.

# Large-Parameter Asymptotics of
Generalized Hasting-McLeod Functions

Kurt Schmidt  Department of Mathematical Sciences
University of Cincinnati
PO Box 210025
Cincinnati, OH 45221<EMAIL_ADDRESS>and  Robert Buckingham  Department
of Mathematical Sciences
University of Cincinnati
PO Box 210025
Cincinnati, OH 45221<EMAIL_ADDRESS>
###### Abstract.

The generalized Hastings-McLeod solutions to the inhomogeneous Painlevé-II
equation arise in multi-critical unitary random matrix ensembles, the chiral
two-matrix model for rectangular matrices, non-intersecting squared Bessel
paths, and non-intersecting Brownian motions on the circle. We establish the
leading-order asymptotic behavior of the generalized Hastings-McLeod functions
as the inhomogeneous parameter approaches infinity using the Deift-Zhou
nonlinear steepest-descent method for Riemann-Hilbert problems. This analysis
is done in both the pole-free region and pole region. The asymptotic formulae
show excellent agreement with numerically computed solutions in both regions.

###### Contents

  1. 1 Introduction
    1. 1.1 Background
      1. 1.1.1 Notation
      2. 1.1.2 Outline of the Paper
    2. 1.2 Acknowledgments
    3. 1.3 Results
  2. 2 The Initial Riemann-Hilbert Problem
  3. 3 Analysis in the Pole-Free Region
    1. 3.1 The g-function
    2. 3.2 Defining the Pole-Free Region
    3. 3.3 short
    4. 3.4 Preliminary Deformations in the Riemann-Hilbert Analysis
      1. 3.4.1
      2. 3.4.2 The Collapsing Deformation
    5. 3.5 Opening of the Lens
    6. 3.6 Applying the g-function
    7. 3.7 The Outer Parametrix
    8. 3.8 The Inner Parametrices
      1. 3.8.1 Local Parametrices at the Endpoints $z=a$ and $z=b$.
    9. 3.9 Error Analysis
      1. 3.9.1 The Global Parametrix
      2. 3.9.2 Tracking the Error Term in the Extraction Formula
  4. 4 Analysis in the Pole Region
    1. 4.1
      1. 4.1.1 Definition and Properties of the G-function
    2. 4.2 Preliminary Deformations in the Riemann-Hilbert Analysis
      1. 4.2.1 Conformal Map Constructions
      2. 4.2.2 The Collapsing Deformation
      3. 4.2.3 Opening Lenses
    3. 4.3 Applying the G-function.
    4. 4.4 Constructing the Outer Parametrix
      1. 4.4.1 The Riemann-Hilbert Problem for the Outer Parametrix
      2. 4.4.2 The Baker-Akhiezer Map
      3. 4.4.3 The Solution to the Model Riemann-Hilbert Problem
    5. 4.5 Constructing the Inner Parametrices
      1. 4.5.1 Local Parametrix at $z=C$
      2. 4.5.2 Local Parametrices at the Endpoints $z=A$, $z=B$, and $z=D$
    6. 4.6 Error Analysis
      1. 4.6.1 The Global Parametrix
      2. 4.6.2 Extraction Formula
  5. 5 Numerical Approximation
    1. 5.1 Spectral Collocation Method
      1. 5.1.1 Collocation Method for Generalized Hastings-McLeod
    2. 5.2 Padé Approximant Method
      1. 5.2.1 Padé Expansion
      2. 5.2.2 Padé Approximant Method for Generalized Hastings-McLeod.
  6. A The Airy Parametrix
  7. B Behavior of Re$(2h+\lambda)$ when $x$ is real.

##  1\. Introduction

###  1.1. Background

We consider the inhomogeneous Painlevé-II equation

(1) $u^{\prime\prime}(y)=2u(y)^{3}+yu(y)-\alpha,$

where $\alpha$ is the inhomogeneity parameter. The well-known Hastings-McLeod
function, $u_{\text{HM}}$, solves (1) when $\alpha=0$ with boundary conditions
$u_{\text{HM}}(y)\sim\sqrt{-y/2}$ as $y\to-\infty$ and
$u_{\text{HM}}(y)\sim\operatorname{Ai}(y)$ as $y\to+\infty$ (where
$\operatorname{Ai}$ is the Airy function) [17]. It arises in the Tracy-Widom
distribution describing a wide range of random phenomena, including, but not
limited to, the largest eigenvalue of a Gaussian Unitary Ensemble random
matrix [20] and the longest increasing subsequence of a permutation [1].

In 2008, Claeys, Kuijlaars, and Vanlessen found that if one perturbs the
Gaussian Unitary Ensemble so that the limiting mean density function of
eigenvalues vanishes quadratically at the origin, then perturbations of the
Hastings-McLeod function arise in a double-scaling limit [10]. These
perturbations solve (1) for $\alpha>-1/2$ with boundary conditions

(2) $\begin{cases}\displaystyle
u_{\text{HM}}^{(\alpha)}(y)=\sqrt{-\frac{y}{2}}\left(1+\mathcal{O}\left(1/(-y)^{3/2}\right)\right)&\text{
as }y\to-\infty\\\\[10.0pt] \displaystyle
u_{\text{HM}}^{(\alpha)}(y)=\frac{\alpha}{y}\left(1+\mathcal{O}\left(1/y^{3}\right)\right)&\text{
as }y\to+\infty\hskip 14.22636pt(\alpha\not=0).\end{cases}$

This family of solutions to (2) is named the generalized Hastings-McLeod
functions. Recently, these functions have been observed in interacting
particle systems [5, 12] and in a different random matrix model [13]. We
establish asymptotic formulae for generalized Hastings-McLeod functions,
$u_{\text{HM}}^{(\alpha)}$, for large, positive half-integers $\alpha$.

The starting point of our analysis is a Riemann-Hilbert representation for
generalized Hastings-McLeod functions discovered by Buckingham and Liechty in
2018 while investigating Dyson Brownian bridges [5]. We apply the Deift-Zhou
method of nonlinear steepest descent developed in [11]. This involves
nontrivial transformations that force the jump discontinuities to be
exponentially small as $\alpha\to\infty$ except at certain points or arcs.
Ignoring these small jumps results in a Riemann-Hilbert problem that is
explicitly solvable and is called a model problem. This provides a rigorous
approximation, with controlled error, to the solution of the original Riemann-
Hilbert problem for the generalized Hastings-McLeod functions.

For large $\alpha$, under the scaling

(3) $x:=-\frac{2^{1/3}}{\left(\alpha-\frac{1}{2}\right)^{2/3}}y,$

$u^{(\alpha)}_{\text{HM}}$ has two distinct behaviors in the $x$-plane. On one
hand, the leading-order asymptotic formula is an analytic function in a
certain region. We call this section of the $x$-plane the pole-free region. On
the other hand, in the complement of the closure of the pole-free region, the
leading-order asymptotic formula is not analytic and has poles at specific
points (depending on $\alpha$). We call this section of the $x$-plane the pole
region. (See Proposition 3.3 for a precise description of the pole-free
region.) Notice, for $\alpha$ large enough the scaled generalized Hastings-
McLeod functions are analytic in the pole-free region and meromorphic in the
pole region. Our Riemann-Hilbert analysis detects this bifurcation. Indeed, in
the pole-free region the model problem is solved in terms of elementary
functions, while in the pole region, it is solved in terms of Riemann
$\Theta$-functions.

####  1.1.1. Notation

Throughout the paper all matrices are denoted by a bold, uppercase letter with
the exception of the Pauli matrix

(4) $\sigma_{3}:=\begin{bmatrix}1&0\\\ 0&-1\end{bmatrix}.$

We also use the following short hand:

(5) $\boldsymbol{U}_{k}(f):=\begin{bmatrix}1&e^{kf}\\\ 0&1\end{bmatrix},\hskip
28.45274pt\boldsymbol{L}_{k}(f):=\begin{bmatrix}1&0\\\
e^{kf}&1\end{bmatrix},\hskip
28.45274pt\boldsymbol{T}(a):=\begin{bmatrix}0&-a^{-1}\\\ a&0\end{bmatrix}.$

Finally, given a smooth, oriented curve $\Sigma$ and a function $f$ that can
be continuously extended to $z_{0}\in\Sigma$, we let $f_{+}(z_{0})$ denote the
non-tangential limit of $f(z)$ as $z$ approaches $z_{0}$ from the left of
$\Sigma$. Similarly, $f_{-}(z_{0})$ denotes the non-tangential limit as $z$
approaches $z_{0}$ from the right of $\Sigma$.

####  1.1.2. Outline of the Paper

First, we state our main results in Theorem 1.1 and Theorem 1.2 and verify
these results by comparing our asymptotic formulae to numerics. Second, in
Section 2 we state the Riemann-Hilbert representation (and its scaled version)
of the generalized Hastings-McLeod functions. Next, in Section 3, we give an
explicit and precise definition for the pole-free region of the $x$-plane.
Then, we proceed to apply the Deift-Zhou steepest-decent method in the pole-
free region, leading to a proof of Theorem 1.1. In Section 4, we carry out
similar analysis in the pole region to prove Theorem (1.2). Finally, in
Section 5, we review the numerical methods we implemented to verify our
asymptotic formulae.

###  1.2. Acknowledgments

The first author was supported by the Charles Phelps Taft Research Center by a
Dissertation Fellowship. The second author was supported by the National
Science Foundation via grant DMS–2108019. The authors also thank Deniz Bilman,
Karl Liechty, Peter Miller, and Andrei Prokhorov for useful discussions.

###  1.3. Results

It is convenient to state our results in terms of the large parameter $k$,
where

(6) $k:=\alpha-\frac{1}{2}.$

######  Theorem 1.1.

Let $x$ be in the pole-free region of the $x$-plane, as defined in Section
3.2. Set $\Sigma_{S}$ to be the union of the closed rays emanating from the
points $x=3e^{\frac{2\pi i}{3}}$ and $x=3e^{-\frac{2\pi i}{3}}$ with arguments
$2\pi/3$ and $-2\pi/3$ respectively. Let $S(x)\equiv S$ denote the solution to
the cubic equation

(7) $S^{3}+xS-2i=0$

cut on $\Sigma_{S}$ which has asymptotic behavior
$S(x)=-i\sqrt{x}+\mathcal{O}(1/x)$ as $x\to+\infty$ along the positive real
line and $S(x)=2i/x+\mathcal{O}\left(1/x^{2}\right)$ as $x\to-\infty$ along
the negative real line. Then, for large $k\in\mathbb{N}$,

(8)
$-(2k)^{-1/3}u^{(k+1/2)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=-i\frac{S(x)}{2}+\mathcal{O}\left(1/k\right).$

######  Theorem 1.2.

Let $x$ be in the pole region of the $x$-plane, i.e. the complement of the
closure of the pole-free region. Suppose $A,B,C,D\in\mathbb{C}$ satisfy the
moment conditions (78) and the Boutroux conditions (81). For large integers
$k$, if $x\in\mathscr{S}_{k}$ (see (133)), then

(9)
$-(2k)^{-1/3}u^{(\alpha)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=i\left(\frac{-[\dot{\boldsymbol{Q}}_{-2}]_{12}+[\dot{\boldsymbol{Q}}_{-1}]_{22}[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}\right),$

where the quantities $[\dot{\boldsymbol{Q}}_{-2}]_{12}$,
$[\dot{\boldsymbol{Q}}_{-1}]_{12}$, and $[\dot{\boldsymbol{Q}}_{-1}]_{22}$ are
defined in (160), (161), and (157). Further, if
$[\dot{\boldsymbol{Q}}_{-1}]_{12}\neq 0$, then the left-hand side of (9) can
be expressed in terms of Riemann $\Theta$-functions. That is,

(10)
$-(2k)^{-1/3}u^{(\alpha)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=i\left(\mathcal{A}_{-1}\cdot(\mathfrak{L}_{22}-\mathfrak{L}_{12})-\frac{B^{2}+D^{2}-A^{2}-C^{2}}{2(B+D-A-C)}\right)+\mathcal{O}(1/k),$

where

(11)
$\mathfrak{L}_{22}=\frac{\Theta^{\prime}(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}+kF_{1}U;\mathcal{B})}{\Theta(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}+kF_{1}U;\mathcal{B})}-\frac{\Theta^{\prime}(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K};\mathcal{B})}{\Theta(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K};\mathcal{B})},$
(12)
$\mathfrak{L}_{12}=\frac{\Theta^{\prime}(\mathcal{A}(\infty)-\mathcal{A}(Q)-\mathcal{K}+kF_{1}U;\mathcal{B})}{\Theta(\mathcal{A}(\infty)-\mathcal{A}(Q)-\mathcal{K}+kF_{1}U;\mathcal{B})}-\frac{\Theta^{\prime}(\mathcal{A}(\infty)-\mathcal{A}(Q)-\mathcal{K};\mathcal{B})}{\Theta(\mathcal{A}(\infty)-\mathcal{A}(Q)-\mathcal{K};\mathcal{B})},$

and $\Theta(\cdot)$, $\mathcal{B}$, $\mathcal{A}_{-1}$, $\mathcal{A}(\infty)$,
$\mathcal{A}(\cdot)$, $Q$, $\mathcal{K}$, $F_{1}$, and $U$ are defined in
(110), (105), (109), (103), (128), (120), (99), and (114).

Here, we compare the results between the asymptotic formulae derived via the
steepest-decent method and the numerical methods discussed in Section 5. In
Figure 1 we take a slice on the real $x$-line. Since the real line is
contained in the pole-free region, the asymptotic formula does not depend on
$k$. So, as $k\to\infty$, $u^{(k+\frac{1}{2})}_{\text{HM}}$ converges (under
the appropriate scaling) to the right-hand side of (8). Notice, we can observe
this convergence even with small values of $k$. In Figure 2 we take a
horizontal slice (namely the slice $\operatorname{Im}(x)=-9i$) through the
pole region of the $x$-plane. For $x$ not in a neighborhood of a pole, we
again note an excellent match between the asymptotics and numerics even for
small $k$.

|
---|---
Figure 1. The left panel is a plot of the left-hand side of (8) evaluated on the real line for $k=1$ (blue), $2$ (orange), $3$ (red) compared to the black curve that is the right-hand side of (8). The right panel is a logarithmic plot of the errors. $|\cdot|$ | $\operatorname{Re}(\cdot)$ | $\operatorname{Im}(\cdot)$
---|---|---
|  |
|  |
|  |
Figure 2. Plots of the left-hand side of (9) evaluated on the slice $\operatorname{Im}(x)=-9$ in the $x$-plane for $k=1$ (top), $2$ (middle), $3$ (bottom). The orange curves show the asymptotic formulae described in Theorems 1.1 and 1.2, while the blue curves are the result of the numerical methods discussed in Section 5. Asymptotic Formula |  | Numerical Plot
---|---|---
|  |
|  |
|  |
Figure 3. Density plots in the $x$-plane of the scaled generalized Hastings-
McLeod functions for $k=1,2,3$. The left column uses the asymptotic formulae
described in Theorems 1.1 and 1.2, while the right column uses the numerical
methods discussed in Section 5.

##  2\. The Initial Riemann-Hilbert Problem

The Painlevé-II equation (1) is equivalent to

(13) $\frac{1}{2}p^{\prime\prime}(s)=p(s)^{3}-sp(s)+\frac{1}{2}+k.$

Indeed, since $u_{\text{HM}}^{(\alpha)}(y)$ satisfies (1) it follows

(14) $p_{k}(s):=-2^{1/3}u_{\text{HM}}^{(k+1/2)}\left(-2^{1/3}s\right)$

satisfies (13) (recall, $k=\alpha+1/2$).

In 2018, Buckingham and Liechty discovered $p_{k}(s)$ can be extracted from
the following $2\times 2$ Riemann-Hilbert problem [5].

Riemann-Hilbert Problem 1:  For each $k\in\mathbb{N}$ and $s\in\mathbb{C}$,
find a $2\times 2$ matrix-valued function
$\boldsymbol{Z}\left(\zeta;s,k\right)\equiv\boldsymbol{Z}(\zeta)$ so that

  1. (1)

Analyticity: The matrix-valued function $\boldsymbol{Z}(\zeta)$ is analytic in
$\zeta$ except on the four rays
$\arg(\zeta)\in\left\\{\pm\frac{\pi}{6},\pm\frac{5\pi}{6}\right\\}$ and is
Hölder continuous up to the boundary in a neighborhood of $\zeta=0$.

  2. (2)

Jump Condition: $\boldsymbol{Z}(\zeta)$ can be continuously extended to the
boundary and the values taken by $\boldsymbol{Z}(\zeta)$ on the four rays of
discontinuity are related by the jump condition
$\boldsymbol{Z}_{+}(\zeta)=\boldsymbol{Z}_{-}(\zeta)\boldsymbol{V}^{(\boldsymbol{Z})}(\zeta)$,
where the jump matrix $\boldsymbol{V}^{(\boldsymbol{Z})}(\zeta)$ is as shown
in Figure 4 and all rays are oriented towards infinity.

$\begin{bmatrix}1&0\\\
e^{2i\left(\frac{4}{3}\zeta^{3}+s\zeta\right)}&1\end{bmatrix}$$\begin{bmatrix}1&0\\\
-e^{2i\left(\frac{4}{3}\zeta^{3}+s\zeta\right)}&1\end{bmatrix}$$\begin{bmatrix}1&e^{-2i\left(\frac{4}{3}\zeta^{3}+s\zeta\right)}\\\
0&1\end{bmatrix}$$\begin{bmatrix}1&-e^{-2i\left(\frac{4}{3}\zeta^{3}+s\zeta\right)}\\\
0&1\end{bmatrix}$$0$ Figure 4. The jumps of $\boldsymbol{Z}(\zeta;s,k)$.

  3. (3)

Normalization: There exist constant (in $\zeta$) matrices
$\boldsymbol{Z}_{-1}(s;k)\equiv\boldsymbol{Z}_{-1}$ and
$\boldsymbol{Z}_{-2}(s;k)\equiv\boldsymbol{Z}_{-2}$ such that, for large
$\zeta$,

(15)
$\boldsymbol{Z}(\zeta)\zeta^{-k\sigma_{3}}=\mathbb{I}+\frac{\boldsymbol{Z}_{-1}}{\zeta}+\frac{\boldsymbol{Z}_{-2}}{\zeta^{2}}+\mathcal{O}(1/\zeta^{3}).$

In particular, the extraction formula is

(16)
$p_{k}(s)=2i\left([\boldsymbol{Z}_{-1}(s;k)]_{22}-\frac{[\boldsymbol{Z}_{-2}(s;k)]_{12}}{[\boldsymbol{Z}_{-1}(s;k)]_{12}}\right).$

We introduce the scalings

(17) $z:=\left(\frac{2}{k}\right)^{1/3}\zeta\hskip 14.22636pt\text{and}\hskip
14.22636ptx:=\left(\frac{2}{k}\right)^{2/3}s$

and define

(18)
$\boldsymbol{M}(z;x,k)\equiv\boldsymbol{M}(z):=\left(\frac{k}{2}\right)^{-\frac{k}{3}\sigma_{3}}\boldsymbol{Z}\left(\left(\frac{k}{2}\right)^{1/3}z;\left(\frac{k}{2}\right)^{2/3}x,k\right).$

$\boldsymbol{M}(z)$ solves the following Riemann-Hilbert problem:

Riemann-Hilbert Problem 2:  For each positive integer $k$, find
$\boldsymbol{M}\left(z;x,k\right)\equiv\boldsymbol{M}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{M}(z)$ is analytic in $z$ off the jumps contours
shown in Figure 5.

  2. (2)

Jump Condition: $\boldsymbol{M}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{M}(z)$ are related by
the jump condition
$\boldsymbol{M}_{+}(z)=\boldsymbol{M}_{-}(z)\boldsymbol{V}^{(\boldsymbol{M})}(z)$,
where $\boldsymbol{V}^{(\boldsymbol{M})}(z)$ is as shown in Figure 5, and

(19) $\theta(z:x):=\frac{4}{3}z^{3}+xz.$

$\boldsymbol{L}_{k}(i\theta)$$\boldsymbol{L}^{-1}_{k}(i\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\boldsymbol{U}_{k}^{-1}(-i\theta$)$0$
Figure 5. The jumps of $\boldsymbol{M}(z)$. Recall, $\boldsymbol{L}_{k}$ and
$\boldsymbol{U}_{k}$ are defined (5).

