# Large n-limit for Random matrices with External Source with three distinct
eigenvalues

Jian Xu  College of Science,
University of Shanghai for Science and Technology,
Shanghai 200093
People’s Republic of China<EMAIL_ADDRESS>or<EMAIL_ADDRESS>,  Engui
Fan  School of Mathematical Sciences, Institute of Mathematics and Key
Laboratory of Mathematics for Nonlinear Science
Fudan University
Shanghai 200433
People’s Republic of China  correspondence<EMAIL_ADDRESS>and  Yang
Chen  Department of Mathematics, University of Macau, Macau, P. R. China
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we analyze the large n-limit for random matrix with external
source with three distinct eigenvalues. And we confine ourselves in the
Hermite case and the three distinct eigenvalues are $-a,0,a$. For the case
$a^{2}>3$, we establish the universal behavior of local eigenvalue
correlations in the limit $n\rightarrow\infty$, which is known from unitarily
invariant random matrix models. Thus, local eigenvalue correlations are
expressed in terms of the sine kernel in the bulk and in terms of the Airy
kernel at the edge of the spectrum. The result can be obtained by analyzing
$4\times 4$ Riemann-Hilbert problem via nonlinear steepest decent method.

###### Key words and phrases:

Riemann-Hilbert problem, Random matrices, Multiple Orthogonal Polynomials,
Large N limit

##  1\. Introduction and Statement of Results

We will consider the ensemble of $n\times n$ Hermitian matrices $M$ with the
density function defined by

$\frac{1}{Z_{n}}e^{-nTr(V(M)-AM)}dM,$ (1.1)

where $V(M)$ is a matrix polynomial with respect to $M$, The number $n$ is a
large parameter in the ensemble, $A$ is a $n\times n$ matrix, and $Z_{n}$ is
the normalization constant,

$Z_{n}=\int{e^{-Tr(V(M)-AM)}dM},$ (1.2)

where the integral is over all $n\times n$ Hermitian matrices. When $A=0$, the
above ensemble is the standard unitary-invariant ensemble in the theory of
random matrices [1]. The case when $A\neq 0$ is called a random matrix model
with external source, and has been studied in, for example, [2, 3, 4, 5, 6, 7,
8, 9, 10]. In a series of paper [2, 3, 4], Bleher and Kuijlaars considered the
large n limit of this model when external source $A$ has two distinct
eigenvalues via nonlinear steepest descent method based on a $3\times 3$
Riemann-Hilbert problem.

In this paper, we consider the case external source $A$ has three distinct
eigenvalues, say $\\{a_{1},a_{2},a_{3}\\}$. It was shown that when $A$ has
three distinct eigenvalues, the correlation kernel function $K_{n}(x,y)$ is
linked to a $4\times 4$ Riemann-Hilbert problem: finding
$Y:{\mathbb{C}}\backslash{\mathbb{R}}\rightarrow{\mathbb{C}}^{4\times 4}$ such
that

  1. (i)

$Y$ is analytic on ${\mathbb{C}}\backslash{\mathbb{R}}$,

  2. (ii)

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

$Y_{+}(x)=Y_{-}(x)\left(\begin{array}[]{cccc}1&\omega_{1}(x)&\omega_{2}(x)&\omega_{3}(x)\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right),$ (1.3)

where $Y_{+}(x)$ and $Y_{-}(x)$ denote the limit of $Y(z)$ as $z\rightarrow x$
from upper and lower half plane,respectively.

  3. (iii)

as $z\rightarrow\infty$, we have

$Y(z)=\left(\mathbb{I}+O(\frac{1}{z})\right)diag{(z^{n},z^{-n_{1}},z^{-n_{2}},z^{-n_{3}})},$
(1.4)

where $\mathbb{I}$ denotes the $4\times 4$ identity matrix.

We consider correlation kernel

$K_{n}(x,y)=\frac{e^{-\frac{1}{2}(V(x)+V(y))}}{2\pi
i(x-y)}(\begin{array}[]{cccc}0&e^{na_{1}y}&e^{na_{2}y}&e^{na_{3}y}\end{array})Y^{-1}(y)Y(x)\left(\begin{array}[]{c}1\\\
0\\\ 0\\\ 0\end{array}\right),$ (1.5)

and assume that the external source $A$ is a fixed diagonal matrix with
$n_{1}$ eigenvalues $a_{1}=a$, $n_{2}$ eigenvalues $a_{2}=0$ and $n_{3}$
eigenvalues $a_{3}=-a$. It is the aim of this paper to analyze the Riemann-
Hilbert problem as $n\rightarrow\infty$, by using the method of nonlinear
steepest descent of Deift and Zhou [11]. And we focus here on Gaussian case
$V(x)=\frac{1}{2}x^{2}$. The main results of this paper are

######  Theorem 1.1.

Let $V(M)=\frac{1}{2}M^{2}$, $n_{1}=n_{3}$, $n_{2}=t$, $a^{2}>3$. Then the
limiting mean density of eigenvalues

$\rho(z)=\lim_{n\rightarrow\infty}\frac{1}{n}K_{n}(x,x),$ (1.6)

exists, and it is supported by three intervals,
$[-z_{3},-z_{2}],[-z_{1},z_{1}]$ and $[z_{2},z_{3}]$.

######  Theorem 1.2.

For every $x_{0}\in(-z_{3},-z_{2})\cup(-z_{1},z_{1})\cup(z_{2},z_{3})$ and
$u,v\in{\mathbb{R}}$, we have

$\lim_{n\rightarrow\infty}\frac{1}{n\rho(x_{0})}\hat{K}_{n}\left(x_{0}+\frac{u}{n\rho(x_{0})},x_{0}+\frac{v}{n\rho(x_{0})}\right)=\frac{sin\pi(u-v)}{\pi(u-v)}.$
(1.7)

######  Theorem 1.3.

We use the same notation as in Theorem 1.2. For every $u,v\in{\mathbb{R}}$, we
have

$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{1}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{3}+\frac{u}{(\rho_{1}n)^{\frac{2}{3}}},z_{3}+\frac{v}{(\rho_{1}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(1.8)
$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{2}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{2}+\frac{u}{(\rho_{2}n)^{\frac{2}{3}}},z_{2}+\frac{v}{(\rho_{2}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(1.9)
$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{3}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{1}+\frac{u}{(\rho_{3}n)^{\frac{2}{3}}},z_{1}+\frac{v}{(\rho_{3}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(1.10)

Similar limits hold near the edge points $-z_{3},-z_{2},-z_{1}$.

##  2\. Differential Equations

In [15], it was shown that the Riemann-Hilbert problem of $Y$ has a unique
solution

$Y=\left(\begin{array}[]{cccc}P_{n_{1},n_{2},n_{3}}&C(P_{n_{1},n_{2},n_{3}}\omega_{1})&C(P_{n_{1},n_{2},n_{3}}\omega_{2})&C(P_{n_{1},n_{2},n_{3}}\omega_{3})\\\
c_{1}P_{n_{1}-1,n_{2},n_{3}}&c_{1}C(P_{n_{1}-1,n_{2},n_{3}}\omega_{1})&c_{1}C(P_{n_{1}-1,n_{2},n_{3}}\omega_{2})&c_{1}C(P_{n_{1}-1,n_{2},n_{3}}\omega_{3})\\\
c_{2}P_{n_{1},n_{2}-1,n_{3}}&c_{2}C(P_{n_{1},n_{2}-1,n_{3}}\omega_{1})&c_{2}C(P_{n_{1},n_{2}-1,n_{3}}\omega_{2})&c_{2}C(P_{n_{1},n_{2}-1,n_{3}}\omega_{3})\\\
c_{3}P_{n_{1},n_{2},n_{3}-1}&c_{3}C(P_{n_{1},n_{2},n_{3}-1}\omega_{1})&c_{3}C(P_{n_{1},n_{2},n_{3}-1}\omega_{2})&c_{3}C(P_{n_{1},n_{2},n_{3}-1}\omega_{3})\end{array}\right)$
(2.1)

where $P_{n_{1},n_{2},n_{3}}(x)=x^{n}+\cdots$ is a monic orthogonal polynomial
of degree $n=n_{1}+n_{2}+n_{3}$ with constants

$c_{1}=-2\pi i(h^{(1)}_{n_{1}-1,n_{2},n_{3}})^{-1},\quad c_{2}=-2\pi
i(h^{(2)}_{n_{1},n_{2}-1,n_{3}})^{-1},\quad c_{3}=-2\pi
i(h^{(3)}_{n_{1},n_{2},n_{3}-1})^{-1}$ (2.2)

and where
$h^{(j)}_{k_{1},k_{2},k_{3}}=\int_{-\infty}^{\infty}P_{k_{1},k_{2},k_{3}}x^{k_{j}}\omega_{j}(x)dx,j=1,2,3$
and $\omega_{j}(x)=e^{-(V(x)-a_{j}x)},\quad j=1,2,3,$ and $Cf$ denotes the
Cauchy transform of $f$, that is,

$Cf(z)=\frac{1}{2\pi i}\int_{{\mathbb{R}}}\frac{f(s)}{s-z}ds.$ (2.3)

Introducing

$\Psi_{n_{1},n_{2},n_{3}}=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&\frac{1}{c_{1}}&0&0\\\ 0&0&\frac{1}{c_{2}}&0\\\
0&0&0&\frac{1}{c_{3}}\end{array}\right)Y_{n_{1},n_{2},n_{3}}\left(\begin{array}[]{cccc}e^{-V(x)}&0&0&0\\\
0&e^{-a_{1}x}&0&0\\\ 0&0&e^{-a_{2}x}&0\\\
0&0&0&e^{-a_{3}x}\end{array}\right).$ (2.4)

we can obtain

  1. (i)

$\Psi_{n_{1},n_{2},n_{3}}$ is analytic on
${\mathbb{C}}\backslash{\mathbb{R}}$,

  2. (ii)

for $x\in{\mathbb{R}}$, we have

$\Psi_{n_{1},n_{2},n_{3},+}(x)=\Psi_{n_{1},n_{2},n_{3},-}(x)\left(\begin{array}[]{cccc}1&1&1&1\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right),$ (2.5)

