数理解析研究所講究録
931 巻 1995 年 1-19

1Connection Formula for Airy-type Equations
Yousuke Ohyama,$7\mathrm{b}$$r_{\backslash }\mathrm{d}\sim$$/_{l}\backslash |$$(f\backslash ?k\mathcal{K}\cdot\gamma \mathrm{g})$\S 0. Introduction
We will study turning point problems for third order equations using technique of WKB
analysis. In the case of second order equations, there are many results which go back to
Liouville. It seems that there are some difficulty to generalize higher-order case ([BNR],[AKT2] . We treat one of the most simple example in this paper.
$)$In$19\mathrm{a}6,$$\mathrm{H}.\mathrm{s}_{\mathrm{C}\mathrm{h}\mathrm{f}\mathrm{f}\acute{\mathrm{e}}}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{u}\mathrm{d}\dot{\mathrm{y}}\mathrm{t}\dot{\mathrm{h}}\mathrm{e}$equations(0.1)$\frac{d^{m}y}{dz^{m}}-z^{n}y=0$,

which he called -equation. He showed that the asymptotic forms of -equation play a key. The asymptotic behavior of -equation isrole in the study of more general equationsof -equation around infinity.
. He calculated Stokes
studied by Turrittin
$t$

$t$

$t$

$([\mathrm{S}])$$\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}_{\mathrm{P}}\mathrm{I}\mathrm{i}\mathrm{e}\mathrm{r}\mathrm{S}$$([\mathrm{T}])$$t$

The Stokes sectors of -equation are divided by $2\langle n+m$ ) lines$t$$\mathrm{a}x\mathrm{g}z=\frac{\pi h}{n+m}$,$h=0,1,$$\cdots,$$2(n+m)-1$ .

But different solutions may have the same asymptotic expansions in some sectors. Hence
we should choose special solutions in order to determine the Stokes multipliers. Turrittin
used Barnes-type integral formula of solutions, and choose an enlarged sector to restore
uniqueness.
-Stark effects and the
Recently Silverstone and others studied the resonances of. In [SNH], they consider Borel summablity of Airy functionenergy eigenvaIue of. The asymptotic expansion of$\mathrm{L}\mathrm{o}\mathrm{S}\mathrm{u}\mathrm{r}\mathrm{d}\mathrm{o}$$H_{2^{+}}$$\mathrm{A}\mathrm{i}(z)$$\mathrm{A}\mathrm{i}(z)$$\mathrm{A}\mathrm{i}(z)\sim\frac{1}{2}\pi^{-\perp}2z^{-_{4}^{\iota}}\exp(-\frac{2}{3}z2)\mathrm{g}\sum_{k=0}^{\infty}(-1)^{k_{\frac{\Gamma(k+\frac{5}{6})\mathrm{r}(k+\frac{1}{6})}{\Gamma(\frac{5}{6})\Gamma(\frac{1}{6})2kk\prime!}}}(-\frac{2}{3}z^{2}2)k$is valid for$|\arg_{\mathcal{Z}}|<\pi$. If we take the Borel resummation of the expansion, we have$\mathrm{A}\mathrm{i}(z)=\frac{1}{2}\pi^{-\frac{1}{2}}z^{-\frac{1}{4}}\exp(-\frac{2}{3}z^{\frac{3}{2}})I^{\infty}\mathrm{o}e^{-}t2F1(\frac{1}{6},$ $\frac{5}{6};1;-\frac{3t}{2z^{\frac{3}{2}}})dt$.

2

By looking at the branch points of the hypergeometric function, they show that thestandard domain of Borel summability is, narrower than the usual sector$|\arg_{Z}|<\pi$ . Borel resummation gives an integral expression of asymptotic
expansions, andwe can understand the valid sectors from the branch points of the integrand.
$| \arg_{Z}|<\frac{2}{3}\pi$$\mathrm{i}$On the other hand, the exact WKB method invented by A.Voros gives a powerful tool inthe study of semi-classical analysis of one-dimensional Shr\"odinger equation. T.Aoki,
T.Kawai and Y.Takei developed the Voros theory, and give exact connection formula ofsecond order equations with regular singularity. In [AKT], they reduce general equations
to the Airy equation near a simple turning point and give the connection formula nearturning point. They also study the case of third order equations. In their theory
the connection problem of the Airy equation is essential and this problem is just the sameas the case of Silverstone.
$([\mathrm{V}])$$([\mathrm{A}\mathrm{K}\mathrm{T}2])$

We will study the connection problem of the Airy-type equation (0.1) when $m=3$ .
The exact WKB analysis of Voros is also useful in this case. In section 1 we review
exact WKB analysis. We must modify Voros’s theory for the third order equations. In
section 2 we discuss WKB solutions of Airy-type equation. The Borel transformations of
WKB solutions to this type of equation can be represented by generalized hypergeometricfunctions $3F2$ . We will give an explicit connection formula of Borel-transform of WKB
solutions. In section 3, using the connection formula of, we study connection problem
of generalized Airy equation around infinity. In section 4, we study connection problem
from infinity to. In both cases, the connection formula of hypergeometric functions$2F1$ plays a key role.
$3F_{2}$$\mathrm{z}\mathrm{e}\mathrm{l}\cdot \mathrm{O}$\S 1. Complex WKB analysis
We first review complex WKB analysis. We follow the notation in [V] and [AKT], although
they discuss second order equations.

We are concerned with an equation of tlue following form:(1.1)$( \frac{d^{3}}{dq^{3}}-xQ3(q))\psi(q, x)=0$

where $Q(q)$ is an analytic function andsolution of the following type:
(1.2)

$x$

is a complex large parameter. We take a formal$\psi(q, x)=\frac{1}{\sqrt{x}}\exp(\int^{q}s(q^{;}, x)dq)/$,$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{l}\cdot \mathrm{e}$(1.3)$S(q, x)=S_{-1}(q)x+S_{0}(q)+S_{0}(q)x-1+\cdots$

3

As a formal power series in(1.4)$x^{-1},$$S(q, x)$ satisfies following nonlinear equation:$3S(q, x) \frac{d}{dq}s(q, x)+\frac{d^{2}}{dq^{2}}S(q, x)+s(q, x)3-XQ3(q)=0$.

Each term $S_{j}(q)$ in (1.3) is uniquely determined if we fix the branch of $S_{-1}=\sqrt[3]{Q(q)}$ .
From (1.4) $S_{j}(q)$ is calculated by the following recursive equation:$3S_{-1}(q)^{2}S_{m+}2+ \frac{d^{2}}{dq^{2}}sm+3\sum S_{i}\frac{d}{dq}S_{j}$$i+j=m$.

$+3S_{-1} \sum_{0\dot{r}+j=m+1i,j\geq},sisj+\sum_{ii+j+k=m,,j,k\geq 0}Sis_{j}S_{k}=0$

Let$S_{(0)}= \sum_{j}S_{3j}x^{3j},$$S_{(}j+12)= \sum_{j}S_{3j2}+X3j+2$$s_{(1})= \sum_{j}s_{3j+1}X^{3},$.

