$\displaystyle\mathbf{\Psi}(\lambda,x)\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-\Theta_{0}\sigma_{3}/2}\mathrm{e}^{\mathrm{i}x\lambda^{-1}\sigma_{3}/2}=\mathbf{\Psi}_{0}^{0}(x)+\mathbf{\Psi}_{1}^{0}(x)\lambda+\mathcal{O}\big{(}\lambda^{2}\big{)},\qquad\lambda\to
0.$

Define the matrices

$\sigma_{+}=\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\qquad\text{and}\qquad\sigma_{-}=\begin{bmatrix}0&0\\\
0&1\end{bmatrix}.$

Following [5], assuming the $(1,1)$ entry of $\mathbf{\Psi}_{0}^{0}(x)$,
denoted ${\Psi}_{0,11}^{0}(x)$, is not identically zero, we consider the
Schlesinger transformation

$\hat{\mathbf{\Psi}}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\big{(}\sigma_{+}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{1/2}+\hat{\mathbf{S}}(x)\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-1/2}\big{)}\mathbf{\Psi}(\lambda,x),$

where

$\hat{\mathbf{S}}(x)\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\Psi_{0,21}^{0}(x)\Psi_{1,12}^{\infty}(x)/\Psi^{0}_{0,11}(x)&-\Psi_{1,12}^{\infty}(x)\vspace{1mm}\\\
-\Psi_{0,21}^{0}(x)/\Psi_{0,11}^{0}(x)&1\end{bmatrix}.$

Since
$\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\pm
1/2}$ has its branch cut along part of the curve $L^{(6)}$, we see that
$\hat{\mathbf{\Psi}}(\lambda,x)$ is analytic in ${\mathbb{C}}\setminus
L^{(6)}$ and, by direct calculation, has the jumps on $L^{(6)}$ summarized by
Figure 3, with the exception of the sign changes

$\mathbf{S}_{2}^{0}e_{0}^{-2\sigma_{3}}\mapsto-\mathbf{S}_{2}^{0}e_{0}^{-2\sigma_{3}}\qquad\text{and}\qquad\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}}\mapsto-\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}}.$

Furthermore, one can verify using the definition of
$\hat{\mathbf{\Psi}}(\lambda,x)$ that

$\hat{\mathbf{\Psi}}(\lambda,x)\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(\Theta_{\infty}-1)\sigma_{3}/2}\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}/2}=\mathbb{I}+\hat{\mathbf{\Psi}}^{\infty}_{1}(x)\lambda^{-1}+\mathcal{O}\big{(}\lambda^{-2}\big{)},\qquad\lambda\to\infty,$
(6.1)

where

$\hat{\mathbf{\Psi}}_{1}^{\infty}(x)\mathrel{\mathop{\mathchar
58\relax}}=\sigma_{+}\mathbf{\Psi}^{\infty}_{1}(x)\sigma_{+}+\sigma_{+}\mathbf{\Psi}_{2}^{\infty}(x)\sigma_{-}+\hat{\mathbf{S}}(x)\sigma_{+}+\hat{\mathbf{S}}(x)\mathbf{\Psi}_{1}^{\infty}(x)\sigma_{-}.$

Similarly, one can check that

$\hat{\mathbf{\Psi}}(\lambda,x)\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-(\Theta_{0}+1)\sigma_{3}/2}\mathrm{e}^{\mathrm{i}x\lambda^{-1}\sigma_{3}/2}=\hat{\mathbf{\Psi}}^{0}_{0}(x)+\hat{\mathbf{\Psi}}^{0}_{1}(x)\lambda+\mathcal{O}\big{(}\lambda^{2}\big{)},\qquad\lambda\to
0,$ (6.2)

where

$\hat{\mathbf{\Psi}}^{0}_{0}(x)=\hat{\mathbf{S}}(x)\mathbf{\Psi}^{0}_{0}(x)\sigma_{-}+\hat{\mathbf{S}}(x)\mathbf{\Psi}^{0}_{1}(x)\sigma_{+}+\sigma_{+}\mathbf{\Psi}^{0}_{0}(x)\sigma_{+}.$

The transformation $\mathbf{\Psi}\mapsto\hat{\mathbf{\Psi}}$ is invertible so
long as $\hat{\Psi}_{0,22}^{0}(x)$ does not identically vanish, and its
inverse is given by

$\mathbf{\Psi}(x,\lambda)\mapsto\check{\mathbf{\Psi}}(x,\lambda)\mathrel{\mathop{\mathchar
58\relax}}=\big{(}\sigma_{-}\lambda^{1/2}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}+\check{\mathbf{S}}(x)\lambda^{-1/2}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}\big{)}\
\mathbf{\Psi}(x,\lambda),$

where

$\check{\mathbf{S}}(x)\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}1&-\Psi_{0,12}^{0}(x)/\Psi_{0,22}^{0}(x)\vspace{1mm}\\\
-\Psi_{1,21}^{\infty}(x)&\Psi_{0,12}^{0}(x)\Psi_{1,21}^{\infty}(x)/\Psi_{0,22}^{0}(x)\end{bmatrix}.$

It follows that $\check{\mathbf{\Psi}}$ satisfies conditions similar to (6.1)
and (6.2) as $\lambda$ approaches $\infty,0$, respectively. That these
operations are inverses of one another is the content of [5, Lemma 1].

In this way, starting with $\mathbf{\Psi}$ and iterating the map
$\mathbf{\Psi\mapsto\hat{\mathbf{\Psi}}}$ (assuming $\Psi^{0}_{0,11}(x)$,
$\Psi^{0}_{0,22}(x)$ do not identically vanish after each step), we may define
the $n$th iterate of this Schlesinger transformation, which we denote
$\mathbf{\Psi}_{n}$. This matrix, if it exists, satisfies the following
Riemann–Hilbert problem.

######  Riemann–Hilbert Problem 6.1.

Fix generic monodromy parameters $(e_{1},e_{2})$, $n\in{\mathbb{Z}}$, and
$x>0$. We seek a $2\times 2$ matrix function
$\lambda\mapsto\mathbf{\Psi}(\lambda,x)$ satisfying:

  * •

Analyticity: $\mathbf{\Psi}_{n}(\lambda,x)$ is analytic in
${\mathbb{C}}\setminus L^{(6)}$, where
$L^{(6)}=\\{|\lambda|=1\\}\cup\mathrm{i}{\mathbb{R}}$ is the jump contour
shown in Figure 3.

  * •

Jump condition: $\mathbf{\Psi}_{n}(\lambda,x)$ has continuous boundary values
on $L^{(6)}\setminus\\{0\\}$ from each component of $\mathbb{C}\setminus
L^{(6)}$, which satisfy

$\mathbf{\Psi}_{n,+}(\lambda,x)=\mathbf{\Psi}_{n,-}(\lambda,x)\mathbf{J}_{\mathbf{\Psi}_{n}}(\lambda),$

where $\mathbf{J}_{\mathbf{\Psi}_{n}}(\lambda)$ is as shown in Figure 3 but
with the modification

$\mathbf{S}_{2}^{0}e_{0}^{-2\sigma_{3}}\mapsto(-1)^{n}\mathbf{S}_{2}^{0}e_{0}^{-2\sigma_{3}}\qquad\text{and}\qquad\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}}\mapsto(-1)^{n}\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}}.$

  * •

Normalization: $\mathbf{\Psi}_{n}(\lambda,x)$ satisfies the asymptotic
conditions

$\boldsymbol{\Psi}_{n}(\lambda,x)=\big{(}\mathbb{I}+\mathbf{\Xi}^{(6)}_{n}(x)+\mathcal{O}\big{(}\lambda^{-2}\big{)}\big{)}\mathrm{e}^{\mathrm{i}x\lambda\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(n-\Theta_{\infty})\sigma_{3}/2}\qquad\text{as}\quad\lambda\to\infty,$
(6.3)

and

$\boldsymbol{\Psi}_{n}(\lambda,x)=\big{(}\mathbf{\Delta}^{(6)}_{n}(x)+\mathcal{O}(\lambda)\big{)}\mathrm{e}^{-\mathrm{i}x\lambda^{-1}\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(\Theta_{0}+n)\sigma_{3}/2}\qquad\text{as}\quad\lambda\to
0,$ (6.4)

where $\mathbf{\Delta}^{(6)}_{n}(x)$ is a matrix determined from
$\mathbf{\Psi}_{n}(\lambda,x)$ having unit determinant.

That $\mathbf{\Psi}_{n}$ solves the above Riemann–Hilbert problem implies the
existence of the limit

$\mathbf{\Xi}^{(6)}_{n}(x)\mathrel{\mathop{\mathchar
58\relax}}=\lim_{\lambda\to\infty}\lambda\big{[}\mathbf{\Psi}_{n}(\lambda,x)\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\Theta_{\infty}\sigma_{3}/2}-\mathbb{I}\big{]}.$
(6.5)

It follows that the function

$u_{n}(x)=\dfrac{-\mathrm{i}\Xi^{(6)}_{n,12}(x)}{\Delta^{(6)}_{n,11}(x)\Delta^{(6)}_{n,12}(x)}$
(6.6)

satisfies PIII($D_{6}$) in the form

$u_{n}^{\prime\prime}=\dfrac{(u_{n}^{\prime})^{2}}{u_{n}}-\dfrac{u_{n}^{\prime}}{x}+\dfrac{4(n+\Theta_{0})u_{n}^{2}}{x}+\dfrac{4(1+n-\Theta_{\infty})}{x}+4u_{n}^{3}-\frac{4}{u_{n}}.$

It was shown in [5, Lemma 2] that if for some $n\in\mathbb{Z}$ the inverse
monodromy problem is solvable for a given $x\in D$, where $D$ is a domain in
${\mathbb{C}}\setminus\\{0\\}$, then $\mathbf{\Psi}_{n}$ satisfies the Lax
pair

$\displaystyle\mathinner{\dfrac{\partial{}\mathbf{\Psi}_{n}}{\partial{\lambda}}}(\lambda,x)$
$\displaystyle=\Bigg{(}\frac{\mathrm{i}x}{2}\sigma_{3}+\frac{1}{2\lambda}\begin{bmatrix}n-\Theta_{\infty}&2y\\\
2v&\Theta_{\infty}-n\end{bmatrix}+\frac{1}{2\lambda^{2}}\begin{bmatrix}\mathrm{i}x-2\mathrm{i}st&2\mathrm{i}s\\\
2\mathrm{i}t(x-st)&2\mathrm{i}st-\mathrm{i}x\end{bmatrix}\Bigg{)}\mathbf{\Psi}_{n}(\lambda,x),$
$\displaystyle\mathinner{\dfrac{\partial{}\mathbf{\Psi}_{n}}{\partial{x}}}(\lambda,x)$
$\displaystyle=\Bigg{(}\dfrac{\mathrm{i}\lambda}{2}\sigma_{3}+\dfrac{1}{x}\begin{bmatrix}0&y\\\
v&0\end{bmatrix}-\dfrac{1}{2\lambda
x}\begin{bmatrix}\mathrm{i}x-2\mathrm{i}st&2\mathrm{i}s\\\
2\mathrm{i}t(x-st)&2\mathrm{i}st-\mathrm{i}x\end{bmatrix}\Bigg{)}\mathbf{\Psi}_{n}(\lambda,x),$

where potentials $s$, $t$, $u$, $v$, $y$ all depend on $x$ and $n$.
Furthermore, in this domain, the functions $\Psi_{0,11}^{0}(x)$,
$\Psi_{0,22}^{0}(x)$ extracted from $\mathbf{\Psi}_{n}(\lambda,x)$ are not
identically zero.888Lemma 2 in [5] was stated for parameters corresponding to
rational solutions of Painlevé-III, but the proof is almost exactly the same
in this case.

One can check that if a solution to Riemann–Hilbert Problem 6.1 exists, it
must be unique, and we attempt to identify this solution as a solution of
Riemann–Hilbert Problem 4.1 with possibly different monodromy data. The
diagonal elements of $\mathbf{S}_{2}^{0}e_{0}^{-2\sigma_{3}}$,
$\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}}$ alternate signs which
implies the change

$e_{0}^{2}\mapsto(-1)^{n}e_{0}^{2},\qquad
e_{\infty}^{2}\mapsto(-1)^{n}e_{\infty}^{2}.$

Furthermore, in view of (4.30), we can write

$(-1)^{n}s_{2}^{\infty}e_{\infty}^{2}=(-1)^{n}\big{(}1-e_{1}^{2}e_{\infty}^{2}\big{)}e_{\infty}^{2}=\big{(}1-(-1)^{n}e_{1}^{2}(-1)^{n}e_{\infty}^{2}\big{)}(-1)^{n}e_{\infty}^{2},$

and

$(-1)^{n}s_{2}^{0}e_{0}^{-2}=(-1)^{n}\big{(}e_{1}^{2}e_{0}^{2}-1\big{)}e_{0}^{-2}=\big{(}(-1)^{n}e_{1}^{2}(-1)^{n}e_{0}^{2}-1\big{)}(-1)^{n}e_{0}^{-2}.$

Combining the above with the fact that $\mathbf{C}_{0\infty}^{\pm}$ remain
invariant under the iterated Schlesinger transformations implies the change in
monodromy data

$\displaystyle e_{1}^{2}\mapsto(-1)^{n}e_{1}^{2}\qquad\text{and}\qquad
e_{2}\mapsto e_{2}.$ (6.7)

Since $e_{1}$, $e_{2}$ are assumed to be nonvanishing, we may write them in
the form (1.16) for some ${\mu,\eta\in{\mathbb{C}}}$ with
$-1<\operatorname{Re}(\mu),\operatorname{Re}(\eta)\leq 1$. Moreover, since the
transformations $e_{1}\mapsto-e_{1}$, ${e_{2}\mapsto-e_{2}}$ preserve the
monodromy data, we can assume
$-\frac{1}{2}<\operatorname{Re}(\eta)\leq\frac{1}{2}$ and
$-\frac{1}{2}<\operatorname{Re}(\mu)\leq\frac{1}{2}$. Equation (6.7) implies
in turn that $\eta$ does not depend on $n\in{\mathbb{Z}}$, while $\mu$ is
replaced with

$\mu\mapsto\mu_{n}\mathrel{\mathop{\mathchar 58\relax}}=\begin{cases}\mu,&n\in
2{\mathbb{Z}},\\\ \mu-\frac{1}{2},&n+1\in 2{\mathbb{Z}},\end{cases}$ (6.8)

This proves Proposition 1.6. We end this section with two important remarks.

######  Remark 6.2.

It was noted in the introduction that one could restrict
$0<\operatorname{Re}(\mu_{n})\leq 1/2$, in which case, the above iterations
interchange the roles of $e_{1}^{2}$, $e_{1}^{-2}$ and we have to perform the
transformation $\mu\to-\mu$, which corresponds to the replacements

$\displaystyle\mathbf{E}^{\infty}\to\left(\sqrt{\frac{e_{\infty}^{2}-e_{1}^{2}}{e_{1}^{2}\big{(}e_{1}^{2}e_{\infty}^{2}-1\big{)}}}\right)^{\sigma_{3}}\mathbf{E}^{\infty}\left(\sqrt{\frac{e_{\infty}^{2}-e_{1}^{2}}{e_{1}^{2}\big{(}e_{1}^{2}e_{\infty}^{2}-1\big{)}}}\right)^{-\sigma_{3}}\sigma_{1}\left(\frac{e_{1}^{2}e_{\infty}^{2}\big{(}1-e_{1}^{2}e_{\infty}^{2}\big{)}}{\big{(}e_{1}^{4}-1\big{)}}\right)^{\sigma_{3}},$
$\displaystyle\mathbf{E}^{0}\to\left(\sqrt{\frac{e_{0}^{2}-e_{1}^{2}}{e_{1}^{2}\big{(}e_{1}^{2}e_{0}^{2}-1\big{)}}}\right)^{\sigma_{3}}\mathbf{E}^{0}\left(\sqrt{\frac{e_{0}^{2}-e_{1}^{2}}{e_{1}^{2}\big{(}e_{1}^{2}e_{0}^{2}-1\big{)}}}\right)^{-\sigma_{3}}\sigma_{1}\left(\frac{e_{1}^{2}\big{(}e_{0}^{2}e_{1}^{2}-1\big{)}}{e_{0}^{2}\big{(}e_{1}^{4}-1\big{)}}\right)^{\sigma_{3}}.$

This gauge transformation then allows us to identify the monodromy parameter
pairs

$\displaystyle(e_{1},e_{2})\sim\left(\frac{1}{e_{1}},\frac{1}{e_{2}e_{0}^{2}e_{\infty}^{2}}\sqrt{\frac{\big{(}e_{1}^{2}-e_{0}^{2}\big{)}\big{(}1-e_{0}^{2}e_{1}^{2}\big{)}}{\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}\big{(}1-e_{1}^{2}e_{\infty}^{2}\big{)}}}\right).$
(6.9)

Therefore, we alternatively can write the monodromy data for the Schlesinger
transformation as

$\displaystyle\mu_{n}=\begin{cases}\mu,&n\in 2{\mathbb{Z}},\\\
\frac{1}{2}-\mu,&n+1\in 2{\mathbb{Z}},\end{cases}$ $\displaystyle
e_{2,n}=\begin{cases}e_{2},&n\in 2{\mathbb{Z}},\\\
\dfrac{1}{e_{2}e_{0}^{2}e_{\infty}^{2}}\sqrt{\dfrac{\big{(}e_{1}^{2}-e_{0}^{2}\big{)}\big{(}1-e_{0}^{2}e_{1}^{2}\big{)}}{\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}\big{(}1-e_{1}^{2}e_{\infty}^{2}\big{)}}},&n+1\in
2{\mathbb{Z}}.\end{cases}$

Furthermore, one can check that $(x_{1},x_{2},x_{3})$ in (1.17)–(1.19) remain
invariant under the map described in (6.9), whereas
$(y_{1},y_{2},y_{3})\mapsto(\pm y_{1},\pm y_{2},y_{3})$ where the sign depends
on the choice of the square root in (6.9) and (7.52) below. In both cases, the
corresponding solution of (1.3) remains invariant, see Remark 5.2.

######  Remark 6.3.

Moving forward, we will slightly abuse notation by suppressing the
$n$-dependence in the parameters

$e_{\infty}=\mathrm{e}^{\pi\mathrm{i}(\Theta_{\infty}-n)/2},\qquad
e_{0}=\mathrm{e}^{\pi\mathrm{i}(\Theta_{0}+n)/2},\qquad
e_{1}=\mathrm{e}^{\pi\mathrm{i}\mu_{n}},\qquad
e_{2}=\mathrm{e}^{\pi\mathrm{i}\eta}.$ (6.10)

##  7 Asymptotics for large $\boldsymbol{n}$ and small $\boldsymbol{x}$
and proof of Theorem 1.4

Let $(e_{1},e_{2})$ be generic monodromy parameters, see Definition 1.3. At
this point, we can see more clearly the meaning of the genericity conditions
formulated there:

  1. (i)

$e_{1}^{4}\neq 1$; this is to guarantee diagonalizability in (4.27),

  2. (ii)

$e_{1}e_{2}\neq 0$; this is to guarantee the unit-determinant condition in
(4.27) and (4.28),

  3. (iii)

$e_{1}^{2}\neq e_{\infty}^{\pm 2}$ and $e_{1}^{2}\neq e_{0}^{\pm 2}$; this, in
particular, implies that the Stokes multipliers (4.30) are nonvanishing.

###  7.1 Opening the lenses

First, we define a new unknown matrix by
$\mathbf{\Phi}_{n}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\boldsymbol{\Psi}_{n}(\lambda,x)\mathbf{L}$ where $\mathbf{L}$ is
the piece-wise constant matrix shown in the left-hand panel of Figure 6. It
follows from (4.28), (4.29) that the resulting jump conditions satisfied by
$\mathbf{\Phi}_{n}(\lambda,x)$ are as shown in the right-hand panel of Figure
6.

Figure 6: Left panel: the definition of the matrix $\mathbf{L}$; the circles
are centered at the origin and have radii $\frac{1}{2}$, $1$, and $2$. Right
panel: the jump contour $\Gamma$ and jump conditions for
$\mathbf{\Phi}_{n}(\lambda,x)$.

######  Remark 7.1.

In the general case $|{\operatorname{Arg}}(x)|<\pi$, the lenses shown in
Figure 6 must be rotated in the manner shown in the left panel of Figure 7.
The resulting jumps follow from the identities (4.33)–(4.34) and are shown in
the right panel of Figure 7.

Figure 7: Analogue of Figure 6 when $\operatorname{Arg}(x)\neq 0$. The thick
line represents the branch cut for the argument chosen as in Remark 4.3.

Figure 8: The contour $\Gamma^{(\infty)}$ includes the circle $|\lambda|=2$.

###  7.2 Parametrix for $\boldsymbol{{\Phi}_{n}(\lambda,x)}$ near
$\boldsymbol{\lambda=\infty}$

By definition, the parametrix
$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ satisfies the following
Riemann–Hilbert problem.

######  Riemann–Hilbert Problem 7.2.

Fix generic monodromy parameters $(e_{1},e_{2})$ determining the Stokes and
connection matrices, $n\in\mathbb{Z}$, and $x>0$. We seek a $2\times 2$ matrix
function $\lambda\mapsto\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
satisfying:

  * •

Analyticity: $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ is analytic
in ${\mathbb{C}}\setminus\Gamma^{(\infty)}$, where

$\Gamma^{(\infty)}=\\{|\lambda|=2\\}\cup(\mathrm{i}{\mathbb{R}}\cap\\{|{\operatorname{Im}}\,\lambda-1|>1\\})$

is the jump contour shown in Figure 8.

  * •

Jump condition: $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ has
continuous boundary values on $\Gamma^{(\infty)}\setminus\\{0\\}$ from each
component of $\mathbb{C}\setminus\Gamma^{(\infty)}$, which satisfy

$\breve{\boldsymbol{\Phi}}_{n,+}^{(\infty)}(\lambda,x)=\breve{\boldsymbol{\Phi}}_{n,-}^{(\infty)}(\lambda,x)\mathbf{J}_{\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}(\lambda),$

where $\mathbf{J}_{\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}(\lambda)$ is as
shown in Figure 8 and where the $+$ $($resp., $-)$ subscript denotes a
boundary value taken from the left $($resp., right$)$ of an arc of
$\Gamma^{(\infty)}$.

