# On the Algebraic Solutions of the Painlevé-III (D7) Equation

R. J. Buckingham and P. D. Miller  Department of Mathematical Sciences,
University of Cincinnati, P. O. Box 210025, Cincinnati, OH 45221
<EMAIL_ADDRESS>Department of Mathematics, University of Michigan, East Hall,
530 Church St., Ann Arbor, MI 48109<EMAIL_ADDRESS>
###### Abstract.

The D7 degeneration of the Painlevé-III equation has solutions that are
rational functions of $x^{1/3}$ for certain parameter values. We apply the
isomonodromy method to obtain a Riemann-Hilbert representation of these
solutions. We demonstrate the utility of this representation by analyzing
rigorously the behavior of the solutions in the large parameter limit.

R. J. Buckingham was supported by National Science Foundation grant
DMS-2108019. P. D. Miller was supported by National Science Foundation grant
DMS-1812625.

##  1\. Introduction

The six Painlevé equations were discovered more than a century ago by Paul
Painlevé in his classification of all second-order first-degree ordinary
differential equations algebraic in the unknown function, rational in its
first derivative, and analytic in the independent variable having the
_Painlevé property_ that all solutions are meromorphic away from certain fixed
singularities whose locations are fully determined from the equation itself.
It turned out that all such equations could be reduced either to previously
known (linear) equations or to one of the equations usually denoted PI, PII,
PIII, PIV, PV, and PVI. See [11, Ch. XIV]. All of these equations except for
PI involve one or more free parameters. As a consequence of the Painlevé
property, solutions of these equations may be regarded as new special
functions, and they occur in many applications. Although typical solutions of
Painlevé equations are highly-transcendental functions (indeed, they are
typically called _Painlevé transcendents_), all of the equations except for PI
also admit, for certain parameter values, solutions in terms of elementary
functions or classical linear special functions (e.g., Airy, Bessel, etc.).
The parameter values for these solutions are related by a certain finitely-
generated group action, and the group acts on the solutions via Bäcklund
transformations that preserve the functional character of the solution
(rational, algebraic, etc.).

From the very beginning, the Painlevé equations were recognized by R. Fuchs
and Garnier as isomonodromic deformations of certain second-order linear
equations (see also [19]). It is convenient to work at the level of first-
order systems instead, in which case each Painlevé equation can be recognized
as the compatibility condition for a certain _Lax pair_ of linear equations
for an auxiliary $2\times 2$ matrix-valued unknown $\boldsymbol{\Psi}$. The
inverse problem of constructing $\boldsymbol{\Psi}$ from its monodromy data
can be usefully formulated as a matrix Riemann-Hilbert problem. The
aforementioned group actions reappear in this context as _Schlesinger
transformations_ , linear gauge transformations acting on the matrix unknown
that preserve the form of the Lax pair as well as the essential monodromy
data, affecting only formal monodromy exponents at various singular points.
This means that the Riemann-Hilbert representation of the whole family of
special solutions can be obtained once it is known for just one particular
choice of parameters and solution, usually called the _seed solution_.
Generally the monodromy data for a given solution cannot be obtained
explicitly; however in the case of elementary-function solutions the direct
problem for the Lax pair can frequently be solved in terms of classical
special functions in which case the monodromy data can be found by applying
known connection formulæ. The isomonodromy method has been successfully
applied to rational solutions of

  * •

The PII equation [5, 6, 7, 10]. Here if one uses the Jimbo-Miwa [12] Lax pair
for the PII equation, the direct problem for the seed solution is solved in
terms of Airy functions. There is another approach for analyzing directly the
Yablonskii-Vorob’ev polynomials that can be used to construct the rational
solutions based on a Hankel determinant representation [1]; this was shown to
be equivalent to the isomonodromy approach in the alternate setting of the
Flaschka-Newell Lax pair in [17].

  * •

The PIII equation (nondegenerate D6 type) [3, 2]. The direct problem for the
seed solution is solved in terms of confluent hypergeometric functions
(Whittaker functions).

  * •

The PIV equation [8]. In this problem there are two distinct families of
rational solutions; the direct problem for the so-called generalized Hermite
rational solutions is solved in terms of elementary functions while that for
the so-called generalized Okamoto rational solutions is solved in terms of
Airy functions. The generalized Hermite solutions can be analyzed by means of
a Hankel determinant representation similarly to the PII case; see [4]. Yet
another approach in this case exploits a connection with the spectral theory
of quantum oscillators in quartic potentials; see [15, 16].

The goal of this paper is to show that the isomonodromy method applies equally
well to the algebraic solutions of the D7 degeneration of the PIII equation.
These are not rational solutions, although they are rational functions of the
cube root of the independent variable.

###  1.1. PIII D7 equation and its algebraic solutions

The Painlevé-III equation of D7 type has the form

(1)
$u^{\prime\prime}=\frac{(u^{\prime})^{2}}{u}-\frac{u^{\prime}}{x}+\frac{\alpha
u^{2}+\beta}{x}+\frac{\delta}{u},\quad u=u(x),\quad\delta\neq 0,$

which is a degenerate case of the general (D6) Painlevé-III equation

(2)
$u^{\prime\prime}=\frac{(u^{\prime})^{2}}{u}-\frac{u^{\prime}}{x}+\frac{\alpha
u^{2}+\beta}{x}+\gamma u^{3}+\frac{\delta}{u}$

in which $\gamma=0$. According to [20, §32.9(i)], if $\alpha=1$, $\beta=2n$
with $n\in\mathbb{Z}$, and $\delta=-1$, the D7 equation (1) has a solution
that is a rational function of $x^{1/3}$. When $n=0$, that solution is
$u(x)=x^{1/3}$. By scaling $u=\sqrt{8}\widetilde{u}$ and
$x=\sqrt{8}\widetilde{x}$ and dropping the tildes, we see that
$u(x)=\tfrac{1}{2}x^{1/3}$ is a solution of (1) with $\alpha=8$, $\beta=0$,
and $\delta=-1$, while for parameters $\alpha=8$, $\beta=2n$, and $\delta=-1$
there is a solution of (1) that is rational in $x^{1/3}$.

If one substitutes into (1) with $\alpha=8$, $\beta=2n$, and $\delta=-1$ the
formal expression $u=\tfrac{1}{2}x^{1/3}v(x^{1/3})$ where
$v(\zeta)=1+v_{1}\zeta^{-1}+v_{2}\zeta^{-2}+\cdots$, then obtains a systematic
recurrence to determine the coefficients $v_{j}$ in order that never requires
division by zero. Hence all coefficients are uniquely determined by the value
of $n$ (and in fact the coefficients of all odd powers of $\zeta$ vanish), so
the solution rational in $\zeta$ is unique for given $n$. We denote this
solution by $u=u_{n}(x)$. There is a sequence
$\\{R_{n}(\zeta)\\}_{n=0}^{\infty}$ of _Ohyama polynomials_ [9, 18] satisfying
a simple recurrence relation such that if $n>0$,
$u_{n}(x)=R_{n+1}(x^{1/3})R_{n-1}(x^{1/3})/R_{n}(x^{1/3})^{2}$. This
representation is consistent with the fact that all poles not at the origin of
solutions of (1) are double poles. As shown in [9, Sec. 3] and Figure 1, when
displayed in the $\zeta=x^{1/3}$ plane, the poles and zeros of $u_{n}(x)$ for
integers $n>0$ form an approximately crystalline pattern confined within two
quasi-triangular regions forming a “bow-tie” shape. One can also deduce from
$u_{0}(x)=\frac{1}{2}x^{1/3}$ and the Bäcklund transformations (131)–(132)
below (see also [9, Eqn. 3.4]) that

(3) $\zeta\mapsto\zeta^{*}\implies u_{n}\mapsto
u_{n}^{*}\quad\text{and}\quad\zeta\mapsto-\zeta\implies u_{n}\mapsto-
u_{n},\quad\text{for all }n\in\mathbb{Z}.$

Figure 1. Complex zeros of the Ohyama polynomials $R_{n}(\zeta)$. Left: $n=5$.
Center: $n=10$. Right: $n=20$.

###  1.2. Lax pair

In [13], the equation (1) is written in the form ($\alpha=-8\epsilon$,
$\beta=2ab$, $\delta=b^{2}$)

(4)
$u^{\prime\prime}=\frac{(u^{\prime})^{2}}{u}-\frac{u^{\prime}}{x}+\frac{-8\epsilon
u^{2}+2ab}{x}+\frac{b^{2}}{u},\quad u=u(x).$

The authors of [13] introduce the following linear equations on an unknown
$\boldsymbol{\Psi}=\boldsymbol{\Psi}(\lambda,x)$:

(5)
$\boldsymbol{\Psi}_{\lambda}=\boldsymbol{\Lambda}(\lambda,x)\boldsymbol{\Psi}\quad\text{and}\quad\boldsymbol{\Psi}_{x}=\mathbf{X}(\lambda,x)\boldsymbol{\Psi}$

where the coefficient matrices are defined by

(6)
$\boldsymbol{\Lambda}(\lambda,x):=-\mathrm{i}x\sigma_{3}-\frac{\mathrm{i}a}{2\lambda}\sigma_{3}-\frac{x}{\lambda}\mathbf{J}(x)+\frac{\mathrm{i}x}{2\lambda^{2}}\mathbf{K}(x)\quad\text{and}\quad\mathbf{X}(\lambda,x):=-\mathrm{i}\lambda\sigma_{3}+\frac{\mathrm{i}a}{2x}\sigma_{3}-\mathbf{J}(x)-\frac{\mathrm{i}}{2\lambda}\mathbf{K}(x),$

with

(7) $\mathbf{J}(x):=\frac{\epsilon x}{4u(x)}\begin{bmatrix}0&p(x)\\\
q(x)&0\end{bmatrix}\quad\text{and}\quad\mathbf{K}(x):=\frac{u(x)}{x}\begin{bmatrix}\epsilon&\mathrm{e}^{\mathrm{i}\varphi(x)}\\\
-\mathrm{e}^{-\mathrm{i}\varphi(x)}&-\epsilon\end{bmatrix}.$

The compatibility condition of the overdetermined system (5) is
$\boldsymbol{\Lambda}_{x}-\mathbf{X}_{\lambda}+[\boldsymbol{\Lambda},\mathbf{X}]=\mathbf{0}$,
which upon separating out the coefficients of $\lambda^{-1}$ and
$\lambda^{-2}$ amounts to the following system of equations:

(8) $\displaystyle
u^{\prime}(x)-\frac{u(x)}{x}-\frac{x}{2}\left[p(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}+q(x)\mathrm{e}^{\mathrm{i}\varphi(x)}\right]$
$\displaystyle=0,$ (9) $\displaystyle-\frac{\epsilon
x^{2}p^{\prime}(x)}{4u(x)}+\frac{\epsilon
x^{2}p(x)u^{\prime}(x)}{4u(x)^{2}}-\frac{\epsilon
xp(x)}{2u(x)}+\frac{\mathrm{i}\epsilon
axp(x)}{2u(x)}-2u(x)\mathrm{e}^{\mathrm{i}\varphi(x)}$ $\displaystyle=0,$ (10)
$\displaystyle-\frac{1}{2}u(x)\varphi^{\prime}(x)\mathrm{e}^{\mathrm{i}\varphi(x)}+\frac{1}{2}\mathrm{i}u^{\prime}(x)\mathrm{e}^{\mathrm{i}\varphi(x)}-\frac{\mathrm{i}}{2x}u(x)\mathrm{e}^{\mathrm{i}\varphi(x)}-\frac{\mathrm{i}\epsilon^{2}}{2}xp(x)$
$\displaystyle=0,$ (11) $\displaystyle-\frac{\epsilon
x^{2}q^{\prime}(x)}{4u(x)}+\frac{\epsilon
x^{2}q(x)u^{\prime}(x)}{4u(x)^{2}}-\frac{\epsilon
xq(x)}{2u(x)}-\frac{\mathrm{i}\epsilon
axq(x)}{2u(x)}-2u(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}$ $\displaystyle=0,$
(12)
$\displaystyle-\frac{1}{2}u(x)\varphi^{\prime}(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}-\frac{1}{2}\mathrm{i}u^{\prime}(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}+\frac{\mathrm{i}}{2x}u(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}+\frac{\mathrm{i}\epsilon^{2}}{2}xq(x)$
$\displaystyle=0.$

We eliminate $p(x)$ and $q(x)$ explicitly using (10) and (12):

(13)
$\begin{split}p(x)&=\frac{1}{\epsilon^{2}}\left[\frac{\mathrm{i}u(x)\varphi^{\prime}(x)\mathrm{e}^{\mathrm{i}\varphi(x)}}{x}+\frac{u^{\prime}(x)\mathrm{e}^{\mathrm{i}\varphi(x)}}{x}-\frac{u(x)\mathrm{e}^{\mathrm{i}\varphi(x)}}{x^{2}}\right]=\frac{1}{\epsilon^{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u(x)\mathrm{e}^{\mathrm{i}\varphi(x)}}{x}\right)\\\
q(x)&=\frac{1}{\epsilon^{2}}\left[-\frac{\mathrm{i}u(x)\varphi^{\prime}(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}}{x}+\frac{u^{\prime}(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}}{x}-\frac{u(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}}{x^{2}}\right]=\frac{1}{\epsilon^{2}}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}}{x}\right).\end{split}$

Then (8) becomes

(14)
$\left(1-\frac{1}{\epsilon^{2}}\right)\left(u^{\prime}(x)-\frac{u(x)}{x}\right)=0,$

and the sum and difference of (9) and (11) become, respectively,

(15) $\displaystyle-8\epsilon
u(x)^{3}-u(x)u^{\prime}(x)+xu^{\prime}(x)^{2}-2au(x)^{2}\varphi^{\prime}(x)+xu(x)^{2}\varphi^{\prime}(x)^{2}-xu(x)u^{\prime\prime}(x)$
$\displaystyle=0$ (16) $\displaystyle
u(x)\varphi^{\prime\prime}(x)+u^{\prime}(x)\varphi^{\prime}(x)-\frac{2au^{\prime}(x)}{x}+\frac{2au(x)}{x^{2}}=0.$

We note that (16) can be written in the form

(17)
$\frac{\mathrm{d}}{\mathrm{d}x}\left(u(x)\varphi^{\prime}(x)-\frac{2au(x)}{x}\right)=0\implies
u(x)\varphi^{\prime}(x)=\frac{2au(x)}{x}+b$

where $b$ is an integration constant. Using this to eliminate
$u(x)\varphi^{\prime}(x)$ from (15) gives the equation (4) on $u(x)$. If we
assume that $\epsilon^{2}=1$, then (14) places no further conditions on $u(x)$
(otherwise conditions on the parameters $\epsilon$, $a$, and $b$ are required
so that (4) admits a solution of the form $u(x)=Ax$ for $A\neq 0$).

Using $\epsilon^{2}=1$, we note that, in terms of the matrix elements of
$\mathbf{J}(x)$ and $\mathbf{K}(x)$, the potentials are given by

(18) $\begin{split}u(x)&=\epsilon xK_{11}(x)=-\epsilon xK_{22}(x)\\\
\mathrm{e}^{\mathrm{i}\varphi(x)}&=\epsilon\frac{K_{12}(x)}{K_{11}(x)}=-\epsilon\frac{K_{12}(x)}{K_{22}(x)}\\\
\mathrm{e}^{-\mathrm{i}\varphi(x)}&=-\epsilon\frac{K_{21}(x)}{K_{11}(x)}=\epsilon\frac{K_{21}(x)}{K_{22}(x)}\\\
p(x)&=4J_{12}(x)K_{11}(x)=-4J_{12}(x)K_{22}(x)\\\
q(x)&=4J_{21}(x)K_{11}(x)=-4J_{21}(x)K_{22}(x).\end{split}$

The equivalence of the two expressions in each case is guaranteed from
$\mathrm{tr}(\mathbf{K}(x))=0$, and the compatibility of the expressions for
$\mathrm{e}^{\pm\mathrm{i}\varphi(x)}$ is implied by $\det(\mathbf{K}(x))=0$.
If we are given the matrices $\mathbf{J}(x)$ (off-diagonal) and
$\mathbf{K}(x)$ (singular and nondiagonalizable), we cannot determine the
value of $\epsilon=\pm 1$. However, from (18) we can see that
$\epsilon\mapsto-\epsilon$ changes the signs of $u(x)$ and
$\mathrm{e}^{\pm\mathrm{i}\varphi(x)}$ but leaves $p(x)$ and $q(x)$ invariant.
It follows that the equations (8)–(12) are invariant under
$\epsilon\mapsto-\epsilon$, and from (17) we see that the integration constant
$b$ is proportional to $\epsilon$. This is then consistent with the obvious
symmetry of (4): $u\mapsto-u$, $\epsilon\mapsto-\epsilon$, $b\mapsto-b$.

Therefore, for the rest of this work, we assume without loss of generality
that $\epsilon=-1$ in the parametrization of the matrices $\mathbf{J}(x)$ and
$\mathbf{K}(x)$ in the Lax pair (5)–(7). Then eliminating $u^{\prime}(x)$
between (10) and (12) allows the integration constant $b$ to be expressed
without derivatives as

(19)
$b=-\frac{2au(x)}{x}-\frac{1}{2}\mathrm{i}xp(x)\mathrm{e}^{-\mathrm{i}\varphi(x)}+\frac{1}{2}\mathrm{i}xq(x)\mathrm{e}^{\mathrm{i}\varphi(x)}.$

###  1.3. Outline of the paper

In Section 2 we show how simultaneous solutions of the Lax pair equations for
the simplest algebraic solution of (4), $u_{0}(x)=\frac{1}{2}x^{1/3}$ for
$\epsilon=-1$, $a=0$, and $b=\mathrm{i}$, can be explicitly obtained in terms
of Airy functions. Then we build canonical bases of simultaneous solutions
near the singular points at $\lambda=0$ and $\lambda=\infty$ and apply
connection formulæ to determine relations among them. Next, in Section 3 we
formulate the inverse monodromy problem for the rational solution $u_{n}(x)$
for $a=-\mathrm{i}n$ and $n\in\mathbb{Z}$ as a Riemann-Hilbert problem. We
demonstrate that the problem is solved for $n=0$ by the seed Lax pair
solutions constructed in Section 2. We then derive differential equations from
its solution in general, recovering the PIII (D7) equation in the form (4) for
$\epsilon=-1$, $a=-\mathrm{i}n$, and $b=\mathrm{i}$. Next we show that the
solutions are related for consecutive values of $n$ by Schlesinger
transformations and hence the solutions of (4) are related by Bäcklund
transformations. Since the latter preserve the algebraic character of the
solution, this shows that the Riemann-Hilbert problem captures the algebraic
solutions of (4); we summarize this result in Theorem 1 below. We conclude
this section by making a convenient transformation of the Riemann-Hilbert
problem that has the effect of simplifying the data for the problem. Then, in
Section 4 we use the Riemann-Hilbert representation of the algebraic solution
$u_{n}(x)$ to consider the asymptotic behavior of the solution for large $n$.
After some preliminary rescaling of the independent variable $x$ and the
spectral variable $\lambda$ to balance exponents, we present for background
some formal arguments applying similar scalings to the PIII (D7) equation
itself. This formal approach suggests two types of approximations: a slowly-
varying “equilibrium” approximation and an approximation based on the
Weierstraß elliptic function. Then, we return to the rescaled Riemann-Hilbert
problem and introduce an appropriate $g$-function via a family of spectral
curves. The family of spectral curves mirrors the family of formally-
approximate solutions of the PIII (D7) equation in a remarkable fashion.
Finally, we carry out a rigorous analysis of the function $u_{n}(x)$ for large
$n$ and rescaled $x>0$ sufficiently large and recover one of the equilibrium
solutions predicted by the formal theory (see Theorem 2 below). With some
extra steps, the method allows this result to be continued to the exterior of
the “bow-tie” domain of the $x^{1/3}$ plane and hence makes rigorous some of
the observations made in [9]. See Theorem 3 below.

###  1.4. Notation

Throughout our paper, square matrices are indicated by boldface capital
letters with the exception of the identity $\mathbb{I}$ and the Pauli
matrices:

(20) $\sigma_{1}:=\begin{bmatrix}0&1\\\
1&0\end{bmatrix},\quad\sigma_{2}:=\begin{bmatrix}0&-\mathrm{i}\\\
\mathrm{i}&0\end{bmatrix},\quad\sigma_{3}:=\begin{bmatrix}1&0\\\
0&-1\end{bmatrix}.$

For a function (matrix or scalar-valued) analytic off an oriented contour arc,
we use subscripts $+$ (resp., $-$) to indicate boundary values at a point of
the arc from the left (resp., right).