By (4) we have
$3 \frac{d}{dq}(s(1)S(0))+3s(2)^{\frac{d}{dq}s\frac{d^{2}}{dq^{2}}}(2)+S_{()()}1+3s_{(0)}2s1+3S^{2}S(2)+1)3S_{()}2S_{(}0(2)=0$,$3 \frac{d}{d_{(\mathit{1}}}(S_{(2)(}S0))+3s_{(1)}\frac{d}{dq}s_{(\rangle}1+\frac{d^{2}}{dq^{2}}s(2)+3S^{2}s(2\rangle+0(1))+3s^{2}s_{(0}3s^{2}S()(2)(1)=0$.

Taking the$\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{l}\cdot \mathrm{e}\mathrm{n}\mathrm{C}\mathrm{e}$of the equations above,$s_{(2)} \frac{d^{2}}{dq^{2}}S(1)-s_{(}1)^{\frac{d^{2}}{dq^{2}}S}(2)+3s^{2})(2\frac{d}{dq}S_{(}2)-3S^{2}(1)\frac{d}{dq}s_{(}1)+3S(2)\frac{d}{dq}(s_{(1})s(0))$$-3s_{(1)} \frac{d}{dq}(S(2)s(0))-3S_{(0)}(S_{(2)}^{3}-s_{(1}3))=0$

.

:

,

$S_{(0)}(s_{(2)}^{3}-^{s+}(1)(2)^{\frac{d}{dq}s_{(1}s_{(}}s)-31)^{\frac{d}{dq}s)}(2)$$\cdot$$+ \frac{1}{3}(3s_{(2}^{2}\frac{d}{dq})S(2)-3S_{(1}^{2})^{\frac{d}{dq}S_{(1}})s(2)^{\frac{d^{2}}{dq^{2}}}(s1)-S_{(1)}\frac{d^{2}}{dq^{2}}S(2))=0$.

Therefore we obtain$S_{(0)}=- \frac{1}{3}\frac{d}{dq}\log(S_{(2}^{3}-^{s_{(1}+^{s\frac{d}{dq}}})3)(2)s(1)-s(1)^{\frac{d}{dq}s)}(2)$Therefore$\mathrm{W}V\mathrm{R}.\mathrm{q}\mathrm{n}\mathrm{l}\mathrm{l}\mathrm{l}\dotplus \mathrm{i}\cap 1\tau.\mathrm{C}\mathrm{i}\mathfrak{m}\mathrm{f}\mathrm{l}\mathrm{v}\mathrm{b}_{P}$.

$\mathrm{w}\Gamma \mathrm{i}+_{l}$},$\mathrm{e}\mathrm{n}$in$\mathrm{t}_{}\mathrm{h}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{o}\mathrm{W}\mathrm{i}\mathrm{n}\rho$.

; form:(1.4)

Although the series (1.3) is divergent, Voros shows that the Borel transform with respect
to the parameter is a ramified analytic function. We will fix the definition of the Borel
transform (see [V], [AKT]):$x$

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DEFINITION. Let $f(x)$ be a formal series$f(x)=e \xi 0xj\sum_{\geq 0}f_{j}x^{-j-}a$$wl_{l}ere$$a$is any complex$\mathrm{n}\mathrm{u}\mathrm{m}ber$.

$Tl_{l}e\mathrm{n}$,its Borel transform$\sum_{j\geq 0}\frac{f_{j}}{\Gamma(j+a+1)}(\xi+\xi 0)^{j}+a$$f_{B}(\xi)$$f(x)=x\mathit{1}_{-}^{\infty}.\epsilon_{0}de-x\xi f_{B}(\xi)\xi$$f_{B}(\xi)$$\frac{d^{3}}{dq^{3}}\psi_{B}(\xi, q)-Q(q)\frac{d^{3}}{d\xi^{3}}\psi_{B(}\xi,$series:.

If $\psi(q, x)$ is a WKB solution of (1.1), it follows from (1.5) that
Bloch equation(1.6)$tl_{l}e$.

At least formally, we can represent $f(x)$ as an integral of(1.5)

is defined by$\psi_{B}(\xi, q)$satisfies the Balian-$q)=0$ .

\S 2. WKB solutions of Airy-type equation
In [AKT], it is shown that the Borel transformed WKB solutions of Airy equation are
represented by Gauss hypergeometric functions. The connection formula is calculatedusing Kummer’s relation for hypergeometric functions. In this section we study following
Airy-type equation:(2.1)$\frac{d^{3}}{dq^{3}}\psi-x^{3n}q\psi_{=}0$.

The WKB solution of (2.1) is as follows:(2.2)$\psi(q, x)=\frac{1}{\sqrt{x}}\exp\int^{q}s(X, q’)dq;$

,

where
(2.3)

$S(q, x)=S_{-1}(q)_{X}+S_{0}(q)+S_{0}(q)x^{-}1+\cdots$

Let $\psi^{(0)}(q, x),$ $\psi(1)(q, x)$ and $\psi^{(-1)}(q, x)$ be the WKB solutions corresponding to theinitial termand, respectively, where. We take a branch$S_{-1}=q^{\frac{n}{3}},$$q^{\frac{n}{3}}\omega$$q^{\frac{n}{3}}\omega^{-1}$$\omega=e^{\frac{2}{3}\pi i}$

5

so that $\Re q^{n/s}>0$ if $q>0$ . We can verify by the induction that the $S_{k}(q)$ has the

of
form

$q^{n/3}$$S_{k}(q)=skq^{-}1-(n+3) \frac{k}{3}$

where

,

is a complex number. For example, if $S_{-1}(q)=\omega^{j/3}q^{n}$ ,$s_{k}$$S_{1}-1(q)=- \frac{n(6+n)\omega^{2}j}{27}q^{-\frac{n+6}{3}},$ $s_{-}2(q)=- \frac{(18n+9n^{2}+n)3\omega^{j}}{81}q-\frac{2n+9}{3}\ldots$$S_{0}(q)=- \frac{n}{3}q,$

Therefore$\psi^{(j)}(q, x)$.

is following:

,

(2.4)

$\psi^{(j)}(q, x)=\exp(\frac{3\omega^{j}}{n+3}q\frac{n+3}{3}x)q^{-}\sum_{0}\frac{n}{3}\tau k(q)k\geq x^{-k}-12$

and $T_{k}(q)$ has also the form
$T_{k}(q)=tkq^{-(+3)\frac{k}{3}}n$where$t_{k}$$\psi^{(j)}(q, x)$,

is a complex number. Especially $t_{0}=1$ . We denote the Borel transform of. By definition, we haveby$\psi_{B}^{(j)}(\xi, q)$$\psi_{B}^{(j)-\frac{n}{3}}(\xi, q)=qk\sum_{\geq 0}\frac{t.’q^{-(n+3)k/}3}{\Gamma(k+\frac{3}{2})}(\xi+\frac{3\omega^{j}}{n+3}q^{\frac{n+3}{3}})k+_{2}^{1}$$=q^{\frac{3-n}{6}\sum_{k\geq}}0 \frac{t_{k}}{\Gamma(k+\frac{3}{2})}(\xi q^{-}+\cdot\frac{3\omega^{j}}{n+3}\frac{n+3}{3})^{k+^{\iota}}2$

(2.5)

where
Let

$=q^{\frac{3-n}{6}} \frac{2}{\sqrt{\pi}}\sum_{\geq k0}\frac{t_{k}}{(\frac{3}{2})_{k}}(\xi q^{-}\frac{n+3}{3}+\frac{3\omega^{j}}{n+3})^{k}+\frac{1}{2}$$(a)_{n}= \frac{\Gamma(a+n)}{\Gamma(a)}$denote$t$$\xi q^{-\frac{n+3}{3}}$

.