  * •

Normalization: $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ satisfies
the asymptotic conditions

$\displaystyle\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=\bigg{(}\mathbb{I}+\dfrac{\mathbf{A}_{n}(x)}{\lambda}+\mathcal{O}\left(\dfrac{1}{\lambda^{2}}\right)\bigg{)}\mathrm{e}^{\mathrm{i}x\lambda\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(n-\Theta_{\infty})\sigma_{3}/2}\qquad\text{as}\quad\lambda\to\infty,$
(7.1)
$\displaystyle\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=(\mathbf{B}_{n}(x)+\mathcal{O}(\lambda))\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\mu_{n}\sigma_{3}}\qquad\text{as}\quad\lambda\to
0,$ (7.2)

where $\mathbf{A}_{n}(x)$ has zero trace and $\mathbf{B}_{n}(x)$ has unit
determinant.

It is easy to see that $\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
necessarily has unit determinant. Furthermore, note that the jump matrix being
$e_{1}^{-2\sigma_{3}}$ across the arc terminating at the origin implies
$e_{1}=\mathrm{e}^{\pi\mathrm{i}\mu_{n}}$, which is consistent with (6.10).

####  7.2.1 Dependence on $\boldsymbol{\lambda}$

It follows from assuming differentiability of the asymptotics in (7.1)–(7.2)
that

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}(\lambda,x)^{-1}$
$\displaystyle\qquad=\left(\mathbb{I}+\dfrac{\mathbf{A}_{n}(x)}{\lambda}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\right)\left(\dfrac{\mathrm{i}x}{2}+\dfrac{n-\Theta_{\infty}}{2\lambda}\right)\sigma_{3}\left(\mathbb{I}+\dfrac{\mathbf{A}_{n}(x)}{\lambda}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\right)^{-1}+\mathcal{O}\big{(}\lambda^{-2}\big{)}$
$\displaystyle\qquad=\dfrac{\mathrm{i}x}{2}\sigma_{3}+\left(\dfrac{\mathrm{i}x}{2}[\mathbf{A}_{n}(x),\sigma_{3}]+\dfrac{n-\Theta_{\infty}}{2}\sigma_{3}\right)\dfrac{1}{\lambda}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\qquad\text{as}\quad\lambda\to\infty,$
(7.3)

and

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}(\lambda,x)^{-1}$
$\displaystyle=\left(\mathbf{B}_{n}(x)+\mathcal{O}(\lambda)\right)\left(\dfrac{\mu_{n}}{\lambda}\sigma_{3}\right)\left(\mathbf{B}_{n}(x)+\mathcal{O}(\lambda)\right)^{-1}+\mathcal{O}(1)$
$\displaystyle=\dfrac{\mu_{n}}{\lambda}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}+\mathcal{O}(1)\qquad\text{as}\quad\lambda\to
0.$ (7.4)

Since the quantity on the left-hand side of (7.3) and (7.4) is otherwise an
analytic function of $\lambda$, it follows from Liouville’s Theorem that

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}=\dfrac{\mathrm{i}x}{2}\sigma_{3}+\dfrac{\mu_{n}}{\lambda}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}\implies$
$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)=\left(\dfrac{\mathrm{i}x}{2}\sigma_{3}+\dfrac{\mu_{n}}{\lambda}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}\right)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x).$
(7.5)

Noting that
$\operatorname{Tr}\big{(}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}\big{)}=0$
and $\det\big{(}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}\big{)}=-1$,
we may write

$\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}=\begin{bmatrix}a_{n}(x)&b_{n}(x)\\\
c_{n}(x)&-a_{n}(x)\end{bmatrix}\qquad\text{subject to}\quad
a_{n}(x)^{2}+b_{n}(x)c_{n}(x)=1$ (7.6)

and use this form in (7.5) to write a coupled scalar system of differential
equations satisfied by the elements $\phi_{1}(\lambda,x)$ and
$\phi_{2}(\lambda,x)$ of the first and second rows, respectively, of any
column of $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$:

$\displaystyle\frac{\partial\phi_{1}}{\partial\lambda}(\lambda,x)$
$\displaystyle=\left(\dfrac{\mathrm{i}x}{2}+\dfrac{\mu_{n}a_{n}(x)}{\lambda}\right)\phi_{1}(\lambda,x)+\dfrac{\mu_{n}b_{n}(x)}{\lambda}\phi_{2}(\lambda,x),$
(7.7) $\displaystyle\frac{\partial\phi_{2}}{\partial\lambda}(\lambda,x)$
$\displaystyle=\dfrac{\mu_{n}c_{n}(x)}{\lambda}\phi_{1}(\lambda,x)-\left(\dfrac{\mathrm{i}x}{2}+\dfrac{\mu_{n}a_{n}(x)}{\lambda}\right)\phi_{2}(\lambda,x).$
(7.8)

Before beginning to solve this system, observe that equating the coefficients
of $\lambda^{-1}$ in (7.3) and (7.4) yields the identity

$\mu_{n}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}=\dfrac{\mathrm{i}x}{2}[\mathbf{A}_{n}(x),\sigma_{3}]+\dfrac{n-\Theta_{\infty}}{2}\sigma_{3}.$
(7.9)

Since $[\mathbf{A}_{n}(x),\sigma_{3}]$ is off-diagonal, we arrive at

$\mu_{n}a_{n}(x)=\frac{n-\Theta_{\infty}}{2}.$ (7.10)

Since $\mu_{n}$ and $n$ are constants, this equation implies that $a_{n}(x)$
is independent of $x$, so we will simply write $a_{n}$ going forward. Now,
solving for $\phi_{1}(\lambda,x)$ in (7.8) and eliminating it from (7.7)
yields (assuming $c_{n}(x)\neq 0$ and using $b_{n}(x)c_{n}(x)=1-a_{n}^{2}$)

$\displaystyle\lambda\frac{\partial^{2}\phi_{2}}{\partial\lambda^{2}}(\lambda,x)+\frac{\partial\phi_{2}}{\partial\lambda}(\lambda,x)+\left[\dfrac{\mathrm{i}x}{2}+\dfrac{x^{2}}{4}\lambda-\mathrm{i}x\mu_{n}a_{n}-\dfrac{\mu_{n}^{2}}{\lambda}\right]\phi_{2}(\lambda,x)=0.$

It is easy to see that the first-order derivative term is removed by the
substitution $\phi_{2}(\lambda,x)=\lambda^{-1/2}w(\lambda,x)$. Indeed,
$w(\lambda,x)$ satisfies

$\displaystyle\frac{\partial^{2}w}{\partial\lambda^{2}}(\lambda,x)+\bigg{[}\dfrac{x^{2}}{4}+\mathrm{i}x\left(\dfrac{1}{2}-\mu_{n}a_{n}\right)\dfrac{1}{\lambda}+\left(\dfrac{1}{4}-\mu_{n}^{2}\right)\dfrac{1}{\lambda^{2}}\bigg{]}w(\lambda,x)$
$\displaystyle=0.$

Finally, the explicit $x$-dependence in the coefficients can be removed by
setting $Z\mathrel{\mathop{\mathchar 58\relax}}=\mathrm{i}x\lambda$ and
writing $w(\lambda,x)=W(Z)$. Note that the notation $W(Z)$ here is not related
to $W(z)$ appearing in Section 5. In this case, $W(Z)$ satisfies the ordinary
differential equation

$W^{\prime\prime}(Z)+\bigg{[}-\dfrac{1}{4}+\left(\dfrac{1}{2}-\mu_{n}a_{n}\right)\dfrac{1}{Z}+\left(\dfrac{1}{4}-\mu_{n}^{2}\right)\dfrac{1}{Z^{2}}\bigg{]}W(Z)=0,$
(7.11)

which is Whittaker’s equation (see [38, Chapter 13]) with parameter

$\kappa=\kappa_{n}\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{2}-\mu_{n}a_{n}=\dfrac{1+\Theta_{\infty}-n}{2}.$ (7.12)

Given $\phi_{2}(\lambda,x)=\lambda^{-1/2}W(Z)$ for $Z=\mathrm{i}x\lambda$, and
a solution $W(Z)$ of (7.11), it follows from (7.8) that the corresponding
first-row entry is

$\phi_{1}(\lambda,x)=\frac{\mathrm{i}x\lambda^{1/2}}{\mu_{n}c_{n}(x)}\bigg{(}W^{\prime}(Z)+\left(\frac{1}{2}-\frac{\kappa_{n}}{Z}\right)W(Z)\bigg{)}.$
(7.13)

A fundamental pair of solutions of (7.11) is given by
$W(Z)=W_{\pm\kappa_{n},\mu_{n}}(\pm Z)$, $\mathrm{arg}(\pm Z)\in(-\pi,\pi)$.
If we take the particular solution
$\phi_{2}(\lambda,x)=\lambda^{-1/2}W_{\kappa_{n},\mu_{n}}(Z)$, then using the
identity

$W^{\prime}_{\kappa,\mu_{n}}(Z)=\left(\dfrac{\kappa}{Z}-\dfrac{1}{2}\right)W_{\kappa,\mu_{n}}(Z)+\bigg{(}\left(\dfrac{1}{2}-\kappa\right)^{2}-\mu_{n}^{2}\bigg{)}\frac{1}{Z}W_{\kappa-1,\mu_{n}}(Z)$

(see [38, equation (13.15.23)]) in (7.13) gives

$\displaystyle\phi_{2}(\lambda,x)=\lambda^{-1/2}W_{\kappa_{n},\mu_{n}}(Z)$
$\displaystyle\implies\phi_{1}(\lambda,x)=\frac{\mathrm{i}x\lambda^{1/2}}{\mu_{n}c_{n}(x)}\bigg{(}\left(\frac{1}{2}-\kappa_{n}\right)^{2}-\mu_{n}^{2}\bigg{)}Z^{-1}W_{\kappa_{n}-1,\mu_{n}}(Z).$

Likewise, if we take the particular solution
$\phi_{2}(\lambda,x)=\lambda^{-1/2}W_{-\kappa_{n},\mu_{n}}(-Z)$, then using
the identity

$W^{\prime}_{\kappa,\mu_{n}}(Z)=\left(\dfrac{1}{2}-\dfrac{\kappa}{Z}\right)W_{\kappa,\mu_{n}}(Z)-\dfrac{1}{Z}W_{\kappa+1,\mu_{n}}(Z)$

(see [38, equation (13.15.26)]) in (7.13) yields

$\phi_{2}(\lambda,x)=\lambda^{-1/2}W_{-\kappa_{n},\mu_{n}}(-Z)\implies\phi_{1}(\lambda,x)=-\frac{\mathrm{i}x\lambda^{1/2}}{\mu_{n}c_{n}(x)}Z^{-1}W_{1-\kappa_{n},\mu_{n}}(-Z).$

Taking linear combinations with coefficients depending generally on the
parameter $x$, the general solution matrix for the system (7.5) can be written
in the form

$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=\widetilde{\boldsymbol{\Phi}}^{(\infty)}_{n}(\lambda,x)\mathbf{K}(x),\qquad\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{H}(\lambda,x)\mathbf{W}(\mathrm{i}\lambda
x;\kappa_{n},\mu_{n}),$ (7.14)

where $\widetilde{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ is a specific
fundamental solution matrix of (7.5) constructed from

$\mathbf{H}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\lambda^{\sigma_{3}/2}\begin{bmatrix}\displaystyle\frac{\mathrm{i}x}{\mu_{n}c_{n}(x)}&0\\\
0&1\end{bmatrix}$ (7.15)

and

$\displaystyle\mathbf{W}(Z;\kappa,\mu_{n})\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z)&-Z^{-1}W_{1-\kappa,\mu_{n}}(-Z)\\\
W_{\kappa,\mu_{n}}(Z)&W_{-\kappa,\mu_{n}}(-Z)\end{bmatrix},$
$\displaystyle\alpha_{\kappa,\mu_{n}}\mathrel{\mathop{\mathchar
58\relax}}=\left(\frac{1}{2}-\kappa\right)^{2}-\mu_{n}^{2},$ (7.16)

in which $\kappa=\kappa_{n}$ and $\mu_{n}$ are given by (7.12) and
$\mathbf{K}(x)$ is a matrix of free coefficients.

####  7.2.2 Dependence on $\boldsymbol{x}$

Going back to (7.1)–(7.2) and now assuming that the asymptotics are
differentiable with respect to $x$,

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}}{\partial
x}(\lambda,x)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}$
$\displaystyle\qquad=\left(\mathbb{I}+\dfrac{\mathbf{A}_{n}(x)}{\lambda}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\right)\left(\dfrac{\mathrm{i}\lambda}{2}\sigma_{3}\right)\left(\mathbb{I}+\dfrac{\mathbf{A}_{n}(x)}{\lambda}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\right)^{-1}+\mathcal{O}\big{(}\lambda^{-1}\big{)}$
$\displaystyle\qquad=\dfrac{\mathrm{i}\lambda}{2}\sigma_{3}+\dfrac{\mathrm{i}}{2}[\mathbf{A}_{n}(x),\sigma_{3}]+\mathcal{O}\big{(}\lambda^{-1}\big{)}\qquad\text{as}\quad\lambda\to\infty,$

and

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}}{\partial
x}(\lambda,x)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}$
$\displaystyle=\mathbf{B}_{n}^{\prime}(x)\mathbf{B}_{n}(x)^{-1}+\mathcal{O}(\lambda)\qquad\text{as}\quad\lambda\to
0.$

So, applying Liouville’s theorem yields

$\frac{\partial\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}}{\partial
x}(\lambda,x)=\left(\dfrac{\mathrm{i}\lambda}{2}\sigma_{3}+\dfrac{\mathrm{i}}{2}[\mathbf{A}_{n}(x),\sigma_{3}]\right)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x),$
(7.17)

and it follows from (7.9) that

$\displaystyle\dfrac{\mathrm{i}}{2}[\mathbf{A}_{n}(x),\sigma_{3}]$
$\displaystyle=\dfrac{1}{x}\left(\mu_{n}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}+\left(\kappa_{n}-\frac{1}{2}\right)\sigma_{3}\right)$
$\displaystyle=\frac{1}{x}\begin{bmatrix}0&\dfrac{\mu_{n}\big{(}1-a_{n}^{2}\big{)}}{c_{n}(x)}\\\
\mu_{n}c_{n}(x)&0\end{bmatrix},$ (7.18)

where $\mu_{n}$, $a_{n}$ are independent of $\lambda$, $x$. To determine the
$x$-dependence of $c_{n}(x)$, we use (7.18) to assemble (7.5) (using also
(7.9) and (7.12)) and (7.17) to give the Lax system

$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)=\breve{\boldsymbol{\Lambda}}(\lambda,x)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x),$
$\displaystyle\breve{\mathbf{\Lambda}}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\dfrac{\mathrm{i}x}{2}\sigma_{3}+\dfrac{1}{\lambda}\begin{bmatrix}-\frac{1}{2}(\Theta_{\infty}-n)&\displaystyle\frac{\mu_{n}\big{(}1-a_{n}^{2}\big{)}}{c_{n}(x)}\\\
\mu_{n}c_{n}(x)&\frac{1}{2}(\Theta_{\infty}-n)\end{bmatrix},$ (7.19)
$\displaystyle\frac{\partial\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial
x}(\lambda,x)=\breve{\mathbf{X}}(\lambda,x)\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x),$
$\displaystyle\breve{\mathbf{X}}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\dfrac{\mathrm{i}\lambda}{2}\sigma_{3}+\dfrac{1}{x}\begin{bmatrix}0&\displaystyle\frac{\mu_{n}\big{(}1-a_{n}^{2}\big{)}}{c_{n}(x)}\\\
\mu_{n}c_{n}(x)&0\end{bmatrix}.$ (7.20)

Since $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ is a simultaneous
fundamental solution matrix for these equations, the Lax system is compatible.
The compatibility condition reads

$\breve{\boldsymbol{\Lambda}}_{x}(\lambda,x)-\breve{\mathbf{X}}_{\lambda}(\lambda,x)+\big{[}\breve{\boldsymbol{\Lambda}}(\lambda,x),\breve{\mathbf{X}}(\lambda,x)\big{]}=\mathbf{0},$

which is equivalent to

$xc_{n}^{\prime}(x)=(1-2\kappa_{n})c_{n}(x)\implies
c_{n}(x)=\gamma_{n}x^{1-2\kappa_{n}},$ (7.21)

for some constant $\gamma_{n}\neq 0$. Thus, the coefficient $c_{n}(x)$ is
determined up to the choice of the constant $\gamma_{n}$. Note also that the
coefficient matrices $\breve{\boldsymbol{\Lambda}}(\lambda,x)$ and
$\breve{\mathbf{X}}(\lambda,x)$ are obviously related by the simple identity

$\breve{\mathbf{X}}(\lambda,x)-\frac{\lambda}{x}\breve{\boldsymbol{\Lambda}}(\lambda,x)=\frac{1}{x}\left(\kappa_{n}-\frac{1}{2}\right)\sigma_{3}.$
(7.22)

Since the fundamental matrix
$\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ defined by (7.14)
satisfies (7.19), then so does

$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)\mathbf{K}(x),$

and $\mathbf{K}(x)$ must now be chosen so that (7.20) is satisfied.
Substituting into (7.20), we obtain an ordinary differential equation on
$\mathbf{K}(x)$:

$\mathbf{K}^{\prime}(x)=\bigg{(}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\mathbf{X}(\lambda,x)\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)-\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial
x}(\lambda,x)\bigg{)}\mathbf{K}(x).$ (7.23)

Now, from the form of
$\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ written in (7.14),
we have both

$\displaystyle\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial
x}(\lambda,x)=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\mathbf{H}}{\partial
x}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\phantom{\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial
x}(\lambda,x)=}{}+\mathrm{i}\lambda\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\mathbf{H}(\lambda,x)\mathbf{W}^{\prime}(Z;\kappa_{n},\mu_{n}),$
$\displaystyle\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\mathbf{H}}{\partial\lambda}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\phantom{\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)=}{}+\mathrm{i}x\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\mathbf{H}(\lambda,x)\mathbf{W}^{\prime}(Z;\kappa_{n},\mu_{n}),$

so it follows that

$\displaystyle\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial
x}(\lambda,x)$
$\displaystyle\quad=\frac{\lambda}{x}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\frac{\partial\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}}{\partial\lambda}(\lambda,x)$
$\displaystyle\quad\phantom{=}{}+\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\left[\frac{\partial\mathbf{H}}{\partial
x}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}-\frac{\lambda}{x}\frac{\partial\mathbf{H}}{\partial\lambda}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}\right]\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\quad=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\left[\frac{\lambda}{x}\breve{\boldsymbol{\Lambda}}(\lambda,x)+\frac{\partial\mathbf{H}}{\partial
x}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}-\frac{\lambda}{x}\frac{\partial\mathbf{H}}{\partial\lambda}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}\right]\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x),$

where we also used (7.19). Using this in (7.23) along with the explicit
definition (7.15) of $\mathbf{H}(\lambda,x)$ and the identities (7.21) and
(7.22) gives

$\displaystyle\mathbf{K}^{\prime}(x)\mathbf{K}(x)^{-1}$
$\displaystyle\quad=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\left[\breve{\mathbf{X}}(\lambda,x)-\frac{\lambda}{x}\breve{\boldsymbol{\Lambda}}(\lambda,x)\right.$
$\displaystyle\left.\hphantom{\quad=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}}{}+\frac{\lambda}{x}\frac{\partial\mathbf{H}}{\partial\lambda}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}-\frac{\partial\mathbf{H}}{\partial
x}(\lambda,x)\mathbf{H}(\lambda,x)^{-1}\right]\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\quad=\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\Bigg{[}\frac{1}{x}\left(\kappa_{n}-\frac{1}{2}\right)\sigma_{3}+\frac{1}{2x}\sigma_{3}+\frac{\mathrm{d}}{\mathrm{d}x}\log\left(\frac{c_{n}(x)}{x}\right)\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\Bigg{]}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\quad=\frac{\kappa_{n}}{x}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)^{-1}\Bigg{[}\sigma_{3}-2\begin{bmatrix}1&0\\\
0&0\end{bmatrix}\Bigg{]}\widetilde{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=-\frac{\kappa_{n}}{x}\mathbb{I}.$

Therefore, the $x$-dependence of the matrix $\mathbf{K}(x)$ is explicitly
given by

$\mathbf{K}(x)=x^{-\kappa_{n}}\mathbf{K},$

where $\mathbf{K}$ is now independent of both $\lambda$ and $x$. However, as
the domain of analyticity of $\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
in the $\lambda$-plane consists of three disjoint regions, we expect to have
to specify a different matrix $\mathbf{K}$ for each. Note also that the
constant $\gamma_{n}$ remains to be determined.

