数理解析研究所講究録1088 巻 1999 年 22-38

22

Exact WKB
for Schr\"odinger
with periodic potentials$\mathrm{a}1’1\mathrm{a}\mathrm{l}\mathrm{y}\mathrm{s}\mathrm{i}\mathrm{s}$$\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}_{0}1\prime 1\mathrm{s}$Tatsuya Koike

1

Introduction.

Let us consider one-dimensional Schr\"odinger equations,$(- \frac{d^{2}}{dz^{2}}+\eta^{2}Q(\cos Z))\phi=0$

,

(1)

and
$(- \frac{d^{2}}{dx^{2}}+\eta^{2}(\frac{Q(x)}{1-x^{2}}-\eta^{-2_{\frac{x^{2}+2}{4(1-x^{2})^{2}}})})\psi=0$,(2)where is a large parameter, and $Q(x)$ is a polynomial in . These equations
are related via the transformation,$x$$\eta$

$x$

$=$

cosz,

$\psi$$=$$\sqrt{\sin z}\phi$

(3)
.

(4)

Because this transformation (3) is a singular one, that is, its first derivative
vanishes at $z=k\pi$ , there appear singularities in (2) at $x=\pm 1$ .
Our interest is to analyze these equations from the viewpoint of exact
WKB analysis, in which the Stokes geometry (turning points, Stokes curves,
singularities, . . .) plays an important role to grasp the global behavior of solutions, and the interesting point in the above example is the Stokes geometryof (1) could not correspond to that of (2) due to the singularities $x=\pm 1$ of(2). To make a further reasoning, we need to clarify the role of $x=\pm 1$ in
WKB analysis. (1Ve already know the resurgent structure of (1) with respect
to due to the work given by E.Delabaere [D].)$\eta$

23

In this note, motivated by this example, we will mainly be concerned with

(5),$(- \frac{d^{2}}{d\overline{x}^{2}}+\eta^{2}(\frac{A(\overline{x})}{\overline{x}}+\eta^{-2_{\frac{B(\overline{x})}{\overline{x}^{2}}}}))\overline{\psi}=0$, wherenear
we also assume$\overline{x}=0$

are holomorphic functions atand. Our purpose here is$B(\overline{x})$$A(\overline{x})$$A(\mathrm{O})\neq 0$$\overline{x}=0$, and(i) to confirm that we must consider $x\sim=0$ as a turning point (Hence one);
Stokes curve emanates from$\overline{x}=0.$(ii) to determine the connection formula at $x=0\sim$ ,from a viewpoint of exact WKB analysis.
Thus $x=\pm 1$ are not only the singular points but also the turning pointsof (2), while the corresponding points $z=k\pi(k\in \mathbb{Z})$ are not. Hence thecorrespondence of the Stokes geometry between (1) and (2) does not hold atall, which leads us to the expectation that the turning points $x=\pm 1$ may beapparent. We can prove this expectation is true as a corollary of our result;
A turning point could become an apparent one if this point is a singularpoint of the subleading term of the potential. This is the second point wewish to emphasize in this note. We will explain its details in the last section,
together with some examples of Stokes curves for (1) and (2).
In what follows we will use the same notations and definitions as in[AKTI], such as WKB solutions, their Borel sum, etc. See [AKTI] or [DP](and, of course, [V]) and the references cited there for the explanation of
exact WKB analysis.

2

A simple pole as a turning point.

2.1

The canonical equation.

Let us start with the following simple, but important equation, which is(just like Airy equation near acalled a canonical equation of (5) at
simple turning point):$\overline{x}=0$$(- \frac{d^{2}}{dx^{2}}+\eta^{2}\frac{1}{x}+\frac{b}{x^{2}})\psi=0$

,

(6)

24

where is some constant, independent of . WKB solutions of this equation$b$$\eta$

are$\psi_{\pm}=\frac{1}{\sqrt{S_{\mathrm{o}\mathrm{d}\mathrm{d}}}}\exp(\pm\int_{0}^{x}S_{\mathrm{o}\mathrm{d}\mathrm{d}}dX)$.

(7), is the odd degree part of thewheresolution $S=\eta S_{-1}+S_{0}+\eta^{-1}S_{1}+\cdots$ , of the Riccati equation$S_{\mathrm{o}\mathrm{d}\mathrm{d}}=\eta S_{-1}+\eta^{-1}S_{1}+\eta^{-3}S_{3}+\cdots$$S^{2}+ \frac{dS}{dx}=\eta^{2_{\frac{1}{x}+}}\frac{b}{x^{2}}$.