Then

$\psi_{B}^{(j\rangle}(\xi, q)$

has the form$\psi_{B}^{(}j)(\xi, q)=q\frac{3-n}{6}h^{(}j)(t)$$h^{(j)}(t)$has a singularity at$h^{(-1)}(t)$,$t=- \frac{\dot{3}\omega^{j}}{n+3}$.

. It follows from (1.6) that $y=h^{(0)}(t),$satisfy the equation$((n+3)^{\mathrm{a}_{t}3}+27) \frac{d^{3}\tau/}{dt^{3}}+\frac{9(3+n)^{3}}{2}t^{2}\frac{d^{2}y}{dt^{2}}$(2.6)$+ \frac{(3+n)(81+78n+13n^{2})}{4}t\frac{d\tau/}{dt}+\frac{(-3+n)(3+n)(9+n)}{8}y=0$$h^{(1)}(t)$,6

The three regular singular points in (2.6) correspond to the choice of. The exponents
at each singularity are. Although the distance of the exponents may be an integer,
any local solution does not have a logarithmic term. The exponent ofisat
the corresponding singular point. We consider analytic continuation of
to anothersingular point.
$S_{-1}$$0,$$\frac{1}{2},1$$h^{(j)}(t)$$\frac{1}{2}$$h^{(j)}(t)$The equation (2.6) is invariant by the$- \frac{(n+3)^{3}}{27}t^{3}$$\frac{2}{3}\pi$-rotation around the origin. Let$s$denote. Then we obtain$s^{2}(_{\mathit{8}}-1) \frac{d^{3}y}{ds^{3}}-(2s-\frac{7}{2}S)2\frac{d^{2}y}{ds^{2}}$(2.7)$-( \frac{2}{9}-\frac{159+114n+19n^{2}}{12(3+n)^{2}}S)\frac{dy}{ds}+\frac{(-3+n)(9+n)}{216(3+n)^{2}}y=0$

We set $p= \frac{1}{n+3}$ . The local solutions of (2.7) at origin are followings:$\phi_{0}^{(0)}(_{S)}=$3$F_{2}( \frac{1}{6}-\frac{p}{2}, \frac{1}{6}+\frac{p}{2}, \frac{1}{6}; \frac{1}{3}, \frac{2}{3};s)$,$\phi_{0}^{(1)\frac{1}{3}}(_{S})=s3F2(\frac{1}{2}-\frac{p}{2}, \frac{1}{2}+\frac{p}{2}, \frac{1}{2};\frac{2}{3}, \frac{4}{3};s)$,$\phi_{0}^{(2)}(S)=s\frac{2}{3}3F2(\frac{5}{6}-\frac{p}{2}, \frac{5}{6}+\frac{p}{2}, \frac{5}{6};\frac{4}{3}, \frac{5}{3};s):.\cdot$

We set$2F_{1}( \frac{1}{12}-\frac{p}{2}, \frac{1}{12}+\frac{p}{2};\frac{2}{3};\mathit{8})$,$u_{2}(S)=S^{\frac{1}{3}}2F1( \frac{5}{12}-\frac{p}{2}, \frac{5}{12}+\frac{p}{2};\frac{4}{3};s)$.

$u_{1}(s)=$Using Clausen’s formula$3F_{2}(2a, a+b, 2b;2a+2b, a+b+ \frac{1}{2};s)=[_{2}F_{1}(a, b;a+b+\frac{1}{2};s)]^{2}$

3 $F_{2}( \frac{1}{2}, a-b+\frac{1}{2}, -a+b+\frac{1}{2}; -a-b+\frac{3}{2}, a+b+\frac{1}{2}; s)$$=_{2}F_{1}(a, b;a+b+ \frac{1}{2}; s)_{2}F_{1}(\frac{1}{2}-a,\frac{1}{2}-b;-a-b+\frac{3}{2};s)$we have$\phi_{0}^{(0)}(S)=u_{1}(s)^{2}$,$\phi_{0}^{(1)}(_{\mathit{8}})=u_{1}(S)u_{2}(S)$$\phi_{0}^{(2)}(s)=u2(S)^{2}$

.

,

,

,

7

.

We set$2F1( \frac{1}{12}-\frac{p}{2}, \frac{1}{12}+\frac{p}{2};\frac{1}{2};1-s)$$u_{3}(s)=$.

$\cdot$,$u_{4}(s)=(1-s)^{1}/2F21( \frac{7}{12}-\frac{p}{2}, \frac{7}{12}+\frac{p}{2};\frac{3}{2};1-s)$.

Then the products$\phi_{1}^{(0})(_{S)=u_{3}}(S)^{2}$,$\phi_{1}^{(1)}(s)=u_{3}(S)u_{4}(_{S)}$$\phi_{1}^{(2)}(S)=u_{4}(S)^{2}$.

are independent solutions of (2.7) at $s=1$ . Since$(1-s)1/2,$ $h^{(j)}(t)$ is equal to$\phi_{1}^{(1)}(-(\frac{t}{3p})^{3})$,$\phi_{1}^{(1)}(s)$has a singularity of the type

up to constant multiplication. We notice thefollowing lemma.

LEMMA.

$c_{on}Si,de\mathrm{r}.thefu\mathrm{n}cti_{ons}1$$\mu(t)=(1+(\frac{t}{3p})^{3})^{\perp}2$

We take$tl_{l}e$following cut$li\mathrm{n}$,$\mu_{j}(t)=(t+3_{P^{\omega}}j)^{\frac{1}{2}}$es in t-plan :$\mu(t)$and:..$\cdot$$\mathrm{e}$$\{t;t+\frac{\omega^{j}}{3p}\in \mathbb{R}_{+}\}$

We take a branch of$\mu_{j}(t)$.

$(j–$ $-1,0.’ 1)$.

as follows.
$\lim_{\epsilon l^{0}}\mu(a+i\epsilon)>0$$\lim_{\epsilon\downarrow 0}\mu(a-\frac{\omega^{j}}{3p}+i\epsilon)>0$

for $a>0$ . Tllen we can take$t=- \frac{\omega^{j}}{3p}s\mathrm{u}d_{l}$$f\mathrm{u}n$,,ctions $A_{j}(t)(j=-1,0,1),$ wllich are holomorphic atthat$\mu(t)=_{l}l_{j}(t)Aj(t)$and$(j=-1,0,1)$$A_{j}(- \frac{\omega^{j}}{3p})=\frac{\omega^{j}}{\sqrt{p}}$,for $j=-1,0,1$ .

The lemma is easily verified by direct calculations.