  3. (3)

Normalization: There exist constant (in $z$) matrices
$\boldsymbol{M}_{-1}(x;k)\equiv\boldsymbol{M}_{-1}$ and
$\boldsymbol{M}_{-2}(x;k)\equiv\boldsymbol{M}_{-2}$ such that, for large $z$

(20)
$\boldsymbol{M}(z)z^{-k\sigma_{3}}=\mathbb{I}+\frac{\boldsymbol{M}_{-1}}{z}+\frac{\boldsymbol{M}_{-2}}{z^{2}}+\mathcal{O}\left(1/z^{3}\right).$

A direct calculation shows that under this scaling the extraction formula (16)
becomes

(21)
$p_{k}\left(\left(\frac{k}{2}\right)^{2/3}x\right)=2i\left(\frac{k}{2}\right)^{1/3}\left([\boldsymbol{M}_{-1}(x;k)]_{22}-\frac{[\boldsymbol{M}_{-2}(x;k)]_{12}}{[\boldsymbol{M}_{-1}(x;k)]_{12}}\right).$

##  3\. Analysis in the Pole-Free Region

In this section we perform the Riemann-Hilbert analysis in the pole-free
region. Our approach is modeled from the methodology presented in Section 3 of
[6]. In the first part, we introduce a technical but standard tool called the
$g$-function. Once that is established, we will be able to define the pole-
free region of the $x$-plane. In the subsequent section, we will use the
$g$-function to reduce Riemann-Hilbert Problem 2 to an explicitly solvable
Riemann-Hilbert problem, thereby giving us a rigorous way to approximate the
left-hand side of (16).

###  3.1. The g-function

We start by constructing the $g$-function for the pole-free region. We define
the functions

(22) $\Delta(x)\equiv\Delta:=\left(-\frac{4i}{S}\right)^{1/2},\hskip
13.00806pta(x)\equiv a:=\frac{S-\Delta}{2},\hskip 13.00806ptb(x)\equiv
b:=\frac{S+\Delta}{2},\hskip 13.00806ptc(x)\equiv c:=-\frac{S}{2}.$

The square root in (22) is principal. This choice guarantees that
$\operatorname{Re}(a)\leq\operatorname{Re}(b)$. Now, let $\Sigma$ denote a
finite, simple, oriented arc that connects $a$ to $b$. (See Remark 3.4 on the
exact placement of $\Sigma$). Given $\Sigma$, let $r(z;x)\equiv r(z)$ be the
function analytic for $z\in\mathbb{C}\setminus\Sigma$ that satisfies the
conditions

(23) $r(z)^{2}=(z-a)(z-b)\hskip 23.84923pt\text{and}\hskip
23.84923ptr(z)=z+\mathcal{O}(1)\text{ as }z\to\infty.$

To define the $g$-function, we first define its derivative and integrate up.
So,

(24) $g^{\prime}(z;x)\equiv
g^{\prime}(z):=\frac{i\theta^{\prime}(z)}{2}-i(S+2z)r(z),$

where $g^{\prime}$ is cut on $\Sigma$. We find that the large-$z$ expansion of
$g^{\prime}(z)$ is

(25)
$g^{\prime}(z)=i\frac{2S^{2}+2x+\Delta^{2}}{4}+i\frac{S\Delta^{2}}{4z}+\mathcal{O}\left(1/z^{2}\right).$

An equivalent way to write (7) and $\Delta^{2}$ is

(26) $2S^{2}+2x+\Delta^{2}=0\hskip 18.77892pt\text{and}\hskip
18.77892pt\Delta^{2}=-\frac{4i}{S}.$

Plugging (26) into (25), we see that $g^{\prime}(z)=1/z+\mathcal{O}(1/z^{2})$
as $z\to\infty$. Thus, $g^{\prime}(z)-1/(z-a)$ is integrable at $z=\infty$ and
we can use the fundamental theorem of calculus to define $g(z)$ as an
antiderivative of $g^{\prime}(z)$. However, we must introduce a logarithmic
cut. Let $L$ denote an oriented, unbounded arc from $z=\infty$ to $z=a$.
Further, suppose that $L$ only intersects $\Sigma$ at $z=a$ and that, for
large enough $z$, $L$ coincides with the negative real numbers. Now, we define
$g(z)$ as an integral where the path of integration can be any path that does
not cross $\Sigma$. Set

(27) $g(z;x)\equiv
g(z):=\log(z-a)+\int_{\infty}^{z}g^{\prime}(w)-\frac{1}{w-a}\text{d}w.$

For notational convenience, we write

(28) $h(z;x)\equiv h(z):=\frac{i\theta(z)}{2}-g(z),\hskip
14.45377ptz\in\mathbb{C}\setminus\Sigma,\hskip
14.45377ptx\in\mathbb{C}\setminus\Sigma_{S}.$

The behavior of $h(z)$ near the points $z=a$ and $z=b$ plays an important role
in the Riemann-Hilbert analysis. Let $\mathbb{D}_{a}$ and $\mathbb{D}_{b}$ be
sufficiently small disks around $z=a$ and $z=b$, respectively. Notice that
$L\cup\Sigma$ divides $\mathbb{D}_{a}$ into two components. Let
$\mathbb{D}_{a}^{\uparrow}$ denote the component that is on the plus side (or
left side) of $L\cup\Sigma$, and let $\mathbb{D}_{a}^{\downarrow}$ denote the
component that is on the minus side (or right side) of $L\cup\Sigma$.

We now list some basic properties of $g$ and $h$ that we will use throughout
the Riemann-Hilbert analysis.

######  Proposition 3.1.

Given that $S,\Delta,a,b$ are defined, guaranteeing that $g(z)$ and $h(z)$ are
well-defined by (27) and (28), respectively, then for each
$x\in\mathbb{C}\setminus\Sigma_{S}$ there exists a complex number
$\lambda(x)\equiv\lambda$ such that for all $z\in\Sigma$ we have that

(29) $h_{+}(z)+h_{-}(z)=i\theta(z)-g_{+}(z)-g_{-}(z)=-\lambda.$

Further, if $x\neq 3e^{\pm 2\pi i/3}$, then under the right perturbation,
$2h+\lambda$ locally behaves like a $3/2$-root function near the endpoints $a$
and $b$. More precisely, if we consider the functions

(30) $J_{a}(z)=\begin{cases}\left(2h(z)+\lambda-2\pi i\right)&\text{if
}z\in\mathbb{D}_{a}^{\uparrow}\\\ \left(2h(z)+\lambda+2\pi i\right)&\text{if
}z\in\mathbb{D}_{a}^{\downarrow}\end{cases}$

and

(31) $J_{b}(z)=2h(z)+\lambda,$

then for each endpoint $p\in\\{a,b\\}$ there exists
$\kappa_{p}\in\mathbb{C}\setminus\\{0\\}$ such that

(32) $J_{p}(z)=\kappa_{p}(z-p)^{3/2}+\mathcal{O}\left((z-p)^{5/2}\right).$

Additionally, when $x\in\mathbb{R}$, for each
$z\in\mathbb{C}\setminus(\Sigma\cup L)$

(33)
$\operatorname{Re}\left(2h\left(-\overline{z};x\right)+\lambda(x)\right)=\operatorname{Re}\left(2h(z;x)+\lambda(x)\right).$

Furthermore, for $x\in\mathbb{R}$,

(34) $\text{there are at most two real }z\text{-values}\text{ such that
}Re\left(2h(z)+\lambda\right)=0.$

Also,

(35) $g_{+}(z)-g_{-}(z)=-2\pi i,\hskip 5.69046ptz\in L.$

Finally, as $z\to\infty$

(36) $g(z)=\log(z)+\mathcal{O}(1/z).$

Most of these statements are standard and follow from direct computation,
however we furnish a proof of statements (33) and (34)in Appendix B. The
function $h(z)$, and the constant $\lambda$ can both be explicitly expressed
in a closed form. Consider

(37)
$\eta(z):=\big{(}(1-z)(1+z)\big{)}^{1/2}=i\big{(}(z-1)(z+1)\big{)}^{1/2},$

where $\eta$ is cut on $[-1,1]$ and as $z\to\infty$,
$\eta(z)=iz+\mathcal{O}(1/z).$ Note that on the cut $[-1,1]$, both the plus
and minus boundary values of $\eta$ are real valued. Now, we let $\mathcal{U}$
denote

(38) $\mathcal{U}(z):=-i\log\left(iz+\eta(z)\right).$

Then,

(39)
$h(z)=\frac{i}{6}r(z)\left(4z^{2}+2Sz+2x\right)-2i\mathcal{U}\left(\left(\frac{a-z}{a-b}\right)^{1/2}\right)+\frac{ixS}{6}+\log\left(\frac{-4}{\Delta}\right)+\frac{1}{3}.$

Moreover, as $\lambda=-(h_{+}+h_{-})$,

(40)
$\lambda=-2\left(\frac{ixS}{6}+\log\left(\frac{-4}{\Delta}\right)+\frac{1}{3}\right).$

###  3.2. Defining the Pole-Free Region

We now have the necessary objects established to define the pole-free region.
In particular, $x$-membership of the pole-free region will depend on the
topology of the zero-level set of $\operatorname{Re}(2h+\lambda)$ in the
$z$-plane (for motivation see Figure 12). The following definition is directly
adapted from Definition 2 in [6].

######  Definition 3.2.

We say $x\in\mathbb{C}$ is in the pole-free region of the complex $x$-plane if
the following five conditions hold:

  1. (a1)

There are exactly three arcs of the zero-level set of
$\operatorname{Re}(2h+\lambda)$ that terminate at the endpoint $z=a$.

  2. (a2)

There exist two curves emanating from $z=a(x)$ that never cross the zero-level
set of $\operatorname{Re}(2h+\lambda)$ and tend to infinity at distinct angles
$5\pi/6$ and $-5\pi/6$.

  3. (a3)

There are exactly three arcs of the zero-level set of
$\operatorname{Re}(2h+\lambda)$ that terminate at the endpoint $z=b$.

  4. (a4)

There exist two curves emanating from $z=b$ that never cross the zero-level
set of $\operatorname{Re}(2h+\lambda)$ and tend to infinity at distinct angles
$\pi/6$ and $-\pi/6$.

  5. (a5)

There exists a finite curve that connects $z=a$ to $z=b$ that never crosses
the zero-level set of $\operatorname{Re}(2h+\lambda)$. We further require that
the curve is below $\Sigma$. That is, except for the endpoints $z=a$ and
$z=b$, the curve is entirely to the right side of $\Sigma$.

As $x$ varies, for $x$ to leave or enter the pole-free region it is necessary
(but not sufficient) for the zero-level set of
$\operatorname{Re}(2h(z;x)+\lambda(x))$ to undergo a sudden topological
change. That is, the zero-level set must intersect itself. Further, the
Cauchy-Riemann equations tell us this can only occur at a point $z_{0}$ where
$h^{\prime}(z_{0};x)=0$ (here the derivative is with respect to $z$). However,
by (24) and (28), for each fixed $x\in\mathbb{C}\setminus\Sigma_{S}$,
$h^{\prime}(z;x)$ can only vanish at $z=a(x)$, $z=b(x)$, or $z=c(x)$. In fact,
since $a(x)$ and $b(x)$ will always lie on the zero-level set (by
construction) and $a(x)\neq b(x)$ (because $\Delta(x)$ never vanishes), our
point of interest is $z=c(x)$. Specifically, we want the $x$-values where
$c(x)$ collides with the zero-level set and so we consider

(41) $\mathfrak{c}(x):=\operatorname{Re}(2h(c(x);x)+\lambda(x)).$

Notice that $g(z)$ (and hence $h(z)$) is continuous at $z=b$. So, at $z=b$
(29) reads $h_{+}(b)+h_{-}(b)=2h(b)=-\lambda$. Thus, by the fundamental
theorem of calculus we have

(42) $\mathfrak{c}(x)=2\operatorname{Re}\int_{b}^{c}h^{\prime}(w)\text{d}w,$

where the path of integration does not cross $\Sigma$. Notice, this implies
that $\mathfrak{c}(x)$ is continuous. The zero-level set of $\mathfrak{c}(x)$
divides the $x$-plane into four connected, open components
$\Omega_{\text{right}}$, $\Omega_{\text{up}}$, $\Omega_{\text{left}}$, and
$\Omega_{\text{down}}$; see Figure 6.

Figure 6. The zero-level set of $\mathfrak{c}(x)$ in the $x$-plane.

######  Proposition 3.3.

The pole-free region of the complex $x$-plane is exactly the union of
$\Omega_{\text{left}}$, $\Omega_{\text{right}}$, and their shared boundary
minus the apex points $x=3e^{\pm 2\pi i/3}$.

###### Proof.

Notice that conditions (a1)–(a5) are preserved as $x$ varies continuously as
long as $x$ does not cross the zero-level set of $\mathfrak{c}(x)$. So, in
order to show that $\Omega_{\text{left}}$ (or $\Omega_{\text{right}}$) is
contained in the pole-free region it suffices to show that at least one of its
points is in the pole-free region.

A direct calculation shows that for, $x\in\mathbb{R}$,

(43)
$\mathfrak{c}(x)=\begin{cases}\displaystyle\frac{\sqrt{2}}{3}(-x)^{3/2}+\mathcal{O}(1)&x\to-\infty\\\\[10.0pt]
\displaystyle-\frac{2}{3}x^{3/2}-\log\left(x^{3/2}\right)+\mathcal{O}(1)&x\to+\infty.\end{cases}$

Given that $\Omega_{\text{left}}$ and $\Omega_{\text{right}}$ are defined by
the zero-level set of $\mathfrak{c}(x)$, (43) tells us that $\mathfrak{c}(x)$
is positive in $\Omega_{\text{left}}$ and negative in $\Omega_{\text{right}}$.

Now, we show that $\Omega_{\text{left}}$ is in the pole-free region. Fix
$x_{\ell}\in\Omega_{\text{left}}\cap\mathbb{R}$. Notice, as $z\to\infty$ the
$i\theta(z)$ part of $h(z)$ dominates because $g(z)=\log(z)+\mathcal{O}(1/z)$
(see (28) and (36)). Thus, at $z=\infty$ we deduce that the zero-level set of
$\operatorname{Re}(2h+\lambda)$ will have exactly six arcs at angles $0$,
$\pi/3$, $2\pi/3$, $\pi$, $-2\pi/3$, and $-\pi/3$. Further, (32) shows that at
$z=b$ there are exactly three arcs of the zero-level set of
$\operatorname{Re}(2h+\lambda)$ emanating from it at angles separated by
$2\pi/3$. Similarly, (32) shows that there are exactly three arcs of the zero-
level set of $\operatorname{Re}(2h+\lambda)$ emanating from $z=a$ at angles
separated by $2\pi/3$.

As $\operatorname{Re}(2h(z)+\lambda)$ is harmonic in $z$ in
$\mathbb{C}\setminus(\Sigma\cup L)$, the aforementioned arcs must connect
pairwise to each other. Further, once connected these six arcs comprise the
entirety of the zero-level set of $\operatorname{Re}(2h+\lambda)$. In
particular, it does not contain a finite arc, nor a closed loop. Indeed, the
former case would violate the fact $\operatorname{Re}(2h+\lambda)$ is
infinitely differentiable, while the latter case would violate the maximum
modulus principle. So, our current task is to determine which configurations
of connections are valid (i.e. connections that are consistent with
Proposition 3.1).

We start by analyzing $2h^{\prime}(z)$ on the imaginary axis in order to
determine how many times the zero-level set intersects the imaginary axis.
Towards that goal we consider the function $f:\mathbb{R}\to\mathbb{C}$ defined
by

(44) $f(v;x)\equiv f(v):=2h(iv)+\lambda.$

A direct calculation shows that when $x$ is real, $a=-\overline{b}$, $c$ is
purely imaginary, and $f^{\prime}$ is a real-valued function with the sign
chart shown in Figure 7.

$\rho$$\operatorname{Im}(c)$$+$$-$$+$$v$ Figure 7. A sign chart for
$f^{\prime}(v)$ (for any $x\in\mathbb{R}$), where $\rho$ is the modulus of the
intersection point of $\Sigma$ and the imaginary axis, i.e. $i\rho\in\Sigma$.

Reading from Figure 7, we see that on the interval
$(\operatorname{Im}(c),\infty)$ $\operatorname{Re}(f)$ has positive slope,
while on $(\rho,\operatorname{Im}(c))$ $F$ has negative slope. Since
$\operatorname{Re}(f(\operatorname{Im}c(x_{\ell})))=\mathfrak{c}(x_{\ell})>0$,
it follows $\operatorname{Re}(f)$ does not vanish on $(\rho,\infty)$ and is
positive.

At $v=\rho$, $f$ has a jump discontinuity because $i\rho\in\Sigma$ and $h$ has
a jump discontinuity on $\Sigma$. In fact, (29) implies that
$f_{+}(\rho)=-f_{-}(\rho)$. Thus, $\operatorname{Re}(f_{-}(\rho))<0$.
Moreover, the fact $f$ has positive slope on $(-\infty,\rho)$ implies
$\operatorname{Re}(f)$ does not vanish on $(-\infty,\rho)$. Therefore,
$\operatorname{Re}(2h+\lambda)$ does not vanish on the imaginary axis. This
means none of the zero-level arcs of $\operatorname{Re}(2h+\lambda)$ cross the
imaginary axis.

Next, since $x_{\ell}\in\mathbb{R}$, the real part of $h(z)$ is symmetric
about the imaginary axis, i.e. (33) holds. Hence, we need only to consider the
left half of the $z$-plane to ascertain how the zero-level arcs of
$\operatorname{Re}(2h+\lambda)$ are connected. Now, label the zero-level arcs
at $\infty$ with angles $2\pi/3$, $\pi$, and $-2\pi/3$ as $\textsc{arc}_{1}$,
$\textsc{arc}_{2}$, and $\textsc{arc}_{3}$. For the arcs emanating from $z=a$,
we label (in a counterclockwise fashion) $\textsc{arc}_{\alpha}$,
$\textsc{arc}_{\beta}$, and $\textsc{arc}_{\gamma}$. Notice, if one of the
arcs at $\infty$ (on the left side of the plane) did not connect to one of the
arcs emanating from $z=a$, then two arcs at $z=a$ would be forced to connect
to each other (because to connect to an arc in the right half of the plane
would require it to cross the imaginary axis). However, this is not a valid
connection because it forms a closed loop. This means we can assume without
loss of generality that $\textsc{arc}_{1}$ is connected to
$\textsc{arc}_{\alpha}$ (otherwise relabel the arcs emanating from $z=a$).
Further, $\textsc{arc}_{2}$ is forced to connect to $\textsc{arc}_{\beta}$.
Otherwise, the connection between $\textsc{arc}_{3}$ and
$\textsc{arc}_{\beta}$ would intersect a zero-level arc; see Figure 8.
Therefore, we have a complete picture of possible configurations of the zero-
level arcs in the left half of the $z$-plane. Indeed, $\textsc{arc}_{1}$ is
connected to $\textsc{arc}_{\alpha}$, $\textsc{arc}_{2}$ is connected to
$\textsc{arc}_{\beta}$, and $\textsc{arc}_{3}$ is connected to
$\textsc{arc}_{\gamma}$. Moreover, we can complete our picture of the entire
$z$-plane using the symmetry (33); see Figure 9. Finally, we see that
$x_{\ell}$ satisfies all of the axioms in the definition of the pole-free
region and therefore $\Omega_{\text{left}}$ is completely contained in the
pole-free region.

$\textsc{arc}_{\gamma}$$\textsc{arc}_{\alpha}$$\textsc{arc}_{\beta}$$a$$\textsc{arc}_{1}$$\textsc{arc}_{2}$$\textsc{arc}_{3}$ | fill |  | fill |
---|---|---|---|---
Figure 8. Finding all of the possible configurations of the zero-level arcs of
$\operatorname{Re}(2h+\lambda)$ in the left half of the $z$-plane, when
$x\in\Omega_{\text{left}}$. The left diagram shows the starting point, the
middle diagram shows a valid configuration, while the right diagram shows an
invalid configuration.

$a$$b$$c$

Figure 9. Valid connections of the zero-level arcs of
$\operatorname{Re}(2h+\lambda)$ for $x\in\Omega_{\text{left}}$.

We now show $\Omega_{\text{right}}$ is contained in the pole-free region. Fix
$x_{r}\in\Omega_{\text{right}}\cap\mathbb{R}$. The same analysis gives us the
same setup as in the previous case. That is, we know that the zero-level set
of $\operatorname{Re}(2h+\lambda)$ will have exactly six unbounded arcs that
tend to $\infty$ at angles $0$, $\pi/3$, $2\pi/3$, $\pi$, $-2\pi/3$, and
$-\pi/3$, and there are exactly three arcs of the zero-level set of
$\operatorname{Re}(2h+\lambda)$ emanating from the points $z=a$ and $z=b$ at
angles separated by $2\pi/3$. We now need to determine the valid connections
between these arcs. Using the same the argument as in the previous case, we
can use Figure 7 to find that the zero-level lines of
$\operatorname{Re}(2h(z)+\lambda)$ cross the imaginary axis in the $z$-plane
at two distinct points.

