Asymptotic expansion of the hard-to-soft edge transition

Luming Yao¹¹1School of Mathematical Sciences, Fudan University, Shanghai 200433, China. E-mail: {lumingyao, lunzhang}@fudan.edu.cn. and Lun Zhang²²footnotemark: 2

Abstract

By showing that the symmetrically transformed Bessel kernel admits a full asymptotic expansion for the large parameter, we establish a hard-to-soft edge transition expansion. This resolves a conjecture recently proposed by Bornemann.

1 Introduction and statement of results

Consideration in this paper is a universal phenomenon arising from the random matrix theory, namely, the hard-to-soft edge transition [8]. As a concrete example, let $X_{1}$ and $X_{2}$ be two $n\times(n+\nu)$ , $\nu\geq 0$ , random matrices, whose element is chosen to be an independent normal random variable. The $n\times n$ complex Wishart matrix $X$ or the Laguerre unitary ensemble (LUE), which plays an important role in statistics and signal processing (cf. [27, 30] and the references therein), is defined to be

X=(X_{1}+\mathrm{i}X_{2})(X_{1}+\mathrm{i}X_{2})^{\ast},

where the superscript ^∗ stands for the operation of conjugate transpose. As $n\to\infty$ with $\nu$ fixed, the smallest eigenvalue of $X$ accumulates near the hard-edge $0$ . After proper scaling, the limiting process is a determinantal point process characterized by the Bessel kernel [15, 18]

K_{\nu}^{\textrm{Bes}}(x,y):=\frac{J_{\nu}(\sqrt{x})\sqrt{y}J_{\nu}^{\prime}(\sqrt{y})-J_{\nu}(\sqrt{y})\sqrt{x}J_{\nu}^{\prime}(\sqrt{x})}{2(x-y)},\qquad x,y>0,

(1.1)

where $J_{\nu}$ is the Bessel function of the first kind of order $\nu$ (cf. [24]). If the parameter $\nu$ grows simultaneously with $n$ in such a way that $\nu/n$ approaches a positive constant, it comes out that the smallest eigenvalue is pushed away from the origin, creating a soft-edge. The fluctuation around the soft-edge, however, is given by the Airy point process [15, 18], which is determined by the Airy kernel

\displaystyle K^{\mathrm{Ai}}(x,y)=\frac{\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)-\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)}{x-y},\qquad x,y\in\mathbb{R},

(1.2)

where $\mathrm{Ai}$ is the standard Airy function. One encounters the same limiting process by considering the scaled cumulative distribution of largest eigenvalues for large random Hermitian matrices with complex Gaussian entries, which is also known as the Tracy-Widom distribution [28].

Besides the above explanation of the hard-to-soft edge transition, the present work is also highly motivated by its connection with distribution of the length of longest increasing subsequences. Let $S_{n}$ be the set of all permutations on $\{1,2,\ldots,n\}$ . Given $\sigma\in S_{n}$ , we denote by $L_{n}(\sigma)$ the length of longest increasing subsequences, which is defined as the maximum of all $k$ such that $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$ with $\sigma(i_{1})<\sigma(i_{2})<\cdots<\sigma(i_{k})$ . Equipped $S_{n}$ with the uniform measure, the question of the distribution of discrete random variable $L_{n}(\sigma)$ for large $n$ was posed by Ulam in the early 1960s [29]. After the efforts of many people (cf. [1, 26] and the references therein), Baik, Deift and Johansson finally answered this question in a celebrated work [2] by showing

\lim_{n\to\infty}\mathbb{P}\left(\frac{L_{n}-2\sqrt{n}}{n^{1/6}}\leq t\right)=F(t),

(1.3)

where

F(t):=\det(I-K^{\mathrm{Ai}})\big{|}_{L^{2}(t,\infty)}

(1.4)

is the aforementioned Tracy-Widom distribution with $K^{\mathrm{Ai}}$ being the Airy kernel in (1.2).

To establish (1.3), a key ingredient of the proof is to introduce the exponential generating function of $L_{n}$ defined by

P(r;l)=e^{-r}\sum_{n=0}^{\infty}\mathbb{P}(L_{n}\leq l)\frac{r^{n}}{n!},\qquad r>0,

which is known as Hammersley’s Poissonization of the random variable $L_{n}$ . The quantity itself can be interpreted as the cumulative distribution of $L(r)$ – the length of longest up/right path from $(0,0)$ to $(1,1)$ with nodes chosen randomly according to a Poisson process of intensity $r$ ; cf. [3, Chapter 2]. By representing $P(r;l)$ as a Toeplitz determinant, it was proved in [2] that

\lim_{r\to\infty}\mathbb{P}\left(\frac{L(r)-2\sqrt{r}}{r^{1/6}}\leq t\right)=F(t).

(1.5)

This, together with Johansson’s de-Poissonization lemma [20], will lead to (1.3).

Alternatively, one has (see [8, 16])

P(r;l)=E_{2}^{\mathrm{hard}}(4r;l),

(1.6)

where

E_{2}^{\mathrm{hard}}(s;\nu):=\det(I-K_{\nu}^{\mathrm{Bes}})\big{|}_{L^{2}(0,s)}

(1.7)

with $K_{\nu}^{\mathrm{Bes}}$ being the Bessel kernel defined in (1.1) is the scaled hard-edge gap probability of LUE over $(0,s)$ . Thus, by showing the hard-to-soft transition

\lim_{\nu\to\infty}E_{2}^{\mathrm{hard}}\left(\left(\nu-t(\nu/2)^{1/3}\right)^{2};\nu\right)=F(t),

(1.8)

Borodin and Forrester reclaimed (1.5) in [8].

An interesting question now is to improve (1.3) and (1.5) by establishing the first few finite-size correction terms or the asymptotic expansion. This is also known as edgeworth expansions in the literature, and we particularly refer to [7, 10, 13, 14, 19, 25] for the relevant results of Laguerre ensembles. In the context of the distribution for the length of longest increasing subsequences, the relationship (1.6) plays an important role in a recent work of Bornemann [5] among various studies toward this aim [4, 6, 17]. Instead of working on the Fredholm determinant directly, the idea in [5] is to establish an expansion between the Bessel kernel and the Airy kernel, which can be lifted to trace class operators. It is the aim of this paper to resolve some conjectures posed therein.

To proceed, we set

h_{\nu}:=2^{-\frac{1}{3}}\nu^{-\frac{2}{3}},

(1.9)

and define, as in [5], the symmetrically transformed Bessel kernel

\displaystyle\hat{K}_{\nu}^{\mathrm{Bes}}(x,y):=\sqrt{\phi_{\nu}^{\prime}(x)\phi_{\nu}^{\prime}(y)}K_{\nu}^{\mathrm{Bes}}(\phi_{\nu}(x),\phi_{\nu}(y)),

(1.10)

where

\phi_{\nu}(t):=\nu^{2}(1-h_{\nu}t)^{2}.

(1.11)

Our main result is stated as follows.

Theorem 1.1.

With $\hat{K}_{\nu}^{\mathrm{Bes}}(x,y)$ defined in (1.10), we have, for any $\mathfrak{m}\in\mathbb{N}$ ,

\displaystyle\hat{K}_{\nu}^{\mathrm{Bes}}(x,y)=K^{\mathrm{Ai}}(x,y)+\sum_{j=1}^{\mathfrak{m}}K_{j}(x,y)h_{\nu}^{j}+h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-(x+y)}\right),\qquad h_{\nu}\to 0^{+},

(1.12)

uniformly valid for $t_{0}\leq x,y<h_{\nu}^{-1}$ with $t_{0}$ being any fixed real number. Preserving uniformity, the expansion can be repeatedly differentiated w.r.t. the variable $x$ and $y$ . Here, $K^{\mathrm{Ai}}$ is the Airy kernel given in (1.2) and

\displaystyle K_{j}(x,y)=\sum_{\kappa,\lambda\in\{0,1\}}p_{j,\kappa\lambda}(x,y)\mathrm{Ai}^{(\kappa)}(x)\mathrm{Ai}^{(\lambda)}(y)

(1.13)

with $p_{j,\kappa\lambda}(x,y)$ being polynomials in $x$ and $y$ . Moreover, we have

	$\displaystyle K_{1}(x,y)$	$\displaystyle=\frac{1}{10}\left(-3(x^{2}+xy+y^{2})\mathrm{Ai}(x)\mathrm{Ai}(y)+2(\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)+\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y))\right.$
		$\displaystyle\left.~{}~{}~{}+3(x+y)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}^{\prime}(y)\right),$		(1.14)

and

$\displaystyle K_{2}(x,y)$	$\displaystyle=\frac{1}{1400}\left((56-235(x^{2}+y^{2})-319xy(x+y))\mathrm{Ai}(x)\mathrm{Ai}(y)\right.$
	$\displaystyle~{}~{}~{}+(63(x^{4}+x^{3}y-x^{2}y^{2}-xy^{3}-y^{4})-55x+239y)\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)$
	$\displaystyle~{}~{}~{}+(63(y^{4}+xy^{3}-x^{2}y^{2}-x^{3}y-x^{4})-55y+239x)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)$
	$\displaystyle\left.~{}~{}+(340(x^{2}+y^{2})+256xy)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}^{\prime}(y)\right).$	(1.15)

Based on a uniform version of transient asymptotic expansion of Bessel functions [23], the above theorem is stated in [5] under the condition that $0\leq\mathfrak{m}\leq\mathfrak{m}_{*}=100$ , where the upper bound $\mathfrak{m}_{*}$ is obtained through a numerical inspection. It is conjectured therein that (1.12) is valid without such a restriction, Theorem 1.1 thus gives a confirm answer to this conjecture.

As long as the Bessel kernel admits an expansion of the form (1.12), it is generally believed that one can lift the expansion to the associated Fredholm determinants. By carefully estimating trace norms in terms of kernel bounds, this is rigorously established in [5, Theorem 2.1] for the perturbed Airy kernel determinants, which allows us to obtain the following hard-to-soft edge transition expansion with the aid of Theorem 1.1.