By (2.5), $h^{(j)}(t)$ has a singularity at$t=- \frac{\omega^{j}}{3p}$such that$h^{(j)}(t)= \frac{2}{\sqrt{\pi}}(t+3p\omega^{j})\frac{1}{2}(1+O(t+\cdot 3\omega^{j}p))$.

8

Therefore it follows from the lemma above that(2.8)$h^{(j)}(t)=2 \omega^{-j}\sqrt{\frac{p}{\pi}}\phi_{1}^{(1})(-(\frac{t}{3p})^{3})$,near $t=-3p\omega^{j}$ .
We will calculate the discontinuity of $h^{(0)}(t)$ at $t=-3\mu v,$ $-3F^{v}-1$ . In the 8-space we(resp.,
should consider the analytic continuation around the origin. We denote
once counter-clockwise (resp., clockwise)as the analytic continuation ofaround the origin.
$\tilde{\phi}_{1}^{(j\rangle}(s)$$\phi_{1}^{(j)}(s)\approx)$$\phi_{1}^{(j)}(S)$PROPOSITION 2.1. We have$\tilde{\phi}_{1}^{(1)}(S)=(2\cos^{2}p\pi-\frac{1}{2})\omega\phi_{1}t1)(S)+$($l_{1O}lomorph\mathrm{j}\mathrm{c}$at $s=1$ ),$\phi_{1}^{(1)}(s\approx)=(2\cos^{2}P^{\pi}-\frac{1}{2})\omega^{2(1}\phi_{1}(_{S})+(holo\mathrm{m}o\mathrm{r}p)l\dot{u}c$near$\mathit{8}=1$

at $s=1$ ),.

PROOF: By Gauss’ connection formula
2 $F1(a, b;C;s)= \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}2F1(a, b;a+b-c+1;1-s)$$+ \frac{\Gamma(c)\Gamma(a+b-C)}{\mathrm{r}(a)\mathrm{r}(l\mathrm{J})}(1-S)^{c}-a-b2F_{1}(c-a, c-b;c-a-b+1;1-s)$

we have(2.9)$u_{1}(s)=Au_{3}(_{S)-}Bu_{4}(S)$,$s^{1/3}u_{2}(_{S)}=^{c()}u3s-Du_{4}(s)$

,

where
$A= \frac{\sqrt{\pi}\Gamma(\frac{2}{3})}{\Gamma(\frac{7}{12}+_{2}^{\mathrm{g}})\mathrm{r}(\frac{7}{12}-^{E}2)}$$C= \frac{\sqrt{\pi}\Gamma(\frac{3}{4})}{\Gamma(\frac{\mathrm{I}1}{12}+_{2}^{2})\Gamma(\frac{11}{12}+_{2}^{\mathrm{z}})}$,,$B= \frac{2\sqrt{\pi}\Gamma(\frac{2}{3})}{\Gamma(\frac{\mathrm{I}}{12}+_{2}\mathrm{r})\Gamma(\frac{1}{12}+E)2}$,$D= \frac{2\sqrt{\pi}\Gamma(\frac{4}{3})}{\Gamma(\frac{5}{12}+^{\mathrm{g}})2\mathrm{r}(\frac{5}{12}+^{\mathrm{g}})2}$.

Therefore(2.10)$(\phi_{0_{2}^{1}}()\phi_{0}^{(0})(\phi_{0}()(s)(s)s))=(_{\phi^{(2}()}^{\phi^{\mathrm{t}}(S)}\phi_{1}^{(}()s)110)1)\mathit{8})$

.

,

9

) as the analytic continuation of(resp.,We denotearoundthe origin. Then we haveclockwise)(resp.,
clockwise$\tilde{\phi}_{1}^{(j)}(S)$$\phi_{1}^{(j)}(s)\approx$$\tilde{\phi}_{0}((i)(js)=\omega^{j}\phi_{0})(_{S})$(2.11)

once counter-$\phi_{0}^{(j)}(s)$,$\phi_{0}^{(j)(j}\approx(s)=\omega 2j\phi 0)(_{S})$.

By direct calculation, we have
$AD= \frac{1}{3}+\frac{2}{3\sqrt{3}}\cos\pi p$,.

$BC=- \frac{1}{3}+\frac{2}{3\sqrt{3}}\cos\pi p$

Hence the inverse transformation of (2.9) is given by$u_{\}(s)= \frac{3}{2}Du_{1}(\mathit{8})-\frac{3}{2}Bu2(S)$,$u_{4}(s)= \frac{3}{2}Cu_{1}(s)-\frac{3}{2}Au_{2}(_{S)}$.

Therefore(2.12)$( \phi_{1}^{(2})(_{\mathit{8})}\phi_{1}^{(1)}\phi^{()}1(0(S)S))=\frac{9}{4}(_{\phi_{0}^{()}(_{\mathit{8})}}^{\phi_{0_{1)}}(}\phi_{0_{2}}\mathrm{t}(0)(S)S))$

into(2.12) is valid if we change all of(2.12), we obtain the proposition 2.1. 1$\phi_{k}^{(j)}$$\tilde{\phi}_{k}^{(j)}$or$\phi_{k}^{(j)}\approx$. Combined with (2.10), (2.11) and

From now on, we will study the analytic continuation from$\mathit{8}=\infty$

to $s=1$ . We s\"et$u_{5}(s)=(-s)- \frac{1}{12}+_{2}\mathrm{z}F21(\frac{1}{12}-\frac{p}{2}, \frac{5}{12}-\frac{p}{2};1-p;s^{-})1$,.

$u_{6}(s)=(-S)- \frac{1}{12}-^{\mathrm{g}}2F21(\frac{1}{12}+\frac{p}{2}, \frac{5}{12}+\frac{p}{2};1+P;s-1)$

The products$\emptyset_{\infty}^{\mathrm{t}0)}(S)=u_{5}(s)^{2}$,$\phi_{\infty}^{(1)}(_{S})=u\mathrm{s}(S).u_{6}(s)$$\phi_{\infty}^{(2)}(_{S})=u_{6}(s)^{2}$,.

are the local sofutions of (2.7) near infinity. By Clausen’s formula we have$\phi_{\infty}^{(0)}(S)=(-S)-\iota+PF_{2}6s(\frac{1}{6}-_{F,p,\frac{5}{6}-}\frac{1}{2}-p;1-p, 1-2p;s-1)$$\phi_{\infty}^{(1)}(_{S})=(-s)^{-\frac{1}{6}}\cdot 3F_{2}(\frac{1}{6}, \frac{1}{2}, \frac{5}{6};1-p, 1+p;\mathit{8}^{-})1$,,$\phi_{\infty}^{(2\rangle}(_{S})=(-s)-\frac{1}{6}-r\mathrm{a}F2(\frac{1}{6}+P, \frac{1}{2}+p, \frac{5}{6}+P\mathrm{i}1+p, 1+2p;S^{-1})$.

