Kapaev’s global asymptotics of the fourth Painlevé transcendents. Elliptic asymptotics

Shun Shimomura Department of Mathematics, Keio University, 3-14-1, Hiyoshi, Kohoku-ku, Yokohama 223-8522 Japan [email protected]

Abstract.

For the fourth Painlevé transcendents we derive elliptic asymptotic representations, which were announced by late Professor Kapaev without proofs. Then we newly obtain related results including the correction function.

2010 Mathematics Subject Classification. 34M55, 34M56, 34M40, 34M60, 33E05.

Key words and phrases. elliptic asymptotic representation; fourth Painlevé transcendents; WKB analysis; isomonodromy deformation; monodromy data.

To the memory of Professor Kapaev

1. Introduction

The fourth Painlevé equation

P_IV

y^{\prime\prime}=\frac{(y^{\prime})^{2}}{2y}+\frac{3}{2}y^{3}+4xy^{2}+(-4\alpha+\beta+2x^{2})y-\frac{\beta^{2}}{2y}

$(y^{\prime}=dy/dx)$ with $\alpha,\beta\in\mathbb{C}$ governs the isomonodromy deformation of the linear system

(1.1)			$\displaystyle\frac{d\Psi}{d\xi}=\Biggl{(}\Bigl{(}\frac{\xi^{3}}{2}+\xi(x+\mathrm{u}\mathrm{v})+\frac{\alpha}{\xi}\Bigr{)}\sigma_{3}+i\begin{pmatrix}0&\xi^{2}\mathrm{u}+2x\mathrm{u}+\mathrm{u}^{\prime}\\ \xi^{2}\mathrm{v}+2x\mathrm{v}-\mathrm{v}^{\prime}&0\end{pmatrix}\Biggr{)}\Psi,$
		$\displaystyle\beta=\mathrm{u}^{\prime}\mathrm{v}-\mathrm{u}\mathrm{v}^{\prime}+2x\mathrm{u}\mathrm{v}-(\mathrm{u}\mathrm{v})^{2},\quad y=\mathrm{u}\mathrm{v}$

by Kitaev [16] (for another isomonodromy system see [10]). Kapaev [13] announced global asymptotic results on solutions of P_IV including elliptic asymptotic representations along generic directions in the complex plane, giving notice of publishing the proofs in [4] written by him in collaboration with Fokas, Its and Novokshenov. The monograph [4], however, contains none of these proofs, and Kapaev passed away without publishing the asymptotics on P_IV except [9] for solutions on the real line and [14]. In Kapaev’s announcement, [13, Theorems 2 and 3] are on elliptic asymptotics described in terms of an elliptic integral as the inverse function; [13, Theorems 4 and 5] on trigonometric asymptotics, and [13, Theorems 6, 7, 8 and 9] on truncated solutions. Asymptotics of P_IV along Stokes rays are also studied by Kitaev [16], [17], [18]. By isomonodromy technique, elliptic asymptotics of general solutions are known for P_I ([15], [20], [21]), P_II ([23], [24], [11], [12], [21]), P ${}_{\mathbf{III}}(D_{6})$ ([25], [26], [28]), P ${}_{\mathbf{III}}(D_{7})$ ([27]) and P_V ([29]), and in each asymptotic formula the phase shift is expressed in terms of the monodromy data. For the complete P_IV the proofs of elliptic asymptotics have not been published, though P_IV with $\alpha=0$ was treated by Vereshchagin [31] for $0<\arg x<\pi/4$ . The present author believes that it is significant to present the proofs of elliptic representations for P_IV in Kapaev’s announcement [13] for the following reasons: (i) in the study of a general solution of P_IV, elliptic asymptotics are crucial information; (ii) as a practical matter, the process of deriving the elliptic representation needs some devices peculiar to P_IV; and (iii) related important materials including the correction function, which appear in the proofs, are not referred to in the announcement.

In this paper we derive the elliptic asymptotics for the complete P_IV by the isomonodromy deformation method along the lines of the arguments in [12], [20], [21] with discussions on the Boutroux equations and on the justification of the asymptotics as a solution of P_IV. Then we newly obtain related results including the correction function $B_{\phi}(t)$ given by (5.1), which contains information on the asymptotics and is essential in the justification procedure as in [21, Section 3].

The main results are stated in Section 2. Theorems 2.1 and 2.2 present elliptic representations of a general solution of P_IV, which correspond to the announced [13, Theorem 2]. These results are also described by an alternative elliptic expression as of Corollary 2.3, which has an advantage in treating in general sectors (cf. Theorem 2.4). For elliptic expressions of Theorems 2.1 and 2.2 in directions neighbouring the positive real axis, degeneration to trigonometric asymptotics may be considered under certain suppositions, and is shown to be consistent with the result of [13, Theorem 4]. This fact supports the validity of signs in Theorems 2.1 and 2.2 contradicting those of [13, Theorem 2] (cf. Remark 2.3). Section 3 summarises necessary facts on the isomonodromy linear system (3.1) and on its monodromy data consisting of Stokes coefficients. Section 4 explains turning points, Stokes graphs and WKB solutions, which are necessary in the WKB analysis. In Section 5 we solve a direct monodromy problem for system (3.1) by the WKB analysis to obtain the key relations consisting of monodromy data and certain integrals (cf. Propositions 5.1 and 5.2). Asymptotics of these key relations are examined in Section 6 by the use of the $\vartheta$ -function. In Section 7 from the formulas thus obtained asymptotic forms of the main theorems are derived by solving an inverse monodromy problem for the prescribed monodromy data. In this process we make technical devices to find necessary special properties of the elliptic function related to our case (Propositions 7.5 and 7.6). The justification as a solution of P_IV is performed along the lines of Kitaev [21] with [19]. The final section is devoted to the proofs of necessary facts on the Boutroux equations summarised in Proposition 8.15, which determine $A_{\phi}$ parametrising the related elliptic function. Furthermore we clarify local structure of Stokes curves near coalescing turning points, which is used in drawing Stokes graphs in Section 4.

Throughout this paper we use the following symbols:

(1) The coefficient $A(\varphi_{0})$ defined by [13, (20)] is denoted by $e^{3i\phi}A_{\phi}$ $(\phi=\varphi_{0})$ ;

(2) $\sigma_{1},$ $\sigma_{2}$ , $\sigma_{3}$ denote the Pauli matrices

\sigma_{1}=\begin{pmatrix}0&1\\ 1&0\end{pmatrix},\quad\sigma_{2}=\begin{pmatrix}0&-i\\ i&0\end{pmatrix},\quad\sigma_{3}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix};

(3) for complex-valued functions $f$ and $g$ , we write $f\ll g$ or $g\gg f$ if $f=O(|g|)$ , and write $f\asymp g$ if $g\ll f\ll g$ .

2. Results

To state the results we explain the monodromy data [13, Section 2], [9, Section 2], and the Boutroux equations [13, Section 3]. For $k\in\mathbb{Z}$ system (1.1) admits the matrix solutions

(2.1)

\Psi_{k}^{\infty}(\xi)=(I+O(\xi^{-1}))\exp((\tfrac{1}{8}\xi^{4}+\tfrac{1}{2}x\xi^{2}+(\alpha-\beta)\ln\xi)\sigma_{3})

as $\xi\to\infty$ through the sector $|\arg\xi+\frac{\pi}{8}-\frac{\pi}{4}k|<\frac{\pi}{4},$ and

\Psi^{0}(\xi)=T_{0}(\xi)\xi^{\alpha\sigma_{3}}(I+J_{0}\ln\xi)e^{\mathrm{int}(x)\sigma_{3}},\quad J_{0}=\begin{pmatrix}0&j_{+}\\ j_{-}&0\end{pmatrix},\quad\mathrm{int}(x)=\int^{x}\mathrm{u}\mathrm{v}\,dx

as $\xi\to 0$ , where $T_{0}(\xi)$ is invertible around $\xi=0$ , and $j_{+}=0$ if $\alpha-\frac{1}{2}\not=0,1,2,\ldots,$ $j_{-}=0$ if $\alpha-\frac{1}{2}\not=-1,-2,-3,\ldots.$ The Stokes matrices

S_{2l-1}=\begin{pmatrix}1&s_{2l-1}\\ 0&1\end{pmatrix},\quad S_{2l}=\begin{pmatrix}1&0\\ s_{2l}&1\end{pmatrix}\quad(l\in\mathbb{Z})

are defined by $\Psi_{k+1}^{\infty}(\xi)=\Psi_{k}^{\infty}(\xi)S_{k}$ , and satisfy $S_{k+4}=e^{-i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}S_{k}\sigma_{3}e^{i\pi(\alpha-\beta)\sigma_{3}},$ $s_{k+4}=-s_{k}e^{(-1)^{k}2\pi i(\alpha-\beta)}.$ For the matrices $M=e^{-\mathrm{int}(x)\sigma_{3}}(I+i\pi J_{0})e^{\mathrm{int}(x)\sigma_{3}}e^{i\pi\alpha\sigma_{3}}$ and $E$ such that $\sigma_{3}\Psi^{0}(e^{i\pi}\xi)\sigma_{3}=\Psi^{0}(\xi)M$ and $\Psi_{1}^{\infty}(\xi)=\Psi^{0}(\xi)E$ , the semi-cyclic relation

S_{1}S_{2}S_{3}S_{4}=E^{-1}\sigma_{3}M^{-1}Ee^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}

holds, and the traces of both sides lead to the surface of the monodromy data

\mathcal{M}_{0}(\alpha,\beta):

((1+s_{1}s_{2})(1+s_{3}s_{4})+s_{1}s_{4})e^{-i\pi(\alpha-\beta)}-(1+s_{2}s_{3})e^{i\pi(\alpha-\beta)}=-2i\sin\pi\alpha.

For each $m\in\mathbb{Z}$ the semi-cyclic relation for $S_{j+m}$ $(1\leq j\leq 4)$ in Proposition 3.1 yields

$\mathcal{M}_{m}(\alpha,\beta):$		$\displaystyle((1+s_{1+m}s_{2+m})$	$\displaystyle(1+s_{3+m}s_{4+m})+s_{1+m}s_{4+m})e^{-i\pi(-1)^{m}(\alpha-\beta)}$
		$\displaystyle-(1+s_{2+m}s_{3+m})e^{i\pi(-1)^{m}(\alpha-\beta)}=-2i(-1)^{m}\sin\pi\alpha.$

Around a nonsingular point on $\mathcal{M}_{m}(\alpha,\beta)$ suitable three of $s_{j+m}$ $(1\leq j\leq 4)$ are independent. For any $c\in\mathbb{C}\setminus\{0\}$ the gauge transformation $\Psi=c^{\sigma_{3}}\tilde{\Psi}$ induces the action

[c]:\,\,\,(S_{1},S_{2},S_{3},S_{4})\mapsto(c^{-\sigma_{3}}S_{1}c^{\sigma_{3}},c^{-\sigma_{3}}S_{2}c^{\sigma_{3}},c^{-\sigma_{3}}S_{3}c^{\sigma_{3}},c^{-\sigma_{3}}S_{4}c^{\sigma_{3}})

on $\mathcal{M}_{0}(\alpha,\beta)$ consistent with the isomonodromy structure of (1.1), and each solution of P_IV corresponds to an orbit, or equivalence class yielded by dividing $\mathcal{M}_{0}(\alpha,\beta)$ by $[c]$ . Thus an orbit passing through a point $(s_{1},s_{2},s_{3},s_{4})\in\mathcal{M}_{0}(\alpha,\beta)$ parametrises a solution of P_IV. Let us call it a solution labelled by $(s_{1},s_{2},s_{3},s_{4})$ . (In [13, Section 2] the gauge symmetry on $(\mathrm{u},\mathrm{v})$ is considered.)

For $0<A<\frac{8}{27}$ let $z=0,$ $x_{1},$ $x_{3},$ $x_{5}$ with $x_{5}<x_{3}<x_{1}<0$ be the zeros of the polynomial $P_{A}(z)=z^{4}+4z^{3}+4z^{2}+4Az$ such that $x_{1}\to 0,$ $x_{3},x_{5}\to-2$ as $A\to 0$ and that $x_{1},x_{3}\to-\frac{2}{3},$ $x_{5}\to-\frac{8}{3}$ as $A\to\frac{8}{27}$ . Let $\Pi_{\pm}$ be two copies of $P^{1}(\mathbb{C})\setminus([x_{5},x_{3}]\cup[x_{1},0])$ . The elliptic curve $w^{2}=P_{A}(z)$ is the two sheeted Riemann surface $\Pi_{A,0}=\Pi_{+}\cup\Pi_{-}$ glued along the cuts $[x_{5},x_{3}]$ , $[x_{1},0]$ . As long as $(\phi,A)\in\mathbb{R}\times\mathcal{D}_{0}$ with $\mathcal{D}_{0}=\mathbb{C}\setminus\{c\leq 0\}\cup\{c\geq\tfrac{8}{27}\}$ , the polynomial $z^{4}+4e^{i\phi}z^{3}+4e^{2i\phi}z^{2}+4e^{3i\phi}Az$ admits the roots $0,$ $z_{j}$ $(j=1,3,5)$ such that $\mathrm{Re\,}e^{-i\phi}z_{5}<\mathrm{Re\,}e^{-i\phi}z_{3}<\mathrm{Re\,}e^{-i\phi}z_{1}$ and that $z_{j}=x_{j}$ if $\phi=0,$ $0<A<\frac{8}{27}$ (cf. Corollary 8.2), and then the elliptic curve

w^{2}=w(A,z)^{2}=z^{4}+4e^{i\phi}z^{3}+4e^{2i\phi}z^{2}+4e^{3i\phi}Az=z(z-z_{1})(z-z_{3})(z-z_{5})

may be considered to be the two sheeted Riemann surface $\Pi_{A,\phi}=\Pi_{+}\cup\Pi_{-}$ that is a continuous modification of $\Pi_{A,0}$ with $\Pi_{\pm}$ glued along cuts $[z_{1},0],$ $[z_{5},z_{3}]$ . Here the branch of $w=\sqrt{z(z-z_{1})(z-z_{3})(z-z_{5})}=z^{2}\sqrt{1-z_{1}z^{-1}}\sqrt{1-z_{3}z^{-1}}\sqrt{1-z_{5}z^{-1}}$ is such that $\sqrt{1-z_{j}z^{-1}}\to 1$ as $z\to\infty$ $(j=1,3,5)$ on the upper sheet $\Pi_{+}$ .

As will be shown in Corollary 8.16 with Remark 8.3, for each $\phi\in\mathbb{R}$ , there exists $A_{\phi}\in\mathbb{C}$ such that, for any cycle $\mathbf{c}$ on $\Pi_{A_{\phi},\phi}$

\mathrm{Re\,}\int_{\mathbf{c}}\frac{w(A_{\phi},z)}{z}dz=0

and that $A_{\phi}$ has the following properties:

(1) for each $\phi\in\mathbb{R}$ , $A_{\phi}$ is uniquely determined;

(2) ${A}_{\phi+\pi/2}={A}_{\phi},$ ${A}_{-\phi}=\overline{{A}_{\phi}}$ ;

(3) $A_{0}=\frac{8}{27},$ $A_{\pm\pi/4}=0$ and $0\leq\mathrm{Re\,}A_{\phi}\leq\tfrac{8}{27};$

(4) $A_{\phi}$ is continuous in $\phi\in\mathbb{R}$ and is smooth in $\phi\in\mathbb{R}\setminus\{\pi k/4\,|\,k\in\mathbb{Z}\}.$

The elliptic curve $\Pi_{A_{\phi},\phi}$ degenerates if and only if $\phi=\pi k/4$ with $k\in\mathbb{Z}.$

2.1. Solutions for $0<|\phi|<\pi/4$ in [13, Theorem 2]

For $0<|\phi|<\pi/4$ and $A\in\mathcal{D}_{0}$ , let the primitive cycles $\mathbf{a}$ and $\mathbf{b}$ on $\Pi_{A,\phi}$ be as described on the upper sheet $\Pi_{+}$ in Figure 2.1. (The cycles $\mathbf{a}$ and $\mathbf{b}$ are consistent with those defined in [13, Section 3].) Then the Boutroux equations

(2.2)

\mathrm{Re\,}\int_{\mathbf{a}}\frac{w(A,z)}{z}dz=\mathrm{Re\,}\int_{\mathbf{b}}\frac{w(A,z)}{z}dz=0

admit a unique solution $A=A_{\phi}$ , which means [13, Theorem 1]. For $0<|\phi|<\pi/4$ the periods of $\Pi_{A_{\phi},\phi}$ along $\mathbf{a}$ and $\mathbf{b}$ are given by

\Omega_{\mathbf{a}}=\Omega_{\mathbf{a}}^{\phi}=\int_{\mathbf{a}}\frac{dz}{w(A_{\phi},z)},\quad\Omega_{\mathbf{b}}=\Omega_{\mathbf{b}}^{\phi}=\int_{\mathbf{b}}\frac{dz}{w(A_{\phi},z)},

which satisfy $\mathrm{Im\,}\Omega_{\mathbf{b}}/\Omega_{\mathbf{a}}>0.$

Figure 2.1. Cycles

\mathbf{a},

\mathbf{b}

\Pi_{A,\phi}=\Pi_{+}\cup\Pi_{-}

Let $\mathrm{P}(u;A)$ denote the elliptic function defined by

\mathrm{P}_{u}^{2}=\mathrm{P}^{4}+4e^{i\phi}\mathrm{P}^{3}+4e^{2i\phi}\mathrm{P}^{2}+4e^{3i\phi}A\mathrm{P},\quad\mathrm{P}(0;A)=0,\quad\text{i.e.}\,\,\,\int^{\mathrm{P}(u;A)}_{0}\frac{dz}{w(A,z)}=u.

Note that $\mathrm{P}(u;A_{\phi})$ does not degenerate as long as $\phi\not=k\pi/4,$ $k\in\mathbb{Z}.$ Then [13, Theorem 2] with $n=0,$ $m=0,1$ may be described as follows.

Let $y=y(\mathbf{s},x)$ denote the solution of P_IV labelled by the monodromy data $\mathbf{s}=(s_{1},s_{2},s_{3},s_{4})\in\mathcal{M}_{0}(\alpha,\beta).$

Theorem 2.1.

Suppose that $-\pi/4<\phi<0$ and that $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0,$ $1+s_{1}s_{2}\not=0.$ Then

	$\displaystyle y(\mathbf{s},x)=e^{-i\phi}x\mathrm{P}(e^{i\phi}t+\chi_{0-}+O(t^{-\delta});A_{\phi}),\quad 2t=(e^{-i\phi}x)^{2},$
	$\displaystyle\chi_{0-}\equiv\frac{\Omega_{\mathbf{a}}}{2\pi i}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)$
	$\displaystyle\phantom{----------}-\frac{\Omega_{\mathbf{b}}}{2\pi i}\ln(1+s_{1}s_{2})+\frac{\Omega_{\mathbf{b}}}{3}(\alpha-\beta)\quad\mod\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}$

as $2t=(e^{-i\phi}x)^{2}\to\infty$ through the cheese-like strip

S_{0}(\phi,t_{\infty},\kappa_{0},\delta_{0})=\{x=e^{i\phi}\sqrt{2t}\,|\,\mathrm{Re\,}t>t_{\infty},\,\,\,|\mathrm{Im\,}t|<\kappa_{0}\}\setminus\bigcup_{\sigma\in\mathcal{P}_{0-}}\{|x-\sigma|<\delta_{0}\}

with $\mathcal{P}_{0-}=\{e^{i\phi}\sqrt{2t_{0-}}\,|\,e^{i\phi}t_{0-}+\chi_{0-}=\pm\tfrac{1}{3}\Omega_{\mathbf{b}}+\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}\}.$ Here $\delta$ is some positive number, $\kappa_{0}$ a given positive number, $\delta_{0}$ a given small positive number, and $t_{\infty}=t_{\infty}(\kappa_{0},\delta_{0})$ a sufficiently large positive number depending on $(\kappa_{0},\delta_{0}).$

Theorem 2.2.

Suppose that $0<\phi<\pi/4$ and that $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0,$ $1+s_{2}s_{3}\not=0.$ Then

	$\displaystyle y(\mathbf{s},x)=e^{-i\phi}x\mathrm{P}(e^{i\phi}t+\chi_{0+}+O(t^{-\delta});A_{\phi}),\quad 2t=(e^{-i\phi}x)^{2},$
	$\displaystyle\chi_{0+}\equiv\frac{\Omega_{\mathbf{a}}}{2\pi i}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)$
	$\displaystyle\phantom{----------}+\frac{\Omega_{\mathbf{b}}}{2\pi i}\ln(1+s_{2}s_{3})+\frac{\Omega_{\mathbf{b}}}{3}(\alpha-\beta)\quad\mod\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}$

as $2t=(e^{-i\phi}x)^{2}\to\infty$ through the cheese-like strip

S_{1}(\phi,t_{\infty},\kappa_{0},\delta_{0})=\{x=e^{i\phi}\sqrt{2t}\,|\,\mathrm{Re\,}t>t_{\infty},\,\,\,|\mathrm{Im\,}t|<\kappa_{0}\}\setminus\bigcup_{\sigma\in\mathcal{P}_{0+}}\{|x-\sigma|<\delta_{0}\}

with $\mathcal{P}_{0+}=\{e^{i\phi}\sqrt{2t_{0+}}\,|\,e^{i\phi}t_{0+}+\chi_{0+}=\pm\tfrac{1}{3}\Omega_{\mathbf{b}}+\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}\}.$

Remark 2.1.

If $A_{\phi}\not=0,\tfrac{8}{27}$ , then $\mathrm{P}(\tau;A_{\phi})=e^{3i\phi}A_{\phi}(\wp(\tau;g_{2},g_{3})-\frac{1}{3}e^{2i\phi})^{-1}$ with $g_{2}=-4e^{4i\phi}(A_{\phi}-\frac{1}{3})$ , $g_{3}=e^{6i\phi}(-A_{\phi}^{2}+\frac{4}{3}A_{\phi}-\frac{8}{27}),$ where $\wp(\tau;g_{2},g_{3})$ is the Weierstrass pe-function satisfying $\wp_{\tau}^{2}=4\wp^{3}-g_{2}\wp-g_{3}.$ Furthermore, $\wp_{\tau}=-e^{3i\phi}A_{\phi}\mathrm{P}^{-2}w(A_{\phi},\mathrm{P}).$

Remark 2.2.

The solutions $y(\mathbf{s},x)$ in Theorems 2.1 and 2.2, and the correction function $B_{\phi}(t)$ in Proposition 7.4 are parametrised by $(\mathfrak{s}_{12},\mathfrak{s}_{23})=(s_{1}s_{2},s_{2}s_{3})$ . The variables $\mathfrak{s}_{12},$ $\mathfrak{s}_{23}$ , and $\mathfrak{s}_{34}=s_{3}s_{4},$ $\mathfrak{s}_{41}=s_{4}s_{1}$ are invariant under the action $[c]$ on $\mathcal{M}_{0}(\alpha,\beta)$ , and hence the family of the orbits generated by $[c]$ on $\mathcal{M}_{0}(\alpha,\beta)$ may be identified with the two-dimensional surface:

$\mathcal{M}^{*}_{0}(\alpha,\beta):$		$\displaystyle\bigl{(}(1+\mathfrak{s}_{12})(1+\mathfrak{s}_{34})+\mathfrak{s}_{41}\bigr{)}e^{i\pi(\beta-\alpha)}-(1+\mathfrak{s}_{23})e^{i\pi(\alpha-\beta)}+2i\sin\pi\alpha=0,$
	$\displaystyle\mathfrak{s}_{12}\mathfrak{s}_{34}-\mathfrak{s}_{23}\mathfrak{s}_{41}=0,$

whose points parametrise solutions of P_IV given in Theorems 2.1, 2.2, and [13, Theorems 2, 4, 5 and 6] with $n=0$ . Singular points on this surface is given by Proposition 3.2. It is easy to see that a point $(\mathfrak{s}_{12},\mathfrak{s}_{23},\mathfrak{s}_{34})\in\mathcal{M}^{*}_{0}(\alpha,\beta)$ satisfying the condition of Theorem 2.1 or 2.2 is nonsingular.

Remark 2.3.

An inconsistency appears between the signs of $\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)$ in [13, Theorem 2] and in our Theorems 2.1 and 2.2. The agreements of the trigonometric asymptotics with [13, Theorem 4] and [9, Theorem 4.1] discussed in Section 2.3 support the correctness of the sign $+$ in our theorems.

2.2. Alternative expression of solutions

Let the elliptic curve $\Pi_{A_{\phi}}^{*}=\Pi^{*}_{+}\cup\Pi_{-}^{*}$ : $v^{2}=v(A_{\phi},\zeta)^{2}=e^{-4i\phi}w(A_{\phi},e^{i\phi}\zeta)^{2}$ be as defined in Section 8, and the cycles $\mathbf{a}^{*}$ and $\mathbf{b}^{*}$ on $\Pi^{*}_{A_{\phi}}$ as drawn in Figure 8.1. These elliptic curve and cycles, which depend on $A_{\phi}$ only, are also the images of $\Pi_{A_{\phi},\phi}=\Pi_{+}\cup\Pi_{-}$ and $\mathbf{a}$ and $\mathbf{b}$ under the mapping $z=e^{i\phi}\zeta$ . Then the Boutroux equation (2.2) is written in the form

(2.3)

\mathrm{Re\,}\,e^{2i\phi}\int_{\mathbf{a}_{*}}\frac{v(A_{\phi},\zeta)}{\zeta}d\zeta=\mathrm{Re\,}\,e^{2i\phi}\int_{\mathbf{b}_{*}}\frac{v(A_{\phi},\zeta)}{\zeta}d\zeta=0,

in which

v(A_{\phi},\zeta):=e^{-2i\phi}w(A_{\phi},e^{i\phi}\zeta)=\sqrt{\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A_{\phi}\zeta},

and for each $\phi\in\mathbb{R},$ $A_{\phi}$ is a unique solution of (2.3) as in Proposition 8.15. Let $\mathfrak{P}(u)=\mathfrak{P}(u,A_{\phi})$ be the elliptic function defined by

\int^{\mathfrak{P}(u,A_{\phi})}_{0}\frac{d\zeta}{v(A_{\phi},\zeta)}=u.

Then it is easy to see that $\mathrm{P}(e^{-i\phi}u;A_{\phi})=e^{i\phi}\mathfrak{P}(u;A_{\phi})$ , and that

\Omega_{\mathbf{a}_{*}}^{*}=\int_{\mathbf{a}_{*}}\frac{d\zeta}{v(A_{\phi},\zeta)}=e^{i\phi}\Omega_{\mathbf{a}},\quad\Omega_{\mathbf{b}_{*}}^{*}=\int_{\mathbf{b}_{*}}\frac{d\zeta}{v(A_{\phi},\zeta)}=e^{i\phi}\Omega_{\mathbf{b}}

are the periods of $\Pi_{A_{\phi}}^{*}$ . Then the solutions given above are also written as follows.

Corollary 2.3.

Suppose that $0<|\phi|<\pi/4$ , and that $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0$ , $\eta_{\phi}(\mathbf{s})\not=0,$ where $\eta_{\phi}(\mathbf{s})=1+s_{1}s_{2}$ if $-\pi/4<\phi<0$ , and $=(1+s_{2}s_{3})^{-1}$ if $0<\phi<\pi/4.$ Then

	$\displaystyle y(\mathbf{s},x)=x\mathfrak{P}(\tfrac{1}{2}x^{2}+\chi_{0}^{*}+O(x^{-\delta});A_{\phi}),$
	$\displaystyle\chi_{0}^{}\equiv\frac{\Omega_{\mathbf{a}_{}}^{*}}{2\pi i}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)$
	$\displaystyle\phantom{----------}-\frac{\Omega_{\mathbf{b}_{}}^{}}{2\pi i}\ln\eta_{\phi}(\mathbf{s})+\frac{\Omega_{\mathbf{b}_{}}^{}}{3}(\alpha-\beta)\mod\Omega_{\mathbf{a}_{}}^{}\mathbb{Z}+\Omega_{\mathbf{b}_{}}^{}\mathbb{Z}$

as $x\to\infty$ through the cheese-like strip

S_{0}^{*}(\phi,t_{\infty},\kappa_{0},\delta_{0})=\{x=e^{i\phi}\sqrt{2t}\,|\,\mathrm{Re\,}t>t_{\infty},\,\,\,|\mathrm{Im\,}t|<\kappa_{0}\}\setminus\bigcup_{\sigma\in\mathcal{P}_{0}^{*}}\{|x-\sigma|<\delta_{0}\}

with $\mathcal{P}_{0}^{*}=\{e^{i\phi}\sqrt{2t_{0}^{*}}\,|\,e^{2i\phi}t^{*}_{0}+\chi^{*}_{0}=\pm\tfrac{1}{3}\Omega^{*}_{\mathbf{b}_{*}}+\Omega^{*}_{\mathbf{a}_{*}}\mathbb{Z}+\Omega^{*}_{\mathbf{b}_{*}}\mathbb{Z}\}.$

The dependence on $\phi$ of $v(A_{\phi},\zeta)$ is on $A_{\phi}$ only such that $A_{\phi+\pi/2}=A_{\phi}$ , and hence so is that of $\Omega_{\mathbf{a}_{*}}^{*}$ , $\Omega_{\mathbf{b}_{*}}^{*}$ and $\mathfrak{P}(u,A_{\phi})$ . Solutions in general directions are given as follows.