##  2\. The direct monodromy problem for the seed solution

###  2.1. Lax pair for the seed solution

To obtain the algebraic solutions of the Painlevé-III D7 equation in the form
(4), we assume that $\epsilon=-1$, $a=-\mathrm{i}n$, and $b=\mathrm{i}$ for
$n\in\mathbb{Z}$. In the case that $n=0$ we have the algebraic solution
$u(x)=\tfrac{1}{2}x^{1/3}$. We call this the _seed solution_ for the algebraic
solution family, and $n=0$ is the _seed parameter value_. Taking $a=0$,
$b=\mathrm{i}$, and $u(x)=\tfrac{1}{2}x^{1/3}$ in (17) gives
$\varphi^{\prime}(x)=2\mathrm{i}x^{-1/3}$ so that
$\varphi(x)=3\mathrm{i}x^{2/3}+\varphi_{0}$ where $\varphi_{0}$ is an
integration constant that we will take to be zero. From (13) and $\epsilon=-1$
we then get

(21)
$p(x)=-\left(\frac{1}{x}+\frac{1}{3}x^{-5/3}\right)\mathrm{e}^{-3x^{2/3}}\quad\text{and}\quad
q(x)=-\left(-\frac{1}{x}+\frac{1}{3}x^{-5/3}\right)\mathrm{e}^{3x^{2/3}}.$

It follows that for the seed solution, the matrices $\mathbf{J}(x)$ and
$\mathbf{K}(x)$ in (7) are given by

(22)
$\mathbf{J}(x)=\begin{bmatrix}0&\left(\tfrac{1}{6}x^{-1}+\tfrac{1}{2}x^{-1/3}\right)\mathrm{e}^{-3x^{2/3}}\\\
\left(\tfrac{1}{6}x^{-1}-\tfrac{1}{2}x^{-1/3}\right)\mathrm{e}^{3x^{2/3}}&0\end{bmatrix},\;\mathbf{K}(x)=\frac{1}{2}x^{-2/3}\begin{bmatrix}-1&\mathrm{e}^{-3x^{2/3}}\\\
-\mathrm{e}^{3x^{2/3}}&1\end{bmatrix}.$

We make a gauge transformation to remove the exponential factors on the off-
diagonal elements of the coefficient matrices:
$\boldsymbol{\Psi}=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\boldsymbol{\Phi}$, which
implies that also
$\boldsymbol{\Psi}_{x}=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}(\boldsymbol{\Phi}_{x}-x^{-1/3}\sigma_{3}\boldsymbol{\Phi})$.
In terms of $\boldsymbol{\Phi}$ then, the (compatible) Lax pair equations for
the seed solution read

(23)
$\boldsymbol{\Phi}_{\lambda}=\left(-\mathrm{i}x\sigma_{3}-\frac{x}{\lambda}\begin{bmatrix}0&\tfrac{1}{6}x^{-1}+\tfrac{1}{2}x^{-1/3}\\\
\tfrac{1}{6}x^{-1}-\tfrac{1}{2}x^{-1/3}&0\end{bmatrix}+\frac{\mathrm{i}x^{1/3}}{4\lambda^{2}}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\right)\boldsymbol{\Phi},$ (24)
$\boldsymbol{\Phi}_{x}=\left(-\mathrm{i}\lambda\sigma_{3}-\begin{bmatrix}-x^{-1/3}&\tfrac{1}{6}x^{-1}+\tfrac{1}{2}x^{-1/3}\\\
\tfrac{1}{6}x^{-1}-\tfrac{1}{2}x^{-1/3}&x^{-1/3}\end{bmatrix}-\frac{\mathrm{i}x^{-2/3}}{4\lambda}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\right)\boldsymbol{\Phi}.$

Our strategy to obtain a fundamental simultaneous solution matrix for
(23)–(24) is to deal first with the equation (24). For this purpose, it is
convenient to simplify the latter equation as much as possible. We observe
that multiples of the same diagonal coefficient matrix appear in the terms
most singular at $\lambda=\infty$ in both equations (23)–(24), and at the same
time the coefficient matrix of the terms most singular at $\lambda=0$ is
nilpotent (that this coefficient matrix is defective for all $x$ is what
distinguishes the D7 Lax pair from the most general D6 Lax pair given by Jimbo
and Miwa) which is a simplification to be retained. Hence we will remove the
leading terms at $\lambda=\infty$ from (24) by introducing the shearing
transformation

(25) $\lambda=X^{-1}\Lambda,\;x=X\iff\Lambda=x\lambda,\;X=x.$

Then the partial derivatives with respect to $\lambda$ and $x$ transform as
follows:

(26)
$\frac{\partial}{\partial\Lambda}=\frac{1}{X}\frac{\partial}{\partial\lambda}\quad\text{and}\quad\frac{\partial}{\partial
X}=\frac{\partial}{\partial
x}-\frac{\Lambda}{X^{2}}\frac{\partial}{\partial\lambda}.$

In terms of the new variables $\Lambda$ and $X$, the compatible system
(23)–(24) becomes

(27)
$\boldsymbol{\Phi}_{\Lambda}=\left(-\mathrm{i}\sigma_{3}-\frac{X}{\Lambda}\begin{bmatrix}0&\tfrac{1}{6}X^{-1}+\tfrac{1}{2}X^{-1/3}\\\
\tfrac{1}{6}X^{-1}-\tfrac{1}{2}X^{-1/3}&0\end{bmatrix}+\frac{\mathrm{i}X^{4/3}}{4\Lambda^{2}}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\right)\boldsymbol{\Phi}$

and

(28)
$\boldsymbol{\Phi}_{X}=\left(X^{-1/3}\sigma_{3}-\frac{\mathrm{i}X^{1/3}}{2\Lambda}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\right)\boldsymbol{\Phi}.$

(We note that the shearing transformation had the added benefit of removing
the off-diagonal terms from the coefficient of $\lambda^{0}$.) To solve (28)
for fixed $\Lambda\in\mathbb{C}$, we make a change of independent variable:

(29)
$Z=X^{1/3}\implies\frac{\mathrm{d}}{\mathrm{d}X}=\frac{1}{3}X^{-2/3}\frac{\mathrm{d}}{\mathrm{d}Z}=\frac{1}{3}Z^{-2}\frac{\mathrm{d}}{\mathrm{d}Z}.$

Then (28) becomes

(30)
$\boldsymbol{\Phi}_{Z}=\left(3Z\sigma_{3}-\frac{3\mathrm{i}Z^{3}}{2\Lambda}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\right)\boldsymbol{\Phi}.$

Introducing the constant gauge transformation

(31)
$\boldsymbol{\Omega}:=\mathbf{G}\boldsymbol{\Phi},\quad\mathbf{G}:=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\\
1&1\end{bmatrix}$

and noting that

(32)
$\mathbf{G}\sigma_{3}\mathbf{G}^{-1}=\sigma_{1}\quad\text{and}\quad\mathbf{G}\begin{bmatrix}-1&1\\\
-1&1\end{bmatrix}\mathbf{G}^{-1}=\begin{bmatrix}0&0\\\ -2&0\end{bmatrix},$

we get a purely off-diagonal system

(33)
$\boldsymbol{\Omega}_{Z}=\left(3Z\sigma_{1}-\frac{3\mathrm{i}Z^{3}}{2\Lambda}\begin{bmatrix}0&0\\\
-2&0\end{bmatrix}\right)\boldsymbol{\Omega}=\begin{bmatrix}0&3Z\\\
3Z+3\mathrm{i}\Lambda^{-1}Z^{3}&0\end{bmatrix}\boldsymbol{\Omega}.$

Therefore, denoting by $D(Z)$ and $S(Z)$ the first-row and second-row elements
respectively of either column of $\boldsymbol{\Omega}$,

(34) $S_{Z}=(3Z+3\mathrm{i}\Lambda^{-1}Z^{3})D\quad\text{and}\quad D_{Z}=3ZS.$

Eliminating $S$ using the second equation gives a second-order equation on
$D$:

(35)
$Z^{2}D_{ZZ}-ZD_{Z}-4Z^{4}(\tfrac{9}{4}+\tfrac{9\mathrm{i}}{4}\Lambda^{-1}Z^{2})D=0.$

It seems like a good idea to consider the substitution
$W:=\tfrac{9}{4}+\tfrac{9\mathrm{i}}{4}\Lambda^{-1}Z^{2}$. Then a simple
computation shows that also

(36)
$\frac{\mathrm{d}}{\mathrm{d}Z}=\frac{9\mathrm{i}Z}{2\Lambda}\frac{\mathrm{d}}{\mathrm{d}W},\quad\frac{\mathrm{d}^{2}}{\mathrm{d}Z^{2}}=\frac{9\mathrm{i}}{2\Lambda}\frac{\mathrm{d}}{\mathrm{d}W}-\frac{81Z^{2}}{4\Lambda^{2}}\frac{\mathrm{d}^{2}}{\mathrm{d}W^{2}}\implies
Z^{2}D_{ZZ}-ZD_{Z}=-\frac{81Z^{4}}{4\Lambda^{2}}D_{WW}.$

Therefore, the equation (35) in fact becomes

(37) $D_{WW}+\frac{16\Lambda^{2}}{81}WD=0$

which is a scaling of Airy’s equation. Its general solution is

(38)
$D=A\mathrm{Ai}\left(\left(-\frac{16\Lambda^{2}}{81}\right)^{1/3}W\right)+B\mathrm{Bi}\left(\left(-\frac{16\Lambda^{2}}{81}\right)^{1/3}W\right)$

where $A,B$ are independent of $W$ (they may depend on $\Lambda$, however).

If $f(\xi)$ is any solution of Airy’s equation $f^{\prime\prime}(\xi)-\xi
f(\xi)=0$, then we have

(39)
$D(Z)=f(\xi),\quad\xi:=\left(\frac{3}{2}\right)^{2/3}(\mathrm{i}\Lambda)^{2/3}\left(1-\frac{Z^{2}}{\mathrm{i}\Lambda}\right),$

and from (34),

(40)
$S(Z)=\frac{D_{Z}(Z)}{3Z}=-\left(\frac{2}{3}\right)^{1/3}(\mathrm{i}\Lambda)^{-1/3}f^{\prime}(\xi).$

So, if $f_{1}(\xi)$ and $f_{2}(\xi)$ form a fundamental pair of solutions of
Airy’s equation $f^{\prime\prime}(\xi)-\xi f(\xi)=0$, then the general
solution of (33) is
$\boldsymbol{\Omega}=\boldsymbol{\Omega}_{0}(Z,\Lambda)\mathbf{H}(\Lambda)$
where $\mathbf{H}(\Lambda)$ is an arbitrary matrix-valued function of
$\Lambda$ only, and where

(41)
$\boldsymbol{\Omega}_{0}(Z,\Lambda):=\boldsymbol{\Delta}(\Lambda)\mathbf{F}(\zeta),\quad\boldsymbol{\Delta}(\Lambda):=\frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\\
0&-\left(\tfrac{2}{3}\right)^{1/3}(\mathrm{i}\Lambda)^{-1/3}\end{bmatrix},\quad\mathbf{F}(\xi):=\begin{bmatrix}f_{1}(\xi)&f_{2}(\xi)\\\
f_{1}^{\prime}(\xi)&f_{2}^{\prime}(\xi)\end{bmatrix},$

and $\xi$ is defined in terms of $Z$ and $\Lambda$ by (39). (The factor of
$1/\sqrt{2}$ is arbitrary and is chosen for later convenience.)

Using (32) and

(42)
$\mathbf{G}\sigma_{1}\mathbf{G}^{-1}=-\sigma_{3}\quad\text{and}\quad\mathbf{G}\begin{bmatrix}0&-1\\\
1&0\end{bmatrix}\mathbf{G}^{-1}=\begin{bmatrix}0&1\\\ -1&0\end{bmatrix},$

we derive from (27) and (31) that

(43)
$\boldsymbol{\Omega}_{\Lambda}=\begin{bmatrix}\displaystyle\frac{1}{6\Lambda}&\displaystyle-\mathrm{i}-\frac{X^{2/3}}{2\Lambda}\vspace{.02in}\\\
\displaystyle-\mathrm{i}+\frac{X^{2/3}}{2\Lambda}-\frac{\mathrm{i}X^{4/3}}{2\Lambda^{2}}&\displaystyle-\frac{1}{6\Lambda}\end{bmatrix}\boldsymbol{\Omega}=\begin{bmatrix}\displaystyle\frac{1}{6\Lambda}&\displaystyle-\mathrm{i}-\frac{Z^{2}}{2\Lambda}\vspace{.02in}\\\
\displaystyle-\mathrm{i}+\frac{Z^{2}}{2\Lambda}-\frac{\mathrm{i}Z^{4}}{2\Lambda^{2}}&\displaystyle-\frac{1}{6\Lambda}\end{bmatrix}\boldsymbol{\Omega}.$

Since the systems (33) and (43) are compatible, it is possible to find
$\mathbf{H}(\Lambda)$ so that
$\boldsymbol{\Omega}=\boldsymbol{\Omega}_{0}(Z,\Lambda)\mathbf{H}(\Lambda)$
solves not only (33) but also (43). Substituting into the latter shows that
$\mathbf{H}(\Lambda)$ must solve the system

(44)
$\mathbf{H}^{\prime}(\Lambda)=\left(\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\begin{bmatrix}\displaystyle\frac{1}{6\Lambda}&\displaystyle-\mathrm{i}-\frac{Z^{2}}{2\Lambda}\vspace{.02in}\\\
\displaystyle-\mathrm{i}+\frac{Z^{2}}{2\Lambda}-\frac{\mathrm{i}Z^{4}}{2\Lambda^{2}}&\displaystyle-\frac{1}{6\Lambda}\end{bmatrix}\boldsymbol{\Omega}_{0}(Z,\Lambda)-\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\frac{\partial\boldsymbol{\Omega}_{0}}{\partial\Lambda}(Z,\Lambda)\right)\mathbf{H}(\Lambda).$

Using (41) shows that

(45)
$\begin{split}\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\frac{\partial\boldsymbol{\Omega}_{0}}{\partial\Lambda}(Z,\Lambda)&=\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\boldsymbol{\Delta}^{\prime}(\Lambda)\boldsymbol{\Delta}(\Lambda)^{-1}\boldsymbol{\Omega}_{0}(Z,\Lambda)+\frac{\partial\xi}{\partial\Lambda}\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\boldsymbol{\Delta}(\Lambda)\mathbf{F}^{\prime}(\xi)\\\
\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\frac{\partial\boldsymbol{\Omega}_{0}}{\partial
Z}(Z,\Lambda)&=\frac{\partial\zeta}{\partial
Z}\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\boldsymbol{\Delta}(\Lambda)\mathbf{F}^{\prime}(\xi).\end{split}$

Eliminating $\mathbf{F}^{\prime}(\xi)$ between these two relations gives

(46)
$\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\frac{\partial\boldsymbol{\Omega}_{0}}{\partial\Lambda}(Z,\Lambda)=\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\boldsymbol{\Delta}^{\prime}(\Lambda)\boldsymbol{\Delta}(\Lambda)^{-1}\boldsymbol{\Omega}_{0}(Z,\Lambda)+\left(\frac{\partial\xi}{\partial
Z}\right)^{-1}\frac{\partial\xi}{\partial\Lambda}\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\frac{\partial\boldsymbol{\Omega}_{0}}{\partial
Z}(Z,\Lambda).$

The $Z$-derivative that remains can be eliminated because
$\boldsymbol{\Omega}_{0}(Z,\Lambda)$ is a solution of (33). Using also that,
from (39),

(47) $\left(\frac{\partial\xi}{\partial
Z}\right)^{-1}\frac{\partial\xi}{\partial\Lambda}=\frac{1}{3\mathrm{i}Z}-\frac{Z}{6\Lambda},$

we therefore find that
$\mathbf{H}^{\prime}(\Lambda)=\boldsymbol{\Omega}_{0}(Z,\Lambda)^{-1}\mathbf{C}(Z,\Lambda)\boldsymbol{\Omega}_{0}(Z,\Lambda)\mathbf{H}(\Lambda)$,
where

(48)
$\mathbf{C}(Z,\Lambda):=\begin{bmatrix}\displaystyle\frac{1}{6\Lambda}&\displaystyle-\mathrm{i}-\frac{Z^{2}}{2\Lambda}\vspace{.02in}\\\
\displaystyle-\mathrm{i}+\frac{Z^{2}}{2\Lambda}-\frac{\mathrm{i}Z^{4}}{2\Lambda^{2}}&\displaystyle-\frac{1}{6\Lambda}\end{bmatrix}-\boldsymbol{\Delta}^{\prime}(\Lambda)\boldsymbol{\Delta}(\Lambda)^{-1}-\left(\frac{1}{3\mathrm{i}Z}-\frac{Z}{6\Lambda}\right)\begin{bmatrix}0&3Z\\\
3Z+3\mathrm{i}\Lambda^{-1}Z^{3}&0\end{bmatrix}.$

Differentiating $\boldsymbol{\Delta}(\Lambda)$ using (41) then shows that a
dramatic simplification occurs:
$\mathbf{C}(Z,\Lambda)=(6\Lambda)^{-1}\mathbb{I}$. Because this is a multiple
of the identity, it follows that the differential equation satisfied by
$\mathbf{H}(\Lambda)$ reads simply

(49)
$\mathbf{H}^{\prime}(\Lambda)=\frac{1}{6\Lambda}\mathbf{H}(\Lambda)\implies\mathbf{H}(\Lambda)=\Lambda^{1/6}\mathbf{H}_{0}$

where $\mathbf{H}_{0}$ is a matrix independent of both $Z$ and $\Lambda$. It
can be absorbed as a change of basis in specifying the fundamental solutions
$f_{1}(\xi)$ and $f_{2}(\xi)$ of Airy’s equation.

Inverting the gauge transformations
$\boldsymbol{\Phi}\mapsto\boldsymbol{\Omega}$ and
$\boldsymbol{\Psi}\mapsto\boldsymbol{\Phi}$ and restoring the original
independent variables by $Z=X^{1/3}=x^{1/3}$ and $\Lambda=x\lambda$, the
general simultaneous solution matrix of the Lax pair for the seed solution is:

(50)
$\boldsymbol{\Psi}(\lambda,x)=(\mathrm{i}x\lambda)^{1/6}\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\mathbf{G}^{-1}\boldsymbol{\Delta}(x\lambda)\mathbf{F}(\xi),\quad\xi:=\left(\frac{3}{2}\right)^{2/3}(\mathrm{i}x\lambda)^{2/3}\left(1-\frac{x^{2/3}}{\mathrm{i}x\lambda}\right),$

where $\mathbf{F}(\xi)$ is built from two independent solutions $f_{1}(\xi)$,
$f_{2}(\xi)$ of Airy’s equation $f^{\prime\prime}(\xi)-\xi f(\xi)=0$ by (41).
The extra factor of $\mathrm{i}^{1/6}$ is included at no cost to give the
simplification

(51)
$(\mathrm{i}x\lambda)^{1/6}\boldsymbol{\Delta}(x\lambda)=\frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\\
0&-(\tfrac{2}{3})^{1/3}\end{bmatrix}(\mathrm{i}x\lambda)^{\sigma_{3}/6}.$

###  2.2. Three particular simultaneous solutions of the Lax pair for the seed

We assume here for simplicity that $x>0$ and that
$(\mathrm{i}x\lambda)^{p}=x^{p}(\mathrm{i}\lambda)^{p}$ for any power $p$. In
particular, $\xi$ is then well-defined as a function on the $\lambda$-plane
once we select the principal branch for $(\mathrm{i}\lambda)^{2/3}$. It has a
branch cut emanating vertically upwards from $\lambda=0$ along the imaginary
axis, and it is positive real on the negative imaginary axis.

We will make use of the fact that for the Airy equation
$f^{\prime\prime}(\xi)-\xi f(\xi)=0$ there are three complementary sectors:
$0\leq\arg(\xi)\leq\tfrac{2}{3}\pi$, $-\tfrac{2}{3}\pi\leq\arg(\xi)\leq 0$,
and $-\tfrac{1}{3}\pi\leq\arg(-\xi)\leq\tfrac{1}{3}\pi$, on each of which
there is a basis of solutions with exponential dichotomy and exhibiting no
Stokes phenomenon as $\xi\to\infty$. Those fundamental pairs are
$(\mathrm{Ai}(\xi),\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi))$,
$(\mathrm{Ai}(\xi),\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi))$, and
$(\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi),\mathrm{Ai}(\mathrm{e}^{-2\pi\i/3}\xi))$,
respectively.