####  7.2.3 The parametrix
$\boldsymbol{\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)}$ on the two
regions with $\boldsymbol{|\lambda|>2}$

To fully specify the parametrix
$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ for $|\lambda|>2$, we
concretely take the jump contours for $|\lambda|>2$ to lie along the real axis
in the $Z$-plane, corresponding to ${\mathbb{R}}_{+}$ and ${\mathbb{R}}_{-}$,
respectively. Thus, the part of the domain of analyticity of
$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ with $|\lambda|>2$ has
two components, corresponding to the upper and lower half $Z$-planes. To
properly define $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ in these
two exterior domains, we firstly take the matrix factor
$\mathbf{H}(\lambda,x)$ defined in (7.15) in the precise form

$\mathbf{H}(\lambda,x)=\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}\begin{bmatrix}\displaystyle\frac{\mathrm{i}x^{2\kappa_{n}}}{\mu_{n}\gamma_{n}}&0\\\
0&1\end{bmatrix}=x^{\kappa_{n}}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n},\qquad\mathbf{D}_{n}\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\displaystyle\frac{\mathrm{i}}{\mu_{n}\gamma_{n}}&0\\\
0&1\end{bmatrix}.$ (7.24)

Then, we assume different constant matrices $\mathbf{K}=\mathbf{K}_{n}^{\pm}$
in the two domains by writing the parametrix for $|\lambda|>2$ and
$\pm\operatorname{Im}(Z)>0$ as

$\displaystyle\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)=\breve{\boldsymbol{\Phi}}_{n}^{(\infty)\pm}(\lambda,x)=x^{-\kappa_{n}}\mathbf{H}(\lambda,x)\mathbf{W}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{\pm}$
$\displaystyle\phantom{\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)}{}=\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{W}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{\pm}.$
(7.25)

We now express the matrices $\mathbf{K}_{n}^{\pm}$ in terms of the remaining
constants $\mu_{n}$ and $\gamma_{n}$ by enforcing the asymptotic condition
(7.1) in each of the two sectors with $|\lambda|>2$. According to [38,
equation (13.19.3)],

$W_{\kappa,\mu_{n}}(Z)=\mathrm{e}^{-Z/2}Z^{\kappa}\big{(}1+\mathcal{O}\big{(}Z^{-1}\big{)}\big{)},\qquad
Z\to\infty,\qquad|\mathrm{arg}(Z)|\leq\frac{3\pi}{2}-\delta$

holds for each $\delta>0$. Hence also

$\displaystyle\begin{split}&\mathbf{W}(Z;\kappa,\mu_{n})=\begin{bmatrix}\alpha_{\kappa,\mu_{n}}Z^{\kappa-2}\big{(}1+\mathcal{O}\big{(}Z^{-1}\big{)}\big{)}&(-Z)^{-\kappa}\big{(}1+\mathcal{O}\big{(}Z^{-1}\big{)}\big{)}\vspace{1mm}\\\
Z^{\kappa}\big{(}1+\mathcal{O}\big{(}Z^{-1}\big{)}\big{)}&(-Z)^{-\kappa}\big{(}1+\mathcal{O}\big{(}Z^{-1}\big{)}\big{)}\end{bmatrix}\mathrm{e}^{-Z\sigma_{3}/2},\\\
&Z\to\infty,\qquad|\mathrm{arg}(Z)|\leq\frac{3\pi}{2}-\delta.\end{split}$

Under the condition given on $\mathrm{arg}(Z)$, we have

$(-Z)^{-\kappa}=Z^{-\kappa}\begin{cases}\mathrm{e}^{\mathrm{i}\pi\kappa},&\text{$\operatorname{Im}(Z)>0$
\ (i.e., $0<\operatorname{Arg}(Z)<\pi$),}\\\
\mathrm{e}^{-\mathrm{i}\pi\kappa},&\text{$\operatorname{Im}(Z)<0$ \ (i.e.,
$-\pi<\operatorname{Arg}(Z)<0$).}\end{cases}$

To calculate $Z^{\pm\kappa}$, we recall $Z=\mathrm{i}x\lambda$ and use [5,
equation (49)]:

$-\dfrac{\pi}{2}-\operatorname{Arg}(x)<\mathrm{arg}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}(\lambda)<\dfrac{3\pi}{2}-\operatorname{Arg}(x),\qquad|\lambda|\to\infty.$

Next,

$\displaystyle\operatorname{Im}(Z)>0$ $\displaystyle\implies
0<\operatorname{Arg}(Z)<\pi\\!\\!\\!\\!$
$\displaystyle\implies\\!-\frac{\pi}{2}-\operatorname{Arg}(x)<\mathrm{arg}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}(\lambda)<\frac{\pi}{2}-\operatorname{Arg}(x),\\!$
$\displaystyle\operatorname{Im}(Z)<0$
$\displaystyle\implies-\pi<\operatorname{Arg}(Z)<0\\!\\!\\!\\!\\!$
$\displaystyle\implies\\!-\frac{3\pi}{2}-\operatorname{Arg}(x)<\mathrm{arg}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}(\lambda)-2\pi<-\dfrac{\pi}{2}-\operatorname{Arg}(x)\\!$

and hence, for any $x\in{\mathbb{C}}\setminus\\{0\\}$ such that
$|{\operatorname{Arg}}(x)|<\pi$,

$Z^{\pm\kappa}=x^{\pm\kappa}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\pm\kappa}\begin{cases}\mathrm{e}^{\pm\mathrm{i}\pi\kappa/2},&\operatorname{Im}(Z)>0,\\\
\mathrm{e}^{\mp 3\mathrm{i}\pi\kappa/2},&\operatorname{Im}(Z)<0.\end{cases}$

Therefore, for $\lambda$ large such that $\operatorname{Im}(Z)>0$,

$\displaystyle\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{W}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})$
$\displaystyle\qquad=\mathbf{D}_{n}\begin{bmatrix}\mathcal{O}(\lambda^{-1})&\mathrm{e}^{\mathrm{i}\pi\kappa_{n}/2}\big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\big{)}\\\
\mathrm{e}^{\mathrm{i}\pi\kappa_{n}/2}\big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\big{)}&\mathcal{O}\big{(}\lambda^{-1}\big{)}\end{bmatrix}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(\kappa_{n}-\frac{1}{2})\sigma_{3}}\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}/2},$

so choosing $\mathbf{K}_{n}^{+}$ so that (7.25) is consistent with (7.1) in
the sector $\operatorname{Im}(Z)>0$ requires that $\mathbf{K}_{n}^{+}$ is an
off-diagonal matrix, namely,

$\mathbf{K}_{n}^{+}\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}0&\mathrm{e}^{-\mathrm{i}\pi\kappa_{n}/2}\\\
-\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa_{n}/2}\mu_{n}\gamma_{n}&0\end{bmatrix}.$

Similarly, for $\lambda$ large such that $\operatorname{Im}(Z)<0$,

$\displaystyle\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{W}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})$
$\displaystyle\qquad=\mathbf{D}_{n}\begin{bmatrix}\mathcal{O}(\lambda^{-1})&\mathrm{e}^{\mathrm{i}\pi\kappa_{n}/2}\big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\big{)}\\\
\mathrm{e}^{-3\mathrm{i}\pi\kappa_{n}/2}\big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\big{)}&\mathcal{O}\big{(}\lambda^{-1}\big{)}\end{bmatrix}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(\kappa_{n}-\frac{1}{2})\sigma_{3}}\mathrm{e}^{-\mathrm{i}x\lambda\sigma_{3}/2},$

so consistency of (7.25) with (7.1) in the sector $\operatorname{Im}(Z)<0$
requires

$\mathbf{K}_{n}^{-}\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}0&\mathrm{e}^{3\mathrm{i}\pi\kappa_{n}/2}\\\
-\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa_{n}/2}\mu_{n}\gamma_{n}&0\end{bmatrix}.$

Some additional useful information can be gleaned by enforcing on
$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ the jump conditions for
$|\lambda|>2$. The jump rays are illustrated in the $Z$-plane with their
orientations in Figure 9.

•$\infty$$\operatorname{Im}(Z)<0$$\operatorname{Im}(Z)>0$$<$$>$$Z\in{\mathbb{R}}_{+}$$Z\in{\mathbb{R}}_{-}$
Figure 9: Jump contour for
$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$ near $\lambda=\infty$.

The Whittaker function $W_{\kappa,\mu_{n}}(Z)$ can be viewed as an analytic
function on the cut plane $|{\operatorname{Arg}}(Z)|<\pi$, and it follows from
the connection formula [38, equation (13.14.13)] that the boundary values on
the negative real axis are related by

$\displaystyle
W_{\kappa,\mu_{n}}(-Z+\mathrm{i}0)=\mathrm{e}^{2\pi\mathrm{i}\kappa}W_{\kappa,\mu_{n}}(-Z-\mathrm{i}0)$
$\displaystyle\hphantom{W_{\kappa,\mu_{n}}(-Z+\mathrm{i}0)=}{}+\frac{2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\pi\kappa}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa\big{)}}W_{-\kappa,\mu_{n}}(Z),\qquad
Z>0.$ (7.26)

Note that the denominators in the second term on the right-hand side of (7.26)
are finite due to condition (iii) in the definition of generic data; see the
beginning of Section 7. Indeed, it follows from (6.10) and (7.12) that

$e_{1}^{\pm 2}=e_{\infty}^{2}\ \Leftrightarrow\
\frac{1}{2}-\kappa_{n}\pm\mu_{n}\in{\mathbb{Z}}.$

On $Z\in{\mathbb{R}}_{-}$ the left ($+$) and right ($-$) boundary values
correspond to limits from ${\operatorname{Im}(Z)<0}$ and
$\operatorname{Im}(Z)>0$, respectively. Therefore, the second column of
$\mathbf{W}(Z;\kappa,\mu_{n})$ is continuous across $\mathbb{R}_{-}$, and from
(7.26) (replacing $Z$ with $-Z$),

$\displaystyle\mathbf{W}_{-}(Z;\kappa,\mu_{n})=\begin{bmatrix}\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z+\mathrm{i}0)&-Z^{-1}W_{1-\kappa,\mu_{n}}(-Z)\vspace{1mm}\\\
W_{\kappa,\mu_{n}}(Z+\mathrm{i}0)&W_{-\kappa,\mu_{n}}(-Z)\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\displaystyle\mathrm{e}^{2\pi\mathrm{i}\kappa}\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z-\mathrm{i}0)-\frac{2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\pi\kappa}Z^{-1}W_{1-\kappa,\mu_{n}}(-Z)}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa\big{)}}&-Z^{-1}W_{1-\kappa,\mu_{n}}(-Z)\vspace{1mm}\\\
\displaystyle\mathrm{e}^{2\pi\mathrm{i}\kappa}W_{\kappa,\mu_{n}}(Z-\mathrm{i}0)+\frac{2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\pi\kappa}W_{-\kappa,\mu_{n}}(-Z)}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa\big{)}}&W_{-\kappa,\mu_{n}}(-Z)\end{bmatrix}$
$\displaystyle=\mathbf{W}_{+}(Z;\kappa,\mu_{n})\begin{bmatrix}\mathrm{e}^{2\pi\mathrm{i}\kappa}&0\vspace{1mm}\\\
\displaystyle\frac{2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\pi\kappa}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa\big{)}}&1\end{bmatrix},\qquad
Z<0.$

Here, on the third line we used the definition (7.16) of
$\alpha_{\kappa,\mu_{n}}$ and the factorial identity
$\Gamma(\diamond+1)=\diamond\Gamma(\diamond)$. Since $\mathbf{H}(\lambda,x)$
is analytic across $\mathrm{i}\mathbb{R}_{+}$, it follows that
$\breve{\mathbf{\Phi}}^{(\infty)}(\lambda)=\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
satisfies the jump condition

$\displaystyle\breve{\boldsymbol{\Phi}}_{+}^{(\infty)}(\lambda)$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\big{[}\mathbf{W}_{-}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{+}\big{]}^{-1}\mathbf{W}_{+}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{-}$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\big{(}\mathbf{K}^{+}_{n}\big{)}^{-1}\begin{bmatrix}\mathrm{e}^{2\pi\mathrm{i}\kappa_{n}}&0\vspace{1mm}\\\
\displaystyle\frac{2\pi\mathrm{i}\mathrm{e}^{\mathrm{i}\pi\kappa_{n}}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa_{n}\big{)}}&1\end{bmatrix}^{-1}\mathbf{K}^{-}_{n}$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\begin{bmatrix}1&\displaystyle\frac{2\pi\mathrm{e}^{\mathrm{i}\pi\kappa_{n}}}{\mu_{n}\gamma_{n}\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa_{n}\big{)}}\\\
0&1\end{bmatrix},\qquad\lambda\in\mathrm{i}\mathbb{R}_{+}.$

Requiring that this matches with the corresponding jump condition in Figure 8
gives the condition

$\frac{2\pi\mathrm{e}^{\mathrm{i}\pi\kappa_{n}}}{\mu_{n}\gamma_{n}\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa_{n}\big{)}}=\frac{e_{\infty}^{2}-e_{1}^{2}}{e_{1}^{2}e_{\infty}^{4}}.$

For $Z\in{\mathbb{R}}_{+}$ the $\pm$ boundary values correspond to the limit
from $\operatorname{Im}(Z)\gtrless 0$. Therefore, now the first column of
$\mathbf{W}(Z;\kappa,\mu_{n})$ is continuous across $\mathbb{R}_{+}$, and from
(7.26),

$\displaystyle\mathbf{W}_{-}(Z;\kappa,\mu_{n})=\begin{bmatrix}\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z)&-Z^{-1}W_{1-\kappa,\mu_{n}}(-Z+\mathrm{i}0)\vspace{1mm}\\\
W_{\kappa,\mu_{n}}(Z)&W_{-\kappa,\mu_{n}}(-Z+\mathrm{i}0)\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\\!\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z)\\!&\displaystyle\\!\\!-\mathrm{e}^{-2\pi\mathrm{i}\kappa}Z^{-1}W_{1-\kappa,\mu_{n}}(-Z\\!-\mathrm{i}0)+\frac{2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa}\alpha_{\kappa,\mu_{n}}Z^{-1}W_{\kappa-1,\mu_{n}}(Z)}{\Gamma\big{(}\frac{1}{2}+\mu_{n}+\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}+\kappa\big{)}}\\!\vspace{1mm}\\\
W_{\kappa,\mu_{n}}(Z)\\!&\\!\displaystyle\mathrm{e}^{-2\pi\mathrm{i}\kappa}W_{-\kappa,\mu_{n}}(-Z\\!-\mathrm{i}0)+\frac{2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa}W_{\kappa,\mu_{n}}(Z)}{\Gamma\big{(}\frac{1}{2}+\mu_{n}+\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}+\kappa\big{)}}\end{bmatrix}$
$\displaystyle=\mathbf{W}_{+}(Z;\kappa,\mu_{n})\begin{bmatrix}1&\displaystyle\frac{2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}+\kappa\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}+\kappa\big{)}}\vspace{1mm}\\\
0&\mathrm{e}^{-2\pi\mathrm{i}\kappa}\end{bmatrix},\qquad Z>0.$

Again here, the finiteness of the denominators is guaranteed by condition
(iii) at the beginning of Section 7. Since $\mathbf{H}(\lambda,x)$ changes
sign across $\mathrm{i}\mathbb{R}_{-}$, we get that
$\breve{\mathbf{\Phi}}^{(\infty)}(\lambda)=\breve{\mathbf{\Phi}}^{(\infty)}_{n}(\lambda,x)$
satisfies the jump condition

$\displaystyle\breve{\boldsymbol{\Phi}}_{+}^{(\infty)}(\lambda)$
$\displaystyle=-\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)[\mathbf{W}_{-}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{-}]^{-1}\mathbf{W}_{+}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{K}_{n}^{+}$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)(\mathbf{K}^{-}_{n})^{-1}\begin{bmatrix}-1&\displaystyle-\frac{2\pi\mathrm{i}\mathrm{e}^{-\mathrm{i}\pi\kappa_{n}}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}+\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}+\kappa_{n}\big{)}}\vspace{1mm}\\\
0&-\mathrm{e}^{-2\pi\mathrm{i}\kappa_{n}}\end{bmatrix}^{-1}\mathbf{K}_{n}^{+}$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\begin{bmatrix}-\mathrm{e}^{2\pi\mathrm{i}\kappa_{n}}&0\vspace{1mm}\\\
\displaystyle\frac{2\pi\mathrm{e}^{-\mathrm{i}\pi\kappa_{n}}\mu_{n}\gamma_{n}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}+\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}+\kappa_{n}\big{)}}&-\mathrm{e}^{-2\pi\mathrm{i}\kappa_{n}}\end{bmatrix}$
$\displaystyle=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\begin{bmatrix}e_{\infty}^{2}&0\vspace{1mm}\\\
e_{\infty}^{2}s_{2}^{\infty}&e_{\infty}^{-2}\end{bmatrix}=\breve{\boldsymbol{\Phi}}_{-}^{(\infty)}(\lambda)\mathbf{S}_{2}^{\infty}e_{\infty}^{2\sigma_{3}},\qquad\lambda\in\mathrm{i}\mathbb{R}_{-}.$

The last two equalities follow by a direct calculation using the definitions
of $s_{2}^{\infty}$, $\kappa_{n}$, $e_{1}$, and $e_{\infty}$ in (4.30),
(7.12), and (6.10), respectively, along with the expression

$\mu_{n}\gamma_{n}=\dfrac{2\pi\mathrm{e}^{\pi\mathrm{i}\kappa_{n}}}{\Gamma\big{(}\frac{1}{2}+\mu_{n}-\kappa_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\mu_{n}-\kappa_{n}\big{)}}\cdot\frac{e_{1}^{2}e_{\infty}^{4}}{e_{\infty}^{2}-e_{1}^{2}},$
(7.27)

and the classical identity

$\Gamma\left(\frac{1}{2}-z\right)\Gamma\left(\frac{1}{2}+z\right)=\dfrac{\pi}{\cos(\pi
z)}.$ (7.28)

####  7.2.4 The parametrix
$\boldsymbol{\breve{\boldsymbol{\Phi}}^{(\infty)}_{n}(\lambda,x)}$ in the
region $\boldsymbol{|\lambda|<2}$

We use the identity [38, equation (13.14.33)] to express the elements of
$\mathbf{W}(Z;\kappa,\mu)$ in terms of the alternative basis of solutions
$M_{-\kappa,\pm\mu}(-Z)$ of Whittaker’s equation with parameters
$(\kappa,\mu)$ that form a numerically satisfactory pair in a neighborhood of
the origin and that are analytic for $\operatorname{Arg}(-Z)\in(-\pi,\pi)$.
Moreover, these functions are the Maclaurin series associated with the regular
singular point at $Z=0$, so they have the property that

$M_{\kappa,\mu}(-Z)(-Z)^{-\frac{1}{2}-\mu}=1+\mathcal{O}(Z)\qquad\text{as}\quad
Z\to 0,$ (7.29)

where the power function denotes the principal branch and where the error term
represents an analytic function of $Z$ vanishing at the origin. To deal with
the first column of $\mathbf{W}(Z;\kappa,\mu)$ we also use the corresponding
identity
$M_{\kappa,\mu}(Z)=\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}+\mu)}M_{-\kappa,\mu}(-Z)$
which holds for $\pm\operatorname{Im}(Z)>0$ (see also [38, equation
(13.14.10)]). Using the above identities, and under the condition
$2\mu\not\in{\mathbb{Z}}$ (which follows from the condition (i) at the
beginning of Section 7 in our case), we can write the elements of
$\mathbf{W}(Z;\kappa,\mu)$ in the form

$\displaystyle
W_{11}(Z;\kappa,\mu)=-Z^{-1}\frac{\Gamma(-2\mu)\Gamma\big{(}\frac{1}{2}-\mu+\kappa\big{)}}{\Gamma\big{(}\frac{1}{2}-\mu-\kappa\big{)}\Gamma\bigl{(}-\frac{1}{2}-\mu+\kappa\bigr{)}}\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}+\mu)}M_{1-\kappa,\mu}(-Z)$
$\displaystyle\hphantom{W_{11}(Z;\kappa,\mu)=}{}-Z^{-1}\frac{\Gamma(2\mu)\Gamma\big{(}\frac{1}{2}+\mu+\kappa\big{)}}{\Gamma\big{(}\frac{1}{2}+\mu-\kappa\big{)}\Gamma\bigl{(}-\frac{1}{2}+\mu+\kappa\bigr{)}}\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}-\mu)}M_{1-\kappa,-\mu}(-Z),\quad\\!\pm\operatorname{Im}(Z)>0,$
$\displaystyle
W_{12}(Z;\kappa,\mu)=-Z^{-1}\frac{\Gamma(-2\mu)}{\Gamma\bigl{(}-\frac{1}{2}-\mu+\kappa\bigr{)}}M_{1-\kappa,\mu}(-Z)-Z^{-1}\frac{\Gamma(2\mu)}{\Gamma\bigl{(}{-}\frac{1}{2}+\mu+\kappa\bigr{)}}M_{1-\kappa,-\mu}(-Z),$
$\displaystyle
W_{21}(Z;\kappa,\mu)=\frac{\Gamma(-2\mu)}{\Gamma\big{(}\frac{1}{2}-\mu-\kappa\big{)}}\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}+\mu)}M_{-\kappa,\mu}(-Z)$
$\displaystyle\hphantom{W_{21}(Z;\kappa,\mu)=}{}+\frac{\Gamma(2\mu)}{\Gamma\big{(}\frac{1}{2}+\mu-\kappa\big{)}}\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}-\mu)}M_{-\kappa,-\mu}(-Z),\qquad\pm\operatorname{Im}(Z)>0,$

and

$W_{22}(Z;\kappa,\mu)=\frac{\Gamma(-2\mu)}{\Gamma\big{(}\frac{1}{2}-\mu+\kappa\big{)}}M_{-\kappa,\mu}(-Z)+\frac{\Gamma(2\mu)}{\Gamma\big{(}\frac{1}{2}+\mu+\kappa\big{)}}M_{-\kappa,-\mu}(-Z).$

These expressions can be usefully combined into a matrix identity:

$\mathbf{W}(Z;\kappa,\mu)=\mathbf{M}(Z;\kappa,\mu)\mathbf{G}^{\pm}_{\kappa,\mu},\qquad\pm\operatorname{Im}(Z)>0,$
(7.30)

where

$\displaystyle\mathbf{M}(Z;\kappa,\mu)\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\big{(}\frac{1}{2}-\kappa+\mu\big{)}Z^{-1}M_{1-\kappa,\mu}(-Z)&\big{(}\frac{1}{2}-\kappa-\mu\big{)}Z^{-1}M_{1-\kappa,-\mu}(-Z)\vspace{1mm}\\\
M_{-\kappa,\mu}(-Z)&M_{-\kappa,-\mu}(-Z)\end{bmatrix},$
$\displaystyle\operatorname{Arg}(-Z)\in(-\pi,\pi),$ (7.31)

and

$\mathbf{G}^{\pm}_{\kappa,\mu}\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\displaystyle\frac{\Gamma(-2\mu)\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}+\mu)}}{\Gamma\big{(}\frac{1}{2}-\mu-\kappa\big{)}}&\displaystyle\frac{\Gamma(-2\mu)}{\Gamma\big{(}\frac{1}{2}-\mu+\kappa\big{)}}\vspace{1mm}\\\
\displaystyle\frac{\Gamma(2\mu)\mathrm{e}^{\pm\mathrm{i}\pi(\frac{1}{2}-\mu)}}{\Gamma\big{(}\frac{1}{2}+\mu-\kappa\big{)}}&\displaystyle\frac{\Gamma(2\mu)}{\Gamma\big{(}\frac{1}{2}+\mu+\kappa\big{)}}\end{bmatrix}.$

To define the parametrix $\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)$
for $|\lambda|<2$, we first introduce a constant matrix by

$\mathbf{J}_{n}\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{G}_{\kappa_{n},\mu_{n}}^{+}\mathbf{K}_{n}^{+}\mathbf{S}_{1}^{\infty}\mathbf{E}^{\infty}=\mathbf{G}_{\kappa_{n},\mu_{n}}^{-}\mathbf{K}_{n}^{-}\mathbf{E}^{\infty}.$
(7.32)

The equality of these two expressions can be seen as follows. First, combining
(7.25) and (7.30), and using the fact that the matrix
$\mathbf{M}(\mathrm{i}x\lambda;\kappa,\mu_{n})$ is analytic in a neighborhood
of $\lambda=2\mathrm{i}$, the jump condition for
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ across the positive
imaginary axis for $|\lambda|>2$ shown in Figure 8 implies the identity
$\mathbf{G}_{\kappa_{n},\mu_{n}}^{+}\mathbf{K}_{n}^{+}\mathbf{S}_{1}^{\infty}=\mathbf{G}_{\kappa_{n},\mu_{n}}^{-}\mathbf{K}_{n}^{-}$,
which yields the desired equality.