Theorem 2.4.

For any $n\in\mathbf{Z}$ , let $\mathbf{s}_{n}=(s_{1+n},s_{2+n},s_{3+n},s_{4+n})$ . Suppose that $0<|\phi-\pi n/2|<\pi/4$ , and that $(1+s_{1+n}s_{2+n})(1+s_{2+n}s_{3+n})-1\not=0,$ $\eta_{\phi}(\mathbf{s}_{n})\not=0,$ where $\eta_{\phi}(\mathbf{s}_{n})=1+s_{1+n}s_{2+n}$ if $-\pi/4<\phi-\pi n/2<0$ , and $=(1+s_{2+n}s_{3+n})^{-1}$ if $0<\phi-\pi n/2<\pi/4.$ Then $\mathrm{P}_{\mathrm{IV}}$ admits a solution represented as follows $:$

	$\displaystyle y(\mathbf{s}_{n},x)=x\mathfrak{P}(\tfrac{1}{2}x^{2}+\chi_{0}^{n}+O(x^{-\delta});A_{\phi}),$
	$\displaystyle\chi_{0}^{n}\equiv\frac{\Omega_{\mathbf{a}_{}}^{}}{2\pi i}\ln((1+s_{1+n}s_{2+n})(1+s_{2+n}s_{3+n})-1)$
	$\displaystyle\phantom{----------}-\frac{\Omega_{\mathbf{b}_{}}^{}}{2\pi i}\ln\eta_{\phi}(\mathbf{s}_{n})+\frac{\Omega_{\mathbf{b}_{}}^{}}{3}(\alpha-\beta)\mod\Omega_{\mathbf{a}_{}}^{}\mathbb{Z}+\Omega_{\mathbf{b}_{}}^{}\mathbb{Z}$

as $x\to\infty$ through the cheese-like strip

S_{0}^{n}(\phi,t_{\infty},\kappa_{0},\delta_{0})=\{x=e^{i\phi}\sqrt{2t}\,|\,\mathrm{Re\,}t>t_{\infty},\,\,\,|\mathrm{Im\,}t|<\kappa_{0}\}\setminus\bigcup_{\sigma\in\mathcal{P}_{0}^{n}}\{|x-\sigma|<\delta_{0}\}

with $\mathcal{P}_{0}^{n}=\{e^{i\phi}\sqrt{2t_{0}^{n}}\,|\,e^{2i\phi}t^{n}_{0}+\chi^{n}_{0}=\pm\tfrac{1}{3}\Omega^{*}_{\mathbf{b}_{*}}+\Omega^{*}_{\mathbf{a}_{*}}\mathbb{Z}+\Omega^{*}_{\mathbf{b}_{*}}\mathbb{Z}\}.$

Remark 2.4.

By the single-valuedness on $\mathbb{C}$ , $y(\mathbf{s}_{n},x)$ coincides with $y(\mathbf{s}_{n+4},x)$ . Indeed $s_{k}s_{k+1}=s_{k+4}s_{k+5}$ for every $k\in\mathbb{Z}.$

2.3. Observation related to trigonometric asymptotics

Let us observe trigonometric asymptotics as degeneration of our elliptic solutions. For elliptic expressions of solutions in Theorems 2.1, 2.2 or Corollary 2.3 we may calculate at least formal leading terms of the analytic continuations to strips lying along the positive real axis, which is expected to behave trigonometrically as in [13, Theorem 4]. Such problems were discussed in [21, Section 4] for the first and the second Painlevé transcendents.

Write the strip $S_{0}(\phi,t_{\infty},\kappa_{0},\delta_{0})$ in Theorem 2.1 in the form

	$\displaystyle S_{0}(\phi,t_{\infty},\kappa_{0},\delta_{0})$	$\displaystyle=e^{i\phi}\sqrt{2T(t_{\infty},\kappa_{0}))}\setminus P_{\phi}=\sqrt{2e^{2i\phi}T(t_{\infty},\kappa_{0}))}\setminus P_{\phi},$
	$\displaystyle T(t_{\infty},\kappa_{0})$	$\displaystyle=\{t\,\|\,\mathrm{Re\,}t>t_{\infty},\,\,\|\mathrm{Im\,}t\|<\kappa_{0}\},$

where $\sqrt{2T}=\{x=\sqrt{2t}\,|\,t\in T\}$ and $P_{\phi}=\bigcup_{\sigma\in\mathcal{P}_{0-}}\{|x-\sigma|<\delta_{0}\}$ . Let us suppose that there exists a strip $\mathcal{S}_{0}$ such that

\mathcal{S}_{0}=\{t\,|\,\mathrm{Re\,}t>t_{\infty}^{0},\,\,-K_{0}<\mathrm{Im\,}t<-K_{1}\}\subset\bigcup_{-\phi_{0}<\phi<0}e^{2i\phi}T(t_{\infty},\kappa_{0}),

where $K_{0}$ , $K_{1}$ are positive constants and $\phi_{0}$ a small positive constant. Then $\sqrt{2\mathcal{S}_{0}}$ is a strip lying along the line $\sqrt{2t^{0}_{\infty}}<x<+\infty$ , and has the properties: for each $\phi$ ,

(1) every $t\in\mathcal{S}_{0}\cap e^{2i\phi}T(t_{\infty},\kappa_{0})$ fulfils $t\phi\asymp 1$ with implied constants independent of $\phi$ and $t$ ;

(2) $\sqrt{2(\mathcal{S}_{0}\cap e^{2i\phi}T(t_{\infty},\kappa_{0}))}\setminus P_{\phi}\subset S_{0}(\phi,t_{\infty},\kappa_{0},\delta_{0}).$

By the property (2) the expression of $y(\mathbf{s},x)$ in Theorem 2.1 is valid in each region $\mathcal{S}_{0}\cap e^{2i\phi}T(t_{\infty},\kappa_{0})$ . By Remark 2.1 this is written in the form

(2.4)

(2t)^{-1/2}y(\mathbf{s},x)=\frac{e^{3i\phi}A_{\phi}}{\wp(\tau;g_{2},g_{3})-\tfrac{1}{3}e^{2i\phi}},\quad\tau=e^{i\phi}t+\chi_{0-}+O(t^{-\delta}).

Note that, by Corollary 8.20 and the property (1),

\Omega_{\mathbf{a}}=e^{-i\phi}\Omega^{*}_{\mathbf{a}_{*}}=\tfrac{1}{2}\sqrt{3}i\ln t\,(1+o(1)),\quad\Omega_{\mathbf{b}}=e^{-i\phi}\Omega^{*}_{\mathbf{b}_{*}}=-\sqrt{3}\pi(1+o(1)),

and $\chi_{0-}=\tfrac{1}{4}\sqrt{3}\pi^{-1}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)\ln t-i\tfrac{1}{2}\sqrt{3}(\ln(1+s_{1}s_{2})-\tfrac{2}{3}\pi i(\alpha-\beta))$ . The periods of $\wp(\tau)$ are given by $(2\omega,2\omega^{\prime})=(\sqrt{3}\pi(1+o(1)),\tfrac{1}{2}\sqrt{3}i\ln t\,(1+o(1))$ such that $\mathrm{Im\,}(\omega^{\prime}/\omega)=(2\pi)^{-1}\ln t\,(1+o(1)).$ In (2.4), we have $\phi=O(t^{-1})$ and $A_{\phi}=\tfrac{8}{27}+O(t^{-1})$ (cf. Proposition 8.19) as $t\to\infty$ , and then $\wp(u)$ degenerates to

(2.5)			$\displaystyle\wp(u)=-\tfrac{1}{3}\hat{\omega}^{-2}+\hat{\omega}^{-2}\sin^{-2}(\hat{\omega}u)+O(h\cos^{2}(2\hat{\omega}u))\quad\text{or}$
(2.6)			$\displaystyle\wp(u)=-\tfrac{1}{3}\hat{\omega}^{2}-8\hat{\omega}^{2}h\cos(2\hat{\omega}(u-\omega^{\prime}))+O(h^{2}\cos^{2}(2\hat{\omega}(u-\omega^{\prime})))$

with $\hat{\omega}=\pi/(2\omega)$ and $h=e^{i\pi{\omega^{\prime}/\omega}}$ [8], [32]. In $\mathcal{S}_{0}\cap(\bigcup_{-\phi_{0}<\phi<0}e^{2i\phi}T(t_{\infty},\kappa_{0}))$ , from (2.5) we have the trigonometric expression

	$\displaystyle\frac{1}{(2t)^{-1/2}y(\mathbf{s},x)+\tfrac{2}{3}}$	$\displaystyle=\frac{1}{2}+\frac{1}{2}(e^{i2\tau/\sqrt{3}}+e^{-i2\tau/\sqrt{3}})(1+O(t^{-\varepsilon})),$
	$\displaystyle{2\tau}/{\sqrt{3}}={2t}/{\sqrt{3}}$	$\displaystyle+({2\pi})^{-1}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)\ln t-\tfrac{2}{3}\pi(\alpha-\beta)+O(1),$

which agrees with [13, Theorem 4, (30)] up to constants. From (2.6) we have

	$\displaystyle y(\mathbf{s},x)+\tfrac{2}{3}$	$\displaystyle=2t^{-1/2}(e^{i2\hat{\tau}/\sqrt{3}}+e^{-i2\hat{\tau}/\sqrt{3}})(1+O(t^{-\varepsilon})),$
	$\displaystyle{2\hat{\tau}}/{\sqrt{3}}$	$\displaystyle={2t}/{\sqrt{3}}+({2\pi})^{-1}\ln(1-(1+s_{1}s_{2})(1+s_{2}s_{3}))\ln t-\tfrac{2}{3}\pi(\alpha-\beta)+O(1),$

which agrees with [13, Theorem 4, (28)] and also with [9, Theorem 1.1, (1.11)] up to constants. In the strip $0<K_{0}<\mathrm{Im\,}t<K_{1}$ a similar argument is possible for the solution in Theorem 2.2. The argument above, though not justified, suggests information about degeneration to the trigonometric asymptotics.

3. Basic facts

3.1. Monodromy data

The monodromy data and the monodromy manifold $\mathcal{M}^{*}_{0}(\alpha,\beta)$ described in Section 2 have the following properties.

Proposition 3.1.

For each $m\in\mathbb{Z}$ ,

	$\displaystyle S_{1+m}S_{2+m}S_{3+m}S_{4+m}e^{-i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}$
	$\displaystyle\phantom{---}=\begin{cases}(ES_{1}\cdots S_{m})^{-1}\sigma_{3}M^{-1}(ES_{1}\cdots S_{m})\quad&\text{if $m\geq 1;$}\\[11.38092pt] (ES_{0}^{-1}\cdots S_{m+1}^{-1})^{-1}\sigma_{3}M^{-1}(ES_{0}^{-1}\cdots S_{m+1}^{-1})\quad&\text{if $m\leq-1.$}\end{cases}$

Proof..

The semi-cyclic condition $S_{1}S_{2}S_{3}S_{4}=E^{-1}\sigma_{3}M^{-1}Ee^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}$ with $S_{k+4}=e^{-i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}S_{k}\sigma_{3}e^{i\pi(\alpha-\beta)\sigma_{3}}$ yields

	$\displaystyle S_{2}S_{3}S_{4}S_{1+4}=$	$\displaystyle S_{1}^{-1}E^{-1}\sigma_{3}M^{-1}Ee^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}e^{-i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}S_{1}\sigma_{3}e^{i\pi(\alpha-\beta)\sigma_{3}}$
	$\displaystyle=$	$\displaystyle(ES_{1})^{-1}\sigma_{3}M^{-1}(ES_{1})e^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3},$
	$\displaystyle S_{0}S_{1}S_{2}S_{3}=$	$\displaystyle S_{0}E^{-1}\sigma_{3}M^{-1}Ee^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}S_{4}^{-1}$
	$\displaystyle=$	$\displaystyle(ES_{0}^{-1})^{-1}\sigma_{3}M^{-1}ES_{0}^{-1}S_{0}e^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}S_{4}^{-1}$
	$\displaystyle=$	$\displaystyle(ES_{0}^{-1})^{-1}\sigma_{3}M^{-1}(ES_{0}^{-1})e^{i\pi(\alpha-\beta)\sigma_{3}}\sigma_{3}.$

Repeating this procedure, we obtain the proposition. ∎

Proposition 3.2.

The surface $\mathcal{M}^{*}_{0}(\alpha,\beta)$ has a singular point $\mathbf{s}_{\mathrm{sing}}$ if and only if $\alpha-\tfrac{1}{2}\in\mathbb{Z}$ , and then $\mathbf{s}_{\mathrm{sing}}=(\mathfrak{s}_{12},\mathfrak{s}_{23},\mathfrak{s}_{34},\mathfrak{s}_{41})=(e^{-i\pi\beta}-1,e^{i\pi\beta}-1,e^{-i\pi\beta}-1,e^{-i\beta\pi}-e^{-2i\beta\pi}).$

Proof..

Write the surface in the form

f=e^{i\pi(\beta-\alpha)}(xz+u-1)-e^{i\pi(\alpha-\beta)}y+2i\sin\pi\alpha=0,\quad(x-1)(z-1)=(y-1)(u-1)

with $(x,y,z,u)=(1+\mathfrak{s}_{12},1+\mathfrak{s}_{23},1+\mathfrak{s}_{34},1+\mathfrak{s}_{41}),$ and examine when $f_{x},$ $f_{y},$ $f_{z}$ , $f_{u}$ and $f$ have a common zero. If, say $y-1\not=0$ , by using $u-1=(x-1)(z-1)/(y-1)$ , we have $\alpha-\tfrac{1}{2}\in\mathbb{Z}$ and $x=z=1/y=e^{-i\pi\beta}.$ ∎

Remark 3.1.

Equation P_IV admits a one-parameter family of classical solutions (respectively, a rational solution) if and only if $\{\alpha-\tfrac{1}{2},\,\tfrac{1}{2}\beta,\,\alpha-\tfrac{1}{2}\beta-\tfrac{1}{2}\}\cap\mathbb{Z}\not=\emptyset$ (respectively, $\{(\alpha-\tfrac{1}{2},\tfrac{1}{2}\beta),(\alpha\pm\tfrac{1}{6},2\alpha-\tfrac{1}{2}\beta)\}\cap\mathbb{Z}^{2}\not=\emptyset$ ) [5], [6], [7, Chap. 6], [22], [30].

3.2. Isomonodromy linear system

Let us transform system (1.1) into a form suitable to our calculation. By

\Psi=\mathrm{u}^{(1/2)\sigma_{3}}x^{-(1/4)\sigma_{3}}\Phi,\quad\mathrm{u}\mathrm{v}=y,\quad\frac{\mathrm{u}_{x}}{\mathrm{u}}=x\mathrm{z},

system (1.1) is changed into

	$\displaystyle\frac{d\Phi}{d\xi}=$	$\displaystyle\Biggl{(}\Bigl{(}\frac{\xi^{3}}{2}+(x+y)\xi+\frac{\alpha}{\xi}\Bigr{)}\sigma_{3}+ix^{1/2}\begin{pmatrix}0&\xi^{2}+x\mathrm{z}+2x\\ x^{-1}(y\xi^{2}-xy\mathrm{z}+y^{2}+\beta)&0\end{pmatrix}\Biggr{)}\Phi,$
	$\displaystyle y^{\prime}=$	$\displaystyle 2xy\mathrm{z}+2xy-y^{2}-\beta.$

The change of variables

\tau=x^{2},\quad y=x\eta,\quad\xi=x^{1/2}\tilde{\xi}

takes this system to

	$\displaystyle\frac{d\Phi}{d\tilde{\xi}}=$	$\displaystyle\tau\Biggl{(}\Bigl{(}\frac{\tilde{\xi}^{3}}{2}+(1+\eta)\tilde{\xi}+\frac{\alpha\tau^{-1}}{\tilde{\xi}}\Bigr{)}\sigma_{3}+i\begin{pmatrix}0&\tilde{\xi}^{2}+\mathrm{z}+2\\ \eta\tilde{\xi}^{2}-\eta\mathrm{z}+\eta^{2}+\beta\tau^{-1}&0\end{pmatrix}\Biggr{)}\Phi,$
	$\displaystyle\tau\frac{d\eta}{d\tau}=$	$\displaystyle\tau\eta(\mathrm{z}+1)-\frac{1}{2}({\tau\eta^{2}}+\eta+\beta).$

The further substitution

\tau=2e^{2i\phi}t,\quad\eta=e^{-i\phi}\psi,\quad e^{i\phi}\mathrm{z}=\mathfrak{z},\quad\lambda=e^{i\phi/2}\tilde{\xi}

leads to

(3.1)			$\displaystyle\frac{d\Phi}{d\lambda}=t\mathcal{B}(t,\lambda)\Phi,\qquad\mathcal{B}(t,\lambda)=b_{3}\sigma_{3}+b_{1}\sigma_{1}+b_{2}\sigma_{2},$
(3.2)			$\displaystyle 2(\mathfrak{z}+e^{i\phi})=e^{-i\phi}\frac{\psi_{t}}{\psi}+\psi+\tfrac{1}{2}(e^{-i\phi}+\beta\psi^{-1})t^{-1}$

with

	$\displaystyle b_{1}=ie^{-i\phi/2}\left((\psi+e^{i\phi})\lambda^{2}-(\psi-e^{i\phi})\mathfrak{z}+\psi^{2}+2e^{2i\phi}+\tfrac{1}{2}\beta t^{-1}\right),$
	$\displaystyle b_{2}=e^{-i\phi/2}\left((\psi-e^{i\phi})\lambda^{2}-(\psi+e^{i\phi})\mathfrak{z}+\psi^{2}-2e^{2i\phi}+\tfrac{1}{2}\beta t^{-1}\right),$
	$\displaystyle b_{3}=\lambda^{3}+2(e^{i\phi}+\psi)\lambda+\alpha t^{-1}\lambda^{-1}$

In (3.2) as a resulting equation, $\psi_{t}$ denotes the derivative $(d/dt)\psi.$ Let us now change the meaning of $\psi_{t}$ in such a way that $\psi_{t}$ is an arbitrary function not necessarily the derivative, and in what follows suppose that system (3.1) is equipped with $\mathfrak{z}$ containing such $\psi_{t}.$ Then the isomonodromy property of (1.1) is converted to that of (3.1).

Proposition 3.3.

The monodromy data of (3.1) is invariant under a small change of $t$ if and only if $\psi_{t}=(d/dt)\psi$ holds in (3.2) and $y(x)=e^{-i\phi}x\psi$ with $2e^{2i\phi}t=x^{2}$ solves $\mathrm{P}_{\mathrm{IV}}$ .

For $k\in\mathbb{Z}$ system (3.1) admits canonical solutions

(3.3)

\Phi_{k}^{\infty}(\lambda)=(I+O(\lambda^{-1}))\exp((\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3})

as $\lambda\to\infty$ through the sector $|\arg(t^{1/4}\lambda)+\tfrac{\pi}{8}-\tfrac{\pi}{4}k|<\tfrac{\pi}{4}$ . The Stokes matrices are defined by $\Phi_{k+1}^{\infty}(\lambda)=\Phi_{k}^{\infty}(\lambda)S_{k}^{*}$ . Recalling the Stokes matrices $S_{k}$ with respect to $\Psi^{\infty}_{k}(\xi)$ solving (1.1), we have the following relation.

Proposition 3.4.

For every $k\in\mathbb{Z}$ ,

S_{k}=\mathrm{u}^{(1/2)\sigma_{3}}x^{-(1/4)\sigma_{3}}(e^{-i\phi/2}x^{1/2})^{-(\alpha-\beta)\sigma_{3}}S_{k}^{*}(e^{-i\phi/2}x^{1/2})^{(\alpha-\beta)\sigma_{3}}\mathrm{u}^{-(1/2)\sigma_{3}}x^{(1/4)\sigma_{3}}.

Proof..

The relation $\Psi^{\infty}_{k}(\xi)=\mathrm{u}^{(1/2)\sigma_{3}}x^{-(1/4)\sigma_{3}}\Phi^{\infty}_{k}(\lambda)(e^{-i\phi/2}x^{1/2})^{(\alpha-\beta)\sigma_{3}}\mathrm{u}^{-(1/2)\sigma_{3}}x^{(1/4)\sigma_{3}}$ yields the conclusion. ∎

The Stokes coefficients $s^{*}_{k}$ for $\Phi^{\infty}_{k}(\lambda)$ are given by

S_{2l-1}^{*}=\begin{pmatrix}1&s^{*}_{2l-1}\\ 0&1\end{pmatrix},\quad S_{2l}^{*}=\begin{pmatrix}1&0\\ s^{*}_{2l}&1\end{pmatrix}\quad(l\in\mathbb{Z}).

Corollary 3.5.

For any $k,j\in\mathbb{Z}$ , $s_{k}^{*}s_{k+2j+1}^{*}=s_{k}s_{k+2j+1}$ .

4. Turning points, Stokes graph, WKB analysis

For system (3.1) we will treat the direct monodromy problem by WKB analysis. Let us start with the characteristic roots $\pm\mu(t,\lambda)$ of $\mathcal{B}(t,\lambda)$ constituting the essential part of the WKB solution (cf. Proposition 4.1). By $\mu(t,\lambda)^{2}=b_{1}^{2}+b_{2}^{2}+b_{3}^{2}$ , we have

(4.1)

\mu(t,\lambda)^{2}=\lambda^{6}+4e^{i\phi}\lambda^{4}+(4e^{2i\phi}+2(\alpha-\beta)t^{-1})\lambda^{2}+4e^{3i\phi}a_{\phi}+\alpha^{2}t^{-2}\lambda^{-2}

with

(4.2)		$\displaystyle 4e^{3i\phi}a_{\phi}=$	$\displaystyle 4e^{3i\phi}a_{\phi}(t)=e^{-2i\phi}{(\psi_{t})^{2}}{\psi^{-1}}-(\psi+2e^{i\phi})^{2}\psi$
		$\displaystyle+(e^{-2i\phi}\psi_{t}+(4\alpha-\beta)\psi+2(2\alpha-\beta)e^{i\phi})t^{-1}+\tfrac{1}{4}(e^{-2i\phi}\psi-\beta^{2}\psi^{-1})t^{-2}.$

To draw Stokes graphs it is necessary to know the location of the turning points. Now we note the following facts on the solution $A_{\phi}$ of the Boutroux equation (2.2) for $|\phi|<\pi/4$ (Proposition 8.15):

(i) $A_{0}=\frac{8}{27},$ and then $w(A_{0},z)=z(z+\frac{2}{3})^{2}(z+\frac{8}{3});$

(ii) $A_{\pm\pi/4}=0,$ and then $w(A_{\pm\pi/4},z)=z^{2}(z+2e^{\pm i\pi/4})^{2};$

(iii) for $0<|\phi|<\pi/4,$ $0<\mathrm{Re\,}A_{\phi}<\frac{8}{27}$ and $w(A_{\phi},z)$ does not degenerate.

Then by Corollary 8.2, for $|\phi|<\pi/4$ , the zeros of $w(A_{\phi},z)$ may be numbered in such a way that $\mathrm{Re\,}e^{-i\phi}z_{5}\leq\mathrm{Re\,}e^{-i\phi}z_{3}\leq\mathrm{Re\,}e^{-i\phi}z_{1}$ and that $(z_{1},z_{3},z_{5})\to(-\frac{2}{3},-\frac{2}{3},-\frac{8}{3})$ as $\phi\to 0$ , and $\to(0,-2e^{\pm i\pi/4},-2e^{\pm i\pi/4})$ as $\phi\to\pm\pi/4$ (cf. Section 2).

Our WKB analysis is carried out under the supposition $a_{\phi}(t)=A_{\phi}+O(t^{-1})$ as $t\to\infty$ , that is, (5.1). The characteristic root satisfies $\lambda\mu(t,\lambda)\to w(A_{\phi},\lambda^{2})$ as $t\to\infty$ . Let $\lambda_{j}$ $(1\leq j\leq 6)$ be turning points of $\mu(t,\lambda)$ such that

	$\displaystyle\lambda_{1}\to z_{1}^{1/2},\,\,\,\lambda_{3}\to z_{3}^{1/2},\,\,\,\lambda_{5}\to z_{5}^{1/2}\,\,\,\text{as}\,\,t\to\infty,$
	$\displaystyle\lambda_{2}=-\lambda_{1},\,\,\,\lambda_{4}=-\lambda_{3},\,\,\,\lambda_{6}=-\lambda_{5},$
	$\displaystyle\|\arg\lambda_{j}-\tfrac{\pi}{2}\|<\tfrac{\pi}{4}\,\,\,(j=1,3,5)\,\,\,\text{for sufficiently large $t$}.$

The algebraic function $\mu(t,\lambda)$ is written in the form

	$\displaystyle\mu(t,\lambda)$	$\displaystyle=\sqrt{\smash{\lambda^{6}}+4e^{i\phi}\lambda^{4}+(4e^{2i\phi}+2(\alpha-\beta)t^{-1})\lambda^{2}+4e^{3i\phi}\smash{a_{\phi}}+\alpha^{2}t^{-2}\lambda^{-2}}$
		$\displaystyle=\lambda^{-1}\sqrt{(\lambda^{2}-\lambda_{1}^{2})(\lambda^{2}-\lambda_{3}^{2})(\lambda^{2}-\lambda_{5}^{2})(\lambda^{2}-\lambda_{0}^{2})}$
		$\displaystyle=\lambda^{-1}\sqrt{(\lambda-\lambda_{1})(\lambda-\lambda_{2})(\lambda-\lambda_{3})(\lambda-\lambda_{4})(\lambda-\lambda_{5})(\lambda-\lambda_{6})(\lambda-\lambda_{0})(\lambda+\lambda_{0})},$
		$\displaystyle\lambda_{0}=O(t^{-1}),$

which is considered on the two sheeted Riemann surface $\mathcal{R}_{t}=\mathcal{R}^{+}_{t}\cup\mathcal{R}^{-}_{t}$ glued along the cuts $[\lambda_{5},\lambda_{3}]$ , $[\lambda_{1},\lambda_{0}],$ $[-\lambda_{0},\lambda_{2}]$ , $[\lambda_{4},\lambda_{6}]$ with $\mathcal{R}^{\pm}_{t}=\mathbb{C}\setminus([\lambda_{5},\lambda_{3}]\cup[\lambda_{1},\lambda_{0}]\cup[-\lambda_{0},\lambda_{2}]\cup[\lambda_{4},\lambda_{6}])$ . The branches of $\mu(\infty,\lambda)$ with $a_{\phi}(\infty)=A_{\phi}$ (by (5.1)) is chosen in such a way that $\mu(\infty,\lambda)=\lambda^{3}(1+O(\lambda^{-2})$ as $\lambda\to\infty$ on the upper sheet $\mathcal{R}^{+}_{t}.$ For $0<|\phi|<\pi/4$ , $w(A_{\phi},z)$ does not degenerate and neither does $\mathcal{R}_{t}$ . Then each turning point satisfies $\lambda_{j}(t)-\lambda_{j}(\infty)=O(t^{-1})$ as $t\to\infty.$

In what follows we treat a Stokes graph with $t=\infty$ , and the limit turning point $\lambda_{j}(\infty)$ is simply denoted by $\lambda_{j}$ . By $z=\lambda^{2}$ the algebraic function $w(A_{\phi},z)$ on $\Pi_{A_{\phi},\phi}$ is mapped to $\lambda\mu(\infty,\lambda)$ on $\mathcal{R}_{\infty}$ , in which

	$\displaystyle\mu(\infty,\lambda)=$	$\displaystyle\sqrt{\lambda^{6}+4e^{i\phi}\lambda^{4}+4e^{2i\phi}\lambda^{2}+4e^{3i\phi}A_{\phi}}=\sqrt{(\lambda^{2}-\lambda_{1}^{2})(\lambda^{2}-\lambda_{3}^{2})(\lambda^{2}-\lambda_{5}^{2})}$
	$\displaystyle=$	$\displaystyle\sqrt{(\lambda^{2}-z_{1})(\lambda^{2}-z_{3})(\lambda^{2}-z_{5})}.$

The Stokes graph on $\mathcal{R}_{\infty}$ consists of vertices and Stokes curves, where the Stokes curve is defined by $\mathrm{Re\,}\int^{\lambda}_{\lambda_{j}}\mu(\infty,\lambda)d\lambda=0,$ and the vertices are turning points and singular points. In our case the turning points and the Stokes curves have the following properties:

(i) if $\phi=0$ , then $\lambda_{1}=\lambda_{3}=\tfrac{1}{3}\sqrt{6}i,$ $\lambda_{5}=\frac{2}{3}\sqrt{6}i,$ that is, $\lambda_{1,3}=\tfrac{1}{3}\sqrt{6}i$ is a double turning point, and if $\phi=\pm\pi/4$ , then $\lambda_{1}=0,$ $\lambda_{3}=\lambda_{5}=\sqrt{2}ie^{\pm i\pi/8}$ , that is, $\lambda_{3,5}=\sqrt{2}ie^{\pm i\pi/8}$ is a double turning point;

(ii) if $\phi$ is close to $0$ , then the double turning point $\lambda_{1,3}$ resolves into

	$\displaystyle\lambda_{2\pm 1}=\tfrac{1}{3}\sqrt{6}i\pm\phi_{}e^{i3\pi/4}+O(\phi_{}^{2})\quad\text{for $\phi>0$},$
	$\displaystyle\lambda_{2\pm 1}=\tfrac{1}{3}\sqrt{6}i\pm\phi_{}e^{i\pi/4}+O(\phi_{}^{2})\quad\text{for $\phi<0$},$

where $\phi_{*}=\phi_{*}(\phi)$ is such that $\phi_{*}\geq 0$ and $\phi_{*}=o(1)$ as $\phi\to 0$ (cf. Propositions 8.17 and 8.18);

(iii) by the Boutroux equations (2.2) with $z=\lambda^{2}$ ,

\mathrm{Re\,}\int_{\lambda_{1}}^{\lambda_{3}}\mu(\infty,\lambda)d\lambda=0,\quad\mathrm{Re\,}\int_{\lambda_{3}}^{\lambda_{5}}\mu(\infty,\lambda)d\lambda=0,

implying the existence of Stokes curves joining $\lambda_{1}$ to $\lambda_{3}$ , and $\lambda_{3}$ to $\lambda_{5}$ ;

(iv) the Stokes curves tending to $\infty$ are asymptotic to the rays $\arg\lambda=\frac{\pi}{8}+\frac{\pi}{4}k$ $(1\leq k\leq 8)$ .