####  2.2.1. Solutions near infinity

Here we consider the domain $|\lambda|>1$ and find simultaneous solutions with
no Stokes phenomenon in two complementary sectors near $\lambda=\infty$ and
that admit a simple normalization as $\lambda\to\infty$ in each of these
sectors. To be precise, we take unbounded domains $D_{\infty}^{\pm}$ defined
by the conditions $|\lambda|>1$ and $\pm\mathrm{Re}(\lambda)>0$. By the
definition of $\xi$ for $x>0$ in terms of the principal branch of
$(\mathrm{i}\lambda)^{2/3}$, in the limit $\lambda\to\infty$ from
$D_{\infty}^{+}$, $\arg(\mathrm{i}\lambda)\in(0,\pi)$ translates into
$\xi\to\infty$ with $\arg(\xi)\in(0,\tfrac{2}{3}\pi)$. Likewise as
$\lambda\to\infty$ from $D_{\infty}^{-}$ we have $\xi\to\infty$ with
$-\tfrac{2}{3}\pi<\arg(\xi)<0$. Therefore, to avoid Stokes phenomenon, we will
take for $f(\xi)$ scalar multiples of the solutions from the fundamental pairs
$(\mathrm{Ai}(\xi),\mathrm{Ai}(\mathrm{e}^{\mp 2\pi\mathrm{i}/3}\xi))$ to
define simultaneous solutions
$\boldsymbol{\Psi}=\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x)$ on the domains
$D_{\infty}^{\pm}$ respectively.

That is, using (51), we take

(52)
$\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x):=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\mathbf{G}^{-1}\frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\\
0&-(\tfrac{2}{3})^{1/3}\end{bmatrix}(\mathrm{i}x\lambda)^{\sigma_{3}/6}\begin{bmatrix}\mathrm{Ai}(\xi)&\mathrm{Ai}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)\\\ \mathrm{Ai}^{\prime}(\xi)&\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)\end{bmatrix}\begin{bmatrix}c_{1}^{\pm}&0\\\
0&c_{2}^{\pm}\end{bmatrix}$

for suitable constants $c_{j}^{\pm}$, $j=1,2$. This can be equivalently
written as

(53)
$\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x)=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\begin{bmatrix}\frac{1}{2}&\frac{1}{2}\vspace{.02in}\\\
-\frac{1}{2}&\frac{1}{2}\end{bmatrix}\\\
{}\cdot\begin{bmatrix}c_{1}^{\pm}(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\xi)&c_{2}^{\pm}(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)\vspace{.02in}\\\
-c_{1}^{\pm}(\frac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{Ai}^{\prime}(\xi)&-c_{2}^{\pm}(\frac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)\end{bmatrix}.$

As $\lambda\to\infty$ from $D_{\infty}^{\pm}$, i.e., with
$\pm\arg(\mathrm{i}x\lambda)\in[0,\pi]$, we have $\xi\to\infty$ with
$\pm\arg(\xi)\in[0,\tfrac{2}{3}\pi]$, and also $\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi\to\infty$ with $\mp\arg(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)\in[0,\tfrac{2}{3}\pi]$. Using the asymptotic formulæ
[20, Eqns. 9.7.5–6]

(54)
$\mathrm{Ai}(z)=\frac{\mathrm{e}^{-2z^{3/2}/3}}{2\sqrt{\pi}z^{1/4}}(1+\mathcal{O}(z^{-3/2}))\;\text{and}\;\mathrm{Ai}^{\prime}(z)=-\frac{\mathrm{e}^{-2z^{3/2}/3}z^{1/4}}{2\sqrt{\pi}}(1+\mathcal{O}(z^{-3/2})),$

valid as $z\to\infty$ with $|\arg(z)|<\pi$ (and hence
$|\arg(z^{3/2})|<\tfrac{3}{2}\pi$ and $|\arg(z^{1/4})|<\tfrac{1}{4}\pi$),
along with
$\xi=(\frac{3}{2})^{2/3}(\mathrm{i}x\lambda)^{2/3}(1-x^{2/3}(\mathrm{i}x\lambda)^{-1})$,
we get

(55)
$(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\xi)=\frac{\mathrm{e}^{3x^{2/3}/2}}{2\sqrt{\pi}}(\tfrac{2}{3})^{1/6}\mathrm{e}^{-\mathrm{i}x\lambda}(1+\mathcal{O}(\lambda^{-1}))$

and

(56)
$-(\tfrac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{Ai}^{\prime}(\xi)=\frac{\mathrm{e}^{3x^{2/3}/2}}{2\sqrt{\pi}}(\tfrac{2}{3})^{1/6}\mathrm{e}^{-\mathrm{i}x\lambda}(1+\mathcal{O}(\lambda^{-1}))$

as $\lambda\to\infty$ in either $D_{\infty}^{+}$ or $D_{\infty}^{-}$.
Likewise,

(57) $(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)=\mathrm{e}^{\pm\mathrm{i}\pi/6}\frac{\mathrm{e}^{-3x^{2/3}/2}}{2\sqrt{\pi}}(\tfrac{2}{3})^{1/6}\mathrm{e}^{\mathrm{i}x\lambda}(1+\mathcal{O}(\lambda^{-1}))$

and

(58) $-(\tfrac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{\mp
2\pi\mathrm{i}/3}\xi)=\mathrm{e}^{\mp
5\pi\mathrm{i}/6}\frac{\mathrm{e}^{-3x^{2/3}/2}}{2\sqrt{\pi}}(\tfrac{2}{3})^{1/6}\mathrm{e}^{\mathrm{i}x\lambda}(1+\mathcal{O}(\lambda^{-1}))$

as $\lambda\to\infty$ from $D_{\infty}^{\pm}$. Therefore, from (53), we get

(59)
$\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x)=\frac{1}{2\sqrt{\pi}}(\tfrac{2}{3})^{1/6}\left(\begin{bmatrix}c_{1}^{\pm}&0\\\
0&-\mathrm{e}^{\pm\mathrm{i}\pi/6}c_{2}^{\pm}\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}}$

in the same limit. Choosing the constants to be

(60) $c_{1}^{\pm}:=2\sqrt{\pi}(\tfrac{3}{2})^{1/6}\quad\text{and}\quad
c_{2}^{\pm}:=-\mathrm{e}^{\mp\mathrm{i}\pi/6}2\sqrt{\pi}(\tfrac{3}{2})^{1/6},$

the simultaneous solutions $\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x)$ are
normalized so that

(61) $\mathop{\lim_{\lambda\to\infty}}_{\lambda\in
D_{\infty}^{\pm}}\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x)\mathrm{e}^{\mathrm{i}x\lambda\sigma_{3}}=\mathbb{I}.$

Since the coefficient matrices $\boldsymbol{\Lambda}(\lambda,x)$ and
$\mathbf{X}(\lambda,x)$ have zero trace, it then follows that
$\det(\boldsymbol{\Psi}_{\infty}^{\pm}(\lambda,x))=1$.

####  2.2.2. Solution near zero

When $\lambda$ is small and $\arg(\mathrm{i}x\lambda)$ varies from $-\pi$ to
$\pi$, $\arg(-\xi)$ varies from $\tfrac{1}{3}\pi$ to $-\tfrac{1}{3}\pi$, so to
avoid Stokes phenomenon and maintain exponential dichotomy we use scalar
multiples of the basis elements
$\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)$ and
$\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)$, so for constants $c^{\pm}$
we use (51) and hence define a solution in the domain $D_{0}$ characterized by
$|\lambda|<1$ and $\arg(\mathrm{i}x\lambda)\in(-\pi,\pi)$ by

(62)
$\boldsymbol{\Psi}_{0}(\lambda,x):=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\mathbf{G}^{-1}\frac{1}{\sqrt{2}}\begin{bmatrix}1&0\\\
0&-(\frac{2}{3})^{1/3}\end{bmatrix}(\mathrm{i}x\lambda)^{\sigma_{3}/6}\\\
{}\cdot\begin{bmatrix}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}\begin{bmatrix}c^{+}&0\\\
0&c^{-}\end{bmatrix},$

which can be equivalently written as

(63)
$\boldsymbol{\Psi}_{0}(\lambda,x)=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\begin{bmatrix}\frac{1}{2}&\frac{1}{2}\\\
-\frac{1}{2}&\frac{1}{2}\end{bmatrix}\\\
{}\cdot\begin{bmatrix}c^{+}(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&c^{-}(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
-c^{+}(\frac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&-c^{-}(\frac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}.$

Using $\mathrm{e}^{\pm
2\pi\mathrm{i}/3}\xi=x^{2/3}(\frac{3}{2})^{2/3}\mathrm{e}^{\mp\mathrm{i}\pi/3}(\mathrm{i}x\lambda)^{-1/3}(1-x^{-2/3}(\mathrm{i}x\lambda))$,
the relevant expansions we need in this situation are then

(64) $(\mathrm{i}x\lambda)^{1/6}\mathrm{Ai}(\mathrm{e}^{\pm
2\pi\mathrm{i}/3}\xi)=\mathrm{e}^{\pm\mathrm{i}\pi/12}\frac{1}{2\sqrt{\pi}}x^{-1/6}(\tfrac{3}{2})^{-1/6}(\mathrm{i}x\lambda)^{1/4}\mathrm{e}^{\pm\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}}(1+\mathcal{O}(\lambda^{1/2})),$

and

(65) $-(\tfrac{2}{3})^{1/3}(\mathrm{i}x\lambda)^{-1/6}\mathrm{e}^{\pm
2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{\pm 2\pi
i/3}\xi)=\mathrm{e}^{\pm
7\pi\mathrm{i}/12}\frac{1}{2\sqrt{\pi}}x^{1/6}(\tfrac{3}{2})^{-1/6}(\mathrm{i}x\lambda)^{-1/4}\mathrm{e}^{\pm\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}}(1+\mathcal{O}(\lambda^{1/2})),$

as $\lambda\to 0$ with $-\pi<\arg(\mathrm{i}x\lambda)<\pi$. It follows that
$\det(\boldsymbol{\Psi}_{0}(\lambda,x))=\tfrac{1}{4\pi\mathrm{i}}c^{+}c^{-}(\tfrac{2}{3})^{1/3}+\mathcal{O}(\lambda^{1/2})$
in the same limit, so by Abel’s theorem
$\det(\boldsymbol{\Psi}_{0}(\lambda,x))=\tfrac{1}{4\pi\mathrm{i}}c^{+}c^{-}(\tfrac{2}{3})^{1/3}$.
We complete the definition of $\boldsymbol{\Psi}_{0}(\lambda,x)$ by choosing

(66) $c^{+}:=2\mathrm{i}\sqrt{\pi}(\tfrac{3}{2})^{1/6}\quad\text{and}\quad
c^{-}:=2\sqrt{\pi}(\tfrac{3}{2})^{1/6}$

which guarantees the identity $\det(\boldsymbol{\Psi}_{0}(\lambda,x))=1$. It
then follows that

(67)
$\boldsymbol{\Psi}_{0}(\lambda,x)\mathrm{e}^{-\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}=\mathrm{e}^{-3x^{2/3}\sigma_{3}/2}\begin{bmatrix}\frac{1}{2}&\frac{1}{2}\\\
-\frac{1}{2}&\frac{1}{2}\end{bmatrix}x^{\sigma_{3}/3}\mathrm{e}^{7\pi\mathrm{i}\sigma_{3}/12}\\\
{}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/4}\begin{bmatrix}1+\mathcal{O}(\lambda^{1/2})&\mathrm{e}^{-2\pi\mathrm{i}/3}(1+\mathcal{O}(\lambda^{1/2}))\\\
\mathrm{e}^{-\mathrm{i}\pi/3}(1+\mathcal{O}(\lambda^{1/2}))&1+\mathcal{O}(\lambda^{1/2})\end{bmatrix},\quad\lambda\to
0.$

Here, the symbol $\mathcal{O}(\lambda^{1/2})$ denotes in each case a
(different) quantity having an asymptotic expansion as $\lambda\to 0$ in
positive integer powers of $\lambda^{1/2}$. This implies the existence of the
limit

(68) $\mathbf{B}_{0}(x):=\lim_{\lambda\to
0}\boldsymbol{\Psi}_{0}(\lambda,x)\mathrm{e}^{-\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/4},\quad\mathbf{E}:=\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{e}^{\mathrm{i}\pi/3}\\\
\mathrm{e}^{2\pi\mathrm{i}/3}&1\end{bmatrix}.$

At this point, we can revisit the expansion of
$\boldsymbol{\Psi}_{0}(\lambda,x)\mathrm{e}^{-\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\mathbf{E}\cdot(\mathrm{i}\lambda/x)^{-\sigma_{3}/4}$
as $\lambda\to 0$, and take advantage of the structure of the complete
asymptotic expansions [20, Eqns. 9.7.5–6] of which the parentheses in (54)
represent just the first explicit terms to deduce that the half-integer powers
all vanish, and in fact we have a full expansion

(69)
$\boldsymbol{\Psi}_{0}(\lambda,x)\mathrm{e}^{-\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/4}\sim\sum_{p=0}^{\infty}\left(\frac{\mathrm{i}\lambda}{x}\right)^{p}\mathbf{B}_{p}(x),\quad\lambda\to
0.$

###  2.3. Jump conditions for the seed

We define (see Figure 2)

Figure 2. Left: the domains $D_{\infty}^{\pm}$ and $D_{0}$ in the
$\lambda$-plane. Right: the arcs of the jump contour for
$\boldsymbol{\Psi}(\lambda,x)$.

(70)
$\boldsymbol{\Psi}(\lambda,x):=\begin{cases}\boldsymbol{\Psi}_{\infty}^{+}(\lambda,x),&\quad\lambda\in
D_{\infty}^{+},\\\ \boldsymbol{\Psi}_{\infty}^{-}(\lambda,x),&\quad\lambda\in
D_{\infty}^{-},\\\ \boldsymbol{\Psi}_{0}(\lambda,x),&\quad\lambda\in
D_{0}.\end{cases}$

The jump matrices are obtained using the connection identity [20, Eqn. 9.2.12]
$\mathrm{Ai}(z)+\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}z)+\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}z)=0$
and its derivative
$\mathrm{Ai}^{\prime}(z)+\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}z)+\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}z)=0$.
We will calculate the jump
$\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{-}(\lambda,x)\mathbf{V}$
across each arc of the jump contour as indicated in the right-hand diagram of
Figure 2 (subscripts $+$/$-$ indicate boundary value from the left/right by
orientation). Note that since in each domain $\boldsymbol{\Psi}(\lambda,x)$ is
a simultaneous fundamental solution matrix with unit determinant for the
equations $\boldsymbol{\Psi}_{x}=\mathbf{X}\boldsymbol{\Psi}$ and
$\boldsymbol{\Psi}_{\lambda}=\boldsymbol{\Lambda\Psi}$, each jump matrix
$\mathbf{V}$ is a constant matrix with $\det(\mathbf{V})=1$.

To compute the jump of $\boldsymbol{\Psi}(\lambda,x)$ across
$\Sigma_{\infty}^{-}$ as well as $C^{\pm}$, we note that $\xi$ and
$(\mathrm{i}x\lambda)^{\sigma_{3}/6}$ are both well-defined on these contours.
So the jump matrix
$\mathbf{V}=\boldsymbol{\Psi}_{-}(\lambda,x)^{-1}\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{\infty}^{+}(\lambda,x)^{-1}\boldsymbol{\Psi}_{\infty}^{-}(\lambda,x)$
on $\Sigma_{\infty}^{-}$ satisfies

(71)
$\begin{bmatrix}\mathrm{Ai}(\xi)&-\mathrm{e}^{\mathrm{i}\pi/6}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)\\\
\mathrm{Ai}^{\prime}(\xi)&-\mathrm{e}^{\mathrm{i}\pi/6}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)\end{bmatrix}=\begin{bmatrix}\mathrm{Ai}(\xi)&-\mathrm{e}^{-\mathrm{i}\pi/6}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
\mathrm{Ai}^{\prime}(\xi)&-\mathrm{e}^{-\mathrm{i}\pi/6}\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}\mathbf{V}.$

Using the connection formulæ to write the second column on the left-hand side
in terms of $\mathrm{Ai}(\xi)$ and
$\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)$ and their
derivatives yields

(72) $\mathbf{V}=\begin{bmatrix}1&-\mathrm{i}\\\
0&1\end{bmatrix},\quad\lambda\in\Sigma_{\infty}^{-}.$

Similarly, the jump matrix on $C^{-}$ is
$\mathbf{V}=\boldsymbol{\Psi}_{-}(\lambda,x)^{-1}\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{\infty}^{-}(\lambda,x)^{-1}\boldsymbol{\Psi}_{0}(\lambda,x)$,
which satisfies

(73)
$\begin{bmatrix}\mathrm{i}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
\mathrm{i}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}=\begin{bmatrix}\mathrm{Ai}(\xi)&-\mathrm{e}^{\mathrm{i}\pi/6}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)\\\
\mathrm{Ai}^{\prime}(\xi)&-\mathrm{e}^{\mathrm{i}\pi/6}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)\end{bmatrix}\mathbf{V}.$

The same connection formulæ applied to the second column of the left-hand side
then yields

(74) $\mathbf{V}=\begin{bmatrix}0&\mathrm{e}^{-\mathrm{i}\pi/3}\\\
\mathrm{e}^{-2\pi\mathrm{i}/3}&\mathrm{e}^{-5\mathrm{i}\pi/6}\end{bmatrix},\quad\lambda\in
C^{-}.$

Likewise, the jump matrix on $C^{+}$ is
$\mathbf{V}=\boldsymbol{\Psi}_{-}(\lambda,x)^{-1}\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{0}(\lambda,x)^{-1}\boldsymbol{\Psi}_{\infty}^{+}(\lambda,x)$,
which satisfies

(75)
$\begin{bmatrix}\mathrm{Ai}(\xi)&-\mathrm{e}^{-\mathrm{i}\pi/6}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
\mathrm{Ai}^{\prime}(\xi)&-\mathrm{e}^{-\mathrm{i}\pi/6}\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}=\begin{bmatrix}\mathrm{i}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\\\
\mathrm{i}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi)&\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi)\end{bmatrix}\mathbf{V}.$

Applying the connection formulæ to the first column of the left-hand side
yields

(76) $\mathbf{V}=\begin{bmatrix}\mathrm{e}^{-5\pi\mathrm{i}/6}&0\\\
\mathrm{e}^{\mathrm{i}\pi/3}&\mathrm{e}^{5\pi\mathrm{i}/6}\end{bmatrix},\quad\lambda\in
C^{+}.$

To compute the jump matrices on $\Sigma_{0}^{+}$ and $\Sigma_{\infty}^{+}$ we
also have to take into account the jump conditions satisfied by $\xi$ and
$(\mathrm{i}x\lambda)^{\sigma_{3}/6}$:

(77)
$\left[(\mathrm{i}x\lambda)^{\sigma_{3}/6}\right]_{+}=\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/3}\left[(\mathrm{i}x\lambda)^{\sigma_{3}/6}\right]_{-}\quad\text{and}\quad\xi_{+}=\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-},\quad\lambda\in\Sigma_{0}^{+}$

and

(78)
$\left[(\mathrm{i}x\lambda)^{\sigma_{3}/6}\right]_{+}=\mathrm{e}^{-\mathrm{i}\pi\sigma_{3}/3}\left[(\mathrm{i}x\lambda)^{\sigma_{3}/6}\right]_{-}\quad\text{and}\quad\xi_{+}=\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-},\quad\lambda\in\Sigma_{\infty}^{+}.$

The difference between these formulæ arises simply from the opposite
orientation of $\Sigma_{0}^{+}$ and $\Sigma_{\infty}^{+}$. So, on
$\Sigma_{0}^{+}$, using (77) on the left-hand side of
$\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{-}(\lambda,x)\mathbf{V}$
(with $\boldsymbol{\Psi}(\lambda,x)=\boldsymbol{\Psi}_{0}(\lambda,x)$), we
have

(79)
$\begin{bmatrix}\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{Ai}(\xi_{-})&\mathrm{e}^{\mathrm{i}\pi/3}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})\\\
\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{Ai}^{\prime}(\xi_{-})&\mathrm{e}^{\mathrm{i}\pi/3}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})\end{bmatrix}=\begin{bmatrix}\mathrm{i}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})&\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\\\
\mathrm{i}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})&\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\end{bmatrix}\mathbf{V}.$

Applying the connection formulæ to the first column of the left-hand side
gives

(80) $\mathbf{V}=\begin{bmatrix}1&\mathrm{e}^{-\mathrm{i}\pi/6}\\\
\mathrm{e}^{-5\pi\mathrm{i}/6}&0\end{bmatrix},\quad\lambda\in\Sigma_{0}^{+}.$

Finally, on $\Sigma_{\infty}^{+}$, we use (78) on the left-hand side of
$\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{-}(\lambda,x)\mathbf{V}$
(this time with
$\boldsymbol{\Psi}_{+}(\lambda,x)=\boldsymbol{\Psi}_{\infty,+}^{-}(\lambda,x)$
and
$\boldsymbol{\Psi}_{-}(\lambda,x)=\boldsymbol{\Psi}_{\infty,-}^{+}(\lambda,x)$),
we find

(81)
$\begin{bmatrix}\mathrm{e}^{-\mathrm{i}\pi/3}\mathrm{Ai}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})&\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\\\
\mathrm{e}^{-\mathrm{i}\pi/3}\mathrm{e}^{2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{2\pi\mathrm{i}/3}\xi_{-})&\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\end{bmatrix}=\begin{bmatrix}\mathrm{Ai}(\xi_{-})&\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{Ai}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\\\
\mathrm{Ai}^{\prime}(\xi_{-})&\mathrm{e}^{5\pi\mathrm{i}/6}\mathrm{e}^{-2\pi\mathrm{i}/3}\mathrm{Ai}^{\prime}(\mathrm{e}^{-2\pi\mathrm{i}/3}\xi_{-})\end{bmatrix}\mathbf{V}.$

Using the connection formulæ on the first column of the left-hand side gives

(82) $\mathbf{V}=\begin{bmatrix}1&0\\\
\mathrm{i}&1\end{bmatrix},\quad\lambda\in\Sigma_{\infty}^{+}.$

This completes the computation of the jump matrices for
$\boldsymbol{\Psi}(\lambda,x)$. It is straightforward to verify that the
cyclic product of jump matrices about either of the two self-intersection
points of the jump contour equals $\mathbb{I}$.