(8)Explicitly we obtain$S= \eta\frac{1}{\sqrt{x}}+\frac{1}{4x}+\eta^{-1}\frac{16b+3}{32\sqrt{x^{3}}}+\eta^{-}2_{\frac{16b+3}{64x^{2}}+}\ldots$.

(9)Note that we can show, by induction, that(10)

$S_{j}=(const.)x- \frac{1}{2}j-1$

at $x=0$ becomes higher andholds, namely, the degree of singularities ofhigher as tends to infinity, which is the common property with “usual”
turning points. This is the intuitive reason why we call $x=0$ as a turning
point.
such thatIn the following we will fix the branch of$S_{j}$$j$$\sqrt{x}$$\sqrt{x}>0$(11)

for $x>0$

holds.
Now we show
Proposition 1 $WKB$ solutionsof (6) are Borel summable except theacross the positive
be analytic continuation of
positive real axis. Letreal axis from the lower half plane to the upper. Then we obtain$\psi_{\pm}$$\hat{\psi}_{\pm}$$\psi_{\pm}$$\hat{\psi}_{+}$$=$$\hat{\psi}_{-}$$=$$\psi_{+}+2i\cos\pi^{\sqrt{1+4b}\psi}-$$\psi_{-}$

.

,

(12)
(13)

We note here that $\sqrt{1+4b}$ is the difference of two characteristic exponents of (6) at $x=0$ .

25

Proof. The proof is a standard one (see [AKT2], [V]), but we shall follow the proof in detail because of its importance to understand the meaningof a turning point and a Stokes curve.
asBecause of (10), we can expand the WKB solutions$\psi_{\pm}$$\psi_{\pm}=\sqrt{x}\exp(\pm\eta\int_{0}^{x}S_{-1}dX)\sum_{j=0}\psi\pm,j(\eta\sqrt{x})\infty-j-1/2$

,

(14)

are constants independent of and . Hence their Borel transform
where
are of the following form:$\psi_{\pm,j}$$\eta$$x$that is,$\psi_{\pm,B}(X, y)$depends on

$x$

and

$s= \frac{1}{4}(\frac{y}{\sqrt{x}}+2)$respectively. In addition,,$\psi_{\pm,B}(x, y)$

$y$

(15)

,

$\psi_{\pm,B}(x, y)=\sum_{j=0}^{\infty}\frac{\psi_{\pm,j}}{\Gamma(j+1/2)}(\frac{y}{\sqrt{x}}\pm 2)^{j-1}/2$

through$1-s= \frac{1}{4}(-\frac{y}{\sqrt{x}}+2)$

,

(16)

satisfy$(- \frac{\partial^{2}}{\partial x^{2}}+\frac{1}{x}\frac{\partial^{2}}{\partial y^{2}}+\frac{b}{x^{2}}\mathrm{I}^{\psi}\pm,B(X, y)=0$

.

(17)

Hence we find that(18)$(s(1-s) \frac{d^{2}}{dx^{2}}+(\frac{3}{2}-3s)\frac{d}{ds}+4b)\psi_{\pm,B}=0$

holds, which enables us to determine$\psi_{\pm,B}$

explicitly as follows:$\psi_{+,B}(x, y)$$=$$\frac{1}{\sqrt{4\pi}}s^{-1/2}F(\alpha-\frac{1}{2},$ $\beta-\frac{1}{2},$$\psi_{-,B}(x, y)$$=$$\frac{1}{\sqrt{-4\pi}}(1-s)^{-1/}2F(\frac{3}{2}-\alpha,$$\frac{1}{2};S\mathrm{I}|_{s=*_{x}+\frac{1}{2}}4$$\frac{3}{2}-\beta,$

,

(19)

$\frac{1}{2};1-S)|_{s=_{\overline{4}\sqrt{x}}}\mathrm{A}-+\frac{1}{2}$

(20)

26

$\llcorner \mathbb{C}_{x}$$O$$\llcorner \mathbb{C}_{y}$${\rm Re} x$$\cross 2\sqrt{x}$$\cross x$