10

PROPOSITION 2.2. The
plane is$\mathrm{a}n\mathrm{a}lyti_{C,1}$

continu. ation of$\phi_{\infty}^{(k)}.(S)$.

along a path in the lower-half$\phi_{\infty}^{(k)}(S)=-\frac{3^{3(k-1})p}{\sqrt{\pi}}e^{i(1k}-\frac{2}{3}+(-)P)\frac{\prod_{j=0}^{2}\Gamma(1+(k-j)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}\pi\phi(1)(1S)$$+$( $hol_{om}orphic$ at $s=1$ )near $s=1$ .
PROOF: When welowing:a pat.. in the lower-half plane, the connection formula is the fol$\mathrm{h}$$\mathrm{t}_{\iota}\mathrm{a}\mathrm{k}\mathrm{e}$$$(-s)^{-a}2F1(a, a+1-c;a+1-b;s^{-1})$$= \frac{\mathrm{r}(_{C+1}-a-b)\mathrm{r}(a+b-c)\mathrm{r}(a+1-b)}{\Gamma(1-b)\Gamma(c-b)\Gamma(a+b+1-C)}e^{-i\pi a_{2}}F1(a, b;a+b-c+1;1-s)$.

$- \frac{\Gamma(a+b-c)\mathrm{r}(a+1-b)}{\Gamma(a)\Gamma(a+1-C)}e^{i\pi(}-\mathrm{c})(b1-S)^{c}-a-bF_{1}2(C-a, C-b;c-a-b+1;1-s)$$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{l}\cdot \mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}$$u_{5}(S)=Eu_{3}(S)+Fu_{4}(s)$ ,(2.13)$u_{6}(s)=Gu3(_{S)H}+u4(s)$ ,where$E=-e^{i\pi(}- \frac{1}{12}+^{\mathrm{g}})_{\frac{\sqrt{\pi}\Gamma(1-p)}{\Gamma(\frac{11}{12}-22)\Gamma(\frac{7}{12}-2\mathrm{g})}}2$$G=-e^{i\pi}(- \frac{1}{12}-^{z_{2}}‘)_{\frac{\sqrt{\pi}\Gamma(1+p)}{\Gamma(\frac{11}{12}+_{2}^{E})\mathrm{r}(\frac{7}{12}+_{2}\mathrm{z})}}$

At first we consider$\phi_{\infty}^{(0\rangle}(\mathit{8})$,,$F=e^{i\pi} \mathit{3}(-\frac{7}{12}+\epsilon)_{\frac{2\sqrt{\pi}\Gamma(1-p):}{\Gamma(\frac{1}{12}-22)\mathrm{r}(\frac{5}{12}-2E)}}2^{\cdot}.,$$f$. By (2.13) we have$u_{5}(s)^{2}=2GHu_{3}(S)u4(s)+$ (Recall the multiplication formula of Ga\’uss(2.14).

$H=e^{i} \pi(-\frac{7}{12}-\mathrm{z}2)_{\frac{2\sqrt{\pi}\Gamma(1+p)}{\Gamma(\frac{1}{12}+_{2}^{\mathrm{g}})\mathrm{r}(\frac{5}{12}+_{2}\mathrm{r})}}$$\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{P}^{\mathrm{h}}\mathrm{i}\mathrm{c}$$\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{L}\mathrm{e}\mathrm{g}\mathrm{e}\check{\mathrm{I}}\mathrm{l}\mathrm{d}\mathrm{r}\mathrm{e}$at $s=1$ ).

:$\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\mathrm{r}(2z)$.

We have$\Gamma(\frac{1}{12}-\frac{p}{2})\Gamma(\frac{7}{12}-\frac{p}{2})=2^{\frac{5}{6}+p}\sqrt{\pi}\mathrm{r}(\frac{1}{6}-p)$,$\Gamma(\frac{5}{12}-\frac{p}{2})\Gamma(\frac{11}{12}-\frac{p}{2})=2^{\frac{1}{6}+p}\sqrt{\pi}\Gamma(\frac{5}{6}-p)$.

11

Therefore we have
$2EF=-4e^{i\pi}(- \frac{2}{3}+p)_{\frac{\Gamma(1-p)^{2}}{2^{1+2p\Gamma}(\frac{1}{6}-p)\mathrm{r}(\frac{5}{6}-p)}}$.

Recall the multiplication formula of the third degree:(2.15)$\Gamma(z)\Gamma(z+\frac{1}{3})\Gamma(z+\frac{2}{3})=\frac{2\sqrt{3}\pi}{3^{3z}}\Gamma(3Z)$.

We have$\Gamma(\frac{1}{6}-p)\Gamma(\frac{1}{2}-p)\Gamma(\frac{5}{6}-p)=2\cdot 3^{3p}\pi \mathrm{r}(\frac{1}{2}-3p)$.

We use (2.14) again and get$\Gamma(\frac{1}{2}-p)\Gamma(1-p)=2^{2_{\sqrt{\pi}\Gamma}}P(1-2p)$.

Hence we obtain.

(2.16)$2EF=- \frac{1}{3^{3p_{\sqrt{\pi}}}}e^{i\pi}(-\frac{2}{3}+p)_{\frac{\Gamma(1-p)\Gamma(1-2p)}{\Gamma(\frac{1}{2}-3p)\}}$In the same way,$u_{6}(s)^{2}=2GHu_{3}(_{S})u_{4}(S)+(\mathrm{h}_{0}..1\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{s}^{r}?;s=1)$.

And we get$2GH=- \frac{3^{3p}}{\sqrt{\pi}}e^{i\pi}(-\frac{2}{3}-p)_{\frac{\Gamma(1+p)\Gamma(1-2p)}{\Gamma(\frac{1}{2}+3p)}}$

We will consider$\phi_{\infty}^{(1)}(S)$.

$\mathrm{B}\mathrm{y}\backslash (2.13)$

we have$u_{5}(s)u_{6}(s)=(EH+FG)u_{3}(S)u4(s)+$ ($\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}$at $s=1$ ).
’.

Since
$\Gamma(\frac{11}{12}-\frac{p}{2})\Gamma(\frac{1}{12}+\frac{p}{2})=\frac{\pi}{\sin(\frac{\pi}{12}+L^{\pi})2}$$\Gamma(\frac{7}{12}-\frac{p}{2})\Gamma(\frac{5}{12}+\frac{p}{2})=\frac{\pi}{\sin(\frac{5\pi}{12}+L^{\pi})2}$

we have,,$\iota$:.

$EH+FG=- \frac{2}{\pi}e^{-\frac{2}{3}i\pi}\mathrm{r}(1-p)\Gamma(1+p)$

.

. .,.

L

$( \sin(\frac{\pi}{12}+\frac{p\pi}{2})\sin(\frac{5\pi}{12}+\frac{p\pi}{2})+\sin(\frac{11\pi}{12}+\frac{p\pi}{2})\sin(\frac{7\pi}{12}+\frac{p\pi}{2}))$$=- \frac{2}{\pi}e^{-\frac{2}{3}i\pi}\Gamma(1-p)\Gamma(1+p)\cos\frac{\pi}{3}$(2.18)’.

(2.17)$=- \frac{1}{\pi}e^{-\frac{2}{3}i\pi}\Gamma(1-p)\mathrm{r}(1+p)$.