Then, we set

$\breve{\boldsymbol{\Phi}}_{n}^{(\infty)}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{J}_{n},\qquad\text{for}\quad|\lambda|<2.$
(7.33)

It is straightforward to then check that, regardless of the choice of
$\mu_{n}$, the matrix $\mathbf{J}_{n}$ defined by (7.32) is diagonal.
Comparing (7.25) and (7.33) shows that the jump conditions for
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ across the arcs of the
circle $|\lambda|=2$ shown in Figure 8 are satisfied. Using (7.29) then proves
that $\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ satisfies the simple
jump condition across the negative imaginary axis with ${|\lambda|<2}$ shown
in Figure 8 and that an expansion of the form shown in (7.2) holds. To check
that the matrix $\mathbf{J}_{n}$ is diagonal and arrive at its final form
below, we use the identity (7.28) to get

$\displaystyle\Gamma\left(\dfrac{1}{2}+\kappa_{n}-\mu_{n}\right)\Gamma\left(\dfrac{1}{2}-\kappa_{n}+\mu_{n}\right)=\dfrac{\pi}{\cos\left(\pi(\kappa_{n}-\mu_{n})\right)}=\dfrac{2\pi\mathrm{i}e_{1}e_{\infty}}{e_{1}^{2}-e_{\infty}^{2}},$
$\displaystyle\Gamma\left(\dfrac{1}{2}+\kappa_{n}+\mu_{n}\right)\Gamma\left(\dfrac{1}{2}-\kappa_{n}-\mu_{n}\right)=\dfrac{\pi}{\cos\left(\pi(\kappa_{n}+\mu_{n})\right)}=\dfrac{2\pi\mathrm{i}e_{1}e_{\infty}}{1-e_{1}^{2}e_{\infty}^{2}}.$
(7.34)

The result is that the diagonal matrix $\mathbf{J}_{n}$ from (7.32) is given
by

$\mathbf{J}_{n}=\begin{bmatrix}\dfrac{\mathrm{e}^{{\mathrm{i}\pi/4}}e_{1}e_{\infty}^{7/2}\Gamma(-2\mu_{n})}{\Gamma\big{(}\frac{1}{2}-\kappa_{n}-\mu_{n}\big{)}}&0\\\
0&\dfrac{\mathrm{e}^{{\mathrm{i}\pi/4}}\big{(}e_{1}^{4}-1\big{)}e_{\infty}^{3/2}\Gamma(2\mu_{n})}{e_{1}\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}\Gamma\big{(}\frac{1}{2}-\kappa_{n}+\mu_{n}\big{)}}\end{bmatrix}.$
(7.35)

###  7.3 Parametrix for $\boldsymbol{{\Phi}_{n}(\lambda,x)}$ near
$\boldsymbol{\lambda=0}$

By definition, the parametrix $\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)$
satisfies the following Riemann–Hilbert problem.

######  Riemann–Hilbert Problem 7.3.

Fix generic monodromy parameters $(e_{1},e_{2})$ determining the Stokes and
connection matrices, $n\in\mathbb{Z}$, and $x>0$. We seek a $2\times 2$ matrix
function $\lambda\mapsto\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)$
satisfying:

  * •

Analyticity: $\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)$ is analytic in
${\mathbb{C}}\setminus\Gamma^{(0)}$, where
$\Gamma^{(0)}=\big{\\{}|\lambda|=\frac{1}{2}\big{\\}}\cup\big{(}\mathrm{i}{\mathbb{R}}\cap\big{\\{}\operatorname{Im}\lambda<\frac{1}{2}\big{\\}}\big{)}$
is the jump contour shown in Figure 10.

  * •

Jump condition: $\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)$ has
continuous boundary values on $\Gamma^{(0)}\setminus\\{0\\}$ from each
component of $\mathbb{C}\setminus\Gamma^{(0)}$, which satisfy

$\breve{\boldsymbol{\Phi}}_{n,+}^{(0)}(\lambda,x)=\breve{\boldsymbol{\Phi}}_{n,-}^{(0)}(\lambda,x)\mathbf{J}_{\breve{\boldsymbol{\Phi}}_{n}^{(0)}}(\lambda),$

where $\mathbf{J}_{\breve{\boldsymbol{\Phi}}_{n}^{(0)}}(\lambda)$ is as shown
in Figure 10 and where the $+$ $($resp., $-)$ subscript denotes a boundary
value taken from the left $($resp., right$)$ of an arc of $\Gamma^{(0)}$.

  * •

Normalization: $\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)$ satisfies the
asymptotic conditions

$\displaystyle\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)=\mathcal{O}(1)\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{{\mu_{n}\sigma_{3}}{}}\qquad\text{as}\quad\lambda\to\infty,$
(7.36)

where $\mathcal{O}(1)$ refers to a function analytic and bounded in a
neighborhood of $\lambda=\infty$ and

$\breve{\boldsymbol{\Phi}}_{n}^{(0)}(\lambda,x)=(\mathbb{I}+\mathcal{O}(\lambda))\mathrm{e}^{-\mathrm{i}x\lambda^{-1}\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{(n+\Theta_{0})\sigma_{3}/2}\qquad\text{as}\quad\lambda\to
0.$ (7.37)

Figure 10: The contour $\Gamma^{(0)}$, which includes the circle
$|\lambda|=\frac{1}{2}$.

We can write down the unique solution
$\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x)$ explicitly in terms of the
parametrix $\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ obtained in
Section 7.2, but taken with the index $1-n$ instead of $n$ and
$\Theta_{\infty}$ replaced by $\Theta_{0}$. If we indicate the dependence of
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ and
$\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x)$ on $\Theta_{\infty}$ and
$\Theta_{0}$ respectively with the notation
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)=\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x,\Theta_{\infty})$
and
$\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x)=\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x,\Theta_{0})$,
then we have the following.

######  Proposition 7.4.

Fix $\Theta_{\infty},\Theta_{0}\in{\mathbb{C}}$ and generic monodromy
parameters $(e_{1},e_{2})$. Then

$\breve{\mathbf{\Phi}}^{(0)}_{n}(\lambda,x,\Theta_{0})=\begin{cases}e_{0}^{3\sigma_{3}}\breve{\mathbf{\Phi}}^{(\infty)}_{-n}\bigl{(}-\lambda^{-1},x,\Theta_{0}\bigr{)}e_{0}^{-2\sigma_{3}}\quad,&|\lambda|<\frac{1}{2},\vspace{1mm}\\\
e_{0}^{3\sigma_{3}}\breve{\mathbf{\Phi}}^{(\infty)}_{-n}\bigl{(}-\lambda^{-1},x,\Theta_{0}\bigr{)}\begin{bmatrix}0&\beta_{n}\\\
-\beta_{n}^{-1}&0\end{bmatrix},&|\lambda|>\frac{1}{2},\end{cases}$ (7.38)

where

$\beta_{n}\mathrel{\mathop{\mathchar
58\relax}}=\dfrac{1-e_{1}^{4}}{e_{1}^{2}\big{(}1-e_{0}^{2}e_{1}^{2}\big{)}}.$
(7.39)

###### Proof.

The mapping $\lambda\mapsto-\lambda^{-1}$ takes the contour $\Gamma^{(0)}$
onto the contour $\Gamma^{(\infty)}$ up to the reversal of orientation of
certain arcs, and swaps the circles centered at the origin of radii
$\frac{1}{2}$ and $2$. Therefore, the domain of analyticity of
$\breve{\mathbf{\Phi}}^{(0)}_{n}(\lambda,x,\Theta_{0})$ is as desired.

Under the map $n\mapsto-n$, $\Theta_{\infty}\mapsto\Theta_{0}$, the
exponentials defined in (6.10) satisfy $e_{0}^{2}\mapsto e_{\infty}^{2}$,
whereas $\mu_{n}=\mu_{-n}$ since this quantity depends only on the parity of
$n$. This implies that the Stokes matrices defined in (4.11)–(5.15) satisfy
the corresponding identities

$\mathbf{S}^{\infty}_{1}\mapsto
e_{0}^{-2\sigma_{3}}\big{(}\mathbf{S}^{0}_{1}\big{)}^{-1}e_{0}^{2\sigma_{3}}\qquad\text{and}\qquad\mathbf{S}^{\infty}_{{2}}e_{\infty}^{2\sigma_{3}}\mapsto
e_{0}^{-2\sigma_{3}}\big{(}\mathbf{S}^{0}_{2}e_{0}^{-2\sigma_{3}}\big{)}^{-1}e_{0}^{2\sigma_{3}}.$

Comparing Figures 8 and 10 then shows that the function defined by (7.38)
satisfied the required jump conditions across the imaginary axis for
$|\lambda|<\frac{1}{2}$. Likewise, the jump condition on the negative
imaginary axis for $|\lambda|>\frac{1}{2}$ is easily verified due to the
identity valid for any $a\neq 0$:

$e_{1}^{-2\sigma_{3}}\begin{bmatrix}0&a\\\
-a^{-1}&0\end{bmatrix}+\begin{bmatrix}0&a\\\
-a^{-1}&0\end{bmatrix}e_{1}^{2\sigma_{3}}=\mathbf{0}.$

Finally, the fact that $\breve{\mathbf{\Phi}}^{(0)}_{n}(\lambda,x,\Theta_{0})$
defined by (7.38) satisfies the required jump conditions across the circle
$|\lambda|=\frac{1}{2}$ follows from the corresponding jump conditions for
$\breve{\mathbf{\Phi}}^{(\infty)}_{1-n}(\lambda,x,\Theta_{0})$ for
$|\lambda|=2$ and the identities

$(\mathbf{E}^{\infty})^{-1}\mapsto\begin{bmatrix}0&\beta_{n}\\\
-\beta_{n}^{-1}&0\end{bmatrix}\big{(}\mathbf{S}_{1}^{0}\mathbf{E}^{0}\big{)}^{-1}e_{0}^{2\sigma_{3}},\qquad(\mathbf{S}_{1}^{\infty}\mathbf{E}^{\infty})^{-1}\mapsto\begin{bmatrix}0&\beta_{n}\\\
-\beta_{n}^{-1}&0\end{bmatrix}\big{(}\mathbf{E}^{0}\big{)}^{-1}e_{0}^{2\sigma_{3}}$

among the matrices $\mathbf{E}^{\infty}$, $\mathbf{E}^{0}$ defined in
(4.31)–(4.32), which hold for the value of $\beta_{n}$ indicated in (7.39).

It only remains to verify the asymptotics in (7.36)–(7.37). However, these
follow from the corresponding formulæ in (7.1)–(7.2) with the help of the
identity

$\lambda^{p}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}=\mathrm{e}^{\mathrm{i}\pi
p}\zeta^{-p}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}},\qquad\zeta\mathrel{\mathop{\mathchar
58\relax}}=-\lambda^{-1},$

which holds for all $\lambda$ not on the negative imaginary axis.

Since the matrix function
$\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x,\Theta_{0})$ defined by
(7.38)–(7.39) satisfies all the required Riemann–Hilbert conditions, and there
is at most one solution of those conditions, as is easily confirmed by a
Liouville argument, the proof is finished. ∎

###  7.4 An equivalent Riemann–Hilbert problem on the unit circle

The parametrix for $\mathbf{\Phi}_{n}(\lambda,x)$ is by definition the
following matrix function:

$\breve{\mathbf{\Phi}}_{n}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\begin{cases}\breve{\mathbf{\Phi}}^{(\infty)}_{n}(\lambda,x,\Theta_{\infty}),&|\lambda|>1,\\\
\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x,\Theta_{0}),&|\lambda|<1.\end{cases}$

This matrix function satisfies exactly the same jump conditions in the domains
$|\lambda|>1$ and ${|\lambda|<1}$ as does $\mathbf{\Phi}_{n}(\lambda,x)$
itself, and it is also consistent with the asymptotics given in (6.3)–(6.4)
(note that $\mathbf{\Psi}_{n}(\lambda,x)=\mathbf{\Phi}_{n}(\lambda,x)$ for
$|\lambda|$ sufficiently large or small). The parametrix has unit determinant,
so the matrix quotient

$\mathbf{Q}_{n}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{\Phi}_{n}(\lambda,x)\breve{\mathbf{\Phi}}_{n}(\lambda,x)^{-1}$

is an analytic function of $\lambda$ except possibly on the jump contour
$\Gamma$ shown in Figure 6 and on the unit circle, where there is a
discontinuity in the definition of $\breve{\mathbf{\Phi}}_{n}(\lambda,x)$.
However, since the jumps of $\breve{\mathbf{\Phi}}_{n}(\lambda,x)$ and
$\mathbf{\Phi}_{n}(\lambda,x)$ agree on $\Gamma$, a Morera argument shows that
$\mathbf{Q}_{n}(\lambda,x)$ is actually analytic both for $|\lambda|>1$ and
for $0<|\lambda|<1$. The asymptotic behavior of the factors in
$\mathbf{Q}_{n}(\lambda,x)$ as $\lambda\to 0$ then shows that any singularity
of $\mathbf{Q}_{n}(\lambda,x)$ at the origin $\lambda=0$ is removable, and the
asymptotic behavior of the same factors in the limit $\lambda\to\infty$ shows
that $\mathbf{Q}_{n}(\lambda,x)\to\mathbb{I}$ as $\lambda\to\infty$.

$\mathbf{Q}_{n}(\lambda,x)$ is therefore characterized by its jump condition
across the unit circle $|\lambda|=1$. Taking counterclockwise orientation for
the circle, the jump condition for $\mathbf{Q}_{n}(\lambda,x)$ reads

$\mathbf{Q}_{n,+}(\lambda,x)=\mathbf{Q}_{n,-}(\lambda,x)\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x,\Theta_{\infty})e_{2}^{\sigma_{3}}\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x,\Theta_{0})^{-1},\qquad|\lambda|=1.$

Using Proposition 7.4, the jump matrix can be written as

$\displaystyle\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x,\Theta_{\infty})e_{2}^{\sigma_{3}}\breve{\mathbf{\Phi}}_{n}^{(0)}(\lambda,x,\Theta_{0})^{-1}$
$\displaystyle\qquad=\breve{\mathbf{\Phi}}^{(\infty)}_{n}(\lambda,x,\Theta_{\infty})e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}\breve{\mathbf{\Phi}}^{(\infty)}_{-n}(-\lambda^{-1},x,\Theta_{0})^{-1}e_{0}^{-3\sigma_{3}},\qquad|\lambda|=1.$
(7.40)

We summarize by writing the Riemann–Hilbert problem for
$\mathbf{Q}_{n}(\lambda,x)$.

######  Riemann–Hilbert Problem 7.5.

Fix generic monodromy parameters $(e_{1},e_{2})$, $n\in\mathbb{Z}$, and
$x\in\mathbb{C}$. Seek a $2\times 2$ matrix function
$\lambda\mapsto\mathbf{Q}_{n}(\lambda,x)$ with the following properties:

  * •

Analyticity: $\mathbf{Q}_{n}(\lambda,x)$ is an analytic function of $\lambda$
for $|\lambda|\neq 1$.

  * •

Jump condition: $\mathbf{Q}_{n}(\lambda,x)$ takes analytic boundary values on
the unit circle from the interior and exterior, denoted
$\mathbf{Q}_{n,+}(\lambda,x)$ and $\mathbf{Q}_{n,-}(\lambda,x)$ for
$|\lambda|=1$ respectively, and they are related by

$\mathbf{Q}_{n,+}(\lambda,x)=\mathbf{Q}_{n,-}(\lambda,x)\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x,\Theta_{\infty})e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}\breve{\mathbf{\Phi}}^{(\infty)}_{-n}\bigl{(}-\lambda^{-1},x,\Theta_{0}\bigr{)}^{-1}e_{0}^{-3\sigma_{3}}.$

  * •

Normalization: $\mathbf{Q}_{n}(\lambda,x)\to\mathbb{I}$ as $\lambda\to\infty$.

Henceforth, to avoid the notation becoming unwieldy, we understand that all
quantities appearing with subscript $n$ are evaluated at parameter
$\Theta_{\infty}$ while quantities appearing with subscript $-n$ are evaluated
at parameter $\Theta_{0}$.

###  7.5 The limit $\boldsymbol{n\to+\infty}$

Having succeeded in removing the problematic jump conditions along rays
emanating from $0$, $\infty$ in the $\lambda$ plane by defining
$\mathbf{Q}_{n}(\lambda,x)$, we would next like to consider the limiting
behavior of this problem as $n\to+\infty$ with $x=z/n$ and $z$ fixed. It is
convenient to first renormalize $\mathbf{Q}_{n}(\lambda,x)$, essentially by a
transformation that diagonalizes the coefficient
$\mu_{n}\mathbf{B}_{n}(x)\sigma_{3}\mathbf{B}_{n}(x)^{-1}$ of $\lambda^{-1}$
in the matrix of the Lax equation (7.5). In other words, in the jump condition
for $\mathbf{Q}_{n}(\lambda,x)$ we prefer to replace
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ with a suitable left-
diagonal multiple of
$\mathbf{B}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$.
Observe that the coefficient $\mathbf{B}_{n}(x)$ is determined up to right-
multiplication by a diagonal matrix by (7.6), in which the second row of the
matrix on the right-hand side is
$(c_{n}(x),-a_{n})=\big{(}\gamma_{n}x^{1-2\kappa_{n}},\mu_{n}^{-1}\big{(}\kappa_{n}-\frac{1}{2}\big{)}\big{)}$,
where we used (7.12) and (7.21). Indeed, the first column
$\mathbf{b}_{n}^{(1)}(x)$ satisfies
$\big{(}\gamma_{n}x^{1-2\kappa_{n}},\mu_{n}^{-1}\big{(}\kappa_{n}-\frac{1}{2}\big{)}-1\big{)}\mathbf{b}_{n}^{(1)}(x)=0$
while the second column $\mathbf{b}_{n}^{(2)}(x)$ satisfies
$\big{(}\gamma_{n}x^{1-2\kappa_{n}},\mu_{n}^{-1}\big{(}\kappa_{n}-\frac{1}{2}\big{)}+1\big{)}\mathbf{b}_{n}^{(2)}(x)=0$.
By selecting specific constant factors for each column, we obtain a matrix
$\mathbf{P}_{n}(x)$ differing from $\mathbf{B}_{n}(x)$ by right-multiplication
by a diagonal matrix, and given explicitly by

$\mathbf{P}_{n}(x)\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}\frac{1}{2}-\kappa_{n}+\mu_{n}&\frac{1}{2}-\kappa_{n}-\mu_{n}\\\
\mu_{n}\gamma_{n}x^{1-2\kappa_{n}}&\mu_{n}\gamma_{n}x^{1-2\kappa_{n}}\end{bmatrix},$

in which the dependence on the index $n$ enters via (7.12) and (7.27). Then to
get the desired modification of the jump matrix we set

$\mathbf{R}_{n}(\lambda,x)\mathrel{\mathop{\mathchar
58\relax}}=\begin{cases}\mathbf{P}_{n}(x)^{-1}\mathbf{Q}_{n}(\lambda,x)\mathbf{P}_{n}(x),&|\lambda|>1,\\\
\mathbf{P}_{n}(x)^{-1}\mathbf{Q}_{n}(\lambda,x)e_{0}^{3\sigma_{3}}\mathbf{P}_{-n}(x)d_{n}(x),&|\lambda|<1,\end{cases}$

where $d_{n}(x)$ is a scalar satisfying

$d_{n}(x)^{2}=\frac{\det(\mathbf{P}_{n}(x))}{\det(\mathbf{P}_{-n}(x))}.$
(7.41)

Then, $\mathbf{R}_{n}(\lambda,x)$ solves the following Riemann–Hilbert
problem.

######  Riemann–Hilbert Problem 7.6.

Fix generic monodromy parameters $(e_{1},e_{2})$, $n\in\mathbb{Z}$, and
$x\in\mathbb{C}$. Seek a $2\times 2$ matrix function
$\lambda\mapsto\mathbf{R}_{n}(\lambda,x)$ with the following properties:

  * •

Analyticity: $\mathbf{R}_{n}(\lambda,x)$ is an analytic function of $\lambda$
for $|\lambda|\neq 1$.