Notice, due to the symmetry condition (33), the topology of the zero-level
lines differs from the previous case. Also, since $\mathfrak{c}(x)$ is
continuous, (43) and the intermediate value theorem imply there exists
$x_{0}\in(x_{\ell},x_{r})$ such that
$\operatorname{Re}(2h(c;x_{0})+\lambda(x_{0}))=0$. Here, we use the assumption
that $\mathfrak{c}(x)$ only vanishes once on the real line, as indicated by
the numerical plots. Indeed, plots suggest this map is monotone.

To determine the valid connections of the zero-level arcs of
$\operatorname{Re}(2h+\lambda)$ at $x_{r}$, we will use the fact that these
arcs deform continuously as $x$ goes from $x_{\ell}$ to $x_{r}$ along the real
line. As $x$ varies, we know that $\mathfrak{c}(x)$ vanishes at $x_{0}$. By
the symmetry condition (33) there are only two mechanisms for
$\mathfrak{c}(x)$ to vanish on the real line. Namely, the arc that starts at
$z=a$ that goes to $\infty$ at an angle of $2\pi/3$ can collide with the arc
that starts at $z=b$ that goes to $\infty$ at an angle of $\pi/3$, or the arc
that starts at $z=a$ that goes to $\infty$ at an angle of $-2\pi/3$ can
collide with the arc that starts at $z=b$ that goes to $\infty$ at an angle of
$-\pi/3$. Any other collision would cause $\mathfrak{c}(x)$ to vanish at least
twice on $\mathbb{R}$. Further, since $\operatorname{Im}(c)>0$ and the
collision must occur at $z=c$, (34) allows us to rule out the latter case.
Therefore, the zero-level arcs of $\operatorname{Re}(2h+\lambda)$ when
$x=x_{0}$ is topologically equivalent to Figure 10.

$a$$b$$c$

Figure 10. The zero-level arcs of $\operatorname{Re}(2h+\lambda)$ when
$x=x_{0}$.

When $x$ is pushed to the right of $x_{0}$ the arcs break apart so that there
is a bounded arc that starts at $z=a$ and goes to $z=b$ and there is an
unbounded arc that tends to $\infty$ in the left half of the $z$-plane at an
angle of $2\pi/3$ and tends to $\infty$ in the right half at an angle of
$\pi/3$. This breaking mechanism is forced because the zero-level set of
$\operatorname{Re}(2h+\lambda)$ intersects the imaginary axis twice. So, we
see that $x_{r}$ satisfies all of the axioms in the definition of the pole-
free region; see Figure 11, and therefore $\Omega_{\text{right}}$ is
completely contained in the pole-free region.

$a$$b$$c$

Figure 11. The zero-level arcs of $\operatorname{Re}(2h+\lambda)$ when
$x\in\Omega_{\text{right}}$.

∎

######  Remark 3.4.

The zero-level set of $\operatorname{Re}(2h+\lambda)$ is independent of the
placement of the band $\Sigma$. That is, the zero-level set of
$\operatorname{Re}(2h+\lambda)$ is invariant with respect to the choice of
$\Sigma$ as long as $\Sigma$ is a finite, simple curve that connects $a$ to
$b$. Therefore, we are at liberty to pick $\Sigma$. For the ease of describing
other objects in the analysis we choose $\Sigma$ such that

  * •

When $x\in\Omega_{\text{left}}$, we require that (i)
$\Sigma\cap\mathbb{D}_{a}$ is contained in the zero-level arc of
$\operatorname{Re}(2h+\lambda)$ that emanates from $z=a$ and goes to $\infty$
at an angle of $2\pi/3$ and (ii) $\Sigma\cap\mathbb{D}_{b}$ is contained in
the zero-level arc of $\operatorname{Re}(2h+\lambda)$ that emanates from $z=b$
and goes to $\infty$ at an angle of $\pi/3$.

  * •

When $x\in\Omega_{\text{right}}$, $\Sigma$ coincides with the zero-level arc
of $\operatorname{Re}(2h+\lambda)$ that connects $z=a$ to $z=b$.

######  Remark 3.5.

The section of the boundary that borders $\Omega_{\text{left}}$ is the two
straight rays that comprise $\Sigma_{S}$ (the branch cut of $S(x)$). The proof
is straightforward and involves evaluating (39) on the boundary.

###  3.3. Surveying the Pole-Free Region

We now survey the pole-free region. See Table 1 for a list of all of the
$x$-values we use to graph the zero-level lines of
$\operatorname{Re}(2h+\lambda)$.

$9e^{2\pi i/3}\approx-4.5+7.794i$ |  |  |  | $\approx 1.5+6.678i$
---|---|---|---|---
$3e^{2\pi i/3}-3\approx-4.5+2.598i$ |  | $3e^{2\pi i/3}\approx-1.5+2.598$ |  | $3e^{2\pi i/3}+3\approx 1.5+2.598i$
$-4.5$ |  | $x_{0}\approx-1.588$ |  | $1.5$
$3e^{-2\pi i/3}-3\approx-4.5-2.598i$ |  | $3e^{-2\pi i/3}\approx-1.5-2.598$ |  | $3e^{-2\pi i/3}+3\approx 1.5-2.598i$
$9e^{-2\pi i/3}\approx-4.5-7.794i$ |  |  |  | $\approx 1.5-6.678i$
Table 1. $x$-values corresponding to Figure 12 |  |
---|---|---
|  |
|  |
|  |
|  |
Figure 12. The zero-level arcs of $\operatorname{Re}(2h+\lambda)$ in the
$z$-plane for various $x$-values as shown in Table 1. Here, the dotted lines
refers to branch cuts that do not coincide with the zero-level set.

Notice the breaking mechanism differs between the left and right side of the
$x$-plane. On one hand, when $x$ leaves the pole-free region to the left of
the vertical line $\operatorname{Re}(x)=-1.5$, two distinct zero-level arcs
collide with each other at $z=c$. In this case, $\operatorname{Re}(c)$ will be
between $\operatorname{Re}(a)$ and $\operatorname{Re}(b)$. The resulting
configuration has a zero-level arc that separates $a$ and $b$. Thus, such a
configuration violates $(\textbf{a5})$. In Figure 12 this corresponds to when
$x=9e^{2\pi i/3}$ or $x=9e^{-2\pi/3}$. Notice that we have an exact formula
for when this phenomenon occurs, $x=\rho e^{\pm 2\pi i/3}$ with $\rho>3$. On
the other hand, when $x$ leaves the pole-free region to the right of
$\operatorname{Re}(x)=-1.5$, two distinct zero-level arcs collide with each
other at $z=c$, but $\operatorname{Re}(c)$ will not be between
$\operatorname{Re}(a)$ and $\operatorname{Re}(b)$. The resulting configuration
violates $(\textbf{a2})$ or $(\textbf{a4})$ depending on if $x$ is in the
upper or lower half of the plane, respectively. The right boundary curves are
defined implicitly by those $x$ satisfying
$\operatorname{Re}\left(2h(c(x);x)+\lambda(x)\right)=0$.

There are two special $x$-values, $x=3e^{\pm 2\pi i/3}$, when an endpoint
collides with $z=c$. These are the apexes of the boundary. When $x=3e^{2\pi
i/3}$, then $a=c$ or, in other words, $3a+b=0$. In fact, $2h(a)+\lambda$ has a
zero of order $5/2$. Hence, (a1) is not satisfied. Similarly, when
$x=3e^{-2\pi i/3}$, (a3) is not satisfied. Finally, we can see that a
topological change is not a sufficient condition to be on the boundary.
Indeed, at $x_{0}$ we see that the zero-level lines undergo a sudden
topological change but of all the axioms are still satisfied. Notice, this
section of the zero-level set of $\mathfrak{c}(x)$ is curved and is not
contained in the vertical line $\operatorname{Re}(x)=-3/2$.

###  3.4. Preliminary Deformations in the Riemann-Hilbert Analysis

For the remainder of Section 3 we fix $x$ in the pole-free region as we
perform the Deift-Zhou nonlinear steepest-descent method. The general line of
attack is to apply a sequence of deformations to $\boldsymbol{M}(z)$ to obtain
a corresponding Riemann-Hilbert problem that can be approximated by a solvable
Riemann-Hilbert problem. Our first step is to collapse the corresponding jump
contours of $\boldsymbol{M}(z)$ in an open bounded set. In order to specify
the exact placement of the deformed jump contours, we need to construct two
conformal maps $\tau_{a}$ and $\tau_{b}$.

####  3.4.1. Conformal Map Constructions

We start with the simpler case and construct $\tau_{b}$. Recall,
$\mathbb{D}_{b}$ is a small disk around $z=b$. Notice, one can write the jump
condition (29) as

(45) $2h_{+}(z)+\lambda=-(2h_{-}(z)+\lambda),\hskip 9.38945ptz\in\Sigma.$

So, (32) and (45) imply that in $\mathbb{D}_{b}$, the map
$z\mapsto(2h(z)+\lambda)^{2}$ is analytic and nonvanishing, except for a
third-order zero at $z=b$. Also, since $\Sigma$ was placed on the zero-level
set of $\operatorname{Re}(2h+\lambda)$, we have that for each
$z\in\Sigma\cap\mathbb{D}_{b}$, $(2h_{\pm}(z)+\lambda)^{2}<0$. With these
considerations in mind, we see that there exists a conformal map $\tau_{b}$
such that

  * •

For each $z\in\mathbb{D}_{b}$, $\tau_{b}(z)^{3}=(2h(z)+\lambda)^{2}$,

  * •

$\tau_{b}$ maps $\mathbb{D}_{b}$ to a small neighborhood of the origin,

  * •

$\tau_{b}$ maps $\mathbb{D}_{b}\cap\Sigma$ to the negative real line.

We need to construct a similar conformal mapping of $\mathbb{D}_{a}$. However,
in this case we need to deal with the logarithmic cut $L$. Notice, one can
write (35) in terms of $h$ as

(46) $2h_{+}(z)-2\pi i=2h_{-}(z)+2\pi i,\hskip 9.38945ptz\in L.$

Consider the map

(47) $z\mapsto\begin{cases}\left(2h(z)+\lambda-2\pi i\right)^{2}&\text{if
}z\in\mathbb{D}_{a}^{\uparrow},\\\ \left(2h(z)+\lambda+2\pi
i\right)^{2}&\text{if }z\in\mathbb{D}_{a}^{\downarrow}.\end{cases}$

Notice, (45) and (46) imply that both the upper and lower functions of the
above map have the same continuous extension on
$(L\cup\Sigma)\cap\mathbb{D}_{a}$. Therefore, the map (47) is analytic in
$\mathbb{D}_{a}$. Moreover, (32) shows that this map is nonvanishing in
$\mathbb{D}_{a}$ except for a third-order zero at $z=a$. Thus, there exists a
conformal map $\tau_{a}$ that maps $\mathbb{D}_{a}$ to a neighborhood of the
origin, that satisfies

(48) $(\tau_{a}(z))^{3}=\begin{cases}\left(2h(z)+\lambda-2\pi
i\right)^{2}&\text{if }z\in\mathbb{D}_{a}^{\uparrow},\\\
\left(2h(z)+\lambda+2\pi i\right)^{2}&\text{if
}z\in\mathbb{D}_{a}^{\downarrow},\end{cases}$

and that maps $\Sigma\cap\mathbb{D}_{a}$ to the negative real numbers.

####  3.4.2. The Collapsing Deformation

We deform jump contours of Riemann-Hilbert problems by the standard
sectionally analytic substitution method (for more details see Section 4.3 in
[4]). Our first deformation is to collapse the four jump rays of
$\boldsymbol{M}(z)$ onto the line segment with endpoints $z=a$ and $z=b$. The
resulting jump contour will be comprised of four unbounded rays and one line
segment from $z=a$ to $z=b$. Two of the unbounded jump rays emanate from $z=a$
and approach $\infty$ at angles $\pm 2\pi/3$, while the other two rays emanate
from $z=b$ and approach $\infty$ at angles $\pm\pi/3$. The jump along the
deformed rays will be the same jump along the parent ray, while the new jump
along the line segment is simply the product

(49) $\begin{bmatrix}1&-e^{-ik\theta}\\\ 0&1\end{bmatrix}\begin{bmatrix}1&0\\\
e^{ik\theta}&1\end{bmatrix}=\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&1\end{bmatrix}.$

Since the deformation occurred on a bounded set, the new resulting Riemann-
Hilbert problem has the same normalization condition as Riemann-Hilbert 2. In
fact, the new Riemann-Hilbert problem is:

Riemann-Hilbert Problem 3:  For each $k\in\mathbb{N}$, find
$\boldsymbol{N}(z;x,k)\equiv\boldsymbol{N}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{N}(z)$ is analytic in $z$ except along the jump
contours shown in Figure 13.

  2. (2)

Jump Condition: $\boldsymbol{N}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{N}(z)$ are related by
the jump condition
$\boldsymbol{N}_{+}(z)=\boldsymbol{N}_{-}(z)\boldsymbol{V}^{(\boldsymbol{N})}(z)$
as shown in Figure 13.

$\boldsymbol{L}_{k}(i\theta)$$\boldsymbol{L}_{k}^{-1}(i\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\boldsymbol{U}_{k}^{-1}(-i\theta)$$\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&1\end{bmatrix}$$a$$b$

Figure 13. The jumps of $\boldsymbol{N}(z)$.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{N}(z)z^{-k\sigma_{3}}=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

Although there is some degree of freedom where we place the deformed jump
rays, we require the following criteria:

  * •

In $\mathbb{C}\setminus(\mathbb{D}_{a}\cup\mathbb{D}_{b})$, the four unbounded
jump contours avoid the zero-level set of $\operatorname{Re}(2h+\lambda)$.
Such a deformation is guaranteed to exist by Axioms (a2) and (a4),

  * •

In $\mathbb{D}_{a}$ the jump contour that has an upper-triangular jump is
contained in $\tau_{a}^{-1}\left(\left(0,\infty\right)\right)$,

  * •

In $\mathbb{D}_{a}$ the jump contour that has a lower-triangular jump is
contained in $\tau_{a}^{-1}\left(\rho e^{-2i/3}\right)$, $\rho>0$,

  * •

In $\mathbb{D}_{b}$ the jump contour that has an upper-triangular jump is
contained $\tau_{b}^{-1}\left(\left(0,\infty\right)\right)$,

  * •

In $\mathbb{D}_{b}$ the jump contour that has a lower-triangular jump is
contained in $\tau_{b}^{-1}\left(\rho e^{2i/3}\right)$, $\rho>0$.

###  3.5. Opening of the Lens

The jump on the line segment with endpoints $z=a$ and $z=b$ has the following
factorization:

(50) $\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&1\end{bmatrix}=\begin{bmatrix}1&0\\\
-e^{ik\theta}&1\end{bmatrix}\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&0\end{bmatrix}.$

As such, we use sectionally analytic substitutions to open a lens. The upper
boundary of the lens, i.e. the jump contour with an off-diagonal jump, is
placed on $\Sigma$. For the lower boundary of the lens, near the endpoints
$z=a$ and $z=b$ we have the following requirements:

  * •

In $\mathbb{D}_{a}$, the lower boundary of the lens, i.e. the jump contour
with a lower-triangular jump, is contained in $\tau_{a}^{-1}\left(\rho e^{2\pi
i/3}\right)$ for $\rho>0$,

  * •

In $\mathbb{D}_{b}$, the lower boundary of the lens is contained in
$\tau_{b}^{-1}\left(\rho e^{-2\pi i/3}\right)$ for $\rho>0$.

Outside of $\mathbb{D}_{a}\cup\mathbb{D}_{b}$ we only require the lower
boundary of the lens does not intersect the zero-level set of
$\operatorname{Re}\left(2h+\lambda\right)$. The existence of such a lens is
guaranteed by Axiom (a5). The opening lens deformation results in another
matrix-valued function $\boldsymbol{O}(z)$. Since this deformation is on a
bounded set, $\boldsymbol{O}(z)$ and $\boldsymbol{N}(z)$ satisfy the same
normalization condition. Thus, $\boldsymbol{O}(z)$ solves the following
Riemann-Hilbert problem:

Riemann-Hilbert Problem 4: For each $k\in\mathbb{N}$, find
$\boldsymbol{O}(z;x,k)\equiv\boldsymbol{O}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{O}(z)$ is analytic in $z$ except along the jump
contours shown in Figure 14.

  2. (2)

Jump Condition: $\boldsymbol{O}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{O}(z)$ are related by
the jump condition
$\boldsymbol{O}_{+}(z)=\boldsymbol{O}_{-}(z)\boldsymbol{V}^{(\boldsymbol{O})}(z)$
as shown in Figure 14.

$\boldsymbol{L}_{k}(i\theta)$$\boldsymbol{L}_{k}^{-1}(i\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\boldsymbol{U}_{k}^{-1}(-i\theta)$$\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&0\end{bmatrix}$$\boldsymbol{L}_{k}^{-1}(i\theta)$$a$$b$

Figure 14. The jumps of $\boldsymbol{O}(z)$.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{O}(z)z^{-k\sigma_{3}}=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

###  3.6. Applying the g-function

We now apply the $g$-function defined in Section 3.1 (see (27)) to the
function $\boldsymbol{O}(z)$. We consider

(51)
$\boldsymbol{P}(z;x,k)\equiv\boldsymbol{P}(z):=e^{-\frac{k\lambda}{2}\sigma_{3}}\boldsymbol{O}(z)e^{-kg(z)\sigma_{3}}e^{\frac{k\lambda}{2}\sigma_{3}},$

where $\lambda$ is defined in (40). Introducing the $g$-function simplifies
both the jump and normalization conditions of the corresponding Riemann-
Hilbert problem. Since $e^{g(z)}$ is analytic everywhere except on $\Sigma$,
the deformation $\boldsymbol{O}(z)\to\boldsymbol{P}(z)$ does not introduce new
jump contours or move any. It does, however, simplify the jump matrices. A
formula for jump matrices of $\boldsymbol{P}(z)$, given the jump matrices of
$\boldsymbol{O}(z)$, is

(52)
$\boldsymbol{V}^{(\boldsymbol{O})}=\begin{bmatrix}v_{11}^{(\boldsymbol{O})}&v_{12}^{(\boldsymbol{O})}\\\
v_{21}^{(\boldsymbol{O})}&v_{22}^{(\boldsymbol{O})}\end{bmatrix}\Longrightarrow\boldsymbol{V}^{(\boldsymbol{P})}=\begin{bmatrix}v_{11}^{(\boldsymbol{O})}e^{-k(g_{+}-g_{-})}&v_{12}^{(\boldsymbol{O})}e^{k(g_{+}+g_{-}-\lambda)}\\\
v_{21}^{(\boldsymbol{O})}e^{-k(g_{+}+g_{-}-\lambda)}&v_{22}^{(\boldsymbol{O})}e^{k(g_{+}-g_{-})}\end{bmatrix}.$

On one hand, $\boldsymbol{V}^{(\boldsymbol{O})}$ is highly oscillatory along
$\Sigma$. On the other hand, due to the jump of $g$ on $\Sigma$ (see (29)) we
see that $\boldsymbol{V}^{(\boldsymbol{P})}$ is a constant off-diagonal
matrix. For all of the other jump matrices, their corresponding jump contour
is placed so that $\boldsymbol{V}^{(\boldsymbol{P})}$ decays to the identity
matrix as $k\to\infty$. Indeed, if $\boldsymbol{V}^{(\boldsymbol{P})}$ is an
upper-triangular matrix, then the exponent of its $12$-entry is
$-(2h+\lambda)$ and its jump contour is completely contained in a region of
the $z$-plane where $\operatorname{Re}(2h+\lambda)>0$. Likewise, if
$\boldsymbol{V}^{(\boldsymbol{P})}$ is a lower-triangular matrix, then the
exponent of its $21$-entry is $2h+\lambda$ and its jump contour is completely
contained in a region of the $z$-plane where $\operatorname{Re}(2h+\lambda)<0$
(see Figure (12) and (15)). Finally, the normalization (36) of $g(z)$
simplifies the large-$z$ behavior of $\boldsymbol{P}(z)$. Therefore,
$\boldsymbol{P}(z)$ is the solution to the following Riemann-Hilbert problem:

Riemann-Hilbert Problem 5:  For each positive integer $k$, find
$\boldsymbol{P}\left(z;x,k\right)\equiv\boldsymbol{P}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{P}(z)$ is analytic in $z$ off the jumps contours
shown in Figure 15.