Figure 1: If is below the positive real axis in -plane (left), the location of
singularities (indicated by x) are as shown in the right figure. The
straight line in the right figure indicates the contour of Borel sum.
$x$

$x$

where $F$ denotes the Gauss hypergeometric function and a,satisfying

Because the singularities of$\alpha+\beta$

$=$

2,

$\alpha\beta$

$=$

$-4b$

$\psi_{\pm,B}(X, y)$$s=0,1$$\Leftrightarrow$$\beta$are constants
(21)
(22)

.

are

$y=-2\sqrt{x},$

,

(23)

$ydy$ ,

(24)

$2\sqrt{x}$its Borel sum$\psi_{\pm}(x, \eta)=\int^{\infty}\mp 2^{\cdot}\sqrt{x}’)e^{-}\psi_{\pm}\eta y(Bx,$where the integral are performed in parallel with the positive real axis, arewell-defined except${\rm Im}(2\sqrt{x})={\rm Im}(-2\sqrt{x})$ ,(25)
namely, except the positive real axis, which is called a Stokes curve (or line)
emanating from a turning point $x=0$ . (See Fig. 1.)
Let be below the positive real axis as shown in the left of Fig.1. The
right of Fig.1 indicates the contour of the Borel sum ofand the singularities of $\psi_{+,B}(x, y)$ in the y-plane.
If moves to the upper half plane across the positive real axis as shownin the left of Fig.2, we must deform the contour of Borel sum to $C$ as shownin the right of Fig.2, because the singularities $y=\pm 2\sqrt{x}$ move as moves.
Hence we obtain
.
(26)
$x$

$\psi_{+}$

$x$

$x$

$\hat{\psi}_{+}(_{X}, \eta)=\int_{C}e^{-\eta y}\psi+,B(X, y)dy$27$|\mathbb{C}_{\underline{y}}--$$\llcorner \mathbb{C}_{x}$.

$\cross\prime r$

Figure 2: If moves as in the left, the singularities also move. We must
deform the contour to .
$x$

$C$

$|\mathbb{C}_{y_{-}}$

Figure 3: The integration along $C$ is equal to sum of the integration along
the contours indicated in the right figure.
This is equal to (See Fig.3.)$\int_{-2\sqrt{x}}^{\infty}e-\eta y\psi_{+},B(X, y)dy+\int_{C’}e^{-\eta y}\psi_{+},B(X, y)dy$

(27)

.

By making use of the connection formula of hypergeometric functions$s^{-1/2}F( \alpha-\frac{1}{2},$ $\beta-\frac{1}{2},$$=$$\frac{1}{2};S\mathrm{I}$$\frac{\Gamma(\frac{1}{2})\Gamma(-\frac{1}{2})}{\Gamma(1-\alpha)\Gamma(1-\beta)}F(\alpha,$$\beta,$$\frac{3}{2};1-S)$$\frac{3}{2}-\beta)\frac{1}{2};1-s)$$+ \frac{\mathrm{I}^{\urcorner}(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(\alpha-\frac{1}{2})\Gamma(\beta-\frac{1}{2})}(1-s)^{-1/}2F(\frac{3}{2}-\alpha,$

,

(28)

28

we obtain$\int_{C’}e^{-\eta y}\psi_{+},B(X, y)$$=$$2i \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(\alpha-\frac{1}{2})\Gamma(\beta-\frac{1}{2})}\int_{2\sqrt{x}}^{\infty}e^{-\eta y}\psi-,B(x, y)dy$$=$$2i\cos\pi^{\sqrt{1+4b}}\psi_{-}(_{X}, \eta)$

This prove the proposition.

.

(29)
(30)

$\square$In the following, for simplicity, we denote (12) and (13) as

2.2
Let

$\psi_{+}$$\mapsto$$\psi_{+}+2_{\dot{i}}\cos\pi\sqrt{1+4b}\psi-$$\psi_{-}$

ト\rightarrow$\psi_{-}$

,

(31)
(32)

.

Connection formulae at a simple pole.
$\tilde{\psi}_{\pm}$

be WKB solutions of (5):
(33)

where
$\overline{S}=\eta\sqrt{\frac{A(\overline{x})}{\overline{x}}}+\frac{1}{4}(\frac{1}{x\sim}-\frac{A’(x\sim)}{A(\overline{x})})+\cdots$

.