Corollary 1.2.

With $E_{2}^{\mathrm{hard}}$ defined in (1.7), we have, for any $\mathfrak{m}\in\mathbb{N}$ ,

\displaystyle E_{2}^{\textrm{hard}}(\phi_{\nu}(t);\nu)=F(t)+\sum_{j=1}^{\mathfrak{m}}F_{j}(t)h_{\nu}^{j}+h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-3t/2}\right),\qquad h_{\nu}\to 0^{+},

(1.16)

uniformly valid for $t_{0}\leq t<h_{\nu}^{-1}$ with $t_{0}$ being any fixed real number. Preserving uniformity, the expansion can be repeatedly differentiated w.r.t. the variable $t$ . Here, $F$ denotes the Tracy-Widom distribution (1.4) and $F_{j}$ are certain smooth functions.

Again the above result is stated in [5] under a restriction on the number of summation but with explicit expressions of $F_{1}$ and $F_{2}$ in terms of the derivatives of $F$ . The expansion (1.16) serves as a preparatory step in establishing the expansion of the limit law (1.3) in [5]. Finally, we also refer to [9] for exponential moments, central limit theorems and rigidity of the hard-to-soft edge transition.

The rest of this paper is devoted to the proof of Theorem 1.1. The difficulty of using transient asymptotic expansion of the Bessel functions for large order to prove Theorem 1.1 lies in checking the divisibility of a certain sequence of polynomials. Indeed, it was commented in [5] that one probably needs some hidden symmetry of the coefficients in the expansion. The approach we adopt here, however, is based on a Riemann-Hilbert (RH) characterization of the Bessel kernel, as described in Section 2. By performing a Deift-Zhou nonlinear steepest descent analysis [12] to the associated RH problem in Section 3, the initial RH problem will be transformed into a small norm problem, for which one can find a uniform estimate. The proof of Theorem 1.1 is an outcome of our asymptotic analysis, which is presented in Section 4.

Notations

Throughout this paper, the following notations are frequently used.

•

If $A$ is a matrix, then $(A)_{ij}$ stands for its $(i,j)$ -th entry. We use $I$ to denote a $2\times 2$ identity matrix.

•

As usual, the three Pauli matrices $\{\sigma_{j}\}_{j=1}^{3}$ are defined by

\sigma_{1}=\begin{pmatrix}0&1\\ 1&0\end{pmatrix},\qquad\sigma_{2}=\begin{pmatrix}0&-\mathrm{i}\\ \mathrm{i}&0\end{pmatrix},\qquad\sigma_{3}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}.

(1.17)

2 An RH characterization of the Bessel kernel

Our starting point is the following Bessel parametrix [21], which particularly characterizes the Bessel kernel $K_{\nu}^{\textrm{Bes}}(x,y)$ in (1.1).

For $z\in\mathbb{C}\setminus[0,+\infty)$ and $\nu>-1$ , we set

\displaystyle\Psi_{\nu}(z)=\sqrt{\pi}e^{-\frac{\pi\mathrm{i}}{4}}\begin{pmatrix}I_{\nu}\left((-z)^{\frac{1}{2}}\right)&-\frac{\mathrm{i}}{\pi}K_{\nu}\left((-z)^{\frac{1}{2}}\right)\\ (-z)^{\frac{1}{2}}I_{\nu}^{\prime}\left((-z)^{\frac{1}{2}}\right)&-\frac{\mathrm{i}}{\pi}(-z)^{\frac{1}{2}}K^{\prime}_{\nu}\left((-z)^{\frac{1}{2}}\right)\end{pmatrix},

(2.1)

where $I_{\nu}$ and $K_{\nu}$ denote the modified Bessel functions of order $\nu$ (see [24]) and define

\displaystyle\Psi(z;\nu)=\Psi_{\nu}(z)\begin{cases}\begin{pmatrix}1&0\\ -e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(0,\frac{\pi}{3}),\\ I,&\qquad\arg z\in(\frac{\pi}{3},\frac{5\pi}{3}),\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(\frac{5\pi}{3},2\pi).\end{cases}

(2.2)

Then $\Psi(z):=\Psi(z;\nu)$ satisfies the following RH problem.

RH problem 2.1.

(a)

$\Psi(z)$ is defined and analytic for $z\in\mathbb{C}\setminus\left\{\cup_{j=1}^{3}\Gamma_{j}\cup\{0\}\right\}$ , where

\displaystyle\Gamma_{1}:=e^{\frac{\pi\mathrm{i}}{3}}(0,+\infty),\qquad\Gamma_{2}:=(0,+\infty),\qquad\Gamma_{3}:=e^{-\frac{\pi\mathrm{i}}{3}}(0,+\infty);

(2.3)

see Figure 1 for an illustration.

(b)

For $z\in\Gamma_{j}$ , $j=1,2,3$ , the limiting values of $\Psi$ exist and satisfy the jump condition

\displaystyle\Psi_{+}(z)=\Psi_{-}(z)\begin{cases}\begin{pmatrix}1&0\\ e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\Gamma_{1},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in\Gamma_{2},\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\Gamma_{3}.\end{cases}

(2.4)

(c)

As $z\to\infty$ , we have

\displaystyle\Psi(z)=\begin{pmatrix}1&0\\ -\frac{4\nu^{2}+3}{8}&1\end{pmatrix}\left(I+\mathcal{O}(z^{-1})\right)(-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{((-z)^{1/2}-\frac{\pi\mathrm{i}}{4})\sigma_{3}},

(2.5)

where $\sigma_{3}$ is defined in (1.17), the branch cuts of $(-z)^{\pm\frac{1}{4}}$ and $(-z)^{\pm\frac{1}{2}}$ are chosen along $[0,+\infty)$ .

(d)

As $z\to 0$ , we have, for $\nu\notin\mathbb{Z}$ ,

\displaystyle\Psi(z)=\widehat{\Psi}(z)(-z)^{\frac{\nu}{2}\sigma_{3}}\begin{pmatrix}1&\frac{\mathrm{i}}{2\sin\pi\nu}\\ 0&1\end{pmatrix}\begin{cases}\begin{pmatrix}1&0\\ -e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(0,\frac{\pi}{3}),\\ I,&\qquad\arg z\in(\frac{\pi}{3},\frac{5\pi}{3}),\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(\frac{5\pi}{3},2\pi),\end{cases}

(2.6)

and for $\nu\in\mathbb{Z}$ ,

\displaystyle\Psi(z)=\widehat{\Psi}(z)(-z)^{\frac{\nu}{2}\sigma_{3}}\begin{pmatrix}1&-\frac{e^{\pi\mathrm{i}\nu}}{2\pi\mathrm{i}}\ln(-z)\\ 0&1\end{pmatrix}\begin{cases}\begin{pmatrix}1&0\\ -e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(0,\frac{\pi}{3}),\\ I,&\qquad\arg z\in(\frac{\pi}{3},\frac{5\pi}{3}),\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad\arg z\in(\frac{5\pi}{3},2\pi),\end{cases}

(2.7)

where $\widehat{\Psi}(z)$ is analytic at $z=0$ and we choose principal branches for $(-z)^{\pm\frac{\nu}{2}}$ and $\ln(-z)$ .

Figure 1: The jump contours of the RH problem for

\Psi

The Bessel kernel then admits the following representation:

\displaystyle K_{\nu}^{\mathrm{Bes}}(x,y)=\frac{1}{2\pi\mathrm{i}(x-y)}\begin{pmatrix}-e^{-\frac{\pi\mathrm{i}\nu}{2}}&e^{\frac{\pi\mathrm{i}\nu}{2}}\end{pmatrix}\Psi_{+}(y)^{-1}\Psi_{+}(x)\begin{pmatrix}e^{\frac{\pi\mathrm{i}\nu}{2}}\\ e^{-\frac{\pi\mathrm{i}\nu}{2}}\end{pmatrix},

(2.8)

where the $+$ sign means taking limits from the upper half-plane.

3 Asymptotic analysis of the RH problem for $\Psi$ with large $\nu$

In this section, we will perform a Deift-Zhou steepest descent analysis [11] for the RH problem for $\Psi$ . It consists of a series of explicit and invertible transformations and the final goal is to arrive at an RH problem tending to the identity matrix as $\nu\to+\infty$ . Without loss of generality, we may assume that $\nu>0$ in what follows.

3.1 First transformation: $\Psi\to Y$

Due to the scaled variable (1.11), the first transformation is a scaling. In addition, we multiply some constant matrices from the left to simplify the asymptotic behavior at infinity.

Define the matrix-valued function

\displaystyle Y(z)=\nu^{\frac{1}{2}\sigma_{3}}\begin{pmatrix}1&0\\ \frac{4\nu^{2}+3}{8}&1\end{pmatrix}\Psi(\nu^{2}z).

(3.1)

It is then readily seen from RH problem 2.1 that $Y$ satisfies the following RH problem.

RH problem 3.1.

(a)

$Y(z)$ is defined and analytic for $z\in\mathbb{C}\setminus\left\{\cup_{j=1}^{3}\Gamma_{j}\cup\{0\}\right\}$ , where $\Gamma_{i}$ , $i=1,2,3$ , is defined in (2.3).

(b)

$Y(z)$ satisfies the jump condition

\displaystyle Y_{+}(z)=Y_{-}(z)\begin{cases}\begin{pmatrix}1&0\\ e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\Gamma_{1},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in\Gamma_{2},\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\Gamma_{3}.\end{cases}

(3.2)

(c)

As $z\to\infty$ , we have

\displaystyle Y(z)=\left(I+\mathcal{O}(z^{-1})\right)(-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{(\nu(-z)^{1/2}-\frac{\pi\mathrm{i}}{4})\sigma_{3}},

(3.3)

where we take the principal branch for fractional exponents.