  3. (iii)

as $z\rightarrow\infty$, we also have

$\Psi(z)=\left(\mathbb{I}+\frac{\Psi^{(1)}_{n_{1},n_{2},n_{3}}}{z}+O(\frac{1}{z^{2}})\right)\left(\begin{array}[]{cccc}z^{n}e^{-V(x)}&0&0&0\\\
0&\frac{1}{c_{1}}z^{-n_{1}}e^{-a_{1}x}&0&0\\\
0&0&\frac{1}{c_{2}}z^{-n_{2}}e^{-a_{2}x}&0\\\
0&0&0&\frac{1}{c_{3}}z^{-n_{3}}e^{-a_{3}x}\end{array}\right)$ (2.6)

where

$\Psi^{(1)}_{n_{1},n_{2},n_{3}}=\left(\begin{array}[]{cccc}p_{n_{1},n_{2},n_{3}}&\frac{h^{(1)}_{n_{1},n_{2},n_{3}}}{h^{(1)}_{n_{1}-1,n_{2},n_{3}}}&\frac{h^{(2)}_{n_{1},n_{2},n_{3}}}{h^{(2)}_{n_{1},n_{2}-1,n_{3}}}&\frac{h^{(3)}_{n_{1},n_{2},n_{3}}}{h^{(3)}_{n_{1},n_{2},n_{3}-1}}\\\
1&\frac{q^{(1)}_{n_{1}-1,n_{2},n_{3}}}{h^{(1)}_{n_{1}-1,n_{2},n_{3}}}&\frac{h^{(2)}_{n_{1}-1,n_{2},n_{3}}}{h^{(2)}_{n_{1},n_{2}-1,n_{3}}}&\frac{h^{(3)}_{n_{1}-1,n_{2},n_{3}}}{h^{(3)}_{n_{1},n_{2},n_{3}-1}}\\\
1&\frac{h^{(1)}_{n_{1},n_{2}-1,n_{3}}}{h^{(1)}_{n_{1}-1,n_{2},n_{3}}}&\frac{q^{(2)}_{n_{1},n_{2}-1,n_{3}}}{h^{(2)}_{n_{1},n_{2}-1,n_{3}}}&\frac{h^{(3)}_{n_{1},n_{2}-1,n_{3}}}{h^{(3)}_{n_{1},n_{2},n_{3}-1}}\\\
1&\frac{h^{(1)}_{n_{1},n_{2},n_{3}-1}}{h^{(1)}_{n_{1}-1,n_{2},n_{3}}}&\frac{h^{(2)}_{n_{1},n_{2},n_{3}-1}}{h^{(2)}_{n_{1},n_{2}-1,n_{3}}}&\frac{q^{(3)}_{n_{1},n_{2},n_{3}-1}}{h^{(3)}_{n_{1},n_{2},n_{3}-1}}\end{array}\right)$
(2.7)

here

$q^{(k)}_{n_{1},n_{2},n_{3}}=\int_{-\infty}^{\infty}P_{n_{1},n_{2},n_{3}}(x)x^{n_{k}+1}\omega_{k}(x)dx,\quad
k=1,2,3.$ (2.8)

and $P_{n_{1},n_{2},n_{3}}(z)=z^{n}+p_{n_{1},n_{2},n_{3}}z^{n-1}+\cdots$ is a
multiple polynomials.

According to the above notations, we have the following property.

######  Proposition 2.1.

We have the differential equation,

$\Psi^{\prime}_{n_{1},n_{2},n_{3}}(z)=N\left(\begin{array}[]{cccc}-z&\frac{n_{1}}{N}&\frac{n_{2}}{N}&\frac{n_{3}}{N}\\\
-1&-a&0&0\\\ -1&0&0&0\\\
-1&0&0&a\end{array}\right)\Psi_{n_{1},n_{2},n_{3}}(z).$ (2.9)

where the prime ′ denotes the derivative with respect to $z$.

###### Proof.

Let

$A_{n_{1},n_{2},n_{3}}=\frac{1}{N}\Psi^{\prime}_{n_{1},n_{2},n_{3}}\Psi_{n_{1},n_{2},n_{3}}$
(2.10)

From (2.5), we know $A_{n_{1},n_{2},n_{3}}$ has no jump, that is to say
$A_{n_{1},n_{2},n_{3}}$ is analytic on the whole complex plane. From (2.6), we
can show as $z\rightarrow\infty$

$A_{n_{1},n_{2},n_{3}}=\left(\mathbb{I}+\frac{\Psi^{(1)}_{n_{1},n_{2},n_{3}}}{+}\dots\right)\left(\begin{array}[]{cccc}-z&0&0&0\\\
0&-a_{1}&0&0\\\ 0&0&-a_{2}&0\\\
0&0&0&-a_{4}\end{array}\right)\left(\mathbb{I}+\frac{\Psi^{(1)}_{n_{1},n_{2},n_{3}}}{+}\dots\right)^{-1}+O(\frac{1}{z}).$
(2.11)

As $A_{n_{1},n_{2},n_{3}}$ is entire function, it implies that

$A_{n_{1},n_{2},n_{3}}=\left(\begin{array}[]{cccc}-z&c_{n_{1},n_{2},n_{3}}&d_{n_{1},n_{2},n_{3}}&e_{n_{1},n_{2},n_{3}}\\\
-1&-a_{1}&0&0\\\ -1&0&-a_{2}&0\\\ -1&0&0&-a_{3}\end{array}\right),$ (2.12)

where $c_{n_{1},n_{2},n_{3}},d_{n_{1},n_{2},n_{3}},e_{n_{1},n_{2},n_{3}}$ are
needed to determine.

From the differential equation we get

$\Psi^{\prime}_{n_{1},n_{2},n_{3}}(z)=NA_{n_{1},n_{2},n_{3}}\Psi_{n_{1},n_{2},n_{3}}(z).$
(2.13)

Let us show how to determine
$c_{n_{1},n_{2},n_{3}},d_{n_{1},n_{2},n_{3}},e_{n_{1},n_{2},n_{3}}$. From the
recursion formula of $\Psi_{n_{1},n_{2},n_{3}}(z)$ [18], say
$\Psi_{n_{1}+1,n_{2},n_{3}}=U_{n_{1},n_{2},n_{3}}\Psi_{n_{1},n_{2},n_{3}}$,
where

$\begin{array}[]{rl}U_{n_{1},n_{2},n_{3}}&=\left(\begin{array}[]{cccc}z-b_{n_{1},n_{2},n_{3}}&-c_{n_{1},n_{2},n_{3}}&-d_{n_{1},n_{2},n_{3}}&-e_{n_{1},n_{2},n_{3}}\\\
1&0&0&0\\\ 1&0&f_{n_{1},n_{2},n_{3}}&0\\\
1&0&0&g_{n_{1},n_{2},n_{3}}\end{array}\right)\end{array}$ (2.14)

we can get

$\Psi^{\prime}_{n_{1}+1,n_{2},n_{3}}=U^{\prime}_{n_{1},n_{2},n_{3}}\Psi_{n_{1},n_{2},n_{3}}+U_{n_{1},n_{2},n_{3}}\Psi^{\prime}_{n_{1},n_{2},n_{3}},$

on the other side, we have

$\Psi^{\prime}_{n_{1}+1,n_{2},n_{3}}(z)=NA_{n_{1}+1,n_{2},n_{3}}\Psi_{n_{1}+1,n_{2},n_{3}}(z).$

This implies

$U^{\prime}_{n_{1},n_{2},n_{3}}+U_{n_{1},n_{2},n_{3}}NA_{n_{1},n_{2},n_{3}}=NA_{n_{1}+1,n_{2},n_{3}}U_{n_{1},n_{2},n_{3}}.$

A simpler calculation shows that

$\begin{array}[]{ll}b_{n_{1},n_{2},n_{3}}=a_{1},&c_{n_{1}+1,n_{2},n_{3}}=c_{n_{1},n_{2},n_{3}}+\frac{1}{N},\\\
d_{n_{1}+1,n_{2},n_{3}}=d_{n_{1},n_{2},n_{3}},&e_{n_{1}+1,n_{2},n_{3}}=e_{n_{1},n_{2},n_{3}},\\\
f_{n_{1},n_{2},n_{3}}=a_{2}-a_{1},&g_{n_{1},n_{2},n_{3}}=a_{3}-a_{1}.\end{array}$
(2.15)

Similarly, we also can show the following formulas via analyze the other
recursion relations of $\Psi_{n_{1},n_{2},n_{3}}(z)$,

$\begin{array}[]{ll}\tilde{b}_{n_{1},n_{2},n_{3}}=a_{2},&\tilde{c}_{n_{1}+1,n_{2},n_{3}}=\tilde{c}_{n_{1},n_{2},n_{3}},\\\
\tilde{d}_{n_{1}+1,n_{2},n_{3}}=\tilde{d}_{n_{1},n_{2},n_{3}}+\frac{1}{N},&\tilde{e}_{n_{1}+1,n_{2},n_{3}}=\tilde{e}_{n_{1},n_{2},n_{3}},\\\
\tilde{f}_{n_{1},n_{2},n_{3}}=a_{1}-a_{2},&\tilde{g}_{n_{1},n_{2},n_{3}}=a_{3}-a_{2}.\end{array}$
(2.16)
$\begin{array}[]{ll}\hat{b}_{n_{1},n_{2},n_{3}}=a_{3},&\hat{c}_{n_{1}+1,n_{2},n_{3}}=\hat{c}_{n_{1},n_{2},n_{3}},\\\
\hat{d}_{n_{1}+1,n_{2},n_{3}}=\hat{d}_{n_{1},n_{2},n_{3}},&\hat{e}_{n_{1}+1,n_{2},n_{3}}=\hat{e}_{n_{1},n_{2},n_{3}}+\frac{1}{N},\\\
\hat{f}_{n_{1},n_{2},n_{3}}=a_{1}-a_{3},&\hat{g}_{n_{1},n_{2},n_{3}}=a_{2}-a_{2}.\end{array}$
(2.17)

By the initial value
$c_{0,n_{2},n_{3}}=d_{n_{1},0,n_{3}}=e_{n_{1},n_{2},0}=0$, we can get

$c_{n_{1},n_{2},n_{3}}=\frac{n_{1}}{N},\quad
d_{n_{1},n_{2},n_{3}}=\frac{n_{2}}{N},\quad
e_{n_{1},n_{2},n_{3}}=\frac{n_{3}}{N}.$ (2.18)

Noticing our assumption $a_{1}=a,a_{2}=0,a_{3}=-a$, we end the proof.

∎

We look for a WKB solution of the differential equation (2.9) of the form

$\Psi_{n_{1},n_{2},n_{3}}(z)=W(z)e^{-N\Lambda(z)},$ (2.19)

where $\Lambda$ is a diagonal matrix. By substituting this form into (2.9) we
obtain the equation,

$-W\Lambda^{\prime}W^{-1}=A+\frac{1}{N}W^{\prime}W^{-1},$ (2.20)

where $A$ is the matrix of coefficients in (2.9). By dropping the last term we
reduce it to the eigenvalue problem,

$W\Lambda^{\prime}W^{-1}=-A.$ (2.21)

The characteristic polynomial is

$\begin{array}[]{rl}det[\xi\mathbb{I}+A]=&{}\left|\begin{array}[]{cccc}\xi-z&t_{1}&t_{2}&t_{3}\\\
-1&\xi-a_{1}&0&0\\\ -1&0&\xi-a_{2}&0\\\ -1&0&0&\xi-a_{3}\end{array}\right|\\\
=&{}\xi^{4}-(z+a_{1}+a_{2}+a_{3})\xi^{3}\\\
&{}+[(t_{1}+t_{2}+t_{3})+a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3}+z(a_{1}+a_{2}+a_{3})]\xi^{2}\\\
&{}-[(a_{1}a_{2}+a_{1}a_{3}+a_{2}a_{3})z+(t_{1}+t_{2})a_{3}+(t_{1}+t_{3})a_{2}+(t_{2}+t_{3})a_{1}+a_{1}a_{2}a_{3}]\xi\\\
&{}+a_{1}a_{2}t_{3}+a_{1}t_{2}a_{3}+t_{1}a_{2}a_{3}+a_{1}a_{2}a_{3}z=0\end{array}$
(2.22)

where $t_{1}=\frac{n_{1}}{N},t_{2}=\frac{n_{2}}{N}$ and
$t_{3}=\frac{n_{3}}{N}$.