  2. (2)

Jump Condition: $\boldsymbol{P}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{P}(z)$ are related by
the jump condition
$\boldsymbol{P}_{+}(z)=\boldsymbol{P}_{-}(z)\boldsymbol{V}^{(\boldsymbol{P})}(z)$,
where $\boldsymbol{V}^{(\boldsymbol{P})}(z)$ is as shown in Figure 15.

$\boldsymbol{L}_{k}(2h+\lambda)$$\boldsymbol{L}_{k}^{-1}(2h+\lambda)$$\boldsymbol{U}_{k}(-(2h+\lambda))$$\boldsymbol{U}_{k}^{-1}(-(2h+\lambda))$$\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}$$\boldsymbol{L}_{k}^{-1}(2h+\lambda)$$a$$b$

Figure 15. Jumps for $\boldsymbol{P}(z)$.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{P}(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

###  3.7. The Outer Parametrix

The phase function was constructed so that if $k$ is large, all of the the
jump matrices (except for the one along $\Sigma$) of $\boldsymbol{P}(z)$ are
close to the identity matrix. Ignoring these jumps we are led to the following
Riemann-Hilbert problem:

Riemann-Hilbert Problem 6:  Determine the matrix-valued function
$\dot{\boldsymbol{P}}^{\text{(out)}}(z;x)\equiv\dot{\boldsymbol{P}}^{\text{(out)}}(z)$
with the following properties:

  1. (1)

Analyticity: $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ is analytic in $z$
except on $\Sigma$. At the endpoints $a$ and $b$, negative one-fourth power
singularities are allowed.

  2. (2)

Jump Condition: $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ can be continuously
extended to the interior of $\Sigma$ and the boundary values taken by
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ are related by the jump condition

(53)
$\dot{\boldsymbol{P}}^{\text{(out)}}_{+}(z)=\dot{\boldsymbol{P}}^{\text{(out)}}_{-}(z)\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}.$

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

.

It is well known that the unique solution to this Riemann-Hilbert problem is

(54)
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)=\begin{bmatrix}\frac{1}{2}\Big{(}\beta(z)+\frac{1}{\beta(z)}\Big{)}&\frac{i}{2}\Big{(}\beta(z)-\frac{1}{\beta(z)}\Big{)}\\\
-\frac{i}{2}\Big{(}\beta(z)-\frac{1}{\beta(z)}\Big{)}&\frac{1}{2}\Big{(}\beta(z)+\frac{1}{\beta(z)}\Big{)}\end{bmatrix},$

where $\beta(z;x)\equiv\beta(z)$ denotes the function analytic for
$z\in\mathbb{C}\setminus\Sigma$ that satisfies the conditions

(55) $\beta(z)^{4}=\left(\frac{z-b}{z-a}\right)\hskip
14.22636pt\text{and}\hskip 14.22636pt\beta(z)=1+\mathcal{O}(1/z)\text{ as
}z\to\infty.$

See, for example, Section 3.3 of [7].

###  3.8. The Inner Parametrices

The deformation $\boldsymbol{P}(z)\to\dot{\boldsymbol{P}}^{\text{(out)}}(z)$
was motivated by the fact that the jump contours we ignored decayed to the
identity as $k\to\infty$. In light of the theory of small-norm Riemann-Hilbert
problems, we need this decay to be uniform (with respect to $z$). However,
near $z=a$ and $z=b$ the convergence is not uniform. We also see that
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ has quarter-root singularities at $a$
and $b$. Therefore, $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ is not a good
approximation of $\boldsymbol{P}(z)$ near $z=a$ and $z=b$, and so we need to
approximate Riemann-Hilbert Problem 3.6 in a different manner. In particular,
in $\mathbb{D}_{a}$ and $\mathbb{D}_{b}$ we construct Airy parametrices that
satisfy the same jump conditions that $\boldsymbol{P}(z)$ satisfies. In
Section 4.5 we give a systematic way to construct Airy parametrices and
provide full details for one example.

####  3.8.1. Local Parametrices at the Endpoints $z=a$ and $z=b$.

Set

(56)
$\boldsymbol{H}^{(a)}(z):=\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(h(z)+\lambda/2)\sigma_{3}}e^{\frac{k}{2}\tau_{a}^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{a}^{-\sigma_{3}/4}$

and

(57)
$\dot{\boldsymbol{P}}^{(a)}:=\boldsymbol{H}^{(a)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{a}(z)\right)e^{-\frac{k}{2}\tau_{a}^{3/2}\sigma_{3}}e^{k(h(z)+\lambda/3)\sigma_{3}},$

where $\boldsymbol{A}(\cdot)$ is the well-known solution to Riemann-Hilbert
Problem A in Appendix A and $\boldsymbol{V}$ is defined in (184). In
$\mathbb{D}_{b}$ set

(58)
$\boldsymbol{H}^{(b)}(z):=\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(h(z)+\lambda/2)\sigma_{3}}e^{\frac{i\pi}{2}\sigma_{3}}e^{\frac{k}{2}\tau_{b}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{b}(z)^{-\sigma_{3}/4}$

and

(59)
$\dot{\boldsymbol{P}}^{(b)}(z):=\boldsymbol{H}^{(b)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{b}(z)\right)e^{-\frac{k}{2}\tau_{b}(z)^{3/2}\sigma_{3}}e^{-\frac{i\pi}{2}\sigma_{3}}e^{k(h(z)+\lambda/2)\sigma_{3}}.$

For $p\in\left\\{a,b\right\\}$, for each $z\in\partial\mathbb{D}_{p}$ as
$k\to\infty$ the mismatch jumps satisfy

(60)
$\dot{\boldsymbol{P}}^{(p)}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1}=\mathbb{I}+\begin{bmatrix}\mathcal{O}\left(k^{-2}\right)&\mathcal{O}\left(k^{-1}\right)\\\
\mathcal{O}\left(k^{-1}\right)&\mathcal{O}\left(k^{-2}\right)\end{bmatrix}.$

###  3.9. Error Analysis

In this section we will quantify the error introduced by ignoring the jumps of
$\boldsymbol{P}(z)$ that decayed to the identity as $k\to\infty$ by analyzing
a small-norm Riemann-Hilbert problem. First, we will compose the global
parametrix out of the local and outer parametrices. Then, we will set up the
Riemann-Hilbert problem that characterizes the error between the global
parametrix and $\boldsymbol{P}(z)$, the actual solution. Finally, using the
theory of small-norm Riemann-Hilbert problems, we will show that our
approximation of the right-hand side of (21) is of order $1/k$.

####  3.9.1. The Global Parametrix

Now that we have constructed a parametrix in all parts of the $z$-plane, we
are ready to define the global parametrix. Naturally, we set

(61)
$\dot{\boldsymbol{P}}(z;x,k)\equiv\dot{\boldsymbol{P}}(z)=\begin{cases}\dot{\boldsymbol{P}}^{(a)}(z)&z\in\mathbb{D}_{a},\\\
\dot{\boldsymbol{P}}^{(b)}(z)&z\in\mathbb{D}_{b},\\\
\dot{\boldsymbol{P}}^{\text{(out)}}(z)&z\in\mathbb{C}\setminus\left(\Sigma\cup\overline{\mathbb{D}_{a}\cup\mathbb{D}_{b}}\right).\end{cases}$

Set the (unknown) error matrix to be the ratio between $\boldsymbol{P}(z)$ and
$\dot{\boldsymbol{P}}(z)$. That is,

(62)
$\boldsymbol{E}(z;x,k)\equiv\boldsymbol{E}(z):=\boldsymbol{P}(z)\dot{\boldsymbol{P}}^{-1}(z).$

Notice, when $z$ is not on the jump contours of $\boldsymbol{P}(z)$ or
$\dot{\boldsymbol{P}}(z)$ then $\boldsymbol{E}(z)$ is a product of analytic
functions, and hence is analytic. Thus, $\boldsymbol{E}(z)$ will have a jump
discontinuity if either $z$ is on a jump contour of $\boldsymbol{P}(z)$ that
is not $\Sigma$ or $z$ is on the boundary of
$\mathbb{D}_{a}\cup\mathbb{D}_{b}$; see Figure 16. Also, since both
$\boldsymbol{P}(z)$ and $\dot{\boldsymbol{P}}(z)$ converge to the identity as
$z\to\infty$, $\boldsymbol{E}(z)$ converges to the identity as $z\to\infty$.
Thus, we find that $\boldsymbol{E}(z)$ is the unique solution to the following
Riemann-Hilbert problem:

Riemann-Hilbert Problem 7:  For each $k\in\mathbb{N}$, find
$\boldsymbol{E}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{E}(z)$ is analytic in $z$ except along the jump
contours shown in Figure 16.

  2. (2)

Jump Condition: $\boldsymbol{E}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{E}(z)$ are related by
the jump condition
$\boldsymbol{E}_{+}(z)=\boldsymbol{E}_{-}(z)\boldsymbol{V}^{(\boldsymbol{E})}(z)$
as shown in Figure 16.

$\dot{\boldsymbol{P}}\boldsymbol{L}_{k}(2h+\lambda)\dot{\boldsymbol{P}}^{-1}$$\dot{\boldsymbol{P}}\boldsymbol{L}_{k}^{-1}(2h+\lambda)\dot{\boldsymbol{P}}^{-1}$$\dot{\boldsymbol{P}}\boldsymbol{U}_{k}(-(2h+\lambda))\dot{\boldsymbol{P}}^{-1}$$\dot{\boldsymbol{P}}\boldsymbol{U}_{k}^{-1}(-(2h+\lambda))\dot{\boldsymbol{P}}^{-1}$$\mathbb{D}_{a}$$\boldsymbol{V}_{a}^{(\boldsymbol{E})}$$\mathbb{D}_{b}$$\boldsymbol{V}_{b}^{(\boldsymbol{E})}$$\dot{\boldsymbol{P}}\boldsymbol{L}_{k}^{-1}(2h+\lambda)\dot{\boldsymbol{P}}^{-1}$
Figure 16. The jumps of $\boldsymbol{E}(z)$.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{E}(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

Notice, not only does $\boldsymbol{E}(z)$ normalize to the identity matrix,
but all of its jumps are close to the identity for large $k$. Indeed, due to
the sign of $\operatorname{Re}(2h+\lambda)$, all of the triangular matrices in
Figure 16 are exponentially close to the identity (and clearly conjugating by
the bounded matrix $\dot{\boldsymbol{P}}$ does not cause growth). Further, the
mismatch jumps on the boundaries of $\mathbb{D}_{a}$ and $\mathbb{D}_{b}$ have
been shown to go like $\mathbb{I}+\mathcal{O}\left(1/k\right)$ as
$k\to\infty$; see (60). The upshot is that, according to the theory of small-
norm Riemann-Hilbert problems, the following proposition is true.

######  Proposition 3.6.

Given that $x$ is in the pole-free region of the $x$-plane, as $k\to\infty$

(63) $\boldsymbol{E}(z;x,k)=\mathbb{I}+\mathcal{O}(1/k).$

####  3.9.2. Tracking the Error Term in the Extraction Formula

We recall the transformations and deformations we employed to arrive at
$\dot{\boldsymbol{P}}$.

(64)
${\boldsymbol{M}(z)}$${\boldsymbol{N}(z)}$${\boldsymbol{O}(z)}$${\boldsymbol{P}(z)}$${\dot{\boldsymbol{P}}(z)}$

We want to approximate $\boldsymbol{M}(z)$ in terms of the explicit function
$\dot{\boldsymbol{P}}(z)$. Towards this goal we start off by observing that
the transformations
$\boldsymbol{M}(z)\to\boldsymbol{N}(z)\to\boldsymbol{O}(z)$ all occur in a
bounded subset of the $z$-plane. Hence, for large enough $z$, we have
$\boldsymbol{M}(z)=\boldsymbol{N}(z)=\boldsymbol{O}(z)$. Thus, inverting (51),
we find for large $z$,

(65)
$\boldsymbol{M}(z)=e^{\frac{k\lambda}{2}\sigma_{3}}\boldsymbol{P}(z)e^{kg(z)\sigma_{3}}e^{-\frac{k\lambda}{2}\sigma_{3}}=e^{\frac{k\lambda}{2}\sigma_{3}}\boldsymbol{E}(z)\dot{\boldsymbol{P}}(z)e^{kg(z)\sigma_{3}}e^{-\frac{k\lambda}{2}\sigma_{3}}.$

Now, write the large-$z$ expansions for $\boldsymbol{E}(z)$ and
$\dot{\boldsymbol{P}}(z)$ as

(66)
$\boldsymbol{E}(z)=\mathbb{I}+\frac{\boldsymbol{E}_{-1}}{z}+\frac{\boldsymbol{E}_{-2}}{z^{2}}+\mathcal{O}\left(\frac{1}{z^{3}}\right)\hskip
7.11317pt\text{and}\hskip
7.11317pt\dot{\boldsymbol{P}}(z)=\mathbb{I}+\frac{\dot{\boldsymbol{P}}_{-1}}{z}+\frac{\dot{\boldsymbol{P}}_{-2}}{z^{2}}+\mathcal{O}\left(\frac{1}{z^{3}}\right).$

Also, using (36) we find there exists some $g_{-1}(x)\equiv
g_{-1}\in\mathbb{C}$ such that as, $z\to\infty$,

(67) $g(z)=\log(z)+\frac{g_{-1}}{z}+\mathcal{O}\left(1/z^{2}\right).$

This implies as, $z\to\infty$,

(68)
$e^{kg(z)\sigma_{3}}z^{-k\sigma_{3}}=\mathbb{I}+\frac{kg_{-1}\sigma_{3}}{z}+\begin{bmatrix}\mathcal{O}\left(1/z^{2}\right)&\mathcal{O}\left(1/z^{3}\right)\\\
\mathcal{O}\left(1/z^{3}\right)&\mathcal{O}\left(1/z^{2}\right)\end{bmatrix}.$

In view of (21), we extract $[\boldsymbol{M}_{-1}]_{22}$,
$[\boldsymbol{M}_{-1}]_{12}$, and $[\boldsymbol{M}_{-2}]_{12}$ and find

(69)
$\displaystyle\begin{split}[\boldsymbol{M}_{-1}]_{22}&=[\boldsymbol{E}_{-1}]_{22}+[\dot{\boldsymbol{P}}_{-1}]_{22}-kg_{-1},\\\
[\boldsymbol{M}_{-1}]_{12}&=\left([\boldsymbol{E}_{-1}]_{12}+[\dot{\boldsymbol{P}}_{-1}]_{12}\right)e^{k\lambda},\\\
[\boldsymbol{M}_{-2}]_{12}&=e^{k\lambda}\Big{(}[\boldsymbol{E}_{-1}]_{11}[\dot{\boldsymbol{P}}_{-1}]_{12}+[\boldsymbol{E}_{-1}]_{12}[\dot{\boldsymbol{P}}_{-1}]_{22}-\\\
&\hskip
64.01869pt\left([\boldsymbol{E}_{-1}]_{12}+[\dot{\boldsymbol{P}}_{-1}]_{12}\right)kg_{-1}+[\boldsymbol{E}_{-2}]_{12}+[\dot{\boldsymbol{P}}_{-2}]_{22}\Big{)}.\end{split}$

Therefore,

(70)
$p_{k}\left(\left(\frac{k}{2}\right)^{2/3}x\right)=2i\left(\frac{k}{2}\right)^{1/3}\Bigg{(}[\boldsymbol{E}_{-1}]_{22}+[\dot{\boldsymbol{P}}_{-1}]_{22}-\\\
\frac{[\boldsymbol{E}_{-1}]_{11}[\dot{\boldsymbol{P}}_{-1}]_{12}+[\boldsymbol{E}_{-1}]_{12}[\dot{\boldsymbol{P}}_{-1}]_{22}+[\boldsymbol{E}_{-2}]_{12}+[\dot{\boldsymbol{P}}_{-2}]_{12}}{[\boldsymbol{E}_{-1}]_{12}+[\dot{\boldsymbol{P}}_{-1}]_{12}}\Bigg{)}.$

A direct calculation finds that

(71)
$[\dot{\boldsymbol{P}}(z)]_{22}=\frac{1}{2}\left(\beta(z)+\frac{1}{\beta(z)}\right)=1+\mathcal{O}\left(z^{-2}\right)$

and

(72)
$[\dot{\boldsymbol{P}}(z)]_{12}=\frac{i}{2}\left(\beta(z)-\frac{1}{\beta(z)}\right)=\frac{i(a-b)}{4z}+\frac{i(a^{2}-b^{2})}{8z^{2}}+\mathcal{O}\left(z^{-3}\right).$

We are now ready to prove Theorem 1.1.

###### Proof of Theorem 1.1.

Using (63), we can write (70) as

(73)
$p_{k}\left(\left(\frac{k}{2}\right)^{2/3}x\right)=2i\left(\frac{k}{2}\right)^{1/3}\left([\dot{\boldsymbol{P}}_{-1}(x)]_{22}-\frac{[\dot{\boldsymbol{P}}_{-2}(x)]_{12}+\mathcal{O}\left(1/k\right)}{[\dot{\boldsymbol{P}}_{-1}(x)]_{12}+\mathcal{O}\left(1/k\right)}\right).$

Further, looking at (72) we have that

(74)
$[\dot{\boldsymbol{P}}_{-1}(x)]_{12}=\frac{i(a(x)-b(x))}{4}=-\frac{i\Delta(x)}{4}.$

Notice that $\Delta(x)=\sqrt{-4i/S(x)}$ never vanishes in the pole-free region
of the $x$-plane because $S(x)$ is analytic (and hence pole free) in the pole-
free region. Therefore, the quotient in (74) never blows up and we can write

(75)
$p_{k}\left(\left(\frac{k}{2}\right)^{2/3}x\right)=2i\left(\frac{k}{2}\right)^{1/3}\left([\dot{\boldsymbol{P}}_{-1}(x)]_{22}-\frac{[\dot{\boldsymbol{P}}_{-2}(x)]_{12}}{[\dot{\boldsymbol{P}}_{-1}(x)]_{12}}+\mathcal{O}(1/k)\right).$

Now, using (71) and (72) to express the right-hand side of (75) in terms of
$S$ we find

(76)
$p_{k}\left(\left(\frac{k}{2}\right)^{2/3}x\right)=-2i\left(\frac{k}{2}\right)^{1/3}\left(\frac{S(x)}{2}+\mathcal{O}(1/k)\right).$

In view of (14) and (17) we obtain,

(77)
$-(2k)^{-1/3}u^{(\alpha)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=-i\frac{S(x)}{2}+\mathcal{O}\left(1/k\right),$

our desired result. ∎

##  4\. Analysis in the Pole Region

We now begin our analysis in the pole region.