3.2 Second transformation: $Y\to T$

In the second transformation we apply contour deformations. The rays $\Gamma_{1}$ and $\Gamma_{3}$ emanating from the origin are replaced by their parallel lines $\widetilde{\Gamma}_{1}$ and $\widetilde{\Gamma}_{3}$ emanating from $1$ . Let I and II be two regions bounded by $\Gamma_{1}\cup\widetilde{\Gamma}_{1}\cup[0,1]$ and $\Gamma_{3}\cup\widetilde{\Gamma}_{3}\cup[0,1]$ , respectively; see Figure 2 for an illustration. We now define

\displaystyle T(z)=Y(z)\begin{cases}\begin{pmatrix}1&0\\ e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\textrm{I},\\ \begin{pmatrix}1&0\\ -e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\textrm{II},\\ I,&\qquad\textrm{elsewhere.}\end{cases}

(3.4)

Figure 2: The jump contour

\Gamma_{T}

of the RH problem for

T

It is easily seen the following RH problem for $T$ .

RH problem 3.2.

(a)

$T(z)$ is defined and analytic in $\mathbb{C}\setminus\Gamma_{T}$ , where

\Gamma_{T}:=\cup_{i=1}^{3}\widetilde{\Gamma}_{i}\cup[0,1],

(3.5)

with

\displaystyle\widetilde{\Gamma}_{1}:=e^{\frac{\pi\mathrm{i}}{3}}(1,+\infty),\qquad\widetilde{\Gamma}_{2}:=(1,+\infty),\qquad\widetilde{\Gamma}_{3}:=e^{-\frac{\pi\mathrm{i}}{3}}(1,+\infty);

(3.6)

see Figure 2.

(b)

$T(z)$ satisfies the jump condition

\displaystyle T_{+}(z)=T_{-}(z)\begin{cases}\begin{pmatrix}1&0\\ e^{-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{1},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{2},\\ \begin{pmatrix}1&0\\ e^{\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{3},\\ \begin{pmatrix}e^{-\pi\mathrm{i}\nu}&1\\ 0&e^{\pi\mathrm{i}\nu}\end{pmatrix},&\qquad z\in(0,1).\end{cases}

(3.7)

(c)

As $z\to\infty$ , we have

\displaystyle T(z)=\left(I+\mathcal{O}(z^{-1})\right)(-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{(\nu(-z)^{1/2}-\frac{\pi\mathrm{i}}{4})\sigma_{3}}.

(3.8)

3.3 Third transformation: $T\to S$

In order to normalize the behavior at infinity, we apply the third transformation $T\to S$ by introducing the so-called $g$ -function:

\displaystyle g(z):=-(1-z)^{\frac{1}{2}}+\frac{1}{2}\ln\left(\frac{1+(1-z)^{1/2}}{1-(1-z)^{1/2}}\right)\pm\frac{\pi\mathrm{i}}{2},\quad\pm\mathrm{Im}\,z>0.

(3.9)

As before, we take a cut along $[1,+\infty)$ for $(1-z)^{\frac{1}{2}}$ . The following proposition of $g$ is immediate from its definition.

Proposition 3.3.

(i)

The function $g(z)$ is analytic in $\mathbb{C}\setminus[0,+\infty)$ .
(ii)

For $z\in(-\infty,0)$ , we have

$\displaystyle g(z)=-\sqrt{1-z}+\frac{1}{2}\ln\left(\frac{1+\sqrt{1-z}}{\sqrt{1-z}-1}\right).$ (3.10)
(iii)

For $z\in(0,1)$ , we have

$\displaystyle g_{\pm}(z)=-\sqrt{1-z}+\frac{1}{2}\ln\left(\frac{1+\sqrt{1-z}}{1-\sqrt{1-z}}\right)\pm\frac{\pi\mathrm{i}}{2}.$ (3.11)

(iv)

For $z\in(1,+\infty)$ , we have

\displaystyle g_{\pm}(z)=\pm\mathrm{i}\sqrt{z-1}\pm\frac{\mathrm{i}}{2}\arg\left(\frac{1-\mathrm{i}\sqrt{z-1}}{1+\mathrm{i}\sqrt{z-1}}\right)\pm\frac{\pi\mathrm{i}}{2}.

(3.12)

(v)

As $z\to\infty$ , we have

$\displaystyle g(z)=-\left(-z\right)^{\frac{1}{2}}+\frac{1}{2}\left(-z\right)^{-\frac{1}{2}}+\mathcal{O}(z^{-1}).$ (3.13)

Refer to caption — Figure 3: Image of $\mathrm{Re}\,g$ : the solid line is the contour of $\mathrm{Re}\,{g(z)}=0$ , the “ $-$ ” sign is the region where $\mathrm{Re}\,{g(z)}<0$ and the “ $+$ ” sign shows the region where $\mathrm{Re}\,{g(z)}>0$ .

By setting

\displaystyle S(z)=T(z)e^{\nu g(z)\sigma_{3}},

(3.14)

it is readily seen from Proposition 3.3 and RH problem 3.2 that $S$ satisfies the RH problem as follows.

RH problem 3.4.

(a)

$S(z)$ is defined and analytic in $\mathbb{C}\setminus\Gamma_{T}$ , where the contour $\Gamma_{T}$ is defined in (3.5).

(b)

$S(z)$ satisfies the jump condition

\displaystyle S_{+}(z)=S_{-}(z)\begin{cases}\begin{pmatrix}1&0\\ e^{2\nu g(z)-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{1},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{2},\\ \begin{pmatrix}1&0\\ e^{2\nu g(z)+\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in\widetilde{\Gamma}_{3},\\ \begin{pmatrix}1&e^{2\nu\left(\sqrt{1-z}-\frac{1}{2}\ln\left(\frac{1+\sqrt{1-z}}{1-\sqrt{1-z}}\right)\right)}\\ 0&1\end{pmatrix},&\qquad z\in(0,1).\end{cases}

(3.15)

(c)

As $z\to\infty$ , we have

\displaystyle S(z)=\left(I+\mathcal{O}(z^{-1})\right)(-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{-\frac{\pi\mathrm{i}}{4}\sigma_{3}}.

(3.16)

3.4 Global parametrix

As $\nu\to+\infty$ , from the image of $\mathrm{Re}\,g$ depicted in Figure 3, we conclude that all the jump matrices of $S$ tend to $I$ exponentially fast except for that along $(1,+\infty)$ . Ignoring the exponential small terms in the jump matrices for $S$ , we come to the following global parametrix.

RH problem 3.5.

(a)

$N(z)$ is defined and analytic in $\mathbb{C}\setminus[1,+\infty)$ .
(b)

$N(z)$ satisfies the jump condition

$\displaystyle N_{+}(z)=N_{-}(z)\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\qquad z\in[1,+\infty).$ (3.17)

(c)

As $z\to\infty$ , we have

\displaystyle N(z)=\left(I+\mathcal{O}(z^{-1})\right)(-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{-\frac{\pi\mathrm{i}}{4}\sigma_{3}}.

(3.18)

An explicit solution to the RH problem for $N$ is given by

\displaystyle N(z)=(1-z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}e^{-\frac{\pi\mathrm{i}}{4}\sigma_{3}}.

(3.19)

3.5 Local parametrix

Since the jump matrices for $S$ and $N$ are not uniformly close to each other near the point $z=1$ , we next construct local parametrix near this point. In a disc $D(1,\varepsilon)$ centered at 1 with certain fixed radius $0<\varepsilon<1$ , we seek a $2\times 2$ matrix-valued function $P(z)$ satisfying an RH problem as follows.

RH problem 3.6.

(a)

$P(z)$ is defined and analytic in $D(1,\varepsilon)\setminus\Gamma_{T}$ , where the contour $\Gamma_{T}$ is defined in (3.5).

(b)

$P(z)$ satisfies the jump condition

\displaystyle P_{+}(z)=P_{-}(z)\begin{cases}\begin{pmatrix}1&0\\ e^{2\nu g(z)-\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in D(1,\varepsilon)\cap\widetilde{\Gamma}_{1},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in D(1,\varepsilon)\cap\widetilde{\Gamma}_{2},\\ \begin{pmatrix}1&0\\ e^{2\nu g(z)+\pi\mathrm{i}\nu}&1\end{pmatrix},&\qquad z\in D(1,\varepsilon)\cap\widetilde{\Gamma}_{3},\\ \begin{pmatrix}1&e^{2\nu\left(\sqrt{1-z}-\frac{1}{2}\ln\left(\frac{1+\sqrt{1-z}}{1-\sqrt{1-z}}\right)\right)}\\ 0&1\end{pmatrix},&\qquad z\in D(1,\varepsilon)\cap(0,1).\end{cases}

(3.20)

(c)

As $\nu\to\infty$ , we have

$\displaystyle P(z)=\left(I+\mathcal{O}(\nu^{-1})\right)N(z),\qquad z\in\partial D(1,\varepsilon),$ (3.21)

where $N$ is given in (3.19).

This local parametrix can be constructed by using the Airy parametrix $\Phi^{(\mathrm{Ai})}$ introduced in Appendix A. To do this, we introduce the function:

	$\displaystyle f(z)$	$\displaystyle=\left(\frac{3g(z)}{2}\mp\frac{3\pi\mathrm{i}}{4}\right)^{\frac{2}{3}},\qquad\pm\mathrm{Im}\,z>0$
		$\displaystyle=-2^{-\frac{2}{3}}(z-1)\left(1-\frac{2}{5}(z-1)+\frac{43}{175}(z-1)^{2}+\mathcal{O}\left((z-1)^{3}\right)\right),\qquad z\to 1.$		(3.22)

It is easily obtained that

\displaystyle f(1)=0,\qquad f^{\prime}(1)=-2^{-\frac{2}{3}}<0.