##  3\. Riemann-Hilbert representation for the algebraic solutions

###  3.1. Riemann-Hilbert problem formulation and basic properties

The algebraic solutions of (1) with $\alpha=8$, $\beta=2n\in 2\mathbb{Z}$, and
$\delta=-1$ (or (4) with $\epsilon=-1$, $a=-\mathrm{i}n$, and $b=\mathrm{i}$)
can be obtained from the following Riemann-Hilbert problem.

######  Riemann-Hilbert Problem 1 (Algebraic solutions of Painlevé-III D7).

Given $x>0$ and $n\in\mathbb{Z}$, seek a $2\times 2$ matrix function
$\lambda\mapsto\mathbf{W}^{(n)}(\lambda,x)$ with the following properties:

  * •

Analyticity: $\mathbf{W}^{(n)}(\lambda,x)$ is analytic for
$\lambda\in\mathbb{C}\setminus(\Sigma_{\infty}^{-}\cup\Sigma_{\infty}^{+}\cup\Sigma_{0}^{+}\cup
C^{+}\cup C^{-})$.

  * •

Jump conditions: $\mathbf{W}^{(n)}(\lambda,x)$ takes continuous boundary
values from each component of its domain of analyticity, and the boundary
values are related by

(83)
$\mathbf{W}^{(n)}_{+}(\lambda,x)=\mathbf{W}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda_{-})^{-1/2})\sigma_{3}}(-1)^{n}\mathbf{V}\mathrm{e}^{\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda_{+})^{-1/2})\sigma_{3}},\quad\lambda\in\Sigma_{0}^{+}\cup\Sigma_{\infty}^{+},$

and

(84)
$\mathbf{W}^{(n)}_{+}(\lambda,x)=\mathbf{W}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda_{-})^{-1/2})\sigma_{3}}\mathbf{V}\mathrm{e}^{\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda_{+})^{-1/2})\sigma_{3}},\quad\lambda\in\Sigma_{\infty}^{-}\cup
C^{+}\cup C^{-},$

where the constant matrix $\mathbf{V}$ is defined on each arc of the jump
contour by (72), (74), (76), (80), and (82).

  * •

Normalization:
$\mathbf{W}^{(n)}(\lambda,x)\left(\frac{\mathrm{i}\lambda}{x}\right)^{n\sigma_{3}/2}\to\mathbb{I}$
as $\lambda\to\infty$.

  * •

Behavior at the origin: the limit

(85) $\mathbf{B}^{(n)}_{0}(x):=\lim_{\lambda\to
0}\mathbf{W}^{(n)}(\lambda,x)\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4}$

exists, where $\mathbf{E}$ is given by (68).

The following lemma is a consequence of the construction given in Section 2.

######  Lemma 1.

The matrix $\mathbf{W}^{(0)}(\lambda,x)$ defined in terms of
$\mathbf{\Psi}(\lambda,x)$ (see (70)) by

(86)
$\mathbf{W}^{(0)}(\lambda,x):=\boldsymbol{\Psi}(\lambda,x)\mathrm{e}^{\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}$

is a solution of Riemann-Hilbert Problem 1 with $n=0$.

######  Lemma 2.

Given $x>0$ and $n\in\mathbb{Z}$, there is at most one solution of Riemann-
Hilbert Problem 1 and any solution has unit determinant.

###### Proof.

Suppose there exist two solutions, denoted $\mathbf{W}(\lambda)$ and
$\widetilde{\mathbf{W}}(\lambda)$. It follows from the conditions of the
problem and Liouville’s theorem that
$\det(\mathbf{W}(\lambda))=\det(\widetilde{\mathbf{W}}(\lambda))\equiv 1$.
Consider the matrix
$\mathbf{R}(\lambda):=\mathbf{W}(\lambda)\widetilde{\mathbf{W}}(\lambda)^{-1}$.
It similarly follows from the conditions of the problem that
$\mathbf{R}(\lambda)$ is entire and tends to the identity matrix as
$\lambda\to\infty$, so by Liouville’s theorem again,
$\mathbf{R}(\lambda)\equiv\mathbb{I}$. ∎

######  Lemma 3.

Given $x>0$ and $n\in\mathbb{Z}$, if the solution
$\mathbf{W}^{(n)}(\lambda,x)$ of Riemann-Hilbert Problem 1 exists, then the
normalization condition holds in the stronger sense that there exist
coefficient matrices $\\{\mathbf{A}_{p}^{(n)}(x)\\}_{p=1}^{\infty}$ such that
the complete asymptotic expansion

(87)
$\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\left(\frac{\mathrm{i}\lambda}{x}\right)^{n\sigma_{3}/2}\sim\mathbb{I}+\sum_{p=1}^{\infty}\left(\frac{\mathrm{i}\lambda}{x}\right)^{-p}\mathbf{A}_{p}^{(n)}(x),\quad\lambda\to\infty$

is uniformly valid with respect to $\arg(\mathrm{i}\lambda)$, and the
expansion is differentiable term-by-term with respect to $\lambda$ and $x$.
Also, the limit in (85) is the leading term in another complete asymptotic
expansion involving other coefficient matrices
$\\{\mathbf{B}_{p}^{(n)}(x)\\}_{p=0}^{\infty}$:

(88)
$\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4}\sim\sum_{p=0}^{\infty}\left(\frac{\mathrm{i}\lambda}{x}\right)^{p}\mathbf{B}_{p}^{(n)}(x),\quad\lambda\to
0$

holding uniformly with respect to $\arg(\mathrm{i}\lambda)$ and enjoying
similar differentiability properties.

###### Proof.

We consider the function $\lambda\mapsto\mathbf{S}(\lambda)$ defined by

(89)
$\mathbf{S}^{(n)}(\lambda):=\begin{cases}\displaystyle\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\left(\frac{\mathrm{i}\lambda}{x}\right)^{n\sigma_{3}/2},&\quad|\lambda|>1,\\\
\displaystyle\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4},&\quad|\lambda|<1.\end{cases}$

Then, $\mathbf{S}^{(n)}(\lambda)$ satisfies the conditions of an equivalent
Riemann-Hilbert problem, and it is easy to check that the jump matrices for
$\mathbf{S}^{(n)}(\lambda)$ tend exponentially rapidly to the identity as
$\lambda\to\infty$ in $\Sigma_{\infty}^{+}\cup\Sigma_{\infty}^{-}$ and as
$\lambda\to 0$ in $\Sigma_{0}^{+}$. This kind of problem is amenable to
analytic (with respect to $x$) Fredholm theory applied to an equivalent
singular integral equation, and the result that $\mathbf{S}^{(n)}(\lambda)$
has asymptotic power series expansions as $\lambda\to\infty$ and $\lambda\to
0$ then follows. ∎

In the next two subsections we show how and why Riemann-Hilbert Problem 1
encodes the algebraic solutions $u=u_{n}(x)$ of (4).

###  3.2. Differential equations satisfied by the solution of Riemann-Hilbert
Problem 1

Suppose $\mathbf{W}^{(n)}(\lambda,x)$ solves Riemann-Hilbert Problem 1, and
consider the related matrix

(90)
$\boldsymbol{\Psi}^{(n)}(\lambda,x):=\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}$

(cf. (86)). It is straightforward to check that
$\boldsymbol{\Psi}^{(n)}(\lambda,x)$ satisfies jump conditions across each arc
of the jump contour that are independent of both $x$ and $\lambda$:
$\boldsymbol{\Psi}^{(n)}_{+}(\lambda,x)=\boldsymbol{\Psi}^{(n)}_{-}(\lambda,x)(-1)^{n}\mathbf{V}$
for $\lambda\in\Sigma_{0}^{+}\cup\Sigma_{\infty}^{+}$ and
$\boldsymbol{\Psi}^{(n)}_{+}(\lambda,x)=\boldsymbol{\Psi}^{(n)}_{-}(\lambda,x)\mathbf{V}$
for $\lambda\in\Sigma_{\infty}^{-}\cup C^{+}\cup C^{-}$. It follows that the
matrices

(91)
$\boldsymbol{\Lambda}^{(n)}(\lambda,x):=\frac{\partial\boldsymbol{\Psi}^{(n)}}{\partial\lambda}(\lambda,x)\boldsymbol{\Psi}^{(n)}(\lambda,x)^{-1}\quad\text{and}\quad\mathbf{X}^{(n)}(\lambda,x):=\frac{\partial\boldsymbol{\Psi}^{(n)}}{\partial
x}(\lambda,x)\boldsymbol{\Psi}^{(n)}(\lambda,x)^{-1}$

are both analytic for $\lambda\in\mathbb{C}\setminus\\{0\\}$. They can be
expressed in terms of the coefficients
$\\{\mathbf{A}_{p}^{(n)}(x)\\}_{p=1}^{\infty}$ and
$\\{\mathbf{B}_{p}^{(n)}(x)\\}_{p=0}^{\infty}$ of Lemma 3 as follows. First,
we write $\boldsymbol{\Psi}^{(n)}(\lambda,x)$ in terms of
$\mathbf{W}^{(n)}(\lambda,x)$ and then the matrix
$\mathbf{S}^{(n)}(\lambda,x)$ defined by (89). Then, assuming that
$|\lambda|>1$, we have

(92)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=\frac{\partial\mathbf{S}^{(n)}}{\partial\lambda}(\lambda,x)\mathbf{S}^{(n)}(\lambda,x)^{-1}-\frac{n}{2\lambda}\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}-\mathrm{i}x\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}\\\
\mathbf{X}^{(n)}(\lambda,x)&=\frac{\partial\mathbf{S}^{(n)}}{\partial
x}(\lambda,x)\mathbf{S}^{(n)}(\lambda,x)^{-1}+\frac{n}{2x}\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}-\mathrm{i}\lambda\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}.\end{split}$

Using (87) from Lemma 3 (the left-hand side of which is exactly
$\mathbf{S}^{(n)}(\lambda,x)$ for $|\lambda|>1$) then gives

(93)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=-\mathrm{i}x\sigma_{3}-\frac{n}{2\lambda}\sigma_{3}-\frac{x^{2}}{\lambda}[\mathbf{A}^{(n)}_{1}(x),\sigma_{3}]+\mathcal{O}(\lambda^{-2}),\quad\lambda\to\infty,\quad\text{and}\\\
\mathbf{X}^{(n)}(\lambda,x)&=-\mathrm{i}\lambda\sigma_{3}+\frac{n}{2x}\sigma_{3}-x[\mathbf{A}^{(n)}_{1}(x),\sigma_{3}]+\mathcal{O}(\lambda^{-1}),\quad\lambda\to\infty.\end{split}$

To analyze the same matrices in the limit $\lambda\to 0$, we use (89) to
obtain, for $|\lambda|<1$,

(94)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=\frac{\partial\mathbf{S}^{(n)}}{\partial\lambda}(\lambda,x)\mathbf{S}^{(n)}(\lambda,x)^{-1}+\frac{(-1)^{n}}{4\lambda}\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}\\\
&\qquad\qquad{}+\frac{1}{2x}\left(\frac{\mathrm{i}\lambda}{x}\right)^{-3/2}\mathbf{S}^{(n)}(\lambda,x)\left(\frac{\mathrm{i}\lambda}{x}\right)^{(-1)^{n}\sigma_{3}/4}\mathbf{E}^{-1}\sigma_{3}\mathbf{E}\left(\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4}\mathbf{S}^{(n)}(\lambda,x)^{-1}\\\
\mathbf{X}^{(n)}(\lambda,x)&=\frac{\partial\mathbf{S}^{(n)}}{\partial
x}(\lambda,x)\mathbf{S}^{(n)}(\lambda,x)^{-1}-\frac{(-1)^{n}}{4x}\mathbf{S}^{(n)}(\lambda,x)\sigma_{3}\mathbf{S}^{(n)}(\lambda,x)^{-1}\\\
&\qquad\qquad{}+\frac{\mathrm{i}}{2x}\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1/2}\mathbf{S}^{(n)}(\lambda,x)\left(\frac{\mathrm{i}\lambda}{x}\right)^{(-1)^{n}\sigma_{3}/4}\mathbf{E}^{-1}\sigma_{3}\mathbf{E}\left(\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4}\mathbf{S}^{(n)}(\lambda,x)^{-1}.\end{split}$

Using (88) from Lemma 3, the left-hand side of which is exactly
$\mathbf{S}^{(n)}(\lambda,x)$ for $|\lambda|<1$, the terms on the first line
of each expression are $\mathcal{O}(\lambda^{-1})$ and $\mathcal{O}(1)$,
respectively, as $\lambda\to 0$. For the terms on the second line of each, we
use the identity

(95)
$\mathbf{E}^{-1}\sigma_{3}\mathbf{E}=\begin{bmatrix}0&\mathrm{e}^{\mathrm{i}\pi/3}\\\
\mathrm{e}^{-\mathrm{i}\pi/3}&0\end{bmatrix}.$

Therefore, using (88) from Lemma 3 again shows that if $n$ is even,

(96)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=-\mathrm{e}^{-\mathrm{i}\pi/3}\frac{x}{2\lambda^{2}}\mathbf{B}^{(n)}_{0}(x)\begin{bmatrix}0&0\\\
1&0\end{bmatrix}\mathbf{B}^{(n)}_{0}(x)^{-1}+\mathcal{O}(\lambda^{-1}),\quad\lambda\to
0,\quad\text{and}\\\
\mathbf{X}^{(n)}(\lambda,x)&=\mathrm{e}^{-\mathrm{i}\pi/3}\frac{1}{2\lambda}\mathbf{B}^{(n)}_{0}(x)\begin{bmatrix}0&0\\\
1&0\end{bmatrix}\mathbf{B}^{(n)}_{0}(x)^{-1}+\mathcal{O}(1),\quad\lambda\to
0,\end{split}$

while if instead $n$ is odd,

(97)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=-\mathrm{e}^{\mathrm{i}\pi/3}\frac{x}{2\lambda^{2}}\mathbf{B}^{(n)}_{0}(x)\begin{bmatrix}0&1\\\
0&0\end{bmatrix}\mathbf{B}^{(n)}_{0}(x)^{-1}+\mathcal{O}(\lambda^{-1}),\quad\lambda\to
0,\quad\text{and}\\\
\mathbf{X}^{(n)}(\lambda,x)&=\mathrm{e}^{\mathrm{i}\pi/3}\frac{1}{2\lambda}\mathbf{B}^{(n)}_{0}(x)\begin{bmatrix}0&1\\\
0&0\end{bmatrix}\mathbf{B}^{(n)}_{0}(x)^{-1}+\mathcal{O}(1),\quad\lambda\to
0.\end{split}$

The Laurent expansions (93) and (96)–(97) then fully determine the matrices
$\boldsymbol{\Lambda}^{(n)}(\lambda,x)$ and $\mathbf{X}^{(n)}(\lambda,x)$:

(98)
$\begin{split}\boldsymbol{\Lambda}^{(n)}(\lambda,x)&=-\mathrm{i}x\sigma_{3}-\frac{n}{2\lambda}\sigma_{3}-\frac{x}{\lambda}\mathbf{J}^{(n)}(x)+\frac{\mathrm{i}x}{2\lambda^{2}}\mathbf{K}^{(n)}(x),\\\
\mathbf{X}^{(n)}(\lambda,x)&=-\mathrm{i}\lambda\sigma_{3}+\frac{n}{2x}\sigma_{3}-\mathbf{J}^{(n)}(x)-\frac{\mathrm{i}}{2\lambda}\mathbf{K}^{(n)}(x),\end{split}$

where

(99)
$\mathbf{J}^{(n)}(x):=x[\mathbf{A}_{1}^{(n)}(x),\sigma_{3}]\quad\text{and}\quad\mathbf{K}^{(n)}(x):=\begin{cases}\mathrm{e}^{\mathrm{i}\pi/6}\mathbf{B}_{0}^{(n)}(x)\begin{bmatrix}0&0\\\
1&0\end{bmatrix}\mathbf{B}_{0}^{(n)}(x)^{-1},&\quad\text{$n$ even},\\\
\mathrm{e}^{5\pi\mathrm{i}/6}\mathbf{B}_{0}^{(n)}(x)\begin{bmatrix}0&1\\\
0&0\end{bmatrix}\mathbf{B}_{0}^{(n)}(x)^{-1},&\quad\text{$n$ odd}.\end{cases}$

Since $\mathbf{J}^{(n)}(x)$ is an off-diagonal matrix and
$\mathbf{K}^{(n)}(x)$ is singular and nondiagonalizable but nonzero, it
follows that whenever $x$ is such that Riemann-Hilbert Problem 1 is solvable,
if potentials $u=u_{n}(x)$,
$\mathrm{e}^{\pm\mathrm{i}\varphi}=\mathrm{e}^{\pm\mathrm{i}\varphi_{n}(x)}$,
$p=p_{n}(x)$, and $q=q_{n}(x)$ are defined in terms of the matrix elements of
$\mathbf{J}=\mathbf{J}^{(n)}(x)$ and $\mathbf{K}=\mathbf{K}^{(n)}(x)$ by (18)
(taking $\epsilon=-1$ by our convention), then in particular $u_{n}(x)$ is a
solution of the Painlevé-III D7 equation in the form (4) for $a=-\mathrm{i}n$
and $b$ determined from (17). Using $\det(\mathbf{B}_{0}^{(n)}(x))=1$ (arising
by combining Lemmas 2 and 3), the formula for $u_{n}(x)$ in terms of the
matrix $\mathbf{B}_{0}^{(n)}(x)$ obtained from $\mathbf{W}^{(n)}(\lambda,x)$
by (88) from Lemma 3 is:

(100)
$u_{n}(x)=\begin{cases}\mathrm{e}^{-5\pi\mathrm{i}/6}xB_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x),&\quad\text{$n$
even,}\\\
\mathrm{e}^{5\pi\mathrm{i}/6}xB_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x),&\quad\text{$n$
odd.}\end{cases}$

###  3.3. Solution of Riemann-Hilbert Problem 1 by Schlesinger transformations

####  3.3.1. Schlesinger transformations for the Painlevé-III (D7) Lax pair

Schlesinger transformations for the Lax pair (5) and their induced Bäcklund
transformations are discussed in [13, Section 6.1]. Define a gauge
transformation matrix $\mathbf{G}(\lambda,x)$ by