By (2.16), (2.17) and (2.18) we obtain the proposition. 1\dag er12

3. Connection formula around infinity

In this section we will calculate the Stokes multipliers around infinity. At first we will
see Stokes regions on (2.1), due to Voros. The Borel transformof the WKB
solution has three branch points$\phi_{B}(\xi, q)$$([\mathrm{V}])$$\xi_{j}(q)=-\frac{3}{n+3}\omega q^{\frac{3}{n+3}}j$,

for $j=-1,0,1$ . The Stokes line is the curve in -space defined by the equation
$q$

(3.1)

$\propto s\xi_{j}(q)=\circ s\xi k(q)$for $j\neq k$ . By (3.1), we have$L_{h}$: $\arg q=\frac{(2h+1)\pi}{2(n+3)}$ .

$(h=0,1,2, \cdots, 2n+5)$

This definition is different from the notation used in introduction.

We se.t$S_{h}$: $\{q;-\frac{(2h-1)\pi}{2(n+3)}<\arg q<\frac{(2h+1)\pi}{2(n+3)}\}$ ,and calculate the Stokes multipliers fromtransform ofwhen $q\in S_{k}$ .

$S_{0}$

to$S_{1}$. We will denote$\psi_{k}^{(j\rangle}(q, x)$$\Re_{\xi_{0}}(q)<\Re_{\xi 1}(q)$.

is the Laplace$\psi_{B}^{(j)}(\xi, q)$On$L_{0}$

we have$\propto s\xi_{0}(q)=s\circ\xi_{1}(q)$,Therefore the Laplace transform$\psi^{(0)}(q, x)=x\int^{\infty}\epsilon 0)e-x\epsilon\psi_{B}((0)\xi,$$qd\xi$is changed whenchanged.

$q$

moves across the Stokes line

$L_{0}$, while$\psi^{(1)}(q, X)$

and$\psi^{(-1)}(q, X)\mathrm{a}\mathrm{l}\mathrm{e}$In [V], Voros showed that the connection formula across the Stokes line is$\psi_{1}^{(1)}(q, x)=\psi_{0}^{(}0)(q, X)+9\psi_{0}(\sim\triangle 1)(q, x)$

Here the complex number$\triangle$

is defined by$\psi_{B}^{(0})(\xi, q)=\triangle\psi^{(}B1)(\xi, q)+$near $\xi=\xi_{1}(q)$ .

.

($\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{P}\mathrm{h}\mathrm{i}_{\mathrm{C}}$at $\xi=\xi_{1}(q)$ ),

not

13

PROPOSITION 3.1. The connection formula from

$S_{0}$

to$S_{1}$

is the followings:$\psi_{1}^{(0)(}(q, x)=\psi 0^{0)2}(q, X)+\omega(4\cos^{2}\frac{\pi}{n+3}-1)\psi 01()(q, x)$$\psi_{1}((1)Xq,)=\psi(1)(\mathrm{o}q, x)$,,$\psi_{1^{-}}^{(\iota)}(q, x)=\psi_{0}(-1)(q, x)$.

PROOF: We should know the behavior of
behavior of$\psi_{B}^{(0)}(\xi, q)$

at $\xi=\xi_{1}(q)$ . By proposition 2.1 the$h^{(0)}(t)=2 \sqrt{\frac{p}{\pi}}\phi_{1}^{\mathrm{t}}1)(-(\frac{t}{3p})^{3})$at $t=-3\mu v$ is$\triangle=2\sqrt{\frac{p}{\pi}}\tilde{\phi}_{1}^{()}1(-(\frac{t}{3p})^{3})|_{t=-3p}\mu$$=2 \sqrt{\frac{p}{\pi}}(2\cos^{2}\pi p-\frac{1}{2})\omega\phi_{1}(1)(-(\frac{t}{3p})^{3})|_{t=-3p\omega}$$= \omega^{2}(,2\cos\pi p-2\frac{1}{2})h(1)(t)|t=-3p‘ d$

Therefore$\triangle=\omega^{2}(2\cos^{2}\pi p-\frac{1}{2})$. I

We will calculate the Stokes multiplier when$\circ s\xi_{2}(q)=\alpha s\xi_{1}(q)$Therefore$\xi^{(0)}(q, x)$,$q$goes across. On$L_{1}$$\Re\xi_{2}(q)<\Re_{\xi 1}(q)$$L_{1}$

we have.

and $\xi^{(1)}(q, X)$ are not changed and$\psi_{2}((-\iota))=\psi_{1}(-q,$$x1)(q, X)+2\Delta/\psi_{1}^{(1)}(q,Here the complex number$\triangle’$X)$ .

is defined by$\psi_{B}^{(2)}(\xi, q)=\Delta’\psi_{B}^{(1)}(\xi, q)+$($\mathrm{h}\mathrm{o}1_{0}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}$at $\xi=\xi_{1}(q)$ ),near $\xi=\xi_{1}(q)$ .
PROPOSITION 3.2. The connection$\mathrm{f}\dot{o}rm\mathfrak{U}\mathit{1}\mathrm{a}$$\psi_{2}^{(0)}(q, X)=\psi 1(0)(q, x)$,$\psi_{2}((1)xq,)=\psi(11\rangle(q, x)$

,

from

$S_{1}$

to$S_{2}$

is the followings:$\psi_{2}^{(-1)}(q, X)=\psi 1^{-}(1)(q, X)+\omega(4\cos 2_{\frac{\pi}{n+3}}-1)\psi_{1}^{(}1\rangle(q, x)$.

14

PROOF: The proof is the same as the proof of proposition 3.1 except that we take.
instead of.
:.
$\tilde{\phi}_{1}^{(1)}$$\phi_{1}^{(1)}\approx$

$1$

Since
$\psi^{(j)}(e^{2i}q, xp\pi)=e^{i\pi(2p-}z_{))}3\psi^{(}j+1(q, X)$,

the other case is deduced from proposition 3.1 and 3.2.
THEOREM 3.3. $Wl_{1}enq$ go es across:and$L_{h}$, we have the connection formulae between$\psi_{h+1}^{(j)}(q, x)$.,$\psi_{6h}^{(0)(0)}+1(q, x)=\psi 6h(q, X)+\omega^{2}\triangle^{\psi^{(}}6h(1)xq,)$,$\psi_{6h+2}(-1)(q, x)=^{\psi_{6}(X)\triangle}(-1h+1)q,+\omega\psi_{6h+}^{(}1)1(q, x)$$\psi_{61_{1+}}^{(}-1)(3C\mathit{1}, X)=\psi_{6}^{(}h-1)(+2q, x)$\dagger$\omega^{2}\triangle\psi^{(}6h+2(0)q,$

,

$x)$

$\psi_{6}^{(1)(1)(0)}h+4(q, x)=\psi_{6h+}3(q, x)+\omega\triangle\psi 6h+3(q, x)$,$\psi_{6h+}^{(1)}5(q, X)=\psi_{6h+}^{(}1)4(q, x)+\omega^{2}\Delta\psi 6h+4((-1)q, x)$$\psi_{6h+6}^{()}(0q, x)=\psi_{6h+}(0\rangle 5(q, X)+\omega\Delta\psi 6h+5((-1)q, x)$

,

,

,

. 1

where $\triangle=4\cos^{2}\frac{\pi}{n+3}-1$ . In

$tl_{l}\mathrm{e}$other$c\mathrm{a}se$

Stokes multipliers are trivial:$\psi_{h+}^{(j\rangle}1(q, x)=\psi_{h}((j)q, x)$.