The above equation defines a Riemann surface, which in the case of interest in
this paper (where $n_{1}=n_{3},a_{1}=-a_{3}=a,a_{2}=0$) reduces to

$\xi^{4}-z\xi^{3}+(1-a^{2})\xi^{2}+a^{2}z\xi-t_{2}a^{2}=0.$ (2.23)

This defines the Riemann surface that will play a central role in the rest of
the paper.

##  3\. Spectral Curve and Riemann Surface

The Riemann surface is given by (2.23) or, if we solve for $z$,

$z=\frac{\xi^{4}+(1-a^{2})\xi^{2}-t_{2}a^{2}}{\xi^{3}-a^{2}\xi}.$ (3.1)

There are four inverse functions to (3.1), which we choose such that as
$z\rightarrow\infty$,

$\begin{array}[]{l}\xi_{1}(z)=z-\frac{1}{z}+O(\frac{1}{z^{3}}),\\\
\xi_{2}(z)=a+\frac{t_{1}}{z}+O(\frac{1}{z^{2}}),\\\
\xi_{3}(z)=\frac{t_{2}}{z}+O(\frac{1}{z^{2}}),\\\
\xi_{4}(z)=-a+\frac{t_{3}}{z}+O(\frac{1}{z^{2}}).\end{array}$ (3.2)

We need to find the sheet structure of the Riemann surface (2.23). The
critical points of $z(\xi)$ satisfy the equation

$\xi^{6}-(1+2a^{2})\xi^{4}+[a^{4}+(3t_{2}-1)a^{2}]\xi^{2}-t_{2}a^{4}=0,$ (3.3)

Let $y=\xi^{2}$, then the above equation becomes

$y^{3}-(1+2a^{2})y^{2}+[a^{4}+(3t_{2}-1)a^{2}]y-t_{2}a^{4}=0$ (3.4)

which is a cubic equation as $y$. The discriminant of this cubic equation is,
and we denote $a^{2}$ as $b$ and $t_{2}$ as $t$,

$\Delta=(1-t)b^{2}\Delta_{c}=(1-t)b^{2}[8b^{3}-(9t+15)b^{2}+(108t^{2}-90t+6)b+(1-9t)]$
(3.5)

and then we write $\Delta_{c}$ as a function of $t$,

$\Delta_{c}(t)=108bt^{2}+(-9b^{2}-90b-9)t+8b^{3}-15b^{2}+6b+1$ (3.6)

This is a quadratic function of $t$. And the discriminant of this quadratic
function is

$\Delta_{q}=-3(b-3)(5b+1)^{3},$ (3.7)

We can obtain that if $b>3$, then $\Delta_{c}>0$ for all $t\in(0,1)$, then we
have $\Delta>0$. That means the cubic equation (3.4)of $y$ has three distinct
real roots. And in here, we can go further beyond this, that is, the three
distinct real roots are all positive. We denote the roots by $y_{1},y_{2}$ and
$y_{3}$ (without loss of generally, we can set $y_{1}<y_{2}<y_{3}$), and then
set

$p=\sqrt{y_{1}},\quad q=\sqrt{y_{2}},\quad r=\sqrt{y_{3}}.$ (3.8)

Then the critical points on the $\xi-$plane are $\pm p,\pm q$ and $\pm r$. The
branch points on the $z-$plane are $\pm z_{1},\pm z_{2}$ and $\pm z_{3}$,
where

$z_{1}=z(\xi=p),\quad z_{2}=z(\xi=q),\quad z_{3}=z(\xi=r),\quad
0<z_{1}<z_{2}<z_{3}.$ (3.9)

We can show that $\xi_{1},\xi_{2},\xi_{3}$ and $\xi_{4}$ have analytic
extensions to
${\mathbb{C}}\backslash([-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}])$,${\mathbb{C}}\backslash[z_{2},z_{3}]$,${\mathbb{C}}\backslash[-z_{1},z_{1}]$
and ${\mathbb{C}}\backslash[-z_{3},-z_{2}]$. On the cut $[z_{2},z_{3}]$,

$\begin{array}[]{l}\xi_{1+}(x)=\overline{\xi_{1-}(x)}=\xi_{2-}(x)=\overline{\xi_{2+}(x)},\quad
z_{2}\leq x\leq z_{3},\\\ \mathrm{Im}{\xi_{1+}}>0,\qquad
z_{2}<x<z_{3},\end{array}$ (3.10)

and $\xi_{3}(x),\xi_{4}(x)$ are real. On the cut $[-z_{1},z_{1}]$,

$\begin{array}[]{l}\xi_{1+}(x)=\overline{\xi_{1-}(x)}=\xi_{3-}(x)=\overline{\xi_{3+}(x)},\quad-
z_{1}\leq x\leq z_{1},\\\ \mathrm{Im}{\xi_{1+}}>0,\qquad-
z_{1}<x<z_{1},\end{array}$ (3.11)

$\xi_{2}(x),\xi_{4}(x)$ are real. On the cut $[-z_{3},-z_{2}]$,

$\begin{array}[]{l}\xi_{1+}(x)=\overline{\xi_{1-}(x)}=\xi_{4-}(x)=\overline{\xi_{4+}(x)},\quad-
z_{3}\leq x\leq-z_{2},\\\ \mathrm{Im}{\xi_{1+}}>0,\qquad-
z_{3}<x<-z_{2},\end{array}$ (3.12)

$\xi_{2}(x),\xi_{3}(x)$ are real. See the figure 1.

Figure 1. Four sheet structure of Riemann surface

We define

$\rho(x)=\frac{1}{\pi}\mathrm{Im}\xi_{1+}(x),\qquad
x\in[-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}],$ (3.13)

then as $x\in(-z_{3},-z_{2})\cup(-z_{1},z_{1})\cup(z_{2},z_{3})$, we have
$\rho(x)>0$ and

$\begin{array}[]{rl}\int_{z_{2}}^{z_{3}}\rho(x)dx&=\int_{z_{2}}^{z_{3}}\frac{1}{\pi}\mathrm{Im}\xi_{1+}(x)dx\\\
{}\hfil&=\int_{z_{2}}^{z_{3}}\frac{1}{2\pi
i}(\xi_{1+}-\overline{\xi_{1+}})(x)dx\\\
{}\hfil&=\int_{z_{2}}^{z_{3}}\frac{1}{2\pi i}(\xi_{2-}-\xi_{2+})(x)dx\\\
{}\hfil&=\frac{1}{2\pi i}\oint\xi_{2}(s)ds=\frac{1-t}{2}.\end{array}$ (3.14)

Similarly, we have

$\int_{-z_{1}}^{z_{1}}\rho(x)dx=t,\quad\int_{-z_{3}}^{-z_{2}}\rho(x)dx=\frac{1-t}{2}.$
(3.15)

And there exists $\rho_{j}>0,j=1,2,3$ such that

$\begin{array}[]{l}\rho(x)=\frac{\rho_{j}}{\pi}|x-z_{j}|^{\frac{1}{2}}\left(1+O(x-z_{j})\right),\quad
x\rightarrow z_{j},x\in(z_{2},z_{3}),\\\
\rho(x)=\frac{\rho_{j}}{\pi}|x+z_{j}|^{\frac{1}{2}}\left(1+O(x+z_{j})\right),\quad
x\rightarrow-z_{j},x\in(-z_{3},-z_{2}),\\\
\rho(x)=\frac{\rho_{j}}{\pi}|x-z_{j}|^{\frac{1}{2}}\left(1+O(x-z_{j})\right),\quad
x\rightarrow z_{j},x\in(-z_{1},z_{1}).\end{array}$ (3.16)

The reason is there exists a constant $\rho_{3}>0$ near the branch point
$z_{3}$ such that as $z\rightarrow z_{3}$,

$\begin{array}[]{l}\xi_{1}(z)=r+\rho_{3}(z-z_{3})^{\frac{1}{2}}+O(z-z_{3}),\\\
\xi_{2}(z)=r-\rho_{3}(z-z_{3})^{\frac{1}{2}}+O(z-z_{3}).\end{array}$ (3.17)

Similarly, there exists a constant $\rho_{2}>0$ near the branch point $z_{2}$
such that as $z\rightarrow z_{2}$,

$\begin{array}[]{l}\xi_{1}(z)=q-\rho_{2}(z-z_{2})^{\frac{1}{2}}+O(z-z_{2}),\\\
\xi_{2}(z)=q+\rho_{2}(z-z_{2})^{\frac{1}{2}}+O(z-z_{2}).\end{array}$ (3.18)

There exists a constant $\rho_{1}>0$ near the branch point $z_{1}$ such that
as $z\rightarrow z_{1}$,

$\begin{array}[]{l}\xi_{1}(z)=p-\rho_{1}(z-z_{1})^{\frac{1}{2}}+O(z-z_{1}),\\\
\xi_{2}(z)=p+\rho_{1}(z-z_{1})^{\frac{1}{2}}+O(z-z_{1}).\end{array}$ (3.19)

Next, we need the integrals of the $\xi-$function,

$\lambda_{j}(z)=\int^{z}\xi_{j}(s)ds,\qquad j=1,2,3,4.$ (3.20)

Here, we choose the integral path so that $\lambda_{1}(z)$ and
$\lambda_{2}(z)$ are analytic at ${\mathbb{C}}\backslash(-\infty,z_{3}]$,
$\lambda_{3}(z)$ is analytic at ${\mathbb{C}}\backslash(-\infty,z_{1}]$ and
$\lambda_{4}(z)$ is analytic at ${\mathbb{C}}\backslash(-\infty,-z_{2}]$. From
(3.2), we know as $z\rightarrow\infty$

$\begin{array}[]{l}\lambda_{1}(z)=\frac{1}{2}z^{2}-\ln{z}+l_{1}+O(\frac{1}{z^{2}}),\\\
\lambda_{2}(z)=az+\frac{1-t}{2}\ln{z}+l_{2}+O(\frac{1}{z}),\\\
\lambda_{3}(z)=t\ln{z}+l_{3}+O(\frac{1}{z}),\\\
\lambda_{4}(z)=-az+\frac{1-t}{2}\ln{z}+l_{4}+O(\frac{1}{z}),\end{array}$
(3.21)

where $l_{1},l_{2},l_{3},l_{4}$ are constants, which we choose as follows. We
choose $l_{1},l_{2}$ such that

$\lambda_{1}(z_{3})=\lambda_{2}(z_{3})=0,$ (3.22)

choose $\l_{3}$ and $l_{4}$ such that

$\lambda_{3}(z_{1})=\lambda_{1+}(z_{1})=\lambda_{1-}(z_{1})-(1-t)\pi i,$
(3.23) $\lambda_{4}(-z_{2})=\lambda_{1+}(-z_{2})=\lambda_{1-}(-z_{2})-(1+t)\pi
i.$ (3.24)

Then we have the following jump relations:

$\begin{array}[]{ll}\lambda_{1+}(x)=\lambda_{2-}(x),\quad\lambda_{1-}(x)=\lambda_{2+}(x),&x\in[z_{2},z_{3}],\\\
\lambda_{2+}(x)-\lambda_{2-}(x)=(1-t)\pi i,&x\in(-\infty,z_{2}],\\\
\lambda_{1+}(x)-\lambda_{1-}(x)=-(1-t)\pi i,&x\in[z_{1},z_{2}],\\\
\lambda_{1+}(x)=\lambda_{3-}(x),\quad\lambda_{1-}(x)-\lambda_{3+}(x)=(1-t)\pi
i,&x\in[-z_{1},z_{1}],\\\ \lambda_{1+}(x)-\lambda_{1-}(x)=-(1+t)\pi
i,\quad\lambda_{3+}(x)-\lambda_{3-}(x)=2t\pi i,&x\in[-z_{2},-z_{1}],\\\
\lambda_{1+}(x)=\lambda_{4-}(x),\quad\lambda_{1-}(x)-\lambda_{4+}(x)=(1+t)\pi
i,&x\in[-z_{3},-z_{2}],\\\ \lambda_{1+}(x)-\lambda_{1-}(x)=-2\pi
i,\quad\lambda_{4+}(x)-\lambda_{4-}(x)=(1-t)\pi
i,&x\in(-\infty,-z_{3}].\end{array}$ (3.25)

For later use, we state the following propositions.