(101)
$\mathbf{G}(\lambda,x):=(\mathrm{i}\lambda)^{1/2}\overline{\mathbf{G}}(x)+(\mathrm{i}\lambda)^{-1/2}\underline{\mathbf{G}}(x),$

where $\overline{\mathbf{G}}(x)$ and $\underline{\mathbf{G}}(x)$ are matrices
to be determined so that when $\boldsymbol{\Psi}$ is a simultaneous
fundamental solution matrix of the Lax pair equations (5), then
$\widetilde{\boldsymbol{\Psi}}:=\mathbf{G}(\lambda,x)\boldsymbol{\Psi}$ is as
well, but with a different value of $a$ and different potentials $u(x)$,
$p(x)$, $q(x)$, and $\varphi(x)$. We also want to normalize
$\mathbf{G}(\lambda,x)$ so that $\det(\mathbf{G}(\lambda,x))\equiv 1$.
Clearly, $\widetilde{\boldsymbol{\Psi}}(\lambda,x)$ is a simultaneous
fundamental solution matrix of
$\widetilde{\boldsymbol{\Psi}}_{\lambda}=\widetilde{\boldsymbol{\Lambda}}(\lambda,x)\widetilde{\boldsymbol{\Psi}}$
and
$\widetilde{\boldsymbol{\Psi}}_{x}=\widetilde{\mathbf{X}}(\lambda,x)\widetilde{\boldsymbol{\Psi}}$,
where

(102)
$\begin{split}\widetilde{\boldsymbol{\Lambda}}(\lambda,x)&:=\widetilde{\boldsymbol{\Psi}}_{\lambda}(\lambda,x)\widetilde{\boldsymbol{\Psi}}(\lambda,x)^{-1}\\\
&=\frac{\partial\mathbf{G}}{\partial\lambda}(\lambda,x)\mathbf{G}(\lambda,x)^{-1}+\mathbf{G}(\lambda,x)\boldsymbol{\Lambda}(\lambda,x)\mathbf{G}(\lambda,x)^{-1}\quad\text{and}\\\
\widetilde{\mathbf{X}}(\lambda,x)&:=\widetilde{\boldsymbol{\Psi}}_{x}(\lambda,x)\widetilde{\boldsymbol{\Psi}}(\lambda,x)^{-1}\\\
&=\frac{\partial\mathbf{G}}{\partial
x}(\lambda,x)\mathbf{G}(\lambda,x)^{-1}+\mathbf{G}(\lambda,x)\mathbf{X}(\lambda,x)\mathbf{G}(\lambda,x)^{-1}.\end{split}$

Writing $a=-\mathrm{i}n$ for $n\in\mathbb{C}$ (in general), we want to pick
the coefficients $\overline{\mathbf{G}}(x)$ and $\underline{\mathbf{G}}(x)$ so
that the matrices

(103)
$\boldsymbol{\Lambda}(\lambda,x):=-\mathrm{i}x\sigma_{3}-\frac{n}{2\lambda}\sigma_{3}-\frac{x}{\lambda}\mathbf{J}(x)+\frac{\mathrm{i}x}{2\lambda^{2}}\mathbf{K}(x)\quad\text{and}\quad\mathbf{X}(\lambda,x):=-\mathrm{i}\lambda\sigma_{3}+\frac{n}{2x}\sigma_{3}-\mathbf{J}(x)-\frac{\mathrm{i}}{2\lambda}\mathbf{K}(x),$

where $\mathbf{J}(x)$ is off-diagonal and $\mathbf{K}(x)$ is non-
diagonalizable with zero eigenvalues, are transformed into corresponding
matrices

(104)
$\widetilde{\boldsymbol{\Lambda}}(\lambda,x):=-\mathrm{i}x\sigma_{3}-\frac{\widetilde{n}}{2\lambda}\sigma_{3}-\frac{x}{\lambda}\widetilde{\mathbf{J}}(x)+\frac{\mathrm{i}x}{2\lambda^{2}}\widetilde{\mathbf{K}}(x)\quad\text{and}\quad\widetilde{\mathbf{X}}(\lambda,x):=-\mathrm{i}\lambda\sigma_{3}+\frac{\widetilde{n}}{2x}\sigma_{3}-\widetilde{\mathbf{J}}(x)-\frac{\mathrm{i}}{2\lambda}\widetilde{\mathbf{K}}(x),$

with $\widetilde{\mathbf{J}}(x)$ off-diagonal and $\widetilde{\mathbf{K}}(x)$
non-diagonalizable with zero eigenvalues, where $\widetilde{n}$ is another
complex parameter. Using (101), (103), and (104), we see that both sides of
each of the equations (equivalent to (102))

(105)
$\begin{split}\lambda^{2}(\mathrm{i}\lambda)^{1/2}\widetilde{\boldsymbol{\Lambda}}(\lambda,x)\mathbf{G}(\lambda,x)&=\lambda^{2}(\mathrm{i}\lambda)^{1/2}\frac{\partial\mathbf{G}}{\partial\lambda}(\lambda,x)+\lambda^{2}(\mathrm{i}\lambda)^{1/2}\mathbf{G}(\lambda,x)\boldsymbol{\Lambda}(\lambda,x)\\\
\lambda(\mathrm{i}\lambda)^{1/2}\widetilde{\mathbf{X}}(\lambda,x)\mathbf{G}(\lambda,x)&=\lambda(\mathrm{i}\lambda)^{1/2}\frac{\partial\mathbf{G}}{\partial
x}(\lambda,x)+\lambda(\mathrm{i}\lambda)^{1/2}\mathbf{G}(\lambda,x)\mathbf{X}(\lambda,x)\end{split}$

are cubic polynomials in $\lambda$. The coefficients of $\lambda^{3}$ from
both equations are balanced exactly when

(106) $\sigma_{3}\overline{\mathbf{G}}(x)=\overline{\mathbf{G}}(x)\sigma_{3},$

i.e., $\overline{\mathbf{G}}(x)$ is a diagonal matrix. The coefficients of
$\lambda^{2}$ give the equations

(107)
$\begin{split}-\frac{\mathrm{i}\widetilde{n}}{2}\sigma_{3}\overline{\mathbf{G}}(x)-\mathrm{i}x\widetilde{\mathbf{J}}(x)\overline{\mathbf{G}}(x)-\mathrm{i}x\sigma_{3}\underline{\mathbf{G}}(x)&=\frac{1}{2}\mathrm{i}\overline{\mathbf{G}}(x)-\frac{\mathrm{i}n}{2}\overline{\mathbf{G}}(x)\sigma_{3}-\mathrm{i}x\overline{\mathbf{G}}(x)\mathbf{J}(x)-\mathrm{i}x\underline{\mathbf{G}}(x)\sigma_{3}\\\
\frac{\mathrm{i}\widetilde{n}}{2x}\sigma_{3}\overline{\mathbf{G}}(x)-\mathrm{i}\widetilde{\mathbf{J}}(x)\overline{\mathbf{G}}(x)-\mathrm{i}\sigma_{3}\underline{\mathbf{G}}(x)&=\mathrm{i}\overline{\mathbf{G}}^{\prime}(x)+\frac{\mathrm{i}n}{2x}\overline{\mathbf{G}}(x)\sigma_{3}-\mathrm{i}\overline{\mathbf{G}}(x)\mathbf{J}(x)-\mathrm{i}\underline{\mathbf{G}}(x)\sigma_{3}.\end{split}$

Using the fact that $\mathbf{J}(x)$ and $\widetilde{\mathbf{J}}(x)$ are off-
diagonal, the diagonal terms of these equations are equivalent to the
equations

(108)
$\begin{split}(\widetilde{n}-n)\sigma_{3}\overline{\mathbf{G}}(x)&=-\overline{\mathbf{G}}(x)\\\
-(\widetilde{n}-n)\sigma_{3}\overline{\mathbf{G}}(x)&=-2x\overline{\mathbf{G}}^{\prime}(x).\end{split}$

Adding these together yields
$2x\overline{\mathbf{G}}^{\prime}(x)=-\overline{\mathbf{G}}(x)$ which implies
that $\overline{\mathbf{G}}(x)=x^{-1/2}\overline{\mathbf{G}}_{0}$ where
$\overline{\mathbf{G}}_{0}$ is a constant diagonal matrix. Then, for $x\neq
0$, the first of these equations is the algebraic relation
$(\widetilde{n}-n)\sigma_{3}\overline{\mathbf{G}}_{0}=-\overline{\mathbf{G}}_{0}$.
Observe that if both diagonal elements of $\overline{\mathbf{G}}_{0}$ are
nonzero, we arrive at the contradiction that both $\widetilde{n}=n+1$ and
$\widetilde{n}=n-1$. Hence nontrivial solutions for $\overline{\mathbf{G}}(x)$
are: either

(109)
$\overline{\mathbf{G}}(x)=\overline{\mathbf{G}}^{\uparrow}(x):=\begin{bmatrix}0&0\\\
0&x^{-1/2}\end{bmatrix}\quad\text{and}\quad\widetilde{n}=n+1$

or

(110)
$\overline{\mathbf{G}}(x)=\overline{\mathbf{G}}^{\downarrow}(x):=\begin{bmatrix}x^{-1/2}&0\\\
0&0\end{bmatrix}\quad\text{and}\quad\widetilde{n}=n-1.$

These are unique up to irrelevant scalings by nonzero constants.

At this point, we enforce $\det(\mathbf{G}(\lambda,x))=1$, which implies two
alternate forms for the gauge transformation $\mathbf{G}(\lambda,x)$:

(111)
$\mathbf{G}(\lambda,x)=\mathbf{G}^{\uparrow}(\lambda,x)=\begin{bmatrix}x^{1/2}(\mathrm{i}\lambda)^{-1/2}&B^{\uparrow}(x)(\mathrm{i}\lambda)^{-1/2}\\\
C^{\uparrow}(x)(\mathrm{i}\lambda)^{-1/2}&x^{-1/2}(\mathrm{i}\lambda)^{1/2}+x^{-1/2}B^{\uparrow}(x)C^{\uparrow}(x)(\mathrm{i}\lambda)^{-1/2}\end{bmatrix}\quad\text{and}\quad\widetilde{n}=n+1$

and

(112)
$\mathbf{G}(\lambda,x)=\mathbf{G}^{\downarrow}(\lambda,x)=\begin{bmatrix}x^{-1/2}(\mathrm{i}\lambda)^{1/2}+x^{-1/2}B^{\downarrow}(x)C^{\downarrow}(x)(\mathrm{i}\lambda)^{-1/2}&B^{\downarrow}(x)(\mathrm{i}\lambda)^{-1/2}\\\
C^{\downarrow}(x)(\mathrm{i}\lambda)^{-1/2}&x^{1/2}(\mathrm{i}\lambda)^{-1/2}\end{bmatrix}\quad\text{and}\quad\widetilde{n}=n-1.$

Using $\det(\mathbf{G}(\lambda,x))=1$, we now solve (105) for
$\widetilde{\boldsymbol{\Lambda}}(\lambda,x)$, which has the form of a Laurent
polynomial in $\lambda$ involving powers ranging from $\lambda^{0}$ through
$\lambda^{-3}$. In order that the coefficient of $\lambda^{0}$ is exactly
$-\mathrm{i}x\sigma_{3}$ as required by the form (104), it is necessary to set

(113) $C^{\uparrow}(x):=\frac{x^{1/2}q(x)}{8u(x)}\quad\text{and}\quad
B^{\downarrow}(x):=-\frac{x^{1/2}p(x)}{8u(x)}.$

In order that the coefficient of $\lambda^{-3}$ vanishes as required by the
form (104) we must then set

(114)
$B^{\uparrow}(x)=-x^{1/2}\mathrm{e}^{\mathrm{i}\varphi(x)}\quad\text{and}\quad
C^{\downarrow}(x)=-x^{1/2}\mathrm{e}^{-\mathrm{i}\varphi(x)}.$

These relations use our convention of $\epsilon=-1$ in the parametrization of
the matrices $\mathbf{J}(x)$ and $\mathbf{K}(x)$ in (7). It is then clear that
if one defines matrices $\widetilde{\mathbf{J}}(x)$ and
$\widetilde{\mathbf{K}}(x)$ from the coefficients of $\lambda^{-1}$ and
$\lambda^{-2}$ respectively after taking the correct incremented/decremented
value of $\widetilde{n}$ in the form of
$\widetilde{\boldsymbol{\Lambda}}(\lambda,x)$ in (104), then
$\widetilde{\mathbf{J}}(x)$ is indeed an off-diagonal matrix, and
$\widetilde{\mathbf{K}}(x)$ is a non-diagonalizable matrix with zero
eigenvalues. Finally, we solve (105) for $\widetilde{\mathbf{X}}(\lambda,x)$
and compare with the form given in (104) for the computed coefficients
$\widetilde{\mathbf{J}}(x)$ and $\widetilde{\mathbf{K}}(x)$. We observe
agreement due to the differential equations (8)–(12).

For the transformation $\mathbf{G}^{\uparrow}(\lambda,x)$, the transformed
coefficients are:

(115)
$\widetilde{\mathbf{J}}(x)=\mathbf{J}^{\uparrow}(x):=\begin{bmatrix}0&2x\mathrm{e}^{\mathrm{i}\varphi(x)}\\\
\frac{x\mathrm{e}^{\mathrm{i}\varphi(x)}q(x)^{2}}{32u(x)^{2}}+\frac{(n+1)q(x)}{8xu(x)}-\frac{\mathrm{e}^{-\mathrm{i}\varphi(x)}u(x)}{2x^{2}}&0\end{bmatrix}$

and

(116)
$\widetilde{\mathbf{K}}(x)=\mathbf{K}^{\uparrow}(x):=\frac{\mathrm{i}bx\mathrm{e}^{\mathrm{i}\varphi(x)}}{64u(x)^{3}}\begin{bmatrix}8u(x)q(x)&-64u(x)^{2}\\\
q(x)^{2}&-8u(x)q(x)\end{bmatrix}.$

For the transformation $\mathbf{G}^{\downarrow}(\lambda,x)$, the transformed
coefficients are:

(117)
$\widetilde{\mathbf{J}}(x)=\mathbf{J}^{\downarrow}(x):=\begin{bmatrix}0&-\frac{x\mathrm{e}^{-\mathrm{i}\varphi(x)}p(x)^{2}}{32u(x)^{2}}+\frac{(n-1)p(x)}{8xu(x)}+\frac{\mathrm{e}^{\mathrm{i}\varphi(x)}u(x)}{2x^{2}}\\\
-2x\mathrm{e}^{-\mathrm{i}\varphi(x)}&0\end{bmatrix}$

and

(118)
$\widetilde{\mathbf{K}}(x)=\mathbf{K}^{\downarrow}(x):=\frac{\mathrm{i}bx\mathrm{e}^{-\mathrm{i}\varphi(x)}}{64u(x)^{3}}\begin{bmatrix}-8u(x)p(x)&-p(x)^{2}\\\
64u(x)^{2}&8u(x)p(x)\end{bmatrix}.$

In these formulæ, $b$ is the constant expressed in terms of the potentials via
(19) with $a=-\mathrm{i}n$.

####  3.3.2. The induction argument

We now show that solutions of Riemann-Hilbert Problem 1 for consecutive
integer values of $n$ are related by suitable Schlesinger transformations.

##### The case of $n$ even.

Suppose that $n$ is even and that for some $x>0$, Riemann-Hilbert Problem 1
has a solution $\mathbf{W}^{(n)}(\lambda,x)$. Let
$\mathbf{G}^{(n)\uparrow}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)$ denote the Schlesinger transformations
associated with the potentials $u=u_{n}(x)$,
$\mathrm{e}^{\pm\mathrm{i}\varphi}=\mathrm{e}^{\pm\mathrm{i}\varphi_{n}(x)}$,
$p=p_{n}(x)$, and $q=q_{n}(x)$ for $\epsilon=-1$. We will now show that
$\mathbf{W}^{(n+1)}(\lambda,x)=\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
and that
$\mathbf{W}^{(n-1)}(\lambda,x)=\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$.
Obviously the transformed matrices are analytic exactly where
$\mathbf{W}^{(n)}(\lambda,x)$ is, and it is easy to see that the jump
conditions are satisfied in both cases because the Schlesinger transformation
induces a sign change on the (positive imaginary) branch cut of
$(\mathrm{i}\lambda)^{\pm 1/2}$ but otherwise leaves the jump conditions
invariant. So it remains to check the normalization condition at
$\lambda=\infty$ and the condition at $\lambda=0$.

We start by combining (18) with (99) to get, for $n$ even,

(119)
$\begin{split}u_{n}(x)=\epsilon\mathrm{e}^{\mathrm{i}\pi/6}xB_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x),&\quad\mathrm{e}^{\pm\mathrm{i}\varphi_{n}(x)}=-\epsilon\left[\frac{B_{0,12}^{(n)}(x)}{B_{0,22}^{(n)}(x)}\right]^{\pm
1},\\\
p_{n}(x)=-8\mathrm{e}^{\mathrm{i}\pi/6}xA_{1,12}^{(n)}(x)B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x),&\quad
q_{n}(x)=8\mathrm{e}^{\mathrm{i}\pi/6}xA_{1,21}^{(n)}(x)B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x).\end{split}$

Using these in (113)–(114) and substituting into (111) and (112) gives

(120) $\mathbf{G}^{(n)\uparrow}(\lambda,x)=\left(\begin{bmatrix}1&0\\\
\epsilon
A_{1,21}^{(n)}(x)&1\end{bmatrix}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\end{bmatrix}\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2},$
(121) $\mathbf{G}^{(n)\downarrow}(\lambda,x)=\left(\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\
0&1\end{bmatrix}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\end{bmatrix}\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}.$

Now using (87) from Lemma 3 gives

(122)
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\left(\frac{\mathrm{i}\lambda}{x}\right)^{(n+1)\sigma_{3}}\\\
\begin{aligned} &=\left(\begin{bmatrix}1&0\\\ \epsilon
A_{1,21}^{(n)}(x)&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\left(\mathbb{I}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\mathbf{A}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{-2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\\\
&=\left(\begin{bmatrix}1&0\\\ \epsilon
A_{1,21}^{(n)}(x)&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\left(\begin{bmatrix}1&0\\\
A_{1,21}^{(n)}(x)&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\\\
&=\mathbb{I}+\mathcal{O}(\lambda^{-1}),\quad\lambda\to\infty,\quad\text{and}\end{aligned}$
(123)
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{\mathrm{i}x(\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\left(\frac{\mathrm{i}\lambda}{x}\right)^{(n-1)\sigma_{3}}\\\
\begin{aligned} &=\left(\begin{bmatrix}1&\epsilon A_{1,12}^{(n)}(x)\\\
0&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\left(\mathbb{I}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\mathbf{A}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{-2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\\\
&=\left(\begin{bmatrix}1&\epsilon A_{1,12}^{(n)}(x)\\\
0&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\left(\begin{bmatrix}1&A_{1,12}^{(n)}(x)\\\
0&1\end{bmatrix}+\mathcal{O}(\lambda^{-1})\right)\\\
&=\mathbb{I}+\mathcal{O}(\lambda^{-1}),\quad\lambda\to\infty,\end{aligned}$

using $\epsilon=-1$. Therefore
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ behave
respectively as $\mathbf{W}^{(n+1)}(\lambda,x)$ and
$\mathbf{W}^{(n-1)}(\lambda,x)$ are required to according to Riemann-Hilbert
Problem 1. On the other hand, using (88) from Lemma 3 for $n$ even gives

(124)
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/4}\\\
\begin{aligned}
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\end{bmatrix}+\begin{bmatrix}1&0\\\
\epsilon A_{1,21}^{(n)}(x)&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\left(\mathbf{B}_{0}^{(n)}(x)+\left(\frac{\mathrm{i}\lambda}{x}\right)\mathbf{B}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{-2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\\\
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,12}^{(n)}(x)B_{0,22}^{(n)}(x)^{-1}\end{bmatrix}+\begin{bmatrix}1&0\\\
\epsilon A_{1,21}^{(n)}(x)&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&B_{0,12}^{(n)}(x)\\\
0&0\end{bmatrix}+\begin{bmatrix}B_{0,11}^{(n)}(x)&B_{1,12}^{(n)}(x)\\\
0&B_{0,22}^{(n)}(x)\end{bmatrix}+\mathcal{O}(\lambda)\right)\\\
&=\mathcal{O}(1),\quad\lambda\to 0,\end{aligned}$

and

(125)
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/4}\\\
\begin{aligned}
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\end{bmatrix}+\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\ 0&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\left(\mathbf{B}_{0}^{(n)}(x)+\left(\frac{\mathrm{i}\lambda}{x}\right)\mathbf{B}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{-2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\\\
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,22}^{(n)}(x)B_{0,12}^{(n)}(x)^{-1}&0\end{bmatrix}+\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\ 0&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&0\\\
0&B_{0,22}^{(n)}(x)\end{bmatrix}+\begin{bmatrix}0&B_{0,12}^{(n)}(x)\\\
B_{0,21}^{(n)}(x)&B_{1,22}^{(n)}(x)\end{bmatrix}+\mathcal{O}(\lambda)\right)\\\
&=\mathcal{O}(1),\quad\lambda\to 0,\end{aligned}$

again using $\epsilon=-1$. This proves that
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ behave,
respectively, as $\mathbf{W}^{(n+1)}(\lambda,x)$ and
$\mathbf{W}^{(n-1)}(\lambda,x)$ are required to according to the conditions of
Riemann-Hilbert Problem 1, taking into account that $n$ is even.