  * •

Jump condition: $\mathbf{R}_{n}(\lambda,x)$ takes analytic boundary values on
the unit circle from the interior and exterior, denoted
$\mathbf{R}_{n,+}(\lambda,x)$ and $\mathbf{R}_{n,-}(\lambda,x)$ for
$|\lambda|=1$ respectively, and they are related by

$\displaystyle\mathbf{R}_{n,+}(\lambda,x)=\mathbf{R}_{n,-}(\lambda,x)\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}$
$\displaystyle\hphantom{\mathbf{R}_{n,+}(\lambda,x)=}{}\times\breve{\mathbf{\Phi}}_{-n}^{(\infty)}\bigl{(}-\lambda^{-1},x\bigr{)}^{-1}\mathbf{P}_{-n}(x)d_{n}(x)$
(7.42)

  * •

Normalization: $\mathbf{R}_{n}(\lambda,x)\to\mathbb{I}$ as $\lambda\to\infty$.

The matrices $\mathbf{\Xi}_{n}^{(6)}(x)$ and $\mathbf{\Delta}_{n}^{(6)}(x)$
defined in (6.5) and (6.4), respectively, can be expressed in terms of
$\mathbf{R}_{n}(\lambda,x)$ as follows:

$\displaystyle\mathbf{\Xi}_{n}^{(6)}(x)=\mathbf{A}_{n}(x)+\mathbf{P}_{n}(x)\big{[}\lim_{\lambda\to\infty}\lambda(\mathbf{R}_{n}(\lambda,x)-\mathbb{I})\big{]}\mathbf{P}_{n}(x)^{-1}-\frac{1}{2}\mathrm{i}x\sigma_{3},$
$\displaystyle\mathbf{\Delta}_{n}^{(6)}(x)=\mathbf{P}_{n}(x)\mathbf{R}_{n}(0,x)\mathbf{P}_{-n}(x)^{-1}d_{n}(x)^{-1}e_{0}^{-3\sigma_{3}}.$
(7.43)

Here, $\mathbf{A}_{n}(x)$ is the matrix coefficient defined in (7.1).

We now show that the jump matrix in (7.42) has explicit limits as
$n\to+\infty$ along even or odd subsequences, with the convergence being
uniform for $|\lambda|=1$ and bounded $z$ where $x=z/n$. To this end, we
compute the asymptotic behavior of
$\mathbf{P}_{n}\big{(}n^{-1}z\big{)}^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}\big{(}\lambda,n^{-1}z\big{)}$
assuming that $|\lambda|=1$. The relevant formula for
$\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$ in this setting is (7.33).
When $|\lambda|=1$,

$\displaystyle\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\qquad=\frac{1}{2\mu_{n}^{2}\gamma_{n}x^{1-2\kappa_{n}}}\begin{bmatrix}\mu_{n}\gamma_{n}x^{1-2\kappa_{n}}&-\frac{1}{2}+\mu_{n}+\kappa_{n}\vspace{1mm}\\\
-\mu_{n}\gamma_{n}x^{1-2\kappa_{n}}&\frac{1}{2}+\mu_{n}-\kappa_{n}\end{bmatrix}\mathbf{D}_{n}x^{\kappa_{n}\sigma_{3}}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{J}_{n}.$

Using (7.24), we see that

$\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)=\frac{1}{2\mu_{n}^{2}\gamma_{n}}x^{\kappa_{n}-\frac{1}{2}}\begin{bmatrix}\mathrm{i}&-\frac{1}{2}+\mu_{n}+\kappa_{n}\vspace{1mm}\\\
-\mathrm{i}&\frac{1}{2}+\mu_{n}-\kappa_{n}\end{bmatrix}x^{\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{J}_{n}.$

Now, for $x>0$, the principal branch power
$(-\mathrm{i}x\lambda)^{\sigma_{3}/2}$ has the same domain of analyticity as
$\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}$,
and these two analytic functions are related by the identity

$\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}=x^{-\sigma_{3}/2}\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/4}(-\mathrm{i}x\lambda)^{\sigma_{3}/2}.$

Therefore,

$\displaystyle\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\qquad=\frac{1}{2\mu_{n}^{2}\gamma_{n}}x^{\kappa_{n}-\frac{1}{2}}\begin{bmatrix}\mathrm{i}&-\frac{1}{2}+\mu_{n}+\kappa_{n}\\\
-\mathrm{i}&\frac{1}{2}+\mu_{n}-\kappa_{n}\end{bmatrix}\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/4}(-\mathrm{i}x\lambda)^{\sigma_{3}/2}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{J}_{n}$
$\displaystyle\qquad=\frac{\mathrm{e}^{-\mathrm{i}\pi/4}}{2\mu_{n}^{2}\gamma_{n}}x^{\kappa_{n}-\frac{1}{2}}\begin{bmatrix}-1&-\frac{1}{2}+\mu_{n}+\kappa_{n}\\\
1&\frac{1}{2}+\mu_{n}-\kappa_{n}\end{bmatrix}(-Z)^{\sigma_{3}/2}\mathbf{M}(Z;\kappa_{n},\mu_{n})\mathbf{J}_{n},\qquad
Z=\mathrm{i}x\lambda.$ (7.44)

Now using (7.31), we have

$\displaystyle(-Z)^{\sigma_{3}/2}\mathbf{M}(Z;\kappa_{n},\mu_{n})$
$\displaystyle\quad=\begin{bmatrix}\bigl{(}-\frac{1}{2}-\mu_{n}+\kappa_{n}\bigr{)}(-Z)^{-\frac{1}{2}}M_{1-\kappa_{n},\mu_{n}}(-Z)&\bigl{(}-\frac{1}{2}+\mu_{n}+\kappa_{n}\bigr{)}(-Z)^{-\frac{1}{2}}M_{1-\kappa_{n},-\mu_{n}}(-Z)\vspace{1mm}\\\
(-Z)^{-\frac{1}{2}}M_{-\kappa_{n},\mu_{n}}(-Z)&(-Z)^{-\frac{1}{2}}M_{-\kappa_{n},-\mu_{n}}(-Z)\end{bmatrix}.$

The diagonal elements in (7.44) can be simplified using the identity (see [38,
equation (13.15.3)])

$\big{(}\kappa-\mu-\tfrac{1}{2}\big{)}M_{\kappa-\frac{1}{2},\mu+\frac{1}{2}}(\diamond)+(1+2\mu)\diamond^{\frac{1}{2}}M_{\kappa,\mu}(\diamond)-\big{(}\kappa+\mu+\tfrac{1}{2}\big{)}M_{\kappa+\frac{1}{2},\mu+\frac{1}{2}}(\diamond)=0$

replacing $\kappa\to\frac{1}{2}-\kappa_{n}$ and $\mu\to\mu_{n}-\frac{1}{2}$
for the $(1,1)$ entry and $\mu\to-\frac{1}{2}-\mu_{n}$ for the $(2,2)$ entry,
and the off-diagonal elements can be simplified using the identity (see [38,
equation (13.15.4)])

$2\mu M_{\kappa-\frac{1}{2},\mu-\frac{1}{2}}(\diamond)-2\mu
M_{\kappa+\frac{1}{2},\mu-\frac{1}{2}}(\diamond)-\diamond^{\frac{1}{2}}M_{\kappa,\mu}(\diamond)=0$

replacing $\kappa\to\frac{1}{2}-\kappa_{n}$ and $\mu\to\frac{1}{2}-\mu_{n}$
for the $(1,2)$ entry and $\mu\to\frac{1}{2}+\mu_{n}$ for the $(2,1)$ entry.
The result is that

$\displaystyle\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)=\frac{\mathrm{e}^{-\mathrm{i}\pi/4}}{2\mu_{n}^{2}\gamma_{n}}x^{\kappa_{n}-\frac{1}{2}}$
$\displaystyle\qquad{}\times\begin{bmatrix}2\mu_{n}M_{\frac{1}{2}-\kappa_{n},\mu_{n}-\frac{1}{2}}(-Z)&\left(\dfrac{\kappa_{n}+\mu_{n}-\frac{1}{2}}{1-2\mu_{n}}\right)M_{\frac{1}{2}-\kappa_{n},\frac{1}{2}-\mu_{n}}(-Z)\\\
\left(\dfrac{\frac{1}{2}-\kappa_{n}+\mu_{n}}{1+2\mu_{n}}\right)M_{\frac{1}{2}-\kappa_{n},\frac{1}{2}+\mu_{n}}(-Z)&2\mu_{n}M_{\frac{1}{2}-\kappa_{n},-\frac{1}{2}-\mu_{n}}(-Z)\end{bmatrix}\mathbf{J}_{n},$
$\displaystyle Z=\mathrm{i}x\lambda.$ (7.45)

We will need the following result for the large $n$ limit of Whittaker
functions appearing here, cf. [38, equation (13.21.1)].

######  Lemma 7.7.

Assume that $\mu$ is fixed with $2\mu\neq-1,-2,-3,\dots$, and let
$f(\zeta;\mu)$ denote the entire function

$f(\zeta;\mu)\mathrel{\mathop{\mathchar
58\relax}}=\sum_{s=0}^{\infty}\frac{(-\zeta)^{s}}{\Gamma(1+2\mu+s)s!}.$ (7.46)

Then the asymptotic formula

$M_{\kappa,\mu}\left(\frac{\zeta}{\kappa}\right)=\Gamma(1+2\mu)\left(\frac{\zeta}{\kappa}\right)^{\mu+1/2}\big{[}f(\zeta;\mu)+\mathcal{O}\big{(}\kappa^{-2}\big{)}\big{]}$

holds uniformly in the limit $\kappa\to\infty$ in any $($possibly complex$)$
direction under the assumption ${\zeta=\mathcal{O}(1)}$.

###### Proof.

We start from the formula [38, equation (13.14.6)] which holds under the
indicated condition on $\mu$:

$M_{\kappa,\mu}\left(\frac{\zeta}{\kappa}\right)=\Gamma(1+2\mu)\left(\frac{\zeta}{\kappa}\right)^{\mu+1/2}\mathrm{e}^{-\zeta/(2\kappa)}\sum_{s=0}^{\infty}\frac{(-\zeta)^{s}}{\Gamma(1+2\mu+s)s!}\prod_{j=0}^{s-1}\left(1-\frac{\mu+1/2+j}{\kappa}\right).$

Clearly,
$\mathrm{e}^{-\zeta/(2\kappa)}=1-\zeta/(2\kappa)+\mathcal{O}\big{(}\kappa^{-2}\big{)}$
as $\kappa\to\infty$ for $\zeta=\mathcal{O}(1)$, and the product in the
summand has the expansion

$\displaystyle\prod_{j=0}^{s-1}\left(1-\frac{\mu+1/2+j}{\kappa}\right)=1-\frac{1}{\kappa}\bigg{[}\left(\mu+\frac{1}{2}\right)s+\frac{1}{2}s(s-1)\bigg{]}+\mathcal{O}\big{(}\kappa^{-2}s^{s}\big{)},\qquad\kappa\to\infty\\!\\!\\!$
(7.47)

uniformly for all indices $s$. This follows from the Fredholm expansion
formula

$\prod_{j=0}^{s-1}(1+r_{j})=1+\sum_{j=0}^{s-1}r_{j}+\sum_{k=2}^{s}\mathop{\sum_{S\subset\mathbb{Z}_{s}}}_{|S|=k}\prod_{l\in
S}r_{l}$

and the estimate

$\Biggl{|}\sum_{\begin{subarray}{c}S\subset\mathbb{Z}_{s}\\\
|S|=k\end{subarray}}\prod_{l\in
S}r_{l}\Biggr{|}\leq\binom{s}{k}R_{s}^{k},\qquad
R_{s}\mathrel{\mathop{\mathchar 58\relax}}=\max_{0\leq l\leq
s-1}\\{|r_{l}|\\}.$

Indeed, with $r_{j}=-\kappa^{-1}(\mu+\frac{1}{2}+j)$, we have

$\sum_{j=0}^{s-1}r_{j}=-\frac{1}{\kappa}\sum_{j=0}^{s-1}\left(\mu+\frac{1}{2}+j\right)=-\frac{1}{\kappa}\left[\left(\mu+\frac{1}{2}\right)s+\frac{1}{2}s(s-1)\right],$

and $R_{s}\leq|\kappa|^{-1}\big{(}\big{|}\mu+\frac{1}{2}\big{|}+(s-1)\big{)}$.
Therefore,
$R_{s}^{k}\leq|\kappa|^{-2}\big{(}\big{|}\mu+\frac{1}{2}\big{|}+(s-1)\big{)}^{k}$
holds for all $k\geq 2$ whenever $|\kappa|\geq 1$. Consequently,

$\displaystyle\left|\prod_{j=0}^{s-1}\left(1-\frac{\mu+1/2+j}{\kappa}\right)-1+\right.\left.\frac{1}{\kappa}\left[\left(\mu+\frac{1}{2}\right)s+\frac{1}{2}s(s-1)\right]\right|$
$\displaystyle\qquad=\Biggl{|}\sum_{k=2}^{s}\mathop{\sum_{S\subset\mathbb{Z}_{s}}}_{|S|=k}\prod_{l\in
S}r_{l}\Biggr{|}\leq\frac{1}{|\kappa|^{2}}\sum_{k=2}^{s}\binom{s}{k}\left(\left|\mu+\frac{1}{2}\right|+(s-1)\right)^{k}1^{s-k}$
$\displaystyle\qquad\leq\frac{1}{|\kappa|^{2}}\sum_{k=0}^{s}\binom{s}{k}\left(\left|\mu+\frac{1}{2}\right|+(s-1)\right)^{k}1^{s-k}=\frac{\big{(}\big{|}\mu+\frac{1}{2}\big{|}+s\big{)}^{s}}{|\kappa|^{2}},$

which proves (7.47). Since the series

$\sum_{s=0}^{\infty}\frac{(-\zeta)^{s}s^{s}}{\Gamma(1+2\mu_{n}+s)s!}$

converges uniformly for $|\zeta|$ bounded, it follows that

$M_{\kappa,\mu}\left(\frac{\zeta}{\kappa}\right)=\Gamma(1+2\mu)\left(\frac{\zeta}{\kappa}\right)^{\mu+1/2}\left[f(\zeta;\mu)-\frac{1}{\kappa}g(\zeta;\mu)+\mathcal{O}\big{(}\kappa^{-2}\big{)}\right]$

holds as $\kappa\to\infty$ in $\mathbb{C}$ with $\zeta=\mathcal{O}(1)$, where

$g(\zeta;\mu)\mathrel{\mathop{\mathchar 58\relax}}=\frac{1}{2}\zeta
f(\zeta;\mu)+\left(\mu+\frac{1}{2}\right)\zeta
f^{\prime}(\zeta;\mu)+\frac{1}{2}\zeta^{2}f^{\prime\prime}(\zeta;\mu).$

Now using the series (7.46) one checks that for all indicated values of $\mu$,
$g(\zeta;\mu)$ vanishes identically, so the proof is complete. ∎

The series in (7.46) defines an entire function of $\zeta$ related to Bessel
functions (see [38, Chapter 10]) in the following way:

$f(\zeta,\mu)=\zeta^{-{\mu}{}}J_{2\mu}\big{(}2\sqrt{\zeta}\big{)}.$ (7.48)

We apply Lemma 7.7 to (7.45) by taking
$\zeta=-\big{(}\frac{1}{2}-\kappa_{n}\big{)}Z=-\big{(}\frac{1}{2}-\kappa_{n}\big{)}\mathrm{i}x\lambda$.
If $x=z/n$ and $z=\mathcal{O}(1)$, then using (7.12), we see that
$\zeta=-\frac{1}{2}\mathrm{i}z\lambda+\mathcal{O}\big{(}n^{-1}\big{)}$ holds
for $|\lambda|=1$. So, (7.45) becomes the statement that

$\displaystyle\mathbf{P}_{n}(x)^{-1}\breve{\mathbf{\Phi}}_{n}^{(\infty)}(\lambda,x)$
$\displaystyle\quad=\frac{\mathrm{e}^{-\mathrm{i}\pi/4}}{2\mu_{n}^{2}\gamma_{n}}x^{\kappa_{n}-\frac{1}{2}}\Bigg{(}\frac{\rho_{\infty}}{2}\begin{bmatrix}\Gamma(2\mu_{n}+1)J_{2\mu_{n}-1}(\rho_{\infty})&-\Gamma(1-2\mu_{n})J_{1-2\mu_{n}}(\rho_{\infty})\vspace{1mm}\\\
\Gamma(2\mu_{n}+1)J_{2\mu_{n}+1}(\rho_{\infty})&-\Gamma(1-2\mu_{n})J_{-1-2\mu_{n}}(\rho_{\infty})\end{bmatrix}+\mathcal{O}\big{(}n^{-1}\big{)}\Bigg{)}$
$\displaystyle\phantom{\quad=}{}\times\left(\dfrac{n}{2}\right)^{-\mu_{n}\sigma_{3}}\mathbf{J}_{n}$

with

$\rho_{\infty}=\rho_{\infty}(\lambda,z)\mathrel{\mathop{\mathchar
58\relax}}=(-2\mathrm{i}z\lambda)^{1/2}\qquad\text{(principal branch)}$ (7.49)

holds in the limit $n\to\infty$ with $x=n^{-1}z$ uniformly for
$z=\mathcal{O}(1)$ and $|\lambda|=1$. Similarly, to study the parametrix near
0, it will be convenient to rewrite formula (7.48) in terms of modified Bessel
functions:

$f(\zeta,\mu)=(-\zeta)^{-\mu}I_{2\mu}\big{(}2\sqrt{-\zeta}\big{)}.$

Replacing $n$ with $-n$, $\Theta_{\infty}$ with $\Theta_{0}$, and $\lambda$
with $-\lambda^{-1}$ and recalling that $\mu_{-n}=\mu_{n}$, gives in the same
limit,

$\displaystyle\mathbf{P}_{-n}(x)^{-1}\breve{\mathbf{\Phi}}_{-n}^{(\infty)}\bigl{(}-\lambda^{-1},x\bigr{)}$
$\displaystyle\quad=\frac{\mathrm{e}^{-\mathrm{i}\pi/4}}{2\mu_{n}^{2}\gamma_{-n}}x^{\kappa_{-n}-\frac{1}{2}}\Bigg{(}\frac{\rho_{0}}{2}\begin{bmatrix}{\Gamma(2\mu_{n}+1)}{}I_{2\mu_{n}-1}(\rho_{0})&{\Gamma(1-2\mu_{n})}{}I_{1-2\mu_{n}}(\rho_{0})\vspace{1mm}\\\
-{\Gamma(2\mu_{n}+1)}{}I_{2\mu_{n}+1}(\rho_{0})&-{\Gamma(1-2\mu_{n})}{}I_{-2\mu_{n}-1}(\rho_{0})\end{bmatrix}+\mathcal{O}\big{(}n^{-1}\big{)}\Bigg{)}$
$\displaystyle\phantom{\quad=}{}\times\left(\frac{n}{2}\right)^{-\mu_{n}\sigma_{3}}\mathbf{J}_{-n}$

with

$\rho_{0}=\rho_{0}(\lambda,z)\mathrel{\mathop{\mathchar
58\relax}}=\big{(}2\mathrm{i}z\lambda^{-1}\big{)}^{1/2}\qquad\text{(principal
branch)}.$ (7.50)

The jump matrix in (7.42) therefore reads

$\displaystyle\mathbf{R}_{n,-}\big{(}\lambda,n^{-1}z\big{)}^{-1}\mathbf{R}_{n,+}\big{(}\lambda,n^{-1}z\big{)}$
$\displaystyle=\frac{\rho_{\infty}}{2}\begin{bmatrix}\Gamma(2\mu_{n}+1)J_{2\mu_{n}-1}(\rho_{\infty})+\mathcal{O}\big{(}n^{-1}\big{)}&-\Gamma(1-2\mu_{n})J_{1-2\mu_{n}}(\rho_{\infty})+\mathcal{O}\big{(}n^{-1}\big{)}\vspace{1mm}\\\
\Gamma(2\mu_{n}+1)J_{2\mu_{n}+1}(\rho_{\infty})+\mathcal{O}\big{(}n^{-1}\big{)}&-\Gamma(1-2\mu_{n})J_{-1-2\mu_{n}}(\rho_{\infty})+\mathcal{O}\big{(}n^{-1}\big{)}\end{bmatrix}$
$\displaystyle\phantom{=}{}\times{}\frac{\gamma_{-n}}{\gamma_{n}}x^{\kappa_{n}-\kappa_{-n}}d_{n}(x)\left(\frac{n}{2}\right)^{-\mu_{n}\sigma_{3}}\mathbf{J}_{n}e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}\mathbf{J}_{-n}^{-1}\left(\frac{n}{2}\right)^{\mu_{n}\sigma_{3}}\frac{2}{\rho_{0}}$
$\displaystyle\phantom{=}{}\times\begin{bmatrix}{\Gamma(2\mu_{n}+1)}{}I_{2\mu_{n}-1}(\rho_{0})+\mathcal{O}\big{(}n^{-1}\big{)}&{\Gamma(1-2\mu_{n})}{}I_{1-2\mu_{n}}(\rho_{0})+\mathcal{O}\big{(}n^{-1}\big{)}\vspace{1mm}\\\
-{\Gamma(2\mu_{n}+1)}{}I_{2\mu_{n}+1}(\rho_{0})+\mathcal{O}\big{(}n^{-1}\big{)}&-{\Gamma(1-2\mu_{n})}{}I_{-2\mu_{n}-1}(\rho_{0})+\mathcal{O}\big{(}n^{-1}\big{)}\end{bmatrix}^{-1}.$
(7.51)