###  4.1. The G-function

Motivated by the bifurcation of the zero-level set of
$\operatorname{Re}(2h+\lambda)$ as $x$ leaves the pole-free region (see Figure
12), we aim to deform Riemann-Hilbert Problem 2 so that its jump matrices
decay to the identity as $k\to\infty$ along all of its jump contours except
for two bands and one gap (instead of one band as in the pole-free case). As
we will have two bands, we will have four endpoints: $A(x)\equiv A$,
$B(x)\equiv B$, $C(x)\equiv C$, and $D(x)\equiv D$. The normalization of the
soon-to-be-defined $G$-function will be dictated by the endpoints. In order to
enforce our desired normalization we require

(78) $\displaystyle\begin{split}\mathfrak{s}_{1}&:=A+B+C+D=0,\\\
\mathfrak{s}_{2}&:=AB+AC+AD+BC+BD+CD=x/2,\\\
\mathfrak{s}_{3}&:=ABC+ACD+BCD=-i.\end{split}$

These conditions give us six real equations to determine the endpoints. The
remaining two real equations will be derived from what are commonly known as
the Boutroux conditions. To state these conditions we need to introduce the
function $R(z;x)$. Temporarily set $\Sigma_{1}$ as the line segment with
endpoints $A$ and $B$, $\Gamma$ as the line segment with endpoints $B$ and
$C$, and $\Sigma_{2}$ as the line segment with endpoints $C$ and $D$. Now, let
$R(z;x)\equiv R(z)$ denote the function analytic for
$z\in\mathbb{C}\setminus(\Sigma_{1}\cup\Sigma_{2})$ that satisfies the
conditions

(79) $R(z)^{2}=(z-A)(z-B)(z-C)(z-D)\hskip 9.38945pt\text{and}\hskip
9.38945ptR(z)=z^{2}+\mathcal{O}(z)\text{ as }z\to\infty.$

Also, the moment conditions (78) imply $R(z)$ has the large-$z$ expansion

(80) $R(z)=z^{2}+\frac{x}{4}+\frac{i}{2z}+\mathcal{O}\left(1/z^{2}\right).$

The Boutroux conditions are

(81) $\text{Im}\left(\int\limits_{\Sigma_{1}}R_{+}(w)\text{d}w\right)=0\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{Im}\left(\int\limits_{\Gamma}R(w)\text{d}w\right)=0.$

####  4.1.1. Definition and Properties of the G-function

As in the pole-free case, we will define the $G$-function as an
antiderivative. Let

(82) $G^{\prime}(z;x)\equiv G^{\prime}(z)=\frac{R(z)}{2\pi
i}\int\limits_{\Sigma_{1}\cup\Sigma_{2}}\frac{i\theta^{\prime}(w)}{R_{+}(w)(w-z)}\text{d}w,$

where the path of integration on $\Sigma_{1}\cup\Sigma_{2}$ goes from $A$ to
$B$ and then from $C$ to $D$. The moment conditions (78) guarantee that
$G^{\prime}(z)=1/z+\mathcal{O}\left(1/z^{2}\right)$ as $z\to\infty$. Thus,
$G^{\prime}(z)-1/(z-A)$ is integrable at $z=\infty$. Integrating the latter
expression introduces a logarithmic cut. Let $L$ denote an oriented, unbounded
arc from $z=\infty$ to $z=A$. We further require that $L$ only intersects
$\Sigma_{1}$ at $z=A$, avoids $\Gamma\cup\Sigma_{2}$, and coincides with the
negative real numbers for large enough $z$. We now define $G(z)$ as an
integral where the path of integration can be any path that does not cross
$\Sigma_{1}\cup\Gamma\cup\Sigma_{2}$. Set

(83) $G(z;x)\equiv
G(z):=\log(z-A)+\int_{\infty}^{z}G^{\prime}(w)-\frac{1}{w-A}\text{d}w.$

For notational convenience, we write

(84) $H(z;x)\equiv H(z):=\frac{i\theta(z)}{2}-G(z).$

Analogously to the pole-free case, the zero-level lines of
$\operatorname{Re}\left(2H(z)+\Lambda\right)$ are independent of the placement
of the curves $\Sigma_{1}$, $\Gamma$, and $\Sigma_{2}$ given the three curves
are finite and simple, and $\Sigma_{1}$ connects $A$ to $B$, $\Gamma$ connects
$B$ to $C$, and $\Sigma_{2}$ connects $C$ to $D$. It is convenient to make the
following selections:

  * •

Let $\Sigma_{1}$ coincide with the zero-level arc of
$\operatorname{Re}\left(2H+\Lambda\right)$ that connects $A$ to $B$.

  * •

Let $\Gamma$ coincide with the zero-level arc of
$\operatorname{Re}\left(2H+\Lambda\right)$ that connects $B$ to $C$.

  * •

Let $\Sigma_{2}$ coincide with the zero-level arc of
$\operatorname{Re}\left(2H+\Lambda\right)$ that connects $C$ to $D$.

In the same vein as in the pole-free case, the leading error in our
approximation will propagate from the endpoints. As such, we will need to do
local analysis at each endpoint. For each $p\in\\{A,B,C,D\\}$, let
$\mathbb{D}_{p}$ denote a sufficiently small disk around $z=p$. For the points
$A,B,C$ we will need to further specify certain regions of $\mathbb{D}_{p}$.
In particular, we make the following partitions (also see Figure 17):

  * •

For $\mathbb{D}_{A}$, notice $L\cup\Sigma_{1}$ divides $\mathbb{D}_{A}$ into
two open components. Let $\mathbb{D}^{\uparrow}_{A}$ denote the open component
that is on the left side of $L\cup\Sigma_{1}$, and set
$\mathbb{D}_{A}^{\downarrow}:=\mathbb{D}_{A}\setminus\overline{\mathbb{D}^{\uparrow}_{A}}$.

  * •

For $\mathbb{D}_{B}$, notice $\Sigma_{1}\cup\Gamma$ divides $\mathbb{D}_{B}$
into two open components (recall that $\Sigma_{1}$ and
$\widetilde{\Sigma}_{1}$ are the same contours with opposite orientations).
Let $\mathbb{D}^{\uparrow}_{B}$ denote the open component that is on the left
side of $\Sigma_{1}\cup\Gamma$, and set
$\mathbb{D}^{\downarrow}_{B}:=\mathbb{D}_{B}\setminus\overline{\mathbb{D}^{\uparrow}_{B}}$.

  * •

For $\mathbb{D}_{C}$, notice $\Gamma\cup\Sigma_{2}$ divides $\mathbb{D}_{C}$
into two components. Let $\mathbb{D}_{C}^{\uparrow}$ denote the open component
that is on the left side of $\Gamma\cup\Sigma_{2}$, and set
$\mathbb{D}_{C}^{\downarrow}:=\mathbb{D}_{C}\setminus\overline{\mathbb{D}^{\uparrow}_{C}}$.

We now list the basic properties of $G$ and $H$ that we will use throughout
the Riemann-Hilbert analysis.

######  Proposition 4.1.

For each $x$ in the pole region of the $x$-plane, given that the moment
conditions (78) and the Boutroux conditions (81) are satisfied, guaranteeing
that $G(z)$ and $H(z)$ are well-defined by (83) and (84) respectively, then
the following statements hold:

  * •

There exists a complex number $\Lambda(x)\equiv\Lambda$ such that

(85) $H_{+}(z)+H_{-}(z)=i\theta(z)-G_{+}(z)-G_{-}(z)=-\Lambda\hskip
14.22636pt\textit{for each }z\in\Sigma_{1}.$

  * •

There exists a real number $\omega(x)\equiv\omega\in\mathbb{R}$ such that

(86) $H_{+}-H_{-}=-(G_{+}-G_{-})=-i\omega\hskip 14.22636pt\textit{for each
}z\in\Gamma.$

  * •

There exists a real number $\Omega(x)\equiv\Omega\in\mathbb{R}$ such that

(87) $H_{+}(z)+H_{-}(z)=i\theta(z)-G_{+}(z)-G_{-}(z)=-\Lambda-i\Omega\hskip
14.22636pt\textit{for each }z\in\Sigma_{2}.$

  * •

On the logarithmic cut $L$,

(88) $G_{+}(z)-G_{-}(z)=-2\pi i.$

  * •

As $z\to\infty$,

(89) $G(z)=\log(z)+\mathcal{O}(1/z).$

  * •

At each endpoint, under the right perturbation, $2H(z)+i\Lambda$ locally looks
like a $3/2$-root function. More precisely, if we consider the functions

(90) $J_{A}(z):=\begin{cases}2H(z)+\Lambda-2\pi
i,&z\in\mathbb{D}_{A}^{\uparrow},\\\ 2H(z)+\Lambda+2\pi
i,&z\in\mathbb{D}_{A}^{\downarrow},\end{cases}$ (91)
$J_{B}(z):=\begin{cases}2H(z)+\Lambda+i\omega,&z\in\mathbb{D}^{\uparrow}_{B},\\\
2H(z)+\Lambda-i\omega,&z\in\mathbb{D}^{\downarrow}_{B},\end{cases}$ (92)
$J_{C}(z):=\begin{cases}2H(z)+\Lambda+i\omega+i\Omega,&z\in\mathbb{D}^{\uparrow}_{C},\\\
2H(z)+\Lambda-i\omega+i\Omega&z\in\mathbb{D}^{\downarrow}_{C},\end{cases}$

and

(93) $J_{D}(z):=2H(z)+\Lambda+i\Omega,$

then for each endpoint $p\in\left\\{A,B,C,D\right\\}$ there exists
$\kappa_{p}\in\mathbb{C}$ such that

(94) $J_{p}(z)=\kappa_{p}(z-p)^{3/2}+\mathcal{O}\left((z-p)^{5/2}\right)\hskip
23.84923pt\text{for }z\in\mathbb{D}_{p}.$

These statements are standard and can be proved by direct computation.

###  4.2. Preliminary Deformations in the Riemann-Hilbert Analysis

As in the pole-free analysis, we deform Riemann-Hilbert Problem 2 in
preparation for introducing the $G$-function. These deformations are motivated
by the sign chart of $\operatorname{Re}(2H+\Lambda)$; see Figure 19. To
specify the exact placement of the deformed jump contours, we will construct
conformal maps at each endpoint. Then, we will collapse contours and open
lenses.

####  4.2.1. Conformal Map Constructions

Using the same argument, as in Section 3.4.1 and Proposition 4.1, we are able
to construct conformal maps $\tau_{A}$, $\tau_{B}$, $\tau_{C}$, and $\tau_{D}$
according to the table below.

$p$ | Conformal Map $\tau_{p}:\mathbb{D}_{p}\to\mathbb{C}$ | $\tau_{p}^{-1}((-\infty,0])\cap\mathbb{D}_{p}$
---|---|---
$z=A$ | $\tau_{A}(z)^{3}=\begin{cases}\left(2H(z)+\Lambda-2\pi i\right)^{2},&z\in\mathbb{D}_{A}^{\uparrow},\\\ \left(2H(z)+\Lambda+2\pi i\right)^{2},&z\in\mathbb{D}_{A}^{\downarrow},\end{cases}$ | $\Sigma_{1}\cap\mathbb{D}_{A}$
$z=B$ | $\tau_{B}(z)^{3}=\begin{cases}(2H(z)+\Lambda+i\omega)^{2},&z\in\mathbb{D}^{\uparrow}_{B},\\\ (2H(z)+\Lambda-i\omega)^{2},&z\in\mathbb{D}^{\downarrow}_{B},\end{cases}$ | $\Sigma_{1}\cap\mathbb{D}_{B}$
$z=C$ | $\tau_{C}(z)^{3}=\begin{cases}(2H(z)+\Lambda+i\omega+i\Omega)^{2},&z\in\mathbb{D}^{\uparrow}_{C},\\\ (2H(z)+\Lambda-i\omega+i\Omega)^{2}&z\in\mathbb{D}^{\downarrow}_{C},\end{cases}$ | $\Gamma\cap\mathbb{D}_{C}$
$z=D$ | $\tau_{D}(z)^{3}=(2H(z)+\Lambda+i\Omega)^{2}$ | $\Sigma_{2}\cap\mathbb{D}_{D}$
Table 2. Conformal maps used in the pole region. $\Sigma_{1}$$L$$A$$\mathbb{D}_{A}^{\downarrow}$$\mathbb{D}_{A}^{\uparrow}$ | fl |  $\widetilde{\Sigma_{1}}$$\Gamma$$B$$\mathbb{D}_{B}^{\downarrow}$$\mathbb{D}_{B}^{\uparrow}$ | fl |  $\widetilde{\Gamma}$$\Sigma_{2}$$C$$\mathbb{D}_{C}^{\downarrow}$$\mathbb{D}_{C}^{\uparrow}$
---|---|---|---|---
Figure 17. The partitions of $\mathbb{D}_{A}$, $\mathbb{D}_{B}$, and
$\mathbb{D}_{C}$. Here, $\widetilde{\Sigma_{1}}$ denotes the same contour
$\Sigma_{1}$ but with reverse orientation. $\widetilde{\Gamma}$ is defined
analogously.

####  4.2.2. The Collapsing Deformation

We collapse the jump contours of $\boldsymbol{M}(z)$ according to Figure 18.
We make the following specifications:

  * •

The jump contour that connects $A$ to $B$ coincides with $\Sigma_{1}$.

  * •

In $\mathbb{D}_{A}$, the jump contour where the jump is upper triangular is
contained in $\tau_{A}^{-1}((0,\infty))$.

  * •

In $\mathbb{D}_{A}$, the jump contour where the jump is lower triangular is
contained in $\tau_{B}^{-1}(\rho e^{2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{B}$, the jump contour where the jump is upper triangular is
contained in $\tau_{B}^{-1}((0,\infty))$ for $\rho>0$.

  * •

In $\mathbb{D}_{B}$, the jump contour that connects $B$ to $C$ is contained in
$\tau_{B}^{-1}(\rho e^{3\pi i/2})$ for $\rho>0$.

  * •

The jump contour that connects $C$ to $D$ coincides with $\Sigma_{2}$.

  * •

In $\mathbb{D}_{C}$, the jump contour that connects $B$ to $C$ is contained in
$\tau_{C}^{-1}(\rho e^{2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{D}$, the jump contour that emanates from $D$ and goes to
$\infty$ is contained $\tau_{D}^{-1}((0,\infty))$.

  * •

For all other jumps contours, we only require that they avoid the zero-level
set of $\operatorname{Re}(2H+\Lambda)$ and do not self-intersect.

Let $\boldsymbol{N}(z)$ denote the solution of the Riemann-Hilbert problem
obtained from the collapsing deformation. More precisely:

Riemann-Hilbert Problem 8:  For each $k\in\mathbb{N}$, find
$\boldsymbol{N}(z;x,k)\equiv\boldsymbol{N}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{N}(z)$ is analytic in $z$ except along the jump
contours shown in Figure 18.

  2. (2)

Jump Condition: $\boldsymbol{N}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{N}(z)$ are related by
the jump condition
$\boldsymbol{N}_{+}(z)=\boldsymbol{N}_{-}(z)\boldsymbol{V}^{(\boldsymbol{N})}(z)$
as shown in Figure 18.

$A$$B$$C$$D$$\boldsymbol{L}_{k}^{-1}(i\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&1\end{bmatrix}$$\boldsymbol{U}_{k}^{-1}(-i\theta)$$\boldsymbol{L}_{k}(i\theta)$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{1}}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{2}}$

Figure 18. The jumps of $\boldsymbol{N}(z)$ in the pole case.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{N}(z)z^{-k\sigma_{3}}=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

Figure 19. The sign chart for $\operatorname{Re}(2H+\Lambda)$ when
$x=-3/2-10i$. The dashed blue-and-orange lines indicate that a zero-level line
is also a cut of $\operatorname{Re}(2H+\Lambda)$.

####  4.2.3. Opening Lenses

We now open lenses on $\Sigma_{1}$ and $\Sigma_{2}$ guided by the sign chart
for $\operatorname{Re}\left(2H+\Lambda\right)$ (see Figure 19).

On $\Sigma_{1}$, we use the same factorization we used in the pole-free
analysis; see (50). The jump matrix on $\Sigma_{2}$ has the factorization

(95) $\begin{bmatrix}1&0\\\
e^{ik\theta}&1\end{bmatrix}=\begin{bmatrix}1&e^{-ik\theta}\\\
0&1\end{bmatrix}\begin{bmatrix}0&-e^{-ik\theta}\\\
e^{ik\theta}&0\end{bmatrix}\begin{bmatrix}1&e^{-ik\theta}\\\
0&1\end{bmatrix}.$

We open three lenses according to Figure 20. We make the following
specifications:

  * •

In $\mathbb{D}_{A}$, the boundary of the lens that is on the right side of
$\Sigma_{1}$ is contained in $\tau^{-1}_{A}(\rho e^{2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{B}$, the boundary of the lens that is on the right side of
$\Sigma_{1}$ is contained in $\tau_{B}^{-1}(\rho e^{-2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{C}$, the boundary of the lens that is on the right side of
$\Sigma_{2}$ is contained in $\tau_{C}^{-1}(\rho e^{-2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{C}$, the boundary of the lens that is on the left side of
$\Sigma_{2}$ is contained in $\tau_{C}^{-1}((0,\infty))$.

  * •

In $\mathbb{D}_{D}$, the boundary of the lens that is on the right side of
$\Sigma_{2}$ is contained in $\tau_{D}^{-1}(\rho e^{-2\pi i/3})$ for $\rho>0$.

  * •

In $\mathbb{D}_{D}$, the boundary of the lens that is on the left side of
$\Sigma_{2}$ is contained in $\tau_{D}^{-1}(\rho e^{2\pi i/3})$ for $\rho>0$.

Let $\boldsymbol{O}(z)$ denote the solution of the Riemann-Hilbert problem
obtained from the opening lenses deformation. Specifically:

Riemann-Hilbert Problem 9: Determine the $2\times 2$ matrix
$\boldsymbol{O}(z;x,k)\equiv\boldsymbol{O}(z)$ with the following properties:

  1. (1)

Analyticity: $\boldsymbol{O}(z)$ is analytic in $z$ except on the jump
contours shown in Figure 20.

  2. (2)

Jump Condition: The boundary values taken by $\boldsymbol{O}(z)$ are related
by the jump condition
$\boldsymbol{O}_{+}(z)=\boldsymbol{O}_{-}(z)\boldsymbol{V}^{(\boldsymbol{O})}(z)$
as shown in Figure 20.

$A$$B$$C$$D$$\boldsymbol{L}_{k}^{-1}(i\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\boldsymbol{L}_{k}^{-1}(i\theta)$$\boldsymbol{T}(e^{ik\theta})$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{1}}$$\boldsymbol{U}_{k}^{-1}(-i\theta)$$\boldsymbol{L}_{k}(i\theta)$$\boldsymbol{L}_{k}(i\theta)$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{2}}$$\boldsymbol{T}(ik\theta)$$\boldsymbol{U}_{k}(-i\theta)$$\boldsymbol{U}_{k}(-i\theta)$

Figure 20. The jumps of $\boldsymbol{O}(z)$ in the pole case.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{O}(z)z^{-k\sigma_{3}}=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

.

###  4.3. Applying the G-function.

Analogously to the pole-free case, we apply the $G$-function defined in
Section 4.1 (see (83)) to the function $\boldsymbol{O}(z)$. We consider

(96)
$\boldsymbol{P}(z;x,k)\equiv\boldsymbol{P}(z):=e^{-\frac{k\Lambda}{2}\sigma_{3}}\boldsymbol{O}(z)e^{-kG(z)\sigma_{3}}e^{\frac{k\Lambda}{2}\sigma_{3}}.$

Then, $\boldsymbol{P}(z)$ is analytic where both $\boldsymbol{O}(z)$ and
$e^{G(z)}$ are (that is, for
$z\in\mathbb{C}\setminus(\Sigma_{1}\cup\Gamma\cup\Sigma_{2})$). Hence, we have
introduced a jump on $\Gamma$. In particular,
$\boldsymbol{V}^{(\boldsymbol{P})}_{\Gamma}=e^{-ki\omega\sigma_{3}}$, where
$\omega$ was introduced in (86). For compensation, both the normalization and
jump conditions for $\boldsymbol{P}(z)$ are now simplified. Indeed, for large
$z$, $\boldsymbol{P}(z)$ converges to $\mathbb{I}$ instead of
$z^{k\sigma_{3}}$.

The jump recipe (52) is valid after replacing the lowercase symbols ‘$g$’ and
‘$\lambda$’ with their corresponding uppercase counterparts. From this, we
glean that the nontrivial exponential entry of the triangular jump matrices is
transformed by the rule $e^{\pm ik\theta}\Rightarrow e^{\pm k(2H+\Lambda)}$.
Further, the sign chart of $\operatorname{Re}(2H+\Lambda)$ implies that all of
the triangular jump matrices for $\boldsymbol{P}(z)$ decay to the identity as
$k\to\infty$. Moreover, the twist jump matrices on $\Sigma_{1}$ and
$\Sigma_{2}$ become constant. Using the jump recipe, (85), and (87), we find

(97) $\boldsymbol{V}^{(\boldsymbol{P})}_{\Sigma_{1}}=\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}\hskip 37.84221pt\text{ and }\hskip
37.84221pt\boldsymbol{V}^{(\boldsymbol{P})}_{\Sigma_{2}}=\begin{bmatrix}0&-e^{ki\Omega}\\\
e^{-ki\Omega}&0\end{bmatrix}.$

Therefore the Riemann-Hilbert problem $\boldsymbol{P}(z)$ solves is:

Riemann-Hilbert Problem 10:  Determine the $2\times 2$ matrix
$\boldsymbol{P}(z;x,k)\equiv\boldsymbol{P}(z)$ with the following properties:

  1. (1)

Analyticity: $\boldsymbol{P}(z)$ is analytic in $z$ except on the jump
contours shown in Figure 21.

  2. (2)

Jump Condition: The boundary values taken by $\boldsymbol{P}(z)$ are related
by the jump condition
$\boldsymbol{P}_{+}(z)=\boldsymbol{P}_{-}(z)\boldsymbol{V}^{(\boldsymbol{P})}(z)$
as shown in Figure 21.