Taking these facts into account, we may draw the limit Stokes graphs for $|\phi|<\pi/4$ as in Figure 4.1, in which the cuts $[\lambda_{5},\lambda_{3}]$ , $[\lambda_{1},\lambda_{2}]$ , $[\lambda_{4},\lambda_{6}]$ are omitted.

Figure 4.1. Limit Stokes graphs on

\mathcal{R}_{\infty}^{+}

An unbounded domain $D\subset\mathcal{R}_{\infty}$ is said to be canonical if, for every $\lambda\in D$ , there exist contours ${C}_{\pm}(\lambda)\subset D$ terminating in $\lambda$ such that

\mathrm{Re\,}\int_{\lambda^{-}_{0}}^{\lambda}\mu(\lambda)d\lambda\to-\infty,\quad\mathrm{Re\,}\int_{\lambda^{+}_{0}}^{\lambda}\mu(\lambda)d\lambda\to+\infty

as $\lambda_{0}^{-}\to\infty$ along $C_{-}(\lambda)$ and as $\lambda_{0}^{+}\to\infty$ along $C_{+}(\lambda)$ , respectively (see [3], [4]). The interior of a canonical domain contains exactly one Stokes curve, and the boundary consists of Stokes curves. System (3.1) admits the following WKB solution ([3], [4, Theorem 7.2], [29, Proposition 3.8]).

Proposition 4.1.

In a canonical domain system (3.1) with $\mathcal{B}(t,\lambda)=b_{1}\sigma_{1}+b_{2}\sigma_{2}+b_{3}\sigma_{3}$ admits an asymptotic solution expressed as

\Psi_{\mathrm{WKB}}(\lambda)=T(I+O(t^{-\delta}))\exp\Bigl{(}\int^{\lambda}_{\lambda_{*}}\Lambda(\tau)d\tau\Bigr{)},\quad T=\begin{pmatrix}1&\frac{b_{3}-\mu}{b_{1}+ib_{2}}\\ \frac{\mu-b_{3}}{b_{1}-ib_{2}}&1\end{pmatrix},

as long as $|b_{1}\pm ib_{2}|^{-1}\ll 1,$ $|\psi|+|\mathfrak{z}|\ll 1,$ $|\lambda-\lambda_{j}|\gg t^{-(2/3)(1-\delta)}$ $(\lambda_{j}:$ a simple turning point $)$ with given implied constants. Here $\delta$ is an arbitrary number such that $0<\delta<1$ , $\lambda_{*}$ is a fixed base point, and

	$\displaystyle\Lambda(\lambda)=t\mu(t,\lambda)\sigma_{3}-\mathrm{diag}T^{-1}T_{\lambda},$
	$\displaystyle\mathrm{diag}T^{-1}T_{\lambda}=\frac{1}{4}\Bigl{(}1-\frac{b_{3}}{\mu}\Bigr{)}\frac{\partial}{\partial\lambda}\ln\frac{b_{1}+ib_{2}}{b_{1}-ib_{2}}\sigma_{3}+\frac{1}{2}\frac{\partial}{\partial\lambda}\ln\frac{\mu}{\mu+b_{3}}I.$

Remark 4.1.

In the WKB solution we write $\Lambda(\lambda)$ in the component-wise form $\Lambda(\lambda)=\Lambda_{3}(\lambda)+\Lambda_{I}(\lambda)$ , $\Lambda_{3}(\lambda)\in\mathbb{C}\sigma_{3},$ $\Lambda_{I}(\lambda)\in\mathbb{C}I$ with

\Lambda_{3}(\lambda)=t\mu(t,\lambda)\sigma_{3}-\mathrm{diag}T^{-1}T_{\lambda}|_{\sigma_{3}}\sigma_{3},\quad\Lambda_{I}(\lambda)=-\mathrm{diag}T^{-1}T_{\lambda}|_{I}I.

The WKB solution fails in expressing asymptotics in a neighbourhood of a turning point. Around a simple turning point equation (3.1) is reduced to the system

(4.3)

\frac{dW}{d\zeta}=\begin{pmatrix}0&1\\ \zeta&0\end{pmatrix}W

having solutions ${}^{T}(\mathrm{Ai}(\zeta),\mathrm{Ai}_{\zeta}(\zeta)),$ ${}^{T}(\mathrm{Bi}(\zeta),\mathrm{Bi}_{\zeta}(\zeta))$ [4, Theorem 7.3], [29, Proposition 3.9], where $\mathrm{Ai}(\zeta)$ is the Airy function and $\mathrm{Bi}(\zeta)=e^{-\pi i/6}\mathrm{Ai}(e^{-2\pi i/3}\zeta)$ [1], [2].

Proposition 4.2.

Let $\lambda_{j}$ be a simple turning point, and write $c_{k}=b_{k}(\lambda_{j}),$ $c^{\prime}_{k}=(b_{k})_{\lambda}(\lambda_{j})$ $(k=1,2,3)$ and $\kappa_{c}=c_{1}c_{1}^{\prime}+c_{2}c_{2}^{\prime}+c_{3}c_{3}^{\prime}.$ Suppose that $c_{k},$ $c_{k}^{\prime}$ are bounded and $c_{1}\pm ic_{2}\not=0.$ Then (3.1) admits a matrix solution of the form

\Phi_{j}(\lambda)=T_{j}(I+O(t^{-\delta^{\prime}}))\begin{pmatrix}1&0\\ 0&\hat{t}^{-1}\end{pmatrix}W(\zeta),\quad T_{j}=\begin{pmatrix}1&-\frac{c_{3}}{c_{1}+ic_{2}}\\ -\frac{c_{3}}{c_{1}-ic_{2}}&1\end{pmatrix},

in which $\hat{t}=2(2\kappa_{c})^{-1/3}(c_{1}-ic_{2})(t/4)^{1/3}$ and $\lambda-\lambda_{j}=(2\kappa_{c})^{-1/3}(t/4)^{-2/3}(\zeta+\zeta_{0})$ with $|\zeta_{0}|\ll t^{-1/3},$ as long as $|\zeta|\ll t^{(2/3-\delta^{\prime})/3},$ that is, $|\lambda-\lambda_{j}|\ll t^{-2/3+(2/3-\delta^{\prime})/3}.$ Here $\delta^{\prime}$ is a given number such that $0<\delta^{\prime}<2/3,$ and $W(\zeta)$ solves (4.3), which admits canonical solutions $W_{\nu}(\zeta)$ $(\nu\in\mathbb{Z})$ such that

W_{\nu}(\zeta)=\zeta^{-(1/4)\sigma_{3}}(\sigma_{3}+\sigma_{1})(I+O(\zeta^{-3/2}))\exp(\tfrac{2}{3}\zeta^{3/2}\sigma_{3})

as $\zeta\to\infty$ through the sector $|\arg\zeta-(2\nu-1)\frac{\pi}{3}|<\frac{2\pi}{3},$ and that $W_{\nu+1}(\zeta)=W_{\nu}(\zeta)G_{\nu}$ with

G_{1}=\begin{pmatrix}1&-i\\ 0&1\end{pmatrix},\quad G_{2}=\begin{pmatrix}1&0\\ -i&1\end{pmatrix},\quad G_{\nu+1}=\sigma_{1}G_{\nu}\sigma_{1}.

Remark 4.2.

The matrix solutions $\Psi_{\mathrm{WKB}}(\lambda)$ and $\Phi_{j}(\lambda)$ in Propositions 4.1 and 4.2 are simultaneously valid in the annulus $\mathcal{A}_{j}:$ $t^{-2/3+\varepsilon}\ll|\lambda-\lambda_{j}|\ll t^{-2/3+2\varepsilon}$ with, say, $\varepsilon=\frac{2}{3}\delta=\frac{1}{6}(\frac{2}{3}-\delta^{\prime})=\frac{1}{10}$ , in which the matching is possible.

5. Direct monodromy problem

By WKB analysis we calculate Stokes matrices of linear system (3.1) as a solution of the direct monodromy problem. Recall $a_{\phi}(t)$ given by (4.2) and a unique solution $A_{\phi}$ of the Boutroux equations (2.2) for $0<|\phi|<\pi/4$ . In our calculation suppose that $\psi$ and $\psi_{t}$ are arbitrary functions and that

(5.1)

a_{\phi}(t)=A_{\phi}+\frac{B_{\phi}(t)}{t},\quad B_{\phi}(t)\ll 1

as $t\to\infty$ in the strip

S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})=\{t\,|\,\mathrm{Re\,}t>t^{\prime}_{\infty},\,\,\,|\mathrm{Im\,}t|<\kappa_{1},\,\,\,|\psi(t)|+|\psi(t)|^{-1}+|\psi_{t}(t)|<\delta_{1}^{-1}\},

where $\kappa_{1}$ is a given number, $\delta_{1}$ a given small number and $t_{\infty}^{\prime}$ a sufficiently large number.

Figure 5.1. Fragments of the Stokes graph for

-\pi/4<\phi<0

Let $-\pi/4<\phi<0.$ Let us calculate the analytic continuations of the matrix solution $\Phi_{1}^{\infty}(\lambda)$ given by (3.3) in the sector $|\arg(t\lambda)-\tfrac{\pi}{8}|<\tfrac{\pi}{4}$ along the fragments of the Stokes graph

\mathfrak{c}_{\pi/8}^{5\pi/8}=\mathbf{c}_{\infty_{1}}^{3}\cup\mathbf{c}_{3}^{5}\cup\mathbf{c}_{5}^{\infty_{3}}\,\,\,\text{and}\,\,\,\mathfrak{c}_{\pi/8}^{7\pi/8}=\mathbf{c}_{\infty_{1}}^{3}\cup\mathbf{c}_{3}^{1}\cup\mathbf{c}_{1}^{\infty_{4}}

with $\mathbf{c}_{\infty_{1}}^{3}=(e^{i\pi/8}\infty,\lambda_{3})^{\sim},$ $\mathbf{c}_{3}^{5}=(\lambda_{3},\lambda_{5})^{\sim},$ $\mathbf{c}_{5}^{\infty_{3}}=(\lambda_{5},e^{5\pi/8})^{\sim},$ $\mathbf{c}_{3}^{1}=(\lambda_{3},\lambda_{1})^{\sim},$ $\mathbf{c}_{1}^{\infty_{4}}=(\lambda_{1},e^{7\pi/8})^{\sim},$ where $(p,q)^{\sim}$ denotes a curve joining $p$ to $q$ , and $\mathbf{c}_{3}^{5}$ lies on the right shore of the cut $[\lambda_{3},\lambda_{5}]$ (cf. Figure 5.1).

The Stokes matrix $S^{*}_{1}S^{*}_{2}=\Phi_{1}^{\infty}(\lambda)^{-1}\Phi_{3}^{\infty}(\lambda)$ follows from the analytic continuation of $\Phi_{1}^{\infty}(\lambda)$ along $\mathfrak{c}^{5\pi/8}_{\pi/8}$ , which is calculated by the matching procedure as below. In the steps (2), (3), (6), (7), analytic continuations are considered in the annulus $\mathcal{A}_{3}$ and $\mathcal{A}_{5}$ , which may be given by $\delta=\tfrac{3}{20},$ $\delta^{\prime}=\tfrac{1}{15}$ as in Remark 4.2. Thus, in what follows we set $\delta=\tfrac{3}{20}.$

(1) For a WKB solution $\Psi^{(\infty)}_{3}(\lambda)$ along $\mathbf{c}_{\infty_{1}}^{3}$ with a base point $\tilde{\lambda}_{3}$ such that $\tilde{\lambda}_{3}-\lambda_{3}\asymp t^{-1}$ , set $\Phi_{1}^{\infty}(\lambda)=\Psi^{(\infty)}_{3}(\lambda)\Gamma_{\infty,3}$ . Then

	$\displaystyle\Gamma_{\infty,3}=$	$\displaystyle\Psi^{(\infty)}_{3}(\lambda)^{-1}\Phi_{1}^{\infty}(\lambda)$
	$\displaystyle=$	$\displaystyle C_{3}(\tilde{\lambda_{3}})c_{I}(\tilde{\lambda_{3}})(I+O(t^{-\delta}))$
		$\displaystyle\times\exp\biggl{(}-\lim_{\begin{subarray}{c}\lambda\to\infty\\[1.42271pt] \lambda\in\mathbf{c}_{\infty_{1}}^{3}\end{subarray}}\biggl{(}\int^{\lambda}_{\lambda_{3}}\Lambda_{3}(\tau)d\tau-(\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3}\biggr{)}\biggr{)},$

where $C_{3}(\tilde{\lambda}_{3})=\exp(\int^{\tilde{\lambda}_{3}}_{\lambda_{3}}\Lambda_{3}(\tau)d\tau),$ $c_{I}(\tilde{\lambda}_{3})=\exp(-\int^{\infty}_{\tilde{\lambda}_{3}}\Lambda_{I}(\tau)d\tau).$

(2) For a canonical solution $\Phi_{3}^{+}(\lambda)$ along $\mathbf{c}^{3}_{\infty_{1}}\cap\mathcal{A}_{3}$ (cf. Remark 4.2), set $\Psi^{(\infty)}_{3}(\lambda)=\Phi_{3}^{+}(\lambda)\Gamma^{+}_{3}$ . Then

\Gamma^{+}_{3}=\Phi_{3}^{+}(\lambda)^{-1}\Psi^{(\infty)}_{3}(\lambda)=(\tilde{\zeta}_{3})^{1/4}(I+O(t^{-\delta}))C_{3}(\tilde{\lambda}_{3})^{-1}\begin{pmatrix}1&0\\ 0&-\frac{c_{1}-ic_{2}}{c_{3}}\end{pmatrix},

where $c_{k}=b_{k}(\lambda_{3})$ and $\tilde{\zeta}_{3}\asymp\tilde{\lambda}_{3}-\lambda_{3}$ is suitably chosen.

(3) For a canonical solution $\Phi_{3}^{-}(\lambda)$ along $\mathbf{c}^{5}_{3}\cap\mathcal{A}_{3}$ , set $\Phi_{3}^{+}(\lambda)=\Phi_{3}^{-}(\lambda)\Gamma_{(3)}$ . Then

\Gamma_{(3)}=\Phi_{3}^{-}(\lambda)^{-1}\Phi_{3}^{+}(\lambda)=G_{1}^{-1}=\begin{pmatrix}1&i\\ 0&1\end{pmatrix}.

(4) For a WKB solution $\Psi_{3}^{(5)}(\lambda)$ along $\mathbf{c}_{3}^{5}$ with a base point $\lambda_{3}^{\prime}$ such that $\lambda_{3}^{\prime}-\lambda_{3}\asymp t^{-1}$ , set $\Phi_{3}^{-}(\lambda)=\Psi_{3}^{(5)}(\lambda)\Gamma_{3}^{-}$ . Then

\Gamma_{3}^{-}=\Psi_{3}^{(5)}(\lambda)^{-1}\Phi_{3}^{-}(\lambda)=(\zeta_{3}^{\prime})^{-1/4}(I+O(t^{-\delta}))C^{\prime}_{3}(\lambda_{3}^{\prime})\begin{pmatrix}1&0\\ 0&-\frac{c_{3}}{c_{1}-ic_{2}}\end{pmatrix},

where $C_{3}^{\prime}(\lambda_{3}^{\prime})=\exp(\int^{\lambda_{3}^{\prime}}_{\lambda_{3}}\Lambda_{3}(\tau)d\tau)$ and $\zeta_{3}^{\prime}\asymp\lambda_{3}^{\prime}-\lambda_{3}$ is suitably chosen.

(5) For a WKB solution $\Psi_{5}^{(3)}(\lambda)$ along $\mathbf{c}_{3}^{5}$ with a base point $\lambda_{5}^{\prime}$ such that $\lambda_{5}^{\prime}-\lambda_{5}\asymp t^{-1}$ , set $\Psi_{3}^{(5)}(\lambda)=\Psi_{5}^{(3)}(\lambda)\Gamma_{3,5}$ . Then

	$\displaystyle\Gamma_{3,5}=$	$\displaystyle\Psi_{5}^{(3)}(\lambda)^{-1}\Psi_{3}^{(5)}(\lambda)$
	$\displaystyle=$	$\displaystyle C_{3}^{\prime}(\lambda^{\prime}_{3})^{-1}C_{3}^{\prime}(\lambda^{\prime}_{5})c_{I}(\lambda_{3}^{\prime},\lambda_{5}^{\prime})(I+O(t^{-\delta}))\exp\biggl{(}\int_{\lambda_{3}}^{\lambda_{5}}\Lambda_{3}(\tau)d\tau\biggr{)},$

where $C_{3}(\lambda_{5}^{\prime})=\exp(\int^{\lambda_{5}^{\prime}}_{\lambda_{5}}\Lambda_{3}(\tau)d\tau)$ and $c_{I}(\lambda^{\prime}_{3},\lambda^{\prime}_{5})=\exp(\int^{\lambda^{\prime}_{5}}_{\lambda^{\prime}_{3}}\Lambda_{I}(\tau)d\tau).$

(6) For a canonical solution $\Phi_{5}^{+}(\lambda)$ along $\mathbf{c}_{3}^{5}\cap\mathcal{A}_{5}$ , set $\Psi_{5}^{(3)}(\lambda)=\Phi_{5}^{+}(\lambda)\Gamma_{5}^{+}.$ Then

\Gamma_{5}^{+}=\Phi_{5}^{+}(\lambda)^{-1}\Psi_{5}^{(3)}(\lambda)=(\zeta_{5}^{\prime})^{1/4}(I+O(t^{-\delta}))C_{3}(\lambda^{\prime}_{5})^{-1}\begin{pmatrix}1&0\\ 0&-\frac{d_{1}-id_{2}}{d_{3}}\end{pmatrix},

where $d_{k}=b_{k}(\lambda_{5})$ and $\zeta^{\prime}_{5}\asymp\lambda^{\prime}_{5}-\lambda_{5}$ is suitably chosen.

(7) For a canonical solution $\Phi_{5}^{-}(\lambda)$ along $\mathbf{c}_{5}^{\infty_{3}}\cap\mathcal{A}_{5}$ , set $\Phi_{5}^{+}(\lambda)=\Phi_{5}^{-}(\lambda)\Gamma_{(5)}$ . Then

\Gamma_{(5)}=\Phi_{5}^{-}(\lambda)^{-1}\Phi_{5}^{+}(\lambda)=(G_{1}G_{2})^{-1}=\begin{pmatrix}1&i\\ i&0\end{pmatrix}.

(8) For a WKB solution $\Psi_{5}^{(\infty)}(\lambda)$ along $\mathbf{c}_{5}^{\infty_{3}}$ with a base point $\tilde{\lambda}_{5}$ such that $\tilde{\lambda}_{5}-\lambda_{5}\asymp t^{-1},$ set $\Phi_{5}^{-}(\lambda)=\Psi_{5}^{(\infty)}(\lambda)\Gamma_{5}^{-}$ . Then

\Gamma_{5}^{-}=\Psi_{5}^{(\infty)}(\lambda)^{-1}\Phi_{5}^{-}(\lambda)=(\tilde{\zeta}_{5})^{-1/4}(I+O(t^{-\delta}))C_{3}(\tilde{\lambda}_{5})\begin{pmatrix}1&0\\ 0&-\frac{d_{3}}{d_{1}-id_{2}}\end{pmatrix},

where $C_{3}(\tilde{\lambda}_{5})=\exp(\int^{\tilde{\lambda}_{5}}_{\lambda_{5}}\Lambda_{3}(\tau)d\tau)$ and $\tilde{\zeta}_{5}\asymp\tilde{\lambda}_{5}-\lambda_{5}$ is suitably chosen.

(9) For $\Phi_{3}^{\infty}(\lambda)$ solving (3.1) in the sector $|\arg\lambda-\frac{5\pi}{8}|<\frac{\pi}{4}$ , set $\Psi_{5}^{(\infty)}(\lambda)=\Phi_{3}^{\infty}(\lambda)\Gamma_{5,\infty}.$ Then

	$\displaystyle\Gamma_{5,\infty}=$	$\displaystyle\Phi_{3}^{\infty}(\lambda)^{-1}\Psi_{5}^{(\infty)}(\lambda)$
	$\displaystyle=$	$\displaystyle C_{3}(\tilde{\lambda}_{5})^{-1}c_{I}(\tilde{\lambda}_{5})^{-1}(I+O(t^{-\delta}))$
		$\displaystyle\times\exp\biggl{(}\lim_{\begin{subarray}{c}\lambda\to\infty\\[1.42271pt] \lambda\in\mathbf{c}_{5}^{\infty_{3}}\end{subarray}}\biggl{(}\int_{\lambda_{5}}^{\lambda}\Lambda_{3}(\tau)d\tau-(\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3}\biggr{)}\biggr{)},$

where $c_{I}(\tilde{\lambda}_{5})=\exp(-\int^{\infty}_{\tilde{\lambda}_{5}}\Lambda_{I}(\tau)d\tau).$

Product of the matrices above along $\mathfrak{c}^{5\pi/8}_{\pi/8}$ yields the Stokes matrix

	$\displaystyle(S_{1}^{}S_{2}^{})^{-1}=$	$\displaystyle\begin{pmatrix}1&-s_{1}^{}\\ -s_{2}^{}&1+s_{1}^{}s_{2}^{}\end{pmatrix}$
	$\displaystyle=$	$\displaystyle\Phi_{3}^{\infty}(\lambda)^{-1}\Phi_{1}^{\infty}(\lambda)$
	$\displaystyle=$	$\displaystyle\Gamma_{5,\infty}\Gamma_{5}^{-}\Gamma_{(5)}\Gamma_{5}^{+}\Gamma_{3,5}\Gamma_{3}^{-}\Gamma_{(3)}\Gamma_{3}^{+}\Gamma_{\infty,3}$
	$\displaystyle=$	$\displaystyle\epsilon_{1}e^{J_{5}^{\infty 3}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-d_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&i\\ i&0\end{pmatrix}\begin{pmatrix}1&0\\ 0&-d_{0}\end{pmatrix}$
		$\displaystyle\times e^{J_{3,5}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-c_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&i\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ 0&-c_{0}\end{pmatrix}e^{-J_{3}^{\infty 1}\sigma_{3}}$
	$\displaystyle=$	$\displaystyle\epsilon_{1}\begin{pmatrix}e^{J_{3,5}+J_{5}^{\infty 3}-J_{3}^{\infty 1}}&-i(c_{0}e^{J_{3,5}}+d_{0}e^{-J_{3,5}})e^{J_{5}^{\infty 3}+J_{3}^{\infty 1}}\\ -id_{0}^{-1}e^{J_{3,5}-J_{5}^{\infty 3}-J_{3}^{\infty 1}}&-c_{0}d_{0}^{-1}e^{J_{3,5}-J_{5}^{\infty 3}+J_{3}^{\infty 1}}\end{pmatrix}$

(up to the multiplier $1+O(t^{-\delta})$ to each entry), in which $\epsilon_{1}^{2}=1,$

	$\displaystyle c_{0}=(c_{1}-ic_{2})/c_{3},\quad d_{0}=(d_{1}-id_{2})/d_{3},\quad c_{k}=b_{k}(\lambda_{3}),\,\,\,d_{k}=b_{k}(\lambda_{5}),$
(5.2)		$\displaystyle J_{3,5}\sigma_{3}=\int^{\lambda_{5}}_{\lambda_{3}}\Lambda_{3}(\tau)d\tau,\quad\text{$\mathbf{c}_{3}^{5}=(\lambda_{3},\lambda_{5})^{\sim}$: on the right shore of the cut $[\lambda_{3},\lambda_{5}]$},$
	$\displaystyle J_{5}^{\infty 3}\sigma_{3}=\lim_{\begin{subarray}{c}\lambda\to\infty\\[1.42271pt] \lambda\in\mathbf{c}_{5}^{\infty_{3}}\end{subarray}}\biggl{(}\int_{\lambda_{5}}^{\lambda}\Lambda_{3}(\tau)d\tau-(\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3}\biggr{)},$
	$\displaystyle J_{3}^{\infty 1}\sigma_{3}=\lim_{\begin{subarray}{c}\lambda\to\infty\\[1.42271pt] \lambda\in\mathbf{c}^{3}_{\infty_{1}}\end{subarray}}\biggl{(}\int_{\lambda_{3}}^{\lambda}\Lambda_{3}(\tau)d\tau-(\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3}\biggr{)}.$

Then the diagonal entries of $(S_{1}^{*}S_{2}^{*})^{-1}$ give $1=\epsilon_{1}e^{J_{3,5}+J_{5}^{\infty 3}-J_{3}^{\infty 1}}(1+O(t^{-\delta})),$ $1+s_{1}^{*}s_{2}^{*}=-\epsilon_{1}c_{0}d_{0}^{-1}e^{J_{3,5}-J_{5}^{\infty 3}+J_{3}^{\infty 1}}(1+O(t^{-\delta})),$ which imply

(5.3)

1+s_{1}^{*}s_{2}^{*}=-c_{0}d_{0}^{-1}e^{2J_{3,5}}(1+O(t^{-\delta})).

Similarly the analytic continuation of $\Phi_{1}^{\infty}(\lambda)$ along $\mathfrak{c}_{\pi/8}^{7\pi/8}$ yields

	$\displaystyle(S^{}_{1}S^{}_{2}S^{*}_{3})^{-1}=$	$\displaystyle\begin{pmatrix}1+s_{2}^{}s_{3}^{}&-s_{1}^{}-s_{3}^{}-s_{1}^{}s_{2}^{}s_{3}^{}\\ -s_{2}^{}&1+s_{1}^{}s_{2}^{}\end{pmatrix}$
	$\displaystyle=$	$\displaystyle\Phi_{4}^{\infty}(\lambda)^{-1}\Phi_{1}^{\infty}(\lambda)$
	$\displaystyle=$	$\displaystyle\epsilon_{2}e^{J_{1}^{\infty 4}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-e_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&i\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ 0&-e_{0}\end{pmatrix}$
		$\displaystyle\times e^{J_{3,1}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-c_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&0\\ -i&1\end{pmatrix}\begin{pmatrix}1&0\\ 0&-c_{0}\end{pmatrix}e^{-J_{3}^{\infty 1}\sigma_{3}}$
	$\displaystyle=$	$\displaystyle\epsilon_{2}\begin{pmatrix}(e^{J_{3,1}}+c_{0}^{-1}e_{0}e^{-J_{3,1}})e^{J_{1}^{\infty 4}-J_{3}^{\infty 1}}&-ie_{0}e^{-J_{3,1}+J_{1}^{\infty 4}+J_{3}^{\infty 1}}\\ ic_{0}^{-1}e^{-J_{3,1}-J_{1}^{\infty 4}-J_{3}^{\infty 1}}&e^{-J_{3,1}-J_{1}^{\infty 4}+J_{3}^{\infty 1}}\end{pmatrix},$

in which $\epsilon_{2}^{2}=1,$

	$\displaystyle e_{0}=(e_{1}-ie_{2})/e_{3},\quad e_{k}=b_{k}(\lambda_{1}),$
(5.4)		$\displaystyle J_{3,1}\sigma_{3}=\int^{\lambda_{1}}_{\lambda_{3}}\Lambda_{3}(\tau)d\tau,$
	$\displaystyle J_{1}^{\infty 4}\sigma_{3}=\lim_{\begin{subarray}{c}\lambda\to\infty\\[1.42271pt] \lambda\in\mathbf{c}_{1}^{\infty_{4}}\end{subarray}}\biggl{(}\int_{\lambda_{1}}^{\lambda}\Lambda_{3}(\tau)d\tau-(\tfrac{1}{4}t\lambda^{4}+e^{i\phi}t\lambda^{2}+(\alpha-\beta)\ln\lambda)\sigma_{3}\biggr{)}.$

Observing the off-diagonal entries, we have $(1+s_{1}^{*}s_{2}^{*})(1+s_{2}^{*}s_{3}^{*})-1=c_{0}^{-1}e_{0}e^{-2J_{3,1}}(1+O(t^{-\delta}))$ . Combining this with (5.3) and using Corollary 3.5, we have the following.