(3.23)

We then set

\displaystyle P(z)=E(z)\Phi^{(\mathrm{Ai})}\left(\nu^{\frac{2}{3}}f(z)\right)e^{\nu\left(g(z)\mp\frac{\pi\mathrm{i}}{2}\right)\sigma_{3}}\sigma_{3}

(3.24)

with

\displaystyle E(z):=N(z)\sigma_{3}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-\mathrm{i}\\ -\mathrm{i}&1\end{pmatrix}f(z)^{\frac{1}{4}\sigma_{3}}\nu^{\frac{1}{6}\sigma_{3}}.

(3.25)

Proposition 3.7.

The matrix-valued function $P(z)$ defined in (3.24) solves RH problem 3.6.

Proof.

We first show the prefactor $E(z)$ is analytic near $z=1$ . According to its definition in (3.25), the only possible jump is on $(1,1+\varepsilon)$ . It follows from (3.17) and (3.5) that, if $z\in(1,1+\varepsilon)$ ,

	$\displaystyle E_{-}(z)^{-1}E_{+}(z)$	$\displaystyle=\nu^{-\frac{1}{6}\sigma_{3}}f_{-}(z)^{-\frac{1}{4}\sigma_{3}}\frac{1}{\sqrt{2}}\begin{pmatrix}1&\mathrm{i}\\ \mathrm{i}&1\end{pmatrix}\sigma_{3}\begin{pmatrix}0&1\\ -1&0\end{pmatrix}\sigma_{3}\frac{1}{\sqrt{2}}\begin{pmatrix}1&-\mathrm{i}\\ -\mathrm{i}&1\end{pmatrix}f_{+}(z)^{\frac{1}{4}\sigma_{3}}\nu^{\frac{1}{6}\sigma_{3}}$
		$\displaystyle=\nu^{-\frac{1}{6}\sigma_{3}}f_{-}(z)^{-\frac{1}{4}\sigma_{3}}\begin{pmatrix}\mathrm{i}&0\\ 0&-\mathrm{i}\end{pmatrix}f_{+}(z)^{\frac{1}{4}\sigma_{3}}\nu^{\frac{1}{6}\sigma_{3}}=I,$		(3.26)

since $f_{+}(z)=e^{-2\pi\mathrm{i}}f_{-}(z)$ for $z\in(1,1+\varepsilon)$ . Thus, $E(z)$ is analytic in $D(1,\varepsilon)\setminus\{1\}$ . Note that, as $z\to 1$ , by using (3.5),

\displaystyle E(z)=(2h_{\nu})^{-\frac{1}{4}\sigma_{3}}e^{-\frac{\pi\mathrm{i}}{4}\sigma_{3}}\sigma_{3}\left(I-\frac{\sigma_{3}}{10}(z-1)+\begin{pmatrix}\frac{13}{280}&0\\ 0&-\frac{51}{1400}\end{pmatrix}(z-1)^{2}+\mathcal{O}((z-1)^{3})\right)

(3.27)

with $h_{\nu}$ given in (1.9), we conclude that $z=1$ is a removable singularity. The jump conditions of $P(z)$ in (3.20) follows directly from the analyticity of $E(z)$ and (A.1). Finally, as $\nu\to+\infty$ , we apply (A.2) and obtain after a straightforward computation that

\displaystyle P(z)N(z)^{-1}=N(z)\sigma_{3}\left(I+\mathcal{O}(\nu^{-1})\right)\sigma_{3}N(z)^{-1}=I+\mathcal{O}(\nu^{-1}),\qquad z\in\partial D(1,\varepsilon).

(3.28)

This completes the proof of Proposition 3.7. ∎

3.6 Final transformation

We define the final transformation

\displaystyle R(z)=\begin{cases}S(z)P(z)^{-1},&\qquad z\in D(1,\varepsilon),\\ S(z)N(z)^{-1},&\qquad\textrm{elsewhere}.\end{cases}

(3.29)

It is then readily seen that $R(z)$ satisfies the following RH problem.

RH problem 3.8.

(a)

$R(z)$ is defined and analytic in $\mathbb{C}\setminus\Gamma_{R}$ , where

$\Gamma_{R}:=\partial D(1,\varepsilon)\cup\Gamma_{T}\setminus D(1,\varepsilon);$

see Figure 4 for an illustration.

(b)

$R(z)$ satisfies the jump condition

\displaystyle R_{+}(z)=R_{-}(z)J_{R}(z),\qquad z\in\Gamma_{R},

(3.30)

where

\displaystyle J_{R}(z)=\begin{cases}P(z)N(z)^{-1},&\qquad z\in\partial D(1,\varepsilon),\\ N(z)S(z)N(z)^{-1},&\qquad z\in\Gamma_{R}\setminus\partial D(1,\varepsilon).\end{cases}

(3.31)

Figure 4: The jump contours of the RH problem for

R

For $z\in\Gamma_{R}\setminus\partial D(1,\varepsilon)$ , we have the estimate

J_{R}(z)=I+\mathcal{O}(e^{-c\nu}),\qquad\nu\to+\infty,

(3.33)

for some constant $c>0$ .

For $z\in\partial D(1,\varepsilon)$ , substituting the full expansion (A.4) of $\Phi^{({\mathrm{Ai}})}$ into (3.24) and (3.31), we have

\displaystyle J_{R}(z)\sim I+\sum_{k=1}^{\infty}J_{k}(z)h_{\nu}^{\frac{3k}{2}},\qquad\nu\to+\infty,

(3.34)

where

\displaystyle J_{k}(z)=\begin{cases}-\frac{3^{k}}{2^{k/2}f(z)^{3k/2}}\begin{pmatrix}0&(1-z)^{-\frac{1}{2}}\mathfrak{u}_{k}\\ (1-z)^{\frac{1}{2}}\mathfrak{v}_{k}&0\end{pmatrix},&\quad\textrm{for odd $k$, }\\ \frac{3^{k}}{2^{k/2}f(z)^{3k/2}}\begin{pmatrix}\mathfrak{u}_{k}&0\\ 0&\mathfrak{v}_{k}\end{pmatrix},&\quad\textrm{for even $k$, }\end{cases}

(3.35)

with $f(z)$ and $\mathfrak{u}_{k}$ , $\mathfrak{v}_{k}$ given in (3.5) and (A.5), respectively. By a standard argument [11, 12], we conclude that, as $\nu\to+\infty$ ,

\displaystyle R(z)\sim I+\sum_{k=1}^{\infty}R_{k}(z)h_{\nu}^{\frac{3k}{2}}

(3.36)

uniformly for $z\in\mathbb{C}\setminus\Gamma_{R}$ . A combination of (3.36) and RH problem 3.8 shows that $R_{1}$ satisfies

RH problem 3.9.

(a)

$R_{1}(z)$ is defined and analytic in $\mathbb{C}\setminus\partial D(1,\varepsilon)$ .

(b)

$R_{1}(z)$ satisfies the jump condition

\displaystyle R_{1,+}(z)=R_{1,-}(z)+J_{1}(z),\qquad z\in\partial D(1,\varepsilon),

(3.37)

where

\displaystyle J_{1}(z)=-\frac{\sqrt{2}}{48f(z)^{3/2}}\begin{pmatrix}0&5(1-z)^{-\frac{1}{2}}\\ -7(1-z)^{\frac{1}{2}}&0\end{pmatrix}.

(3.38)

From the local behavior of $f(z)$ near $z=1$ given in (3.5), we obtain that

\displaystyle J_{1}(z)=\frac{1}{(z-1)^{2}}\begin{pmatrix}0&\frac{5\sqrt{2}}{24}\\ 0&0\end{pmatrix}+\frac{1}{z-1}\begin{pmatrix}0&\frac{\sqrt{2}}{8}\\ -\frac{7\sqrt{2}}{24}&0\end{pmatrix}-\begin{pmatrix}0&\frac{\sqrt{2}}{70}\\ \frac{7\sqrt{2}}{40}&0\end{pmatrix}+\mathcal{O}(z-1).

(3.39)

By Cauchy’s residue theorem, we have

	$\displaystyle R_{1}(z)$	$\displaystyle=\frac{1}{2\pi\mathrm{i}}\oint_{\partial D(1,\varepsilon)}\frac{J_{1}(s)}{z-s}\,\mathrm{d}s$
		$\displaystyle=\begin{cases}\frac{1}{(z-1)^{2}}\begin{pmatrix}0&\frac{5\sqrt{2}}{24}\\ 0&0\end{pmatrix}+\frac{1}{z-1}\begin{pmatrix}0&\frac{\sqrt{2}}{8}\\ -\frac{7\sqrt{2}}{24}&0\end{pmatrix},\quad&z\in\mathbb{C}\setminus D(1,\varepsilon),\\ \frac{1}{(z-1)^{2}}\begin{pmatrix}0&\frac{5\sqrt{2}}{24}\\ 0&0\end{pmatrix}+\frac{1}{z-1}\begin{pmatrix}0&\frac{\sqrt{2}}{8}\\ -\frac{7\sqrt{2}}{24}&0\end{pmatrix}-J_{1}(z),\quad&z\in D(1,\varepsilon).\end{cases}$		(3.40)

Similarly, $R_{2}$ satisfies the following RH problem.

RH problem 3.10.

(a)

$R_{2}(z)$ is defined and analytic in $\mathbb{C}\setminus\partial D(1,\varepsilon)$ .
(b)

$R_{2}(z)$ satisfies the jump condition

$\displaystyle R_{2,+}(z)=R_{2,-}(z)+R_{1,-}(z)J_{1}(z)+J_{2}(z),\qquad z\in\partial D(1,\varepsilon),$ (3.41)

where

$\displaystyle J_{2}(z)=\frac{9}{2f(z)^{3}}\begin{pmatrix}\mathfrak{u}_{2}&0\\ 0&\mathfrak{v}_{2}\end{pmatrix}$ (3.42)

with $\mathfrak{u}_{2}$ and $\mathfrak{v}_{2}$ given in (A.5).
(c)

As $z\to\infty$ , we have $R_{2}(z)=\mathcal{O}(z^{-1})$ .