Therefore if $\mathbf{W}^{(n)}(\lambda,x)$ is the (unique, by Lemma 2)
solution of Riemann-Hilbert Problem 1 for $n$ even, then
$\mathbf{W}^{(n+1)}(\lambda,x):=\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
and
$\mathbf{W}^{(n-1)}(\lambda,x):=\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
satisfy all the conditions of the same problem for $n\mapsto n+1$ and
$n\mapsto n-1$ respectively, so they are the unique solutions of those
problems.

##### The case of $n$ odd.

Now suppose that $n$ is odd and that for some $x>0$, Riemann-Hilbert Problem 1
has a solution $\mathbf{W}^{(n)}(\lambda,x)$. Again let
$\mathbf{G}^{(n)\uparrow}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)$ denote the Schlesinger transformations
associated with the potentials $u=u_{n}(x)$,
$\mathrm{e}^{\pm\mathrm{i}\varphi}=\mathrm{e}^{\pm\mathrm{i}\varphi_{n}(x)}$,
$p=p_{n}(x)$, and $q=q_{n}(x)$ for $\epsilon=-1$. As before, we just have to
check that the transformed matrices
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ behave as
required as $\lambda\to\infty$ and $\lambda\to 0$. Combining (18) with (99)
for $n$ odd gives

(126)
$\begin{split}u_{n}(x)=-\epsilon\mathrm{e}^{5\pi\mathrm{i}/6}xB_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x),&\quad\mathrm{e}^{\pm\mathrm{i}\varphi_{n}(x)}=-\epsilon\left[\frac{B^{(n)}_{0,11}(x)}{B^{(n)}_{0,21}(x)}\right]^{\pm
1},\\\
p_{n}(x)=8\mathrm{e}^{5\pi\mathrm{i}/6}xA_{1,12}^{(n)}(x)B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x),&\quad
q_{n}(x)=-8\mathrm{e}^{5\pi\mathrm{i}/6}xA_{1,21}^{(n)}(x)B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x).\end{split}$

Then using (111)–(114) gives

(127) $\mathbf{G}^{(n)\uparrow}(\lambda,x)=\left(\begin{bmatrix}1&0\\\
\epsilon
A_{1,21}^{(n)}(x)&1\end{bmatrix}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\end{bmatrix}\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}$

and

(128) $\mathbf{G}^{(n)\downarrow}(\lambda,x)=\left(\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\
0&1\end{bmatrix}+\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\end{bmatrix}\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}.$

Since the Schlesinger transformations for $n$ odd agree with those for $n$
even in their leading terms for large $\lambda$, exactly the same calculations
(122)–(123) apply for $n$ odd (although the Laudau symbols stand for different
expressions in the even and odd cases). Therefore using $\epsilon=-1$,
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ behave
respectively as $\mathbf{W}^{(n+1)}(\lambda,x)$ and
$\mathbf{W}^{(n-1)}(\lambda,x)$ are required to in the limit
$\lambda\to\infty$ according to Riemann-Hilbert Problem 1. Now using (88) from
Lemma 3 for $n$ odd gives

(129)
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/4}\\\
\begin{aligned}
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\end{bmatrix}+\begin{bmatrix}1&0\\\
\epsilon A_{1,21}^{(n)}(x)&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\left(\mathbf{B}_{0}^{(n)}(x)+\left(\frac{\mathrm{i}\lambda}{x}\right)\mathbf{B}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{-2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\\\
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&\epsilon
B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\\\
0&A_{1,21}^{(n)}(x)B_{0,11}^{(n)}(x)B_{0,21}^{(n)}(x)^{-1}\end{bmatrix}+\begin{bmatrix}1&0\\\
\epsilon A_{1,21}^{(n)}(x)&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}B_{0,11}^{(n)}(x)&0\\\
0&0\end{bmatrix}+\begin{bmatrix}B_{1,11}^{(n)}(x)&B_{0,12}^{(n)}(x)\\\
B_{0,21}^{(n)}(x)&0\end{bmatrix}+\mathcal{O}(\lambda)\right)\\\
&=\mathcal{O}(1),\quad\lambda\to 0,\quad\text{and}\end{aligned}$ (130)
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda}\mathbf{E}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/4}\\\
\begin{aligned}
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\end{bmatrix}+\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\ 0&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\frac{\mathrm{i}\lambda}{x}\right)^{\sigma_{3}/2}\left(\mathbf{B}_{0}^{(n)}(x)+\left(\frac{\mathrm{i}\lambda}{x}\right)\mathbf{B}_{1}^{(n)}(x)+\mathcal{O}(\lambda^{2})\right)\left(\frac{\mathrm{i}\lambda}{x}\right)^{-\sigma_{3}/2}\\\
&=\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}A_{1,12}^{(n)}(x)B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\\\
\epsilon
B_{0,21}^{(n)}(x)B_{0,11}^{(n)}(x)^{-1}&0\end{bmatrix}+\begin{bmatrix}1&\epsilon
A_{1,12}^{(n)}(x)\\\ 0&1\end{bmatrix}\right)\\\
&\quad\quad\quad{}\cdot\left(\left(\frac{\mathrm{i}\lambda}{x}\right)^{-1}\begin{bmatrix}0&0\\\
B_{0,21}^{(n)}(x)&0\end{bmatrix}+\begin{bmatrix}B_{0,11}^{(n)}(x)&0\\\
B_{1,21}^{(n)}(x)&B_{0,22}^{(n)}(x)\end{bmatrix}+\mathcal{O}(\lambda)\right)\\\
&=\mathcal{O}(1),\quad\lambda\to 0,\end{aligned}$

using $\epsilon=-1$. So,
$\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$ and
$\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
respectively behave the same as $\lambda\to 0$ as
$\mathbf{W}^{(n+1)}(\lambda,x)$ and $\mathbf{W}^{(n-1)}(\lambda,x)$. Along
with the behavior as $\lambda\to\infty$ and the analyticity and jump
properties, we conclude that
$\mathbf{W}^{(n+1)}(\lambda,x):=\mathbf{G}^{(n)\uparrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
and
$\mathbf{W}^{(n-1)}(\lambda,x):=\mathbf{G}^{(n)\downarrow}(\lambda,x)\mathbf{W}^{(n)}(\lambda,x)$
are the unique solutions of Riemann-Hilbert Problem 1 for $n\mapsto n+1$ and
for $n\mapsto n-1$ respectively.

####  3.3.3. Bäcklund transformations

Since for any $n\in\mathbb{Z}$,
$\mathbf{J}^{(n+1)}(x)=\mathbf{J}^{(n)\uparrow}(x)$,
$\mathbf{K}^{(n+1)}(x)=\mathbf{K}^{(n)\uparrow}(x)$,
$\mathbf{J}^{(n-1)}(x)=\mathbf{J}^{(n)\downarrow}(x)$, and
$\mathbf{K}^{(n-1)}(x)=\mathbf{K}^{(n)\downarrow}(x)$ we derive from (18) with
$\epsilon=-1$ and (115)–(116) the explicit Bäcklund transformations

(131)
$\begin{split}u_{n+1}(x)&=-\frac{\mathrm{i}b_{n}x^{2}\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}q_{n}(x)}{8u_{n}(x)^{2}},\quad\mathrm{e}^{\pm\mathrm{i}\varphi_{n+1}(x)}=\left[\frac{8u_{n}(x)}{q_{n}(x)}\right]^{\pm
1},\\\
p_{n+1}(x)&=\frac{\mathrm{i}b_{n}x^{2}\mathrm{e}^{2\mathrm{i}\varphi_{n}(x)}q_{n}(x)}{u_{n}(x)^{2}},\\\
q_{n+1}(x)&=\frac{\mathrm{i}b_{n}x\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}q_{n}(x)}{2u_{n}(x)^{2}}\left[\frac{x\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}q_{n}(x)^{2}}{32u_{n}(x)^{2}}+\frac{(n+1)q_{n}(x)}{8xu_{n}(x)}-\frac{\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}u_{n}(x)}{2x^{2}}\right],\end{split}$

and using (117)–(118) instead of (115)–(116) we get

(132)
$\begin{split}u_{n-1}(x)&=\frac{\mathrm{i}b_{n}x^{2}\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}p_{n}(x)}{8u_{n}(x)^{2}},\quad\mathrm{e}^{\pm\mathrm{i}\varphi_{n-1}(x)}=-\left[\frac{p_{n}(x)}{8u_{n}(x)}\right]^{\pm
1},\\\
p_{n-1}(x)&=\frac{\mathrm{i}b_{n}x\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}p_{n}(x)}{2u_{n}(x)^{2}}\left[\frac{x\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}p_{n}(x)^{2}}{32u_{n}(x)^{2}}-\frac{(n-1)p_{n}(x)}{8xu_{n}(x)}-\frac{\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}u_{n}(x)}{2x^{2}}\right],\\\
q_{n-1}(x)&=\frac{\mathrm{i}b_{n}x^{2}\mathrm{e}^{-2\mathrm{i}\varphi_{n}(x)}p_{n}(x)}{u_{n}(x)^{2}}.\end{split}$

In these expressions, $b_{n}$ is the constant given, for any $n\in\mathbb{Z}$,
by

(133)
$b_{n}=\frac{2\mathrm{i}nu_{n}(x)}{x}-\frac{1}{2}\mathrm{i}xp_{n}(x)\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}+\frac{1}{2}\mathrm{i}xq_{n}(x)\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}.$

It then follows from (131) that $b_{n+1}=b_{n}$ and from (132) that
$b_{n-1}=b_{n}$ as well. Since $u_{0}(x)=\tfrac{1}{2}x^{1/3}$ satisfies (4)
with $\epsilon=-1$, $a=-\mathrm{i}n$ for $n=0$ and $b=b_{0}=\mathrm{i}$, it is
then clear that the function $u_{n}(x)$ extracted from Riemann-Hilbert Problem
1 using (100) satisfies (4) with $\epsilon=-1$, $a=-\mathrm{i}n$, and
$b=b_{n}=\mathrm{i}$ for all $n\in\mathbb{Z}$.

Now, when $n=0$, $u_{0}(x)$ is obviously a rational function of $x^{1/3}$,
while from (21) we see that
$\mathrm{e}^{\mathrm{i}\varphi_{0}(x)}\mathrm{e}^{3x^{2/3}}$,
$\mathrm{e}^{-\mathrm{i}\varphi_{0}(x)}\mathrm{e}^{-3x^{2/3}}$,
$p_{0}(x)\mathrm{e}^{3x^{2/3}}$, and $q_{0}(x)\mathrm{e}^{-3x^{2/3}}$ are
rational in $x^{1/3}$ as well. It follows inductively from (131) and (132)
that for all $n\in\mathbb{Z}$, $u_{n}(x)$,
$\mathrm{e}^{\mathrm{i}\varphi_{n}(x)}\mathrm{e}^{3x^{2/3}}$,
$\mathrm{e}^{-\mathrm{i}\varphi_{n}(x)}\mathrm{e}^{-3x^{2/3}}$,
$p_{n}(x)\mathrm{e}^{3x^{2/3}}$, and $q_{n}(x)\mathrm{e}^{-3x^{2/3}}$ are
rational functions of $x^{1/3}$.

We summarize the results of Sections 3.2–3.3 in the following theorem.

######  Theorem 1.

Suppose that Riemann-Hilbert Problem 1 is solvable for given $n\in\mathbb{Z}$
and $x>0$. Then the function $u_{n}(x)$ defined by (100) is the unique
solution of (4) with $\epsilon=-1$, $a=-\mathrm{i}n$, and $b=\mathrm{i}$ that
is a rational function of $x^{1/3}$.

###  3.4. Change of branch cut

Although the principal branch of $(\mathrm{i}\lambda)^{p}$ for various powers
$p$ (cut on the positive imaginary axis) is most convenient for describing the
canonical solutions in different Stokes sectors at $\lambda=0$ and
$\lambda=\infty$, for subsequent asymptotic analysis in the limit of large $n$
it will be better to reformulate a version of Riemann-Hilbert Problem 1
involving power functions with branch cuts on the negative imaginary axis
instead. To this end, we start with $\mathbf{W}^{(n)}(\lambda,x)$ solving
Riemann-Hilbert Problem 1 and, letting $D_{0}^{\pm}\subset D_{0}$ denote the
region between $C^{\pm}$ and the imaginary axis, we set

(134)
$\mathbf{Y}^{(n)}(\lambda,x):=\mathrm{i}^{n\sigma_{3}}\mathbf{W}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}(x\lambda-x(\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\begin{cases}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},&\quad\lambda\in
D_{\infty}^{+},\\\
(-1)^{n}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},&\quad\lambda\in
D_{\infty}^{-},\\\
\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\mathrm{e}^{-5\pi\mathrm{i}\sigma_{3}/6},&\quad\lambda\in
D_{0}^{+},\\\
(-1)^{n}(-\mathrm{i}\sigma_{2})\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/3},&\quad\lambda\in
D_{0}^{-}.\end{cases}$

Letting $\Sigma_{0}^{-}$ denote the segment of the imaginary axis between
$\lambda=0$ and $\lambda=-\mathrm{i}$, oriented downwards,

(135)
$\mathbf{Y}^{(n)}_{+}(\lambda,x)=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\begin{bmatrix}1&0\\\
\mathrm{i}&1\end{bmatrix}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},\quad\lambda\in\Sigma_{\infty}^{+},$
(136)
$\mathbf{Y}^{(n)}_{+}(\lambda,x)=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda_{-})^{-1/2})\sigma_{3}}(-1)^{n}\begin{bmatrix}1&-\mathrm{i}\\\
0&1\end{bmatrix}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda_{+})^{-1/2})\sigma_{3}},\quad\lambda\in\Sigma_{\infty}^{-},$
(137)
$\mathbf{Y}^{(n)}_{+}(\lambda,x)=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\begin{bmatrix}1&0\\\
\mathrm{i}&1\end{bmatrix}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},\quad\lambda\in\Sigma_{0}^{+},$
(138)
$\begin{split}\mathbf{Y}^{(n)}_{+}(\lambda,x)&=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda_{-})^{-1/2})\sigma_{3}}(-1)^{n}\mathrm{i}\sigma_{1}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda_{+})^{-1/2})\sigma_{3}}\\\
&=\mathbf{Y}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}}(-1)^{n}\mathrm{i}\sigma_{1}\mathrm{e}^{\mathrm{i}x\lambda\sigma_{3}},\quad\lambda\in\Sigma_{0}^{-},\end{split}$
(139)
$\mathbf{Y}^{(n)}_{+}(\lambda,x)=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\begin{bmatrix}1&0\\\
-\mathrm{i}&1\end{bmatrix}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},\quad\lambda\in
C^{+},$ (140)
$\mathbf{Y}^{(n)}_{+}(\lambda,x)=\mathbf{Y}^{(n)}_{-}(\lambda,x)\mathrm{e}^{-(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}}\begin{bmatrix}1&0\\\
-\mathrm{i}&1\end{bmatrix}\mathrm{e}^{(\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2})\sigma_{3}},\quad\lambda\in
C^{-}.$

Using the identity relating principal branches:

(141) $(\mathrm{i}\lambda)^{p}=\mathrm{e}^{\pm\mathrm{i}\pi
p}(-\mathrm{i}\lambda)^{p},\quad\pm\mathrm{Im}(\mathrm{i}\lambda)>0,$

we see from (87) that regardless of whether $\lambda\to\infty$ from
$D_{\infty}^{+}$ or $D_{\infty}^{-}$,

(142)
$\mathbf{Y}^{(n)}(\lambda,x)\mathrm{e}^{x(-\mathrm{i}x\lambda)^{-1/2}\sigma_{3}}\left(-\frac{\mathrm{i}\lambda}{x}\right)^{n\sigma_{3}/2}\sim\mathbb{I}+\sum_{p=1}^{\infty}\left(-\frac{\mathrm{i}\lambda}{x}\right)^{-p}(-1)^{p}\mathrm{i}^{n\sigma_{3}}\mathbf{A}_{p}^{(n)}(x)\mathrm{i}^{-n\sigma_{3}},\quad\lambda\to\infty.$

Similarly, using (88), regardless of whether $\lambda\to 0$ from $D_{0}^{+}$
or $D_{0}^{-}$,

(143)
$\mathbf{Y}^{(n)}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}}\widetilde{\mathbf{E}}\cdot\left(-\frac{\mathrm{i}\lambda}{x}\right)^{-(-1)^{n}\sigma_{3}/4}\sim\sum_{p=0}^{\infty}\left(-\frac{\mathrm{i}\lambda}{x}\right)^{p}(-1)^{p}\mathrm{i}^{n\sigma_{3}}\mathbf{B}_{p}^{(n)}(x)\mathrm{e}^{\mathrm{i}\pi(-1)^{n}\sigma_{3}/4},\quad\lambda\to
0,\quad\text{where}$ (144)
$\widetilde{\mathbf{E}}:=\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}\mathbf{E}=\mathrm{e}^{-\mathrm{i}\pi\sigma_{3}/3}\mathrm{i}\sigma_{2}\mathbf{E}\mathrm{i}^{\sigma_{3}}=\frac{1}{\sqrt{2}}\begin{bmatrix}\mathrm{e}^{5\pi\mathrm{i}/6}&\mathrm{e}^{-5\pi\mathrm{i}/6}\\\
\mathrm{e}^{-\mathrm{i}\pi/6}&\mathrm{e}^{-5\pi\mathrm{i}/6}\end{bmatrix}.$

Along with the analyticity of $\mathbf{Y}^{(n)}(\lambda,x)$ for
$\lambda\in\mathbb{C}\setminus(\Sigma_{\infty}^{+}\cup\Sigma_{\infty}^{-}\cup\Sigma_{0}^{+}\cup\Sigma_{0}^{-}\cup
C^{+}\cup C^{-})$ and continuity of boundary values implied by the
substitution (134), these conditions amount to an equivalent Riemann-Hilbert
problem for the matrix $\mathbf{Y}^{(n)}(\lambda,x)$.

##  4\. Application: asymptotic analysis of the algebraic solution $u_{n}(x)$
for large $n$

###  4.1. Rescaling

Note that $u\mapsto\mathrm{i}u$ and $x\mapsto\mathrm{i}x$ takes a solution of
(4) to a solution of the same equation for the same values of $b$ and
$\epsilon$, but with $a\mapsto-a$. Hence it is sufficient to assume that $n$
is a large positive integer. We initially scale the variables in Riemann-
Hilbert Problem 1 with $n>0$ as

(145) $x=n^{3/2}y\quad\text{and}\quad\lambda=n^{-1/2}\nu.$

The two terms in the exponent in the jump matrices then balance with the
exponent in the normalization condition as $\lambda\to\infty$, which is
proportional to $n$:

(146)
$\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2}=n(\mathrm{i}y\nu-y(-\mathrm{i}y\nu)^{-1/2}).$

It is convenient to further scale the spectral parameter by the rescaled
parameter $y$, to make the dominant term near $\infty$ independent of $y$.
Thus for $y>0$, we set $\nu=y^{-1}\eta$ and obtain

(147)
$\mathrm{i}x\lambda-x(-\mathrm{i}x\lambda)^{-1/2}=n\Phi(\eta,y),\quad\Phi(\eta,y):=\mathrm{i}\eta-y(-\mathrm{i}\eta)^{-1/2}.$

The quantity $\mathrm{i}\lambda/x$ appearing in the normalization condition
then reads

(148)
$\frac{\mathrm{i}\lambda}{x}=\frac{1}{n^{2}}\frac{\mathrm{i}\nu}{y}=\frac{1}{n^{2}y^{2}}\mathrm{i}\eta,\quad\text{so}\quad\left(-\frac{\mathrm{i}\lambda}{x}\right)^{n\sigma_{3}/2}=(ny)^{-n\sigma_{3}}(-\mathrm{i}\eta)^{n\sigma_{3}/2}.$

We define a new unknown that is a function of the variables $y,\eta$ instead
of $x,\lambda$ by setting

(149)
$\mathbf{Z}^{(n)}(\eta,y):=(ny)^{-n\sigma_{3}}\mathbf{Y}^{(n)}(n^{3/2}y,n^{-1/2}y^{-1}\eta).$

Then, after an unimportant rescaling of the jump contour for
$\mathbf{Y}^{(n)}(\lambda,x)$ to fix it in the $\eta$-plane,
$\mathbf{Z}^{(n)}(\eta,y)$ satisfies the following conditions:

  * •

$\mathbf{Z}^{(n)}(\eta,y)$ is analytic for
$\eta\in\mathbb{C}\setminus(\Sigma_{\infty}^{+}\cup\Sigma_{\infty}^{-}\cup\Sigma_{0}^{+}\cup\Sigma_{0}^{-}\cup
C^{+}\cup C^{-})$.