Proposition 5.1.

Suppose that $-\pi/4<\phi<0.$ Then

	$\displaystyle 1+s_{1}s_{2}=-c_{0}d_{0}^{-1}e^{2J_{3,5}}(1+O(t^{-\delta})),$
	$\displaystyle(1+s_{1}s_{2})(1+s_{2}s_{3})-1=c_{0}^{-1}e_{0}e^{-2J_{3,1}}(1+O(t^{-\delta})),$

where $J_{3,5}$ and $J_{3,1}$ are integrals given by (5.2) and (5.4).

In the case where $0<\phi<\pi/4$ , we calculate the analytic continuations of

\Phi_{2}^{\infty}(\lambda)\,\,\,\text{along}\,\,\hat{\mathfrak{c}}_{3\pi/8}^{7\pi/8}=\mathbf{c}_{\infty_{2}}^{5}\cup{\mathbf{c}}_{5}^{3-}\cup\mathbf{c}_{3}^{\infty_{4}},\quad\Phi_{1}^{\infty}(\lambda)\,\,\,\text{along}\,\,\hat{\mathfrak{c}}_{\pi/8}^{7\pi/8}=\mathbf{c}_{\infty_{1}}^{1}\cup\mathbf{c}_{1}^{3}\cup\mathbf{c}_{3}^{\infty_{4}}

with $\mathbf{c}_{\infty_{1}}^{1}=(e^{\pi i/8}\infty,\lambda_{1})^{\sim}$ , $\mathbf{c}_{1}^{3}=(\lambda_{1},\lambda_{3})^{\sim}$ , $\mathbf{c}_{3}^{\infty_{4}}=(\lambda_{3},e^{7\pi i/8}\infty)^{\sim}$ , $\mathbf{c}^{5}_{\infty_{2}}=(e^{3\pi i/8}\infty,\lambda_{5})^{\sim}$ , ${\mathbf{c}}_{5}^{3-}=(\lambda_{5},\lambda_{3})^{\sim}$ , where ${\mathbf{c}}_{5}^{3-}$ lies on the left shore of the cut $[\lambda_{3},\lambda_{5}]$ (cf. Figure 5.2).

Figure 5.2. Fragments of the Stokes graph for

0<\phi<\pi/4

This procedure results in

	$\displaystyle(S_{2}^{}S_{3}^{})^{-1}=$	$\displaystyle\begin{pmatrix}1+s_{2}^{}s_{3}^{}&-s_{3}^{}\\ -s_{2}^{}&1\end{pmatrix}$
	$\displaystyle=$	$\displaystyle\Phi_{4}^{\infty}(\lambda)^{-1}\Phi_{2}^{\infty}(\lambda)$
	$\displaystyle=$	$\displaystyle\tilde{\epsilon}_{1}e^{J_{3}^{\infty 4}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-c_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&i\\ 0&{1}\end{pmatrix}\begin{pmatrix}1&0\\ 0&-c_{0}\end{pmatrix}$
		$\displaystyle\times e^{J_{5,3}^{-}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-d_{0}^{-1}\end{pmatrix}\begin{pmatrix}0&i\\ i&{1}\end{pmatrix}\begin{pmatrix}1&0\\ 0&-d_{0}\end{pmatrix}e^{-J_{5}^{\infty 2}\sigma_{3}}$
	$\displaystyle=$	$\displaystyle\tilde{\epsilon}_{1}\begin{pmatrix}-c_{0}d_{0}^{-1}e^{-J^{-}_{5,3}+J_{3}^{\infty 4}-J_{5}^{\infty 2}}&-i(d_{0}e^{J^{-}_{5,3}}+c_{0}e^{-J^{-}_{5,3}})e^{J_{3}^{\infty 4}+J_{5}^{\infty 2}}\\ -id_{0}^{-1}e^{-J^{-}_{5,3}-J_{3}^{\infty 4}-J_{5}^{\infty 2}}&e^{-J^{-}_{5,3}-J_{3}^{\infty 4}+J_{5}^{\infty 2}}\end{pmatrix}$

with $\tilde{\epsilon}_{1}^{2}=1,$ $J_{3}^{\infty 4}\sigma_{3}=J_{1}^{\infty 4}\sigma_{3}|_{\mathbf{c}_{1}^{\infty_{4}}\mapsto\mathbf{c}_{3}^{\infty_{4}},\,\,\lambda_{1}\mapsto\lambda_{3}}$ , $J_{5}^{\infty 2}\sigma_{3}=J_{3}^{\infty 1}\sigma_{3}|_{\mathbf{c}^{3}_{\infty_{1}}\mapsto\mathbf{c}^{5}_{\infty_{2}},\,\,\lambda_{3}\mapsto\lambda_{5}}$ and

(5.5)

J^{-}_{5,3}\sigma_{3}=\int^{\lambda_{3},-}_{\lambda_{5}}\Lambda_{3}(\tau)d\tau,\quad\text{$\mathbf{c}_{5}^{3-}=(\lambda_{5},\lambda_{3})^{\sim}$: on the left shore of the cut $[\lambda_{3},\lambda_{5}]$},

and

	$\displaystyle(S_{1}^{}S_{2}^{}S_{3}^{*})^{-1}=$	$\displaystyle\Phi_{4}^{\infty}(\lambda)^{-1}\Phi_{1}^{\infty}(\lambda)$
	$\displaystyle=$	$\displaystyle\tilde{\epsilon}_{2}e^{J_{3}^{\infty 4}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-c_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&0\\ -i&{1}\end{pmatrix}\begin{pmatrix}1&0\\ 0&-c_{0}\end{pmatrix}$
		$\displaystyle\times e^{J_{1,3}\sigma_{3}}\begin{pmatrix}1&0\\ 0&-e_{0}^{-1}\end{pmatrix}\begin{pmatrix}1&i\\ 0&{1}\end{pmatrix}\begin{pmatrix}1&0\\ 0&-e_{0}\end{pmatrix}e^{-J_{1}^{\infty 1}\sigma_{3}}$
	$\displaystyle=$	$\displaystyle\tilde{\epsilon}_{2}\begin{pmatrix}e^{J_{1,3}+J_{3}^{\infty 4}-J_{1}^{\infty 1}}&-ie_{0}e^{J_{1,3}+J_{3}^{\infty 4}+J_{1}^{\infty 1}}\\ ic_{0}^{-1}e^{J_{1,3}-J_{3}^{\infty 4}-J_{1}^{\infty 1}}&(c_{0}^{-1}e_{0}e^{J_{1,3}}+e^{-J_{1,3}})e^{-J_{3}^{\infty 4}+J_{1}^{\infty 1}}\end{pmatrix}$

with $\tilde{\epsilon}_{2}^{2}=1,$ $J_{1}^{\infty 1}\sigma_{3}=J_{3}^{\infty 1}\sigma_{3}|_{\mathbf{c}^{3}_{\infty_{1}}\mapsto\mathbf{c}^{1}_{\infty_{1}},\,\,\lambda_{3}\mapsto\lambda_{1}}$ and $J_{1,3}\sigma_{3}=-J_{3,1}\sigma_{3}.$ Thus we have the following.

Proposition 5.2.

Suppose that $0<\phi<\pi/4.$ Then

	$\displaystyle 1+s_{2}s_{3}=-c_{0}d_{0}^{-1}e^{-2J^{-}_{5,3}}(1+O(t^{-\delta})),$
	$\displaystyle(1+s_{1}s_{2})(1+s_{2}s_{3})-1=c_{0}^{-1}e_{0}e^{2J_{1,3}}(1+O(t^{-\delta})),$

where $J^{-}_{5,3}$ is given by (5.5) and $J_{1,3}=-J_{3,1}$ .

6. Asymptotics of monodromy data

Recall that $\lambda\mu(\infty,\lambda)=w(A_{\phi},\lambda^{2})$ is considered on the Riemann surface $\mathcal{R}_{\infty}$ and that $w(A_{\phi},z)$ defines the elliptic curve $\Pi_{A_{\phi},\phi}$ with the primitive cycles $\mathbf{a}$ and $\mathbf{b}$ described as in Figure 2.1. Note that, by $z=\lambda^{2}$ , $\mathcal{R}_{\infty}$ is mapped to $\Pi_{A_{\phi},\phi}$ . Let $\hat{\mathbf{a}},$ $\hat{\mathbf{b}}\subset\mathcal{R}_{\infty}$ denote the inverse images of $\mathbf{a},$ $\mathbf{b}$ , respectively, such that $\hat{\mathbf{a}}\subset\mathcal{R}^{+}_{\infty}$ surrounds the cut $[\lambda_{5},\lambda_{3}]$ and that $\hat{\mathbf{b}}$ links with $\hat{\mathbf{a}}$ . For $\lambda_{j}(t)\in\mathcal{R}_{t}$ and $\lambda_{j}=\lambda_{j}(\infty)\in\mathcal{R}_{\infty}$ , we have $\lambda_{j}(t)=\lambda_{j}(\infty)+O(t^{-1})$ if $0<|\phi|<\pi/4,$ and hence the cycles $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ may be regarded to be those on $\mathcal{R}_{t}$ as well for sufficiently large $t$ . Furthermore $\mathbf{a},$ $\mathbf{b}\subset\Pi_{A_{\phi},\phi}$ may be regarded to be the primitive cycles on $\Pi_{a_{\phi}(t),\phi}$ by (5.1).

6.1. Integrals

We would like to calculate the asymptotics of $\int_{\hat{\mathbf{a}},\,\,\hat{\mathbf{b}}}\Lambda_{3}(\lambda)\lambda$ . By (4.1)

	$\displaystyle\mu(t,\lambda)$	$\displaystyle=\lambda^{-1}\left(\lambda^{8}+4e^{i\phi}\lambda^{6}+4e^{2i\phi}\lambda^{4}+4e^{3i\phi}a_{\phi}\lambda^{2}+2(\alpha-\beta)t^{-1}\lambda^{4}+\alpha^{2}t^{-2}\right)^{1/2}$
		$\displaystyle=\lambda^{-1}{\mathbf{w}(\lambda)}\Bigl{(}1+2(\alpha-\beta)t^{-1}\frac{\lambda^{4}}{\mathbf{w}(\lambda)^{2}}+O(t^{-2}\mathbf{w}(\lambda)^{-2})\Bigr{)}^{1/2}$
		$\displaystyle=\frac{\mathbf{w}(\lambda)}{\lambda}+(\alpha-\beta)t^{-1}\frac{\lambda^{3}}{\mathbf{w}(\lambda)}+O(t^{-2})$

along $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ , where

\mathbf{w}(\lambda)=w(a_{\phi},\lambda^{2}),\quad w(a_{\phi},z)=\sqrt{z^{4}+4e^{i\phi}z^{3}+4e^{2i\phi}z^{2}+4e^{3i\phi}a_{\phi}z}.

Substitution $\lambda^{2}=z$ yields

(6.1)		$\displaystyle\int_{\hat{\mathbf{a}},\,\,\hat{\mathbf{b}}}\mu(t,\lambda)d\lambda=$	$\displaystyle\frac{1}{2}\int_{\mathbf{a},\,\,\mathbf{b}}\frac{w(a_{\phi},z)}{z}dz+\frac{1}{2}(\alpha-\beta)t^{-1}\int_{\mathbf{a},\,\,\mathbf{b}}\frac{z}{w(a_{\phi},z)}dz+O(t^{-2})$
	$\displaystyle=$	$\displaystyle e^{4i\phi}a_{\phi}\int_{{\mathbf{a}},\,\,{\mathbf{b}}}\frac{dz}{zw(a_{\phi},z)}+\frac{1}{2}(\alpha-\beta)t^{-1}\int_{\mathbf{a},\,\,\mathbf{b}}\frac{z}{w(a_{\phi},z)}dz$
		$\displaystyle+\frac{3e^{3i\phi}a_{\phi}}{2}\omega_{\mathbf{a},\,\,\mathbf{b}}+O(t^{-2}),\quad\,\,\omega_{\mathbf{a},\,\,\mathbf{b}}=\int_{\mathbf{a},\,\,\mathbf{b}}\frac{dz}{w(a_{\phi},z)},$

in which the second equality is obtained by using

\int w(a_{\phi},z)\frac{dz}{z}=\frac{w(a_{\phi},z)}{2}+e^{i\phi}\frac{w(a_{\phi},z)}{z}+3e^{3i\phi}a_{\phi}\int\frac{dz}{w(a_{\phi},z)}+2e^{4i\phi}a_{\phi}\int\frac{dz}{zw(a_{\phi},z)}.

Let us calculate

\mathrm{diag}\,T^{-1}T_{\lambda}|_{\sigma_{3}}d\lambda=\frac{1}{4}\Bigl{(}1-\frac{b_{3}}{\mu}\Bigl{)}\frac{d}{d\lambda}\ln\frac{b_{1}+ib_{2}}{b_{1}-ib_{2}}d\lambda.

Recalling (3.1) and (3.2), and setting $\lambda^{2}=z$ , we have

(6.2)			$\displaystyle b_{1}-ib_{2}=2ie^{i\phi/2}(z-z_{+}),\quad b_{1}+ib_{2}=2ie^{-i\phi/2}(z-z_{-}),$
		$\displaystyle z_{+}:=-\frac{e^{-i\phi}}{2}\frac{\psi_{t}}{\psi}-\frac{\psi}{2}-e^{i\phi}+O(t^{-1}),\quad z_{-}:=\frac{e^{-i\phi}}{2}\frac{\psi_{t}}{\psi}-\frac{\psi}{2}-e^{i\phi}+O(t^{-1}),$

with $(b_{1}-ib_{2})(z_{+})=0,$ $(b_{1}+ib_{2})(z_{-})=0.$ Furthermore

\frac{b_{3}}{\mu(t,\lambda)}=z(z+2(e^{i\phi}+\psi))\Bigl{(}\frac{1}{w(a_{\phi},z)}+O(t^{-1})\Bigr{)},

which satisfies $(b_{3}/\mu)(z_{\pm})=1$ on the upper sheet $\Pi_{+}$ of $\Pi_{a_{\phi},\phi}$ . Then it follows that

(6.3)		$\displaystyle\mathrm{diag}\,T^{-1}T_{\lambda}\|_{\sigma_{3}}d\lambda$	$\displaystyle=\frac{1}{4}\Bigl{(}\frac{1}{z-z_{-}}-\frac{1}{z-z_{+}}\Bigr{)}dz$
		$\displaystyle+\frac{1}{4}\Bigl{(}z_{+}-z_{-}+\frac{w(a_{\phi},z_{+})}{z-z_{+}}-\frac{w(a_{\phi},z_{-})}{z-z_{-}}\Bigr{)}\frac{dz}{w(a_{\phi},z)}+O(t^{-1})dz,$

and that

(6.4)

\begin{split}&\frac{1}{2}\int^{z_{3}}_{z_{5}}\Bigl{(}\frac{1}{z-z_{+}}-\frac{1}{z-z_{-}}\Bigr{)}dz=\frac{1}{2}\ln\frac{(\lambda_{3}^{2}-z_{+})(\lambda_{5}^{2}-z_{-})}{(\lambda_{3}^{2}-z_{-})(\lambda_{5}^{2}-z_{+})}=\ln(c_{0}d_{0}^{-1}),\\ &\frac{1}{2}\int^{z_{1}}_{z_{3}}\Bigl{(}\frac{1}{z-z_{+}}-\frac{1}{z-z_{-}}\Bigr{)}dz=\frac{1}{2}\ln\frac{(\lambda_{1}^{2}-z_{+})(\lambda_{3}^{2}-z_{-})}{(\lambda_{1}^{2}-z_{-})(\lambda_{3}^{2}-z_{+})}=\ln(c_{0}^{-1}e_{0}).\end{split}

Suppose that $-\pi/4<\phi<0$ , and recall Proposition 5.1. Note that

J_{3,5}=\int_{\lambda_{3}}^{\lambda_{5}}\Bigl{(}t\mu(t,\lambda)-\mathrm{diag}T^{-1}T_{\lambda}|_{\sigma_{3}}\Bigr{)}d\lambda,

along $\mathbf{c}_{3}^{5}=(\lambda_{3},\lambda_{5})^{\sim}$ in (5.2), and that $\mathbf{c}_{3}^{5}$ is $\frac{1}{2}(-\hat{\mathbf{a}})$ , which is the image of $\frac{1}{2}(-\mathbf{a})$ of Figure 2.1 under the map $\lambda^{2}=z.$ The integral $J_{3,1}$ of (5.4) is along $\mathbf{c}_{3}^{1}$ , which is the image of $\tfrac{1}{2}\mathbf{b}$ . Then by (6.1), (6.3) and (6.4), we have the following proposition, in which

(6.5)			$\displaystyle W(z)=\Bigl{(}z_{+}-z_{-}+\frac{w(a_{\phi},z_{+})}{z-z_{+}}-\frac{w(a_{\phi},z_{-})}{z-z_{-}}\Bigr{)}\frac{1}{w(a_{\phi},z)},$
(6.6)			$\displaystyle\omega_{\mathbf{a},\,\,\mathbf{b}}=\int_{\mathbf{a},\,\,\mathbf{b}}\frac{dz}{w(a_{\phi},z)},\qquad C_{\alpha,\beta}(\mathbf{a},\,\,\mathbf{b})=\frac{1}{2}(\alpha-\beta)\int_{\mathbf{a},\,\,\mathbf{b}}\frac{zdz}{w(a_{\phi},z)}.$

Proposition 6.1.

Suppose that $-\pi/4<\phi<0$ . Then

	$\displaystyle\ln(1+s_{1}s_{2})=$	$\displaystyle-\frac{t}{2}\int_{\mathbf{a}}\frac{w(a_{\phi},z)}{z}dz+\frac{1}{4}\int_{\mathbf{a}}W(z)dz-C_{\alpha,\beta}({\mathbf{a}})+\pi i+O(t^{-\delta}),$
	$\displaystyle\ln((1+s_{1}s_{2})(1$	$\displaystyle+s_{2}s_{3})-1)$
	$\displaystyle=$	$\displaystyle-\frac{t}{2}\int_{\mathbf{b}}\frac{w(a_{\phi},z)}{z}dz+\frac{1}{4}\int_{\mathbf{b}}W(z)dz-C_{\alpha,\beta}({\mathbf{b}})+O(t^{-\delta}).$

From Proposition 5.2 we have the following.

Proposition 6.2.

Suppose that $0<\phi<\pi/4.$ Then

	$\displaystyle\ln(1+s_{2}s_{3})=$	$\displaystyle\frac{t}{2}\int_{\mathbf{a}}\frac{w(a_{\phi},z)}{z}dz-\frac{1}{4}\int_{\mathbf{a}}W(z)dz+C_{\alpha,\beta}(\mathbf{a})+\pi i+O(t^{-\delta}),$
	$\displaystyle\ln((1+s_{1}s_{2})($	$\displaystyle 1+s_{2}s_{3})-1)$
	$\displaystyle=$	$\displaystyle-\frac{t}{2}\int_{\mathbf{b}}\frac{w(a_{\phi},z)}{z}dz+\frac{1}{4}\int_{\mathbf{b}}W(z)dz-C_{\alpha,\beta}(\mathbf{b})+O(t^{-\delta}).$

Remark 6.1.

In the propositions above,

\frac{1}{2}\int_{\mathbf{a},\,\,\mathbf{b}}\frac{w(a_{\phi},z)}{z}dz=e^{4i\phi}a_{\phi}\int_{\mathbf{a},\,\,\mathbf{b}}\frac{dz}{zw(a_{\phi},z)}+\frac{3}{2}e^{3i\phi}a_{\phi}\omega_{\mathbf{a},\,\,\mathbf{b}}.

6.2. Theta-function

Further calculation needs the theta-function

\vartheta(z,\tau)=\sum_{n=-\infty}^{\infty}e^{\pi i\tau n^{2}+2\pi izn},\quad\tau=\frac{\omega_{\mathbf{b}}}{\omega_{\mathbf{a}}},\quad\mathrm{Im\,}\tau>0,\quad\nu=\frac{1+\tau}{2}.

For $z,$ $\tilde{z}\in\Pi_{a_{\phi},\phi}=\Pi_{+}\cup\Pi_{-}$ , set

F(\tilde{z},z)=\frac{1}{\omega_{\mathbf{a}}}\int^{z}_{\tilde{z}}\frac{dz}{w(a_{\phi},z)}.

For $z_{0}\in\Pi_{a_{\phi},\phi}=\Pi_{+}\cup\Pi_{-}$ write $z_{0}^{+}=(z_{0},w(a_{\phi},z_{0}^{+})),$ $z_{0}^{-}=(z_{0},-w(a_{\phi},z_{0}^{+})$ .

Proposition 6.3.

For any $z_{0}\in\Pi_{a_{\phi},\phi},$ $w(z)=w(a_{\phi},z)$ fulfils

	$\displaystyle\frac{dz}{(z-z_{0})w(z)}$	$\displaystyle=\frac{1}{w(z_{0}^{+})}d\ln\frac{\vartheta(F(z_{0}^{+},z)+\nu,\tau)}{\vartheta(F(z_{0}^{-},z)+\nu,\tau)}-g_{0}(z_{0})\frac{dz}{w(z)},$
	$\displaystyle g_{0}(z_{0})$	$\displaystyle=\frac{w^{\prime}(z^{+}_{0})}{2w(z^{+}_{0})}-\frac{1}{\omega_{\mathbf{a}}w(z_{0}^{+})}\Bigl{(}\pi i+\frac{\vartheta^{\prime}}{\vartheta}(F(z^{-}_{0},z^{+}_{0})+\nu,\tau)\Bigr{)}.$

Proof..

Let $z=\varphi(u)$ be an elliptic function such that $\varphi_{u}^{2}=w(\varphi)^{2}$ and $z_{0}=\varphi(u_{0})$ . Then $(z-z_{0})^{-1}w(z)^{-1}dz=(\varphi(u)-\varphi(u_{0}))^{-1}du.$ Write $\varphi(u_{0}^{\pm})=z_{0}^{\pm},$ $\varphi_{u}(u^{\pm}_{0})=\pm w(z_{0}^{+}).$ Around each $u=u_{0}^{\pm}$ ,

(\varphi(u)-\varphi(u_{0}^{\pm}))^{-1}=\pm\frac{(u-u_{0}^{\pm})^{-1}}{w(z_{0}^{+})}-\frac{w^{\prime}(z_{0}^{+})}{2w(z_{0}^{+})}+O(u-u_{0}^{\pm}),

and hence we have

	$\displaystyle(\varphi(u)-\varphi(u_{0}^{\pm}))^{-1}=$	$\displaystyle\frac{1}{\omega_{\mathbf{a}}w(z_{0})}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}(\omega_{\mathbf{a}}^{-1}(u-u_{0}^{+})+\nu,\tau)-\frac{\vartheta^{\prime}}{\vartheta}(\omega_{\mathbf{a}}^{-1}(u-u_{0}^{-})+\nu,\tau)\Bigr{)}+C_{0}$
	$\displaystyle=$	$\displaystyle\frac{1}{\omega_{\mathbf{a}}w(z_{0})}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}(F(z_{0}^{+},z)+\nu,\tau)-\frac{\vartheta^{\prime}}{\vartheta}(F(z_{0}^{-},z))+\nu,\tau)\Bigr{)}+C_{0}$

for some constant $C_{0}$ . Passage to the limit $z\to z^{+}_{0}$ , i.e. $u\to u_{0}^{+}$ leads to

C_{0}=\frac{1}{\omega_{\mathbf{a}}w(z_{0}^{+})}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}(F(z_{0}^{-},z_{0}^{+})+\nu,\tau)-\frac{\vartheta^{\prime\prime}(\nu)}{2\vartheta^{\prime}(\nu)}\Bigr{)}-\frac{w^{\prime}(z^{+}_{0})}{2w(z^{+}_{0})},

which implies the required formula. ∎

Corollary 6.4.

For any $z_{0}\in\Pi_{a_{\phi},\phi}$ , and for $w(z)=w(a_{\phi},z)$ ,

(6.7)			$\displaystyle\int_{\mathbf{a}}\frac{dz}{(z-z_{0})w(z)}=-g_{0}(z_{0})\omega_{\mathbf{a}},$
(6.8)			$\displaystyle\Bigl{(}\int_{\mathbf{b}}-\tau\int_{\mathbf{a}}\Bigr{)}\frac{dz}{(z-z_{0})w(z)}=\frac{2\pi i}{w(z_{0}^{+})}F(z_{0}^{-},z_{0}^{+}),$
(6.9)			$\displaystyle\Bigl{(}\int_{\mathbf{b}}-\tau\int_{\mathbf{a}}\Bigr{)}\frac{dz}{zw(z)}=\frac{2\pi i}{a_{\phi}\omega_{\mathbf{a}}}e^{-3i\phi}.$

Corollary 6.5.

For $z_{+},$ $z_{-}$ and $W(z)$ given by (6.2) and (6.5), and for $w(z)=w(a_{\phi},z)$ , we have, on $\Pi_{a_{\phi},\phi}$ ,

	$\displaystyle\int_{\mathbf{a}}W(z)dz=$	$\displaystyle\Bigl{(}z_{+}-\frac{w^{\prime}(z_{+}^{+})}{2}\Bigr{)}\omega_{\mathbf{a}}+\frac{\vartheta^{\prime}}{\vartheta}(F(z_{+}^{-},z_{+}^{+})+\nu,\tau)$
		$\displaystyle\phantom{--------}-\Bigl{(}z_{-}-\frac{w^{\prime}(z_{-}^{+})}{2}\Bigr{)}\omega_{\mathbf{a}}-\frac{\vartheta^{\prime}}{\vartheta}(F(z_{-}^{-},z_{-}^{+})+\nu,\tau),$
	$\displaystyle\Bigl{(}\int_{\mathbf{b}}-\tau\int_{\mathbf{a}}\Bigr{)}$	$\displaystyle W(z)dz=2\pi i(F(z_{+}^{-},z_{+}^{+})-F(z^{-}_{-},z_{-}^{+})).$

Another expression of $\int_{\mathbf{a}}W(\lambda)d\lambda$ is derived by using more information on the poles of $\mathrm{P}(u;a_{\phi})$ (cf. Proposition 7.6).

Proposition 6.6.

Under the same condition as above

\int_{\mathbf{a}}W(z)dz=2\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}(\tfrac{1}{2}F(z_{+}^{-},z_{+}^{+})+\tfrac{1}{2}+\tfrac{\tau}{6},\tau)-\frac{\vartheta^{\prime}}{\vartheta}(\tfrac{1}{2}F(z_{-}^{-},z_{-}^{+})+\tfrac{1}{2}+\tfrac{\tau}{6},\tau)\Bigr{)}.

Proof..

Let $z_{0}\in\Pi_{a_{\phi},\phi}=\Pi_{+}\cup\Pi_{-}.$ Note that $z_{0}-\tfrac{1}{2}w^{\prime}(a_{\phi},z_{0})$ is holomorphic around $z_{0}^{+}=\infty^{+}\in\Pi_{+}$ and admits a pole at $z_{0}^{+}=\infty^{-}\in\Pi_{-}$ with the residue $2$ . Then, we have

(z_{0}-\tfrac{1}{2}w^{\prime}(a_{\phi},z^{+}_{0}))\omega_{\mathbf{a}}+(\vartheta^{\prime}/\vartheta)(F(z_{0}^{-},z_{0}^{+})+\nu,\tau)=2(\vartheta^{\prime}/\vartheta)(\tfrac{1}{2}F(z_{0}^{-},z_{0}^{+})+\tfrac{1}{2}+\tfrac{\tau}{6},\tau)+c_{0}

for some constant $c_{0}.$ Indeed, as shown later by Proposition 7.6, the elliptic function $\varphi_{0}(u)$ defined by $u=\int_{0}^{\varphi_{0}}w(a_{\phi},z)^{-1}dz$ has poles with the residue $\mp 1$ at $\pm\tfrac{1}{3}\omega_{\mathbf{b}}+\omega_{\mathbf{a}}\mathbb{Z}+\omega_{\mathbf{b}}\mathbb{Z}$ , which implies $\omega_{\mathbf{a}}^{-1}\int^{\infty^{+}}_{0}w(a_{\phi},z)^{-1}dz+\tfrac{1}{2}+\tfrac{\tau}{6}=\tfrac{1}{2}-\tfrac{\tau}{6}$ for $z_{0}^{+}=\infty^{+}\in\Pi_{+},$ and $\omega_{\mathbf{a}}^{-1}\int^{\infty^{-}}_{0}w(a_{\phi},z)^{-1}dz+\tfrac{1}{2}+\tfrac{\tau}{6}=\tfrac{1}{2}+\tfrac{\tau}{2}=\nu$ for $z_{0}^{-}=\infty^{-}\in\Pi_{-}$ ; and furthermore, at the branch points $z_{0}=0,z_{1},z_{3},z_{5}$ , the leading terms of $-\tfrac{1}{2}w^{\prime}(a_{\phi},z^{+}_{0})\omega_{\mathbf{a}}$ and $(\vartheta^{\prime}/\vartheta)(F(z_{0}^{-},z_{0}^{+})+\nu,\tau)$ are cancelled out (cf. Subsection 6.3). Putting $z_{0}=z_{\pm}$ and using Corollary 6.5, we obtain the proposition. ∎

6.3. Expression of $B_{\phi}(t)$

Recall that our calculations are carried out under the supposition (5.1) in the strip $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1}).$ Let

\Omega_{\mathbf{a},\,\,\mathbf{b}}:=\int_{\mathbf{a},\,\,\mathbf{b}}\frac{dz}{w(A_{\phi},z)},\qquad\mathcal{J}_{\mathbf{a},\,\,\mathbf{b}}:=\int_{\mathbf{a},\,\,\mathbf{b}}\frac{w(A_{\phi},z)}{z}dz

for $\mathbf{a},$ $\mathbf{b}$ on $\Pi_{A_{\phi},\phi}=\Pi_{+}\cup\Pi_{-}=\lim_{a_{\phi}\to A_{\phi}}\Pi_{a_{\phi},\phi},$ i.e. $\Pi_{A_{\phi},\phi}=\lim_{t\to\infty}\Pi_{a_{\phi}(t),\phi}$ . We would like to express $B_{\phi}(t)$ defined by (5.1) in terms of these quantities.