From (3.6), (3.38), (3.42) and Cauchy’s residue theorem, it follows that

\displaystyle R_{2}(z)=\frac{1}{2\pi\mathrm{i}}\oint_{\partial D(1,\varepsilon)}\frac{R_{1,-}(s)J_{1}(s)+J_{2}(s)}{z-s}\,\mathrm{d}s

(3.43)

is a diagonal matrix. For general $k\geq 3$ , the functions $R_{k}$ are analytic in $\mathbb{C}\setminus\partial D(1,\varepsilon)$ with asymptotic behavior $\mathcal{O}(1/z)$ as $z\to\infty$ , and satisfy

\displaystyle R_{k,+}(z)=R_{k,-}(z)+\sum_{l=1}^{k}R_{k-l,-}(z)J_{l}(z),\qquad z\in\partial D(1,\varepsilon),

(3.44)

where the functions $J_{k}(z)$ are given in (3.35). By Cauchy’s residue theorem, we have

R_{k}(z)=\frac{1}{2\pi\mathrm{i}}\oint_{\partial D(1,\varepsilon)}\sum_{l=1}^{k}R_{k-l,-}(s)J_{l}(s)\frac{\,\mathrm{d}s}{z-s}.

(3.45)

One can check that, by the structure of $J_{k}(z)$ and mathematical induction, each $R_{k}$ takes the following structure:

R_{k}(z)=\begin{cases}\begin{pmatrix}0&\left(R_{k}(z)\right)_{12}\\ \left(R_{k}(z)\right)_{21}&0\end{pmatrix},&\textrm{for odd $k$,}\\ \begin{pmatrix}\left(R_{k}(z)\right)_{11}&0\\ 0&\left(R_{k}(z)\right)_{22}\end{pmatrix},&\textrm{for even $k$.}\end{cases}

(3.46)

This, together with (3.36), gives us

	$\displaystyle h_{\nu}^{\frac{1}{4}\sigma_{3}}R(z)h_{\nu}^{-\frac{1}{4}\sigma_{3}}$	$\displaystyle\sim I+\sum_{k=0}^{\infty}\left(h_{\nu}^{3k+1}\begin{pmatrix}0&0\\ \left(R_{2k+1}(z)\right)_{21}&0\end{pmatrix}+h_{\nu}^{3k+2}\begin{pmatrix}0&\left(R_{2k+1}(z)\right)_{12}\\ 0&0\end{pmatrix}\right.$
		$\displaystyle~{}~{}~{}\left.+h_{\nu}^{3k+3}\begin{pmatrix}\left(R_{2k+2}(z)\right)_{11}&0\\ 0&\left(R_{2k+2}(z)\right)_{22}\end{pmatrix}\right),\qquad\nu\to+\infty.$		(3.47)

We are now ready to prove our main result.

4 Proof of Theorem 1.1

Recall the RH characterization of the Bessel kernel given in (2.8), we then follow the series of transformations $\Psi\to Y\to T\to S$ in (3.1) (3.4) and (3.14) to obtain that

\nu^{2}K_{\nu}^{\mathrm{Bes}}(\nu^{2}u,\nu^{2}v)=\frac{1}{2\pi\mathrm{i}(u-v)}\begin{pmatrix}0&1\end{pmatrix}e^{\nu(g_{+}(v)-\pi\mathrm{i}/2)\sigma_{3}}S_{+}(v)^{-1}S_{+}(u)\\ \times e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)\sigma_{3}}\begin{pmatrix}1\\ 0\end{pmatrix},\qquad u,v>0,

(4.1)

where $g$ is given in (3.9). In what follows, we split our discussions into different cases based on different ranges of $u$ and $v$ .

If $u\in(0,1-\varepsilon]$ and $v\in(1-\varepsilon,1+\varepsilon)$ , applying the final transformation (3.29) and (3.24) shows that

		$\displaystyle\nu^{2}K_{\nu}^{\mathrm{Bes}}(\nu^{2}u,\nu^{2}v)$
		$\displaystyle=\frac{1}{2\pi\mathrm{i}(u-v)}\begin{pmatrix}0&1\end{pmatrix}e^{\nu(g_{+}(v)-\pi\mathrm{i}/2)\sigma_{3}}P_{+}(v)^{-1}R(v)^{-1}R_{+}(u)N(u)e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)\sigma_{3}}\begin{pmatrix}1\\ 0\end{pmatrix}$
		$\displaystyle=-\frac{e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)}}{\mathrm{i}\sqrt{2\pi}(u-v)}\begin{pmatrix}\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f(v))&\mathrm{Ai}(\nu^{\frac{2}{3}}f(v))\end{pmatrix}E(v)^{-1}R(v)^{-1}R_{+}(u)N(u)\begin{pmatrix}1\\ 0\end{pmatrix},$		(4.2)

where the functions $E,R$ and $N$ are given in (3.25), (3.29) and (3.19), respectively. By (3.11), it is readily seen that

g_{+}^{\prime}(x)=-\frac{\sqrt{1-x}}{2x}<0,\qquad x\in(0,1-\varepsilon].

(4.3)

Thus, for $u\in(0,1-\varepsilon]$ ,

e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)}\leq e^{-\nu(g_{+}(1-\varepsilon)-\pi\mathrm{i}/2)}=e^{-\nu\left(\frac{1}{2}\ln\frac{1+\sqrt{\varepsilon}}{1-\sqrt{\varepsilon}}-\sqrt{\varepsilon}\right)}<e^{-\frac{3}{2}h_{\nu}^{-1}},\qquad\nu\to+\infty,

(4.4)

and other terms in (4) are bounded for large positive $\nu$ , which follow from (3.19), (3.25) and (3.36). By taking

u=(1-xh_{\nu})^{2},\qquad v=(1-yh_{\nu})^{2}

(4.5)

in (4), where

(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}\leq x<h_{\nu}^{-1},\qquad t_{0}\leq y<(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}

(4.6)

with $t_{0}$ being any fixed real number, we have, for any $\mathfrak{m}\in\mathbb{N}$ ,

	$\displaystyle\sqrt{\phi_{\nu}^{\prime}(x)\phi_{\nu}^{\prime}(y)}K_{\nu}^{\mathrm{Bes}}(\phi_{\nu}(x),\phi_{\nu}(y))<e^{-\frac{3}{2}h_{\nu}^{-1}}\cdot\mathcal{O}(e^{-y})$
	$\displaystyle<e^{-\frac{1}{2}h_{\nu}^{-1}}\cdot\mathcal{O}\left(e^{-(x+y)}\right)=h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-(x+y)}\right),\qquad h_{\nu}\to 0^{+}.$		(4.7)

Here, $\phi_{\nu}$ is defined in (1.11) and the error term $\mathcal{O}(e^{-y})$ in the first inequality comes from the estimate [24, Formula 9.7.15]

|p(\zeta)|\cdot\max\left(|\mathrm{Ai}(\zeta)|,|\mathrm{Ai}^{\prime}(\zeta)|\right)\leq c_{p}e^{-\zeta},\qquad\zeta\in\mathbb{R},

(4.8)

where $p$ is an arbitrary polynomial and the constant $c_{p}$ only depends on $p$ . As a consequence, the transformed Bessel kernel (1.10) will be absorbed into the error term of (1.12) in this case. As for the expansion terms in (1.12), note that $x$ is large as $\nu\to+\infty$ , we obtain again from [24, Formula 9.7.15] that, for any arbitrary polynomials $q$ ,

q(x)\cdot\mathrm{Ai}(x)\leq q(x)\cdot\frac{e^{-\frac{2}{3}x^{3/2}}}{2\sqrt{\pi}x^{1/4}}<e^{-\frac{3}{2}h_{\nu}^{-1}}<h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-x}\right),

(4.9)

and

q(x)\cdot|\mathrm{Ai}^{\prime}(x)|\leq q(x)\cdot\frac{x^{1/4}e^{-\frac{2}{3}x^{3/2}}}{2\sqrt{\pi}}\left(1+\frac{7}{48x^{3/2}}\right)<e^{-\frac{3}{2}h_{\nu}^{-1}}<h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-x}\right).

(4.10)

This, together with the estimate (4.8) for $\mathrm{Ai}(y)$ and $\mathrm{Ai}^{\prime}(y)$ , implies that the expansion terms in (1.12) is also absorbed into the error term, which shows that (1.12) is valid under the condition (4.6).

A similar argument holds if $u\in(1-\varepsilon,1+\varepsilon)$ and $v\in(0,1-\varepsilon]$ , which implies (1.12) for $t_{0}\leq x<(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}$ and $(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}\leq y<h_{\nu}^{-1}$ .

If $0<u,v\leq 1-\varepsilon$ , we obtain from (3.29) that

		$\displaystyle\nu^{2}K_{\nu}^{\mathrm{Bes}}(\nu^{2}u,\nu^{2}v)$
		$\displaystyle=\frac{1}{2\pi\mathrm{i}(u-v)}\begin{pmatrix}0&1\end{pmatrix}e^{\nu(g_{+}(v)-\pi\mathrm{i}/2)\sigma_{3}}N(v)^{-1}R_{+}(v)^{-1}R_{+}(u)N(u)e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)\sigma_{3}}\begin{pmatrix}1\\ 0\end{pmatrix}$
		$\displaystyle=\frac{e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)}e^{-\nu(g_{+}(v)-\pi\mathrm{i}/2)}}{2\pi\mathrm{i}(u-v)}\begin{pmatrix}0&1\end{pmatrix}N(v)^{-1}R_{+}(v)^{-1}R_{+}(u)N(u)\begin{pmatrix}1\\ 0\end{pmatrix}.$		(4.11)

From (4.3), it follows that

e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)}e^{-\nu(g_{+}(v)-\pi\mathrm{i}/2)}\leq e^{-2\nu(g_{+}(1-\varepsilon)-\pi\mathrm{i}/2)}\\ =e^{-\nu\left(\ln\frac{1+\sqrt{\varepsilon}}{1-\sqrt{\varepsilon}}-2\sqrt{\varepsilon}\right)}<e^{-3h_{\nu}^{-1}},\qquad\nu\to+\infty,

(4.12)

and other terms in (4) are bounded for large positive $\nu$ , as can be seen from their definitions in (3.19) and (3.36). Substituting (4.5) into (4) with

(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}\leq x,y<h_{\nu}^{-1},

(4.13)

we have, for any $\mathfrak{m}\in\mathbb{N}$ ,

	$\displaystyle\sqrt{\phi_{\nu}^{\prime}(x)\phi_{\nu}^{\prime}(y)}K_{\nu}^{\mathrm{Bes}}(\phi_{\nu}(x),\phi_{\nu}(y))<e^{-3h_{\nu}^{-1}}\cdot\mathcal{O}(1)$
	$\displaystyle<e^{-h_{\nu}^{-1}}\cdot\mathcal{O}\left(e^{-(x+y)}\right)=h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-(x+y)}\right),\qquad h_{\nu}\to 0^{+}.$		(4.14)

The transformed Bessel kernel is again absorbed into the error term completely in this case. The removability of the singularities at $x=y$ comes from the symmetric structure of the expansion (4). Since the Airy functions in the expansion (1.12) related to $x$ and $y$ are superexponential decay as $\nu\to+\infty$ (see (4.9) and (4.10)), we conclude (1.12) under the condition (4.13).