######  Lemma 3.1.

(a)The open interval $(z_{2},z_{3})$ has a neighborhood $U_{1}$ in the complex
plane such that when $z\in U_{1}\backslash(z_{2},z_{3})$, the real part
$\mathrm{Re}\lambda_{2}$ is the biggest one among $\lambda_{j},j=1,2,3,4$;

(b)The open interval $(-z_{1},z_{1})$ has a neighborhood $U_{2}$ in the
complex plane such that when $z\in U_{1}\backslash(-z_{1},z_{1})$, the real
part $\mathrm{Re}\lambda_{3}$ is the biggest one among
$\lambda_{j},j=1,2,3,4$;

(c)The open interval $(-z_{3},-z_{2})$ has a neighborhood $U_{3}$ in the
complex plane such that when $z\in U_{1}\backslash(-z_{3},-z_{2})$, the real
part $\mathrm{Re}\lambda_{4}$ is the biggest one among
$\lambda_{j},j=1,2,3,4$.

###### Proof.

We just prove (a), the following (b) and (c) are similarly.

First, we have
$\mathrm{Re}\lambda_{3+}<\mathrm{Re}\lambda_{1-},\mathrm{Re}\lambda_{4+}<\lambda_{1-}$
when $z\in U_{1}\backslash[z_{2},z_{3}]$. Then, define
$F(x)=\lambda_{2+}(x)-\lambda_{1+}(x)$, we have
$F(x)=\overline{\lambda_{1+}(x)}-\lambda_{1+}(x)$. That is to say, $F(x)$ is
pure image on the interval $(z_{2},z_{3})$. As we have
$F^{\prime}(x)=\xi_{2+}(x)-\xi_{1+}(x)=\xi_{2+}(x)-\xi_{2-}(x)=-2\pi
i\rho(x)$, then $\mathrm{Im}F^{\prime}(x)<0$. By Cauchy-Riemann equation,
$\mathrm{Re}F(x)>0$ as we move from the interval $(z_{2},z_{3})$ into the
upper half-plane. $\mathrm{Re}F(x)>0$ as we move from the interval
$(z_{2},z_{3})$ into the lower half-plane. ∎

##  4\. Transformations of the Riemann-Hilbert Problem

###  4.1. First Transform

Using the functions $\lambda_{j}$ and constants $l_{j},j=1,2,3,4$, which are
defined in the previous section, we define

$T(z)=diag\left(e^{-nl_{1}},e^{-nl_{2}},e^{-nl_{3}},e^{-nl_{4}}\right)Y(z)diag\left(e^{n(\lambda_{1}(z)-\frac{1}{2}z^{2})},e^{n(\lambda_{2}(z)-az)},e^{n\lambda_{3}(z)},e^{n(\lambda_{4}(z)+az)}\right),$
(4.1)

Then by (1.3) and (4.1), we have $T_{+}(x)=T_{-}(x)J_{T}(x),x\in{\mathbb{R}}$,
where

$J_{T}(x)=\left(\begin{array}[]{cccc}e^{n(\lambda_{1+}-\lambda_{1-})}&e^{n(\lambda_{2+}-\lambda_{1-})}&e^{n(\lambda_{3+}-\lambda_{1-})}&e^{n(\lambda_{4+}-\lambda_{1-})}\\\
0&e^{n(\lambda_{2+}-\lambda_{2-})}&0&0\\\
0&0&e^{n(\lambda_{3+}-\lambda_{3-})}&0\\\
0&0&0&e^{n(\lambda_{4+}-\lambda_{4-})}\end{array}\right)$ (4.2)

Be more exact, on $[z_{2},z_{3}]$

$J_{T}(x)=\left(\begin{array}[]{cccc}e^{n(\lambda_{1+}-\lambda_{2+})}&1&e^{n(\lambda_{3}-\lambda_{1-})}&e^{n(\lambda_{4}-\lambda_{1-})}\\\
0&e^{n(\lambda_{1-}-\lambda_{2-})}&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right);$ (4.3)

on $[-z_{1},z_{1}]$

$J_{T}(x)=\left(\begin{array}[]{cccc}e^{n(\lambda_{1+}-\lambda_{3+})}&e^{n(\lambda_{2+}-\lambda_{1-})}&1&e^{n(\lambda_{4}-\lambda_{1-})}\\\
0&1&0&0\\\ 0&0&e^{n(\lambda_{1-}-\lambda_{3-})}&0\\\
0&0&0&1\end{array}\right);$ (4.4)

on $[-z_{3},-z_{2}]$

$J_{T}(x)=\left(\begin{array}[]{cccc}e^{n(\lambda_{1+}-\lambda_{1-})}&e^{n(\lambda_{2+}-\lambda_{1-})}&e^{n(\lambda_{3+}-\lambda_{1-})}&1\\\
0&1&0&0\\\ 0&0&1&0\\\
0&0&0&e^{n(\lambda_{4+}-\lambda_{4-})}\end{array}\right);$ (4.5)

on
${\mathbb{R}}\backslash([z_{2},z_{3}]\cup[-z_{1},z_{1}]\cup[-z_{3},-z_{2}])$

$J_{T}(x)=\left(\begin{array}[]{cccc}1&e^{n(\lambda_{2+}-\lambda_{1-})}&e^{n(\lambda_{3+}-\lambda_{1-})}&e^{n(\lambda_{4+}-\lambda_{1-})}\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right).$ (4.6)

By (1.4), (3.21) and (4.1), we know the asymptotics of $T$ is

$T(z)=\mathbb{I}+O(\frac{1}{z}),\qquad z\rightarrow\infty.$ (4.7)

Thus $T$ solves the Riemann-Hilbert problem

  * •

$T$ is analytic on ${\mathbb{C}}\backslash{\mathbb{R}}$,

  * •

$T_{+}(x)=T_{-}(x)J_{T}(x),\quad x\in{\mathbb{R}},$ (4.8)

  * •

As $z\rightarrow\infty$,

$T(z)=\mathbb{I}+O(\frac{1}{z}).$ (4.9)

Inserting (4.1) into (1.5), we see that the kernel $K_{n}$ can be expressed in
terms of $T$ as follows:

$K_{n}(x,y)=\frac{e^{\frac{1}{4}n(x^{2}-y^{2})}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}0&e^{n\lambda_{2+}(y)}&e^{n\lambda_{3+}(y)}&e^{n\lambda_{4+}(y)}\end{array}\right)T_{+}^{-1}(y)T_{+}(x)\left(\begin{array}[]{c}e^{-n\lambda_{1+}(x)}\\\
0\\\ 0\\\ 0\end{array}\right).$ (4.10)

###  4.2. Second Transform

The second transformation of the Rirmann-Hilbert problem is opening of lenses.
Consider a lens with vertices $z_{2},z_{3}$, see figure 2.

Figure 2. The lens of vertices of $\pm z_{j},j=1,2,3$.

The lens is contained in the neighborhood $U_{1}$ of $(z_{2},z_{3})$. We have
the factorization,

$\begin{array}[]{l}\left(\begin{array}[]{cccc}e^{n(\lambda_{1+}-\lambda_{2+})}&1&e^{n(\lambda_{3}-\lambda_{1-})}&e^{n(\lambda_{4}-\lambda_{1-})}\\\
0&e^{n(\lambda_{1-}-\lambda_{2-})}&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right)\\\ =\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})_{-}}&1&-e^{n(\lambda_{3}-\lambda_{2})_{-}}&-e^{n(\lambda_{4}-\lambda_{2})_{-}}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right)\left(\begin{array}[]{cccc}0&1&0&0\\\
-1&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right)\\\
\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})_{+}}&1&e^{n(\lambda_{3}-\lambda_{2})_{+}}&e^{n(\lambda_{4}-\lambda_{2})_{+}}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right).\end{array}$ (4.11)

Set

$S(z)=\left\\{\begin{array}[]{l}T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\
-e^{n(\lambda_{1}-\lambda_{2})}&1&-e^{n(\lambda_{3}-\lambda_{2})}&-e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right)\mbox{ ͸ ϱ }\\\
T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&-e^{n(\lambda_{3}-\lambda_{2})}&-e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right)\mbox{ ͸ \textpm }\end{array}\right.$
(4.12)

Then (4.22) and (4.12) imply that

$S_{+}(x)=S_{-}(x)J_{S}(x),\quad
J_{S}(x)=\left(\begin{array}[]{cccc}0&1&0&0\\\ -1&0&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right),\quad x\in[z_{2},z_{3}].$ (4.13)

Similarly, consider a lens with vertices $-z_{1},z_{1}$, that is contained in
$U_{2}$ and set

$S(z)=\left\\{\begin{array}[]{l}T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\
-e^{n(\lambda_{1}-\lambda_{3})}&-e^{n(\lambda_{2}-\lambda_{3})}&1&-e^{n(\lambda_{4}-\lambda_{3})}\\\
0&0&0&1\end{array}\right)\mbox{ ͸ ϱ }\\\
T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\
e^{n(\lambda_{1}-\lambda_{3})}&-e^{n(\lambda_{2}-\lambda_{3})}&1&-e^{n(\lambda_{4}-\lambda_{3})}\\\
0&0&0&1\end{array}\right)\mbox{ ͸ \textpm }\end{array}\right.$ (4.14)

Then (4.22) and (4.14) imply that

$S_{+}(x)=S_{-}(x)J_{S}(x),\quad
J_{S}(x)=\left(\begin{array}[]{cccc}0&0&1&0\\\ 0&1&0&0\\\ -1&0&0&0\\\
0&0&0&1\end{array}\right),\quad x\in[-z_{1},z_{1}].$ (4.15)