(34)

Here, for brevity, we use the following notation:$\int_{0}^{\overline{x}}=\frac{1}{2}\int_{C_{\overline{x}}}$where$C_{\overline{x}}$

,

(35)

is the contour in -plane as shown in Fig.4. (Note that$\overline{x}$$S_{\mathrm{o}\mathrm{d}\mathrm{d},j(^{\sim})}x=R_{j}(\overline{X})^{\sim}X^{-\frac{1}{2}}j-1$

holds, where$R_{j}(^{\sim}X)$

is a holomorphic function at $x\sim=0.$ )

(36)

29
$\llcorner \mathbb{C}_{\overline{x}}$Figure 4: The contour$S_{\mathrm{o}\mathrm{d}\mathrm{d}}$Let$\Gamma$$C_{x}$. We place a cut ($\overline{1}\mathrm{n}\mathrm{d}\mathrm{l}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}$by the wiggly line) for.

be a (real one-dimensional) curve defined by(37)

We call this curve as a Stokes curve emanating fromwe show the following:$\overline{x}=0$. In this subsection

Borel sum of $WKB$ solutionsof (5) are well-defined nearin a counterclockwise manner with
$x=0$ except . Crossing this curve, we obtainrespect to the center$\overline{\psi}_{\pm}$$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}-2$$\Gamma$$\Gamma$$\overline{x}=0$$\overline{\psi}_{+}$$\overline{\psi}_{-}$$\mapsto$$\overline{\psi}_{+}+2i\cos\pi\sqrt{1+4B(0)}\overline{\psi}-$$\overline{\psi}_{-}$$\mapsto$

,

(38)

(39)

,

when

.

${\rm Re} \int_{0}^{\overline{x}}\sqrt{\frac{A(\overline{x})}{\overline{x}}}d\overline{x}>0$(In this case we call$\overline{\psi}_{+}$

is dominant on$\Gamma.$) In the case,${\rm Re} \int_{0}^{\overline{x}}\sqrt{\frac{A(\overline{x})}{\overline{x}}}d\overline{x}<0$

we replace$\psi_{\pm}$to$\psi_{\mp}$(40)

in (38) and (39) respectively.

(41)

30

We note here that

. Roughly
is independent of(i) the Stokes multiplierof the potential determines the
speaking, the leading term
detershape of Stokes geometry, while the lower order term
mines its Stokes multiplier;$2_{\dot{i}\mathrm{c}\mathrm{o}}\mathrm{s}\pi\sqrt{1+4B(0)}$$A(\overline{x})$$A(\overline{x})/x$$B(\overline{x})/\overline{x}^{2}$(ii) $\sqrt{1+4B(0)}$ is the difference of two characteristic exponents at$\overline{x}=0$.

We give the proof of Theorem 2 based on the transformation theory developed in [AKT2]. The first step in the proof is to show the following
proposition:
Proposition 3 There exists an infinite series
$x$

$=$

$x(\overline{x}, \eta)$$=$$x_{0}(\overline{x})+\eta^{-1_{X_{1}}}(\overline{x})+\cdots$

(42)

,

such that(i) each$x_{j}(\overline{x})$$x_{0}’(0)\neq 0$

are holomorphic at$\overline{x}=0$, satisfying $x_{0}(0)=0$ and;(ii) The following relation holds:$\frac{A(\overline{x})}{\overline{x}}+\eta^{-2}\frac{B(\overline{x})}{x^{2}\sim}=(\frac{\partial x}{\partial\overline{x}})^{2}(\frac{1}{x}+\eta^{-2_{\frac{B(0)}{x^{2}}}})-\frac{1}{2}\eta^{-2}\{x;\overline{X}\}$

.

(43)

Here ’ denotes the differentiation with respect to , and$\overline{x}$$\{_{X;\overline{X}}\}=\frac{x’’’}{x’}-\frac{3}{2}(\frac{x’’}{x’})^{2}$the Schwartzian derivative.
Moreover we can show that$x_{j}(0)=0$ holds for $j=1,2,$.

$x=x(\overline{x}, \eta)$

(44)

,

is $pre- B_{ore}l_{-}summab\iota e$ and$\cdots$

This proposition enables us to reduce (5) to (6), e.g., we find$\overline{\psi}_{\pm}(\overline{x}, \eta)=(\frac{\partial x}{\partial\overline{x}})^{-1/2}\psi\pm(X(\overline{x}, \eta),$$\eta)$

.