$\psi_{h}^{(j)}(q, x)$

15

4. Connection formula from zero to infinity(2.1) has solutions(4.1)$y_{k}(q, x)=x \frac{3k}{n+3}q^{k}m\sum_{=0}^{\infty}\frac{p^{3m}}{\prod_{s=0}^{2}\Gamma(1+(k-S)p+m)}xq3m(n+3)m$,$(k=0,,1,2)\backslash$near $q=0$ . (4.1) converge for $|q|<\infty$ . In this section we study connection formula
between $y_{k}(q, x)$ and $\psi(q, x)$ . This connection formula gives asymptotic expansions of$y_{k}(q, x)$ , whenis in the large.
$q$

The independent solutions of (2.6) near $t=\infty$ are given by the following:(4.2)$\phi_{\infty}^{(k)}(-(\frac{t}{3p})^{3})=(3p)^{\frac{1}{2}}+3(k-1)p\pi i(-\frac{1}{6}+(1-k)P)(e-t)^{-\frac{1}{2}-}3(k-1)_{P}\sum_{m=0}^{\infty}c^{(k}m)(-\frac{3p}{t})^{3m}$,for $k=0,1,2$ , Here$c_{m}^{(k)}= \frac{(\frac{1}{6}+(k^{\wedge-1)p)(+(k}m\frac{1}{2}-1)p)m(\frac{5}{6}+(k-1)p)_{m}}{(1+kp)m(1+(k-1)p)m(1+(k-2)p)_{m}}$.

In (4.2), we take a sector$- \frac{2}{3}\pi<\arg(-t)<0$,

which corresponds to the sector$-\pi<\arg(-s)<\pi$ ,by the transform$-S^{-=}e \pi i(\frac{-t}{3p})^{3}$

The Borel transform$\psi_{B}(\xi, q)$(4.3)

is a sum of the following functions near $\xi=\infty$ :$\psi_{\infty}^{(k)}(\xi, q)=q-\mathrm{g}\phi^{(k}16\infty)(\xi q^{-\perp_{p}}3)$.

, turns
We will take Laplace transform of (4.3). Let be a curve which starts. And Let $C_{j}(j=0,1,2)$
counter-clockwise and returns
aroundreturns to, turns aroundbe a curve which starts from.
. Consider the Laplace integral$C$$\xi_{0}(q),$$\xi 1(q),$ $\xi_{-1}(q)$$\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}+\infty$$\mathrm{t}\mathrm{o}+\infty$$+\infty$.and$\xi_{j}(q)\mathrm{c}.\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}- \mathrm{C}\mathrm{l}\mathrm{o}\mathrm{C}\mathrm{k}_{\mathrm{W}\mathrm{i}_{\mathrm{S}}}\mathrm{e}‘$$+\infty$$y_{\infty}^{(k)}=x \int_{C}e^{-x\xi}\psi(\infty(\xi, q)dk)\xi$.

16Then we have$y_{\infty}^{(k)}=x \sum_{1j=-}\triangle\int(j\psi^{(}B(\xi, q)de^{-}\xi 1k)c_{i}x\epsilon j)$

where
is the discontinuity of
at $\xi=\xi_{j}(q)$ . Since
square root type at $\xi=\xi_{j}(q)$ , we have$\psi_{\infty}^{(k)}$$\triangle_{j}(k)$$y_{\infty}^{(k)}=x.

,$\psi_{B}^{(j)}(\xi, q)$\sum 12\triangle_{j}\int_{\xi(q}^{\infty}(k)x\epsilon\psi_{B}^{(}j)(\xi, q)de^{-}\xi j)$

has a singularity of’$J^{=-1}$$= \sum 12\triangle_{j}\psi(j)((k)q, x)$

(4.4)

.

$j=-1$In the following, we set$r_{k}=(3p)^{\frac{1}{2}}+3(k-1)_{P}\pi ie(-^{1}6+(1-k)p)$PROPOSITION 4.1. we.

$(k=0,1,2)$$h\mathrm{a}ve$.

$y_{\infty}^{(k)}(q, x)=2 \pi e\pi i(\frac{1}{3}+(1-k)p)_{(3}p)^{\frac{1}{2}+}3(k-1)p\frac{1}{2}-3p\frac{\prod_{s=0}^{2}\Gamma(1+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}xyk(q, x)$

PROOF: We can develop$\psi_{\infty}^{(k)}(\xi)$near $\xi=-\infty$ as follows:.

$\psi_{\infty}^{(k)}(\xi)=r_{kq^{k}\sum q^{\frac{m}{p}(}}m\infty=0(3P)3mc(mk)-\xi)^{-\iota}2-3(k-1)p-3m$

Since$\int_{c^{e(}}-x\epsilon-\xi)\alpha_{d\xi}=\frac{2\pi i}{\Gamma(-\alpha)}x-1-\alpha$,..

we have$y_{\infty}^{(k)()p}=2 \pi ir_{k}X^{\frac{1}{2}}-q\sum_{m}^{\infty}+3k1k=0\frac{(3p)3mcm(k\rangle}{\Gamma(\frac{1}{2}+3(k-1)p+3n)}xq+3m\frac{m}{p}$.

By (.2.15) we get
$ti.$

.

$v$

‘

$c_{m}^{(k)}= \frac{\prod_{s^{--0}}^{2}\Gamma(\frac{1}{6}+(k--1)p+m+\frac{s}{3})}{\prod_{s=0}^{2}\mathrm{r}(\frac{1}{6}+(k-1)p+\frac{s}{3})}\frac{\prod_{s=0}^{2}\Gamma(1+(k-s)p)}{\prod_{s=0}^{2}\Gamma(1+(k-\mathit{8})p+m)}$$= \frac{\Gamma(\frac{1}{2}+3(k-1)p+3m)}{3^{3m}\Gamma(\frac{1}{2}+3\langle k-1)p)}\frac{\prod^{2}=0\Gamma(\backslash ’+1(k-s)p)}{\prod_{s=0^{\Gamma(+}}^{2}1(k-\mathit{8})P+m)}$.

17

Therefore we obtain that$y_{\infty}^{(k)}(q, x)= \frac{2\pi ir_{k}}{\Gamma(\frac{1}{2}+3(k-1)p)}x^{\frac{1}{2}+3(k-}q^{k}q\sum_{m}1)_{P}\frac{p^{3m}\prod_{s=}^{2}0\Gamma(1+(k-s)p)}{\prod_{s=0^{\Gamma}}^{2}(1+(k-S)p+m)}xq\infty=03m\frac{m}{\mathrm{p}}$$= \frac{2\pi ir_{k^{X}}\frac{1}{2}-3pkq\prod_{S-}^{2}-0^{\Gamma(1}+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}y^{(k)}(q, x)$.

$=2 \pi e^{\pi i}(\frac{1}{3}+(1-k)p)(3p)\frac{1}{2}+3(k-1)p\frac{1}{2}x-3p\frac{\prod_{s=0}^{2}\Gamma(1+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}yk(q, x)$. .