Consider a lens with vertices $-z_{3},-z_{2}$, that is contained in $U_{3}$
and set

$S(z)=\left\\{\begin{array}[]{l}T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\ 0&0&1&0\\\
-e^{n(\lambda_{1}-\lambda_{4})}&-e^{n(\lambda_{2}-\lambda_{4})}&-e^{n(\lambda_{3}-\lambda_{4})}&1\end{array}\right)\mbox{
͸ ϱ }\\\ T(z)\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
e^{n(\lambda_{1}-\lambda_{4})}&-e^{n(\lambda_{2}-\lambda_{4})}&-e^{n(\lambda_{3}-\lambda_{4})}&1\end{array}\right)\mbox{
͸ \textpm }\end{array}\right.$ (4.16)

Then (4.22) and (4.16) imply that

$S_{+}(x)=S_{-}(x)J_{S}(x),\quad
J_{S}(x)=\left(\begin{array}[]{cccc}0&0&0&1\\\ 0&1&0&0\\\ 0&0&1&0\\\
-1&0&0&0\end{array}\right),\quad x\in[-z_{3},-z_{2}].$ (4.17)

And set

$S(z)=T(z)\qquad\mbox{outside of the lens region}.$ (4.18)

Then we have jumps on the boundary of the lenses,

$S_{+}(z)=S_{-}(z)J_{S}(z)$ (4.19)

where the contours are oriented from left to right (that is, from $-z_{3}$ to
$-z_{2}$, or from $-z_{1}$ to $z_{1}$, or from $z_{2}$ to $z_{3}$). The jump
matrix in (4.19) is

$\begin{array}[]{l}J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&e^{n(\lambda_{3}-\lambda_{2})}&e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\mbox{on the upper boundary of the
$[z_{2},z_{3}]-$lens}\\\ J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&-e^{n(\lambda_{3}-\lambda_{2})}&-e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\mbox{on the lower boundary of the
$[z_{2},z_{3}]-$lens}\\\ J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\
e^{n(\lambda_{1}-\lambda_{3})}&e^{n(\lambda_{2}-\lambda_{3})}&1&e^{n(\lambda_{4}-\lambda_{3})}\\\
0&0&0&1\end{array}\right),\mbox{on the upper boundary of the
$[-z_{1},z_{1}]-$lens}\\\ J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\
e^{n(\lambda_{1}-\lambda_{3})}&-e^{n(\lambda_{2}-\lambda_{3})}&1&-e^{n(\lambda_{4}-\lambda_{3})}\\\
0&0&0&1\end{array}\right),\mbox{on the lower boundary of the
$[-z_{1},z_{1}]-$lens}\\\ J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&0&0\\\ 0&0&1&0\\\
e^{n(\lambda_{1}-\lambda_{4})}&e^{n(\lambda_{2}-\lambda_{4})}&e^{n(\lambda_{3}-\lambda_{4})}&1\end{array}\right),\mbox{on
the upper boundary of the $[-z_{3},-z_{2}]-$lens}\\\
J_{S}(z)=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
e^{n(\lambda_{1}-\lambda_{4})}&-e^{n(\lambda_{2}-\lambda_{4})}&-e^{n(\lambda_{3}-\lambda_{4})}&1\end{array}\right),\mbox{on
the lower boundary of the $[-z_{3},-z_{2}]-$lens}\end{array}$ (4.20)

And

$S_{+}(x)=S_{-}(x)J_{S}(x),\qquad J_{S}(x)=J_{T}(x),\quad
x\in(-\infty,-z_{3}]\cup[-z_{2},-z_{1}]\cup[z_{1},z_{2}]\cup[z_{3},\infty).$
(4.21)

Thus $S$ solves the Riemann-Hilbert problem

  * •

$S$ is analytic on ${\mathbb{C}}\backslash({\mathbb{R}}\cup\Gamma)$, where
$\Gamma$ denotes the boundary of the lens,

  * •

$S_{+}(z)=S_{-}(z)J_{S}(z),\quad z\in{\mathbb{R}}\cup\Gamma,$ (4.22)

  * •

As $z\rightarrow\infty$,

$S(z)=\mathbb{I}+O(\frac{1}{z}).$ (4.23)

The kernel $K_{n}(x,y)$ is expressed in terms of $S$ as follows. By (4.10) and
the definitions (4.12), (4.14), (4.16), for $x,y$ in $(z_{2},z_{3})$, we have

$K_{n}(x,y)=\frac{e^{\frac{n}{4}(x^{2}-y^{2})}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{n\lambda_{1+}(y)}&e^{n\lambda_{2+}(y)}&0&0\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-n\lambda_{1+}(x)}\\\
e^{-n\lambda_{2+}(x)}\\\ 0\\\ 0\end{array}\right),$ (4.24)

for $x,y$ in $(-z_{1},z_{1})$, we have

$K_{n}(x,y)=\frac{e^{\frac{n}{4}(x^{2}-y^{2})}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{n\lambda_{1+}(y)}&0&e^{n\lambda_{3+}(y)}&0\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-n\lambda_{1+}(x)}\\\
0\\\ e^{-n\lambda_{3+}(x)}\\\ 0\end{array}\right),$ (4.25)

while for $x,y$ in $(-z_{3},-z_{2})$, we have

$K_{n}(x,y)=\frac{e^{\frac{n}{4}(x^{2}-y^{2})}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{n\lambda_{1+}(y)}&0&0&e^{n\lambda_{4+}(y)}\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-n\lambda_{1+}(x)}\\\
0\\\ 0\\\ e^{-n\lambda_{4+}(x)}\end{array}\right).$ (4.26)

Since $\lambda_{1+}$ and $\lambda_{2+}$ are complex conjugates on
$(z_{2},z_{3})$, we can rewrite (4.24) for $x,y$ in $(z_{2},z_{3})$ as

$K_{n}(x,y)=\frac{e^{n(h(y)-h(x))}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{ni\mathrm{Im}\lambda_{1+}(y)}&e^{-ni\mathrm{Im}\lambda_{1+}(y)}&0&0\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-ni\mathrm{Im}\lambda_{1+}(x)}\\\
e^{ni\mathrm{Im}\lambda_{1+}(x)}\\\ 0\\\ 0\end{array}\right),$ (4.27)

where $h(x)=-\frac{1}{4}x^{2}+\mathrm{Re}\lambda_{1+}(x)$. Similarly, for
$x,y$ in $(-z_{1},z_{1})$, we have

$K_{n}(x,y)=\frac{e^{n(h(y)-h(x))}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{ni\mathrm{Im}\lambda_{1+}(y)}&0&e^{-ni\mathrm{Im}\lambda_{1+}(y)}&0\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-ni\mathrm{Im}\lambda_{1+}(x)}\\\
0\\\ e^{ni\mathrm{Im}\lambda_{1+}(x)}\\\ 0\end{array}\right),$ (4.28)

for $x,y$ in $(-z_{3},-z_{2})$, we have

$K_{n}(x,y)=\frac{e^{n(h(y)-h(x))}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{ni\mathrm{Im}\lambda_{1+}(y)}&0&0&e^{-ni\mathrm{Im}\lambda_{1+}(y)}\end{array}\right)S_{+}^{-1}(y)S_{+}(x)\left(\begin{array}[]{c}e^{-ni\mathrm{Im}\lambda_{1+}(x)}\\\
0\\\ 0\\\ e^{ni\mathrm{Im}\lambda_{1+}(x)}\end{array}\right).$ (4.29)

###  4.3. Model Riemann-Hilbert problem

As $n\rightarrow\infty$, the jump matrix $J_{S}(z)$ is exponentially close to
the identity matrix at every $z$ outside of
$[-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}]$. This follows from (4.20)
and Lemma 3.1 for $z$ on the boundary of the lenses, and (4.21), (4.6), Lemma
3.1 for $z$ on the intervals
$(-\infty,-z_{3}]\cup[-z_{2},-z_{1}]\cup[z_{1},z_{2}]\cup[z_{3},\infty)$.

In this subsection, we solve the following model Riemann-Hilbert problem,
where we ignore the exponentially small jumps: find
$M:{\mathbb{C}}\backslash([-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}])\rightarrow{\mathbb{C}}^{4\times
4}$ such that

  * •

$M$ is analytic on
${\mathbb{C}}\backslash([-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}])$,

  * •

$M_{+}(x)=M_{-}(x)J_{S}(x),\quad
x\in(-z_{3},-z_{2})\cup(-z_{1},z_{1})\cup(z_{2},z_{3}),$ (4.30)

  * •

As $z\rightarrow\infty$,

$M(z)=\mathbb{I}+O(\frac{1}{z})$ (4.31)

This problem is similar to the RH problem considered in [22, Sect. 6.1].We
also follow a similar method to solve it.

We lift the model Riemann-Hilbert problem to the Riemann surface of (2.23)
with the sheet structure as in figure 1. Consider

$\begin{array}[]{l}\Omega_{1}=\xi_{1}({\mathbb{C}}\backslash([-z_{3},-z_{2}]\cup[-z_{1},z_{1}]\cup[z_{2},z_{3}])),\\\
\Omega_{2}=\xi_{2}({\mathbb{C}}\backslash[z_{2},z_{3}]),\\\
\Omega_{3}=\xi_{3}({\mathbb{C}}\backslash[-z_{1},z_{1}]),\\\
\Omega_{4}=\xi_{4}({\mathbb{C}}\backslash[-z_{3},-z_{2}]).\end{array}$ (4.32)

Then $\Omega_{1},\Omega_{2},\Omega_{3},\Omega_{4}$ give a partition of the
complex plane into four regions, see figure 3.