(45)

31

Proof of the .roposition. First, we construct two infinite series (42) and$\mathrm{p}$$b=b_{0}+\eta^{-1}b_{1}+\cdots$

where$b_{j}$(46),

are some constants, satisfying the condition (i) and$\frac{A(\overline{x})}{\overline{x}}+\eta^{-2}\frac{B(\overline{x})}{\overline{x}^{2}}=(\frac{\partial x}{\partial\overline{x}})^{2}(\frac{1}{x}+\eta-2\frac{b}{x^{2}})-\frac{1}{2}\eta^{-2}\{X;\overline{x}\}$

.

(47)

By substituting (42) into (47), and comparing both sides degree by degree,
we obtain(48)$\frac{A(\overline{x})}{\overline{x}}=(\frac{\partial x_{0}}{\partial\overline{x}})^{2}\frac{1}{x_{0}}$for the O-th degree, and$( \frac{2x_{0}’}{x_{0}}\frac{d}{d\overline{x}}-(\frac{x_{0}’}{x_{0}})^{2})Xj(_{\overline{X})=F_{j()-}}\overline{x}(\frac{x_{0}’}{x_{0}})^{2}b_{j-2}, (49.j)$for the $j(\geq 1)$ -th degree, where $b_{-1}=0$ and(50),$F_{1}(x)$

$=$

$0$

$F_{2}(x)$$=$$\frac{B(\overline{x})}{\overline{x}}-\frac{(X_{0}’X_{1})^{2}}{x_{0^{3}}}+2\frac{X^{J}X’x_{1}01}{x_{0^{2}}}+\frac{1}{2}\{_{X}0;x\}\sim$

,

(51)

and
$F_{n}(.x)$$=$$k_{1,\mu_{1^{+}\mu}}+k.2..+ \mu\sum_{\mu+\iota^{=}}(-1)^{l}+1X+\iota=n\frac{X\cdots X\mu_{1}+1\mu\iota+1}{x_{0^{l+1}}}JJk_{1}^{X_{k}}2$$+ \sum_{\mu\mu 1+\cdots\mu l=}(-1)^{l}+1(\iota+1k_{1}+k_{2}+m+\mu=n-2)b_{m^{X’}}k_{1}^{X^{J}}k2\frac{x_{\mu_{1+}1}\cdots x\mu\iota+1}{x_{0^{l+2}}}$$+ \frac{1}{2}\{_{X;\overline{x}}\}l+2$

Here we denote by$\{x;\overline{x}\}_{n}$.

$(52.n)$

the n-th degree part of$\{x;\overline{x}\}=\sum_{n=0}\eta\infty-n\{X;\overline{x}\}_{n}$.

$\{x;\overline{x}\}$

:

(53)

32

The holomorphic solutions of (48) and (49.1) are$x_{0}(\overline{x})$(54)$\frac{1}{4}(\int_{0}^{x}\sqrt{\frac{A(\overline{x})}{\overline{x}}}\sim d_{X}^{\sim})^{2}$$=$$A(0)_{X}+\cdots$$=$,$x_{1}(_{\overline{X})} =(55)
0,$(56)

which satisfy the condition (i) and (ii).
The holomorphic solution of (49.2) is(57)$x_{2}(_{\overline{X})}= \sqrt{x_{0}}\int_{0}^{\overline{x}}\frac{\sqrt{x_{0}}}{2x_{0}},(\frac{B(\overline{x})}{\overline{x}^{2}}-(\frac{x_{0}’}{x_{0}})^{2}b_{0}+\frac{1}{2}\{X0;X\}\sim \mathrm{I}d_{X}^{\sim}.$does not vanish at
We chose $b_{0}=B(0)$ to ensure $x_{2}(0)=0$ ; Otherwise. In this case) has a pole of degree(or, at least,, and(49.4) (or (49.6)) does note have a solution which satisfies (i).
’ can solve $(49.j)(j\geq 3)$ recursively in a similar fashion:$x_{2}(\overline{x})$$\overline{x}=0$$\geq 3$

$F_{6}(X)\sim$$F_{4}(\overline{x})$$)\eta$$x_{j}( \overline{x})=\sqrt{x_{0}}\int_{0}^{\overline{x}}\frac{\sqrt{x_{0}}}{2x_{0}},(F_{j}(\overline{x})-(\frac{x_{0}’}{x_{0}})^{2}bj-2)d_{X}^{\sim}$

and$b_{j-2}=\overline{x}^{2}F_{j}(\overline{x})|_{\overline{x}=0}$,$(58.j)$.