1
THEOREM 4.2. $y_{j}(q, x)h$ as an asymptotic expansion in$S_{0}$

in the form

$y_{k}(q, x)= \frac{p^{3(1-k)_{P^{-}}1}}{2\sqrt{3}\pi}x^{-}\iota 2^{+p}3\sum_{j=-1}e^{-}(\frac{2}{3}+2(k-1)p)\psi j\pi\dot{\cdot}(j)(1q, X)$

PROOF: We will consider the behavior ofhas the formthat$\phi_{\infty}^{(k)}(t)$.

at $t=-3p\omega^{j}$ for $j=-1,0,1$ . We assume$\phi_{\infty}^{(k)}(t)$$d^{(k)}(t+3p) \frac{1}{2}(1+O(t+3p))+$ ($\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}$near $t=-.3p$ . Since$\phi_{\infty}^{(k})(t)=e^{\pi i(+2()}\frac{1}{3}k-1p)\phi^{(k}\infty()\omega t)$$\phi_{\infty}^{(k)}(t)$at $t=-3p$ ),has the folm$e^{\pi i(-1} \frac{1}{3}+2(k)p)d(k)(\omega t+3p)^{\frac{1}{2}}(1+O(t+3\mu_{U^{-1}))}+$($\mathrm{h}\mathrm{o}1_{0}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}$$=e( \frac{2}{3}+2(k-1)p)d\pi i(k)(t+3p\omega-1)\frac{1}{2}(1+o(t+3p\omega-1))+$neal’ $t=-3\mu v^{-1}$(

at $t=-3\mu_{U^{-1}}$ )$\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{p}\mathrm{h}\mathrm{i}\mathrm{c}$at)$t=-3_{P^{\omega^{-1}}}$.

In the same way$\phi_{\infty}^{(k\rangle}(t)$has the form

( lomorphic at $t=-3\mu v$ ).
near $t=-3_{l}w$ . Therefore we should know only the discontinuity at $t=-3p$ .
$e^{\pi}i(-_{3^{-}}^{2,2(1}k-)p)d(k)(t+3p\omega)^{\frac{1}{2}(}1+O(t$ $+\backslash 3p\omega))+$We will take a path from infinity to$t\overline{\mapsto}..-3p$$\mathrm{h}o$in the sector $- \frac{2}{3}\pi<\arg(-t)<0$ , In-space, this path goes from infinity to $s=1$ in the lower space. By proposition 2.2
has the form$s$$- \frac{3^{3(k-1)p}}{\sqrt{\pi}}e^{i\pi(+(1k))_{\frac{\prod_{s=0}^{2}\Gamma(1+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}}}-\frac{2}{3}-P\phi 1((1)-(\frac{t}{3p})^{3})$

(4.5)
$+$

’(

1

$\mathrm{h}\mathrm{o}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{P}^{\mathrm{h}}\mathrm{i}\mathrm{c}$at $t=-3p$ ).

$\phi_{\infty}^{(k)}(t)$18

By (2.8) the singular part of (4.5) is$- \frac{3^{3(k-1)p}}{2\sqrt{p}}e-\frac{2}{3}+(1-k)p\frac{\prod_{s=0}^{2}\Gamma(1+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}i\pi()h(0)(t)$.

Therefore(4.6)$\triangle_{0}(k)=-\frac{3^{3(k-1})p}{2\sqrt{p}}e^{i\pi}(-\frac{2}{3}+(1-k)_{P})\frac{\prod_{s=0}^{2}\mathrm{r}(1+(k-s)p)}{\Gamma(\frac{1}{2}+3(k-1)p)}$in (4.4). By proposition 4.1, we obtain the proposition 4.2. 1
We considereda variable$x$

is a large parameter, and(4.7)
and consider$q$

is a finite number. From now on, we take

$z=x^{3p}q$$z$,

in the large. By (4.7), the equation (2.1) is(4.8)$\frac{d^{3}}{dz^{3}}\psi-z^{n}\psi=^{\mathrm{o}}$.

The local solution (4.1) is$\tilde{?/}k(_{Z)}=z\sum_{m=0}k\infty\frac{p^{3m}}{\prod_{s=0^{\Gamma}}^{2}(1+(k-s)P+m)}z(n+3)m$.

And the WKB solution (2.4) is$\psi^{(j)},(.Z)=x\frac{1}{2}-3PZ^{-\frac{n}{3}}\exp(3\mu\backslash \backslash \tilde{q})j\frac{1}{3\mathrm{p}}\sum t_{k}k\geq 0z\frac{1}{3p}$We set$\tilde{\psi}^{(j)}(- z)=\tilde{q}^{-}\frac{n}{3}\exp(3\mu vjZ\frac{1}{3p})k\sum_{\geq 0}t_{k}z^{\frac{1}{\epsilon_{p}}}$.

.

By theorem 4.2 we have asymptotic expansion of solutions of (4.8).

THEOREM 4.3. In $tlle$ sector $| \arg_{Z}|<\frac{\pi}{2(n+3)}$ , we$(n+1)^{\frac{3(k-1)}{3+n}+1}$$h\mathrm{a}ve$

an asymptotic expansion

1$\tilde{l/}k(z)=\sum_{j=-1}e\overline{2^{\sqrt{3}\pi}}-j\pi i(\mathrm{g}3^{+}\frac{2(k-1)}{n+3})_{\tilde{\psi}(}(j)z)$.

REFERENCES[AKT] T.Aoki, T.Kawai and Y.Takei, The Bender-Wu analysis and the Voros theory, ICM-90 Satellite Conference Proceedings, Special Functions, Springer-Verlag, 1991, pp.1-29.

19[AKT2] T.Aoki, T.Kawai and Y.Takei, New turning points in the exact WKB analysis for higher-order ordinary
Analysis and Singular perturbations”, 1994.
differential equations, Proc.
(($\mathrm{A}\mathrm{l}\mathrm{g}\mathrm{e}\mathrm{b}_{\Gamma}\mathrm{a}\mathrm{i}\mathrm{c}$.

[ [BNR] H.L.Berk, W.M.Nevins and K.V.Roberts, New Stokes’ line in WKB theory, J. Math. Phys., Math. Soc.,.’
23(1982), 988-1002. .,$\cdot$[S] H.Scheff\’e, Asymptotic solutions of certain linear differential equations in which the coefficients of themay have a zero, Trans. Amer. Math. Soc., 40(1936), 127-154.
$\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{e},\mathrm{t}\mathrm{e}\mathrm{r}$

[SNH]
[T]

.Harris, Observations on the summability of confluent hypergeometric.Silverstone, S. Nakai andmechanics, Phys. Rev. 32(1985), 1341-1345.
quantum
semiclassicalonandfunctions$\mathrm{H}.\mathrm{J}$$\mathrm{J}.\mathrm{G}$.Turittin, Stokes multipliers for asymptotic solutions of a certain differential equation, Trans. Amer.
Math. Soc., 68 (1950), 304-329$\mathrm{H}.\mathrm{L}$[V] A.Voros, The return of the39 (1983), 211-338.

$\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{i},\mathrm{c}$oscillator. The complex WKB method. Ann. Inst. Henri Poincare,