Figure 3. Partition of the complex $\xi$ plane

The regions $\Omega_{2},\Omega_{3},\Omega_{4}$ are bounded, $a\in\Omega_{2}$,
$0\in\Omega_{3}$, $-a\in\Omega_{4}$, with the symmetry conditions

$\bar{\Omega}_{2}=\Omega_{2},\quad\bar{\Omega}_{4}=\Omega_{4},\quad\bar{\Omega}_{3}=\Omega_{3},\quad\Omega_{2}=-\Omega_{4},\quad\Omega_{3}=-\Omega_{3}.$
(4.33)

Denote by $\Gamma_{k}$ the boundary of $\Omega_{k},k=2,3,4$, see figure 3.
Then we have

$\begin{array}[]{l}\xi_{1+}([z_{2},z_{3}])=\xi_{2-}([z_{2},z_{3}])=\Gamma^{+}_{2}=\Gamma_{2}\cap\\{\mathrm{Im}z\geq
0\\},\\\
\xi_{1-}([z_{2},z_{3}])=\xi_{2+}([z_{2},z_{3}])=\Gamma^{-}_{2}=\Gamma_{2}\cap\\{\mathrm{Im}z\leq
0\\},\\\
\xi_{1+}([-z_{1},z_{1}])=\xi_{3-}([-z_{1},z_{1}])=\Gamma^{+}_{3}=\Gamma_{3}\cap\\{\mathrm{Im}z\geq
0\\},\\\
\xi_{1-}([-z_{1},z_{1}])=\xi_{3+}([-z_{1},z_{1}])=\Gamma^{-}_{3}=\Gamma_{3}\cap\\{\mathrm{Im}z\leq
0\\},\\\
\xi_{1+}([-z_{3},-z_{2}])=\xi_{4-}([-z_{3},-z_{2}])=\Gamma^{+}_{4}=\Gamma_{4}\cap\\{\mathrm{Im}z\geq
0\\},\\\
\xi_{1-}([-z_{3},-z_{2}])=\xi_{4+}([-z_{3},-z_{2}])=\Gamma^{-}_{4}=\Gamma_{4}\cap\\{\mathrm{Im}z\leq
0\\}.\end{array}$ (4.34)

We are looking for a solution $M$ in the following form:

$M(z)=\left(\begin{array}[]{cccc}M_{1}(\xi_{1}(z))&M_{1}(\xi_{2}(z))&M_{1}(\xi_{3}(z))&M_{1}(\xi_{4}(z))\\\
M_{2}(\xi_{1}(z))&M_{2}(\xi_{2}(z))&M_{2}(\xi_{3}(z))&M_{2}(\xi_{4}(z))\\\
M_{3}(\xi_{1}(z))&M_{3}(\xi_{2}(z))&M_{3}(\xi_{3}(z))&M_{3}(\xi_{4}(z))\\\
M_{4}(\xi_{1}(z))&M_{4}(\xi_{2}(z))&M_{4}(\xi_{3}(z))&M_{4}(\xi_{4}(z))\\\
\end{array}\right),$ (4.35)

where $M_{1}(\xi),M_{2}(\xi),M_{3}(\xi),M_{4}(\xi)$ are four analytic
functions on ${\mathbb{C}}\backslash(\Gamma_{2}\cup\Gamma_{3}\cup\Gamma_{4})$.
To satisfy the jump condition (4.30), we need the following relations for
$k=1,2,3,4$,

$\begin{array}[]{l}M_{k+}(\xi)=-M_{k-}(\xi),\qquad\xi\in\Gamma_{2}^{+}\cup\Gamma_{3}^{+}\cup\Gamma_{4}^{+},\\\
M_{k+}(\xi)=M_{k-}(\xi),\qquad\xi\in\Gamma_{2}^{-}\cup\Gamma_{3}^{-}\cup\Gamma_{4}^{-}.\end{array}$
(4.36)

Since
$\xi_{1}(\infty)=\infty,\xi_{2}(\infty)=a,\xi_{3}(\infty)=0,\xi_{4}(\infty)=-a$,
then to satisfy (4.31), we have

$\begin{array}[]{llll}M_{1}(\infty)=1,&M_{1}(a)=0,&M_{1}(0)=0,&M_{1}(-a)=0,\\\
M_{2}(\infty)=0,&M_{2}(a)=1,&M_{2}(0)=0,&M_{2}(-a)=0,\\\
M_{3}(\infty)=0,&M_{3}(a)=0,&M_{3}(0)=1,&M_{4}(-a)=0,\\\
M_{4}(\infty)=0,&M_{4}(a)=0,&M_{3}(0)=0,&M_{4}(-a)=1.\end{array}$ (4.37)

Equation (4.36) and (4.37) have the following solution

$\begin{array}[]{ll}M_{1}(\xi)=\frac{\xi(\xi^{2}-a^{2})}{\sqrt{(\xi^{2}-p^{2})(\xi^{2}-q^{2})(\xi^{2}-r^{2})}},&M_{2}(\xi)=c_{2}\frac{\xi(\xi+a)}{\sqrt{(\xi^{2}-p^{2})(\xi^{2}-q^{2})(\xi^{2}-r^{2})}},\\\
M_{3}(\xi)=c_{3}\frac{\xi^{2}-a^{2}}{\sqrt{(\xi^{2}-p^{2})(\xi^{2}-q^{2})(\xi^{2}-r^{2})}},&M_{4}(\xi)=c_{4}\frac{\xi(\xi-a)}{\sqrt{(\xi^{2}-p^{2})(\xi^{2}-q^{2})(\xi^{2}-r^{2})}}.\end{array}$
(4.38)

Here the constants $c_{2},c_{3},c_{4}$ are defined by
$M_{2}(a)=1,M_{3}(0)=1,M_{4}(-a)=1$. Notice

$(\xi^{2}-p^{2})(\xi^{2}-q^{2})(\xi^{2}-r^{2})=\xi^{6}-(1+2a^{2})\xi^{4}+[a^{4}+(3t-1)a^{2}]\xi^{2}-ta^{4},$
(4.39)

hence

$M_{2}(a)=c_{2}\frac{2a^{2}}{\sqrt{2(t-1)a^{4}}}.$ (4.40)

By taking into account the cuts of $M_{2}(\xi)$, we have

$M_{2}(a)=c_{2}i\sqrt{\frac{2}{1-t}},$ (4.41)

hence

$c_{2}=-i\sqrt{\frac{1-t}{2}}.$ (4.42)

Similarly, we have

$c_{3}=-i\sqrt{t},\qquad c_{4}=-i\sqrt{\frac{1-t}{2}}.$ (4.43)

Thus, the solution to the model Riemann-Hilbert is given as following,

$\begin{array}[]{l}M(z)=\\\
\left(\begin{array}[]{cccc}\frac{\xi_{1}(\xi_{1}^{2}-a^{2})}{\sqrt{(\xi_{1}^{2}-p^{2})(\xi_{1}^{2}-q^{2})(\xi_{1}^{2}-r^{2})}}&\frac{\xi_{2}(\xi_{2}^{2}-a^{2})}{\sqrt{(\xi_{2}^{2}-p^{2})(\xi_{2}^{2}-q^{2})(\xi_{2}^{2}-r^{2})}}&\frac{\xi_{3}(\xi_{3}^{2}-a^{2})}{\sqrt{(\xi_{3}^{2}-p^{2})(\xi_{3}^{2}-q^{2})(\xi_{3}^{2}-r^{2})}}&\frac{\xi_{4}(\xi_{4}^{2}-a^{2})}{\sqrt{(\xi_{4}^{2}-p^{2})(\xi_{4}^{2}-q^{2})(\xi_{4}^{2}-r^{2})}}\\\
-i\frac{\sqrt{1-t}\xi_{1}(\xi_{1}+a)}{\sqrt{2(\xi_{1}^{2}-p^{2})(\xi_{1}^{2}-q^{2})(\xi_{1}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{2}(\xi_{2}+a)}{\sqrt{2(\xi_{2}^{2}-p^{2})(\xi_{2}^{2}-q^{2})(\xi_{2}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{3}(\xi_{3}+a)}{\sqrt{2(\xi_{3}^{2}-p^{2})(\xi_{3}^{2}-q^{2})(\xi_{3}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{4}(\xi_{4}+a)}{\sqrt{2(\xi_{4}^{2}-p^{2})(\xi_{4}^{2}-q^{2})(\xi_{4}^{2}-r^{2})}}\\\
-i\frac{\sqrt{t}(\xi_{1}^{2}-a^{2})}{\sqrt{(\xi_{1}^{2}-p^{2})(\xi_{1}^{2}-q^{2})(\xi_{1}^{2}-r^{2})}}&-i\frac{\sqrt{t}(\xi_{2}^{2}-a^{2})}{\sqrt{(\xi_{2}^{2}-p^{2})(\xi_{2}^{2}-q^{2})(\xi_{2}^{2}-r^{2})}}&-i\frac{\sqrt{t}(\xi_{3}^{2}-a^{2})}{\sqrt{(\xi_{3}^{2}-p^{2})(\xi_{3}^{2}-q^{2})(\xi_{3}^{2}-r^{2})}}&-i\frac{\sqrt{t}(\xi_{4}^{2}-a^{2})}{\sqrt{(\xi_{4}^{2}-p^{2})(\xi_{4}^{2}-q^{2})(\xi_{4}^{2}-r^{2})}}\\\
-i\frac{\sqrt{1-t}\xi_{1}(\xi_{1}-a)}{\sqrt{2(\xi_{1}^{2}-p^{2})(\xi_{1}^{2}-q^{2})(\xi_{1}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{2}(\xi_{2}-a)}{\sqrt{2(\xi_{2}^{2}-p^{2})(\xi_{2}^{2}-q^{2})(\xi_{2}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{3}(\xi_{3}-a)}{\sqrt{2(\xi_{3}^{2}-p^{2})(\xi_{3}^{2}-q^{2})(\xi_{3}^{2}-r^{2})}}&-i\frac{\sqrt{1-t}\xi_{4}(\xi_{4}-a)}{\sqrt{2(\xi_{4}^{2}-p^{2})(\xi_{4}^{2}-q^{2})(\xi_{4}^{2}-r^{2})}}\\\
\end{array}\right)\end{array}$ (4.44)

The solution $M$ to the model Riemann-Hilbert problem will be used to
construct a parametrix for the Riemann-Hilbert problem for $S$ outside of a
small neighborhood of the edge points. Namely, we will fix some $r>0$ and
consider the disks of radius $r$ around the edge points. At the edge points
$M$ is not analytic and in a neighborhood of the edge points the parametrix is
constructed differently.

###  4.4. Parametrix at Edge Points

We consider small disks $D(\pm z_{j},r),j=1,2,3$ with radius $r>0$ and
centered at the edge points, and look for a local parametrix $P$ defined on
the union of the six disks such that

  * •

$P$ is analytic at $D(\pm z_{j},r)\backslash({\mathbb{R}}\cup\Gamma)$,

  * •

$P_{+}(z)=P_{-}(z)J_{S}(z),\qquad z\in({\mathbb{R}}\cup\Gamma)\cap D(\pm
z_{j},r),$ (4.45)

  * •

As $n\rightarrow\infty$,

$P(z)=\left(\mathbb{I}+O(\frac{1}{z})\right)M(z),\quad\mbox{uniformly for
$z\in\partial D(\pm z_{j},r)$}.$ (4.46)

We consider here the edge point $z_{3}$ in detail. We note that as
$z\rightarrow z_{3}$,

$\begin{array}[]{l}\lambda_{1}(z)=r(z-z_{3})+\frac{2\rho_{1}}{3}(z-z_{3})^{\frac{3}{2}}+O(z-z_{3})^{2},\\\
\lambda_{2}(z)=r(z-z_{3})-\frac{2\rho_{1}}{3}(z-z_{3})^{\frac{3}{2}}+O(z-z_{3})^{2},\end{array}$
(4.47)

thus, we have as $z\rightarrow z_{3}$,

$\lambda_{1}(z)-\lambda_{2}(z)=\frac{4\rho_{1}}{3}(z-z_{3})^{\frac{3}{2}}+O(z-z_{3})^{\frac{5}{2}}.$
(4.48)

Hence

$\beta(z)=\left[\frac{3}{4}(\lambda_{1}(z)-\lambda_{2}(z))\right]^{\frac{2}{3}}$
(4.49)

is analytic at $z_{3}$, real-valued on the real axis near $z_{3}$, and
$\beta^{\prime}(z)=\rho_{1}^{\frac{2}{3}}>0$. So $\beta$ is a conformal map
from $D(z_{3},r)$ to a convex neighborhood of the origin, if $r$ is
sufficiently small. We take $\Gamma$ near $z_{3}$ such that

$\beta(\Gamma\cap D(z_{3},r))\subset\\{z|arg(z)=\pm\frac{2\pi}{3}\\}.$ (4.50)

Thus, $\Gamma$ and ${\mathbb{R}}$ divide the disk $D(z_{3},r)$ into four
regions numbered $\@slowromancap i@,\@slowromancap ii@,\@slowromancap iii@$
and $\@slowromancap iv@$ such that
$0<arg\beta(z)<\frac{2\pi}{3},\frac{2\pi}{3}<\arg\beta(z)<\pi,-\pi<arg\beta(z)<-\frac{2\pi}{3}$
and $-\frac{2\pi}{3}<arg\beta(z)<0$ for $z$ in regions $\@slowromancap
i@,\@slowromancap ii@,\@slowromancap iii@$ and $\@slowromancap iv@$,
respectively.