Expanding (7.41) for large $n>0$ gives

$\displaystyle
d_{n}(x)^{2}=\frac{\gamma_{n}}{\gamma_{-n}}x^{2\kappa_{-n}-2\kappa_{n}}=\dfrac{e_{\infty}^{5}}{e_{0}^{5}}\dfrac{e_{0}^{2}-e_{1}^{2}}{e_{\infty}^{2}-e_{1}^{2}}\dfrac{\Gamma\big{(}\frac{1}{2}-\kappa_{-n}-\mu_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\kappa_{-n}+\mu_{n}\big{)}}{\Gamma\big{(}\frac{1}{2}-\kappa_{n}-\mu_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\kappa_{n}+\mu_{n}\big{)}}x^{2\kappa_{-n}-2\kappa_{n}}$
$\displaystyle\hphantom{d_{n}(x)^{2}}{}=\dfrac{e_{\infty}^{5}e_{1}^{2}}{e_{0}^{3}\big{(}1-e_{1}^{2}e_{0}^{2}\big{)}\big{(}e_{\infty}^{2}-e_{1}^{2}\big{)}}$
$\displaystyle\hphantom{d_{n}(x)^{2}=}{}\times\dfrac{4\pi^{2}x^{2\kappa_{-n}-2\kappa_{n}}}{\Gamma\big{(}\frac{1}{2}+\kappa_{-n}-\mu_{n}\big{)}\Gamma\big{(}\frac{1}{2}+\kappa_{-n}+\mu_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\kappa_{n}+\mu_{n}\big{)}\Gamma\big{(}\frac{1}{2}-\kappa_{n}-\mu_{n}\big{)}}$
$\displaystyle\hphantom{d_{n}(x)^{2}}{}=4\dfrac{e_{\infty}^{4}}{e_{0}^{4}}\dfrac{e_{0}e_{\infty}e_{1}^{2}}{\big{(}1-e_{1}^{2}e_{0}^{2}\big{)}\big{(}e_{\infty}^{2}-e_{1}^{2}\big{)}}x^{2\kappa_{-n}-2\kappa_{n}}n^{-2n-\Theta_{0}+\Theta_{\infty}}\mathrm{e}^{2n}2^{2n+\Theta_{0}-\Theta_{\infty}-2}\big{(}1+\mathcal{O}\big{(}n^{-1}\big{)}\big{)},$
$\displaystyle n\to+\infty.$

We now properly define $d_{n}(x)$ for large $n$ by selecting a definite value
for the square root of

$\sqrt{\dfrac{1-e_{1}^{2}e_{0}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}$ (7.52)

after which $d_{n}(x)$ has the asymptotic expansion

$\displaystyle
d_{n}(x)=\dfrac{e_{1}e_{\infty}^{5/2}}{e_{0}^{3/2}}\dfrac{1}{\big{(}1-e_{1}^{2}e_{0}^{2}\big{)}}\sqrt{\dfrac{1-e_{1}^{2}e_{0}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}x^{\kappa_{-n}-\kappa_{n}}n^{-n+\frac{1}{2}(-\Theta_{0}+\Theta_{\infty})}\mathrm{e}^{n}2^{n+\frac{1}{2}(\Theta_{0}-\Theta_{\infty})}$
$\displaystyle\hphantom{d_{n}(x)=}{}\times\big{(}1+\mathcal{O}\big{(}n^{-1}\big{)}\big{)},\qquad
n\to+\infty.$

Then, by definition, we have

$\displaystyle\dfrac{\gamma_{-n}}{\gamma_{n}}d_{n}(x)x^{\kappa_{n}-\kappa_{-n}}={}$
$\displaystyle\dfrac{e_{0}^{3/2}}{e_{1}e_{\infty}^{5/2}}{\big{(}1-e_{1}^{2}e_{0}^{2}\big{)}}\left(\sqrt{\dfrac{1-e_{1}^{2}e_{0}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}\right)^{-1}n^{n-\frac{1}{2}(-\Theta_{0}+\Theta_{\infty})}\mathrm{e}^{-n}$
$\displaystyle\times
2^{-n-\frac{1}{2}(\Theta_{0}-\Theta_{\infty})}\big{(}1+\mathcal{O}\big{(}n^{-1}\big{)}\big{)},\qquad
n\to+\infty.$

Furthermore, using identities (7.28), (7.34), and Stirling’s formula yields

$\displaystyle-\left(\frac{2}{n}\right)^{2\mu_{n}}\beta_{n}\frac{J_{n,11}}{J_{-n,22}}=-\left(\frac{2}{n}\right)^{2\mu_{n}}\frac{e_{\infty}^{7/2}\big{(}e_{1}^{2}-e_{0}^{2}\big{)}\Gamma(-2\mu_{n})\Gamma\big{(}\frac{1}{2}-\kappa_{-n}+\mu_{n}\big{)}}{e_{0}^{3/2}\big{(}e_{0}^{2}e_{1}^{2}-1\big{)}\Gamma(2\mu_{n})\Gamma\big{(}\frac{1}{2}-\kappa_{n}-\mu_{n}\big{)}}$
$\displaystyle\hphantom{-\left(\frac{2}{n}\right)^{2\mu_{n}}\beta_{n}\frac{J_{n,11}}{J_{-n,22}}}{}=\frac{e_{\infty}^{7/2}}{e_{0}^{3/2}}\dfrac{\mathrm{i}e_{0}e_{1}}{\big{(}1-e_{0}^{2}e_{1}^{2}\big{)}}\dfrac{\Gamma(-2\mu_{n})}{\Gamma(2\mu_{n})}n^{-n+\frac{1}{2}(-\Theta_{0}+\Theta_{\infty})}\mathrm{e}^{n}$
$\displaystyle\hphantom{-\left(\frac{2}{n}\right)^{2\mu_{n}}\beta_{n}\frac{J_{n,11}}{J_{-n,22}}=}{}\times
2^{n+\frac{1}{2}(\Theta_{0}-\Theta_{\infty})}\big{(}1+\mathcal{O}\big{(}n^{-1}\big{)}\big{)},\qquad
n\to+\infty,$

and similarly,

$\displaystyle\left(\frac{n}{2}\right)^{2\mu_{n}}\frac{J_{n,22}}{\beta_{n}J_{-n,11}}=\left(\frac{n}{2}\right)^{2\mu_{n}}\frac{e_{\infty}^{3/2}\big{(}e_{0}^{2}e_{1}^{2}-1\big{)}\Gamma(2\mu_{n})\Gamma\big{(}\frac{1}{2}-\kappa_{-n}-\mu_{n}\big{)}}{e_{0}^{7/2}\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}\Gamma(-2\mu_{n})\Gamma\big{(}\frac{1}{2}-\kappa_{n}+\mu_{n}\big{)}}$
$\displaystyle\hphantom{\left(\frac{n}{2}\right)^{2\mu_{n}}\frac{J_{n,22}}{\beta_{n}J_{-n,11}}}{}=\frac{e_{\infty}^{3/2}}{e_{0}^{7/2}}\dfrac{\mathrm{i}e_{0}e_{1}}{\big{(}e_{\infty}^{2}-e_{1}^{2}\big{)}}\dfrac{\Gamma(2\mu_{n})}{\Gamma(-2\mu_{n})}n^{-n+\frac{1}{2}(-\Theta_{0}+\Theta_{\infty})}\mathrm{e}^{n}$
$\displaystyle\hphantom{\left(\frac{n}{2}\right)^{2\mu_{n}}\frac{J_{n,22}}{\beta_{n}J_{-n,11}}=}{}\times
2^{n+\frac{1}{2}(\Theta_{0}-\Theta_{\infty})}\big{(}1+\mathcal{O}\big{(}n^{-1}\big{)}\big{)},\qquad
n\to+\infty.$

Therefore, the central factor on the right-hand side of (7.51) satisfies

$\displaystyle\frac{\gamma_{-n}}{\gamma_{n}}x^{\kappa_{n}-\kappa_{-n}}d_{n}(x)\left(\frac{n}{2}\right)^{-\mu_{n}\sigma_{3}}\mathbf{J}_{n}e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}\mathbf{J}_{-n}^{-1}\left(\frac{n}{2}\right)^{\mu_{n}\sigma_{3}}$
$\displaystyle\qquad=\begin{bmatrix}0&-\dfrac{e_{0}e_{2}e_{\infty}}{\mathrm{i}}\dfrac{\Gamma(-2\mu_{n})}{\Gamma(2\mu_{n})}\left(\sqrt{\dfrac{1-e_{0}^{2}e_{1}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}\right)^{-1}\\\
\dfrac{\mathrm{i}}{e_{0}e_{2}e_{\infty}}\dfrac{\Gamma(2\mu_{n})}{\Gamma(-2\mu_{n})}\sqrt{\dfrac{1-e_{0}^{2}e_{1}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}&0\end{bmatrix}+\mathcal{O}\big{(}n^{-1}\big{)}.$

The leading term is independent of $n\pmod{2}$ and has unit determinant. This
proves the following.

######  Proposition 7.8.

Define the constant matrix which depends only on the even/odd parity of $n$
via $\mu_{n}$, $e_{1}$, $e_{0}^{2}$, and $e_{\infty}^{2}$:

$\mathbf{V}^{\mathrm{even/odd}}\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}0&-\dfrac{e_{0}e_{2}e_{\infty}}{\mathrm{i}}\left(\sqrt{\dfrac{1-e_{0}^{2}e_{1}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}\right)^{-1}\\\
\dfrac{\mathrm{i}}{e_{0}e_{2}e_{\infty}}\sqrt{\dfrac{1-e_{0}^{2}e_{1}^{2}}{e_{\infty}^{2}-e_{1}^{2}}}&0\end{bmatrix}.$
(7.53)

Then the following asymptotic formula holds uniformly for $|\lambda|=1$ and
$z$ bounded:

$\displaystyle\mathbf{R}_{n,-}(\lambda,z/n)^{-1}\mathbf{R}_{n,+}(\lambda,z/n)={\rho_{\infty}}\begin{bmatrix}J_{2\mu_{n}-1}(\rho_{\infty})&-J_{1-2\mu_{n}}(\rho_{\infty})\\\
J_{2\mu_{n}+1}(\rho_{\infty})&-J_{-1-2\mu_{n}}(\rho_{\infty})\end{bmatrix}\cdot\mathbf{V}^{\mathrm{even/odd}}$
$\displaystyle\hphantom{\mathbf{R}_{n,-}(\lambda,z/n)^{-1}\mathbf{R}_{n,+}(\lambda,z/n)=}{}\times\dfrac{1}{\rho_{0}}\begin{bmatrix}I_{2\mu_{n}-1}(\rho_{0})&I_{1-2\mu_{n}}(\rho_{0})\\\
-I_{2\mu_{n}+1}(\rho_{0})&-I_{-1-2\mu_{n}}(\rho_{0})\end{bmatrix}^{-1}+\mathcal{O}\big{(}n^{-1}\big{)},$

as $n\to\infty$ along even/odd subsequences.

Proposition 7.8 suggests defining the following limiting Riemann–Hilbert
problem.

######  Riemann–Hilbert Problem 7.9 (limiting problem, even/odd subsequences
of $n$).

Fix generic monodromy parameters $(e_{1},e_{2})$, and $z\in\mathbb{C}$ with
$|{\operatorname{Arg}}(z)|<\pi$. Seek a $2\times 2$ matrix function
$\lambda\mapsto\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$ with the
following properties:

  * •

Analyticity: $\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$ is an analytic
function of $\lambda$ for $|\lambda|\neq 1$.

  * •

Jump condition: $\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$ takes
analytic boundary values on the unit circle from the interior and exterior,
denoted $\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)$ and
$\hat{\mathbf{R}}^{\mathrm{even/odd}}_{-}(\lambda,z)$ for $|\lambda|=1$
respectively, and they are related by

$\displaystyle\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)=\hat{\mathbf{R}}^{\mathrm{even/odd}}_{-}(\lambda,z){\rho_{\infty}}\begin{bmatrix}J_{2\mu_{n}-1}(\rho_{\infty})&-J_{1-2\mu_{n}}(\rho_{\infty})\\\
J_{2\mu_{n}+1}(\rho_{\infty})&-J_{-1-2\mu_{n}}(\rho_{\infty})\end{bmatrix}$
$\displaystyle\hphantom{\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)=}{}\times\mathbf{V}^{\mathrm{even/odd}}\cdot\dfrac{1}{\rho_{0}}\begin{bmatrix}I_{2\mu_{n}-1}(\rho_{0})&I_{1-2\mu_{n}}(\rho_{0})\\\
-I_{2\mu_{n}+1}(\rho_{0})&-I_{-1-2\mu_{n}}(\rho_{0})\end{bmatrix}^{-1}.$
(7.54)

  * •

Normalization: $\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)\to\mathbb{I}$
as $\lambda\to\infty$.

Note that the Bessel functions $J_{\nu}(\rho_{\infty})$ and
$I_{\nu}(\rho_{0})$ appearing in the jump matrix in (7.54) are analytic on the
unit circle $|\lambda|=1$ except at the point
$\lambda=\lambda_{\mathrm{c}}\mathrel{\mathop{\mathchar
58\relax}}=-\mathrm{i}\mathrm{e}^{-\mathrm{i}\operatorname{Arg}(z)}$. However,
from the identities

$\left.J_{\nu}(\rho_{\infty})\right|_{\lambda=\lambda_{\mathrm{c}}\mathrm{e}^{-\mathrm{i}0}}=\mathrm{e}^{\mathrm{i}\pi\nu}\left.J_{\nu}(\rho_{\infty})\right|_{\lambda=\lambda_{\mathrm{c}}\mathrm{e}^{\mathrm{i}0}}\qquad\text{and}\qquad\left.I_{\nu}(\rho_{0})\right|_{\lambda=\lambda_{\mathrm{c}}\mathrm{e}^{-\mathrm{i}0}}=\mathrm{e}^{-\mathrm{i}\pi\nu}\left.I_{\nu}(\rho_{0})\right|_{\lambda=\lambda_{\mathrm{c}}\mathrm{e}^{\mathrm{i}0}}$

and the fact that the indices $\nu$ in each column of the Bessel matrix
factors in (7.54) differ by $2$, combined with the fact that
$\mathbf{V}^{\mathrm{even/odd}}$ is an off-diagonal matrix, one sees easily
that

$\begin{bmatrix}J_{2\mu_{n}-1}(\rho_{\infty})&-J_{1-2\mu_{n}}(\rho_{\infty})\\\
J_{2\mu_{n}+1}(\rho_{\infty})&-J_{-1-2\mu_{n}}(\rho_{\infty})\end{bmatrix}\cdot\mathbf{V}^{\mathrm{even/odd}}\cdot\begin{bmatrix}I_{2\mu_{n}-1}(\rho_{0})&I_{1-2\mu_{n}}(\rho_{0})\\\
-I_{2\mu_{n}+1}(\rho_{0})&-I_{-1-2\mu_{n}}(\rho_{0})\end{bmatrix}^{-1}$

is continuous at $\lambda=\lambda_{\mathrm{c}}$ and hence is an analytic
function of $\lambda$ on the unit circle. The scalar factor
$\rho_{\infty}/\rho_{0}$ is also analytic for $|\lambda|=1$, and therefore the
jump matrix in (7.54) is an analytic function of $\lambda$ when $|\lambda|=1$.
At this stage, the existence of a matrix function
$\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$ satisfying Riemann–Hilbert
Problem 7.9 is not clear. However, it turns out that there exists a discrete
set $\Sigma^{\mathrm{even/odd}}\subset{\mathbb{C}}$ such that for
$z\in{\mathbb{C}}\setminus\Sigma^{\mathrm{even/odd}}$, such a matrix does
exist and is in fact a meromorphic function of $z$, see Section 9.1 below.

######  Lemma 7.10.

Let $(e_{1},e_{2})$ be generic monodromy parameters and take
$z\in{\mathbb{C}}\setminus\Sigma^{\mathrm{even/odd}}$. Then,

$\displaystyle\mathop{\lim_{n\to\infty}}_{\text{$n$
even/odd}}u_{n}\big{(}n^{-1}z;m\big{)}={}$
$\displaystyle-\dfrac{8\mu_{n}^{2}}{z}\big{[}\hat{R}_{11}^{\mathrm{even/odd}}(0,z)+\hat{R}_{21}^{\mathrm{even/odd}}(0,z)$
$\displaystyle-\hat{R}_{12}^{\mathrm{even/odd}}(0,z)+\hat{R}_{22}^{\mathrm{even/odd}}(0,z)\big{]}^{-2}.$
(7.55)

###### Proof.

Noting that $\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$ necessarily has
unit determinant, we form the matrix quotient

$\mathbf{E}_{n}(\lambda,z)\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{R}_{n}\big{(}\lambda,n^{-1}z\big{)}\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)^{-1},\qquad|\lambda|\neq
1.$

Clearly, $\mathbf{E}_{n}(\lambda,z)$ is analytic as a function of $\lambda$ in
the domain of definition, and for each fixed $n$ it tends to $\mathbb{I}$ as
$\lambda\to\infty$ as this is true for both
$\mathbf{R}_{n}\big{(}\lambda,n^{-1}z\big{)}$ and
$\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$. Across the unit circle, the
boundary values of $\mathbf{E}_{n}(\lambda,z)$ are related by

$\displaystyle\mathbf{E}_{n,+}(\lambda,z)=\mathbf{E}_{n,-}(\lambda,z)\hat{\mathbf{R}}_{-}^{\mathrm{even/odd}}(\lambda,z)\big{[}\mathbf{R}_{n,-}\big{(}\lambda,n^{-1}z\big{)}^{-1}\mathbf{R}_{n,+}\big{(}\lambda,n^{-1}z\big{)}\big{]}$
$\displaystyle\hphantom{\mathbf{E}_{n,+}(\lambda,z)=}{}\times\big{[}\hat{\mathbf{R}}^{\mathrm{even/odd}}_{-}(\lambda,z)^{-1}\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)\big{]}^{-1}\hat{\mathbf{R}}^{\mathrm{even/odd}}_{-}(\lambda,z)^{-1},\qquad|\lambda|=1.$

Thus, the jump matrix for $\mathbf{E}_{n}(\lambda,z)$ is the conjugation, by a
unit-determinant matrix function of $\lambda$ independent of $n$, of the
matrix ratio of the jump matrices for
$\mathbf{R}_{n}\big{(}\lambda,n^{-1}z\big{)}$ and for
$\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$. But by Proposition 7.8, the
latter ratio is $\mathbb{I}+\mathcal{O}\big{(}n^{-1}\big{)}$ uniformly on the
unit circle as $n\to\infty$ along even or odd subsequences. The conjugating
factors exist and are uniformly bounded for $z$ in compact subsets of
${\mathbb{C}}\setminus\Sigma(y_{1},y_{2},y_{3})$. It follows that in this
limit,
$\mathbf{E}_{n,+}(\lambda,z)=\mathbf{E}_{n,-}(\lambda,z)\big{(}\mathbb{I}+\mathcal{O}\big{(}n^{-1}\big{)}\big{)}$
uniformly for $|\lambda|=1$ and $z$ in compact subsets of
${\mathbb{C}}\setminus\Sigma(y_{1},y_{2},y_{3})$ as $n\to\infty$. By standard
small-norm theory, $\mathbf{E}_{n}(\lambda,z)$ exists for large enough even or
odd $n$, and tends to the identity as $n\to\infty$, in particular in the sense
that

$\lim_{\lambda\to\infty}\lambda(\mathbf{E}_{n}(\lambda,z)-\mathbb{I})\to\mathbf{0}\qquad\text{and}\qquad\mathbf{E}_{n}(0,z)\to\mathbb{I}$

as $n\to\infty$ along even/odd subsequences. By the definition of
$\mathbf{E}_{n}(\lambda,z)$ it follows that in the same limit

$\displaystyle\lim_{\lambda\to\infty}\lambda\big{(}\mathbf{R}_{n}\big{(}\lambda,n^{-1}z\big{)}-\mathbb{I}\big{)}\to\lim_{\lambda\to\infty}\lambda\big{(}\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)-\mathbb{I}\big{)},\qquad\text{and}$
$\displaystyle\mathbf{R}_{n}\big{(}0,n^{-1}z\big{)}\to\hat{\mathbf{R}}^{\mathrm{even/odd}}(0,z).$

Combining (6.6) with (7.43) then shows (7.55). Partly, this works because the
dominant term in $\Xi_{n,12}^{(6)}\big{(}n^{-1}z;m\big{)}$ is
$A_{n,12}\big{(}n^{-1}z\big{)}$. ∎

###  7.6 Transformations of the limiting Riemann–Hilbert problem

In this section we transform Riemann–Hilbert Problem 7.9 to match the form of
Riemann–Hilbert Problem 5.1. To this end, using [38, equations (10.4.4) and
(10.4.6)] to express the Bessel function $J_{\nu}(\diamond)$ in terms of the
Hankel functions $H_{\nu}^{(1)}(\diamond)$, $H_{\nu}^{(2)}(\diamond)$ and the
relations [38, equation (10.4.4) and (10.4.6)],

$\displaystyle H_{-\nu}^{(1)}(\diamond)$
$\displaystyle=\mathrm{e}^{\pi\mathrm{i}\nu}H_{\nu}^{(1)}(\diamond),\qquad
H_{-\nu}^{(2)}(\diamond)=\mathrm{e}^{-\pi\mathrm{i}\nu}H_{\nu}^{(2)}(\diamond),$

we arrive at the identity:

$\displaystyle\rho_{\infty}\begin{bmatrix}J_{2\mu_{n}-1}(\rho_{\infty})&-J_{1-2\mu_{n}}(\rho_{\infty})\\\
J_{2\mu_{n}+1}(\rho_{\infty})&-J_{-1-2\mu_{n}}(\rho_{\infty})\end{bmatrix}$
$\displaystyle\qquad=\dfrac{\rho_{\infty}}{2}\begin{bmatrix}H_{2\mu_{n}-1}^{(1)}(\rho_{\infty})&H_{1-2\mu_{n}}^{(2)}(\rho_{\infty})\\\
H^{(1)}_{2\mu_{n}+1}(\rho_{\infty})&H_{-1-2\mu_{n}}^{(2)}(\rho_{\infty})\end{bmatrix}\begin{bmatrix}1&\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}&-1\end{bmatrix}.$ (7.56)

To obtain appropriate asymptotic formulæ for the matrix on the right-hand side
of (7.56), we first apply the identity [38, equation (10.6.1)]