By Corollary 6.5 the integral $\int_{\mathbf{a}}W(z)dz$ is a linear combination of $z_{\pm}$ , $w^{\prime}(z_{\pm}^{+})$ and $(\vartheta^{\prime}/\vartheta)(F(z_{\pm}^{-},z_{\pm}^{+})+\nu,\tau)$ . In $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ , the functions $\psi$ , $1/\psi$ and $\psi^{\prime}$ has no poles, and hence $z_{\pm}=z_{\pm}(t)$ are bounded in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ and so are $w^{\prime}(z^{+}_{\pm})$ except for neighbourhoods of the zeros of $w(z)$ , i.e., $0,z_{1},z_{3},z_{5}$ . Furthermore, around these points, the leading terms of $-\tfrac{1}{2}w^{\prime}(z^{+}_{\pm})\omega_{\mathbf{a}}$ and $(\vartheta^{\prime}/\vartheta)(F(z_{\pm}^{-},z_{\pm}^{+})+\nu,\tau)$ are cancelled out. Indeed we have, say, around $z^{+}_{\pm}=0$ , $-\tfrac{1}{2}w^{\prime}(z_{\pm}^{+})\omega_{\mathbf{a}}=-\tfrac{1}{2}e^{3i\phi/2}A_{\phi}^{1/2}\omega_{\mathbf{a}}(z_{\pm}^{+})^{-1/2}+O(1)$ and $F(z_{\pm}^{-},z_{\pm}^{+})^{-1}=\tfrac{1}{2}e^{3i\phi/2}A_{\phi}^{1/2}\omega_{\mathbf{a}}(z_{\pm}^{+})^{-1/2}+O(1)$ . Since $z_{\pm}=z_{\pm}(t)$ moves on $\Pi_{a_{\phi},\phi}$ crossing $\mathbf{a}$ - and $\mathbf{b}$ -cycles, $\omega_{\mathbf{a}}F(z_{\pm}^{-},z_{\pm}^{+})=2p_{\pm}(t)\omega_{\mathbf{a}}+2q_{\pm}(t)\omega_{\mathbf{b}}+O(1)$ with $p_{\pm}(t),$ $q_{\pm}(t)\in\mathbb{Z},$ which implies the boundedness of $\mathrm{Re\,}(\vartheta^{\prime}/\vartheta)(F(z^{-}_{\pm},z_{\pm}^{+})+\nu,\tau)$ . Thus we have verified that $\mathrm{Re\,}\int_{\mathbf{a}}W(z)dz$ is bounded in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1}).$

Suppose that $-\pi/4<\phi<0$ . By (5.1)

\frac{1}{z}(w(a_{\phi},z)-w(A_{\phi},z))=\frac{2e^{3i\phi}t^{-1}B_{\phi}(t)}{w(A_{\phi},z)}(1+O(t^{-1}B_{\phi}(t))).

By using this with $B_{\phi}(t)\ll 1$ the first formula in Proposition 6.1 is written in the form

\ln(1+s_{1}s_{2})=-\frac{t}{2}\int_{\mathbf{a}}\Bigl{(}\frac{w(A_{\phi},z)}{z}+\frac{2e^{3i\phi}t^{-1}B_{\phi}(t)}{w(A_{\phi},z)}\Bigr{)}dz+\frac{1}{4}\int_{\mathbf{a}}W(z)dz-C_{\alpha,\beta}(\mathbf{a})+\pi i+O(t^{-\delta}),

that is,

t\mathcal{J}_{\mathbf{a}}+2e^{3i\phi}\Omega_{\mathbf{a}}B_{\phi}(t)=\frac{1}{2}\int_{\mathbf{a}}W(z)dz-2C_{\alpha,\beta}(\mathbf{a})+2\pi i-2\ln(1+s_{1}s_{2})+O(t^{-\delta}).

In Proposition 6.1, $(\ln(1+s_{1}s_{2}),\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1))$ , generally depending on $t$ , is a solution of the direct monodromy problem. Suppose that

(6.10)

\ln(1+s_{1}s_{2})\ll 1,\quad\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)\ll 1\,\,\,\text{in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$.}

By the Boutroux equations (2.2), we have $\mathrm{Re\,}e^{3i\phi}\Omega_{\mathbf{a}}B_{\phi}(t)\ll 1$ as $t\to\infty$ in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ . From the second formula of Proposition 6.1 we similarly derive

t\mathcal{J}_{\mathbf{b}}+2e^{3i\phi}\Omega_{\mathbf{b}}B_{\phi}(t)=\frac{1}{2}\int_{\mathbf{b}}W(z)dz-2C_{\alpha,\beta}(\mathbf{b})-2\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)+O(t^{-\delta}),

in which $\int_{\mathbf{b}}W(z)dz$ is expressed by $\vartheta(z,\hat{\tau})$ with $\hat{\tau}=(-\omega_{\mathbf{a}})/\omega_{\mathbf{b}},$ and this formula yields $\mathrm{Re\,}e^{3i\phi}\Omega_{\mathbf{b}}B_{\phi}(t)\ll 1$ . These two estimates leads to the inequality $|B_{\phi}(t)|\leq C_{0}$ in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ for some $C_{0}>0,$ while the implied constant of (5.1) may be supposed to be $2C_{0}$ if $t^{\prime}_{\infty}$ is taken sufficiently large. Hence the boundedness of $B_{\phi}(t)$ may be derived under (6.10) independently of (5.1). The case $0<\phi<\pi/4$ is discussed in the same way by using the Boutroux equations (2.2).

Remark 6.2.

Under a relaxed condition, say $B_{\phi}(t)\ll\ln|t|$ instead of $\ll 1$ of (5.1), each turning point is located within the distance $O(t^{-1}\ln|t|)$ from the limit one, and Propositions 6.1 and 6.2 are obtained by choosing a slightly smaller $\delta$ . Then the equivalence between (6.10) and the boundedness of $B_{\phi}(t)$ may be proved as above.

Proposition 6.7.

Suppose that $0<|\phi|<\pi/4$ and let $\mathfrak{l}(\mathbf{s},\phi)=\ln(1+s_{1}s_{2})$ if $-\pi/4<\phi<0,$ and $=-\ln(1+s_{2}s_{3})$ if $0<\phi<\pi/4.$ In $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ , we have

\mathfrak{l}(\mathbf{s},\phi)\ll 1,\quad\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)\ll 1

if and only if $B_{\phi}(t)\ll 1$ , and then

t\mathcal{J}_{\mathbf{a}}+2e^{3i\phi}\Omega_{\mathbf{a}}B_{\phi}(t)=\frac{1}{2}\int_{\mathbf{a}}W(z)dz-2C_{\alpha,\beta}(\mathbf{a})+2\pi i-2\mathfrak{l}(\mathbf{s},\phi)+O(t^{-\delta}).

The following fact guarantees the possibility of the limit $a_{\phi}\to A_{\phi}$ under integration.

Proposition 6.8.

Suppose that $0<|\phi|<\pi/4.$ Then, in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ ,

\biggl{(}\int^{z^{+}_{+}}_{z^{-}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\frac{dz}{w(a_{\phi},z)}=\biggl{(}\int^{z^{+}_{+}}_{z^{-}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\frac{dz}{w(A_{\phi},z)}+O(t^{-1}).

Proof..

By (5.1) it is easy to see $\omega_{\mathbf{a},\,\,\mathbf{b}}=\Omega_{\mathbf{a},\,\,\mathbf{b}}+O(t^{-1}).$ Suppose that $-\pi/4<\phi<0$ . By Proposition 6.1, Remark 6.1, Corollary 6.5 and (6.9),

	$\displaystyle\ln((1+s_{1}s_{2})$	$\displaystyle(1+s_{2}s_{3})-1)-\tau\ln(1+s_{1}s_{2})$
	$\displaystyle=$	$\displaystyle-a_{\phi}e^{4i\phi}t\Bigl{(}\int_{\mathbf{b}}-\tau\int_{\mathbf{a}}\Bigr{)}\frac{dz}{zw(a_{\phi},z)}+\frac{1}{4}\Bigl{(}\int_{\mathbf{b}}-\tau\int_{\mathbf{a}}\Bigr{)}W(z)dz$
		$\displaystyle-(C_{\alpha,\beta}(\mathbf{b})-\tau C_{\alpha,\beta}(\mathbf{a}))-\pi i\tau+O(t^{-\delta})$
	$\displaystyle=$	$\displaystyle-\frac{2\pi ie^{i\phi}t}{\omega_{\mathbf{a}}}+\frac{\pi i}{2}(F(z^{-}_{+},z^{+}_{+})-F(z^{-}_{-},z^{+}_{-}))-C_{\alpha,\beta}^{*}-\pi i\tau+O(t^{-\delta})$

with $C_{\alpha,\beta}^{*}=C_{\alpha,\beta}(\mathbf{b})-\tau C_{\alpha,\beta}(\mathbf{a}).$ This implies

-2e^{i\phi}t+\frac{1}{2}\biggl{(}\int_{z^{-}_{+}}^{z^{+}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\frac{dz}{w(a_{\phi},z)}=O(1).

Write $\omega_{\mathbf{a}}F(z_{\pm}^{-},z_{\pm}^{+})=2p_{\pm}(t)\omega_{\mathbf{a}}+2q_{\pm}(t)\omega_{\mathbf{b}}+O(1)$ with $p_{\pm}(t),$ $q_{\pm}(t)\in\mathbb{Z}$ , and set $-i\mathcal{J}_{\mathbf{a}}t/2+\pi q(t)=X,$ $-i\mathcal{J}_{\mathbf{b}}t/2-\pi p(t)=Y,$ where $p(t)=p_{+}(t)-p_{-}(t),$ $q(t)=q_{+}(t)-q_{-}(t).$ By the Boutroux equations (2.2), $\mathrm{Im\,}X$ and $\mathrm{Im\,}Y$ are bounded. Note that $\omega_{\mathbf{b}}\mathcal{J}_{\mathbf{a}}-\omega_{\mathbf{a}}\mathcal{J}_{\mathbf{b}}=\Omega_{\mathbf{b}}\mathcal{J}_{\mathbf{a}}-\Omega_{\mathbf{a}}\mathcal{J}_{\mathbf{b}}+O(t^{-1})=-4\pi ie^{i\phi}+O(t^{-1})$ by Corollary 8.10. Then the estimate above is

		$\displaystyle-2e^{i\phi}t+\omega_{\mathbf{a}}p(t)+\omega_{\mathbf{b}}q(t)+O(1)$
	$\displaystyle=$	$\displaystyle-2e^{i\phi}t+\pi^{-1}\omega_{\mathbf{a}}(-Y-i\mathcal{J}_{\mathbf{b}}t/2)+\pi^{-1}\omega_{\mathbf{b}}(X+i\mathcal{J}_{\mathbf{a}}t/2)+O(1)$
	$\displaystyle=$	$\displaystyle-2e^{i\phi}t+\pi^{-1}it(\omega_{\mathbf{b}}\mathcal{J}_{\mathbf{a}}-\omega_{\mathbf{a}}\mathcal{J}_{\mathbf{b}})/2+\pi^{-1}(\omega_{\mathbf{b}}X-\omega_{\mathbf{a}}Y)+O(1)$
	$\displaystyle=$	$\displaystyle\pi^{-1}(\omega_{\mathbf{b}}X-\omega_{\mathbf{a}}Y)+O(1)\ll 1$

with $\mathrm{Im\,}(\omega_{\mathbf{b}}/\omega_{\mathbf{a}})>0$ uniformly. This implies $X$ , $Y\ll 1$ , and hence

\pi p(t)=-i\mathcal{J}_{\mathbf{b}}t/2+O(1),\quad\pi q(t)=i\mathcal{J}_{\mathbf{a}}t/2+O(1).

Observing that $w(a_{\phi},z)^{-1}-w(A_{\phi},z)^{-1}=-2ze^{3i\phi}B_{\phi}(t)t^{-1}w(A_{\phi},z)^{-3}+O(t^{-2}),$ we have

	$\displaystyle\biggl{\|}\biggl{(}\int^{z^{+}_{+}}_{z^{-}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\Bigl{(}\frac{1}{w(a_{\phi},z)}-\frac{1}{w(A_{\phi},z)}\Bigr{)}dz\biggr{\|}\ll\biggl{\|}\biggl{(}\int^{z^{+}_{+}}_{z^{-}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\frac{zB_{\phi}(t)t^{-1}}{w(A_{\phi},z)^{3}}dz\biggr{\|}+O(t^{-1})$
	$\displaystyle\ll\biggl{\|}t^{-1}\biggl{(}\int^{z^{+}_{+}}_{z^{-}_{+}}-\int^{z^{+}_{-}}_{z^{-}_{-}}\biggr{)}\frac{zdz}{w(A_{\phi},z)^{3}}\biggr{\|}+O(t^{-1})\ll\|t^{-1}(p(t)j_{\mathbf{a}}+q(t)j_{\mathbf{b}})\|+O(t^{-1})$
	$\displaystyle=\|\mathcal{J}_{\mathbf{b}}j_{\mathbf{a}}-\mathcal{J}_{\mathbf{a}}j_{\mathbf{b}}\|+O(t^{-1})=\tfrac{1}{2}\|(\partial/\partial A_{\phi})(\mathcal{J}_{\mathbf{b}}\Omega_{\mathbf{a}}-\mathcal{J}_{\mathbf{a}}\Omega_{\mathbf{b}})\|+O(t^{-1})\ll t^{-1}$

with $j_{\mathbf{a},\,\,\mathbf{b}}=\int_{\mathbf{a},\,\,\mathbf{b}}zw(A_{\phi},z)^{-3}dz$ . This completes the proof. ∎

7. Proofs of the main results

7.1. Derivation of $\psi(t)$

Suppose that $-\pi/4<\phi<0.$ Let $\mathbf{s}=(s_{1},s_{2},s_{3},s_{4})$ with $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0,$ $1+s_{1}s_{2}\not=0$ be a solution of the direct monodromy problem discussed above in which $(\psi,\psi_{t})$ be such that (5.1) is valid in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ . As in the proof of Proposition 6.8 we set

	$\displaystyle\ln((1+s_{1}s_{2})$	$\displaystyle(1+s_{2}s_{3})-1)-\tau\ln(1+s_{1}s_{2})$
	$\displaystyle=$	$\displaystyle-\frac{2\pi ie^{i\phi}t}{\omega_{\mathbf{a}}}+\frac{\pi i}{2}(F(z^{-}_{+},z^{+}_{+})-F(z^{-}_{-},z^{+}_{-}))-C_{\alpha,\beta}^{*}-\pi i\tau+O(t^{-\delta})$

with $C_{\alpha,\beta}^{*}=C_{\alpha,\beta}(\mathbf{b})-\tau C_{\alpha,\beta}(\mathbf{a}).$ Then by Proposition 6.8 we have

(7.1)

\biggl{(}\int^{z^{+}_{+}}_{0}-\int^{z^{+}_{-}}_{0}\biggr{)}\frac{dz}{w(A_{\phi},z)}=2e^{i\phi}t+2\tilde{\chi}_{0}+\Omega_{\mathbf{b}}+O(t^{-\delta}),

in which

	$\displaystyle 2\tilde{\chi}_{0}=\frac{1}{\pi i}(\Omega_{\mathbf{a}}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)-\Omega_{\mathbf{b}}\ln(1+s_{1}s_{2}))+\Gamma_{\alpha,\beta},$
	$\displaystyle\Gamma_{\alpha,\beta}=\frac{(\alpha-\beta)}{2\pi i}\Bigl{(}\Omega_{\mathbf{a}}\int_{\mathbf{b}}\frac{zdz}{w(A_{\phi},z)}-\Omega_{\mathbf{b}}\int_{\mathbf{a}}\frac{zdz}{w(A_{\phi},z)}\Bigr{)}.$

Let us derive the asymptotic form as in Theorem 2.1 from (7.1). Set

u=-\int^{z^{+}_{+}}_{0}\frac{dz}{w(A_{\phi},z)},\quad v=-\int^{z^{+}_{-}}_{0}\frac{dz}{w(A_{\phi},z)}.

By the change of variables $z=e^{3i\phi}A_{\phi}(\wp-\tfrac{1}{3}e^{2i\phi})^{-1}$ (cf. Remark 2.1),

	$\displaystyle-\int^{z^{+}_{\pm}}_{0}\frac{dz}{w(A_{\phi},z)}=\int^{\wp_{\pm}}_{\infty}\frac{d\wp}{\sqrt{4\wp^{3}-g_{2}\wp-g_{3}}},$
	$\displaystyle\wp_{\pm}=\tfrac{1}{3}{e^{2i\phi}}+e^{3i\phi}{A_{\phi}}z_{\pm}^{-1},\quad g_{2}=e^{4i\phi}(\tfrac{4}{3}-4A_{\phi}),\quad g_{3}=e^{6i\phi}(-\tfrac{8}{27}+\tfrac{4}{3}A_{\phi}-A_{\phi}^{2}),$

and hence $\wp(u)=\wp_{+},$ $\wp(v)=\wp_{-}$ . Observing that, by (4.2), (5.1) and (6.2),

	$\displaystyle z_{+}+z_{-}=-\psi-2e^{i\phi}+O(t^{-1}),\quad z_{+}z_{-}=-e^{3i\phi}A_{\phi}\psi^{-1}+O(t^{-1}),$
	$\displaystyle w(A_{\phi},z_{\pm})=\lambda b_{3}(\lambda)\|_{\lambda^{2}=z_{\pm}}=z_{\pm}^{2}+2(e^{i\phi}+\psi)z_{\pm}+O(t^{-1}),$

we have, up to the error term $+O(t^{-1})$ ,

	$\displaystyle\wp(u)+\wp(v)=$	$\displaystyle\wp_{+}+\wp_{-}=\tfrac{2}{3}e^{2i\phi}+e^{3i\phi}A_{\phi}(z_{+}+z_{-})(z_{+}z_{-})^{-1}=\tfrac{2}{3}e^{2i\phi}+\psi^{2}+2e^{i\phi}\psi,$
	$\displaystyle-(\wp^{\prime}(u)-\wp^{\prime}(v))=$	$\displaystyle e^{3i\phi}A_{\phi}(z_{+}^{-2}w(A_{\phi},z_{+})-z_{-}^{-2}w(A_{\phi},z_{-}))=2(e^{i\phi}+\psi)e^{3i\phi}A_{\phi}(z_{+}^{-1}-z_{-}^{-1})$
	$\displaystyle=$	$\displaystyle 2(e^{i\phi}+\psi)(\wp_{+}-\wp_{-})=2(e^{i\phi}+\psi)(\wp(u)-\wp(v)).$

Then by the addition theorem

\wp(u+v)=-\wp(u)-\wp(v)+\frac{1}{4}\Bigl{(}\frac{\wp^{\prime}(u)-\wp^{\prime}(v)}{\wp(u)-\wp(v)}\Bigr{)}^{2}=\frac{e^{2i\phi}}{3}+O(t^{-1}),

which implies, by Proposition 7.6,

	$\displaystyle\biggl{(}\int^{z_{+}^{+}}_{0}+\int^{z_{-}^{+}}_{0}\biggr{)}\frac{dz}{w(A_{\phi},z)}=$	$\displaystyle-(u+v)=-\int_{\infty}^{e^{2i\phi}/3}\frac{d\wp}{\sqrt{4\wp^{3}-g_{2}\wp-g_{3}}}$
	$\displaystyle=$	$\displaystyle\int^{\infty}_{0}\frac{dz}{w(A_{\phi},z)}=\Omega_{0}\in\{\pm\tfrac{1}{3}\Omega_{\mathbf{b}}\}.$

This combined with (7.1) leads to

(7.2)

\begin{split}&\int^{z^{+}_{+}}_{0}\frac{dz}{w(A_{\phi},z)}=e^{i\phi}t+\tilde{\chi}_{0}+\frac{1}{2}(\Omega_{\mathbf{b}}+\Omega_{0})+O(t^{-\delta}),\\ &\int^{z^{+}_{-}}_{0}\frac{dz}{w(A_{\phi},z)}=-e^{i\phi}t-\tilde{\chi}_{0}-\frac{1}{2}(\Omega_{\mathbf{b}}-\Omega_{0})+O(t^{-\delta}).\end{split}

Remark 7.1.

A similar argument with use of the addition theorem leads to

	$\displaystyle\biggl{(}\int^{z_{+}^{+}}_{0}+\int^{z_{-}^{+}}_{0}\biggr{)}\frac{dz}{w(a_{\phi},z)}=-\int_{\infty}^{e^{2i\phi}/3}\frac{d\wp}{\sqrt{4\wp^{3}-\tilde{g}_{2}\wp-\tilde{g}_{3}}}+O(t^{-1})$
	$\displaystyle=-\int_{\infty}^{e^{2i\phi}/3}\frac{d\wp}{\sqrt{4\wp^{3}-{g}_{2}\wp-{g}_{3}}}+O(t^{-1})=\biggl{(}\int^{z_{+}^{+}}_{0}+\int^{z_{-}^{+}}_{0}\biggr{)}\frac{dz}{w(A_{\phi},z)}+O(t^{-1}),$

where $\tilde{g}_{2}=e^{4i\phi}(\tfrac{4}{3}-4a_{\phi}),$ $\tilde{g}_{3}=e^{6i\phi}(-\tfrac{27}{8}+\tfrac{4}{3}a_{\phi}-a_{\phi}^{2}).$ Combining this with Proposition 6.8 we have

\int^{z^{+}_{\pm}}_{0}\frac{dz}{w(a_{\phi},z)}=\int^{z^{+}_{\pm}}_{0}\frac{dz}{w(A_{\phi},z)}+O(t^{-1}).

Recall that $\mathrm{P}(u;A_{\phi})$ solves $\mathrm{P}_{u}^{2}=w(A_{\phi},\mathrm{P})^{2}.$ By Proposition 7.6,

(7.3)

\mathrm{P}(u;A_{\phi})=\frac{1}{\Omega_{\mathbf{a}}}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}+\frac{\tau^{*}}{3}+\nu,\tau^{*}\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}-\frac{\tau^{*}}{3}+\nu,\tau^{*}\Bigr{)}+C_{\mathrm{P}}\Bigr{)}

with $C_{\mathrm{P}}=-(\vartheta^{\prime}/\vartheta)(\tfrac{1}{3}\tau^{*}+\nu,\tau^{*})+(\vartheta^{\prime}/\vartheta)(-\tfrac{1}{3}\tau^{*}+\nu,\tau^{*})$ and $\tau^{*}=\Omega_{\mathbf{b}}/\Omega_{\mathbf{a}}$ . Then we have

	$\displaystyle\psi(t)=$	$\displaystyle-z_{+}-z_{-}-2e^{i\phi}+O(t^{-1})$
	$\displaystyle=$	$\displaystyle-\mathrm{P}(\tilde{t}_{\phi}+\tfrac{1}{2}(\Omega_{\mathbf{b}}+\Omega_{0});A_{\phi})-\mathrm{P}(\tilde{t}_{\phi}+\tfrac{1}{2}(\Omega_{\mathbf{b}}-\Omega_{0});A_{\phi})-2e^{i\phi}+O(t^{-\delta})$
	$\displaystyle=$	$\displaystyle-\frac{1}{\Omega_{\mathbf{a}}}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}-\frac{\tau^{}}{6}\pm\frac{\tau^{}}{6}+\nu\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}+\frac{\tau^{}}{6}\pm\frac{\tau^{}}{6}+\nu\Bigr{)}$
		$\displaystyle+\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}-\frac{\tau^{}}{6}\mp\frac{\tau^{}}{6}+\nu\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}+\frac{\tau^{}}{6}\mp\frac{\tau^{}}{6}+\nu\Bigr{)}\Bigr{)}+C_{\phi}+O(t^{-\delta})$
	$\displaystyle=$	$\displaystyle\frac{1}{\Omega_{\mathbf{a}}}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}+\frac{\tau^{}}{3}+\nu\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{\tilde{t}_{\phi}}{\Omega_{\mathbf{a}}}-\frac{\tau^{}}{3}+\nu\Bigr{)}\Bigr{)}+C_{\phi}+O(t^{-\delta})$
	$\displaystyle=$	$\displaystyle\mathrm{P}(\tilde{t}_{\phi};A_{\phi})+C_{\phi,0}+O(t^{-\delta}),$

where $C_{\phi},C_{\phi,0}\in\mathbb{C},$ and $\tilde{t}_{\phi}=e^{i\phi}t+\tilde{\chi}_{0}$ . If $\psi_{t}=(d/dt)\psi+O(t^{-1})$ , then by (4.2) $e^{-2i\phi}((d/dt)\psi)^{2}=w(A_{\phi},\psi)^{2}+O(t^{-1})$ , which implies $C_{\phi,0}=0.$ In the case $0<\phi<\pi/4$ , for $\mathbf{s}$ such that $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0,$ $1+s_{2}s_{3}\not=0$ , the same argument is possible. Thus we have the following.

Proposition 7.1.

If $\psi_{t}=d\psi/dt+O(t^{-1}),$ then equation (7.1) in $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1})$ implies

\psi(t)=\mathrm{P}(e^{i\phi}t+\tilde{\chi}_{0};A_{\phi})+O(t^{-\delta}),

in which

2\tilde{\chi}_{0}=\frac{1}{\pi i}(\Omega_{\mathbf{a}}\ln((1+s_{1}s_{2})(1+s_{2}s_{3})-1)-\Omega_{\mathbf{b}}\mathfrak{l}(\mathbf{s},\phi))+\Gamma_{\alpha,\beta},

with $\mathfrak{l}(\mathbf{s},\phi)=\ln(1+s_{1}s_{2})$ if $-\pi/4<\phi<0,$ and $=-\ln(1+s_{2}s_{3})$ if $0<\phi<\pi/4$ .

Let us calculate the value $\Gamma_{\alpha,\beta}$ in $\tilde{\chi}_{0}.$ The substitution $z=\mathrm{P}(u;A_{\phi})$ with (7.3) leads to

	$\displaystyle\int_{\mathbf{b}}\frac{zdz}{w(A_{\phi},z)}=$	$\displaystyle\int^{\Omega_{\mathbf{b}}}_{0}\mathrm{P}(u;A_{\phi})du$
	$\displaystyle=$	$\displaystyle\int^{\Omega_{\mathbf{b}}}_{0}\frac{1}{\Omega_{\mathbf{a}}}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}+\frac{\tau^{}}{3}+\nu\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}-\frac{\tau^{}}{3}+\nu\Bigr{)}+C_{\mathrm{P}}\Bigr{)}du$
	$\displaystyle=$	$\displaystyle\Bigl{[}\ln\vartheta(\Omega_{\mathrm{a}}^{-1}u+\tfrac{1}{3}\tau^{}+\nu)-\ln\vartheta(\Omega_{\mathrm{a}}^{-1}u-\tfrac{1}{3}\tau^{}+\nu)\Bigr{]}_{0}^{\Omega_{\mathbf{b}}}+C_{\mathrm{P}}\tau^{*}$
	$\displaystyle=$	$\displaystyle-\pi i(\tau^{}+2(\tfrac{1}{3}\tau^{}+\nu))+\pi i(\tau^{}+2(-\tfrac{1}{3}\tau^{}+\nu))+C_{\mathrm{P}}\tau^{*}$
	$\displaystyle=$	$\displaystyle(-\tfrac{4}{3}\pi i+C_{\mathrm{P}})\tau^{*},$

and

\int_{\mathbf{a}}\frac{zdz}{w(A_{\phi},z)}=\int^{\Omega_{\mathbf{a}}}_{0}\frac{1}{\Omega_{\mathbf{a}}}\Bigl{(}\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}+\frac{\tau^{*}}{3}+\nu\Bigr{)}-\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{u}{\Omega_{\mathbf{a}}}-\frac{\tau^{*}}{3}+\nu\Bigr{)}+C_{\mathrm{P}}\Bigr{)}du=C_{\mathrm{P}}.