Recall that the jumps $J_{S}$ near $z_{3}$ are given by

$\begin{array}[]{l}J_{S}=\left(\begin{array}[]{cccc}0&1&0&0\\\ -1&0&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\quad x\in[z_{3}-r,z_{3}),\\\
J_{S}=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&e^{n(\lambda_{3}-\lambda_{2})}&e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\mbox{on the upper boundary of the lens
in $D(z_{3},r)$}\\\ J_{S}=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&-e^{n(\lambda_{3}-\lambda_{2})}&-e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\mbox{on the lower boundary of the lens
in $D(z_{3},r)$}\\\
J_{S}=\left(\begin{array}[]{cccc}1&e^{n(\lambda_{2}-\lambda_{1})}&e^{n(\lambda_{3}-\lambda_{1})}&e^{n(\lambda_{4}-\lambda_{1})}\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right),\quad
x\in(z_{3},z_{3}+r].\end{array}$ (4.51)

We write

$\tilde{P}=\left\\{\begin{array}[]{ll}P\left(\begin{array}[]{cccc}1&0&0&0\\\
0&1&-e^{n(\lambda_{3}-\lambda_{2})}&-e^{n(\lambda_{4}-\lambda_{2})}\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),&\mbox{in regions $\@slowromancap
i@,\@slowromancap iv@$},\\\ P,&\mbox{in regions $\@slowromancap
ii@,\@slowromancap <EMAIL_ADDRESS>(4.52)

Then the jumps $\tilde{P}$ are $\tilde{P}_{+}=\tilde{P}_{-}J_{\tilde{P}}$,
where

$\begin{array}[]{l}J_{\tilde{P}}=\left(\begin{array}[]{cccc}0&1&0&0\\\
-1&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right),\quad x\in[z_{3}-r,z_{3}),\\\
J_{\tilde{P}}=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right),\mbox{on the upper boundary of the lens in
$D(z_{3},r)$}\\\ J_{\tilde{P}}=\left(\begin{array}[]{cccc}1&0&0&0\\\
e^{n(\lambda_{1}-\lambda_{2})}&1&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right),\mbox{on the lower boundary of the lens in
$D(z_{3},r)$}\\\
J_{\tilde{P}}=\left(\begin{array}[]{cccc}1&e^{n(\lambda_{2}-\lambda_{1})}&0&0\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{array}\right),\quad
x\in(z_{3},z_{3}+r].\end{array}$ (4.53)

The Riemann-Hilbert problem for $\tilde{P}$ is essentially a $2\times 2$
problem, since the jumps (4.53) are non-trivial only in the upper $2\times 2$
block. A solution can be constructed in a standard way out of Airy functions.
The Airy function $Ai(z)$ solves the differential equation

$y^{{}^{\prime\prime}}=zy.$ (4.54)

For any $\varepsilon>0$, in the sector $\pi+\varepsilon\leq
argz\leq\pi-\varepsilon$, it has the asymptotics as $z\rightarrow\infty$,

$Ai(z)=\frac{1}{2\sqrt{\pi}z^{\frac{1}{4}}}e^{-\frac{2}{3}z^{\frac{3}{2}}}\left(1+O(z^{-\frac{3}{2}})\right).$
(4.55)

The functions $Ai(\omega z),Ai(\omega^{2}z)$, where $\omega=e^{\frac{2\pi
i}{3}}$, also solve the equation (4.54). And we have the linear relation,

$Ai(z)+\omega Ai(\omega z)+\omega^{2}Ai(\omega^{2}z)=0.$ (4.56)

Write

$y_{0}(z)=Ai(z),\quad y_{1}(z)=\omega Ai(\omega z),\quad
y_{2}(z)=\omega^{2}Ai(\omega^{2}z),$ (4.57)

using these functions to define

$\phi(z)=\left\\{\begin{array}[]{ll}\left(\begin{array}[]{cccc}y_{0}(z)&-y_{2}(z)&0&0\\\
y^{\prime}_{0}(z)&-y^{\prime}_{2}(z)&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right)&\mbox{ ~{}$0<argz<\frac{2\pi}{3}$},\\\
\left(\begin{array}[]{cccc}-y_{1}(z)&-y_{2}(z)&0&0\\\
-y^{\prime}_{1}(z)&-y^{\prime}_{2}(z)&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right)&\mbox{ ~{}$\frac{2\pi}{3}<argz<\pi$},\\\
\left(\begin{array}[]{cccc}-y_{2}(z)&y_{1}(z)&0&0\\\
-y^{\prime}_{2}(z)&y^{\prime}_{1}(z)&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right)&\mbox{ ~{}$-\pi<argz<-\frac{2\pi}{3}$},\\\
\left(\begin{array}[]{cccc}y_{0}(z)&y_{1}(z)&0&0\\\
y^{\prime}_{0}(z)&y^{\prime}_{1}(z)&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right)&\mbox{
~{}$-\frac{2\pi}{3}<argz<0$}.\end{array}\right.$ (4.58)

To match the asymptotic condition as $z\rightarrow\infty$, we should have

$\tilde{P}=\left(\mathbb{I}+O(\frac{1}{n})\right)M(z),\quad\mbox{uniformly in
$z\in\partial D(z_{3},r)$}.$ (4.59)

Then

$\tilde{P}(z)=E_{n}(z)\phi(n^{\frac{2}{3}}\beta(z))diag\left(e^{\frac{1}{2}n(\lambda_{1}(z)-\lambda_{2}(z))},e^{-\frac{1}{2}n(\lambda_{1}(z)-\lambda_{2}(z))},1,1\right),$
(4.60)

where

$E_{n}=\sqrt{\pi}M\left(\begin{array}[]{cccc}1&-1&0&0\\\ -i&-i&0&0\\\
0&0&1&0\\\
0&0&0&1\end{array}\right)\left(\begin{array}[]{cccc}n^{\frac{1}{6}}\beta^{\frac{1}{4}}&0&0&0\\\
0&n^{-\frac{1}{6}}\beta^{-\frac{1}{4}}&0&0\\\ 0&0&1&0\\\
0&0&0&1\end{array}\right).$ (4.61)

A similar construction works for a parametrix $P$ around the other edge
points.

###  4.5. Third Transformation

Let

$\begin{array}[]{ll}R(z)=S(z)M(z)^{-1}&\mbox{$z$ outside of the disks $D(\pm
z_{j},r),j=1,2,3$},\\\ R(z)=S(z)P(z)^{-1}&\mbox{$z$ inside of the
disks}.\end{array}$ (4.62)

Then we have $R(z)$ is analytic on ${\mathbb{C}}\backslash\Gamma_{R}$, where
$\Gamma_{R}$ consists of the six circles $\partial D(\pm z_{j},r),j=1,2,3$,
the parts of $\Gamma$ outside of the six disks, and the real intervals
$(-\infty,-z_{3}-r),(-z_{2}+r,-z_{1}-r),(z_{1}+r,z_{2}-r),(z_{3}+r,\infty)$.

The jump relations for $R(z)$ are

$R_{+}=R_{-}J_{R},$ (4.63)

where

$\begin{array}[]{ll}J_{R}=MP^{-1},&\mbox{on the circles, oriented
counterclockwise},\\\ J_{R}=MJ_{S}M^{-1},&\mbox{on the remaining parts of
$\Gamma_{R}$}.\end{array}$ (4.64)

From (4.46) it follows that $J_{R}=\mathbb{I}+O(\frac{1}{n})$ uniformly on the
circles, and from (4.20), (4.21), (4.2) and lemma 3.1 it follows that
$J_{R}=\mathbb{I}+O(e^{-cn})$ for $c>0$ as $n\rightarrow\infty$, uniformly on
the remaining parts of $\Gamma_{R}$. So we can conclude

$J_{R}(z)=\mathbb{I}+O(\frac{1}{n}),\quad\mbox{as $n\rightarrow\infty$,
uniformly on $\Gamma_{R}$}.$ (4.65)

As $z\rightarrow\infty$, we have

$R(z)=\mathbb{I}+O(\frac{1}{z}).$ (4.66)

From (4.63), (4.65), (4.66) and and the fact that we can deform the contours
in any desired direction, it follows that

$R(z)=\mathbb{I}+O(\frac{1}{n|z|+1}),\qquad
n\rightarrow\infty,\quad\mbox{uniformly for
$z\in{\mathbb{R}}\backslash\Gamma_{R}$},$ (4.67)

see [17].

By Cauchy theorem, we then also have

$R^{\prime}(z)=O(\frac{1}{n|z|+1}),$ (4.68)

thus,

$R^{-1}(y)R(x)=\mathbb{I}+R^{-1}(y)(R(x)-R(y))=\mathbb{I}+O(\frac{x-y}{n}).$
(4.69)

##  5\. Proofs of the Results

######  Theorem 5.1.

Let $V(M)=\frac{1}{2}M^{2}$, $n_{1}=n_{3}$, $n_{2}=t$, $a^{2}>3$. Then the
limiting mean density of eigenvalues

$\rho(z)=\lim_{n\rightarrow\infty}\frac{1}{n}K_{n}(x,x),$ (5.1)

exists, and it is supported by three intervals,
$[-z_{3},-z_{2}],[-z_{1},z_{1}]$ and $[z_{2},z_{3}]$.