It remains to consider the final case, namely, $1-\varepsilon<u,v<1+\varepsilon$ , which corresponds to

t_{0}\leq x,y<(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}

(4.15)

through (4.5). To proceed, we again observe from (3.29) and (3.24) that

		$\displaystyle\nu^{2}K_{\nu}^{\mathrm{Bes}}(\nu^{2}u,\nu^{2}v)$
		$\displaystyle=\frac{1}{2\pi\mathrm{i}(u-v)}\begin{pmatrix}0&1\end{pmatrix}e^{\nu(g_{+}(v)-\pi\mathrm{i}/2)\sigma_{3}}P_{+}(v)^{-1}R(v)^{-1}R(u)P_{+}(u)e^{-\nu(g_{+}(u)-\pi\mathrm{i}/2)\sigma_{3}}\begin{pmatrix}1\\ 0\end{pmatrix}$
		$\displaystyle=-\frac{1}{\mathrm{i}(u-v)}\begin{pmatrix}\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f(v))&\mathrm{Ai}(\nu^{\frac{2}{3}}f(v))\end{pmatrix}E(v)^{-1}R(v)^{-1}R(u)E(u)\begin{pmatrix}\mathrm{Ai}(\nu^{\frac{2}{3}}f(u))\\ -\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f(u))\end{pmatrix}.$		(4.16)

Inserting (4.5) into the above formula gives us that

	$\displaystyle\sqrt{\phi_{\nu}^{\prime}(x)\phi_{\nu}^{\prime}(y)}K_{\nu}^{\mathrm{Bes}}(\phi_{\nu}(x),\phi_{\nu}(y))$
	$\displaystyle=\frac{\sqrt{(1-h_{\nu}x)(1-h_{\nu}y)}}{x-y-\frac{h_{\nu}}{2}(x^{2}-y^{2})}\begin{pmatrix}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))&-\mathrm{i}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))\end{pmatrix}E((1-h_{\nu}y)^{2})^{-1}$
	$\displaystyle\quad\times R((1-h_{\nu}y)^{2})^{-1}R((1-h_{\nu}x)^{2})E((1-h_{\nu}x)^{2})\begin{pmatrix}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\\ -\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\end{pmatrix}.$		(4.17)

We now show expansions of different parts on the right-hand side of the above formula. According to [5, Lemma 3.1], one has

\frac{\sqrt{(1-h_{\nu}x)(1-h_{\nu}y)}}{x-y-\frac{h_{\nu}}{2}(x^{2}-y^{2})}=\frac{1}{x-y}-(x-y)\sum_{j=2}^{\infty}r_{j}(x,y)h_{\nu}^{j},

(4.18)

where each $r_{j}(x,y)$ is certain polynomial of degree $j-2$ .

Next, it is observed that if $x=y$ ,

E((1-h_{\nu}y)^{2})^{-1}R((1-h_{\nu}y)^{2})^{-1}R((1-h_{\nu}x)^{2})E((1-h_{\nu}x)^{2})=I.

(4.19)

This, together with (3.6) and the fact that $E(z)$ is an analytic function in $D(1,\varepsilon)$ , implies that, as $\nu\to+\infty$ ,

E((1-h_{\nu}y)^{2})^{-1}R((1-h_{\nu}y)^{2})^{-1}R((1-h_{\nu}x)^{2})E((1-h_{\nu}x)^{2})\\ \sim I+(x-y)\sum_{j=1}^{\infty}e_{j}(x,y)h_{\nu}^{j},

(4.20)

where $e_{j}(x,y)$ are certain matrices with all the entries being polynomials in $x$ and $y$ . Indeed, by (3.27), (3.6) and (3.6), it follows that, as $h_{\nu}\to 0^{+}$ ,

	$\displaystyle E((1-h_{\nu}x)^{2})$	$\displaystyle=(2h_{\nu})^{-\frac{1}{4}\sigma_{3}}e^{-\frac{\pi\mathrm{i}}{4}\sigma_{3}}\sigma_{3}\left(I+\frac{x}{5}\sigma_{3}h_{\nu}+\begin{pmatrix}\frac{3}{35}x^{2}&0\\ 0&-\frac{8}{175}x^{2}\end{pmatrix}h_{\nu}^{2}+\mathcal{O}(h_{\nu}^{3})\right),$		(4.21)
	$\displaystyle R((1-h_{\nu}x)^{2})$	$\displaystyle=h_{\nu}^{-\frac{1}{4}\sigma_{3}}\left(I+\begin{pmatrix}0&0\\ \frac{7\sqrt{2}}{40}&0\end{pmatrix}h_{\nu}+\begin{pmatrix}0&\frac{\sqrt{2}}{70}\\ \frac{\sqrt{2}}{25}x&0\end{pmatrix}h_{\nu}^{2}+\mathcal{O}(h_{\nu}^{3})\right)h_{\nu}^{\frac{1}{4}\sigma_{3}}.$		(4.22)

Thus, we have

\displaystyle e_{1}(x,y)=\frac{1}{5}\sigma_{3},\qquad e_{2}(x,y)=\begin{pmatrix}\frac{15x+8y}{175}&0\\ \frac{\mathrm{i}}{25}&-\frac{8x+15y}{175}\end{pmatrix}.

(4.23)

Then, we consider the parts involving the Airy functions. From the expansion of $f$ near $z=1$ given in (3.5), we have

\displaystyle\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2})=x+\sum_{j=1}^{\infty}p_{1,j}(x)h_{\nu}^{j},\qquad\textrm{$\nu\to+\infty$},

(4.24)

where $p_{1,j}(x)$ are certain polynomials of degree $j+1$ with

\displaystyle p_{1,1}(x)=\frac{3}{10}x^{2},\qquad p_{1,2}(x)=\frac{32}{175}x^{3}.

(4.25)

Moreover, by writing

\displaystyle\frac{1}{m!}\left(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2})-x\right)^{m}=\sum_{j=m}^{\infty}p_{m,j}(x)h_{\nu}^{j},\qquad h_{\nu}\to 0^{+},

(4.26)

it is easily seen that

\displaystyle p_{m,j}(x)=\frac{(m-n)!n!}{m!}\sum_{k=m-n}^{j-n}p_{m-n,k}(x)p_{n,j-k}(x)

(4.27)

for $1<n<m\leq j$ . We then obtain from the analyticity of Airy function and (4.26) that

\displaystyle\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))=\mathrm{Ai}(x)+\sum_{j=1}^{\infty}\sum_{m=1}^{j}p_{m,j}(x)\mathrm{Ai}^{(m)}(x)h_{\nu}^{j},\qquad\nu\to+\infty,

(4.28)

where $\mathrm{Ai}^{(m)}(x)$ denotes the $m$ -th derivative of $\mathrm{Ai}(x)$ with respect to $x$ . From the differential equation

\frac{\,\mathrm{d}^{2}\mathrm{Ai}(z)}{\,\mathrm{d}z^{2}}=z\mathrm{Ai}(z)

(4.29)

satisfied by Airy function, a direct calculation gives us

\displaystyle\mathrm{Ai}^{(m)}(x)=P_{m}(x)\mathrm{Ai}(x)+Q_{m}(x)\mathrm{Ai}^{\prime}(x),

(4.30)

where $P_{m}(x)$ and $Q_{m}(x)$ are polynomials with $P_{0}(x)=1$ and $Q_{0}(x)=0$ . They satisfy the recurrence relations (cf. [22])

\displaystyle P_{n+1}(x)=P_{n}^{\prime}(x)+xQ_{n}(x),\qquad Q_{n+1}(x)=Q_{n}^{\prime}(x)+P_{n}(x).