From these quantities, the required constant follows.

Proposition 7.2.

We have

	$\displaystyle\Gamma_{\alpha,\beta}=\Gamma_{\alpha,\beta,\mathbf{b}}-\Gamma_{\alpha,\beta,\mathbf{a}}=-\frac{2}{3}(\alpha-\beta)\Omega_{\mathbf{b}},$
	$\displaystyle\Gamma_{\alpha,\beta,\mathbf{b}}=\Gamma_{\alpha,\beta}+\frac{1}{2\pi i}(\alpha-\beta)C_{\mathrm{P}}\Omega_{\mathbf{b}},\quad\Gamma_{\alpha,\beta,\mathbf{a}}=\frac{1}{2\pi i}(\alpha-\beta)C_{\mathrm{P}}\Omega_{\mathbf{b}}$

with $C_{\mathrm{P}}=-(\vartheta^{\prime}/\vartheta)(\tfrac{1}{3}\tau^{*}+\nu,\tau^{*})+(\vartheta^{\prime}/\vartheta)(-\tfrac{1}{3}\tau^{*}+\nu,\tau^{*})$ .

By Propositions 7.1 and 7.2 we may derive from (7.1) asymptotic forms as in Theorems 2.1 and 2.2.

Proposition 7.3.

In (7.2), $\Omega_{0}=-\tfrac{1}{3}\Omega_{\mathbf{b}}$ .

Proof..

Note that $\psi(t)=\mathrm{P}(e^{i\phi}t+\tilde{\chi}_{0};A_{\phi})+O(t^{-\delta})$ . By Proposition 7.6, if $\psi(t)=e^{3i\phi}A_{\phi}(e^{i\phi}t-t_{0})^{2}(1+o(1))$ around $t=t_{0}$ , then $\psi(t)\sim(e^{i\phi}t-t_{0}^{-})^{-1}-e^{i\phi}+o(1)$ and $\psi(t)\sim-(e^{i\phi}t-t_{0}^{+})^{-1}-e^{i\phi}+o(1)$ around each of the points $t_{0}^{\pm}:=t_{0}\pm\tfrac{1}{3}\Omega_{\mathbf{b}}$ up to $O(t^{-\delta}).$ Then, by $z_{\pm}(t)=\tfrac{1}{2}(\mp e^{-i\phi}\psi_{t}\psi^{-1}-\psi-2e^{i\phi})+O(t^{-1})$ , it follows that

	$\displaystyle z_{+}(t)=O(e^{i\phi}t-t_{0}^{-}),\quad z_{+}(t)=-(e^{i\phi}t-t_{0})^{-1}+O(1),\quad z_{+}(t)=(e^{i\phi}t-t_{0}^{+})^{-1}+O(1),$
	$\displaystyle z_{-}(t)=-(e^{i\phi}t-t_{0}^{-})^{-1}+O(1),\quad z_{-}(t)=(e^{i\phi}t-t_{0})^{-1}+O(1),\quad z_{-}(t)=O(e^{i\phi}t-t_{0}^{+})$

around each point. Since $z_{\pm}(t)=\mathrm{P}(e^{i\phi}t+\tilde{\chi}_{0}+\tfrac{1}{2}(\Omega_{\mathbf{b}}\pm\Omega_{0});A_{\phi})+O(t^{-\delta})$ we conclude that $\Omega_{0}=-\tfrac{1}{3}\Omega_{\mathbf{b}}.$ ∎

7.2. Asymptotic representation of $B_{\phi}(t)$

Recalling Propositions 6.6 and 6.7, and applying Remark 7.1, we have

	$\displaystyle t\mathcal{J}_{\mathbf{a}}+2e^{3i\phi}\Omega_{\mathbf{a}}B_{\phi}(t)=$	$\displaystyle\frac{\vartheta^{\prime}}{\vartheta}(\tfrac{1}{2}F_{}(z^{-}_{+},z^{+}_{+})+\tfrac{1}{2}+\tfrac{\tau^{}}{6},\tau^{})-\frac{\vartheta^{\prime}}{\vartheta}(\tfrac{1}{2}F_{}(z^{-}_{-},z^{+}_{-})+\tfrac{1}{2}+\tfrac{\tau^{}}{6},\tau^{})$
		$\displaystyle-(\alpha-\beta)C_{\mathrm{P}}+2\pi i-2\mathfrak{l}(\mathbf{s},\phi)+O(t^{-\delta}),$

where

F_{*}(z^{-}_{\pm},z^{+}_{\pm})=\frac{1}{\Omega_{\mathbf{a}}}\int_{z^{-}_{\pm}}^{z^{+}_{\pm}}\frac{dz}{w(A_{\phi},z)}=\frac{2}{\Omega_{\mathbf{a}}}\int_{0}^{z^{+}_{\pm}}\frac{dz}{w(A_{\phi},z)}.

By Proposition 7.3, insertion of (7.2) with $\Omega_{0}=-\tfrac{1}{3}\Omega_{\mathbf{b}}$ yields the asymptotic representation of the correction function $B_{\phi}(t),$ which plays an essential role in the justification of our asymptotic solution.

Proposition 7.4.

In $S_{\phi}(t^{\prime}_{\infty},\kappa_{1},\delta_{1}),$

t\mathcal{J}_{\mathbf{a}}+2e^{3i\phi}\Omega_{\mathbf{a}}B_{\phi}(t)=2\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{e^{i\phi}t+\tilde{\chi_{0}}}{\Omega_{\mathbf{a}}}+\nu,\tau^{*}\Bigr{)}-(\alpha-\beta)C_{\mathrm{P}}+2\pi i-2\mathfrak{l}(\mathbf{s},\phi)+O(t^{-\delta}),

where $\mathfrak{l}(\mathbf{s},\phi)$ and $C_{\mathrm{P}}$ are constants given in Propositions 7.1 and 7.2.

7.3. Proofs of Theorems 2.1 and 2.2

For a prescribed monodromy data $\mathbf{s}$ , the asymptotic expression of $\psi(t)$ given in Proposition 7.1 is, at least formally, a solution of the inverse monodromy problem. To prove Theorems 2.1 and 2.2 let us make the justification for $y=e^{-i\phi}x\psi(t)$ as a solution of $\mathrm{P}_{\mathrm{IV}}$ along the lines in [21, pp. 105–106, pp. 120–121]. Suppose that $-\pi/4<\phi<0.$ Let $\mathbf{s}=(s_{1},s_{2},s_{3},s_{4})$ with $(1+s_{1}s_{2})(1+s_{2}s_{3})-1\not=0,$ $1+s_{1}s_{2}\not=0$ be a given point on the monodromy manifold $\mathcal{M}_{0}(\alpha,\beta)$ for isomonodromy system (3.1). Set

	$\displaystyle\psi_{\mathrm{as}}=\psi_{\mathrm{as}}(\mathbf{s},t):=$	$\displaystyle\mathrm{P}(e^{i\phi}t+\tilde{\chi}_{0};A_{\phi}),$
	$\displaystyle(B_{\phi})_{\mathrm{as}}=(B_{\phi})_{\mathrm{as}}(\mathbf{s},t):=$	$\displaystyle\frac{e^{-3i\phi}}{2\Omega_{\mathbf{a}}}\Bigl{(}2\frac{\vartheta^{\prime}}{\vartheta}\Bigl{(}\frac{e^{i\phi}t+\tilde{\chi_{0}}}{\Omega_{\mathbf{a}}}+\nu,\tau^{*}\Bigr{)}-t\mathcal{J}_{\mathbf{a}}+C_{\alpha,\beta,\mathbf{s},\phi}\Bigr{)},$
	$\displaystyle C_{\alpha,\beta,\mathbf{s},\phi}$	$\displaystyle=2\pi i-(\alpha-\beta)C_{\mathrm{P}}-2\mathfrak{l}(\mathbf{s},\phi),$

which are leading term expressions of $\psi(t)$ and $B_{\phi}(t)$ without $O(t^{-\delta})$ in Propositions 7.1 and 7.4. Taking (4.2) and (5.1) into account, we set

	$\displaystyle e^{-i\phi}\psi^{*}_{\mathrm{as}}=$	$\displaystyle-\tfrac{1}{2}e^{-i\phi}\psi_{\mathrm{as}}t^{-1}+\sqrt{\psi_{\mathrm{as}}((\psi_{\mathrm{as}}+2e^{i\phi})^{2}\psi_{\mathrm{as}}+4e^{3i\phi}A_{\phi})+\Delta(t,\psi_{\mathrm{as}},(B_{\phi})_{\mathrm{as}})t^{-1}},$
	$\displaystyle\Delta(t,\psi_{\mathrm{as}},$	$\displaystyle(B_{\phi})_{\mathrm{as}})=\psi_{\mathrm{as}}(4(B_{\phi})_{\mathrm{as}}-(4\alpha-\beta)\psi_{\mathrm{as}}-2(2\alpha-\beta)e^{i\phi})+\tfrac{1}{4}\beta^{2}t^{-1},$

where the branch of the square root is chosen in such a way that $\psi^{*}_{\mathrm{as}}$ is compatible with $(d/dt)\psi_{\mathrm{as}}$ , that is, $\psi^{*}_{\mathrm{as}}=(d/dt)\psi_{\mathrm{as}}+O(t^{-1}).$ Then for $a_{\phi}(t,\psi_{\mathrm{as}},\psi^{*}_{\mathrm{as}})$ and for $(B_{\phi})_{\mathrm{as}}$ inequality (5.1) is valid in the domain

\tilde{S}(\phi,t_{\infty},\kappa_{0},\delta_{2})=\{t\,|\,\mathrm{Re\,}t>t_{\infty},\,|\mathrm{Im\,}t|<\kappa_{0}\}\setminus\bigcup_{t_{*}\in Z_{\phi}}\{|t-t_{*}|<\delta_{2}\},\quad Z_{\phi}=Z_{\phi}^{\infty}\cup Z_{\phi}^{0},

where $Z_{\phi}^{\infty}=\{t_{*}\,|\,e^{i\phi}t_{*}+\tilde{\chi}_{0}=\pm\tfrac{1}{3}\Omega_{\mathbf{b}}+\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}\},$ $Z_{\phi}^{0}=\{t_{*}\,|\,e^{i\phi}t_{*}+\tilde{\chi}_{0}\in\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}\}.$ Consider system (3.1) with $\mathcal{B}(t,\lambda)$ containing $(\psi_{\mathrm{as}},\psi_{\mathrm{as}}^{*})=(\psi_{\mathrm{as}}(\mathbf{s},t),\psi_{\mathrm{as}}^{*}(\mathbf{s},t))$ . Then the direct monodromy problem for this system by the WKB analysis results in the monodromy data $(\mathbf{s})_{\mathrm{as}}(t)$ such that $\|(\mathbf{s})_{\mathrm{as}}(t)-\mathbf{s}\|\leq C|t|^{-\delta}$ for $|t|>t_{\infty}(\mathbf{s})$ , in which $C$ and $\delta$ are some constant independent of $\mathbf{s}$ . Then the justification scheme of Kitaev [19] applies to our case. By the maximal modulus principle the excluded disc around each point in $Z_{\phi}^{0}$ is removed. Thus we obtain Theorems 2.1 and 2.2.

7.4. Proof of Theorem 2.4

For given $n\in\mathbb{Z}$ the substitution

	$\displaystyle\phi=\tfrac{1}{2}\pi n+\tilde{\phi},\quad x=e^{i\pi n/2}\tilde{x},\quad\xi=e^{i\pi n/4}\tilde{\xi},\quad\mathrm{u}=e^{i\pi n/4}\tilde{\mathrm{u}}_{(n)},\quad\mathrm{v}=e^{i\pi n/4}\tilde{\mathrm{v}}_{(n)},$
	$\displaystyle\alpha=e^{i\pi n}\tilde{\alpha},\quad\beta=e^{i\pi n}\tilde{\beta},\quad\psi=e^{i\pi n/2}\tilde{\psi},\quad y=e^{i\pi n/2}\tilde{y},$

where $(\tilde{\mathrm{u}}_{(n)},\tilde{\mathrm{v}}_{(n)})=(\tilde{\mathrm{v}},\tilde{\mathrm{u}})$ if $n$ is odd, and $=(\tilde{\mathrm{u}},\tilde{\mathrm{v}})$ if $n$ is even, changes isomonodromy system (1.1) to

(7.4)		$\displaystyle\frac{d\tilde{\Psi}}{d\tilde{\xi}}=$	$\displaystyle\sigma_{2}^{n}\Biggl{(}\Bigl{(}\frac{\tilde{\xi}^{3}}{2}+\tilde{\xi}(\tilde{x}+\tilde{\mathrm{u}}_{(n)}\tilde{\mathrm{v}}_{(n)})+\frac{\tilde{\alpha}}{\tilde{\xi}}\Bigr{)}\sigma_{3}$
		$\displaystyle\phantom{--}+i\begin{pmatrix}0&\tilde{\xi}^{2}\tilde{\mathrm{u}}_{(n)}+2\tilde{x}\tilde{\mathrm{u}}_{(n)}+(\tilde{u}_{(n)})_{\tilde{x}}\\ \tilde{\xi}^{2}\tilde{\mathrm{v}}_{(n)}+2\tilde{x}\tilde{\mathrm{v}}_{(n)}-(\tilde{v}_{(n)})_{\tilde{x}}&0\end{pmatrix}\Biggr{)}\sigma_{2}^{n}\tilde{\Psi},$
	$\displaystyle\tilde{\beta}=$	$\displaystyle(\tilde{\mathrm{u}}_{(n)})_{\tilde{x}}\tilde{\mathrm{v}}_{(n)}-\tilde{\mathrm{u}}_{(n)}(\tilde{\mathrm{v}}_{(n)})_{\tilde{x}}+2\tilde{x}\tilde{\mathrm{u}}_{(n)}\tilde{\mathrm{v}}_{(n)}-(\tilde{\mathrm{u}}_{(n)}\tilde{\mathrm{v}}_{(n)})^{2},\quad\tilde{y}=\tilde{\mathrm{u}}_{(n)}\tilde{\mathrm{v}}_{(n)},$

and $y=e^{-i\phi}x\psi$ to $\tilde{y}=e^{-i\tilde{\phi}}\tilde{x}\tilde{\psi}.$ For $k\in\mathbb{Z}$ system (7.4) admits the canonical solutions

\sigma_{2}^{n}\tilde{\Psi}^{\infty}_{k}(\tilde{\xi})=\Psi_{k}^{\infty}(\tilde{\xi})=(I+O(\tilde{\xi}^{-1}))\exp((\tfrac{1}{8}\tilde{\xi}^{4}+\tfrac{1}{2}\tilde{x}\tilde{\xi}^{2}+(\tilde{\alpha}-\tilde{\beta})\ln\tilde{\xi})\sigma_{3})

as $\tilde{\xi}\to\infty$ through the sector $|\arg\tilde{\xi}+\tfrac{\pi}{8}-\tfrac{\pi}{4}k|<\tfrac{\pi}{4}$ , where $\Psi^{\infty}_{k}(\xi)$ are solutions of (1.1) given by (2.1). The Stokes matrices $\tilde{S}_{k}$ $(k\in\mathbb{Z})$ with

\tilde{S}_{2l-1}=\begin{pmatrix}1&\tilde{s}_{2l-1}\\ 0&1\end{pmatrix},\quad\tilde{S}_{2l}=\begin{pmatrix}1&0\\ \tilde{s}_{2l}&1\end{pmatrix}\quad(l\in\mathbb{Z})

for system (7.4) are defined by $\Psi_{k+1}^{\infty}(\tilde{\xi})=\Psi_{k}^{\infty}(\tilde{\xi})\tilde{S}_{k}.$ Observing that

	$\displaystyle\Psi^{\infty}_{k}(\tilde{\xi})=$	$\displaystyle(I+O(\xi^{-1}))\exp((-1)^{n}(\tfrac{1}{8}\xi^{4}+\tfrac{1}{2}x\xi^{2}+(\alpha-\beta)\ln\xi)\sigma_{3})e^{-(-1)^{n}(\alpha-\beta)(i\pi n/4)\sigma_{3}}$
	$\displaystyle=$	$\displaystyle\sigma_{2}^{n}\Bigl{(}(I+O(\xi^{-1}))\exp((\tfrac{1}{8}\xi^{4}+\tfrac{1}{2}x\xi^{2}+(\alpha-\beta)\ln\xi)\sigma_{3})e^{-(\alpha-\beta)(i\pi n/4)\sigma_{3}}\Bigr{)}\sigma_{2}^{n}$

in $|\arg\xi+\tfrac{\pi}{8}-\tfrac{\pi}{4}(k+n)|<\tfrac{\pi}{4}$ , we have $\Psi_{k}^{\infty}(\tilde{\xi})=\sigma_{2}^{n}\Psi^{\infty}_{k+n}(\xi)e^{-(\alpha-\beta)(i\pi n/4)\sigma_{3}}\sigma_{2}^{n}.$ This relation immediately leads to

\tilde{S}_{k}=\sigma_{2}^{n}e^{(\alpha-\beta)(i\pi n/4)\sigma_{3}}S_{k+n}e^{-(\alpha-\beta)(i\pi n/4)\sigma_{3}}\sigma_{2}^{n}

(cf. [13, (13)]), which implies $\tilde{s}_{k}\tilde{s}_{k+1}=s_{k+n}s_{k+1+n}$ . Thus system (1.1) for $0<|\phi-\pi n/2|<\pi/4$ is converted to (7.4) for $0<|\tilde{\phi}|<\pi/4$ with the monodromy data $\tilde{\mathbf{s}}=(\tilde{s}_{1},\tilde{s}_{2},\tilde{s}_{3},\tilde{s}_{4})=\mathbf{s}_{n}=(s_{1+n},s_{2+n},s_{3+n},s_{4+n})$ . Then application of Corollary 2.3 yields the theorem.

7.5. Properties of the elliptic function $\mathrm{P}(u;A)$

The elliptic function $p=\mathrm{P}(u;A)$ is a solution of

(7.5)

4e^{3i\phi}A=\frac{(p^{\prime})^{2}}{p}-p(p+2e^{i\phi})^{2}\qquad(p^{\prime}=dp/du).

About this equation relations in (7.2) suggest the following interesting fact.

Proposition 7.5.

Let $p=\eta=\eta(u)$ be a given solution of (7.5). Then the functions $\chi_{\pm}=\mp\tfrac{1}{2}\eta^{\prime}\eta^{-1}-\tfrac{1}{2}\eta-e^{i\phi}$ also solves (7.5).

Proof..

Note that

(\chi_{\pm}+\tfrac{1}{2}\eta+e^{i\phi})^{2}=\tfrac{1}{4}(\eta^{\prime})^{2}\eta^{-2}=\tfrac{1}{4}\eta^{-1}(4e^{3i\phi}A+\eta(\eta+2e^{i\phi})^{2})=e^{3i\phi}A\eta^{-1}+\tfrac{1}{4}(\eta+2e^{i\phi})^{2},

which implies that $\chi_{\pm}$ and $\eta$ satisfy the relation

(7.6)

\eta^{2}\chi_{\pm}+\eta\chi_{\pm}^{2}+2e^{i\phi}\eta\chi_{\pm}-e^{3i\phi}A=0.

Since $\eta$ solves (7.5), using (7.6) we have

	$\displaystyle\mp\chi^{\prime}_{\pm}$	$\displaystyle=\tfrac{1}{2}(\eta^{\prime\prime}\eta^{-1}-(\eta^{\prime})^{2}\eta^{-2})\pm\tfrac{1}{2}\eta^{\prime}$
		$\displaystyle=\tfrac{1}{2}\eta^{-1}(\eta^{2}(\eta+2e^{i\phi})-2e^{3i\phi}A)\pm\tfrac{1}{2}\eta^{\prime}$
		$\displaystyle=\eta(\pm\tfrac{1}{2}\eta^{\prime}\eta^{-1}+\tfrac{1}{2}\eta+e^{i\phi})-\chi_{\pm}^{2}-(\eta+2e^{i\phi})\chi_{\pm}$
		$\displaystyle=-2\eta\chi_{\pm}-\chi_{\pm}(\chi_{\pm}+2e^{i\phi}),$

and hence $\eta=\pm\tfrac{1}{2}\chi^{\prime}_{\pm}\chi_{\pm}^{-1}-\tfrac{1}{2}\chi_{\pm}-e^{i\phi}$ . Then by (7.6)

	$\displaystyle(\chi^{\prime}_{\pm})^{2}\chi_{\pm}^{-1}=$	$\displaystyle\chi_{\pm}(2\eta+(\chi_{\pm}+2e^{i\phi}))^{2}$
	$\displaystyle=$	$\displaystyle\chi_{\pm}(\chi_{\pm}+2e^{i\phi})^{2}+4(\eta^{2}\chi_{\pm}+\eta\chi_{\pm}(\chi_{\pm}+2e^{i\phi}))$
	$\displaystyle=$	$\displaystyle\chi_{\pm}(\chi_{\pm}+2e^{i\phi})^{2}+4e^{3i\phi}A,$

which implies $\chi_{\pm}$ solves (7.5). ∎

Recall that $\mathrm{P}(u;A)$ is a solution of (7.5) such that $\mathrm{P}(0;A)=0.$ Using Proposition 7.5 we have the following.

Proposition 7.6.

The elliptic function $\mathrm{P}(u;A)$ has simple poles with residue $-1$ at $u=\tfrac{1}{3}\Omega_{\mathbf{b}}+\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}$ and simple poles with residue $1$ at $u=-\tfrac{1}{3}\Omega_{\mathbf{b}}+\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}$ .

Figure 7.1. Cycle

\mathbf{b}=\mathbf{l}_{-1}\cup\mathbf{l}_{-2}\cup\mathbf{l}_{2}\cup\mathbf{l}_{1}

Proof..

Consider the loop on $\Pi_{A,\phi}=\Pi_{+}\cup\Pi_{-}$ defined by $\mathbf{l}_{-1}\cup\mathbf{l}_{-2}\cup\mathbf{l}_{2}\cup\mathbf{l}_{1}$ homotopic to the cycle $\mathbf{b}$ , in which $\mathbf{l}_{-1}=[0,\infty^{-}],$ $\mathbf{l}_{-2}=[\infty^{-},z_{3}]\subset\Pi_{-}$ , and $\mathbf{l}_{2}=[z_{3},\infty^{+}],$ $\mathbf{l}_{1}=[\infty^{+},0]\subset\Pi_{+}$ as in Figure 7.1. Let

\gamma_{-}:=\int_{\mathbf{l}_{-1}}\frac{dz}{w(A,z)},\quad\gamma_{+}:=\Omega_{\mathbf{b}}-\gamma_{-}=\int_{\mathbf{l}_{-1}\cup\mathbf{l}_{-2}\cup\mathbf{l}_{2}}\frac{dz}{w(A,z)}=\Bigl{(}\int_{\mathbf{b}}-\int_{\mathbf{l}_{1}}\Bigr{)}\frac{dz}{w(A,z)},

with $\int_{\mathbf{l}_{1}}w(A,z)^{-1}dz=\int_{\mathbf{l}_{-1}}w(A,z)^{-1}dz.$ Then $\mathrm{P}(u)=\mathrm{P}(u;A)$ has simple poles at $u=u_{\pm}\equiv\gamma_{\pm}\mod\Omega_{\mathbf{a}}\mathbb{Z}+\Omega_{\mathbf{b}}\mathbb{Z}$ , and double zeros at $u=u_{0}\equiv 0,$ and around these poles and zeros, $\mathrm{P}(u)=\pm(u-u_{\pm})^{-1}-e^{i\phi}+o(1)$ and $\mathrm{P}(u)=e^{3i\phi}A(u-u_{0})^{2}(1+o(1)).$ Let

\mathrm{P}_{-}(u)=\tfrac{1}{2}\mathrm{P}^{\prime}(u)\mathrm{P}(u)^{-1}-\tfrac{1}{2}\mathrm{P}(u)-e^{i\phi}.

It is easy to see that $\mathrm{P}_{-}(u)$ has simple poles with residue $1$ at $u=u_{0}\equiv 0$ , with residue $-1$ at $u=u_{+}\equiv\gamma_{+}$ , and at least vanishes at $u=u_{-}\equiv\gamma_{-}$ . By Proposition 7.5, $\mathrm{P}_{-}(u)$ also solves (7.5), and hence $\mathrm{P}_{-}(u+\gamma_{-})=\mathrm{P}(u)$ . Then the string of poles and zeros of $\mathrm{P}(u)$ given by $\mathcal{S}:=\Omega_{\mathbf{b}}\mathbb{Z}\cup(\gamma_{-}+\Omega_{\mathbf{b}}\mathbb{Z})\cup(\gamma_{+}+\Omega_{\mathbf{b}}\mathbb{Z})$ coincides with the shifted one $\mathcal{S}+\gamma_{-}$ . From this fact we derive $2\gamma_{-}=\gamma_{+}=\Omega_{\mathbf{b}}-\gamma_{-},$ which implies $\gamma_{-}=\tfrac{1}{3}\Omega_{\mathbf{b}}.$ Thus the proposition is obtained. ∎

8. Boutroux equations

We show basic facts on the Boutroux equations used in the preceding sections.

8.1. Basic facts

The definitions of the cycles $\mathbf{a},$ $\mathbf{b}$ , $\mathbf{a}_{*},$ $\mathbf{b}_{*}$ are based on the following stable configuration of zeros.

Lemma 8.1.

As long as $A\in\mathcal{D}_{0}:=\mathbb{C}\setminus(\{c\leq 0\}\cup\{c\geq\frac{8}{27}\})$ , the polynomial $\zeta^{3}+4\zeta^{2}+4\zeta+4A$ has zeros $\zeta_{1},$ $\zeta_{3}$ , $\zeta_{5}$ with the properties $:$

$(1)$ $\zeta_{j}=\zeta_{j}(A)$ $(j=1,3,5)$ are continuous in $A;$

$(2)$ $\mathrm{Re\,}\zeta_{5}<\mathrm{Re\,}\zeta_{3}<\mathrm{Re\,}\zeta_{1};$

$(3)$ $(\zeta_{5},\zeta_{3},\zeta_{1})\to(-2,-2,0)$ as $A\to 0$ , and $(\zeta_{5},\zeta_{3},\zeta_{1})\to(-\frac{8}{3},-\frac{2}{3},-\frac{2}{3})$ as $A\to\frac{8}{27}.$

Proof..

Suppose that $A\in\mathcal{D}_{0}$ . The substitution $\zeta=\frac{4}{3}(\xi-1)$ leads to the polynomial

\varpi(\xi)=4\xi^{3}-3\xi-c_{0},\quad c_{0}=1-\tfrac{27}{4}A\in\mathcal{D}_{1}=\mathbb{C}\setminus(\{c\leq-1\}\cup\{c\geq 1\}),

which has multiple zeros if and only if $c_{0}=\pm 1.$ The zeros of $\varpi(\xi)$ may be numbered in such a way that $-1<\xi_{5}<\xi_{3}<\xi_{1}<1$ for $-1<c_{0}<1$ , and clearly $\xi_{j}$ $(j=1,3,5)$ move continuously in $c_{0}\in\mathcal{D}_{1}.$ Then to verify (2) for $\zeta_{j}=\frac{4}{3}(\xi_{j}-1)$ $(j=1,3,5)$ , it is sufficient to show that $\xi_{j}-\xi_{k}\not\in i\mathbb{R}$ as long as $c_{0}\in\mathcal{D}_{1}$ if $j\not=k$ . By Cardano’s formula $\xi_{j},$ $\xi_{k}\in\{\gamma_{0},\gamma_{\pm 1}\},$ in which $2\gamma_{0}=u^{1/3}+u^{-1/3},$ $2\gamma_{\pm 1}=\omega^{\pm 1}u^{1/3}+\omega^{\mp 1}u^{-1/3}$ with $\omega=e^{2\pi i/3}$ and $u=c_{0}+\sqrt{c_{0}^{2}-1}$ . Suppose that $2(\gamma_{0}-\gamma_{1})=(1-\omega)u^{1/3}-(1-\omega^{-1})u^{-1/3}=ir$ with $r\in\mathbb{R}.$ Then $v=(1-\omega)u^{1/3}$ fulfils $v^{2}-irv+3=0$ , which implies $(1-\omega)u^{1/3}\in i\mathbb{R}$ with $1-\omega=i\sqrt{3}\omega^{2}.$ Hence $u=c_{0}+\sqrt{c_{0}^{2}-1}=r_{0}\in\mathbb{R}$ , that is, $c_{0}=(r_{0}+r_{0}^{-1})/2\in\{c\leq-1\}\cup\{c\geq 1\}$ , contradicting $c_{0}\in\mathcal{D}_{1}.$ Supposition $\gamma_{1}-\gamma_{-1}\in i\mathbb{R}$ implies $u^{1/3}\in\mathbb{R}$ leading to the same contradiction. Thus the property (2) is verified. It is easy to see that (1) and (3) are fulfilled. ∎

The following corollary is obtained by setting $\zeta=e^{-i\phi}z.$

Corollary 8.2.