###### Proof.

Consider $x\in(z_{2},z_{3})$. We may assume that the circles around the edge
points are such that $x$ is outside of the six disks. From (4.62) it follows
that $S(x)=R(x)M(x)$. And from (4.69)and $M_{+}$ is real analytic in a
neighborhood of $x$, then

$S^{-1}_{+}(y)S_{+}(x)=\mathbb{I}+O(x-y),\quad\mbox{ }\quad y\rightarrow x,$
(5.2)

uniformly for all $n$. Thus, we have

$\begin{array}[]{rl}K_{n}(x,y)&=\frac{e^{n(h(y)-h(x))}}{2\pi
i(x-y)}\left(\begin{array}[]{cccc}-e^{ni\mathrm{Im}\lambda_{1+}(y)}&e^{-ni\mathrm{Im}\lambda_{1+}(y)}&0&0\end{array}\right)(\mathbb{I}+O(x-y))\left(\begin{array}[]{c}e^{-ni\mathrm{Im}\lambda_{1+}(x)}\\\
e^{ni\mathrm{Im}\lambda_{1+}(x)}\\\ 0\\\ 0\end{array}\right)\\\
{}\hfil&=e^{n(h(y)-h(x))}\left(\frac{-e^{ni(\mathrm{Im}\lambda_{1+}(y)-\mathrm{Im}\lambda_{1+}(x))}+e^{-ni(\mathrm{Im}\lambda_{1+}(y)-\mathrm{Im}\lambda_{1+}(x))}}{2\pi
i(x-y)}+O(1)\right)\\\
{}\hfil&=e^{n(h(y)-h(x))}\left(\frac{sin{(n\mathrm{Im}(\lambda_{1+}(x)-\lambda_{1+}(y)))}}{\pi(x-y)}+O(1)\right),\end{array}$
(5.3)

Letting $y\rightarrow x$ and noting that

$\frac{d}{dy}\mathrm{Im}\lambda_{1+}(y)=\mathrm{Im}\xi_{1+}(y)=\pi\rho(y),$
(5.4)

By L’Hospital rule, we have

$K_{n}(x,x)=n\rho(x)+O(1).$ (5.5)

Similarly, for $x$ in $(-z_{3},-z_{2}),(-z_{1},z_{1})$

For
$x\in(-\infty,-z_{3})\cup(-z_{2},-z_{1})\cup(z_{1},z_{2})\cup(z_{3},\infty)$,
we have $K_{n}(x,x)$ decreases exponentially fast. This implies that

$\lim_{n\rightarrow\infty}\frac{1}{n}K_{n}(x,x)=0.$ (5.6)

If x is one of the edge points, it is shown in the proof of 5.3 that as
$n\rightarrow\infty$,

$\frac{1}{n}K_{n}(x,x)=O\left(\frac{1}{n^{\frac{1}{3}}}\right).$ (5.7)

∎

Let

$\hat{K}_{n}(x,y)=e^{n(h(x)-h(y))}K_{n}(x,y),$ (5.8)

where for $x\in(-z_{3},-z_{2})\cup(-z_{1},z_{1})\cup(z_{2},z_{3})$,

$h(x)=-\frac{1}{4}x^{2}+\mathrm{Re}\lambda_{1+}(x),\quad\lambda_{1+}(x)=\int_{z_{1}}^{x}\xi_{1+}(s)ds.$
(5.9)

######  Theorem 5.2.

For every $x_{0}\in(-z_{3},-z_{2})\cup(-z_{1},z_{1})\cup(z_{2},z_{3})$ and
$u,v\in{\mathbb{R}}$, we have

$\lim_{n\rightarrow\infty}\frac{1}{n\rho(x_{0})}\hat{K}_{n}\left(x_{0}+\frac{u}{n\rho(x_{0})},x_{0}+\frac{v}{n\rho(x_{0})}\right)=\frac{sin\pi(u-v)}{\pi(u-v)}.$
(5.10)

###### Proof.

We just prove $x\in(z_{2},z_{3})$, similarly for the rest. Letting

$x=x_{0}+\frac{u}{n\rho(x_{0})},\quad y=x_{0}+\frac{v}{n\rho(x_{0})}.$ (5.11)

By the definition of $\hat{K}_{n}$, we have

$\frac{1}{n\rho(x_{0})}\hat{K}_{n}(x,y)=\frac{sin(n\mathrm{Im}(\lambda_{1+}(x)-\lambda_{1+}(y)))}{\pi(u-v)}+O(\frac{1}{n}).$
(5.12)

By the mean value theorem, we have

$\mathrm{Im}(\lambda_{1+}(x)-\lambda_{1+}(y))=(x-y)\pi\rho(t),$ (5.13)

for some $t$ between $x$ and $y$. We also can know that $t=x_{0}+O(1/n)$ and

$n\mathrm{Im}(\lambda_{1+}(x)-\lambda_{1+}(y))=\pi(u-v)\frac{\rho(t)}{\rho(x_{0})}=\pi(u-v)\left(1+O(\frac{1}{n})\right).$
(5.14)

Then we have

$\frac{1}{n\rho(x_{0})}\hat{K}_{n}(x,y)=\frac{sin\pi(u-v)}{\pi(u-v)}+O(\frac{1}{n}).$
(5.15)

This is proven 5.2 ∎

######  Theorem 5.3.

We use the same notation as in Theorem 5.2. For every $u,v\in{\mathbb{R}}$, we
have

$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{1}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{3}+\frac{u}{(\rho_{1}n)^{\frac{2}{3}}},z_{3}+\frac{v}{(\rho_{1}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(5.16)
$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{2}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{2}+\frac{u}{(\rho_{2}n)^{\frac{2}{3}}},z_{2}+\frac{v}{(\rho_{2}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(5.17)
$\lim_{n\rightarrow\infty}\frac{1}{(\rho_{3}n)^{\frac{2}{3}}}\hat{K}_{n}\left(z_{1}+\frac{u}{(\rho_{3}n)^{\frac{2}{3}}},z_{1}+\frac{v}{(\rho_{3}n)^{\frac{2}{3}}}\right)=\frac{Ai(u)Ai^{\prime}(v)-Ai^{\prime}(u)Ai(v)}{u-v},$
(5.18)

Similar limits hold near the edge points $-z_{3},-z_{2},-z_{1}$.

###### Proof.

We just prove the first formula, similarly for the rest. Noting that
$\beta^{\prime}(z_{3})=\rho_{1}^{\frac{2}{3}}$.

Choosing that $u,v\in{\mathbb{R}}$ and letting

$x=z_{3}+\frac{u}{(\rho_{1}n)^{\frac{2}{3}}},\quad
y=z_{3}+\frac{v}{(\rho_{1}n)^{\frac{2}{3}}}$ (5.19)

Suppose that $u,v<0$, so we can use the formula (4.27) for $K_{n}(x,y)$. Thus,
we have as $n$ goes to infinity, $x$ is inside of $D(z_{3},r)$, from (4.48),
(4.60), (4.62), we have

$\begin{array}[]{rl}S_{+}(x)=&R(x)P_{+}(x)=R(x)\tilde{P}(x)\\\
{}=&R(x)E_{n}(x)\Phi_{+}(n^{\frac{2}{3}}\beta(x))diag\left(e^{\frac{1}{2}n(\lambda_{1}(z)-\lambda_{2}(z))},e^{-\frac{1}{2}n(\lambda_{1}(z)-\lambda_{2}(z))},1,1\right)\\\
{}=&R(x)E_{n}(x)\Phi_{+}(n^{\frac{2}{3}}\beta(x))diag\left(e^{ni\mathrm{Im}\lambda_{1+}(x)},e^{-ni\mathrm{Im}\lambda_{1+}(x)},1,1\right),\end{array}$
(5.20)

$S_{+}(y)$ has similar equation hold. Thus,

$\begin{array}[]{rl}\frac{1}{(\rho_{1}n)^{2/3}}\hat{K}_{n}(x,y)=&\frac{1}{2\pi
i(u-v)}\left(\begin{array}[]{cccc}-1&1&0&0\end{array}\right)\Phi^{-1}_{+}(n^{2/3}\beta(y))E^{-1}_{n}(y)R^{-1}(y)\\\
{}\hfil&{}\times
R(x)E_{n}(x)\Phi_{+}(n^{2/3}\beta(x))\left(\begin{array}[]{c}1\\\ 1\\\ 0\\\
0\end{array}\right).\end{array}$ (5.21)

Since $\rho^{2/3}=\beta^{\prime}(z_{3})$, as $n\rightarrow\infty$,

$n^{2/3}\beta(x)=n^{2/3}\beta\left(z_{3}+\frac{u}{(\rho_{1}n)^{2/3}}\right)\rightarrow
u,$ (5.22)

Then $\Phi_{+}(n^{2/3}\beta(x))\rightarrow\Phi_{+}(u)$. We use the second
formula of (4.58) to compute $\Phi_{+}(u)$, we have

$\lim_{n\rightarrow\infty}\Phi_{+}(n^{2/3}\beta(x))\left(\begin{array}[]{c}1\\\
1\\\ 0\\\ 0\end{array}\right)=\left(-y_{1}(u)-y_{2}(u)\\\
-y^{\prime}_{1}(u)-y^{\prime}_{2}(u)\\\ 0\\\
0\right)=\left(\begin{array}[]{c}y_{0}(u)\\\ y^{\prime}_{0}(u)\\\ 0\\\
0\end{array}\right).$ (5.23)

Similary, we have

$\lim_{n\rightarrow\infty}\left(\begin{array}[]{cccc}-1&1&0&0\end{array}\right)\Phi^{-1}_{+}(n^{2/3}\beta(y))=-2\pi
i\left(\begin{array}[]{cccc}-y^{\prime}_{0}(v)&y_{0}(v)&0&0\end{array}\right).$
(5.24)

Noting that $R^{-1}(y)R(x)=\mathbb{I}+O(\frac{x-y}{n})$, then

$R^{-1}(y)R(x)=\mathbb{I}+O(\frac{1}{n^{5/3}}).$ (5.25)

From the explicit formula of $E_{n}$,

$E_{n}(x)=O(n^{1/6}),\quad E^{-1}_{n}(y)=O(n^{1/6}),\quad
E^{-1}_{n}(y)E_{n}(x)=\mathbb{I}+O(\frac{1}{n^{1/3}}).$ (5.26)

we have

$\lim_{n\rightarrow\infty}E^{-1}_{n}(y)R^{-1}(y)R(x)E_{n}(x)=\mathbb{I}.$
(5.27)

Thus, we have

$\begin{array}[]{rl}\lim_{n\rightarrow\infty}\frac{1}{cn^{2/3}}\hat{K}_{n}(x,y)=&\frac{1}{2\pi
i(u-v)}\times(-2\pi
i)\left(\begin{array}[]{cccc}-y^{\prime}_{0}(v)&y_{0}(v)&0&0\end{array}\right)\left(\begin{array}[]{c}y_{0}(u)\\\
y^{\prime}_{0}(u)\\\ 0\\\ 0\end{array}\right)\\\
{}=&\frac{y_{0}(u)y^{\prime}_{0}(v)-y^{\prime}_{0}(u)y_{0}(v)}{u-v}.\end{array}$
(5.28)

Since $y_{0}=Ai$, then we prove the first equation as $u,v<0$. The rest is
similarly, we just use the another relation between $P$ and $\tilde{P}$. ∎

Acknowledgements This work of Xu was supported by Shanghai Sailing Program
supported by Science and Technology Commission of Shanghai Municipality under
Grant NO. 15YF1408100 and the Hujiang Foundation of China (B14005). Most of
this work was done during the author(Xu) visited Indiana University-Purdue
University, Indianapolis. And the author also would like to thank Prof. Bleher
for his advise and help and arranging this visit. Fan was support by grants
from the National Science Foundation of China (Project No. 11271079). Yang
thank the Macau Science and Technology Development Fund for generous support:
MYRG2014-00011-FST from Macau

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