$H^{(k)}_{\nu-1}(\diamond)+H^{(k)}_{\nu+1}(\diamond)=\dfrac{2\nu}{\diamond}H^{(k)}_{\nu}(\diamond),\qquad
k=1,2,$

which gives

$\displaystyle\rho_{\infty}^{-\sigma_{3}/2}\begin{bmatrix}1&0\\\
1&1\end{bmatrix}\dfrac{\rho_{\infty}}{2}\begin{bmatrix}H_{2\mu_{n}-1}^{(1)}(\rho_{\infty})&H_{1-2\mu_{n}}^{(2)}(\rho_{\infty})\vspace{1mm}\\\
H^{(1)}_{2\mu_{n}+1}(\rho_{\infty})&H_{-1-2\mu_{n}}^{(2)}(\rho_{\infty})\end{bmatrix}$
$\displaystyle\qquad=\dfrac{\sqrt{\rho_{\infty}}}{2}\begin{bmatrix}H_{2\mu_{n}-1}^{(1)}(\rho_{\infty})&H_{1-2\mu_{n}}^{(2)}(\rho_{\infty})\vspace{1mm}\\\
4\mu_{n}H_{2\mu_{n}}^{(1)}(\rho_{\infty})&-4\mu_{n}H_{-2\mu_{n}}^{(2)}(\rho_{\infty})\end{bmatrix}.$
(7.57)

The matrix on the right-hand side is amenable to asymptotic analysis as
$\rho_{\infty}\to\infty$; using the asymptotics of Hankel functions [38,
equations (10.17.5) and (10.17.6)] and (7.57) yields

$\displaystyle\dfrac{\rho_{\infty}}{2}\begin{bmatrix}H_{2\mu_{n}-1}^{(1)}(\rho_{\infty})&H_{1-2\mu_{n}}^{(2)}(\rho_{\infty})\vspace{1mm}\\\
H^{(1)}_{2\mu_{n}+1}(\rho_{\infty})&H_{-1-2\mu_{n}}^{(2)}(\rho_{\infty})\end{bmatrix}$
$\displaystyle\qquad=\begin{bmatrix}1&\dfrac{1}{2}-\dfrac{\mu_{n}}{2}-\dfrac{3}{32\mu_{n}}\vspace{1mm}\\\
-1&\dfrac{1}{2}+\dfrac{\mu_{n}}{2}+\dfrac{3}{32\mu_{n}}\end{bmatrix}\Biggl{(}\mathbb{I}+\dfrac{1}{128\rho_{\infty}^{2}}$
$\displaystyle\phantom{\qquad=}{}\times\begin{bmatrix}\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}&\dfrac{\big{(}16\mu_{n}^{2}-13\big{)}\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}}{48\mu_{n}}\vspace{1mm}\\\
64\mu_{n}\big{(}16\mu_{n}^{2}-1\big{)}&-\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}\end{bmatrix}+\mathcal{O}\big{(}\rho_{\infty}^{-4}\big{)}\Biggr{)}\rho_{\infty}^{\sigma_{3}/2}$
$\displaystyle\phantom{\qquad=}{}\times(2\sqrt{\mu_{n}})^{\mathbb{I}-\sigma_{3}}{\dfrac{\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}}{\sqrt{2\pi}}}{}{}\mathrm{e}^{\pi\mathrm{i}\sigma_{3}/4}\begin{bmatrix}1&\mathrm{i}\\\
1&-\mathrm{i}\end{bmatrix}\mathrm{e}^{\mathrm{i}\rho_{\infty}\sigma_{3}},\qquad\operatorname{Arg}(\rho_{\infty})\in(-\pi,\pi).$
(7.58)

We turn to analogously treating the final factor of the jump of
$\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$; using [38, equation
(10.27.7)] and the above relations, we have

$\displaystyle{\rho_{0}}\begin{bmatrix}I_{2\mu_{n}-1}(\rho_{0})&I_{1-2\mu_{n}}(\rho_{0})\vspace{1mm}\\\
-I_{2\mu_{n}+1}(\rho_{0})&-I_{-1-2\mu_{n}}(\rho_{0})\end{bmatrix}$
$\displaystyle\qquad=\dfrac{\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}}{2}\begin{bmatrix}H^{(1)}_{2\mu_{n}-1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}&H^{(2)}_{1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\vspace{1mm}\\\
H^{(1)}_{1+2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}&H^{(2)}_{-1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\end{bmatrix}\begin{bmatrix}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}&\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{3\pi\mathrm{i}\mu_{n}}&-\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}\end{bmatrix}.$

This allows us to find the following large-$\rho_{0}$ asymptotics:

$\displaystyle\dfrac{\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}}{2}\begin{bmatrix}H^{(1)}_{2\mu_{n}-1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}&H^{(2)}_{1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\vspace{1mm}\\\
H^{(1)}_{2\mu_{n}+1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}&H^{(2)}_{-1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\end{bmatrix}$
$\displaystyle\qquad=\begin{bmatrix}1&\dfrac{1}{2}-\dfrac{\mu_{n}}{2}-\dfrac{3}{32\mu_{n}}\vspace{1mm}\\\
-1&\dfrac{1}{2}+\dfrac{\mu_{n}}{2}+\dfrac{3}{32\mu_{n}}\end{bmatrix}\Biggl{(}\mathbb{I}-\dfrac{1}{128\rho_{0}^{2}}$
$\displaystyle\phantom{\qquad=}{}\times\begin{bmatrix}\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}&\dfrac{\big{(}16\mu_{n}^{2}-13\big{)}\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}}{48\mu_{n}}\vspace{1mm}\\\
64\mu_{n}\big{(}16\mu_{n}^{2}-1\big{)}&-\big{(}16\mu_{n}^{2}-9\big{)}\big{(}16\mu_{n}^{2}-1\big{)}\end{bmatrix}+\mathcal{O}\big{(}\rho_{0}^{-4}\big{)}\Biggr{)}\rho_{0}^{\sigma_{3}/2}$
$\displaystyle\phantom{\qquad=}{}\times\big{(}2\sqrt{\mu_{n}}\big{)}^{\mathbb{I}-\sigma_{3}}{\dfrac{\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}}{\sqrt{2\pi}}}{}{}\begin{bmatrix}1&\mathrm{i}\\\
1&-\mathrm{i}\end{bmatrix}\mathrm{e}^{\rho_{0}\sigma_{3}},\qquad\operatorname{Arg}(\rho_{0})\in\left(-\dfrac{\pi}{2},\dfrac{3\pi}{2}\right).$
(7.59)

For convenience, we introduce the notation

$\mathbf{H}_{n}(\diamond)\mathrel{\mathop{\mathchar
58\relax}}=\sqrt{\dfrac{\pi}{4\mu_{n}}}\mathrm{e}^{\pi\mathrm{i}/4}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\cdot\dfrac{\diamond}{2}\begin{bmatrix}H^{(1)}_{2\mu_{n}-1}(\diamond)&H^{(2)}_{1-2\mu_{n}}(\diamond)\vspace{1mm}\\\
H^{(1)}_{1+2\mu_{n}}(\diamond)&H^{(2)}_{-1-2\mu_{n}}(\diamond)\end{bmatrix},$
(7.60)

with a fixed determination of the square root; this choice of prefactor
guarantees that we have $\det(\mathbf{H}_{n})=1$ identically. Using the
identity [38, equation (10.11.4)], we note that $\mathbf{H}_{n}$ satisfies

$\displaystyle\mathbf{H}_{n}(\mathrm{e}^{\pi\mathrm{i}}\diamond)=\mathbf{H}_{n}(\diamond)\begin{bmatrix}0&-1\\\
1&2\cos(2\pi\mu_{n})\end{bmatrix}.$ (7.61)

We can now rewrite the jump condition (7.54) as

$\displaystyle\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)=\hat{\mathbf{R}}^{\mathrm{even/odd}}_{-}(\lambda,z)\mathbf{H}_{n}(\rho_{\infty})\begin{bmatrix}1&\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}&-1\end{bmatrix}\mathbf{V}^{\mathrm{even/odd}}$
$\displaystyle\hphantom{\hat{\mathbf{R}}^{\mathrm{even/odd}}_{+}(\lambda,z)=}{}\times\begin{bmatrix}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}&\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{3\pi\mathrm{i}\mu_{n}}&-\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}\end{bmatrix}^{-1}\mathbf{H}_{n}^{-1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}.$

Next, define

$\displaystyle\mathbf{\Omega}^{\mathrm{even/odd}}(\lambda,z)\mathrel{\mathop{\mathchar
58\relax}}=\begin{bmatrix}1&\dfrac{1}{2}-\dfrac{\mu_{n}}{2}-\dfrac{3}{32\mu_{n}}\vspace{1.5mm}\\\
-1&\dfrac{1}{2}+\dfrac{\mu_{n}}{2}+\dfrac{3}{32\mu_{n}}\end{bmatrix}^{-1}\big{(}2\sqrt{\mu_{n}}\big{)}^{\sigma_{3}}$
$\displaystyle\hphantom{\mathbf{\Omega}^{\mathrm{even/odd}}(\lambda,z)\mathrel{\mathop{\mathchar
58\relax}}=}{}\times\hat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)\begin{cases}\mathbf{H}_{n}(\rho_{\infty}),&|\lambda|>1,\\\
\mathbf{H}_{n}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)},&|\lambda|<1.\end{cases}$
(7.62)

Then, $\mathbf{\Omega}^{\mathrm{even/odd}}$ satisfies

$\displaystyle\mathbf{\Omega}^{\mathrm{even/odd}}_{+}(\lambda,z)=\mathbf{\Omega}_{-}^{\mathrm{even/odd}}(\lambda,z)\begin{bmatrix}1&\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}&-1\end{bmatrix}\cdot\mathbf{V}^{\mathrm{even/odd}}$
$\displaystyle\hphantom{\mathbf{\Omega}^{\mathrm{even/odd}}_{+}(\lambda,z)=}{}\times\begin{bmatrix}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}&\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{3\pi\mathrm{i}\mu_{n}}&-\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}\end{bmatrix}^{-1},\qquad|\lambda|=1,$
(7.63)

where the jump depends only on the parity of $n$. Furthermore, since
$\rho_{\infty}$ and $\rho_{0}$ change signs across the negative imaginary
axis, we may use (7.61) to find

$\mathbf{\Omega}_{+}^{\mathrm{even/odd}}(\lambda,z)=\mathbf{\Omega}_{-}^{\mathrm{even/odd}}(\lambda,z)\begin{bmatrix}0&-1\\\
1&2\cos(2\pi\mu_{n})\end{bmatrix},$ (7.64)

for $\lambda$ on the negative imaginary axis with $|\lambda|>1$, oriented
towards the origin and

$\mathbf{\Omega}_{+}^{\mathrm{even/odd}}(\lambda,z)=\mathbf{\Omega}_{-}^{\mathrm{even/odd}}(\lambda,z)\begin{bmatrix}0&-1\\\
1&2\cos(2\pi\mu_{n})\end{bmatrix},$ (7.65)

for $\lambda$ on the negative imaginary axis with $|\lambda|<1$, oriented away
from the origin.

It follows from Riemann–Hilbert Problem 7.9, (7.62), and (7.58) that
$\mathbf{\Omega}^{\mathrm{even/odd}}$ has the following asymptotic behavior as
$\lambda\to\infty$:

$\displaystyle\mathbf{\Omega}^{\mathrm{even/odd}}(\lambda,z)=\big{(}\mathbb{I}+\mathbf{\Xi^{\mathrm{even/odd}}}(z)\lambda^{-1}+\mathcal{O}\big{(}\lambda^{-2}\big{)}\big{)}\rho_{\infty}^{\sigma_{3}/2}{\dfrac{1}{\sqrt{2}}}\begin{bmatrix}\mathrm{i}&-1\\\
1&-\mathrm{i}\end{bmatrix}\mathrm{e}^{\mathrm{i}\rho_{\infty}\sigma_{3}},$
(7.66)

where the $\mathcal{O}\big{(}\lambda^{-2}\big{)}$ represents an asymptotic
series that is differentiable term-by-term with respect to both $\lambda$ and
$z$. Analogously, we have

$\displaystyle\mathbf{\Omega}^{\mathrm{even/odd}}(\lambda,z)=\mathbf{\Delta}^{\mathrm{even/odd}}(z)\big{(}\mathbb{I}+\mathbf{\Pi}^{\mathrm{even/odd}}(z)\lambda+\mathcal{O}\big{(}\lambda^{2}\big{)}\big{)}\rho_{0}^{\sigma_{3}/2}\mathrm{e}^{-\pi\mathrm{i}\sigma_{3}/4}$
$\displaystyle\hphantom{\mathbf{\Omega}^{\mathrm{even/odd}}(\lambda,z)=}{}\times{\dfrac{1}{\sqrt{2}}}{}{}\begin{bmatrix}\mathrm{i}&-1\\\
1&-\mathrm{i}\end{bmatrix}\mathrm{e}^{\rho_{0}\sigma_{3}},\qquad\lambda\to 0,$
(7.67)

where $\mathcal{O}\big{(}\lambda^{2}\big{)}$ represents an asymptotic series
at the origin $\lambda=0$ which is similarly term-by-term differentiable.
Notice that we can now relate the limiting formula from Lemma 7.10 to
$\mathbf{\Omega}^{\mathrm{even/odd}}$ using definitions (7.62) and (7.60) to
find that

$\displaystyle\hat{R}^{\mathrm{even/odd}}_{11}(0,z)+\hat{R}^{\mathrm{even/odd}}_{21}(0,z)-\hat{R}^{\mathrm{even/odd}}_{12}(0,z)-\hat{R}^{\mathrm{even/odd}}_{22}(0,z)$
$\displaystyle\qquad=\sqrt{\frac{\pi}{4\mu_{n}}}\mathrm{e}^{\pi\mathrm{i}/4}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\frac{\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}}{2}\big{[}\big{(}H^{(2)}_{-1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}+H^{(2)}_{1-2\mu_{n}}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\big{)}\Omega^{\mathrm{even/odd}}_{21}(\lambda,z)$
$\displaystyle\phantom{\qquad=}{}-\big{(}H^{(1)}_{2\mu_{n}-1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}+H^{(1)}_{2\mu_{n}+1}\big{(}\mathrm{e}^{-\pi\mathrm{i}/2}\rho_{0}\big{)}\big{)}\Omega^{\mathrm{even/odd}}_{22}(\lambda,z)\big{]}_{\lambda=0}.$

Then, using (7.67) and (7.59) yields

$\displaystyle\mathop{\lim_{n\to\infty}}_{\text{$n$
even/odd}}u_{n}\big{(}n^{-1}z\big{)}=U^{\mathrm{even/odd}}(z)\mathrel{\mathop{\mathchar
58\relax}}=-\frac{1}{2z\Delta^{\mathrm{even/odd}}_{21}(z)^{2}}.$

To extract the monodromy parameters of $U(z)$ from
$\mathbf{\Omega}(\lambda,z)$, we notice that it solves Riemann–Hilbert Problem
5.1 with

$t_{1}^{\infty}=t_{0}^{0}=-2\cos(2\pi\mu_{n})=-\left(e_{1}^{2}+\frac{1}{e_{1}^{2}}\right),$
(7.68)

and

$\mathbf{C}_{0\infty}=\begin{bmatrix}1&\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{2\pi\mathrm{i}\mu_{n}}&-1\end{bmatrix}\cdot\mathbf{V}^{\mathrm{even/odd}}\cdot\begin{bmatrix}\mathrm{e}^{\pi\mathrm{i}\mu_{n}}&\mathrm{e}^{\pi\mathrm{i}\mu_{n}}\\\
-\mathrm{e}^{3\pi\mathrm{i}\mu_{n}}&-\mathrm{e}^{-\pi\mathrm{i}\mu_{n}}\end{bmatrix}^{-1}.$
(7.69)

Since $\mu_{n}$ and $\mathbf{V}^{\mathrm{even/odd}}$ depend only on the parity
of $n$ (see (6.8), (7.53), respectively), and
${e_{1}=e_{1,n}=\mathrm{e}^{\pi\mathrm{i}\mu_{n}}}$ (see Remark 6.3), it
follows that (7.68)–(7.69) depend only on the parity of $n$. Recalling the
formulæ for $y_{i}$ in Section 5.4, one immediately arrives at formulæ
(1.20)–(1.22).

##  8 Small $\boldsymbol{x}$ asymptotics and proof of Proposition 1.5

Inspired by [33], see also [22, Theorem 3.2], the goal of this section is to
compute the asymptotics as $x\to 0$ of the Bäcklund iterates $u_{n}(x)$ for
fixed $n$ and, by evaluating at $n=0$, arriving at the asymptotic behavior of
a generic solution of PIII($D_{6}$) in this limit. Observe that the matrices
$\mathbf{\Xi}^{(6)}_{n}(x)$ and $\mathbf{\Delta}^{(6)}_{n}(x)$ defined in
(6.5) and (6.4), respectively, can be expressed in terms of
$\mathbf{Q}_{n}(\lambda,x)$ as follows:

$\mathbf{\Xi}^{(6)}_{n}(x)=\mathbf{A}_{n}(x)+\lim_{\lambda\to\infty}\lambda(\mathbf{Q}_{n}(\lambda,x)-\mathbb{I})-\frac{1}{2}\mathrm{i}x\sigma_{3},\qquad\mathbf{\Delta}^{(6)}_{n}(x)=\mathbf{Q}_{n}(0,x).$

Using (6.6) then gives

$u_{n}(x)=\frac{\displaystyle-\mathrm{i}A_{n,12}(x)-\mathrm{i}\lim_{\lambda\to\infty}\lambda
Q_{n,12}(\lambda,x)}{Q_{n,11}(0,x)Q_{n,12}(0,x)}.$ (8.1)

Also, combining (7.18) and (7.21), we have

$A_{n,12}(x)=\frac{\mathrm{i}\mu_{n}\big{(}1-a_{n}^{2}\big{)}}{\gamma_{n}x^{2-2\kappa_{n}}}$
(8.2)

in which all $x$-dependence is explicit. To analyze $u_{n}(x)$ for small $x$
it therefore remains to obtain asymptotics of $\mathbf{Q}_{n}(\lambda,x)$ as
$x\to 0$. To do this, let $\mathbf{V}_{\mathbf{Q}}$ denote the jump matrix
(7.40). Then, denoting the off-diagonal constant matrix

$\mathbf{O}_{n}\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{J}_{n}e_{2}^{\sigma_{3}}\begin{bmatrix}0&-\beta_{n}\\\
\beta_{n}^{-1}&0\end{bmatrix}\mathbf{J}_{-n}^{-1},$ (8.3)

we arrive at

$\displaystyle\mathbf{V}_{\mathbf{Q}}(\lambda,x)=\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})\mathbf{O}_{n}\mathbf{M}\bigl{(}-\mathrm{i}x\lambda^{-1};\kappa_{-n},\mu_{n}\bigr{)}^{-1}$
$\displaystyle\hphantom{\mathbf{V}_{\mathbf{Q}}(\lambda,x)=}{}\times\mathbf{D}_{-n}^{-1}x^{-\kappa_{-n}\sigma_{3}}\left(-\frac{1}{\lambda}\right)_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-\sigma_{3}/2}e_{0}^{-3\sigma_{3}},\qquad|\lambda|=1,$

where matrices $\mathbf{D}_{n}$, $\mathbf{M}$ are defined in (7.24) and
(7.31), respectively. To consider the limit $x\to 0$, we start with the
Whittaker matrix $\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})$. Using
[38, equation (13.14.6)], we obtain

$M_{\kappa,\mu}(z)=z^{\frac{1}{2}+\mu}\left(1-\frac{\kappa}{1+2\mu}z+\mathcal{O}\big{(}z^{2}\big{)}\right),\qquad
z\to 0.$

Therefore, using the definition (7.31) of $\mathbf{M}(Z;\kappa,\mu)$ we have

$\displaystyle(-\mathrm{i}x\lambda)^{\sigma_{3}/2}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})(-\mathrm{i}x\lambda)^{-\mu_{n}\sigma_{3}}$
$\displaystyle\qquad=\begin{bmatrix}\kappa_{n}-\frac{1}{2}-\mu_{n}&\kappa_{n}-\frac{1}{2}+\mu_{n}\\\\[5.69046pt]
1&1\end{bmatrix}+\mathcal{O}(\lambda x)=\mathrel{\mathop{\mathchar
58\relax}}\mathbf{M}_{\infty}(\lambda x),$

in the limit $x\to 0$ uniformly for $|\lambda|=1$. Similarly, in the same
limit,

$\displaystyle\big{(}\mathrm{i}x\lambda^{-1}\big{)}^{\sigma_{3}/2}\mathbf{M}\bigl{(}-\mathrm{i}x\lambda^{-1};\kappa_{-n},\mu_{-n}\bigr{)}\big{(}\mathrm{i}x\lambda^{-1}\big{)}^{-\mu_{-n}\sigma_{3}}$
$\displaystyle\qquad=\begin{bmatrix}\kappa_{-n}-\frac{1}{2}-\mu_{-n}&\kappa_{-n}-\frac{1}{2}+\mu_{-n}\\\
1&1\end{bmatrix}+\mathcal{O}\big{(}x\lambda^{-1}\big{)}$
$\displaystyle\qquad=\mathrel{\mathop{\mathchar
58\relax}}\mathbf{M}_{0}\big{(}x\lambda^{-1}\big{)}=\mathbf{M}_{0}(0)+\mathcal{O}\big{(}x\lambda^{-1}\big{)}.$

For $x>0$, the functions
$\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{p}$,
$\bigl{(}-\lambda^{-1}\bigr{)}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{p}$,
$(-\mathrm{i}x\lambda)^{p}$, and $\big{(}\mathrm{i}x\lambda^{-1}\big{)}^{p}$
(the latter two being principal branches) all have the same branch cut, namely
$\mathrm{i}\mathbb{R}_{-}$. One has the following identities:

$(-\mathrm{i}x\lambda)^{p}=\mathrm{e}^{-\mathrm{i}\pi
p/2}x^{p}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{p},\qquad\big{(}\mathrm{i}x\lambda^{-1}\big{)}^{p}=\mathrm{e}^{\mathrm{i}\pi
p/2}x^{p}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-p},\qquad\big{(}-\lambda^{-1}\big{)}^{p}_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}=\mathrm{e}^{\mathrm{i}\pi
p}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-p}.$