$A$$B$$C$$D$$\boldsymbol{L}_{k}^{-1}$$\boldsymbol{U}_{k}$$\boldsymbol{L}^{-1}_{k}$$\boldsymbol{T}(1)$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{1}}$$\boldsymbol{U}_{k}^{-1}$$\boldsymbol{L}_{k}$$\boldsymbol{L}_{k}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Sigma_{2}}$$\boldsymbol{T}(e^{-ik\Omega})$$\boldsymbol{U}_{k}$$\boldsymbol{U}_{k}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Gamma}$$e^{-ik\omega\sigma_{3}}$

Figure 21. The jumps of $\boldsymbol{P}(z)$ for $x$ in the pole region. Here,
if the jump matrix is upper triangular then its exponent is $-(2H+\Lambda)$,
while if the jump matrix is lower triangular then its exponent is
$2H+\Lambda$.

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{P}(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

###  4.4. Constructing the Outer Parametrix

In this section we explicitly construct the outer parametrix for the pole
region. See [2] for a similar construction. We start by introducing the
Riemann-Hilbert problem corresponding to the outer parametrix. Next, we deform
the Riemann-Hilbert problem so that both of its jumps on $\Sigma_{1}$ and
$\Sigma_{2}$ are the twist matrix $\boldsymbol{T}(1)$. Then, we recall and
implement the necessary machinery (the Abel map, the Riemann
$\Theta$-function, and holomorphic differentials) in order to construct a
$2\times 2$ matrix-valued function whose jumps on $\Sigma_{1}\cup\Sigma_{2}$
ultimately amounts to switching the columns. Finally, we use algebraic
functions to complete our construction of the outer parametrix.

####  4.4.1. The Riemann-Hilbert Problem for the Outer Parametrix

Disregarding the jumps of $\boldsymbol{P}(z)$ that decay to the identity
matrix we obtain the following Riemann-Hilbert problem:

Riemann-Hilbert Problem 11:  For each $k\in\mathbb{N}$, find the $2\times 2$
matrix-valued function
$\dot{\boldsymbol{P}}^{\text{(out)}}(z;x,k)\equiv\dot{\boldsymbol{P}}^{\text{(out)}}(z)$
that satisfies the following properties:

  1. (1)

Analyticity: $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ is analytic in the
complex $z$-plane except on $\Sigma_{1}\cup\Gamma\cup\Sigma_{2}$.

  2. (2)

Jump Condition: The jumps of $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ are
topologically equivalent to those shown in Figure 22.

$\begin{bmatrix}0&-1\\\ 1&0\end{bmatrix}$$B$$\begin{bmatrix}e^{-ik\omega}&0\\\
0&e^{ik\omega}\end{bmatrix}$$C$$\begin{bmatrix}0&-e^{ik\Omega}\\\
e^{-ik\Omega}&0\end{bmatrix}$$A$$D$

Figure 22. A topologically equivalent version of the jump contours for
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$. Here, $\omega,\Omega\in\mathbb{R}$,
and $\Sigma_{1}$ corresponds to the line segment $AB$, $\Gamma$ corresponds to
the line segment $BC$, and $\Sigma_{2}$ corresponds to the line segment $CD$.

  3. (3)

Normalization: As $z\to\infty$,
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)=\mathbb{I}+\mathcal{O}(1/z)$.

Our next step is to move the dependence on $\omega$ and $\Omega$ from the jump
matrices to the normalization condition. We consider

(98) $F(z;x)\equiv F(z):=\frac{R(z)}{2\pi
i}\left(-\int\limits_{\Gamma}\frac{i\omega}{R(w)(w-z)}\text{d}w-\int\limits_{\Sigma_{2}}\frac{i\Omega}{R_{+}(w)(w-z)}\text{d}w\right).$

Notice $F$ satisfies the following jumps:

  * •

For $z\in\Sigma_{1}$, $F_{+}(z)+F_{-}(z)=0$.

  * •

For $z\in\Gamma$, $F_{+}(z)-F_{-}(z)=-i\omega$.

  * •

For $z\in\Sigma_{2}$, $F_{+}(z)+F_{-}(z)=-i\Omega$.

Also, as $z\to\infty$, $F(z)=F_{1}z+F_{0}+\mathcal{O}(1/z)$, where

(99) $F_{1}:=\frac{1}{2\pi
i}\left(\int\limits_{\Gamma}\frac{i\omega}{R(w)}\text{d}w+\int\limits_{\Sigma_{2}}\frac{i\Omega}{R_{+}(w)}\text{d}w\right)$

and

(100) $F_{0}:=\frac{1}{2\pi i}\left(\int\limits_{\Gamma}\frac{i\omega
w}{R(w)}\text{d}w+\int\limits_{\Sigma_{2}}\frac{i\Omega
w}{R_{+}(w)}\text{d}w\right).$

Set

(101)
$\dot{\boldsymbol{Q}}(z;x,k)\equiv\dot{\boldsymbol{Q}}(z):=e^{kF_{0}\sigma_{3}}\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-kF(z)\sigma_{3}}.$

Then $\dot{\boldsymbol{Q}}$ solves the following Riemann-Hilbert problem:

Riemann-Hilbert Problem 12:  For each $k\in\mathbb{N}$, find the $2\times 2$
matrix-valued function
$\dot{\boldsymbol{Q}}(z;x,k)\equiv\dot{\boldsymbol{Q}}(z)$ that satisfies the
following properties:

  1. (1)

Analyticity: $\dot{\boldsymbol{Q}}(z)$ is analytic in the complex $z$-plane
except on $\Sigma_{1}\cup\Sigma_{2}$.

  2. (2)

Jump Condition: The boundary values taken by $\dot{\boldsymbol{Q}}$ on
$\Sigma_{1}\cup\Sigma_{2}$ are related by the jump condition
$\dot{\boldsymbol{Q}}_{+}=\dot{\boldsymbol{Q}}_{-}\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}$.

  3. (3)

Normalization: As $z\to\infty$,

(102)
$\dot{\boldsymbol{Q}}(z)e^{kF_{1}z\sigma_{3}}=\mathbb{I}+\frac{\dot{\boldsymbol{Q}}_{-1}}{z}+\frac{\dot{\boldsymbol{Q}}_{-2}}{z^{2}}+\mathcal{O}(1/z^{3}).$

Notice we eliminated the jump on $\Gamma$.

####  4.4.2. The Baker-Akhiezer Map

Recall the scalar function $R(z)$ is cut on $\Sigma_{1}\cup\Sigma_{2}$. We
consider $\mathbb{T}$, the genus-one Riemann surface defined by $R(z)$. For
our purposes it is convenient to view $\mathbb{T}$ as two copies of the
complex $z$-plane cut on $\Sigma_{1}\cup\Sigma_{2}$. Given a complex point
$z$, we relate the two copies or sheets by the identity
$R\left(z^{+}\right)=-R\left(z^{-}\right)$. We say $z^{+}$ lives on the plus
sheet of $\mathbb{T}$, while $z^{-}$ lives on the minus sheet of $\mathbb{T}$.
In our discussion unless otherwise specified we assume the inputs of
$\mathcal{A}(\cdot)$ are on the plus sheet. We now introduce a basis of
homology cycles $\\{\mathfrak{a},\mathfrak{b}\\}$ on $\mathbb{T}$ as shown in
Figure 23. Here integration on the second sheet is accomplished by replacing
$R(z)$ by $-R(z)$.

$B$$A$$C$$D$$\mathfrak{a}$$\mathfrak{b}$

Figure 23.  The homology basis $\left\\{\mathfrak{a},\mathfrak{b}\right\\}$ of
$\mathbb{T}$. Thick solid curves lie on the plus sheet while the dotted curve
lies on the minus sheet.

As $\\{\mathfrak{a},\mathfrak{b}\\}$ is a homology basis, we see that any map
defined via path integration on this genus-one surface will be well-defined
modulo the periods defined by integrating along $\mathfrak{a}$ and
$\mathfrak{b}$. In particular, in order for the Abel map,

(103) $\mathcal{A}(z;x)\equiv\mathcal{A}(z):=\frac{2\pi
i}{\ointctrclockwise_{\mathfrak{a}}\frac{\text{d}w}{R(w)}}\int_{A}^{z}\frac{\text{d}w}{R(w)},$

to be well defined we mod out the value obtained from integrating along
$\mathfrak{a}$,

(104) $\frac{2\pi
i}{\ointctrclockwise_{\mathfrak{a}}\frac{\text{d}w}{R(w)}}\ointctrclockwise_{\mathfrak{a}}\frac{\text{d}w}{R(w)}=2\pi
i,$

and the value obtained from integrating along $\mathfrak{b}$,

(105) $\mathcal{B}(x)\equiv\mathcal{B}:=\frac{2\pi
i}{\ointctrclockwise_{\mathfrak{a}}\frac{\text{d}w}{R(w)}}\ointclockwise_{\mathfrak{b}}\frac{\text{d}w}{R(w)}.$

We now list the basic properties of $\mathcal{A}(z)$.

######  Proposition 4.2.

Given the lattice $\mathcal{L}:=\left\\{j_{1}2\pi
i+j_{2}\mathcal{B}:j_{1},j_{2}\in\mathbb{Z}\right\\}$, the boundary values
taken by the Abel map, $\mathcal{A}(z)$, on the cuts on the plus sheet are
related by the jump conditions

(106)
$\mathcal{A}_{+}\left(z^{+}\right)-\mathcal{A}_{-}\left(z^{+}\right)=0\mod\mathcal{L}\hskip
28.45274pt\text{for }z^{+}\in\Sigma_{1},$ (107)
$\mathcal{A}_{+}\left(z^{+}\right)-\mathcal{A}_{-}\left(z^{+}\right)=-2\pi
i\mod\mathcal{L}\hskip 28.45274pt\text{for }z^{+}\in\Gamma,$

and

(108)
$\mathcal{A}_{+}\left(z^{+}\right)-\mathcal{A}_{-}\left(z^{+}\right)=-\mathcal{B}\mod\mathcal{L}\hskip
28.45274pt\text{for }z^{+}\in\Sigma_{2}.$

Also, as $z\to\infty$ on the plus sheet,

(109)
$\mathcal{A}(z)=\mathcal{A}(\infty)+\frac{\mathcal{A}_{-1}}{z}+\mathcal{O}\left(1/z^{2}\right),\hskip
28.45274pt\text{where }\mathcal{A}_{-1}(x)\equiv\mathcal{A}_{-1}:=-\frac{2\pi
i}{\oint_{\mathfrak{a}}\frac{\text{d}w}{R(w)}}.$

Next, we use $\mathcal{A}(z)$ to construct a function (the Baker-Akhiezer
function), defined on $\mathbb{T}$, that is path independent without modding.
To do this, we employ Riemann $\Theta$-functions. Recall, given
$M\in\mathbb{C}$, with $\operatorname{Re}(M)<0$, then for each
$z\in\mathbb{C}$,

(110)
$\Theta(z;M)\equiv\Theta(z):=\sum\limits_{k\in\mathbb{Z}}\exp\left(kz+\frac{1}{2}Mk^{2}\right).$

It is well known (see [18]) that $\Theta(z;M)$ satisfies the following
properties:

  * •

Periodicity: $\Theta(z+2\pi i)=\Theta(z)$.

  * •

Quasiperiodicity: $\Theta(z+M;M)=e^{-\frac{M}{2}}e^{-z}\Theta(z;M)$. More
generally, for $n\in\mathbb{Z}$
$\Theta(z+nM;M)=e^{-Mn^{2}/2}e^{-nz}\Theta(z;M)$.

  * •

Evenness: $\Theta(-z)=\Theta(z)$.

  * •

All of the zeros of $\Theta$ are simple, and $\Theta(z_{0})=0$ if and only if
there exist integers $j_{1},j_{2}$ such that

(111) $z_{0}=\left(\pi i+M/2\right)+j_{1}2\pi i+j_{2}M.$

We see that the $\Theta$-function gives us a handle on our goal of
constructing a well-defined function on $\mathbb{T}$ from $\mathcal{A}(z)$.
Naturally, we consider $\Theta(z;\mathcal{B})$. Notice by using the $2\pi
i$-periodicity of the $\Theta$-function, $\Theta\left(\mathcal{A}(z)\right)$
is independent of the number of $\mathfrak{a}$-cycles used in the path that
connects $A$ to $z$ (or any cycle in the same homology class as
$\mathfrak{a}$). We now develop the machinery that allows us to use the
$\mathcal{B}$-quasiperiodicity of $\Theta$ to construct a function that is
also independent of the number of $\mathfrak{b}$-cycles used in the path that
connects $A$ to $z$. We start with the observation that the
$\mathcal{B}$-quasiperiodicity of $\Theta$ allows us to construct a function
whose $\mathfrak{b}$-cycle dependence is multiplicative, not additive. Indeed,
for complex constants $C_{1},C_{2}\in\mathbb{C}$ and $n\in\mathbb{Z}$, notice

(112)
$\displaystyle\begin{split}\frac{\Theta\left(\mathcal{A}(z)+C_{1}+C_{2}+n\mathcal{B}\right)}{\Theta\left(\mathcal{A}(z)+C_{1}+n\mathcal{B}\right)}&=\frac{\Theta\left(\mathcal{A}(z)+C_{1}+C_{2}\right)}{\Theta\left(\mathcal{A}(z)+C_{1}\right)}e^{-nC_{2}}.\end{split}$

We see that the above quotient “counts” (in the negative direction in the
exponent with weight $C_{2}$) the number of $\mathfrak{b}$-cycles used to
connect $A$ to $z$. So, if we multiply it with a function that “counts” in the
positive direction, then the product will be independent of the number of
$\mathfrak{b}$-cycles.

We define the differentials

(113) $\Phi:=\frac{2\pi
i}{\ointctrclockwise_{\mathfrak{a}}\frac{\text{d}w}{R(w)}}\frac{\text{d}w}{R(w)},\hskip
18.77892pt\Upsilon_{0}:=\frac{w^{2}}{R(w)}\text{d}w,\hskip 18.77892pt\text{
and }\hskip 18.77892pt\Upsilon:=\Upsilon_{0}-\left(\frac{1}{2\pi
i}\ointctrclockwise_{\mathfrak{a}}\Upsilon_{0}\right)\Phi.$

Notice that $\Upsilon$ vanishes when it is integrated against an
$\mathfrak{a}$-cycle. We set the constant

(114) $U(x)\equiv U:=\ointclockwise_{\mathfrak{b}}\Upsilon.$

We now list some of the basic properties of the differential $\Upsilon$.

######  Proposition 4.3.

The boundary values taken by the map

(115) $z\mapsto\int_{A}^{z}\Upsilon$

on $\Sigma_{1}\cup\Sigma_{2}$ in the plus-sheet are related by the jump
conditions

(116)
$\left(\int_{A}^{z^{+}}\Upsilon\right)_{+}+\left(\int_{A}^{z^{+}}\Upsilon\right)_{-}=0\mod
U\mathbb{Z}\hskip 28.45274pt\text{for }z^{+}\in\Sigma_{1}$

and

(117)
$\left(\int_{A}^{z^{+}}\Upsilon\right)_{+}+\left(\int_{A}^{z^{+}}\Upsilon\right)_{-}=-U\mod
U\mathbb{Z}\hskip 28.45274pt\text{for }z^{+}\in\Sigma_{2}.$

Also, for some constants $\Upsilon_{0},\Upsilon_{-1}\in\mathbb{C}$, as
$z\to\infty$

(118)
$z-\int_{A}^{z}\Upsilon=\Upsilon_{0}+\frac{\Upsilon_{-1}}{z}+\mathcal{O}\left(1/z^{2}\right).$

Let $P$ be a path that connects $A$ to $z$ using exactly $n$
$\mathfrak{b}$-cycles, for some $n\in\mathbb{Z}$. Notice

(119)
$\displaystyle\begin{split}\exp\left(kF_{1}\int_{P}\Upsilon\right)&=\exp\left(kF_{1}n\ointclockwise_{\mathfrak{b}}\Upsilon\right)\exp\left(kF_{1}\int_{A}^{z}\Upsilon\right)\\\
&=\exp(nkF_{1}U)\exp\left(kF_{1}\int_{A}^{z}\Upsilon\right).\end{split}$

Looking at (112) and (119), we see that if we choose $C_{2}=kF_{1}U$ and if
the paths in the Abel map and the differential are consistent, then their
product is path independent on $\mathbb{T}$.

We also need to specify the constant $C_{1}$. As we are trying to solve a
Riemann-Hilbert problem with no poles, we need to control where the
denominator of (112) vanishes. Thus, we set

(120) $\mathcal{K}(x)\equiv\mathcal{K}:=i\pi+\frac{\mathcal{B}}{2}$

to be the Riemann constant and

(121) $C_{1}:=-\mathcal{A}(Q)-\mathcal{K},$

where $Q\in\mathbb{C}$ is specified in (128). Now, the Baker-Akhiezer function
is given as

(122) $q(z;x,k)\equiv
q(z):=\frac{\Theta\left(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K}-kF_{1}U\right)}{\Theta\left(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K}\right)}e^{-kF_{1}\int_{A}^{z}\Upsilon}.$

Notice $q$ has a simple pole at $z=Q$.

####  4.4.3. The Solution to the Model Riemann-Hilbert Problem

We start with the observation that by permuting the signs of $C_{1}$ and
$F_{1}U$ in $q(z)$ we can construct a matrix-valued function that is analytic
in $\mathbb{C}\setminus(\Sigma_{1}\cup\Sigma_{2})$ with a simple jump
discontinuity. Define

(123) $\dot{\boldsymbol{R}}(z;x,k)\equiv\dot{\boldsymbol{R}}(z):=\\\
\begin{bmatrix}\displaystyle\frac{\Theta(\mathcal{A}(z)+\mathcal{A}(Q)+\mathcal{K}-kF_{1}U)}{\Theta(\mathcal{A}(z)+\mathcal{A}(Q)+\mathcal{K})}e^{-kF_{1}\int_{A}^{z}\Upsilon}&\displaystyle\frac{\Theta(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K}+kF_{1}U)}{\Theta(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K})}e^{kF_{1}\int_{A}^{z}\Upsilon}\\\
\displaystyle\frac{\Theta(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K}-kF_{1}U)}{\Theta(\mathcal{A}(z)-\mathcal{A}(Q)-\mathcal{K})}e^{-kF_{1}\int_{A}^{z}\Upsilon}&\displaystyle\frac{\Theta(\mathcal{A}(z)+\mathcal{A}(Q)+\mathcal{K}+kF_{1}U)}{\Theta(\mathcal{A}(z)+\mathcal{A}(Q)+\mathcal{K})}e^{kF_{1}\int_{A}^{z}\Upsilon}\end{bmatrix}.$

The jump conditions in Propositions 4.2 and 4.3 imply

(124)
$\dot{\boldsymbol{R}}_{+}(z)=\dot{\boldsymbol{R}}_{-}(z)\begin{bmatrix}0&1\\\
1&0\end{bmatrix}\hskip 28.45274pt\text{ for }z\in\Sigma_{1}\cup\Sigma_{2}.$

We will construct $\dot{\boldsymbol{Q}}(z)$ out of $\dot{\boldsymbol{R}}(z)$.
Our first step is to introduce a sign change into the jump of
$\dot{\boldsymbol{R}}$. In order to do so, we define

(125) $\gamma(z):=\left(\frac{(z-A)(z-C)}{(z-B)(z-D)}\right)^{1/4},$

where $\gamma$ is cut on $\Sigma_{1}\cup\Sigma_{2}$ and as $z\to\infty$,
$\gamma(z)=1+\mathcal{O}(1/z)$. We also define

(126) $f^{\text{D}}(z):=\frac{\gamma(z)+\gamma^{-1}(z)}{2}\hskip
49.79231pt\text{ and }\hskip
49.79231ptf^{OD}(z):=\frac{\gamma(z)-\gamma^{-1}(z)}{2i}.$

A direct calculation shows that $\gamma_{+}=\gamma_{-}e^{-i\pi/2}$ on
$\Sigma_{1}\cup\Sigma_{2}$. Hence,

(127) $f^{D}_{+}(z)=f^{OD}_{-}(z)\hskip 28.45274pt\text{and}\hskip
28.45274ptf^{OD}_{+}(z)=-f^{D}_{-}(z)\hskip 28.45274pt\text{ for
}z\in\Sigma_{1}\cup\Sigma_{2}.$

Also, a straightforward computation shows that

(128) $Q(x)\equiv Q:=\frac{BD-AC}{B+D-A-C}$

is the unique solution to the equation $f^{\text{D}}(z)f^{\text{OD}}(z)=0$.
Further, it is a simple zero and either $f^{\text{D}}(Q)=0$ or
$f^{\text{OD}}(Q)=0$. It has been shown in [3] that while the construction of
the parametrix does depend on whether $Q$ is a zero of $f^{\text{D}}$ or
$f^{\text{OD}}$, the final answer (i.e. the formula we extract from
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$; see (10)) is the same regardless of
which function vanishes at $Q$. Numerically, we see that $f^{\text{OD}}(Q)=0$
for various $x$-values in the pole region. Thus, we assume that $Q$ is the
unique simple zero of $f^{\text{OD}}(z)$, and we proceed accordingly.