As long as $(\phi,A)\in\mathbb{R}\times\mathcal{D}_{0}$ , the polynomial $z^{3}+4e^{i\phi}z^{2}+4e^{2i\phi}z+4e^{3i\phi}A$ has zeros $z_{1},$ $z_{3}$ , $z_{5}$ with the properties $:$

$(1)$ $z_{j}=z_{j}(\phi,A)$ $(j=1,3,5)$ are continuous in $(\phi,A);$

$(2)$ $\mathrm{Re\,}e^{-i\phi}z_{5}<\mathrm{Re\,}e^{-i\phi}z_{3}<\mathrm{Re\,}e^{-i\phi}z_{1};$

$(3)$ $(z_{5},z_{3},z_{1})\to(-2e^{i\phi},-2e^{i\phi},0)$ as $A\to 0$ , and $(z_{5},z_{3},z_{1})\to(-\frac{8}{3}e^{i\phi},-\frac{2}{3}e^{i\phi},-\frac{2}{3}e^{i\phi})$ as $A\to\frac{8}{27}.$

Let us define the elliptic curve $\Pi_{A}^{*}=\Pi_{+}^{*}\cup\Pi_{-}^{*}$ and the cycles $\mathbf{a}_{*}$ and $\mathbf{b}_{*}$ on $\Pi^{*}_{A}$ . As long as $A\in\mathcal{D}_{0}$ , by Lemma 8.1, the elliptic curve

v(A,\zeta)^{2}=\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A\zeta=\zeta(\zeta-\zeta_{1})(\zeta-\zeta_{3})(\zeta-\zeta_{5}),

is the two sheeted Riemann surface $\Pi^{*}_{A}=\Pi^{*}_{+}\cup\Pi^{*}_{-}$ glued along the cuts $[\zeta_{5},\zeta_{3}],$ $[\zeta_{1},0]$ , where $\Pi^{*}_{\pm}$ are two copies of $P^{1}(\mathbb{C})\setminus([\zeta_{5},\zeta_{3}]\cup[\zeta_{1},0])$ . Then on $\Pi^{*}_{A}$ the cycles $\mathbf{a}_{*}$ and $\mathbf{b}_{*}$ are defined as in Figure 8.1. Note that $\Pi^{*}_{A}$ and $(\mathbf{a}_{*},\mathbf{b}_{*})$ are parametrised by $A\in\mathcal{D}_{0}$ . Let the elliptic curve $\Pi_{A,\phi}=\Pi_{+}\cup\Pi_{-}$ and the pair of cycles $(\mathbf{a},\mathbf{b})$ be the images of $\Pi^{*}_{A}$ and $(\mathbf{a}_{*},\mathbf{b}_{*})$ under the map $z=e^{i\phi}\zeta$ . Then $\Pi_{A,\phi}$ is given by $w(A,z)^{2}=z^{4}+4e^{i\phi}z^{3}+4e^{2i\phi}z^{2}+4e^{3i\phi}Az$ as in Section 2, and the cycles $\mathbf{a}$ and $\mathbf{b}$ are as in Figure 2.1 for $0<|\phi|<\pi/4$ .

Figure 8.1. Cycles

\mathbf{a}_{*},

\mathbf{b}_{*}

\Pi^{*}_{A}=\Pi^{*}_{+}\cup\Pi^{*}_{-}

8.2. Uniqueness

Let us treat the integrals

	$\displaystyle J_{\mathbf{a}_{}}(A)=\int_{\mathbf{a}_{}}\frac{v(A,\zeta)}{\zeta}d\zeta,\qquad J_{\mathbf{b}_{}}(A)=\int_{\mathbf{b}_{}}\frac{v(A,\zeta)}{\zeta}d\zeta,$
	$\displaystyle v(A,\zeta)=\sqrt{\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A\zeta}=e^{-2i\phi}w(A,z).$

Then the Boutroux equations (2.2) become (2.3), that is,

(BE)_ϕ

\mathrm{Re\,}e^{2i\phi}\int_{\mathbf{a}_{*}}\frac{v(A_{\phi},\zeta)}{\zeta}d\zeta=\mathrm{Re\,}e^{2i\phi}\int_{\mathbf{b}_{*}}\frac{v(A_{\phi},\zeta)}{\zeta}d\zeta=0.

Example 8.1.

Let $\phi=0,$ $A=\tfrac{8}{27}.$ Then $\zeta_{1}=\zeta_{3}=-\tfrac{2}{3},$ $\zeta_{5}=-\tfrac{8}{3},$ and hence

J_{\mathbf{a}_{*}}(\tfrac{8}{27})=2\int^{-2/3}_{-8/3}(\zeta+\tfrac{2}{3})\sqrt{\zeta(\zeta+\smash{\tfrac{8}{3})}}\frac{d\zeta}{\zeta}=2e^{i\pi/2}\int^{2/3}_{8/3}(\tfrac{2}{3}-t)\sqrt{\frac{\smash{\tfrac{8}{3}}-t}{t}}dt=\frac{4i}{\sqrt{3}}

and $J_{\mathbf{b}_{*}}(\tfrac{8}{27})=0.$ This implies $A_{0}=\tfrac{8}{27}$ is a solution of the Boutroux equations (BE)_ϕ=0.

In accordance with Kitaev [20, Section 7] we would like to show the uniqueness of a solution of (BE)_ϕ=0. To do so we begin with the following.

Proposition 8.3.

If $\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)=\mathrm{Re\,}J_{\mathbf{b}_{*}}(A)=0,$ then $A\in\mathbb{R}.$

Proof..

The condition $\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)=0$ implies

(8.1)		$\displaystyle 0=$	$\displaystyle J_{\mathbf{a}_{}}(A)+\overline{J_{\mathbf{a}_{}}(A)}=J_{\mathbf{a}_{}}(A)+J_{\overline{\mathbf{a}_{}}}(\overline{A})=J_{\mathbf{a}_{}}(A)-J_{\mathbf{a}_{}}(\overline{A})=4(A-\overline{A})I_{0},$
	$\displaystyle I_{0}:=$	$\displaystyle\int_{\mathbf{a}_{*}}\frac{d\zeta}{v(A,\zeta)+v(\overline{A},\zeta)}.$

Let us derive $A\in\mathbb{R}$ by supposing the contrary $A-\overline{A}\not=0.$ Then, by Lemma 8.1, it may be supposed that $\mathrm{Re\,}\zeta_{5}<\mathrm{Re\,}\zeta_{3}<\mathrm{Re\,}\zeta_{1}$ , which is divided into two cases according to the location of $0.$

(i) Case $\mathrm{Re\,}\zeta_{3}\leq 0$ . It is sufficient to show that $I_{0}\not=0$ in (8.1). The algebraic function $v(A,\zeta)+v(\overline{A},\zeta)$ is considered on the two sheeted Riemann surface $\Pi_{A}^{*0}$ glued along the cuts $[\zeta_{1},0]$ , $[\overline{\zeta_{1}},0]$ , $[\zeta_{5},\zeta_{3}]$ , $[\overline{\zeta_{5}},\overline{\zeta_{3}}],$ which is constructed by adding to $\Pi^{*}_{A}$ the new two cuts $[\overline{\zeta_{1}},0]$ , $[\overline{\zeta_{5}},\overline{\zeta_{3}}]$ and gluing along them. Choose the cycle $\mathbf{a}_{*}$ on $\Pi_{A}^{*0}$ in such a way that $\mathbf{a}_{*}$ surrounds the cuts $[\zeta_{5},\zeta_{3}]$ and $[\overline{\zeta_{5}},\overline{\zeta_{3}}]$ as in Figure 8.2, (a), and modify $\mathbf{a}_{*}$ as in Figure 8.2, (b), where $\zeta_{5}=a+ib$ , $\zeta_{3}=c+ib^{\prime}$ , $a<c\leq 0$ , $b,b^{\prime}\in\mathbb{R}.$

Figure 8.2. Cycle

\mathbf{a}_{*}

and modification in the case

\mathrm{Re\,}\zeta_{3}\leq 0

To simplify the description we write $v_{A}(\zeta)=v(A,\zeta),$ $v_{\overline{A}}(\zeta)=v(\overline{A},\zeta)$ and $(v_{A}\pm v_{\overline{A}})(\zeta)=v(A,\zeta)\pm v(\overline{A},\zeta)$ . Let us take the contour of $I_{0}$ to be the modified $\mathbf{a}_{*}$ starting from and terminating in the point $\zeta=\mathrm{Re\,}\zeta_{5}+=a+$ on the upper shore of the cut $[a,c]$ . Then $I_{0}$ is decomposed into the left- and right-vertical, and the horizontal parts:

	$\displaystyle I_{0}=$	$\displaystyle I_{\mathrm{hor}}+I_{\mathrm{right\text{-}v}}+I_{\mathrm{left\text{-}v}},$
	$\displaystyle I_{\mathrm{hor}}=$	$\displaystyle 2\int^{c}_{a}\frac{ds}{(v_{A}+v_{\overline{A}})(s)},$
	$\displaystyle I_{\mathrm{right\text{-}v}}=$	$\displaystyle\int_{0}^{b^{\prime}}\frac{idt}{(v_{A}+v_{\overline{A}})(c+it)}+\int_{b^{\prime}}^{-b^{\prime}}\frac{idt}{(-v_{A}+v_{\overline{A}})(c+it)}+\int^{0}_{-b^{\prime}}\frac{idt}{(-v_{A}-v_{\overline{A}})(c+it)},$
	$\displaystyle-I_{\mathrm{left\text{-}v}}=$	$\displaystyle\int_{0}^{-b}\frac{idt}{(v_{A}+v_{\overline{A}})(a+it)}+\int_{-b}^{b}\frac{idt}{(v_{A}-v_{\overline{A}})(a+it)}+\int_{b}^{0}\frac{idt}{(-v_{A}-v_{\overline{A}})(a+it)}.$

Here

	$\displaystyle I_{\mathrm{right\text{-}v}}=$	$\displaystyle\frac{i}{4(A-\overline{A})}\biggl{(}\int_{0}^{b^{\prime}}\frac{(v_{A}-v_{\overline{A}})(c+it)}{c+it}dt+\int_{0}^{b^{\prime}}\frac{(v_{A}+v_{\overline{A}})(c+it)}{c+it}dt\biggr{)}$
		$\displaystyle-\frac{i}{4(A-\overline{A})}\biggl{(}\int_{0}^{-b^{\prime}}\frac{(v_{A}+v_{\overline{A}})(c+it)}{c+it}dt-\int_{0}^{-b^{\prime}}\frac{(v_{A}-v_{\overline{A}})(c+it)}{c+it}dt\biggr{)}$
	$\displaystyle=$	$\displaystyle\frac{i}{2(A-\overline{A})}\biggl{(}\int_{0}^{b^{\prime}}\frac{v_{A}(c+it)}{c+it}dt+\int_{0}^{b^{\prime}}\frac{v_{\overline{A}}(c-it)}{c-it}dt\biggr{)}\in\mathbb{R},$

and similarly $I_{\mathrm{left\text{-}v}}\in\mathbb{R}.$ Let $A=\kappa+i\kappa^{\prime}$ with $\kappa^{\prime}\not=0.$ If, say $\mathrm{Im\,}\zeta_{5}>0$ , along the upper shore of the cut $[a,c]$ , $v_{A}(s)=e^{-\pi i/2}\sqrt{-\varphi(s)-4i\kappa^{\prime}s}$ , $v_{\overline{A}}(s)=-e^{\pi i/2}\sqrt{-\varphi(s)+4i\kappa^{\prime}s}$ , where $\varphi(s)=s^{4}+4s^{3}+4s^{2}+4\kappa s<0$ , i.e. $-\varphi(s)>0$ for $a<s<c\leq 0.$ Then the horizontal integral along the upper shore is

\int^{c}_{a}\frac{ds}{(v_{A}+v_{\overline{A}})(s)}=\frac{i}{\sqrt{2}}\int^{c}_{a}\frac{ds}{\sqrt{-\varphi(s)+\sqrt{\varphi(s)^{2}+16(\kappa^{\prime})^{2}s^{2}}}}=i\gamma_{0}

with some $\gamma_{0}>0$ . Thus we have $I_{\mathrm{hor}}\in\mathbb{C}\setminus\mathbb{R},$ and hence $I_{0}\not=0,$ which contradicts (8.1). This implies $A\in\mathbb{R}$ in this case.

(ii) Case $\mathrm{Re\,}\zeta_{3}>0$ . Let $\Pi^{**}_{A}=\Pi^{**}_{+}\cup\Pi^{**}_{-}$ be a two sheeted Riemann surface glued along the cuts $[\zeta_{3},\zeta_{1}]$ , $[\zeta_{5},0]$ with $\Pi^{**}_{\pm}=P^{1}(\mathbb{C})\setminus([\zeta_{3},\zeta_{1}]\cup[\zeta_{5},0])$ . Instead of $\Pi^{*}_{A}$ and $\mathbf{a}_{*}$ of the case (i) we treat $\Pi^{**}_{A}$ and a cycle on it. Draw a closed curve $\mathbf{b}_{0}$ on the upper sheet $\Pi^{**}_{+}\subset\Pi^{**}_{A}$ surrounding the cut $[\zeta_{3},\zeta_{1}]$ clockwise as in Figure 8.3.

Figure 8.3. Cycle

\mathbf{b}_{0}

\Pi_{A}^{**}

in the case

\mathrm{Re\,}\zeta_{3}>0

Consider the algebraic function $v^{*}(A,\zeta)=\sqrt{\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A\zeta}$ on $\Pi^{**}_{A}$ , where the branch is chosen in such a way that $v^{*}(A,\zeta)$ coincides with $v(A,\zeta)$ along the upper shore of the cut $[\zeta_{3},\zeta_{1}]$ . The cycle $\mathbf{b}_{0}$ on $\Pi^{**}_{A}$ is a substitute for $\mathbf{b}_{*}$ on $\Pi_{A}^{*}.$ Indeed

J_{\mathbf{b}_{0}}^{*}(A):=\int_{\mathbf{b}_{0}}\frac{v^{*}(A,\zeta)}{\zeta}d\zeta=2\int^{\zeta_{1}}_{\zeta_{3}}\frac{v(A,\zeta)}{\zeta}d\zeta=J_{\mathbf{b}_{*}}(A).

Hence $\mathrm{Re\,}J_{\mathbf{b}_{*}}(A)=0$ is equivalent to $\mathrm{Re\,}J^{*}_{\mathbf{b}_{0}}(A)=0.$ Under this supposition,

	$\displaystyle 0=$	$\displaystyle J^{}_{\mathbf{b}_{0}}(A)+\overline{J^{}_{\mathbf{b}_{0}}(A)}=J^{}_{\mathbf{b}_{0}}(A)+J^{}_{\overline{\mathbf{b}_{0}}}(\overline{A})=J^{}_{\mathbf{b}_{0}}(A)-J^{}_{\mathbf{b}_{0}}(\overline{A})=4(A-\overline{A})I_{0}^{*},$
	$\displaystyle I_{0}^{*}:=$	$\displaystyle\int_{\mathbf{b}_{0}}\frac{d\zeta}{v^{}(A,\zeta)+v^{}(\overline{A},\zeta)}$

instead of (8.1). As in the case (i), to prove $A\in\mathbb{R}$ it is sufficient to show $I_{0}^{*}\not=0.$ The algebraic function $v^{*}(A,\zeta)+v^{*}(\overline{A},\zeta)$ is considered on the two sheeted Riemann surface $\Pi_{A}^{**0}$ glued along the cuts $[\zeta_{3},\zeta_{1}]$ , $[\overline{\zeta_{3}},\overline{\zeta_{1}}]$ , $[\zeta_{5},0]$ , $[\overline{\zeta_{5}},0]$ , and the cycle $\mathbf{b}_{0}$ may be taken in such a way that $\mathbf{b}_{0}$ surrounds the cuts $[\zeta_{3},\zeta_{1}]$ and $[\overline{\zeta_{3}},\overline{\zeta_{1}}]$ as in Figure 8.4, (a).

Figure 8.4. Modification of

\mathbf{b}_{0}

in the case

\mathrm{Re}\,\zeta_{3}>0

The cycle $\mathbf{b}_{0}$ is modified as in Figure 8.4, (b), in which $\zeta_{1}=a+ib,$ $\zeta_{3}=c+ib^{\prime}$ , $a>c>0,$ $b,b^{\prime}\in\mathbb{R}.$ The contour of the integral $I^{*}_{0}$ is taken to be the modified $\mathbf{b}_{0}$ that starts from and terminates in the point $\zeta=\mathrm{Re\,}\zeta_{3}+=c+$ on the upper shore of the cut $[c,a].$ Dividing $I^{*}_{0}$ into the horizontal, and the left- and right-vertical parts, we may show $I^{*}_{0}\not=0$ by the same arguments as in the case (i). Thus the proposition is proved. ∎

As in Example 8.1, $\mathrm{Re\,}J_{\mathbf{a}_{*}}(\tfrac{8}{27})=\mathrm{Re\,}J_{\mathbf{b}_{*}}(\tfrac{8}{27})=0$ . Let us examine $\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)$ or $\mathrm{Re\,}J_{\mathbf{b}_{*}}(A)$ for $A\in\mathbb{R}\setminus\{\tfrac{8}{27}\}.$

Consider the polynomial $f(\zeta)=v(A,\zeta)^{2}=\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A\zeta$ for $A\in\mathbb{R}$ on the real line $(-\infty,+\infty)\subset\Pi_{+}^{*}.$ The zeros of $f(\zeta)$ are located as follows, in which (z.3) is given by Lemma 8.1 and (z.4) is treated in Example 8.1:

(z.1) if $A<0$ , then $0<\zeta_{1},$ and $\zeta_{5},\zeta_{3}\not\in\mathbb{R};$

(z.2) if $A=0$ , then $\zeta_{5}=\zeta_{3}<\zeta_{1}=0;$

(z.3) if $0<A<\tfrac{8}{27}$ , then $\zeta_{5}<\zeta_{3}<\zeta_{1}<0;$

(z.4) if $A=\tfrac{8}{27}$ , then $\zeta_{5}<\zeta_{3}=\zeta_{1}<0;$

(z.5) if $A>\tfrac{8}{27}$ , then $\zeta_{5}<0$ , and $\zeta_{3},\zeta_{1}\not\in\mathbb{R}.$

Figure 8.5.

f(\zeta)

(-\infty,+\infty)

Let us show that $\mathrm{Re\,}J_{\mathbf{a}^{*}}(A)\not=0$ or $\mathrm{Re\,}J_{\mathbf{b}^{*}}(A)\not=0$ for $A\in\mathbb{R}\setminus\{\tfrac{8}{27}\}.$ In each case, $f(\zeta)$ behaves as in Figure 8.5: say, in the case (z.3), $f(\zeta)>0$ on $(-\infty,\zeta_{5})\cup(\zeta_{3},\zeta_{1})\cup(0,+\infty)$ , and $f(\zeta)<0$ on $(\zeta_{5},\zeta_{3})\cup(\zeta_{1},0)$ . It is easy to see that, in (z.2) or (z.3),

J_{\mathbf{b}_{*}}(A)=2\int_{\zeta_{3}}^{\zeta_{1}}\frac{v(A,\zeta)}{\zeta}d\zeta=2\int_{\zeta_{3}}^{\zeta_{1}}\frac{-\sqrt{f(\zeta)}}{\zeta}d\zeta<0,

and that, in (z.2), $J_{\mathbf{a}_{*}}(0)=0.$ In (z.1), we write $\zeta_{5}=c+ib,$ $\zeta_{3}=c-ib$ with $c<0,$ $b>0$ such that $2c+\zeta_{1}=-4.$ Then

J_{\mathbf{b}_{*}}(A)=2\int_{\zeta_{3}}^{\zeta_{1}}\frac{v(A,\zeta)}{\zeta}d\zeta=2\int_{c}^{0}\frac{-\sqrt{f(\zeta)}}{\zeta}d\zeta<0.

It remains to discuss the case (z.5). Set $\zeta_{1}=c-ib$ and $\zeta_{3}=c+ib$ with $b>0.$ The relation $2c+\zeta_{5}=-4$ yields $A=2(c+2)(c+1)^{2}$ and $b=\sqrt{(3c+2)(c+2)}.$ Then, according to the location of $c$ we consider two cases: (z.5a) $-\tfrac{2}{3}<c<0$ if $\tfrac{8}{27}<A<4$ ; and (z.5b) $c\geq 0$ if $A\geq 4$ . First consider the case (z.5a) as in Figure 8.5. Note that

\frac{\partial}{\partial A}\int^{\zeta_{1}}_{0}\frac{v(A,\zeta)}{\zeta}d\zeta=2\int^{\zeta_{1}}_{0}\frac{d\zeta}{v(A,\zeta)}+(\zeta_{1})_{A}\frac{v(A,\zeta_{1})}{\zeta_{1}}=2\int^{\zeta_{1}}_{0}\frac{d\zeta}{v(A,\zeta)},

since $v(A,\zeta_{1})=0.$ Then, by $f(\zeta)<0$ on $(c,0)$ ,

\frac{\partial}{\partial A}\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)=2\frac{\partial}{\partial A}\mathrm{Re\,}\int^{\zeta_{1}}_{0}\frac{v(A,\zeta)}{\zeta}d\zeta=4\mathrm{Re\,}\int^{\zeta_{1}}_{0}\frac{d\zeta}{v(A,\zeta)}=4\mathrm{Re\,}\int^{\zeta_{1}}_{c}\frac{d\zeta}{v(A,\zeta)},

in which the contour is taken along the upper shore of the cut $[0,\zeta_{1}].$ Using

	$\displaystyle f(c+it)=g(t)+ih(t),$
	$\displaystyle g(t)=t^{4}-2(3c^{2}+6c+2)t^{2}+f(c),$
	$\displaystyle h(t)=-4(c+1)t^{3}+4(c+1)(c+2)(3c+2)t$

fulfilling $g(t)<0$ and $h(t)<0$ for $-\sqrt{(3c+2)(c+2)}<t<0$ , and $v(A,c+it)=e^{\pi i/2}\sqrt{{p}(t)+i{q}(t)}$ with ${p}(t)=-g(t)$ , ${q}(t)=-h(t)$ , we have, for $\tfrac{8}{27}<A<4,$

\frac{\partial}{\partial A}\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)=4\mathrm{Re\,}\int^{-b}_{0}\frac{dt}{\sqrt{{p}(t)+i{q}(t)}}=-2\sqrt{2}\int^{0}_{-b}\frac{\sqrt{\sqrt{{p}(t)^{2}+{q}(t)^{2}}+{p}(t)}}{\sqrt{{p}(t)^{2}+{q}(t)^{2}}}dt<0.

From this combined with $\mathrm{Re\,}J_{\mathbf{a}_{*}}(\tfrac{8}{27})=0$ , it follows that $\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)<0$ in the case (z.5a).

In the case (z.5b) in Figure 8.5, $f(\zeta)>0$ on $(0,c)$ , and hence

	$\displaystyle-\frac{\partial}{\partial A}\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)=$	$\displaystyle 2\frac{\partial}{\partial A}\mathrm{Re\,}\int^{\zeta_{1}}_{0}\frac{v(A,\zeta)}{\zeta}d\zeta=4\mathrm{Re\,}\int^{\zeta_{1}}_{0}\frac{d\zeta}{v(A,\zeta)}=4\mathrm{Re\,}\int^{\zeta_{1}}_{c}\frac{d\zeta}{v(A,\zeta)}+I(c),$
	$\displaystyle I(c)=$	$\displaystyle 4\int_{0}^{c}\frac{d\zeta}{v(A,\zeta)}\geq 0,$

in which the contour is taken along the upper shore of the cut $[\zeta_{1},0]$ . Note that $g(t)>0$ for $-\sqrt{c(3c+4)}<t<0$ and $h(t)<0$ for $-\sqrt{(3c+2)(c+2)}<t<0$ , and set $q(t)=-h(t).$ Then

	$\displaystyle\mathrm{Re\,}\int^{\zeta_{1}}_{c}\frac{d\zeta}{v(A,\zeta)}$	$\displaystyle=\mathrm{Re\,}\int^{-b}_{0}\frac{idt}{v(A,c+it)}=\mathrm{Im\,}\int^{0}_{-b}\frac{dt}{\sqrt{g(t)-iq(t)}}$
		$\displaystyle=\frac{1}{\sqrt{2}}\int_{-b}^{0}\frac{\sqrt{\sqrt{g(t)^{2}+q(t)^{2}}-g(t)}}{\sqrt{g(t)^{2}+q(t)^{2}}}dt>0,$

which implies $(\partial/\partial A)\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)<0$ for $A\geq 4.$ This fact combined with $\mathrm{Re\,}J_{\mathbf{a}_{*}}(4)<0$ leads to $\mathrm{Re\,}J_{\mathbf{a}_{*}}(A)<0$ for $A\geq 4.$

Thus we have shown the following uniqueness.

Proposition 8.4.

The Boutroux equations $(\mathrm{BE})_{\phi=0}$ admit a unique solution $A_{0}=\tfrac{8}{27}$ .

Corollary 8.5.

For every $A\in\mathbb{C}$ , $(J_{\mathbf{a}_{*}}(A),J_{\mathbf{b}_{*}}(A))\not=(0,0).$

Proposition 8.6.

Suppose that $0<|\phi|\leq\pi/4.$ Let $A_{\phi}$ solve $(\mathrm{BE})_{\phi}$ , and let the elliptic curve $v^{2}=\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A_{\phi}\zeta$ be degenerate. Then $\phi=\pm\pi/4$ and $A_{\pm\pi/4}=0.$

Proof..

The degeneration occurs only when $A_{\phi}=0,$ $\tfrac{8}{27}$ . By Example 8.1 $A=\tfrac{8}{27}$ does not solve $(\mathrm{BE})_{\phi}$ for $0<|\phi|\leq\pi/4.$ As shown in the case (z.2) above $J_{\mathbf{a}_{*}}(0)=0$ and $J_{\mathbf{b}_{*}}(0)<0$ , which implies $A=0$ solves $(\mathrm{BE})_{\phi}$ only when $\phi=\pm\pi/4.$ This completes the proof. ∎

8.3. Trajectory

The ratio $I(A)={J_{\mathbf{b}_{*}}(A)}/{J_{\mathbf{a}_{*}}(A)}$ by Novokshenov [24, Appendix I] is useful in examining the Boutroux equations. The following is easily verified.

Proposition 8.7.

$(1)$ If $I(A)\in\mathbb{R}$ , then $A$ solves $(\mathrm{BE})_{\phi}$ for some $\phi\in\mathbb{R}.$

$(2)$ If $A$ solves $(\mathrm{BE})_{\phi}$ for some $\phi\in\mathbb{R}$ and $J_{\mathbf{a}_{*}}(A)\not=0,$ then $I(A)\in\mathbb{R}.$

Remark 8.1.

By Corollary 8.5 if $J_{\mathbf{a}_{*}}(A)=0$ then $I(A)^{-1}={J_{\mathbf{a}_{*}}(A)}/{J_{\mathbf{b}_{*}}(A)}=0$ , and around such a point $I(A)^{-1}$ has the same property as in Proposition 8.7.

Let $D_{0}=\{A\in\mathbb{C}\,|\,\text{$A$ solves $(\mathrm{BE})_{\phi}$ for some $\phi$}\}$ .

Proposition 8.8.

The set $D_{0}$ is bounded.

Proof..

Let us calculate $\lim_{A\to\infty}I(A)$ . The zeros of $\zeta^{3}+4\zeta^{2}+4\zeta+4A=0$ are asymptotically expressed as $\zeta_{j}\sim e^{(1-2j)\pi i/3}(4A)^{1/3}$ $(j=1,3,5)$ as $A\to\infty$ . Then, by $\zeta=e^{\pi i/3}(4A)^{1/3}z$ ,

	$\displaystyle J_{\mathbf{a}_{*}}(A)=$	$\displaystyle 2\int^{\zeta_{3}}_{\zeta_{5}}\frac{v(A,\zeta)}{\zeta}d\zeta\sim 2e^{2\pi i/3}(4A)^{2/3}\int^{1}_{e^{2\pi i/3}}\frac{\sqrt{z^{4}-z}}{z}dz,$
	$\displaystyle J_{\mathbf{b}_{*}}(A)=$	$\displaystyle 2\int^{\zeta_{1}}_{\zeta_{3}}\frac{v(A,\zeta)}{\zeta}d\zeta\sim 2e^{2\pi i/3}(4A)^{2/3}\int_{1}^{e^{-2\pi i/3}}\frac{\sqrt{z^{4}-z}}{z}dz.$

Since

\int\frac{\sqrt{z^{4}-z}}{z}dz=\frac{1}{2}\sqrt{z^{4}-z}-\frac{3}{4}\int\frac{dz}{\sqrt{z^{4}-z}},

$\lim_{A\to\infty}I(A)$ is a ratio of periods of the elliptic curve $v^{2}=z^{4}-z,$ which implies $\mathrm{Im\,}\lim_{A\to\infty}I(A)\not=0.$ By Proposition 8.7 the set $D_{0}$ is bounded. ∎

Recall the periods $\Omega^{*}_{\mathbf{a}_{*}}=\int_{\mathbf{a}_{*}}v(A,\zeta)^{-1}d\zeta$ and $\Omega^{*}_{\mathbf{b}_{*}}=\int_{\mathbf{b}_{*}}v(A,\zeta)^{-1}d\zeta$ of $\Pi^{*}_{A}$ given in Section 2.2. To examine the conformality of $I(A)$ we need the following (see also [31, Lemma 6.1]).