  * •

$\mathbf{Z}^{(n)}(\eta,y)$ takes continuous boundary values on the jump
contour related by the jump conditions

(150)
$\mathbf{Z}^{(n)}_{+}(\eta,y)=\mathbf{Z}^{(n)}_{-}(\eta,y)\mathrm{e}^{-n\Phi(\eta,y)\sigma_{3}}\begin{bmatrix}1&0\\\
\mathrm{i}&1\end{bmatrix}\mathrm{e}^{n\Phi(\eta,y)\sigma_{3}},\quad\eta\in\Sigma_{\infty}^{+}\cup\Sigma_{0}^{+},$
(151)
$\mathbf{Z}_{+}^{(n)}(\eta,y)=\mathbf{Z}^{(n)}_{-}(\eta,y)\mathrm{e}^{-n\Phi(\eta,y)\sigma_{3}}\begin{bmatrix}1&0\\\
-\mathrm{i}&1\end{bmatrix}\mathrm{e}^{n\Phi(\eta,y)\sigma_{3}},\quad\eta\in
C^{+}\cup C^{-},$ (152)
$\mathbf{Z}_{+}^{(n)}(\eta,y)=\mathbf{Z}^{(n)}_{-}(\eta,y)\mathrm{e}^{-n\Phi_{-}(\eta,y)\sigma_{3}}(-1)^{n}\begin{bmatrix}1&-\mathrm{i}\\\
0&1\end{bmatrix}\mathrm{e}^{n\Phi_{+}(\eta,y)\sigma_{3}},\quad\eta\in\Sigma_{\infty}^{-},\quad\text{and}$
(153)
$\mathbf{Z}_{+}^{(n)}(\eta,y)=\mathbf{Z}^{(n)}_{-}(\eta,y)\mathrm{e}^{-n\Phi_{-}(\eta,y)\sigma_{3}}(-1)^{n}\mathrm{i}\sigma_{1}\mathrm{e}^{n\Phi_{+}(\eta,y)\sigma_{3}},\quad\eta\in\Sigma_{0}^{-}.$

  * •

$\mathbf{Z}^{(n)}(\eta,y)(-\mathrm{i}\eta)^{n\sigma_{3}/2}\to\mathbb{I}$ as
$\eta\to\infty$.

  * •

The limit of
$\mathbf{Z}^{(n)}(\eta,y)\mathrm{e}^{-\mathrm{i}n\eta\sigma_{3}}\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}$
as $\eta\to 0$ exists. In terms of the leading coefficient in the expansion
(88) (and see also (143)),

(154) $\lim_{\eta\to
0}\mathbf{Z}^{(n)}(\eta,y)\mathrm{e}^{-\mathrm{i}n\eta\sigma_{3}}\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}=(-\mathrm{i}ny)^{-n\sigma_{3}}\mathbf{B}_{0}^{(n)}(n^{3/2}y)(-\mathrm{i}ny)^{-(-1)^{n}\sigma_{3}/2}.$

This may be regarded as the definition of $\mathbf{B}_{0}^{(n)}(n^{3/2}y)$ in
terms of $\mathbf{Z}^{(n)}(\eta,y)$.

###  4.2. Aside: scaling formalism

We perturb the basic scaling $y=n^{-3/2}x$ designed to balance the terms in
the exponents in Riemann-Hilbert Problem 1 by writing $x=n^{3/2}(y+n^{-p}z)$
for some $p>0$ to be determined. Then we also scale $u=n^{q}U$ and consider
what the Painlevé-III (D7) equation (4) on $u(x)$ with $\epsilon=-1$,
$a=-\mathrm{i}n$, and $b=\mathrm{i}$ implies for $U=U(z)$ when $y\neq 0$ is a
fixed parameter:

(155)
$n^{q-3+2p}U^{\prime\prime}(z)=n^{q-3+2p}\frac{U^{\prime}(z)^{2}}{U(z)}-n^{q-3+p}\frac{U^{\prime}(z)}{y+n^{-p}z}+n^{2q-3/2}\frac{8U(z)^{2}}{y+n^{-p}z}+n^{-1/2}\frac{2}{y+n^{-p}z}-n^{-q}\frac{1}{U(z)}.$

Because $p>0$, the second term on the right-hand side is negligible compared
with the left-hand side and the preceding term. All remaining terms can be
balanced with these by choosing $p=1$ and $q=\frac{1}{2}$. Then (155) becomes

(156)
$U^{\prime\prime}(z)=\frac{U^{\prime}(z)^{2}}{U(z)}+\frac{8}{y}U(z)^{2}+\frac{2}{y}-\frac{1}{U(z)}+\mathcal{O}(n^{-1}).$

The _approximating equation_ is obtained by dropping the formally-small
$\mathcal{O}(n^{-1})$ error term. It is an autonomous nonlinear second order
equation on $U(z)$ with parameter $y\neq 0$. It turns out that under these
scalings, the algebraic solutions $u_{n}(x)$ of (4) behave for large $n$ like
one or the other of two types of solutions of the approximating equation,
depending on the value of the parameter $y$. Although the motivation comes
from the simple scaling formalism above, these are correspondences that are
proved using techniques of analysis of Riemann-Hilbert problems.

####  4.2.1. Equilibrium solutions

If $U$ is independent of $z$, the approximating equation yields equilibrium
solutions solving the cubic equation

(157) $8U^{3}+2U-y=0.$

For large $y$, the solutions are $U\approx\frac{1}{2}y^{1/3}$ and
$U\approx\frac{1}{2}\mathrm{e}^{\pm 2\pi\mathrm{i}/3}y^{1/3}$. Reversing the
scalings by $U=n^{-1/2}u$ and $y=n^{-3/2}x$ (neglecting $n^{-1}z$) the
large-$y$ equilibrium solution $U\approx\frac{1}{2}y^{1/3}$ reads
$u\approx\frac{1}{2}x^{1/3}$, which is exactly the seed solution for $n=0$.
The three distinct solutions for large $y$ can coalesce at branch points that
can be found by equating to zero the discriminant of the cubic (157) with
respect to $U$, which is $-64(27y^{2}+4)$.

####  4.2.2. Non-equilibrium solutions

Multiplying the approximating equation through by $U^{\prime}(z)/U(z)^{2}$ we
obtain

(158)
$\frac{U^{\prime}(z)U^{\prime\prime}(z)}{U(z)^{2}}-\frac{U^{\prime}(z)^{3}}{U(z)^{3}}-\frac{8}{y}U^{\prime}(z)-\frac{2}{y}\frac{U^{\prime}(z)}{U(z)^{2}}+\frac{U^{\prime}(z)}{U(z)^{3}}=0$

or, equivalently,

(159)
$\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{U^{\prime}(z)^{2}}{2U(z)^{2}}-\frac{8}{y}U(z)+\frac{2}{y}\frac{1}{U(z)}-\frac{1}{2}\frac{1}{U(z)^{2}}\right]=0.$

Letting $E$ denote the implied integration constant,

(160) $U^{\prime}(z)^{2}=\frac{16}{y}U(z)^{3}+2EU(z)^{2}-\frac{4}{y}U(z)+1.$

Setting

(161) $U(z)=\frac{1}{4}yw(z)-\frac{1}{24}yE,$

$w(z)$ solves the Weierstraß equation
$w^{\prime}(z)^{2}=4w(z)^{3}-g_{2}w(z)-g_{3}$ (see [14, Ch. III.5] or [20, Ch.
23]) with invariants

(162) $g_{2}=\frac{16}{y^{2}}+\frac{E^{2}}{3};\quad
g_{3}=-\frac{16}{y^{2}}-\frac{8E}{3y^{2}}-\frac{E^{3}}{27}.$

Therefore, the general solution of the approximating equation can be written
in terms of the Weierstraß elliptic function $\wp(z)$ with these invariants as

(163) $U(z)=\frac{1}{4}y\wp(z-z_{0})-\frac{1}{24}yE.$

The fact that all poles of $\wp(z)$ are double is consistent with the fact
that all nonzero poles of solutions of (4) are also double.

###  4.3. $g$-function and spectral curves

Returning to the large-$n$ analysis of $\mathbf{Z}^{(n)}(\eta,y)$, we now
introduce a $g$-function $\eta\mapsto g(\eta,y)$ with parameter $y$ with the
following properties:

  * •

$g$ is analytic for
$\eta\in\mathbb{C}\setminus(\Sigma^{+}_{\infty}\cup\Sigma^{-}_{\infty}\cup\Sigma_{0}\cup
C^{+}\cup C^{-})$ and is bounded for bounded $\eta$.

  * •

$g$ takes continuous boundary values on the jump contour with the property
that

(164) $\eta\mapsto F(\eta,y):=\left(\frac{\partial
g}{\partial\eta}(\eta,y)-\frac{\partial\Phi}{\partial\eta}(\eta,y)\right)^{2}$

is continuous except at $\eta=0$.

  * •

There is a quantity $g_{0}(y)$ such that
$g(\eta,y)=-\tfrac{1}{2}\log(-\mathrm{i}\eta)+g_{0}(y)+o(1)$ as
$\eta\to\infty$.

  * •

$g$ is independent of $n$.

Then we use such a function $g(\eta,y)$ to define a new unknown by

(165)
$\mathbf{M}^{(n)}(\eta,y):=\mathrm{e}^{ng_{0}(y)\sigma_{3}}\mathbf{Z}^{(n)}(\eta,y)\mathrm{e}^{-ng(\eta,y)\sigma_{3}}.$

Then $\eta\mapsto\mathbf{M}^{(n)}(\eta,y)$ is analytic where
$\mathbf{Z}^{(n)}(\eta,y)$ is, satisfies the simplified normalization
condition that $\mathbf{M}^{(n)}(\eta,y)\to\mathbb{I}$ as $\eta\to\infty$, has
the property that
$\mathbf{M}^{(n)}(\eta,y)\widetilde{\mathbf{E}}^{\pm}(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}$
has a common limit as $\eta\to 0$ from $D_{0}^{\pm}$, and has jump conditions
explicitly related to those satisfied by $\mathbf{Z}^{(n)}(\eta,y)$ that
involve the boundary values of $g$. We will explain these jump conditions
later.

But first, we consider the function $F(\eta,y)$ defined by (164). Obviously,
this function is not only continuous for $\eta\neq 0$, but it is also analytic
for $\eta\in\mathbb{C}\setminus\\{0\\}$. It is therefore determined by its
asymptotic behavior as $\eta\to 0$ and $\eta\to\infty$. Since then $\partial
g/\partial\eta$ has to have Laurent expansions in both limits in powers of
$(-\mathrm{i}\eta)^{1/2}$ for $F$ to be analytic, we interpret the conditions
on $g(\eta,y)$ near $\eta=0,\infty$ in terms of $\partial g/\partial\eta$ as
follows:

(166) $\frac{\partial
g}{\partial\eta}(\eta,y)=\begin{cases}-\frac{1}{2\eta}+a_{1}(-\mathrm{i}\eta)^{-3/2}+a_{2}(-\mathrm{i}\eta)^{-2}+\cdots,&\quad\eta\to\infty\\\
b_{1}(-\mathrm{i}\eta)^{-1/2}+b_{2}+\cdots,&\quad\eta\to 0.\end{cases}$

It then follows from (147) and (164) that

(167)
$F(\eta,y)=\begin{cases}\left(-\mathrm{i}-\frac{1}{2\eta}+\mathcal{O}(\eta^{-3/2})\right)^{2}=-1+\frac{\mathrm{i}}{\eta}+\mathcal{O}(\eta^{-2}),&\quad\eta\to\infty,\\\
\left(\frac{1}{2}\mathrm{i}y(-\mathrm{i}\eta)^{-3/2}+\mathcal{O}(\eta^{-1/2})\right)^{2}=-\frac{y^{2}}{4}(-\mathrm{i}\eta)^{-3}+\mathcal{O}(\eta^{-2}),&\quad\eta\to
0.\end{cases}$

Therefore, $F(\eta,y)$ is a Laurent polynomial of the form

(168)
$F(\eta,y)=-1+\frac{1}{\mu}+\frac{c}{\mu^{2}}-\frac{y^{2}}{4\mu^{3}},\quad\mu:=-\mathrm{i}\eta,$

involving an undetermined coefficient $c$. Writing $F(\eta,y)$ in the form
$F(\eta,y)=P(-\mathrm{i}\eta,y)(-\mathrm{i}\eta)^{-3}$, where $P(\mu,y)$ is
the cubic polynomial111Note that this polynomial is closely related to
$\frac{16}{y}U^{3}+2EU^{2}-\frac{4}{y}U+1$ appearing in (160). Indeed,
$-\frac{64U^{3}}{y^{3}}P\left(\frac{y}{2U};y\right)=\frac{16}{y}U^{3}-\frac{16c}{y^{2}}U^{2}-\frac{4}{y}U+1$
which matches the target form if we identify the integration constants by
$y^{2}E=-8c$.

(169) $P(\mu,y):=-\mu^{3}+\mu^{2}+c\mu-\frac{y^{2}}{4}.$

We now assume that $c$ is chosen so that $P(\mu,y)$ has a double root $\mu=d$
and a simple root $\mu=s$. Expanding $P(\mu,y)=-(\mu-d)^{2}(\mu-s)$ and
matching the coefficients with (169) gives three equations:

(170) $2d+s=1,\quad d^{2}+2ds=-c,\quad\text{and}\quad
d^{2}s=-\tfrac{1}{4}y^{2}.$

Combining the first equation with the last, one eliminates $d$ and obtains a
cubic equation for $s$: $s(s-1)^{2}=-y^{2}$. Given any solution of this
equation, $d$ and $c$ are determined explicitly from the first and second
equations. It is clear that as long as $y>0$, there exists one real negative
root and a complex-conjugate pair of roots. We select the real solution (a
guess based on symmetry to be justified later); thus $s$ is decreasing from
$s=0$ as $y$ increases from $y=0$, with asymptotic behavior
$s=-y^{2/3}(1+o(1))$ as $y\to+\infty$. Then from $2d+s=1$ we have that $d$ is
increasing from $d=\tfrac{1}{2}$ as $y$ increases from $y=0$, with asymptotic
behavior $d=\frac{1}{2}y^{2/3}(1+o(1))$ as $y\to+\infty$. We observe that the
discriminant of the cubic $s(s-1)^{2}=-y^{2}$ with respect to $s$ is

(171) $y^{2}(27y^{2}+4)^{3}=0,$

which should be compared with the discriminant of (157).

Now we discuss the jump conditions satisfied by $\mathbf{M}^{(n)}(\eta,y)$.
These take the form shown in (150)–(153) with the only change being that the
function $\Phi(\eta,y)$ in the exponents is replaced with $-h(\eta,y)$, where

(172) $h(\eta,y):=g(\eta,y)-\Phi(\eta,y).$

The effect of the conjugation of the constant jump matrices involving the
exponent $h(\eta,y)$ depends on the sign of $\mathrm{Re}(h(\eta,y))$. Under
the assumption that $P(\mu,y)=-(\mu-d)^{2}(\mu-s)$, with $s=s(y)<0$ and
$d=(1-s)/2>0$, we have from (164) that

(173) $\frac{\partial
h}{\partial\eta}(\eta,y)=\sqrt{f(\eta,y)}=-(\eta-\mathrm{i}d)(-\mathrm{i}\eta)^{-3/2}(-\mathrm{i}\eta-s)^{1/2}.$

Here, all fractional powers refer to the principal branch, and the sign of the
square root was chosen to match the asymptotic behavior at $\eta=\infty$
according to the definitions of $h(\eta,y)$, $\Phi(\eta,y)$, and the
large-$\eta$ behavior of $g(\eta,y)$. This function is analytic except on the
negative imaginary axis between $\eta=0$ and $\eta=\mathrm{i}s$, because the
sign changes of the latter two factors cancel for $\eta$ below $\mathrm{i}s$
on the imaginary axis.

Integrating using $d=(1-s)/2$,

(174)
$h(\eta,y)=(-\mathrm{i}\eta-s+1)(-\mathrm{i}\eta-s)^{1/2}(-\mathrm{i}\eta)^{-1/2}+\frac{1}{2}\log\left(\frac{(-\mathrm{i}\eta-s)^{1/2}-(-\mathrm{i}\eta)^{1/2}}{(-\mathrm{i}\eta-s)^{1/2}+(-\mathrm{i}\eta)^{1/2}}\right)+C$

where $C$ is an integration constant, which we take to be $C=0$. Then one can
check that $h_{+}(\eta,y)+h_{-}(\eta,y)\equiv 0$ for $\eta$ on the segment
between $\eta=0$ and $\eta=\mathrm{i}s$. Also,
$g(\eta,y)=h(\eta,y)+\Phi(\eta,y)$, so

(175)
$g(\eta,y)=(-\mathrm{i}\eta-s+1)(-\mathrm{i}\eta-s)^{1/2}(-\mathrm{i}\eta)^{-1/2}+\frac{1}{2}\log\left(\frac{(-\mathrm{i}\eta-s)^{1/2}-(-\mathrm{i}\eta)^{1/2}}{(-\mathrm{i}\eta-s)^{1/2}+(-\mathrm{i}\eta)^{1/2}}\right)+\mathrm{i}\eta-y(-\mathrm{i}\eta)^{-1/2},$

which indeed has the expansion

(176)
$g(\eta,y)=-\frac{1}{2}\log(-\mathrm{i}\eta)+g_{0}(y)+\mathcal{O}(\eta^{-1/2}),\quad\eta\to\infty,\quad
g_{0}(y):=1-\frac{3}{2}s+\frac{1}{2}\ln\left(-\frac{1}{4}s\right),\quad
s=s(y)<0.$

Also, using $s<0$ and $d^{2}s=-\tfrac{1}{4}y^{2}$ shows that

(177)
$g(\eta,y)=-\frac{1+3s}{2(-s)^{1/2}}(-\mathrm{i}\eta)^{1/2}+\mathcal{O}(\eta),\quad\eta\to
0.$

Assuming that $y>0$ is sufficiently large, the function
$\mathrm{Re}(h(\eta,y))$ is continuous except at $\eta=0$, harmonic except on
the negative imaginary segment connecting the origin with $\eta=\mathrm{i}s$,
and has a zero level set consisting of that same segment and two curves
emanating from $\eta=\mathrm{i}s$ into the left and right half-planes and
extending to $\infty$. These curves are reflections of each other in the
imaginary axis, and all three branches emanating from $\eta=\mathrm{i}s$ are
separated by angles of $2\pi/3$ at that junction point. We have
$\mathrm{Re}(h(\eta,y))>0$ above the two curves (a region containing the
positive imaginary axis) and $\mathrm{Re}(h(\eta,y))<0$ below them. See the
left-hand panel of Figure 5 below.