We now modify $\dot{\boldsymbol{R}}(z)$ by setting

(129)
$\dot{\boldsymbol{Q}}(z;x,k)\equiv\dot{\boldsymbol{Q}}(z)=\begin{bmatrix}C_{11}&0\\\
0&C_{22}\end{bmatrix}\begin{bmatrix}f^{\text{D}}(z)[\dot{\boldsymbol{R}}(z)]_{11}&f^{\text{OD}}(z)[\dot{\boldsymbol{R}}(z)]_{12}\\\
-f^{\text{OD}}(z)[\dot{\boldsymbol{R}}(z)]_{21}&f^{\text{D}}(z)[\dot{\boldsymbol{R}}(z)]_{22}\end{bmatrix},$

where $C_{11}$ and $C_{22}$ are normalizing constants specified in (130) and
(131). Now, $\dot{\boldsymbol{Q}}(z)$ satisfies the jump condition
$\dot{\boldsymbol{Q}}_{+}(z)=\dot{\boldsymbol{Q}}_{-}(z)\boldsymbol{T}(1)$ for
$z\in z\in\Sigma_{1}\cup\Sigma_{2}$.

It is well known that the Abel map $\mathcal{A}(z)$ with domain $\mathbb{T}$
is injective (see for example the discussion in Section 4.5.2 in [6]). Since
$\mathcal{A}(Q^{-})+\mathcal{A}(Q)=0$ (here $Q$ is on the plus sheet and
$Q^{-}$ is the equivalent point on the minus sheet), it follows that for each
$z$ on the plus sheet, $\mathcal{A}(z)+\mathcal{A}(Q)\neq 0$, otherwise
$\mathcal{A}(z)$ would not be injective. Hence, the diagonal entries of
$\dot{\boldsymbol{R}}(z)$ are analytic in
$\mathbb{C}\setminus(\Sigma_{1}\cup\Sigma_{2})$. Further, $Q$ was selected so
that in the off-diagonal entries of $\dot{\boldsymbol{R}}(z)$ the simple pole
$z=Q$ is canceled by the simple zero of $f^{\text{OD}}(z)$. Therefore, the
off-diagonal entries of $\dot{\boldsymbol{Q}}(z)$ are analytic in
$\mathbb{C}\setminus(\Sigma_{1}\cup\Sigma_{2})$.

Finally, to deduce the values of $C_{11},C_{22}$, we notice as $z\to\infty$,
$f^{\text{D}}(z)=1+\mathcal{O}(1/z)$ and $f^{\text{OD}}(z)=\mathcal{O}(1/z)$,
thus we find the limits of $[\dot{\boldsymbol{R}}]_{11}$ and
$[\dot{\boldsymbol{R}}]_{22}$ as $z\to\infty$ and take the reciprocal to find

(130) $C_{11}(x;k)\equiv
C_{11}:=\frac{\Theta\left(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}\right)}{\Theta\left(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}-kF_{1}U\right)}e^{-kF_{1}\Upsilon_{0}}$

and

(131) $C_{22}(x;k)\equiv
C_{22}:=\frac{\Theta\left(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}\right)}{\Theta\left(\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}+kF_{1}U\right)}e^{kF_{1}\Upsilon_{0}}.$

Here, $C_{11}$ and $C_{22}$ are not guaranteed to exist. Indeed, their
denominators can vanish. In other words, $C_{11}$ and $C_{22}$ do not exist if
and only if

(132) $\mathcal{A}(\infty)+\mathcal{A}(Q)+\mathcal{K}=\pm
kF_{-1}U\mod\left\\{\mathcal{K}+j_{1}2\pi
i+j_{2}\mathcal{B}:j_{1},j_{2}\in\mathbb{Z}\right\\}.$

When $k$ is large enough, if the latter equality holds for some $x$ in the
pole region, then the left-hand side of (1.1) has a pole near
$-k^{2/3}2^{1/3}x$.

For each $k\in\mathbb{N}$, let $\mathscr{P}_{k}$ denote all of the solutions
in the $x$-plane of (132). Notice, $\mathscr{P}_{k}$ is the set of exceptional
points for which Riemann-Hilbert Problem 4.4.1 does not admit a solution. Now,
let $\delta>0$ be a fixed small number. For each $k\in\mathbb{N}$ let

(133) $\mathscr{S}_{k}:=\left\\{x\text{ in the pole region
}:\text{dist}(x,\mathscr{P}_{k})>\frac{\delta}{k^{2/3}}\right\\}.$

That is, $\mathscr{S}_{k}$ is the result of excising a neighborhood of each
exceptional point from the pole region.

We have completed the construction of $\dot{\boldsymbol{Q}}(z)$, and thus
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ with the help of (101).

###  4.5. Constructing the Inner Parametrices

In this section, we construct the inner parametrices around the points $A$,
$B$, $C$, and $D$. In particular, we go over in detail the construction of the
local parametrix at $z=C$. For the other three we just give the resulting
formulae. Our approach is modeled from the methodology presented in Section
7.6.2 in [8].

####  4.5.1. Local Parametrix at $z=C$

Our first step is to massage the jumps of $\boldsymbol{P}(z)$ so that they
match the jumps of $\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)$, where
$\boldsymbol{A}(\cdot)$ is the solution to Riemann-Hilbert Problem A. We give
a systematic approach to do this.

  1. (1)

Clear the $z$-dependence. Multiply (on the right) by
$e^{-k(H(z)+\Lambda/2)\sigma_{3}}$ to obtain the “core” jumps of
$\boldsymbol{P}(z)$.

  2. (2)

Mold the core jumps into the shape of the Airy jump matrices. In this case, we
move (by sectionally analytic substitutions) $\Sigma_{2}$ onto $\Gamma$. This
deformation flips the triangularity of the jump matrix whose jump contour is
contained in $\mathbb{D}_{C}^{\downarrow}$.

  3. (3)

Insert $\tau$ dependence. Multiply by
$e^{\frac{k}{2}\tau_{C}^{3/2}\sigma_{3}}$. Here, $\tau_{C}(z)^{3/2}$ is cut on
$\widetilde{\Gamma}$ with
$\left(\tau_{C}(z)^{3/2}\right)_{+}=-\left(\tau_{C}(z)^{3/2}\right)_{-}$.

In Figure 24 we track the jumps of $\boldsymbol{P}(z)$ as we apply the above
procedure. Following it yields

(134)
$\boldsymbol{Q}^{(C)}(z;x,k)\equiv\boldsymbol{Q}^{(C)}(z):=\displaystyle\begin{cases}\boldsymbol{P}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}e^{\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}},&z\in\mathbb{D}_{C}^{\uparrow},\\\
\boldsymbol{P}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}\boldsymbol{T}(1)e^{\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}},&z\in\mathbb{D}_{C}^{\downarrow}.\end{cases}$

$e^{ik\omega\sigma_{3}}$$\boldsymbol{T}(e^{-ik\Omega})$$\boldsymbol{U}_{k}(-\chi)$$\boldsymbol{L}_{k}^{-1}(\chi)$$\boldsymbol{U}_{k}(-\chi)$$C$$z$ | fill me up | $\boldsymbol{T}(1)$$\boldsymbol{U}_{0}$$\boldsymbol{L}_{0}^{-1}$$\boldsymbol{U}_{0}$$C$$z$$(1)$
---|---|---
s | s | s
$\boldsymbol{T}(1)$$\boldsymbol{L}_{0}^{-1}$$\boldsymbol{L}_{0}^{-1}$$\boldsymbol{U}_{0}$$C$$z$$(2)$ | fill me up | $\boldsymbol{U}_{k}(-\tau_{C}^{3/2})$$\boldsymbol{T}(1)$$\boldsymbol{L}_{k}^{-1}(\tau_{C}^{3/2})$$\boldsymbol{L}_{k}^{-1}(\tau_{C}^{3/2})$$0$$\tau_{C}$$(3)$
Figure 24. The top-left panel shows the jumps of $\boldsymbol{P}(z)$ in
$\mathbb{D}_{C}$, where $\chi:=\operatorname{Re}(2H+\Lambda)$. The top-right
panel shows the jumps of $\boldsymbol{P}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}$.
The bottom-left panel shows the jumps of
$\boldsymbol{Q}^{(C)}(z)e^{-\frac{k}{2}\tau_{C}(z)^{3/2}}$. The bottom-right
panel shows the jumps of $\boldsymbol{Q}^{(C)}(z)$ in the $\tau_{C}$-plane.

Comparing the jumps of $\boldsymbol{Q}^{(C)}(z)$ to the jumps of
$\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)$ (see Appendix A) we see that
they agree. Notice, if we replace $\boldsymbol{P}(z)$ with
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ in (134), then the resulting function
will only have a jump of $\boldsymbol{T}(1)$ on $\widetilde{\Gamma}$. Set

(135)
$\boldsymbol{H}^{(C)}(z):=\displaystyle\begin{cases}\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}e^{\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{C}^{-\sigma_{3}/4},&z\in\mathbb{D}_{C}^{\uparrow},\\\
\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}\boldsymbol{T}(1)e^{\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{C}^{-\sigma_{3}/4},&z\in\mathbb{D}_{C}^{\downarrow},\end{cases}$

where $\boldsymbol{V}$ is defined in (184). Seeing as
$\tau_{C}(z)^{-\sigma_{3}/4}\boldsymbol{V}$ also has a jump of
$\boldsymbol{T}(1)$ on $\widetilde{\Gamma}$ (because $\tau_{C}$ maps
$\widetilde{\Gamma}$ to the negative real numbers and the $3/2$-root function
is principal), it follows that $\boldsymbol{H}^{(C)}(z)$ is continuous across
$\widetilde{\Gamma}$. Further, as $z\to C$, the largest blowup in (135) is a
square-root blowup (coming from the quarter-root blowups of
$\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ and $\tau_{C}(z)^{-\sigma_{3}/4}$).
This implies that $\boldsymbol{H}^{(C)}(z)$ is analytic in $\mathbb{D}_{C}$
because $\boldsymbol{H}^{(C)}(z)$ has no square-root jumps in $\mathbb{D}_{C}$
as it is continuous in $\mathbb{D}_{C}$.

Now, since $\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)$ and
$\boldsymbol{Q}^{(C)}(z)$ satisfy the same jump conditions, and
$\boldsymbol{H}^{(C)}$ is analytic in $\mathbb{D}_{C}$, it follows that
$\boldsymbol{H}^{(C)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)\boldsymbol{Q}^{(C)}(z)^{-1}$
is analytic in $\mathbb{D}_{C}$. This in turn implies

(136)
$\dot{\boldsymbol{P}}^{(C)}(z;x,k)\equiv\dot{\boldsymbol{P}}^{(C)}(z):=\boldsymbol{H}^{(C)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)\boldsymbol{Q}^{(C)}(z)^{-1}\boldsymbol{P}(z)$

satisfies the same jump conditions as $\boldsymbol{P}(z)$. The upshot is that
we have an explicit formula for $\dot{\boldsymbol{P}}^{(C)}(z)$ because we can
solve for $\boldsymbol{Q}^{(C)}(z)^{-1}\boldsymbol{P}(z)$ in (134). Indeed,

(137)
$\dot{\boldsymbol{P}}^{(C)}(z)=\begin{cases}\boldsymbol{H}^{(C)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)e^{-\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}}e^{k(H(z)+\Lambda/2)\sigma_{3}},&z\in\mathbb{D}_{C}^{\uparrow},\\\
\boldsymbol{H}^{(C)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{C}(z)\right)e^{-\frac{k}{2}\tau_{C}(z)^{3/2}\sigma_{3}}\boldsymbol{T}(-1)e^{k(H(z)+\Lambda/2)\sigma_{3}},&z\in\mathbb{D}_{C}^{\downarrow}.\end{cases}$

Also, we can express $\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1}$ in terms of
$\boldsymbol{H}^{(C)}(z)$ to find that

(138)
$\dot{\boldsymbol{P}}^{(C)}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1}=\boldsymbol{H}^{(C)}(z)k^{-\sigma_{3}/6}\left(\boldsymbol{A}\left(k^{2/3}\tau_{C}\right)\boldsymbol{V}^{-1}\left(k^{2/3}\tau_{C}(z)\right)^{-\sigma_{3}/4}\right)k^{\sigma_{3}/6}\boldsymbol{H}^{(C)}(z)^{-1}.$

Therefore, using the normalization of $\boldsymbol{A}$ (see (183)), on
$\partial\mathbb{D}_{C}$, as $k\to\infty$,

(139)
$\dot{\boldsymbol{P}}^{(C)}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1}=\mathbb{I}+\begin{bmatrix}\mathcal{O}\left(k^{-2}\right)&\mathcal{O}\left(k^{-1}\right)\\\
\mathcal{O}\left(k^{-1}\right)&\mathcal{O}\left(k^{-2}\right)\end{bmatrix}.$

####  4.5.2. Local Parametrices at the Endpoints $z=A$, $z=B$, and $z=D$

We now construct the other parametrices. In $\mathbb{D}_{A}$, after clearing
the $z$-dependence, the core jumps of $\boldsymbol{P}(z)$ are already in the
correct shape. Thus, set

(140)
$\boldsymbol{H}^{(A)}(z):=\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}e^{\frac{k}{2}\tau_{A}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{A}(z)^{-\sigma_{3}/4}$

and

(141)
$\dot{\boldsymbol{P}}^{(A)}(z):=\boldsymbol{H}^{(A)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{A}(z)\right)e^{-\frac{k}{2}\tau_{A}(z)^{3/2}\sigma_{3}}e^{k(H(z)+\Lambda/2)\sigma_{3}}.$

In $\mathbb{D}_{B}$, after clearing the $z$-dependence, the core jumps of
$\boldsymbol{P}(z)$ are all off by a sign in the off-diagonal entries. So, set

(142)
$\boldsymbol{H}^{(B)}(z):=\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}e^{\frac{i\pi}{2}\sigma_{3}}e^{\frac{k}{2}\tau_{B}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{B}(z)^{-\sigma_{3}/4}$

and

(143)
$\dot{\boldsymbol{P}}^{(B)}(z):=\boldsymbol{H}^{(B)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{B}(z)\right)e^{-\frac{k}{2}\tau_{B}(z)^{3/2}\sigma_{3}}e^{-\frac{i\pi}{2}\sigma_{3}}e^{k(H(z)+\Lambda/2)\sigma_{3}}.$

In $\mathbb{D}_{D}$, after clearing the $z$-dependence, the core triangular
jumps of $\boldsymbol{P}(z)$ are all the wrong triangularity. Hence, set

(144)
$\boldsymbol{H}^{(D)}(z):=\dot{\boldsymbol{P}}^{\text{(out)}}(z)e^{-k(H(z)+\Lambda/2)\sigma_{3}}\boldsymbol{T}(1)e^{\frac{i\pi}{2}\sigma_{3}}e^{\frac{k}{2}\tau_{D}(z)^{3/2}\sigma_{3}}\boldsymbol{V}^{-1}\tau_{D}(z)^{-\sigma_{3}/4}$

and

(145)
$\dot{\boldsymbol{P}}^{(D)}(z):=\boldsymbol{H}^{(D)}(z)k^{-\sigma_{3}/6}\boldsymbol{A}\left(k^{2/3}\tau_{D}(z)\right)e^{-\frac{k}{2}\tau_{D}(z)^{3/2}\sigma_{3}}e^{-\frac{i\pi}{2}\sigma_{3}}\boldsymbol{T}(-1)e^{k(H(z)+\Lambda/2)\sigma_{3}}.$

Further, for each endpoint the local parametrix matches well with the outer
parametrix. That is, for $p\in\left\\{A,B,C,D\right\\}$, for each
$z\in\partial\mathbb{D}_{p}$, as $k\to\infty$, the mismatch jumps satisfy

(146)
$\dot{\boldsymbol{P}}^{(p)}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1}=\mathbb{I}+\begin{bmatrix}\mathcal{O}\left(k^{-2}\right)&\mathcal{O}\left(k^{-1}\right)\\\
\mathcal{O}\left(k^{-1}\right)&\mathcal{O}\left(k^{-2}\right)\end{bmatrix}.$

###  4.6. Error Analysis

In this section we quantify the error introduced by omitting the triangular
jumps that decay to the identity.

####  4.6.1. The Global Parametrix

We are now ready to define the global parametrix. In
$\mathbb{C}\setminus(\Sigma_{1}\cup\Gamma\cup\Sigma_{2})$ set

(147)
$\dot{\boldsymbol{P}}(z;x,k)\equiv\dot{\boldsymbol{P}}(z)=\begin{cases}\dot{\boldsymbol{P}}^{(A)}(z),&z\in\mathbb{D}_{A},\\\
\dot{\boldsymbol{P}}^{(B)}(z),&z\in\mathbb{D}_{B},\\\
\dot{\boldsymbol{P}}^{(C)}(z),&z\in\mathbb{D}_{C},\\\
\dot{\boldsymbol{P}}^{(D)}(z),&z\in\mathbb{D}_{D},\\\
\dot{\boldsymbol{P}}^{\text{(out)}}(z),&\text{otherwise}.\end{cases}$

Notice, for $z$ large enough,
$\dot{\boldsymbol{P}}(z)=\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ and we can
use the normalization condition of $\dot{\boldsymbol{P}}^{\text{(out)}}(z)$ to
write

(148)
$\dot{\boldsymbol{P}}(z)=\mathbb{I}+\frac{\dot{\boldsymbol{P}}_{-1}}{z}+\frac{\dot{\boldsymbol{P}}_{-2}}{z^{2}}+\mathcal{O}(1/z^{3}),\hskip
28.45274ptz\to\infty.$

We introduce the error as

(149)
$\boldsymbol{E}(z;x,k)\equiv\boldsymbol{E}(z):=\boldsymbol{P}(z)\dot{\boldsymbol{P}}^{-1}(z).$

As in the pole-free case, when $z$ is not on a jump contour of
$\boldsymbol{P}(z)$ or $\dot{\boldsymbol{P}}(z)$ then $\boldsymbol{E}(z)$ is
the product of analytic functions and thus is analytic. Further, for each
$p\in\left\\{A,B,C,D\right\\}$, $\boldsymbol{P}(z)$ and
$\dot{\boldsymbol{P}}(z)$ have exactly the same jump conditions inside
$\mathbb{D}_{p}$, and thus $\boldsymbol{E}(z)$ is analytic in
$\mathbb{D}_{p}$. Similarly, $\boldsymbol{P}(z)$ and $\dot{\boldsymbol{P}}(z)$
satisfying the same jump conditions on $\Sigma_{1}\cup\Gamma\cup\Sigma_{2}$
implies that $\boldsymbol{E}(z)$ is analytic on
$\Sigma_{1}\cup\Gamma\cup\Sigma_{2}$. Therefore, the only jumps of
$\boldsymbol{E}(z)$ are the boundary of each disk and the jump contours of
$\boldsymbol{P}(z)$ that are bounded away from the endpoints $A$, $B$, $C$,
and $D$.

Riemann-Hilbert Problem 13: For each $k\in\mathbb{N}$, find
$\boldsymbol{E}(z)$ so that

  1. (1)

Analyticity: $\boldsymbol{E}(z)$ is analytic in $z$ except along the jump
contours described above.

  2. (2)

Jump Condition: $\boldsymbol{E}(z)$ can be continuously extended to the
boundary and the boundary values taken by $\boldsymbol{E}(z)$ are related by
the jump condition
$\boldsymbol{E}_{+}(z)=\boldsymbol{E}_{-}(z)\boldsymbol{V}^{\boldsymbol{(E)}}(z)$,
where

(150)
$\boldsymbol{V}^{\boldsymbol{(E)}}(z)=\begin{cases}\dot{\boldsymbol{P}}^{(p)}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1},&z\in\partial\mathbb{D}_{p}\text{
for }p\in\\{A,B,C,D\\},\\\
\dot{\boldsymbol{P}}^{\text{(out)}}(z)\boldsymbol{V}^{(\boldsymbol{P})}(z)\dot{\boldsymbol{P}}^{\text{(out)}}(z)^{-1},&\text{otherwise}.\end{cases}$

  3. (3)

Normalization: As $z\to\infty$, the matrix
$\boldsymbol{E}(z)=\mathbb{I}+\mathcal{O}\left(1/z\right)$.