$(59.j)$,has a (at most) double pole at the origin because
Note that; Oth.
vanish atchoose $(59.j)$ to ensure
vanish atand (49) does note have a solution
has a pole of degree
erwisewhich satisfies (i).
and as desired.
We have thus constructed
Next, by multiplying both side of (47) by , and taking the limit tends
to , we obtain$F_{j}(x)$$\overline{x}=0$$x_{0}(\overline{x}),$$\mathrm{V}^{\gamma}\mathrm{e}$$x_{j}(\overline{x})$$\cdots$$x_{j-1}(\overline{x})$$\overline{x}=0$$\geq 3$

$F_{j+2}(\overline{x})$$b$$x(\overline{x}, \eta)$$\overline{x}^{2}$$\overline{x}$$0$$\eta^{-2}B(0)$$=$$\frac{1}{x}\mathrm{i}\mathrm{m}_{0}arrow(\frac{\partial x}{\partial\overline{x}})^{2}(\overline{\frac{x}{x}})^{2}\eta^{-2}b$$=$$\eta^{-2}b$

(60)

(61)

,

because $x_{j}(0)=0$ . Hence we obtain$b_{0}=B(0),$ $b_{j}=0$

for $j\geq 1$ .

(62)

33

of (42) we have conWhat remains is to show thestructed so far, and we can show that there exist constants $C>0$ suchthat$\mathrm{p}\mathrm{r}\mathrm{e}- \mathrm{B}_{0}\mathrm{r}\mathrm{e}1_{-}\mathrm{S}\mathrm{u}\mathrm{m}\mathrm{m}\mathrm{a}\mathrm{b}\mathrm{i}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}$$r,$$|^{\sim} \sup_{x|\leq r-\epsilon}|\frac{x_{n}(_{\overline{X})}}{x_{0}(\overline{x})}|\leq n!c^{n-1}\epsilon^{-n}$$| \overline{x}|\leq r-\sup|X_{n}(\overline{X}\epsilon)|,$for any$\epsilon$$| \sup_{\overline{x}|\leq r-\epsilon}|X_{n}(/\overline{X})|,$,$(63.n)$satisfying $0<\epsilon<r$ , but we omit the details here. (Cf. [AKT2].)$\square$We give one remark that we can show, by induction, that$x_{2j+1}(_{\overline{X}})\equiv 0$

holds for $j=0,1,2,$$\cdots$

(64)

,

.

By expanding the right hand side of (45), we obtain(65)$\overline{\psi}_{\pm}(_{\overline{X}}, \eta)=(\frac{\partial x}{\partial\overline{x}})^{-1/}\sum_{n}^{2}\infty=0\frac{1}{n!}(_{j=}\sum_{1}^{\infty}xj\eta^{-})^{n}j(\frac{\partial}{\partial x})^{n}\psi_{\pm}(x, \eta)$$x=x\mathrm{o}(\overline{x})$Hence, after the Borel transform, we obtain,$\overline{\psi}_{\pm,B}(\overline{x}, y)$$=$$( \frac{\partial x(\overline{x},\partial_{y})}{\partial\overline{x}})^{-1/2}$$\cross\sum_{n=0}^{\infty}\frac{1}{n!}(_{j}=\sum_{1}^{\infty}Xj(\frac{\partial}{\partial y})^{-j}\mathrm{I}n(\frac{\partial}{\partial x})^{n}\psi_{\pm,B}(x, y)|_{x=x_{0}}(\overline{x})(66)$-summay of
By using this relation (together with theand Proposition 1, we can prove Theorem 2. (We omit the details here.)$\mathrm{P}^{\mathrm{r}\mathrm{e}- \mathrm{B}}\mathrm{o}\mathrm{r}\mathrm{e}$3$x=x(\overline{x},$ $\eta)$)

Application to the periodic potential.

Let us now return to analyze (1) and (2) by making use of the result obtainedso far. In the following we will assume that $Q(\pm 1)\neq 0$ holds to ensure that$x=\pm 1$ are simple poles of the leading term of the potential.