Lemma 8.9.

$\Omega^{*}_{\mathbf{b}_{*}}J_{\mathbf{a}_{*}}(A)-\Omega^{*}_{\mathbf{a}_{*}}J_{\mathbf{b}_{*}}(A)=-4\pi i.$

Proof..

Let $\zeta(t)$ be the elliptic function given by $(\zeta^{\prime})^{2}=v(A,\zeta)^{2}=\zeta^{4}+4\zeta^{3}+4\zeta^{2}+4A\zeta,$ and let $P,Q,R,S$ be the vertices of its period parallelogram such that $Q=P+\Omega^{*}_{\mathbf{a}_{*}}$ and $S=P+\Omega^{*}_{\mathbf{b}_{*}}$ . Then the function ${g}(u)=\int^{u}_{P}{\zeta^{\prime}(t)^{2}}{\zeta(t)^{-1}}dt$ fulfils

{g}(u+\Omega^{*}_{\mathbf{b}_{*}})-g(u)=\int^{u+\Omega^{*}_{\mathbf{b}_{*}}}_{u}\frac{\zeta^{\prime}(t)^{2}}{\zeta(t)}dt=\int_{\mathbf{b}_{*}}\frac{v(A,\zeta)}{\zeta}d\zeta=J_{\mathbf{b}_{*}}(A),

and hence

\biggl{(}\int_{P}^{Q}+\int_{R}^{S}\biggr{)}g(u)du=\int_{P}^{Q}(g(u)-g(u+\Omega^{*}_{\mathbf{b}_{*}}))du=\int_{P}^{Q}(-J_{\mathbf{b}_{*}}(A))du=-\Omega^{*}_{\mathbf{a}_{*}}J_{\mathbf{b}_{*}}(A).

Combining this with $(\int_{Q}^{R}+\int_{S}^{P})g(u)du=\Omega^{*}_{\mathbf{b}_{*}}J_{\mathbf{a}_{*}}(A)$ similarly obtained, we have

\Omega^{*}_{\mathbf{b}_{*}}J_{\mathbf{a}_{*}}(A)-\Omega^{*}_{\mathbf{a}_{*}}J_{\mathbf{b}_{*}}(A)=\int_{(PQRSP)}g(u)du=2\pi i(\mathrm{Res}(u_{\infty}^{-})+\mathrm{Res}(u_{\infty}^{+}))=-4\pi i,

where $u_{\infty}^{\pm}$ are poles of $g(u)$ in the periodic parallelogram. ∎

As an immediate corollary for $\Omega_{\mathbf{a},\,\mathbf{b}}$ and $\mathcal{J}_{\mathbf{a},\,\mathbf{b}}$ (cf. Section 6.3) we have the following.

Corollary 8.10.

$\Omega_{\mathbf{b}}\mathcal{J}_{\mathbf{a}}-\Omega_{\mathbf{a}}\mathcal{J}_{\mathbf{b}}=-4\pi ie^{i\phi}.$

Observing that $(\partial/\partial A)J_{\mathbf{a}_{*},\,\mathbf{b}_{*}}(A)=2\Omega^{*}_{\mathbf{a}_{*},\,\mathbf{b}_{*}},$ we have the following.

Proposition 8.11.

$I^{\prime}(A)=-8\pi iJ_{\mathbf{a}_{*}}(A)^{-2}$ and $(1/I)^{\prime}(A)=8\pi iJ_{\mathbf{b}_{*}}(A)^{-2}$ .

Remark 8.2.

By Example 8.1 and the proposition above, $I(A)$ is conformal around $A=A_{0}=\tfrac{8}{27}$ , and given by

I(A)=\tfrac{3}{2}\pi i(A-\tfrac{8}{27})(1+o(1)).

By Proposition 8.7 the inverse image of the interval $(-\varepsilon,\epsilon)\subset\mathbb{R}$ under $I(A)$ is a local trajectory consisting of $A_{\phi}$ for $|\phi|<\varepsilon_{0}$ , each $A_{\phi}$ solving (BE)_ϕ, where $\varepsilon$ and $\varepsilon_{0}$ are sufficiently small. This local trajectory is expressed as

A_{\phi}=\tfrac{8}{27}+i\rho(\phi),\quad\tfrac{3}{2}\pi\mathrm{Re\,}\rho(\phi)\in(-\varepsilon,\varepsilon),\quad\text{$\mathrm{Im\,}\rho(\phi)=o(\mathrm{Re\,}\rho(\phi))$ as $\phi\to 0$.}

Similarly, there exists a local trajectory for $|\phi\pm\pi/4|<\varepsilon_{0}$ (cf. Proof of Proposition 8.6).

Suppose that $0<|\phi|<\pi/4$ . Write

J_{\mathbf{a}_{*}}(A)=u(A)+iv(A),\quad J_{\mathbf{b}_{*}}(A)=U(A)+iV(A),\quad A=x+iy.

Then $A$ solves (BE)_ϕ if and only if

(8.2)

u(A)-v(A)\tan 2\phi=0,\quad U(A)-V(A)\tan 2\phi=0,

which define the trajectory of $A_{\phi}$ . The Jacobian for (8.2) around $A=A_{\phi}$ is

	$\displaystyle\det J(x,y)=$	$\displaystyle\det\begin{pmatrix}u_{x}-v_{x}\tan 2\phi&u_{y}-v_{y}\tan 2\phi\\ U_{x}-V_{x}\tan 2\phi&U_{y}-V_{y}\tan 2\phi\end{pmatrix}=(1+\tan^{2}2\phi)(v_{x}V_{y}-v_{y}V_{x})$
	$\displaystyle=$	$\displaystyle-4(1+\tan^{2}2\phi)\mathrm{Im\,}\overline{\Omega^{}_{\mathbf{a}_{}}}\Omega^{}_{\mathbf{b}_{}}=-4(1+\tan^{2}2\phi)\|\Omega^{}_{\mathbf{a}_{}}\|^{2}\mathrm{Im\,}\frac{\Omega^{}_{\mathbf{a}_{}}}{\Omega^{}_{\mathbf{b}_{}}}\not=0$

by Proposition 8.6. This fact implies that a given local trajectory for $|\phi-\phi_{0}|<\varepsilon_{0}$ with $|\phi_{0}|<\pi/4$ is extended smoothly for $0<|\phi|<\pi/4$ and continuously for $|\phi|\leq\pi/4$ , and so are the trajectories described in Remark 8.2. For such an extended trajectory, which is bounded by Proposition 8.8, let $A_{\phi_{n}}$ with $\phi_{n}\to 0$ be a given sequence. By the boundedness there exists a subsequence convergent to some $A_{0}^{*}\in\mathbb{C}$ solving (BE)_ϕ=0, and then $A_{0}^{*}=\tfrac{8}{27},$ which implies the uniqueness of the trajectory for $|\phi|\leq\pi/4.$ Thus we have the following.

Proposition 8.12.

There exists a trajectory $A=A_{\phi}$ for $|\phi|\leq\pi/4$ with the properties $:$

$(1)$ for each $\phi$ , $A_{\phi}$ is a unique solution of $(\mathrm{BE})_{\phi};$

$(2)$ $A_{\phi}$ is smooth in $\phi$ for $0<|\phi|<\pi/4,$ and continuous in $\phi$ for $|\phi|\leq\pi/4;$

$(3)$ $A_{0}=\tfrac{8}{27},$ $A_{\pm\pi/4}=0.$

For any $\phi\in\mathbb{R}$ we have

e^{2i\phi}J_{\mathbf{a}_{*},\,\mathbf{b}_{*}}(A_{\phi})=-e^{2i(\phi\pm\pi/2)}J_{\mathbf{a}_{*},\,\mathbf{b}_{*}}(A_{\phi}),\quad\overline{e^{2i\phi}J_{\mathbf{a}_{*},\,\mathbf{b}_{*}}(A_{\phi})}=e^{-2i\phi}J_{\overline{\mathbf{a}_{*}},\,\overline{\mathbf{b}_{*}}}(\overline{A_{\phi}}),

which leads to the following.

Proposition 8.13.

$A_{\phi\pm\pi/2}=A_{\phi}$ and $A_{-\phi}=\overline{A_{\phi}}.$

To know the shape of the trajectory $A=A_{\phi}$ it is sufficient to examine it for $|\phi|\leq\pi/4$ . For $0<|\phi|<\pi/4$ the derivative of (8.2) along $A=A_{\phi}$ with respect to $t=\tan 2\phi$ is written in the form $J(x,y){}^{\mathrm{T}}\!(x^{\prime}(t),y^{\prime}(t))-{}^{\mathrm{T}}\!(v(A_{\phi}),V(A_{\phi}))=\mathbf{o}$ with $A_{\phi}=x(t)+iy(t),$ where $J(x,y)$ is the Jacobi matrix above. Then, for $0<|\phi|<\pi/4$

(8.3)

(x^{\prime}(t),y^{\prime}(t))\not=(0,0),\quad(d/d\phi)A_{\phi}=2(x^{\prime}(t)+iy^{\prime}(t))\cos^{-2}2\phi\not=0.

By Propositions 8.7 and 8.11, for every $\phi$ such that $|\phi|\leq\pi/4$ ,

\frac{d}{dt}I(A_{\phi})=-\frac{8\pi i(x^{\prime}(t)+iy^{\prime}(t))}{J_{\mathbf{a}_{*}}(A_{\phi})^{2}}\,\,\,\text{or}\,\,\,\frac{d}{dt}(1/I)(A_{\phi})=\frac{8\pi i(x^{\prime}(t)+iy^{\prime}(t))}{J_{\mathbf{b}_{*}}(A_{\phi})^{2}}\in\mathbb{R},

where $(J_{\mathbf{a}_{*}}(A_{\phi}),J_{\mathbf{b}_{*}}(A_{\phi}))\not=(0,0)$ by Corollary 8.5. Setting $J_{\mathbf{a}_{*}}(A_{\phi})^{-1}$ or $J_{\mathbf{b}_{*}}(A_{\phi})^{-1}=P+iQ=i(Q-iP)$ with $P^{2}+Q^{2}>0$ , and observing $\mathrm{Im\,}(d/dt)I(A_{\phi})^{\pm 1}=0$ , we have

x^{\prime}(t)(P^{2}-Q^{2})-2y^{\prime}(t)PQ=0.

For $0<|\phi|<\pi/4,$ by (8.3), $x^{\prime}(t)\not=0$ , and if $y^{\prime}(t)=0,$ then $P=\pm Q$ , implying $2\phi=\mp\pi/4.$ For $-\pi/4<\phi<0$ , $x^{\prime}(t)>0$ , and for $0<\phi<\pi/4$ , $x^{\prime}(t)<0$ , since $A_{\pm\pi/4}=0$ and $A_{0}=\tfrac{8}{27}$ . If $-\pi/8<\phi<0$ (respectively, $0<\phi<\pi/8$ ) then $y^{\prime}(t)<0$ , since $|P|<|Q|$ , $PQ>0$ (respectively, $PQ<0$ ).

Proposition 8.14.

Let $A_{\phi}=x(t)+iy(t)$ with $t=\tan 2\phi.$ Then, for $|\phi|\leq\pi/4,$

$(1)$ $x^{\prime}(t)>0$ for $-\pi/4<\phi<0,$ $x^{\prime}(t)<0$ for $0<\phi<\pi/4;$

$(2)$ $y^{\prime}(t)<0$ for $0<|\phi|<\pi/4,$ $y^{\prime}(t)>0$ for $\pi/4<|\phi|<\pi/2;$

$(3)$ $x^{\prime}(0)=x^{\prime}(\pm\tan(\pi/2))=0,$ $y^{\prime}(\pm\tan(\pi/4))=0.$

By this proposition and Remark 8.2 the trajectory $A_{\phi}$ for $|\phi|\leq\pi/4$ is roughly drawn as in Figure 8.6.

Figure 8.6. Trajectory

A_{\phi}

for

|\phi|\leq\pi/4

Proposition 8.15.

There exists a trajectory $A=A_{\phi}$ for $\phi\in\mathbb{R}$ with the properties $:$

$(1)$ for each $\phi$ , $A_{\phi}$ is a unique solution of $(\mathrm{BE})_{\phi};$

$(2)$ $A_{\phi+\pi/2}=A_{\phi},$ $A_{-\phi}=\overline{A_{\phi}};$

$(3)$ $A_{0}=\tfrac{8}{27},$ $A_{\pm\pi/4}=0$ and $0\leq\mathrm{Re\,}A_{\phi}\leq\tfrac{8}{27};$

$(4)$ $A_{\phi}$ is continuous in $\phi\in\mathbb{R}$ , and smooth in $\phi\in\mathbb{R}\setminus\{m\pi/4\,|\,m\in\mathbb{Z}\}.$

As in the derivation of Corollary 8.2 by the change of variables $\zeta=e^{-\phi}z$ Proposition 8.15 is converted to the results on the trajectory for the Boutroux equations (2.2) with the cycles $\mathbf{a}$ and $\mathbf{b}$ on the elliptic curve $\Pi_{A,\phi}:$ $w(A,z)^{2}=z^{4}+4e^{i\phi}z^{3}+4e^{2i\phi}z^{2}+4e^{3i\phi}Az.$

Corollary 8.16.

The trajectory $A=A_{\phi}$ in Proposition 8.15 fulfils

$(1^{\prime})$ for each $\phi\in\mathbb{R}$ , $A_{\phi}$ is a unique solution of (2.2).

Remark 8.3.

In the corollary above $(1^{\prime})$ may be replaced with

$(1^{\prime\prime})$ for each $\phi$ , there exists $A_{\phi}$ uniquely such that, for every cycle $\mathbf{c}$ on $\Pi_{A_{\phi},\phi}$ ,

\mathrm{Re\,}\int_{\mathbf{c}}\frac{w(A_{\phi},z)}{z}dz=0.

8.4. Coalescing turning points

Let us observe coalescing turning points as an application of Proposition 8.14. On the trajectory $A_{\phi}$ , at $A_{\phi=0}=\tfrac{8}{27}$ and $A_{\phi=\pm\pi/4}=0$ , coalescences of, respectively, $\zeta_{3}$ and $\zeta_{1},$ and $\zeta_{5}$ and $\zeta_{3}$ occur, and then the elliptic curve $\Pi^{*}_{A}$ degenerates.

For $A_{\phi}=\tfrac{8}{27}+\varepsilon$ the zeros of $v(A_{\phi},\zeta)^{2}$ are denoted by $\zeta_{5}=-\tfrac{8}{3}+\delta_{5},$ $\zeta_{3,1}=-\tfrac{2}{3}+\delta_{3,1}$ , where $\delta$ and $\delta_{j}$ $(j=1,3,5)$ are small. Then $\delta_{3,1}=\pm i(2\varepsilon)^{1/2}+O(\varepsilon),$ $\delta_{5}=-\varepsilon+O(\varepsilon^{2}).$ By Proposition 8.14, $\varepsilon$ may be written in the form $2\varepsilon=e^{-i\theta(\phi)}\rho(\phi)(1+o(1))$ , where $\theta(\phi)=\pi/2$ if $\phi>0$ and $=-\pi/2$ if $\phi<0$ , and $\rho(0)=0$ and $\rho(\phi)>0$ for $|\phi|>0$ around $\phi=0$ . Hence $\zeta_{3,1}=-\tfrac{2}{3}(1\mp\tfrac{3}{2}ie^{-i\theta(\phi)/2}\rho(\phi)^{1/2}(1+o(1)))$ , $\zeta_{5}=-\tfrac{8}{3}(1+\tfrac{3}{16}e^{-i\theta(\phi)}\rho(\phi)(1+o(1)))$ as $\phi\to 0,$ where each of $\mp$ is chosen in such a way that $\mathrm{Re\,}\zeta_{3}<\mathrm{Re\,}\zeta_{1}.$ By $\lambda^{2}=z=e^{i\phi}\zeta$ , the turning points $\lambda_{j}$ $(j=1,3,5)$ in Section 4 have the following property.

Proposition 8.17.

Around $\phi=0$ ,

	$\displaystyle\lambda_{5}$	$\displaystyle=i\sqrt{\tfrac{8}{3}}+\hat{\varrho}(\phi)(1+o(1)),\quad$
	$\displaystyle\lambda_{3}$	$\displaystyle=i\sqrt{\tfrac{2}{3}}+e^{3\pi i/4}\varrho(\phi)(1+o(1)),\quad\lambda_{1}=i\sqrt{\tfrac{2}{3}}+e^{-\pi i/4}\varrho(\phi)(1+o(1))$

as $\phi\to 0+,$ and

	$\displaystyle\lambda_{5}$	$\displaystyle=i\sqrt{\tfrac{8}{3}}-\hat{\varrho}(\phi)(1+o(1)),\quad$
	$\displaystyle\lambda_{3}$	$\displaystyle=i\sqrt{\tfrac{2}{3}}+e^{\pi i/4}\varrho(\phi)(1+o(1)),\quad\lambda_{1}=i\sqrt{\tfrac{2}{3}}+e^{-3\pi i/4}\varrho(\phi)(1+o(1))$

as $\phi\to 0-,$ where $\varrho(0)=0$ , $\varrho(\phi)>0$ for $|\phi|>0$ , and $\hat{\varrho}(\phi)\asymp\varrho(\phi)^{2}.$

For $A_{\phi}=\varepsilon$ near $\phi=\pm\pi/4$ , the zeros of $v(A,\zeta)^{2}$ are $\zeta_{5,3}=-2+\delta_{5,3},$ $\zeta_{1}=\delta_{1}$ with $\delta_{5,3}=\pm(2\varepsilon)^{1/2}+O(\varepsilon),$ $\delta_{1}=-\varepsilon+O(\varepsilon^{2}).$ Writing $2\varepsilon=e^{i\theta(\phi^{\prime})}\tilde{\rho}(\theta^{\prime})$ , where $\phi=\pm\pi/4\mp\phi^{\prime}$ , $\phi^{\prime}\geq 0$ , and $\tilde{\rho}(0)=0,$ $\tilde{\rho}(\phi^{\prime})\geq 0$ , we have $\zeta_{5,3}=-2\pm e^{i\theta(\phi^{\prime})/2}\tilde{\rho}(\phi^{\prime})^{1/2}(1+o(1))$ , $\zeta_{1}=-\tfrac{1}{2}e^{i\theta(\phi^{\prime})}\tilde{\rho}(\phi^{\prime})(1+o(1))$ .

Proposition 8.18.

Around $\phi=\pm\pi/4$ ,

	$\displaystyle\lambda_{5}$	$\displaystyle=e^{-\pi i/8}(i\sqrt{2}+e^{3\pi i/4}\varrho(\phi^{\prime})(1+o(1))),\quad\lambda_{3}=e^{-\pi i/8}(i\sqrt{2}+e^{-\pi i/4}\varrho(\phi^{\prime})(1+o(1))),$
	$\displaystyle\lambda_{1}$	$\displaystyle=\tfrac{1}{2}e^{-\pi i/8}e^{3\pi i/4}{\varrho}(\phi^{\prime})(1+o(1))),\quad$

as $\phi^{\prime}=\phi+\pi/4\to 0+,$ and

	$\displaystyle\lambda_{5}$	$\displaystyle=e^{\pi i/8}(i\sqrt{2}+e^{\pi i/4}\varrho(\phi^{\prime})(1+o(1))),\quad\lambda_{3}=e^{\pi i/8}(i\sqrt{2}+e^{-3\pi i/4}\varrho(\phi^{\prime})(1+o(1))),$
	$\displaystyle\lambda_{1}$	$\displaystyle=\tfrac{1}{2}e^{\pi i/8}e^{\pi i/4}{\varrho}(\phi^{\prime})(1+o(1))),$

as $\phi^{\prime}=\phi-\pi/4\to 0-,$ where ${\varrho}(0)=0$ , and ${\varrho}(\phi^{\prime})>0$ for $|\phi^{\prime}|>0$ .

The quantities above are also written as functions of $\phi.$ We have the following, say, around $\phi=0.$

Proposition 8.19.

As $\phi\to 0-$ $($ respectively, $\phi\to 0+$ $)$ , $\zeta_{p}=-\tfrac{2}{3}-i^{p}\sqrt{2\varepsilon_{\phi}}+O(\varepsilon_{\phi})$ $($ respectively, $=-\tfrac{2}{3}+i^{p}\sqrt{2\varepsilon_{\phi}}+O(\varepsilon_{\phi})$ $)$ for $p=1,3$ , and $\zeta_{5}=-\tfrac{8}{3}-\varepsilon_{\phi}+O(\varepsilon_{\phi}^{2}),$ where $A_{\phi}=\tfrac{8}{27}+\varepsilon_{\phi}$ with

\varepsilon_{\phi}=\frac{8i}{3}\frac{\phi}{\ln\phi}\Bigl{(}1+O\Bigl{(}\frac{\ln\ln\phi}{\ln\phi}\Bigr{)}\Bigr{)}.

Proof..

In the case $\phi\to 0-,$ set $A_{\phi}=\tfrac{8}{27}+\varepsilon$ , and $\zeta_{1}=-\tfrac{2}{3}-i\sqrt{2\varepsilon}+O(\varepsilon),$ $\zeta_{3}=-\tfrac{2}{3}+i\sqrt{2\varepsilon}+O(\varepsilon),$ $\zeta_{5}=-\tfrac{8}{3}-\varepsilon+O(\varepsilon^{2}),$ where $\varepsilon=i|\varepsilon|(1+o(1))$ . Then

	$\displaystyle\tfrac{1}{2}J_{\mathbf{a}_{*}}(A_{\phi})$	$\displaystyle=i^{3}\int^{\zeta_{3}}_{\zeta_{5}}\sqrt{(\zeta-\zeta_{5})(\zeta_{3}-\zeta)(\zeta_{1}-\zeta)(-\zeta)}\frac{d\zeta}{\zeta}$
		$\displaystyle=-i\int^{-2/3+i\sqrt{2\varepsilon}}_{-8/3}\sqrt{(\zeta+\tfrac{8}{3})(-\tfrac{2}{3}+i\sqrt{2\varepsilon}-\zeta)(-\tfrac{2}{3}-i\sqrt{2\varepsilon}-\zeta)(-\zeta)}\frac{d\zeta}{\zeta}+O(\varepsilon)$
		$\displaystyle=i\int^{2+i\sqrt{2\varepsilon}}_{0}\sqrt{\frac{t}{\frac{8}{3}-t}}\sqrt{(2-t)^{2}+2\varepsilon}\,dt+O(\varepsilon)\qquad(\zeta+\tfrac{8}{3}=t)$
		$\displaystyle=i\int^{2+i\sqrt{2\varepsilon}}_{0}\sqrt{\frac{t}{\frac{8}{3}-t}}\Bigl{(}2-t+{\varepsilon}(2-t)^{-1}+O(\varepsilon^{2}(2-t)^{-3})\Bigr{)}dt+O(\varepsilon)$
		$\displaystyle=i(\tfrac{2}{3}\sqrt{3}-\tfrac{1}{2}\sqrt{3}\varepsilon\ln\varepsilon+O(\varepsilon)),$

and

	$\displaystyle\tfrac{1}{2}J_{\mathbf{b}_{*}}(A_{\phi})$	$\displaystyle=i^{2}\int^{\zeta_{1}}_{\zeta_{3}}\sqrt{(\zeta-\zeta_{5})(\zeta-\zeta_{3})(\zeta_{1}-\zeta)(-\zeta)}\frac{d\zeta}{\zeta}$
		$\displaystyle=-\int^{-2/3-i\sqrt{2\varepsilon}}_{-2/3+i\sqrt{2\varepsilon}}\sqrt{(\zeta+\tfrac{8}{3})(\zeta+\tfrac{2}{3}-i\sqrt{2\varepsilon})(-\tfrac{2}{3}-i\sqrt{2\varepsilon}-\zeta)(-\zeta)}\frac{d\zeta}{\zeta}+O(\varepsilon^{3/2})$
		$\displaystyle=\sqrt{3}\int^{-i\sqrt{2\varepsilon}}_{i\sqrt{2\varepsilon}}\sqrt{-2\varepsilon-t^{2}}\,dt+O(\varepsilon^{3/2})\qquad(\zeta+\tfrac{2}{3}=t)$
		$\displaystyle=-\sqrt{3}\pi+O(\varepsilon^{3/2}).$

Thus we have $J_{\mathbf{a}_{*}}(A_{\phi})=i\sqrt{3}(\tfrac{4}{3}-\varepsilon\ln\varepsilon+O(\varepsilon))$ and $J_{\mathbf{b}_{*}}(A_{\phi})=-2\sqrt{3}\pi\varepsilon+O(\varepsilon^{3/2})$ . For small $\phi$ , the Boutroux equations $\mathrm{Re\,}e^{2i\phi}J_{\mathbf{a}_{*}}(A_{\phi})=\mathrm{Re\,}r^{2i\phi}J_{\mathbf{b}_{*}}(A_{\phi})=0$ yield $\varepsilon=\varepsilon_{\phi}$ as in the proposition. ∎

By $2\Omega^{*}_{\mathbf{a}_{*},\,\mathbf{b}_{*}}=(\partial/\partial\varepsilon_{\phi})J_{\mathbf{a}_{*},\,\mathbf{b}_{*}}(A_{\phi})$ we have the following corollary.

Corollary 8.20.

Around $\phi=0$ ,

	$\displaystyle J_{\mathbf{a}_{}}(A_{\phi})=\tfrac{4}{3}i{\sqrt{3}}(1-2i\phi(1+O(\delta_{\phi}))),\quad J_{\mathbf{b}_{}}(A_{\phi})=-\tfrac{16}{3}\pi i{\sqrt{3}}\,{\phi}\,({\ln\phi})^{-1}(1+O(\delta_{\phi})),$
	$\displaystyle\Omega_{\mathbf{a}_{}}^{}=-\tfrac{1}{2}i\sqrt{3}\ln\phi\,(1+O(\delta_{\phi})),\quad\Omega_{\mathbf{b}_{}}^{}=-{\sqrt{3}\pi}+O({\phi}^{1/2}),$

where $\delta_{\phi}=\ln\ln\phi\,(\ln\phi)^{-1}$ as $\phi\to 0.$

Acknowledgements. The author is grateful to Professor Yousuke Ohyama for a stimulating conversation informing circumstances of Kapaev’s announcement [13] and inspiring the author to tackle this work.

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Kapaev’s global asymptotics of the fourth Painlevé transcendents. Elliptic asymptotics

Abstract.

1. Introduction

2. Results

2.1. Solutions for 0<|ϕ|<π/40<|\phi|<\pi/4 in [13, Theorem 2]

Theorem 2.1.

Theorem 2.2.

Remark 2.1.

Remark 2.2.

Remark 2.3.

2.2. Alternative expression of solutions

Corollary 2.3.

Theorem 2.4.

Remark 2.4.

2.3. Observation related to trigonometric asymptotics

3. Basic facts

3.1. Monodromy data

Proposition 3.1.

Proof..

Proposition 3.2.

Proof..

Remark 3.1.

3.2. Isomonodromy linear system

Proposition 3.3.

Proposition 3.4.

Proof..

Corollary 3.5.

4. Turning points, Stokes graph, WKB analysis

Proposition 4.1.

Remark 4.1.

Proposition 4.2.

Remark 4.2.

5. Direct monodromy problem

Proposition 5.1.

Proposition 5.2.

6. Asymptotics of monodromy data

6.1. Integrals

Proposition 6.1.

Proposition 6.2.

Remark 6.1.

6.2. Theta-function

Proposition 6.3.

Proof..

Corollary 6.4.

Corollary 6.5.

Proposition 6.6.

Proof..

6.3. Expression of Bϕ​(t)B_{\phi}(t)

Remark 6.2.

Proposition 6.7.

Proposition 6.8.

Proof..

7. Proofs of the main results

7.1. Derivation of ψ​(t)\psi(t)

Remark 7.1.

Proposition 7.1.

Proposition 7.2.

Proposition 7.3.

Proof..

7.2. Asymptotic representation of Bϕ​(t)B_{\phi}(t)

Proposition 7.4.

7.3. Proofs of Theorems 2.1 and 2.2

7.4. Proof of Theorem 2.4

7.5. Properties of the elliptic function P​(u;A)\mathrm{P}(u;A)

Proposition 7.5.

Proof..

Proposition 7.6.

Proof..

8. Boutroux equations

8.1. Basic facts

Lemma 8.1.

Proof..

Corollary 8.2.

8.2. Uniqueness

Example 8.1.

Proposition 8.3.

Proof..

Proposition 8.4.

Corollary 8.5.

Proposition 8.6.

2.1. Solutions for $0<|\phi|<\pi/4$ in [13, Theorem 2]

6.3. Expression of $B_{\phi}(t)$

7.1. Derivation of $\psi(t)$

7.2. Asymptotic representation of $B_{\phi}(t)$

7.5. Properties of the elliptic function $\mathrm{P}(u;A)$