###  4.4. Parametrix construction and error analysis

To make use of this structure, we assume that the jump contour is deformed so
that $\eta=\mathrm{i}s$ is the junction point of $C^{+}$, $C^{-}$,
$\Sigma_{0}^{-}$, and $\Sigma_{\infty}^{-}$, and so that $\Sigma_{0}^{-}$
coincides with the segment joining $\eta=\mathrm{i}s$ with the origin. Then,
since the sum of the boundary values of $\partial h/\partial\eta(\eta,y)$
vanishes and $\mathrm{Re}(h(\eta,y))=0$ on $\Sigma_{0}^{-}$, the jump
condition of $\mathbf{M}^{(n)}(\eta,y)$ on this arc reads

(178)
$\mathbf{M}^{(n)}_{+}(\eta,y)=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}0&\mathrm{e}^{\mathrm{i}\phi}\\\
-\mathrm{e}^{-\mathrm{i}\phi}&0\end{bmatrix},\quad\phi:=\frac{\pi}{2}+n\pi-\mathrm{i}n(h_{+}(\eta,y)+h_{-}(\eta,y))=\frac{\pi}{2}+n\pi,\quad\eta\in\Sigma_{0}^{-},$

and here, $\phi$ is a real phase that is constant along $\Sigma_{0}^{-}$.
Because $h(\eta,y)$ is analytic on all other arcs of the jump contour except
for $\Sigma_{\infty}^{-}$, along which
$h_{+}(\eta,y)-h_{-}(\eta,y)=-\mathrm{i}\pi$, all other jump matrices for
$\mathbf{M}^{(n)}(\eta,y)$ are exponentially small perturbations of the
identity matrix when $n\to+\infty$, estimates that are uniform with respect to
$\eta$ except in a neighborhood of $\eta=\mathrm{i}s$. Here it is possible to
install a standard Airy parametrix to solve the jump conditions exactly
locally. To construct an outer parametrix, we solve the jump condition (178)
exactly and build in suitable singularities near the two endpoints of
$\Sigma_{0}^{-}$ to allow matching onto the Airy parametrix at
$\eta=\mathrm{i}s$ and to match the required behavior of
$\mathbf{M}^{(n)}(\eta,y)$ near the origin.

####  4.4.1. Outer parametrix

By definition, the outer parametrix is the matrix
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$ with the following properties:

  * •

$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$ is analytic for
$\eta\in\mathbb{C}\setminus\Sigma_{0}^{-}$, taking continuous boundary values
from each side except at the endpoints.

  * •

The boundary values are related by the jump condition (178).

  * •

$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\to\mathbb{I}$ as $\eta\to\infty$.

  * •

$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)=\mathcal{O}(1+|\eta|^{-1/4}+|\eta-\mathrm{i}s|^{-1/4})$.

We construct $\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$ explicitly by first
removing the phase factors $\mathrm{e}^{\pm\mathrm{i}\phi}$ from the jump
condition without changing any of the other conditions; we write
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\mathbf{N}(\eta,y)\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2}$
so that (178) for $\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$ implies

(179)
$\mathbf{N}_{+}(\eta,y)=\mathbf{N}_{-}(\eta,y)\mathrm{i}\sigma_{2},\quad\eta\in\Sigma_{0}^{-}.$

Next, we diagonalize the jump matrix $\mathrm{i}\sigma_{2}$ by the
substitution (again, not changing any of the other properties)

(180) $\mathbf{N}(\eta,y)=\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\mathbf{O}(\eta,y)\frac{1}{\sqrt{2}}\begin{bmatrix}1&-\mathrm{i}\\\
-\mathrm{i}&1\end{bmatrix}.$

Then $\mathbf{O}(\eta,y)$ is analytic for
$\eta\in\mathbb{C}\setminus\Sigma_{0}^{-}$, takes continuous boundary values
except at the endpoints where $-1/4$ power singularities are admissible, tends
to $\mathbb{I}$ as $\eta\to\infty$, and satisfies the diagonal jump condition
$\mathbf{O}_{+}(\eta,y)=\mathbf{O}_{-}(\eta,y)\mathrm{i}^{\sigma_{3}}$ for
$\eta\in\Sigma_{0}^{-}$ (note that the dependence on $y>0$ enters via the
moving endpoint $\eta=\mathrm{i}c$). The unique solution of these conditions
is the diagonal matrix

(181)
$\mathbf{O}(\eta,y):=(-\mathrm{i}(\eta-\mathrm{i}s))^{\sigma_{3}/4}(-\mathrm{i}\eta)^{-\sigma_{3}/4}=\left(\frac{\eta-\mathrm{i}s}{\eta}\right)^{\sigma_{3}/4}$

where in each case the principal branch power is intended. Inverting the
transformations
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\mapsto\mathbf{N}(\eta,y)\mapsto\mathbf{O}(\eta,y)$
gives the explicit formula for the outer parametrix as

(182)
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y):=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\left(\frac{\eta-\mathrm{i}s}{\eta}\right)^{\sigma_{3}/4}\frac{1}{\sqrt{2}}\begin{bmatrix}1&-\mathrm{i}\\\
-\mathrm{i}&1\end{bmatrix}\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2}.$

A calculation shows that
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}$
is analytic at $\eta=0$. Indeed, letting $\mathbf{C}^{(n)}$ denote the matrix

(183) $\mathbf{C}^{(n)}:=\begin{cases}-\mathrm{i}\sigma_{1},&\quad\text{$n$
even,}\\\ \mathbb{I},&\quad\text{$n$ odd,}\end{cases}$

we see that

(184)
$\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}\\\
\begin{aligned}
&=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\left(\frac{\eta-\mathrm{i}s}{\eta}\right)^{\sigma_{3}/4}\mathbf{C}^{(n)}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2}\\\
&=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}(-\mathrm{i}(\eta-\mathrm{i}s))^{\sigma_{3}/4}(-\mathrm{i}\eta)^{-\sigma_{3}/4}\mathbf{C}^{(n)}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2}\\\
&=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}(-\mathrm{i}(\eta-\mathrm{i}s))^{\sigma_{3}/4}\mathbf{C}^{(n)}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2},\end{aligned}$

and having a vertical branch cut emanating downward from $\eta=\mathrm{i}s$
with $s<0$, the diagonal matrix
$(-\mathrm{i}(\eta-\mathrm{i}s))^{\sigma_{3}/4}$ is analytic at $\eta=0$.
Since $\mathbf{C}^{(n)}$ is either diagonal or off-diagonal, we can further
simplify the resulting formula as follows:

(185)
$\begin{split}\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}&=\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\mathbf{C}^{(n)}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}\mathrm{e}^{-\mathrm{i}\phi\sigma_{3}/2}(-\mathrm{i}(\eta-\mathrm{i}s))^{-(-1)^{n}\sigma_{3}/4}\\\
&=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\\
-1&1\end{bmatrix}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}(-\mathrm{i}(\eta-\mathrm{i}s))^{-(-1)^{n}\sigma_{3}/4},\end{split}$

where we also used the $n$-dependent definition of $\phi$ in (178). In
particular, this implies that

(186) $\lim_{\eta\to
0}\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\widetilde{\mathbf{E}}\cdot(-\mathrm{i}\eta)^{-(-1)^{n}\sigma_{3}/4}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\\
-1&1\end{bmatrix}\mathrm{e}^{5\pi\mathrm{i}\sigma_{3}/6}(-s)^{-(-1)^{n}\sigma_{3}/4}.$

####  4.4.2. Airy parametrix

Since $h_{+}(\mathrm{i}s,y)+h_{-}(\mathrm{i}s,y)=0$,
$h_{+}(\eta,y)+h_{-}(\eta,y)$ is real-valued and strictly decreasing in the
negative imaginary direction along $\Sigma_{\infty}^{-}$, and
$h^{\prime}(\eta,y)$ is an analytic function on this contour arc that vanishes
like a square root at $\eta=\mathrm{i}s$, we may introduce a conformal
coordinate $\eta\mapsto W(\eta)$ on a neighborhood $D_{\mathrm{i}s}$ of
$\eta=\mathrm{i}s$ with the properties that $W(\mathrm{i}s)=0$, $W(\eta)>0$
for $\eta\in\Sigma_{\infty}^{-}\cap D_{\mathrm{i}s}$, and
$W(\eta)=-(h_{+}(\eta,y)+h_{-}(\eta,y))^{2/3}$ for
$\eta\in\Sigma_{\infty}^{-}\cap D_{\mathrm{i}s}$. Using
$h_{+}(\eta,y)-h_{-}(\eta,y)=-\mathrm{i}\pi$ for $\eta\in\Sigma_{\infty}^{-}$,
we see that on $C^{\pm}\cap D_{\mathrm{i}s}$, the exponent function
$2h(\eta,y)$ is the analytic continuation from $\Sigma_{\infty}^{-}\cap
D_{\mathrm{i}s}$ through $D_{\pm}^{\infty}\cap D_{\mathrm{i}s}$ of
$2h_{\mp}(\eta,y)=(h_{+}(\eta,y)+h_{-}(\eta,y))\mp(h_{+}(\eta,y)-h_{-}(\eta,y))=h_{+}(\eta,y)+h_{-}(\eta,y)\pm\mathrm{i}\pi$.
Rescaling the conformal coordinate by $\zeta=n^{2/3}W(\eta)$, the jump
conditions for $\mathbf{M}^{(n)}(\eta,y)$ within $D_{\mathrm{i}s}$ can be
written as follows:

(187)
$\begin{split}\mathbf{M}^{(n)}_{+}(\eta,y)&=\mathbf{M}^{(n)}_{-}(\eta,y)\mathrm{e}^{nh_{-}(\eta,y)\sigma_{3}}\begin{bmatrix}(-1)^{n}&-\mathrm{i}(-1)^{n}\\\
0&(-1)^{n}\end{bmatrix}\mathrm{e}^{-nh_{+}(\eta,y)\sigma_{3}}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\mathrm{e}^{n(h_{+}(\eta,y)+h_{-}(\eta,y))\sigma_{3}/2}\mathrm{e}^{\mathrm{i}n\pi\sigma_{3}/2}\begin{bmatrix}(-1)^{n}&-\mathrm{i}(-1)^{n}\\\
0&(-1)^{n}\end{bmatrix}\mathrm{e}^{\mathrm{i}n\pi\sigma_{3}/2}\mathrm{e}^{-n(h_{+}(\eta,y)+h_{-}(\eta,y))\sigma_{3}/2}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}1&-\mathrm{i}(-1)^{n}\mathrm{e}^{n(h_{+}(\eta,y)+h_{-}(\eta,y))}\\\
0&1\end{bmatrix}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}1&-\mathrm{i}(-1)^{n}\mathrm{e}^{-\zeta^{3/2}}\\\
0&1\end{bmatrix}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\mathrm{e}^{\mathrm{i}(-1)^{n}\pi\sigma_{3}/4}\begin{bmatrix}1&-\mathrm{e}^{-\zeta^{3/2}}\\\
0&1\end{bmatrix}\mathrm{e}^{-\mathrm{i}(-1)^{n}\pi\sigma_{3}/4},\quad\eta\in\Sigma_{\infty}^{-}\cap
D_{\mathrm{i}s};\end{split}$ (188)
$\begin{split}\mathbf{M}^{(n)}_{+}(\eta,y)&=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}1&0\\\
-\mathrm{i}\mathrm{e}^{-n(h_{+}(\eta,y)+h_{-}(\eta,y))}&1\end{bmatrix}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}1&0\\\
-\mathrm{i}(-1)^{n}\mathrm{e}^{\zeta^{3/2}}&1\end{bmatrix}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\mathrm{e}^{\mathrm{i}(-1)^{n}\pi\sigma_{3}/4}\begin{bmatrix}1&0\\\
\mathrm{e}^{\zeta^{3/2}}&1\end{bmatrix}\mathrm{e}^{-\mathrm{i}(-1)^{n}\pi\sigma_{3}/4},\quad\eta\in
C^{\pm}\cap D_{\mathrm{i}s};\quad\text{and}\end{split}$ (189)
$\begin{split}\mathbf{M}^{(n)}_{+}(\eta,y)&=\mathbf{M}^{(n)}_{-}(\eta,y)\begin{bmatrix}0&\mathrm{i}(-1)^{n}\\\
\mathrm{i}(-1)^{n}&0\end{bmatrix}\\\
&=\mathbf{M}^{(n)}_{-}(\eta,y)\mathrm{e}^{\mathrm{i}(-1)^{n}\pi\sigma_{3}/4}\begin{bmatrix}0&1\\\
-1&0\end{bmatrix}\mathrm{e}^{-\mathrm{i}(-1)^{n}\pi\sigma_{3}/4},\quad\eta\in\Sigma_{0}^{-}\cap
D_{\mathrm{i}s}.\end{split}$

We deform the contours $C^{\pm}$ within $D_{\mathrm{i}s}$ to lie along the
rays $\arg(\zeta)=\pm 2\pi/3$ as shown in the left-hand panel of Figure 3
below. Then one can check that
$\mathbf{M}(\eta,y)\mathrm{e}^{\mathrm{i}(-1)^{n}\pi\sigma_{3}/4}\mathbf{A}(n^{2/3}W(\eta))^{-1}$
is holomorphic in $D_{\mathrm{i}s}$, where $\mathbf{A}(\zeta)$ is the standard
Airy parametrix as defined (for instance) by the solution of [2, Riemann-
Hilbert Problem 4]. Defining a matrix holomorphic in $D_{\mathrm{i}s}$ and
uniformly bounded as $n\to\infty$ by

(190)
$\mathbf{H}(\eta):=\mathrm{e}^{\mathrm{i}(-1)^{n}\pi\sigma_{3}/4}\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\left(\frac{\eta-\mathrm{i}s}{\eta}\right)^{\sigma_{3}/4}W(\eta)^{-\sigma_{3}/4},\quad\eta\in
D_{\mathrm{i}s},$

we take as the inner parametrix

(191)
$\dot{\mathbf{M}}^{(n),\mathrm{in}}(\eta,y):=\mathbf{H}(\eta)n^{-\sigma_{3}/6}\mathbf{A}(n^{2/3}W(\eta))\mathrm{e}^{-\mathrm{i}(-1)^{n}\pi\sigma_{3}/4},\quad\eta\in
D_{\mathrm{i}s}.$

Then we can compare the inner and outer parametrices on $\partial
D_{\mathrm{i}s}$:

(192)
$\dot{\mathbf{M}}^{(n),\mathrm{in}}(\eta,y)\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)^{-1}=\mathbf{H}(\eta)n^{-\sigma_{3}/6}\mathbf{A}(\zeta)\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\zeta^{-\sigma_{3}/4}\cdot
n^{\sigma_{3}/6}\mathbf{H}(\eta)^{-1},\quad\eta\in\partial D_{\mathrm{i}s},$

where $\zeta=n^{2/3}W(\eta)$ and we used the fact that
$\mathrm{e}^{-\mathrm{i}(-1)^{n}\sigma_{3}/4}\mathrm{e}^{\mathrm{i}\phi\sigma_{3}/2}$
is a multiple (by $\pm 1$) of the identity. Since $\zeta$ is uniformly large
when $\eta\in\partial D_{\mathrm{i}s}$, we use the large-$\zeta$ asymptotic of
$\mathbf{A}(\zeta)$ (see for instance [2, Eqn. (113)]):

(193) $\mathbf{A}(\zeta)\frac{1}{\sqrt{2}}\begin{bmatrix}1&\mathrm{i}\\\
\mathrm{i}&1\end{bmatrix}\zeta^{-\sigma_{3}/4}=\mathbb{I}+\begin{bmatrix}\mathcal{O}(\zeta^{-3})&\mathcal{O}(\zeta^{-1})\\\
\mathcal{O}(\zeta^{-2})&\mathcal{O}(\zeta^{-3})\end{bmatrix},\quad\zeta\to\infty,$

which implies that

(194) $\sup_{\eta\in\partial
D_{\mathrm{i}s}}\|\dot{\mathbf{M}}^{(n),\mathrm{in}}(\eta,y)\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)^{-1}-\mathbb{I}\|=\mathcal{O}(n^{-1}),\quad
n\to\infty.$

####  4.4.3. Global parametrix and error estimation

We define a global parametrix for $\mathbf{M}(\eta,y)$ by

(195)
$\dot{\mathbf{M}}^{(n)}(\eta,y):=\begin{cases}\dot{\mathbf{M}}^{(n),\mathrm{in}}(\eta,y),&\quad\eta\in
D_{\mathrm{i}s}\\\
\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y),&\quad\eta\in\mathbb{C}\setminus\overline{D_{\mathrm{i}s}}.\end{cases}$

The mismatch between the global parametrix and $\mathbf{M}^{(n)}(\eta,y)$ is
defined as

(196)
$\mathbf{D}^{(n)}(\eta,y):=\mathbf{M}^{(n)}(\eta,y)\dot{\mathbf{M}}^{(n)}(\eta,y)^{-1}.$

Since the inner parametrix is an exact solution of the jump conditions for
$\mathbf{M}^{(n)}(\eta,y)$ within $D_{\mathrm{i}s}$,
$\mathbf{D}^{(n)}(\eta,y)$ may be regarded as being analytic for $\eta\in
D_{\mathrm{i}s}$. Similarly, because the outer parametrix is an exact solution
of the jump condition for $\mathbf{M}^{(n)}(\eta,y)$ on the arc
$\Sigma_{0}^{-}$, $\mathbf{D}^{(n)}(\eta,y)$ has no jump across this contour
either. Therefore $\mathbf{D}^{(n)}(\eta,y)$ is analytic in the complement of
the jump contour illustrated in the right-hand panel of Figure 3.

Figure 3. Left: the jump contour near $\eta=\mathrm{i}s$ in the $\zeta$-plane.
Right: the jump contour for $\mathbf{D}^{(n)}(\eta,y)$

On all arcs of the jump contour except for $\partial D_{\mathrm{i}s}$ we have
$\dot{\mathbf{M}}^{(n)}(\eta,y)=\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$,
and the outer parametrix $\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)$ is
analytic; therefore on those arcs,

(197)
$\mathbf{D}^{(n)}_{+}(\eta,y)=\mathbf{D}^{(n)}_{-}(\eta,y)\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)\cdot[\mathbf{M}^{(n)}_{-}(\eta,y)^{-1}\mathbf{M}^{(n)}_{+}(\eta,y)]\cdot\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)^{-1}.$

Because $\mathrm{Re}(h(\eta,y))\geq\delta>0$ holds on the parts of $C^{\pm}$,
$\Sigma_{0}^{+}$, and $\Sigma_{\infty}^{+}$ outside of $D_{\mathrm{i}s}$,
while $\mathrm{Re}(h(\eta,y))\leq-\delta<0$ holds on the part of
$\Sigma_{\infty}^{-}$ outside of $D_{\mathrm{i}s}$, we have
$\mathbf{M}^{(n)}_{-}(\eta,y)^{-1}\mathbf{M}^{(n)}_{+}(\eta,y)-\mathbb{I}$
decays exponentially to zero as $n\to\infty$ on these arcs, so since the outer
parametrix has unit determinant and is bounded on these contours, uniformly so
as $n\to\infty$, we have
$\mathbf{D}^{(n)}_{+}(\eta,y)=\mathbf{D}^{(n)}_{-}(\eta,y)(\mathbb{I}+\text{exponentially
small})$ as $n\to\infty$, where the exponentially small term is measured in
both $L^{\infty}$ and $L^{2}$. On $\partial D_{\mathrm{i}s}$ taken with
clockwise orientation, we have

(198)
$\mathbf{D}^{(n)}_{+}(\eta,y)=\mathbf{D}^{(n)}_{-}(\eta,y)\dot{\mathbf{M}}^{(n),\mathrm{in}}(\eta,y)\dot{\mathbf{M}}^{(n),\mathrm{out}}(\eta,y)^{-1},\quad\eta\in\partial
D_{\mathrm{i}s}$

because $\mathbf{M}^{(n)}(\eta,y)$ has no jump across this contour. Therefore
from (194) it follows that
$\mathbf{D}^{(n)}_{+}(\eta,y)=\mathbf{D}^{(n)}_{-}(\eta,y)(\mathbb{I}+\mathcal{O}(n^{-1}))$
holds uniformly on $\partial D_{\mathrm{i}s}$ as $n\to\infty$. Finally, we
note that both factors in the definition of $\mathbf{D}^{(n)}(\eta,y)$ tend to
the identity as $\eta\to\infty$, so $\mathbf{D}^{(n)}(\eta,y)$ does as well.

The matrix $\mathbf{D}^{(n)}(\eta,y)$ therefore satisfies the conditions of a
Riemann-Hilbert problem of small-norm type, and the key implication we need of
this fact is that $\mathbf{D}^{(n)}(\eta,y)$ is continuous at $\eta=0$ and
$\mathbf{D}^{(n)}(\eta,y)=\mathbb{I}+\mathcal{O}(n^{-1})$ holds uniformly in
the $\eta$-plane, and in particular in a neighborhood of $\eta=0$.

###  4.5. Asymptotic formula for $u_{n}(n^{3/2}y)$ with $y>0$ sufficiently
large

According to (154),

(199)