34

Figure 5: The Stokes curves for (1) (left) and for (2) (right) when $Q(x)=$x.These Stokes curves are degenerate.

(2).

are

First of all we will see some examples of the Stokes geometry of (1) andAs the first example, we take $Q(x)=x$ . In this case turning points of (1)$z=(k+ \frac{1}{2})\pi$: simple turning points(67)

where $k\in$ Z. See the left of Fig.5 for the Stokes curves of (1), which aredefined by${\rm Im} \int_{a}^{z}\sqrt{\cos z}dZ=0$

,

(68)

where is a turning point.
On the other hand, turning points of (2) are$a$$x=\{\pm 10$.
.

ing point,$\mathrm{S}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{e}\mathrm{p}\mathrm{o}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{m}_{11}\mathrm{p}1\mathrm{e}\mathrm{t}\mathrm{u}\mathrm{r}.\mathrm{n}$$\cdot$(69)

See the right of Fig.5 for Stokes curves of (2), which are defined by${\rm Im} \int_{a}^{x}\sqrt{\frac{x}{1-x^{2}}}dx=0$

,

(70)

35

Figure 6: We resolve the degeneracy in Fig.4 by changing$\eta$

to$e^{i\theta}\eta$for$\theta=$$-1/2$ .

where is a turning point.
These Fig.5 are, however, degenerate, namely, there are Stokes curvesconnecting two turning points. (In this case, there appear singularities on
the contour of the Borel sum. See [DP].) We resolve this degeneracy by, where $\theta<0$ is sufficiently small. Then Stokes curves arechanging to
as in Fig.6.
By the transformation $x=\cos z$ , simple turning points of (1) are mapped
to a simple turning point (2), while the normal points $z=k\pi(k\in \mathbb{Z})$ of (1)are mapped to $x=\pm 1$ . Hence the correspondence of the Stokes geometry
does note hold, as we observed in Introduction.
$a$$e^{i\theta}\eta$$\eta$As the second example, we take $Q(x)=x+t$ , where $t>1$ . In this case,turning points of (1) are$z=\pm\tau+(2k+1)\pi$: simple turning points(71)

where $t=\cos\tau$ and $k\in$ Z. See the left of Fig.7 for the Stokes curves.
Turning points of (2) are36

Figure 7: The Stokes curves for (1) (left) and for (2) (right) when $Q(x)=$Stokes curves are degenerate.
$x+2.\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{e}$$x=\{$$-t$$\pm 1$: a simple turning pints,: simples poles,(72)

and their Stokes curves are as in the right of Fig.7. These Stokes curves
are also degenerate. We try to resolve this degeneracy in the same manneras in the above example. For generic $\theta<0$ , we find that we can resolve
this degeneracy of the Stokes geometry. But, what is interesting is that the
Stokes geometry becomes degenerate infinitely many times, as tends to .
(See Fig.8, 9.)$\theta$$0$

In any example, the correspondence of the Stokes geometry does note
hold between (1) and (2).

37

Figure 8: We resolve the degeneracy in Fig.6 by changing$-1$ .

$\eta$

to$e^{i\theta}\eta$for$\theta=$for
toFigure 9: We resolve the degeneracy in Fig.6 by changing$\theta=-1/2$ .
By comparing this figure with Fig.7, we find that
the Stokes geometry becomes degenerate between-l $<\theta<-1/2$ .
Furthermore, we find that the Stokes geometry becomes degenerate
infinitely many times as tends to .
$e^{i\theta}\eta$$\eta$$\theta$$0$

38

Now we apply Theorem 2 to (2) at $x=\pm 1$ . We must note that(i) the connection formula (2) depends only on the lower degree part ofthe potential;(ii) The lower degree part of the potential are independent of $Q(x)$ .

Thus the connection formulae at $x=\pm 1$ do not depend on $Q(x)$ .
Characteristic exponents of (2) at $x=\pm 1$ are$\frac{3}{4}$$\frac{1}{4}$

.

(73)

Hence the Stokes multiplier which arises when we cross the Stokes curves
emanating from $x=\pm 1$ vanishes because$2i \cos\pi(\frac{3}{4}-\frac{1}{4})=0$(74)

holds, which implies the turning points $x=\pm 1$ are apparent, as we expected.

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