(4.31)

With the aids of the expansions (4.28) and (4.30), we obtain

		$\displaystyle\begin{pmatrix}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))&-\mathrm{i}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))\end{pmatrix}\begin{pmatrix}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\\ -\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\end{pmatrix}$
		$\displaystyle=\left(\mathrm{Ai}(x)+\sum_{j=1}^{\infty}\sum_{m=1}^{j}p_{m,j}(x)\mathrm{Ai}^{(m)}(x)h_{\nu}^{j}\right)\cdot\left(\mathrm{Ai}^{\prime}(y)+\sum_{j=1}^{\infty}\sum_{m=1}^{j}p_{m,j}(y)\mathrm{Ai}^{(m+1)}(y)h_{\nu}^{j}\right)$
		$\displaystyle\quad-\left(\mathrm{Ai}^{\prime}(x)+\sum_{j=1}^{\infty}\sum_{m=1}^{j}p_{m,j}(x)\mathrm{Ai}^{(m+1)}(x)h_{\nu}^{j}\right)\cdot\left(\mathrm{Ai}(y)+\sum_{j=1}^{\infty}\sum_{m=1}^{j}p_{m,j}(y)\mathrm{Ai}^{(m)}(y)h_{\nu}^{j}\right)$
		$\displaystyle=\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)-\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)+\sum_{N=1}^{\infty}\left(a_{N,00}(x,y)\mathrm{Ai}(x)\mathrm{Ai}(y)+a_{N,01}(x,y)\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)\right.$
		$\displaystyle\quad\left.-a_{N,01}(y,x)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)+a_{N,11}(x,y)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}^{\prime}(y)\right)h_{\nu}^{N},$		(4.32)

where

	$\displaystyle a_{N,00}(x,y)=\sum_{n=1}^{N}\left(p_{n,N}(y)P_{n+1}(y)-p_{n,N}(x)P_{n+1}(x)\right)$
	$\displaystyle\quad+\sum_{\begin{subarray}{c}j+k=N\\ j\geq 1,k\geq 1\end{subarray}}\sum_{m=1}^{j}\sum_{n=1}^{k}\left(p_{m,j}(x)p_{n,k}(y)P_{m}(x)P_{n+1}(y)-p_{m,j}(y)p_{n,k}(x)P_{m}(y)P_{n+1}(x)\right),$		(4.33)
	$\displaystyle a_{N,01}(x,y)=\sum_{n=2}^{N}\left(p_{n,N}(y)Q_{n+1}(y)+p_{n,N}(x)P_{n}(x)\right)$
	$\displaystyle\quad+\sum_{\begin{subarray}{c}j+k=N\\ j\geq 1,k\geq 1\end{subarray}}\sum_{m=1}^{j}\sum_{n=1}^{k}p_{m,j}(x)p_{n,k}(y)\left(P_{m}(x)Q_{n+1}(y)-P_{m+1}(x)Q_{n}(y)\right),$		(4.34)

and

	$\displaystyle a_{N,11}(x,y)=\sum_{n=1}^{N}\left(p_{n,N}(x)Q_{n}(x)-p_{n,N}(y)Q_{n}(y)\right)$
	$\displaystyle\quad+\sum_{\begin{subarray}{c}j+k=N\\ j\geq 1,k\geq 1\end{subarray}}\sum_{m=1}^{j}\sum_{n=1}^{k}\left(p_{m,j}(x)p_{n,k}(y)Q_{m}(x)Q_{n+1}(y)-p_{m,j}(y)p_{n,k}(x)Q_{m}(y)Q_{n+1}(x)\right),$		(4.35)

are polynomials in $x$ and $y$ . Since the polynomials $a_{N,00}(x,y)$ and $a_{N,11}(x,y)$ are anti-symmetric in $x$ , $y$ , they must have the form

(x-y)\times(\textrm{polynomials in $x$ and $y$}).

(4.36)

We would like to show the polynomials $a_{N,01}(x,y)$ admit the same structure. In other words, we want to show

a_{N,01}(x,x)=0,

(4.37)

or equivalently,

		$\displaystyle\sum_{n=2}^{N}p_{n,N}(x)\left(Q_{n+1}(x)+P_{n}(x)\right)$
		$\displaystyle+\sum_{\begin{subarray}{c}j+k=N\\ j\geq 1,k\geq 1\end{subarray}}^{N}\sum_{m=1}^{j}\sum_{n=1}^{k}\left(p_{m,j}(x)p_{n,k}(x)(P_{m}(x)Q_{n+1}(x)-P_{m+1}(x)Q_{n}(x)\right)=0.$		(4.38)

To prove the above equality, we need the following lemma.

Lemma 4.1.

With polynomials $P_{m}$ and $Q_{m}$ defined through (4.30), we have

\displaystyle\sum_{j=0}^{N}\frac{1}{j!(N-j)!}\left(P_{j}(x)Q_{N+1-j}(x)-Q_{j}(x)P_{N+1-j}(x)\right)=0,\qquad N\geq 1.

(4.39)

Proof.

We use the method of mathematical induction to prove the above identity. It is clear that (4.39) holds for $N=1$ . Assume that (4.39) is valid for $N=k>1$ , i.e.,

\displaystyle\sum_{j=0}^{k}\frac{1}{j!(k-j)!}\left(P_{j}(x)Q_{k+1-j}(x)-Q_{j}(x)P_{k+1-j}(x)\right)=0.

(4.40)

After taking derivative on both sides with respected to $x$ , it follows that

\displaystyle\sum_{j=0}^{k}\frac{1}{j!(k-j)!}\left(P_{j}^{\prime}(x)Q_{k+1-j}(x)+P_{j}(x)Q_{k+1-j}^{\prime}(x)-Q_{j}^{\prime}(x)P_{k+1-j}(x)-Q_{j}(x)P_{k+1-j}^{\prime}(x)\right)=0,

Applying the recurrence relations (4.31) to the above formula, we arrive at

	$\displaystyle\sum_{j=0}^{k}\frac{1}{j!(k-j)!}\left(P_{j+1}(x)Q_{k+1-j}(x)-Q_{j+1}(x)P_{k+1-j}(x)+P_{j}(x)Q_{k+2-j}(x)-Q_{j}(x)P_{k+2-j}(x)\right)$
	$\displaystyle=\sum_{j=0}^{k+1}\frac{j}{j!(k+1-j)!}\left(P_{j}(x)Q_{k+2-j}(x)-Q_{j}(x)P_{k+2-j}(x)\right)$
	$\displaystyle\qquad+\sum_{j=0}^{k+1}\frac{k+1-j}{j!(k+1-j)!}\left(P_{j}(x)Q_{k+2-j}(x)-Q_{j}(x)P_{k+2-j}(x)\right)$
	$\displaystyle=\sum_{j=0}^{k+1}\frac{k+1}{j!(k+1-j)!}\left(P_{j}(x)Q_{k+2-j}(x)-Q_{j}(x)P_{k+2-j}(x)\right)=0,$

which is (4.39) with $N=k+1$ . ∎

Using(4.27) and (4.39), we could rewrite the left-hand side of (4) as

	$\displaystyle\sum_{n=2}^{N}p_{n,N}(x)\left(Q_{n+1}(x)+P_{n}(x)\right)+\sum_{\begin{subarray}{c}m+n=2\\ m\geq 1,n\geq 1\end{subarray}}^{N}\sum_{\begin{subarray}{c}j+k=N\\ j\geq m,k\geq n\end{subarray}}\left(p_{m,j}(x)p_{n,k}(x)(P_{m}(x)Q_{n+1}(x)-P_{m+1}(x)Q_{n}(x))\right)$
	$\displaystyle=\sum_{t=2}^{N}\left(p_{t,N}(x)\left(Q_{t+1}(x)+P_{t}(x)\right)+\sum_{\begin{subarray}{c}m+n=t\\ m\geq 1,n\geq 1\end{subarray}}\sum_{\begin{subarray}{c}j+k=N,\\ j\geq m,k\geq n\end{subarray}}\left(p_{m,j}(x)p_{n,k}(x)(P_{m}(x)Q_{n+1}(x)-P_{m+1}(x)Q_{n}(x))\right)\right)$
	$\displaystyle=\sum_{t=2}^{N}\left(p_{t,N}(x)\left(Q_{t+1}(x)+P_{t}(x)\right)+\sum_{\begin{subarray}{c}m+n=t\\ m\geq 1,n\geq 1\end{subarray}}\left(\frac{t!}{m!n!}p_{t,N}(x)(P_{m}(x)Q_{n+1}(x)-P_{m+1}(x)Q_{n}(x))\right)\right)$
	$\displaystyle=\sum_{t=2}^{N}p_{t,N}(x)\left(Q_{t+1}(x)+P_{t}(x)+t!\sum_{m=1}^{t-1}\left(\frac{1}{m!(t-m)!}(P_{m}(x)Q_{t+1-m}(x)-P_{t+1-m}(x)Q_{m}(x))\right)\right)$
	$\displaystyle=\sum_{t=2}^{N}t!p_{t,N}(x)\left(\sum_{m=0}^{t}\left(\frac{1}{m!(t-m)!}(P_{m}(x)Q_{t+1-m}(x)-P_{t+1-m}(x)Q_{m}(x))\right)\right)=0,$		(4.41)

as required. Since the polynomials $a_{N,00},a_{N,01}$ and $a_{N,11}$ all take the same structure (4.36), we obtain from (4) and the estimates of Airy functions in (4.8) that, for any $\mathfrak{m}\in\mathbb{N}$ ,

		$\displaystyle\frac{1}{x-y}\begin{pmatrix}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))&-\mathrm{i}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}y)^{2}))\end{pmatrix}\begin{pmatrix}\mathrm{Ai}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\\ -\mathrm{i}\mathrm{Ai}^{\prime}(\nu^{\frac{2}{3}}f((1-h_{\nu}x)^{2}))\end{pmatrix}$
		$\displaystyle=\frac{\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)-\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)}{x-y}+\sum_{N=1}^{\mathfrak{m}}\left(\tilde{a}_{N,00}(x,y)\mathrm{Ai}(x)\mathrm{Ai}(y)+\tilde{a}_{N,01}(x,y)\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)\right.$
		$\displaystyle\quad\left.+\tilde{a}_{N,01}(y,x)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)+\tilde{a}_{N,11}(x,y)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}^{\prime}(y)\right)h_{\nu}^{N}+h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-(x+y)}\right),$		(4.42)

where $\tilde{a}_{N,00}(x,y)$ , $\tilde{a}_{N,01}(x,y)$ and $\tilde{a}_{N,11}(x,y)$ are certain polynomials in $x$ and $y$ .