Just like in the pole-free case, since $\boldsymbol{E}(z)$ normalizes to the
identity and all of its jump matrices are close to the identity matrix we have
the following proposition:

######  Proposition 4.4.

Let $x$ be in the pole region of the $x$-plane. For large $k\in\mathbb{N}$, if
$x\in\mathscr{S}_{k}$ then

(151) $\boldsymbol{E}(z)=\mathbb{I}+\mathcal{O}(1/k).$

####  4.6.2. Extraction Formula

The formula (70) is still valid in the pole region given that
$x\in\mathscr{S}_{k}$. Moreover, Proposition 4.4 implies (70) can be written
as

(152)
$p_{k}(s)=2i\left(\frac{k}{2}\right)^{1/3}\left(\frac{-[\dot{\boldsymbol{P}}_{-2}]_{12}+[\dot{\boldsymbol{P}}_{-1}]_{22}[\dot{\boldsymbol{P}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{P}}_{-1}]_{12}+\mathcal{O}(1/k)}\right).$

A direct computation shows that

(153)
$\frac{-[\dot{\boldsymbol{P}}_{-2}]_{12}+[\dot{\boldsymbol{P}}_{-1}]_{22}[\dot{\boldsymbol{P}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{P}}_{-1}]_{12}+\mathcal{O}(1/k)}=\frac{-[\dot{\boldsymbol{Q}}_{-2}]_{12}+[\dot{\boldsymbol{Q}}_{-1}]_{22}[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)},$

where the quantities in the right-hand side are defined in (102). We now
recover the large-$z$ expansion coefficients that are present in the right-
hand side of (153). First, using (109) and the fact $\Theta(z)$ is entire we
have for any constant $v$, as $z\to\infty$,

(154)
$\Theta(\mathcal{A}(z)+v)=\Theta(\mathcal{A}(\infty)+v)+\Theta^{\prime}(\mathcal{A}(\infty)+v)\frac{\mathcal{A}_{-1}}{z}+\mathcal{O}\left(1/z^{2}\right).$

Recall

(155)
$[\dot{\boldsymbol{Q}}]_{22}e^{-kF_{1}z}=C_{22}\frac{\Theta(\mathcal{A}(z)-C_{1}+kF_{1}U)}{\Theta(\mathcal{A}(z)-C_{1})}e^{kF_{1}\int_{A}^{z}\Upsilon-z}f^{\text{D}}(z),$

where $C_{1}=-\mathcal{A}(Q)-\mathcal{K}$. Using (131), (154), and (118) we
can the above equation as

(156)
$\displaystyle\begin{split}[\dot{\boldsymbol{Q}}]_{22}e^{-kF_{1}z}&=\left(\frac{\Theta(\mathcal{A}(\infty)-C_{1})}{\Theta(\mathcal{A}(\infty)-C_{1}+kF_{1}U)}\frac{\Theta(\mathcal{A}(z)-C_{1}+kF_{1}U)}{\Theta(\mathcal{A}(z)-C_{1})}\right)\left(e^{kF_{1}\left(\int_{A}^{z}\Upsilon-z+\Upsilon_{0}\right)}\right)f^{\text{D}}(z)\\\\[5.0pt]
&=\left(1+\frac{\mathfrak{L}_{22}\mathcal{A}_{-1}}{z}+\mathcal{O}(1/z^{2})\right)\left(1-\frac{kF_{1}\Upsilon_{-1}}{z}+\mathcal{O}(1/z^{2})\right)(1+\mathcal{O}(1/z^{2}))\\\\[5.0pt]
&=1+\frac{\mathfrak{L}_{22}\mathcal{A}_{-1}-kF_{1}\Upsilon_{-1}}{z}+\mathcal{O}(1/z^{2}),\end{split}$

where $\mathfrak{L}_{22}(x;k)\equiv\mathfrak{L}_{22}$ is defined in (11).
Therefore, we find that

(157)
$[\dot{\boldsymbol{Q}}_{-1}]_{22}=\mathcal{A}_{-1}\mathfrak{L}_{22}-kF_{1}\Upsilon_{-1}.$

Next, we find the quotient
$[\dot{\boldsymbol{Q}}_{-2}]_{12}/[\dot{\boldsymbol{Q}}_{-1}]_{12}$. We write

(158)
$\displaystyle\begin{split}[\dot{\boldsymbol{Q}}]_{12}e^{-kF_{1}z}&=C_{11}\frac{\Theta(\mathcal{A}(z)+C_{1}+kF_{1}U)}{\Theta(\mathcal{A}(z)+C_{1})}f^{\text{OD}}(z)e^{kF_{1}\left(\int_{A}^{z}\Upsilon-z\right)}\\\\[5.0pt]
&=\mu
e^{-kF_{1}\Upsilon_{0}}\left(1+\frac{\mathfrak{L}_{12}\mathcal{A}_{-1}-kF_{1}\Upsilon_{-1}}{z}+\mathcal{O}(1/z^{2})\right)\\\
&\hskip
9.38945pt\cdot\left(\frac{-i(B+D-A-C)}{4z}-\frac{i(B^{2}+D^{2}-A^{2}-C^{2})}{8z^{2}}+\mathcal{O}(1/z^{3})\right),\end{split}$

where

(159)
$\mu(x;k)\equiv\mu:=C_{11}\frac{\Theta(\mathcal{A}(\infty)+C_{1}+kF_{1}U)}{\Theta(\mathcal{A}(\infty)+C_{1})}$

and $\mathfrak{L}_{12}(x;k)\equiv\mathfrak{L}_{12}$ is defined in (12). Thus,
we have that

(160) $[\dot{\boldsymbol{Q}}_{-1}]_{12}=\mu
e^{-kF_{1}\Upsilon_{0}}\left(-i\frac{B+D-A-C}{4}\right)$

and

(161) $\displaystyle\begin{split}&[\dot{\boldsymbol{Q}}_{-2}]_{12}=\\\ &\mu
e^{-kF_{1}\Upsilon_{0}}\left(-\frac{i(B+D-A-C)(\mathfrak{L}_{12}\mathcal{A}_{-1}-kF_{1}\Upsilon_{-1})}{4}-\frac{i(B^{2}+D^{2}-A^{2}-C^{2})}{8}\right).\end{split}$

Therefore,

(162)
$\frac{[\dot{\boldsymbol{Q}}_{-2}]_{12}}{[\dot{\boldsymbol{Q}}_{-1}]_{12}}=\mathcal{A}_{-1}\mathfrak{L}_{12}-kF_{1}\Upsilon_{-1}+\frac{B^{2}+D^{2}-A^{2}-C^{2}}{2(B+D-A-C)}.$

We are now ready to prove Theorem 1.2.

###### Proof of Theorem 1.2.

Notice (152) and (153) give

(163)
$p_{k}(s)=2i\left(\frac{k}{2}\right)^{1/3}\left(\frac{-[\dot{\boldsymbol{Q}}_{-2}]_{12}+[\dot{\boldsymbol{Q}}_{-1}]_{22}[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}\right).$

Writing $p_{k}$ in terms of $u^{(k+1/2)}_{\text{HM}}$ using (14) and (17)
results in

(164)
$-(2k)^{-1/3}u^{(k+1/2)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=i\left(\frac{-[\dot{\boldsymbol{Q}}_{-2}]_{12}+[\dot{\boldsymbol{Q}}_{-1}]_{22}[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}{[\dot{\boldsymbol{Q}}_{-1}]_{12}+\mathcal{O}(1/k)}\right).$

This is the first statement of the theorem. Now, assume that
$[\dot{\boldsymbol{Q}}_{-1}]_{12}\neq 0$. In this case, we can geometrically
expand the denominator of the above equation to find

(165)
$-(2k)^{-1/3}u^{(k+1/2)}_{\text{HM}}\left(-\frac{k^{2/3}}{2^{1/3}}x\right)=-i\left([\dot{\boldsymbol{Q}}_{-1}]_{22}-\frac{[\dot{\boldsymbol{Q}}_{-2}]_{12}}{[\dot{\boldsymbol{Q}}_{-1}]_{12}}+\mathcal{O}\left(1/k\right)\right).$

Finally, (157) and (162) give our desired result. ∎

##  5\. Numerical Approximation

We use numerical methods to obtain approximations of the exact generalized
Hastings-McLeod functions. However, since $u_{\text{HM}}^{(\alpha)}$ has poles
in the pole region of the plane, we will employ two distinct solvers: a
spectral collocation boundary condition solver to handle the smooth, pole-free
region, and a pole-vaulting method for the pole region developed by Fornberg
and Weideman [15]. This numerical scheme has been used to survey the pole
fields of various Painlevé transcendents; see [14, 16, 19]. In this section we
will give a brief description of each method.

###  5.1. Spectral Collocation Method

For $N\in\mathbb{N}$, we consider the interpolation nodes

(166) $t_{k}=\cos\left(\frac{(k-1)\pi}{N-1}\right),\hskip
28.45274ptk=1,\dots,N,$

which are Chebyshev points of the second kind, i.e. the extreme points on
$[-1,1]$ of $T_{N-1}(t)$, the Chebyshev polynomial of degree $N-1$. For a
smooth function $f$ on $[-1,1]$ we set the interpolant $P_{N-1}(t)$ to be a
linear combination of Lagrangian interpolating polynomials. So,

(167) $f(t)\approx P_{N-1}(t)=\sum\limits_{j=1}^{N}\varphi_{j}(t)f_{j},$

where

(168) $f_{j}:=f(t_{j})\hskip 18.77892pt\text{and}\hskip
18.77892pt\varphi_{j}(t):=\prod\limits_{k=1,k\neq
j}^{N}\frac{(t-t_{k})}{(t_{j}-t_{k})}.$

In fact, since $t_{2},\dots,t_{N-1}$ are the critical points of $T_{N-1}(t)$,

(169)
$\varphi_{j}(t)=\frac{(-1)^{j}}{c_{j}}\frac{1-t^{2}}{(N-1)^{2}}\frac{T_{N-1}^{\prime}(t)}{t-t_{j}},$

where $c_{1}=c_{N}=2$ and $c_{2},\dots,c_{N-1}=1$. Notice, on the set of
interpolation nodes $\\{t_{1},\dots,t_{N}\\}$, $\varphi_{j}(t_{k})$ acts as a
Kronecker delta function of $j$ and $k$. That is,

(170) $\varphi_{j}\left(t_{k}\right)=\delta_{jk}=\begin{cases}0&\text{if
}j\neq k,\\\ 1,&\text{otherwise}.\end{cases}$

We now define a collocation derivative operator, $\boldsymbol{D}$, on the
interpolant $P_{N-1}(t)$ by

(171)
$\boldsymbol{D}p_{N-1}(t):=\left[P^{\prime}_{N-1}\left(t_{1}\right),\dots,P^{\prime}_{N-1}\left(t_{N}\right)\right]^{T}.$

It was shown in [9, p. 69] that

(172)
$[\boldsymbol{D}]_{k,j}=\begin{cases}\displaystyle\frac{c_{k}(-1)^{j+k}}{c_{j}(t_{k}-t_{j})},&j\neq
k,\\\\[10.0pt] \displaystyle\frac{2(N-1)^{2}+1}{6},&j=k=1,\\\\[10.0pt]
\displaystyle-\frac{2(N-1)^{2}+1}{6},&j=k=N,\\\
\displaystyle\frac{-t_{k}}{2(1-t_{k}^{2})},&\text{otherwise.}\end{cases}.$

Furthermore, in our context, taking $k$ derivatives of $P_{N-1}(t)$ and
evaluating at the nodes $\\{t_{1},\dots,t_{N}\\}$ can be accomplished by the
operator $\boldsymbol{D}^{(k)}$ where

(173) $\boldsymbol{D}^{(k)}:=\left(\boldsymbol{D}\right)^{k}.$

####  5.1.1. Collocation Method for Generalized Hastings-McLeod

We now turn our attention to applying this method to the generalized Hastings-
McLeod function for fixed $\alpha>-1/2$. First, we fix complex numbers $y_{1}$
and $y_{2}$ in the $y$-plane with
$\operatorname{Im}\left(y_{1}\right)=\operatorname{Im}\left(y_{2}\right)$,
$\operatorname{Re}\left(y_{1}\right)\ll
0\ll\operatorname{Re}\left(y_{2}\right)$, so that the asymptotics (2) are good
approximations for both $u(y_{1})$ and $u(y_{2})$. We also require that, after
the appropriate scaling, the horizontal line segment connecting $x_{1}$ to
$x_{2}$ (which we will denote as $[x_{1},x_{2}]$) is contained in the pole-
free region of the $x$-plane. Next, we transform the ODE (1) from
$[y_{1},y_{2}]$, to $[-1,1]$. In particular, we use the linear transformation

(174) $f(t):=\frac{y_{1}+y_{2}+\left(y_{2}-y_{1}\right)t}{2}.$

Then, if $u$ satisfies (1), then $v:[-1,1]\to\mathbb{C}$ defined by
$v(t):=u(f(t))$ satisfies

(175)
$\frac{\text{d}^{2}v}{\text{d}t^{2}}(t)=\frac{\left(y_{2}-y_{1}\right)^{2}}{4}\left(2v(t)^{3}+f(t)v(t)-\alpha\right).$

We interpolate $v(t)$ as in (167). That is, for $N$ unknown values
$v_{1},\dots,v_{N}$ we set

(176) $P_{N-1}(t):=\sum\limits_{j=1}^{N}\varphi_{j}(t)v_{j}.$

Our aim is for $v_{1},\dots,v_{N}$ to approximate the values
$v(t_{1}),\dots,v(t_{N})$ respectively. Then, $v(t)\approx P_{N-1}(t)$. Now,
we apply the discrete derivative operator (173) on the left-hand side of (176)
and plug in $P_{N-1}$ for $v$ in the right-hand side of (175). As a result, we
obtain an $N\times N$ system of equations. We can enforce the boundary
conditions of the generalized Hastings-McLeod solutions (2) by setting

(177) $v_{N}:=f\left(\sqrt{-a/2}\right),$

since $x_{1}$ was chosen so that (2) holds. Hence,
$f\left(\sqrt{-a/2}\right)\approx
f\left(u^{(\alpha)}_{\text{HM}}\left(x_{1}\right)\right)\approx v_{N}$.
Similarly, we set $v_{1}:=f\left(\alpha/2\right)$. Finally, we use a Newton
solver on the remaining $N-2$ equations and variables.

###  5.2. Padé Approximant Method

The collocation method is a boundary condition solver and hence assumes the
function is smooth between the two boundary conditions. However, this is not
always the case for the generalized Hastings-McLeod functions. Indeed, we know
they blow up in the pole region of the $x$-plane. Therefore, we use a Padé
approximant method due to Fornberg and Weideman, which is an initial-value
problem solver, to handle this case. We now give a brief overview of Padé
approximants and apply this method to the generalized Hastings-McLeod
functions.

####  5.2.1. Padé Expansion

Let $U$ be an open set in $\mathbb{C}$. Say $f$ is a meromorphic function in
$U$ and, for some fixed $x_{0}\in U$, $f$ is analytic at $x_{0}$. As such, $f$
has the following Taylor expansion about $x_{0}$:

(178)
$f(x_{0}+h)=\sum\limits_{k=0}^{n}c_{k}h^{k}+\mathcal{O}\left(h^{n+1}\right).$

However, this approximation fails in a neighborhood of a pole. To obtain a
more robust approximation that allows for some poles, we convert the Taylor
expansion into a Padé expansion. That is, instead of approximating $f$ as a
polynomial, we approximate $f$ as a nontrivial rational function. So, we need
to find complex coefficients $a_{0},\dots,a_{\nu_{1}},b_{1},\dots b_{\nu_{2}}$
such that

(179)
$f(x_{0}+h)=\frac{a_{0}+a_{1}h+\cdots+a_{\nu_{1}}h^{\nu_{1}}}{1+b_{1}h+\cdots+b_{\nu_{2}}h^{\nu_{2}}}+\mathcal{O}\left(h^{n+1}\right),$

where $\nu_{1}+\nu_{2}=n$. In our problem we take $n$ to be even and set
$\nu=\nu_{1}=\nu_{2}:=n/2$. The neighborhood in which the Padé approximation
(179) is decent is larger than the corresponding neighborhood of the Taylor
approximation (178) because if $x_{*}$ is a pole, then the denominator of
(179) can vanish near $x_{*}$. As the denominator has $\nu_{2}$ zeros, we see
that the closure of the domain of (179) can contain $\nu_{2}$ poles before we
lose control of the error term.

####  5.2.2. Padé Approximant Method for Generalized Hastings-McLeod.

Fix $\alpha>-1/2$ and let $u(y)$ be the generalized Hastings-McLeod solution
to (1). In order to obtain an approximation of $u$ in the form of (179) we
will need a pair of initial values $u(y_{0})$ and $u^{\prime}(y_{0})$. First,
we use the forward Euler numerical scheme to obtain a partial Taylor series
approximation of $u$ about $y_{0}$ for some even integer $n$ (in our problem
we set $n=24$). That is, we set $c_{0}=u(y_{0})$ and
$c_{1}=u^{\prime}\left(y_{0}\right)$ and write

(180)
$u(y_{0}+h)=\sum\limits_{k=0}^{n}c_{k}h^{k}+\mathcal{O}\left(h^{n+1}\right)$

for $n-1$ unknown complex values $c_{2},\dots,c_{n}$. To solve for
$c_{2},\dots,c_{n}$ we plug (180) into the ODE (1) and find that

$\sum\limits_{k=2}^{n}k(k-1)c_{k}h^{k-2}+\mathcal{O}\left(h^{n-1}\right)=2\left(\sum_{k=0}^{n}c_{k}h^{k}\right)^{3}+(y_{0}+h)\left(\sum\limits_{k=0}^{n}c_{k}h^{k}\right)-\alpha+\mathcal{O}\left(h^{n-1}\right).$

So, we can compare the coefficients of $1,h,\dots,h^{n-2}$ and obtain an
$(n-1)\times(n-1)$ system of equations (that can be solved recursively) for
the variables $c_{2},\dots,c_{n}$. To convert (180) into a Padé approximant,
with $n+1$ unknown complex values $a_{0},\dots,a_{\nu},b_{1},\dots b_{\nu}$ we
write

(181)
$\frac{a_{0}+\cdots+a_{\nu}h^{\nu}}{1+b_{1}h+\cdots+b_{\nu}h^{\nu}}=\sum\limits_{k=0}^{n}c_{k}h^{k}+\mathcal{O}\left(h^{n+1}\right).$

Now, by multiplying both sides of (181) by $1+b_{1}h+\cdots+b_{\nu}h^{\nu}$,
we can compare the coefficients of the $n+1$ monomials $1,h,\dots,h^{n}$ and
obtain an $(n+1)\times(n+1)$ system of equations that once again can be solved
recursively. Hence, we find that

(182)
$u(y_{0}+h)=\frac{a_{0}+\cdots+a_{\nu}h^{\nu}}{1+b_{1}h+\cdots+b_{\nu}h^{\nu}}+\mathcal{O}\left(h^{n+1}\right)$

in a neighborhood of $y_{0}$ for which $\nu$ poles have been excised.

Notice, in order to obtain an approximation resilient to $\nu$ poles, we just
needed the initial values $u(y_{0})$ and $u^{\prime}(y_{0})$. Further, (182)
can generate decent approximations of $u$ (and hence $u^{\prime}$) at points
that are not $y_{0}$. Thus, by choosing a point $y_{1}$ that is close enough
to $y_{0}$ such that (182) is still a good approximation of $u(y_{1})$, we are
able to extend the region of the $y$-plane where we can approximate $u$. As
such, given $U$, a “pole filled” region of the complex $y$-plane and $y_{0}\in
U$ such that the values $u(y_{0})$ and $u^{\prime}(y_{0})$ are known, we
create a coarse grid of node points filling out $U$, with the aim of finding a
Padé approximant near each node point. Fornberg and Weideman first implemented
such an algorithm in [15].

Their idea is to start at $y_{0}$ and to generate integration paths that
branch out so that each node point is near a path. Further, to preserve
accuracy by minimizing numerical cancellations, we want to avoid integrating
near a pole. Therefore, we want to choose paths along which the solution
remains small in magnitude. This is accomplished by running the following
routine.