It follows that

$\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\sigma_{3}/2}x^{\kappa_{n}\sigma_{3}}\mathbf{D}_{n}\mathbf{M}(\mathrm{i}x\lambda;\kappa_{n},\mu_{n})=x^{(\kappa_{n}-\frac{1}{2})\sigma_{3}}\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/4}\mathbf{D}_{n}\mathbf{M}_{\infty}(\lambda
x)x^{\mu_{n}\sigma_{3}}\mathrm{e}^{-\mathrm{i}\pi\mu_{n}\sigma_{3}/2}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\mu_{n}\sigma_{3}},$

and

$\displaystyle\mathbf{M}\bigl{(}-\mathrm{i}x\lambda^{-1};\kappa_{-n},\mu_{-n}\bigr{)}^{-1}\mathbf{D}_{-n}^{-1}x^{-\kappa_{-n}\sigma_{3}}\left(-\frac{1}{\lambda}\right)_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{-\sigma_{3}/2}$
$\displaystyle\qquad=\mathrm{e}^{-\mathrm{i}\pi\mu_{-n}\sigma_{3}/2}x^{-\mu_{-n}\sigma_{3}}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\mu_{-n}\sigma_{3}}\mathbf{M}_{0}\big{(}x\lambda^{-1}\big{)}^{-1}\mathrm{e}^{-\mathrm{i}\pi\sigma_{3}/4}\mathbf{D}_{-n}^{-1}x^{(\frac{1}{2}-\kappa_{-n})\sigma_{3}}.$

Because $\mathbf{O}_{n}$ is off-diagonal, the central factors in
$\mathbf{V}_{\mathbf{Q}}(\lambda,x)$ simplify as follows:

$x^{\mu_{n}\sigma_{3}}\mathrm{e}^{-\frac{1}{2}\mathrm{i}\pi\mu_{n}\sigma_{3}}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\mu_{n}\sigma_{3}}\mathbf{O}_{n}\mathrm{e}^{-\frac{1}{2}\mathrm{i}\pi\mu_{n}\sigma_{3}}x^{-\mu_{n}\sigma_{3}}\lambda_{\vbox{\hbox{\includegraphics[width=4.20004pt]{Figures/squarell.pdf}}}}^{\mu_{n}\sigma_{3}}=\mathbf{O}_{n}x^{-2\mu_{n}\sigma_{3}}.$

Consequently, we have

$\displaystyle\mathbf{V}_{\mathbf{Q}}(\lambda,x)=x^{(\kappa_{n}-\frac{1}{2})\sigma_{3}}\mathrm{e}^{\mathrm{i}\pi\sigma_{3}/4}\mathbf{D}_{n}{}\cdot\mathbf{M}_{\infty}(\lambda
x)\cdot\mathbf{O}_{n}x^{-2\mu_{n}\sigma_{3}}$
$\displaystyle\hphantom{\mathbf{V}_{\mathbf{Q}}(\lambda,x)=}{}\times\mathbf{M}_{0}\big{(}x\lambda^{-1}\big{)}^{-1}\mathrm{e}^{-\mathrm{i}\pi\sigma_{3}/4}\mathbf{D}_{-n}^{-1}x^{(\frac{1}{2}-\kappa_{-n})\sigma_{3}}e_{0}^{-3\sigma_{3}}.$
(8.4)

The matrix $\mathbf{V}_{\mathbf{Q}}(\lambda,x)$ does not possess a finite
limit as $x\to 0$ due to the factors $x^{(\kappa_{n}-\frac{1}{2})\sigma_{3}}$,
$x^{(\frac{1}{2}-\kappa_{-n})\sigma_{3}}$; this can be handled by introducing
the following transformation. Let $\varsigma\mathrel{\mathop{\mathchar
58\relax}}=\lambda x$ and

$\displaystyle\widetilde{\mathbf{Q}}_{n}(\varsigma,x)\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{D}_{n}^{-1}\mathrm{e}^{-\frac{1}{4}\mathrm{i}\pi\sigma_{3}}x^{(\frac{1}{2}-\kappa_{n})\sigma_{3}}$
$\displaystyle\quad\times\begin{cases}\mathbf{Q}_{n}(\frac{\varsigma}{x},x)x^{(-\frac{1}{2}+\kappa_{n})\sigma_{3}}\mathrm{e}^{\frac{1}{4}\mathrm{i}\pi\sigma_{3}}\mathbf{D}_{n},&|\varsigma|>1,\\\
\mathbf{Q}_{n}(\frac{\varsigma}{x},x)\mathbf{V}_{\mathbf{Q}}(\frac{\varsigma}{x},x)e_{0}^{3\sigma_{3}}x^{(\kappa_{-n}-\frac{1}{2})\sigma_{3}}\mathbf{D}_{-n}\mathrm{e}^{\frac{1}{4}\mathrm{i}\pi\sigma_{3}}\\\
\quad\times\mathbf{M}_{0}(0)x^{2\mu_{n}\sigma_{3}}\mathbf{O}_{n}^{-1}\mathbf{M}_{\infty}(\varsigma)^{-1},&|x|<|\varsigma|<1,\\\
\mathbf{Q}_{n}(\frac{\varsigma}{x},x)e_{0}^{3\sigma_{3}}x^{(\kappa_{-n}-\frac{1}{2})\sigma_{3}}\mathbf{D}_{-n}\mathrm{e}^{\frac{1}{4}\mathrm{i}\pi\sigma_{3}}\mathbf{M}_{0}(0)x^{2\mu_{n}\sigma_{3}}\mathbf{O}_{n}^{-1}\mathbf{M}_{\infty}(\varsigma)^{-1},&|\varsigma|<|x|.\end{cases}$
(8.5)

It follows that $\widetilde{\mathbf{Q}}_{n}$ is analytic as a function of
$\varsigma$ for $\varsigma\in{\mathbb{C}}\setminus\\{|\varsigma|=1\\}$ and
satisfies

$\lim_{\varsigma\to\infty}\widetilde{\mathbf{Q}}_{n}(\varsigma,x)=\mathbb{I}.$

Furthermore, on the circle $|\varsigma|=1$, the jump condition
$\widetilde{\mathbf{Q}}_{n,+}(\varsigma,x)=\widetilde{\mathbf{Q}}_{n,-}(\varsigma,x)\mathbf{V}_{\widetilde{\mathbf{Q}}}(\varsigma,x)$
holds, where the jump contour has counterclockwise orientation, and

$\displaystyle\mathbf{V}_{\widetilde{\mathbf{Q}}}(\varsigma,x)\mathrel{\mathop{\mathchar
58\relax}}=\mathbf{D}_{n}^{-1}\mathrm{e}^{-\pi\mathrm{i}\sigma_{3}/4}x^{(\frac{1}{2}-\kappa_{n})\sigma_{3}}\mathbf{V}_{\mathbf{Q}}\left(\frac{\varsigma}{x},x\right)e_{0}^{3\sigma_{3}}x^{(\kappa_{-n}-\frac{1}{2})\sigma_{3}}\mathrm{e}^{\pi\mathrm{i}\sigma_{3}/4}$
$\displaystyle\hphantom{\mathbf{V}_{\widetilde{\mathbf{Q}}}(\varsigma,x)\mathrel{\mathop{\mathchar
58\relax}}=}{}\times\mathbf{D}_{-n}\mathbf{M}_{0}(0)x^{2\mu_{n}\sigma_{3}}\mathbf{O}_{n}^{-1}\mathbf{M}_{\infty}^{-1}(\varsigma).$

Using (8.4) immediately yields

$\mathbf{V}_{\widetilde{\mathbf{Q}}}(\varsigma,x)=\mathbf{M}_{\infty}(\varsigma)\mathbf{O}_{n}x^{-2\mu_{n}\sigma_{3}}\mathbf{M}_{0}\big{(}{x^{2}}\varsigma^{-1}\big{)}^{-1}\mathbf{M}_{0}(0)x^{2\mu_{n}\sigma_{3}}\mathbf{O}_{n}^{-1}\mathbf{M}_{\infty}(\varsigma)^{-1}.$

Therefore, $\widetilde{\mathbf{Q}}_{n}(\varsigma,x)$ solves the following
Riemann–Hilbert problem.

######  Riemann–Hilbert Problem 8.1.

Fix generic monodromy parameters $(e_{1},e_{2})$, and $x\in\mathbb{C}$. Seek a
$2\times 2$ matrix function
$\varsigma\mapsto\widetilde{\mathbf{Q}}_{n}(\varsigma,x)$ with the following
properties:

  * •

Analyticity: $\widetilde{\mathbf{Q}}_{n}(\varsigma,x)$ is an analytic function
of $\varsigma$ for $|\varsigma|\neq 1$.

  * •

Jump condition: $\widetilde{\mathbf{Q}}_{n}(\varsigma,x)$ takes analytic
boundary values on the unit circle from the interior and exterior, denoted
$\widetilde{\mathbf{Q}}_{n,+}(\varsigma,x)$ and
$\widetilde{\mathbf{Q}}_{n,-}(\varsigma,x)$ for $|\varsigma|=1$ respectively,
and they are related by

$\widetilde{\mathbf{Q}}_{n,+}(\varsigma,x)=\widetilde{\mathbf{Q}}_{n,-}(\varsigma,x)\mathbf{M}_{\infty}(\varsigma)\mathbf{O}_{n}x^{-2\mu_{n}\sigma_{3}}\mathbf{M}_{0}\big{(}{x^{2}}\varsigma^{-1}\big{)}^{-1}\mathbf{M}_{0}(0)x^{2\mu_{n}\sigma_{3}}\mathbf{O}_{n}^{-1}\mathbf{M}_{\infty}(\varsigma)^{-1}.$

  * •

Normalization: $\widetilde{\mathbf{Q}}_{n}(\varsigma,x)\to\mathbb{I}$ as
$\varsigma\to\infty$.

The jump $\mathbf{V}_{\widetilde{\mathbf{Q}}}$ has a limit as $x\to 0$,
uniformly for $|\lambda|=1$ for $|{\operatorname{Re}}\,\mu_{n}|<\frac{1}{2}$,
and satisfies the estimate

$\mathbf{V}_{\widetilde{\mathbf{Q}}}(\varsigma,x)=\mathbb{I}+\mathcal{O}\big{(}x^{2-|4\operatorname{Re}\mu_{n}|}\big{)}.$

By the standard theory of small-norm Riemann–Hilbert problems, we arrive at

$\widetilde{\mathbf{Q}}_{n}(\varsigma,x)=\mathbb{I}+\mathcal{O}\big{(}x^{2-|4\operatorname{Re}\mu_{n}|}\big{)}\qquad\text{as}\quad
x\to 0$

uniformly for $\varsigma$ sufficiently small, and

$\lim_{\varsigma\to\infty}\varsigma\big{(}\widetilde{\mathbf{Q}}_{n}(\varsigma,x)-\mathbb{I}\big{)}=\lim_{\varsigma\to\infty}\varsigma\big{(}\widetilde{\mathbf{Q}}_{n}(\varsigma,0)-\mathbb{I}\big{)}+\mathcal{O}\big{(}x^{2-|4\operatorname{Re}\mu_{n}|}\big{)}=\mathcal{O}\big{(}x^{2-|4\operatorname{Re}\mu_{n}|}\big{)}.$

We can now use the above estimate and expressions (8.1) and (8.2) to compute
the asymptotic behavior of $u_{n}(x)$ as $x\to 0$. To this end, note that by
(8.5) and the definition of $\mathbf{D}_{n}$ and $\varsigma$,

$\lim_{\lambda\to\infty}\lambda
Q_{n,12}(\lambda,x)=-\frac{1}{\mu_{n}\gamma_{n}}x^{2\kappa_{n}-2}\lim_{\varsigma\to\infty}\varsigma\widetilde{Q}_{n,12}(\varsigma,x),\qquad|\varsigma|>1,$

and so

$\lim_{\lambda\to\infty}\lambda
Q_{n,12}(\lambda,x)=\mathcal{O}\big{(}x^{2\kappa_{n}-|4\operatorname{Re}\mu_{n}|}\big{)},\qquad
x\to 0.$ (8.6)

Likewise, (8.5) gives

$\displaystyle Q_{n,11}(0,x)Q_{n,12}(0,x)$
$\displaystyle\qquad=-\mathrm{i}\frac{D_{n,11}^{2}}{4\mu_{n}^{2}D_{-n,11}D_{-n,22}}\begin{cases}{O_{n,21}^{2}\big{(}\kappa_{n}-\tfrac{1}{2}+\mu_{n}\big{)}^{2}\big{(}\kappa_{-n}-\tfrac{1}{2}+\mu_{n}\big{)}}{}x^{2\kappa_{n}-1-4\mu_{n}}\\\
\quad{}+\mathcal{O}\big{(}x^{2\kappa_{n}-|4\operatorname{Re}\mu_{n}|}\big{)},\quad\operatorname{Re}\mu_{n}>0,\\\
{O_{n,12}^{2}\big{(}\kappa_{n}-\tfrac{1}{2}-\mu_{n}\big{)}^{2}\big{(}\kappa_{-n}-\tfrac{1}{2}-\mu_{n}\big{)}}{}x^{2\kappa_{n}-1+4\mu_{n}}\\\
\quad{}+\mathcal{O}\big{(}x^{2\kappa_{n}-|4\operatorname{Re}\mu_{n}|}\big{)},\quad\operatorname{Re}\mu_{n}<0.\end{cases}$
(8.7)

Using (8.6), (8.7) in (8.1) yields

$u_{n}(x)=\dfrac{4\mathrm{i}\mu_{n}^{3}\big{(}1-a_{n}^{2}\big{)}D_{-n,11}D_{-n,22}}{\gamma_{n}D_{n,11}^{2}O_{n,21}^{2}\big{(}\kappa_{n}-\tfrac{1}{2}+\mu_{n}\big{)}^{2}\big{(}\kappa_{-n}-\tfrac{1}{2}+\mu_{n}\big{)}}x^{4\mu_{n}-1}\big{(}1+\mathcal{O}\big{(}x^{\delta}\big{)}\big{)}\qquad\text{as}\quad
x\to 0,$

when $\operatorname{Re}\mu_{n}>0$ and

$\displaystyle
u_{n}(x)=\dfrac{4\mathrm{i}\mu_{n}^{3}\big{(}1-a_{n}^{2}\big{)}D_{-n,11}D_{-n,22}}{\gamma_{n}D_{n,11}^{2}O_{n,12}^{2}\big{(}\kappa_{n}-\tfrac{1}{2}-\mu_{n}\big{)}^{2}\big{(}\kappa_{-n}-\tfrac{1}{2}-\mu_{n}\big{)}}x^{-4\mu_{n}-1}\big{(}1+\mathcal{O}\big{(}x^{\delta}\big{)}\big{)}\qquad\text{as}\quad
x\to 0,$

when $\operatorname{Re}\mu_{n}<0$, where
$\delta=\min(1,2-|4\operatorname{Re}(\mu_{n})|)$ in both cases. Using (8.3),
(7.39), (7.35), (7.27), (7.24), and (7.10) gives the expression999The case
$\operatorname{Re}(\mu_{n})=0$ can be treated similarly, and produces a
leading term that is a combination of both leading terms, which we omit for
brevity.

$\displaystyle
u_{n}(x)=-\frac{\Gamma(1-2\epsilon_{n}\mu_{n})^{2}\Gamma\bigl{(}-\frac{n}{2}+\epsilon_{n}\mu_{n}-\frac{\Theta_{0}}{2}\bigr{)}\Gamma\big{(}\frac{n}{2}+\epsilon_{n}\mu_{n}-\frac{\Theta_{\infty}}{2}+1\big{)}}{\Gamma(2\epsilon_{n}\mu_{n})^{2}\Gamma\bigl{(}-\frac{n}{2}-\epsilon_{n}\mu_{n}-\frac{\Theta_{0}}{2}+1\bigr{)}\Gamma\big{(}\frac{n}{2}-\epsilon_{n}\mu_{n}-\frac{\Theta_{\infty}}{2}+1\big{)}}x^{4\epsilon_{n}\mu_{n}-1}$
$\displaystyle\hphantom{u_{n}(x)=}{}\times\big{(}1+\mathcal{O}(x^{\delta})\big{)}\begin{cases}\dfrac{e_{0}^{2}e_{2}^{2}e_{\infty}^{2}\big{(}e_{0}^{2}-e_{1}^{2}\big{)}\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}}{\big{(}e_{0}^{2}e_{1}^{2}-1\big{)}^{2}},&\operatorname{Re}\mu_{n}>0,\\\
\dfrac{\big{(}e_{0}^{2}e_{1}^{2}-1\big{)}^{2}}{e_{0}^{2}e_{2}^{2}e_{\infty}^{2}\big{(}e_{0}^{2}-e_{1}^{2}\big{)}\big{(}e_{1}^{2}-e_{\infty}^{2}\big{)}},&\operatorname{Re}\mu_{n}<0,\end{cases}$
(8.8)

where $\epsilon_{n}=\operatorname{sgn}(\operatorname{Re}\mu_{n})$. The
concerned reader may note that the leading coefficient in (8.8) is finite due
to the genericity conditions on $(e_{1},e_{2})$ (see the beginning of Section
7). Indeed, assumption (i) guarantees that $2\mu_{n}\not\in{\mathbb{Z}}$,
condition (ii) requires $e_{1}e_{2}\neq 0$, and condition (iii) guarantees
that

$\dfrac{n}{2}\pm\mu_{n}+\dfrac{\Theta_{0}}{2}\not\in{\mathbb{Z}}\qquad\text{and}\qquad\dfrac{n}{2}\pm\mu_{n}+\dfrac{\Theta_{\infty}}{2}\not\in{\mathbb{Z}}.$

Evaluating the above at $n=0$ yields (1.24) and finishes the proof of
Proposition 1.5.

One notable application of this is to the family of rational solutions of
Painlevé-III already discussed at the end of Section 4.2. This corresponds to
the choice $m=\Theta_{0}=\Theta_{\infty}-1$ and ${\mu_{0}=1/4}$. It follows
from (8.8) that $u_{n}(x;m)$ has a well-defined value at $x=0$ which is given
by (2.20), (2.21) in the case where $n$ is even or odd, respectively. We can
verify that these values are consistent with (8.8) by noting that $e_{1}$,
$e_{2}$, $e_{0}^{2}$, $e_{\infty}^{2}$ are invariant under an even increment
$n\mapsto n+2$, and so we have the general formulæ

$\displaystyle\dfrac{u_{2k+2}(0)}{u_{2k}(0)}$
$\displaystyle=\frac{(2k+2\mu_{2k+2}+\Theta_{0})(2k+2\mu_{2k+2}+1-\Theta_{\infty})}{(2+2k-2\mu_{2k}+\Theta_{0})(2+2k-2\mu_{2k}-\Theta_{\infty})},$
$\displaystyle\dfrac{u_{2k+1}(0)}{u_{2k-1}(0)}$
$\displaystyle=\frac{(2k-1+2\mu_{2k+1}+\Theta_{0})(2k+1+2\mu_{2k+1}-\Theta_{\infty})}{(2k+1-2\mu_{2k-1}+\Theta_{0})(+2k+1-2\mu_{2k-1}-\Theta_{\infty})}.$

Plugging in the specialized values of the parameters and using the known
values of $u_{0}(0;m)$, $u_{1}(0;m)$ yields the equality of the expression in
(8.8) with the product formulæ (2.20)–(2.21).

##  9 Alternative Riemann–Hilbert problem for Painlevé-
III($\boldsymbol{D_{8}}$)

###  9.1 Fabry-type transformation and existence of
$\boldsymbol{\widehat{R}^{\mathrm{even/odd}}(\lambda,z)}$

The Lax pair (5.25) is unusual in that its coefficient matrices have non-
diagonalizable leading terms at both of its singular points $\lambda=0$ and
$\lambda=\infty$, i.e., the coefficients of $\lambda^{0}$ and $\lambda^{-2}$
in (5.24) are not diagonalizable. To deduce the existence of the matrix
functions ${\mathbf{\Omega}}^{\mathrm{even/odd}}(\lambda,z)$ and
$\widehat{\mathbf{R}}^{\mathrm{even/odd}}(\lambda,z)$, we identify this Lax
pair with ones appearing in the literature by considering the following Fabry-
type transformation

$\displaystyle\mathbf{S}(\xi,z)\mathrel{\mathop{\mathchar
58\relax}}={\dfrac{1}{\sqrt{2}}}\begin{bmatrix}-\mathrm{i}&1\\\
-1&\mathrm{i}\end{bmatrix}(2z)^{-\sigma_{3}/4}\xi^{-\sigma_{3}/2}$
$\displaystyle\hphantom{\mathbf{S}(\xi,z)\mathrel{\mathop{\mathchar
58\relax}}=}{}\times\begin{cases}\mathbf{\Omega}\big{(}\xi^{2}\mathrm{e}^{\frac{\mathrm{i}\pi}{2}},z\big{)},&-\frac{\pi}{2}<\operatorname{Arg}(\xi)<\frac{\pi}{2},\\\
\mathbf{\Omega}\big{(}\xi^{2}\mathrm{e}^{-\frac{3\pi\mathrm{i}}{2}},z\big{)}(-\mathrm{i}\sigma_{2}),&\frac{\pi}{2}<\operatorname{Arg}(\xi)<\pi,\\\
\mathbf{\Omega}\big{(}\xi^{2}\mathrm{e}^{\frac{5\pi\mathrm{i}}{2}},z\big{)}\mathrm{i}\sigma_{2},&-\pi<\operatorname{Arg}(\xi)<-\frac{\pi}{2},\end{cases}$
(9.1)

when $|\xi|>1$, and

$\displaystyle\mathbf{S}(\xi,z)\mathrel{\mathop{\mathchar
58\relax}}={\dfrac{1}{\sqrt{2}}}\begin{bmatrix}-\mathrm{i}&1\\\