Combining (4.18), (4.20) and (4) together gives us that, under the condition (4.15),

	$\displaystyle\sqrt{\phi_{\nu}^{\prime}(x)\phi_{\nu}^{\prime}(y)}K_{\nu}^{\mathrm{Bes}}(\phi_{\nu}(x),\phi_{\nu}(y))$
	$\displaystyle=\frac{\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)-\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)}{x-y}+\sum_{j=1}^{\mathfrak{m}}\left(p_{j,00}(x,y)\mathrm{Ai}(x)\mathrm{Ai}(y)+p_{j,01}(x,y)\mathrm{Ai}(x)\mathrm{Ai}^{\prime}(y)\right.$
	$\displaystyle\quad\left.+p_{j,10}(x,y)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}(y)+p_{j,11}(x,y)\mathrm{Ai}^{\prime}(x)\mathrm{Ai}^{\prime}(y)\right)h_{\nu}^{j}+h_{\nu}^{\mathfrak{m}+1}\cdot\mathcal{O}\left(e^{-(x+y)}\right),$		(4.43)

as $h_{\nu}\to 0^{+}$ , where $p_{j,\kappa\lambda}(x,y)$ , $\kappa,\lambda\in\{0,1\}$ , are polynomials in $x$ and $y$ . Additionally, it is worth noting that even though the calculations involved are inherently complicated, it is possible to derive precise formulaes for $p_{j,\kappa\lambda}$ utilizing the explicit expressions (3.5), (3.27), (3.6) and (4.18). The first few polynomials are hereby presented:

	$\displaystyle p_{1,00}(x,y)$	$\displaystyle=-\frac{3}{10}(x^{2}+xy+y^{2}),$
	$\displaystyle p_{1,01}(x,y)$	$\displaystyle=p_{1,10}(x,y)=\frac{1}{5},$
	$\displaystyle p_{1,11}(x,y)$	$\displaystyle=\frac{3}{10}(x+y),$

and

	$\displaystyle p_{2,00}(x,y)$	$\displaystyle=\frac{56-235(x^{2}+y^{2})-319xy(x+y)}{1400},$
	$\displaystyle p_{2,01}(x,y)$	$\displaystyle=p_{2,10}(y,x)=\frac{63(x^{4}+x^{3}y-x^{2}y^{2}-xy^{3}-y^{4})-55x+239y}{1400},$
	$\displaystyle p_{2,11}(x,y)$	$\displaystyle=\frac{340(x^{2}+y^{2})+256xy}{1400},$

which lead to (1.1) and (1.1).

Finally, since all the terms $\mathrm{Ai}(\cdot)$ , $\mathrm{Ai}^{\prime}(\cdot)$ , $E(\cdot)$ and $R(\cdot)$ are analytic functions for $t_{0}\leq x,y<(1-\sqrt{1-\varepsilon})h_{\nu}^{-1}$ , all the expansions can repeatedly be differentiated w.r.t. the variables $x$ and $y$ while preserving the uniformity, which holds for the kernel expansion.

This completes the proof of Theorem 1.1. ∎

Appendix A The Airy parametrix

The Airy parametrix $\Phi^{({\mathrm{Ai}})}$ is the unique solution of the following RH problem.

RH problem A.1.

(a)

$\Phi^{(\mathrm{Ai})}(z)$ is analytic in $\mathbb{C}\setminus\{\cup_{j=1}^{4}\Sigma_{j}\cup\{0\}\}$ , where the contours $\Sigma_{j}$ , $j=1,2,3,4$ , are indicated in Figure 5.

(b)

$\Phi^{(\mathrm{Ai})}(z)$ satisfies the jump condition

\displaystyle\Phi^{({\mathrm{Ai}})}_{+}(z)=\Phi^{({\mathrm{Ai}})}_{-}(z)\begin{cases}\begin{pmatrix}1&1\\ 0&1\end{pmatrix},&\qquad z\in\Sigma_{1},\\ \begin{pmatrix}1&0\\ 1&1\end{pmatrix},&\qquad z\in\Sigma_{2}\cup\Sigma_{4},\\ \begin{pmatrix}0&1\\ -1&0\end{pmatrix},&\qquad z\in\Sigma_{3}.\end{cases}

(A.1)

(c)

As $z\to\infty$ , we have

\displaystyle\Phi^{({\mathrm{Ai}})}(z)=\frac{1}{\sqrt{2}}\begin{pmatrix}z^{-\frac{1}{4}}&0\\ 0&z^{\frac{1}{4}}\end{pmatrix}\begin{pmatrix}1&\mathrm{i}\\ \mathrm{i}&1\end{pmatrix}\left(I+\mathcal{O}(z^{-\frac{3}{2}})\right)e^{-\frac{2}{3}z^{3/2}\sigma_{3}}.

(A.2)

(d)

$\Phi^{({\mathrm{Ai}})}(z)$ is bounded near the origin.

Figure 5: The jump contours of the RH problem for

\Phi^{({\mathrm{Ai}})}

Denote $\omega:=e^{2\pi\mathrm{i}/3}$ , the unique solution is given by (cf. [11])

\displaystyle\Phi^{({\mathrm{Ai}})}(z)=\sqrt{2\pi}\begin{cases}\begin{pmatrix}\mathrm{Ai}(z)&-\omega^{2}\mathrm{Ai}(\omega^{2}z)\\ -\mathrm{i}\mathrm{Ai}^{\prime}(z)&\mathrm{i}\omega\mathrm{Ai}^{\prime}(\omega^{2}z)\end{pmatrix},&\arg z\in\left(0,\frac{3\pi}{4}\right),\\ \begin{pmatrix}-\omega\mathrm{Ai}(\omega z)&-\omega^{2}\mathrm{Ai}(\omega^{2}z)\\ \mathrm{i}\omega^{2}\mathrm{Ai}^{\prime}(\omega z)&\mathrm{i}\omega\mathrm{Ai}^{\prime}(\omega^{2}z)\end{pmatrix},&\arg z\in\left(\frac{3\pi}{4},\pi\right),\\ \begin{pmatrix}-\omega^{2}\mathrm{Ai}(\omega^{2}z)&\omega\mathrm{Ai}(\omega z)\\ \mathrm{i}\omega\mathrm{Ai}^{\prime}(\omega^{2}z)&-\mathrm{i}\omega^{2}\mathrm{Ai}^{\prime}(\omega z)\end{pmatrix},&\arg z\in\left(-\pi,-\frac{3\pi}{4}\right),\\ \begin{pmatrix}\mathrm{Ai}(z)&\omega\mathrm{Ai}(\omega z)\\ -\mathrm{i}\mathrm{Ai}^{\prime}(z)&-\mathrm{i}\omega^{2}\mathrm{Ai}^{\prime}(\omega z)\end{pmatrix},&\arg z\in\left(-\frac{3\pi}{4},0\right).\end{cases}

(A.3)

Furthermore, applying the asymptotics of $\mathrm{Ai}$ and $\mathrm{Ai}^{\prime}$ in [24, Chapter 9], we have, as $z\to\infty$ ,

\displaystyle\Phi^{({\mathrm{Ai}})}(z)\sim\frac{1}{2\sqrt{2}}\begin{pmatrix}z^{-\frac{1}{4}}&0\\ 0&z^{\frac{1}{4}}\end{pmatrix}\begin{pmatrix}1&\mathrm{i}\\ \mathrm{i}&1\end{pmatrix}\begin{pmatrix}\sum\limits_{k=0}^{\infty}\left(-\frac{3}{2}\right)^{k}\frac{\mathfrak{u}_{k}+\mathfrak{v}_{k}}{z^{3k/2}}&\mathrm{i}\sum\limits_{k=0}^{\infty}\left(\frac{3}{2}\right)^{k}\frac{\mathfrak{u}_{k}-\mathfrak{v}_{k}}{z^{3k/2}}\\ -\mathrm{i}\sum\limits_{k=0}^{\infty}\left(-\frac{3}{2}\right)^{k}\frac{\mathfrak{u}_{k}-\mathfrak{v}_{k}}{z^{3k/2}}&\sum\limits_{k=0}^{\infty}\left(\frac{3}{2}\right)^{k}\frac{\mathfrak{u}_{k}+\mathfrak{v}_{k}}{z^{3k/2}}\end{pmatrix}e^{-\frac{2}{3}z^{3/2}\sigma_{3}},

(A.4)

where $\mathfrak{u}_{0}=\mathfrak{v}_{0}=1$ and

\displaystyle\mathfrak{u}_{k}=\frac{(6k-5)(6k-3)(6k-1)}{216(2k-1)k}\mathfrak{u}_{k-1},\qquad\mathfrak{v}_{k}=\frac{6k+1}{1-6k}\mathfrak{u}_{k}.

(A.5)

Acknowledgements

We thank Folkmar Bornemann for helpful comments on the preliminary version of this paper. This work was partially supported by National Natural Science Foundation of China under grant numbers 12271105, 11822104, and “Shuguang Program” supported by Shanghai Education Development Foundation and Shanghai Municipal Education Commission.

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Asymptotic expansion of the hard-to-soft edge transition

Abstract

1 Introduction and statement of results

Theorem 1.1.

Corollary 1.2.

Notations

2 An RH characterization of the Bessel kernel

RH problem 2.1.

3 Asymptotic analysis of the RH problem for Ψ\Psi with large ν\nu

3.1 First transformation: Ψ→Y\Psi\to Y

RH problem 3.1.

3.2 Second transformation: Y→TY\to T

RH problem 3.2.

3.3 Third transformation: T→ST\to S

Proposition 3.3.

RH problem 3.4.

3.4 Global parametrix

RH problem 3.5.

3.5 Local parametrix

RH problem 3.6.

Proposition 3.7.

Proof.

3.6 Final transformation

RH problem 3.8.

RH problem 3.9.

RH problem 3.10.

4 Proof of Theorem 1.1

Lemma 4.1.

Proof.

Appendix A The Airy parametrix

RH problem A.1.

Acknowledgements

References

3 Asymptotic analysis of the RH problem for $\Psi$ with large $\nu$

3.1 First transformation: $\Psi\to Y$

3.2 Second transformation: $Y\to T$

3.3 Third transformation: $T\to S$