数理解析研究所講究録 1433 巻 2005 年 9-26

9On a connection problem of simple
pole type operators of second order inexact WKB analysis
京都大学理学研究科 小池 達也 (KOIKE, Tatsuya)
Department of Mathematics, Kyoto University\S 0. Introduction
In our exact WKB theoretic study of simple pole type operators ([AKKT2]),
we announced that the connection problemfor WKB solutions of any simple
pole type operators can be reduced to that of a second order equation of the
form
(0.1)

where
(0.2)

$(- \frac{d^{2}}{dx^{2}}+\eta^{2}V(x, \eta))\psi$$\eta$$=0$,denotes a large parameter,$V(x, \eta)=\frac{V_{0}(x)}{x}+\eta^{-1}\frac{V_{1}(x)}{x}+\eta^{-2}\frac{V_{2}(x)}{x^{2}}+\sum_{j\geq 3}\eta^{-j}\frac{V_{j}(x)}{x}$,. Actually, thisare holomorphic functions near the origin, andequation is obtained if we eliminate the first order termof simple pole type
$V_{0}(0)\neq 0$$\{V_{j}\}$operator of the second order by changing the unknown functions.
In [K1] and [K2] the case where $V_{1}=0$ and $V_{j}=0$ for $j\geq 3$ wasconsidered. There it was shown that although the origin is a regular singularpoint of (0.1), it plays the same role as a turning point. For example thelogarithmic derivative $S(x, \eta)=\eta S_{-1}+S_{0}+\eta^{-1}\mathit{3}_{1}+\cdots$ of WKB solutions
behaves like $S_{j}=O(x^{-j/2-1})$ near the origin. That is, the order of singularityof becomes worse and worse as increases. This is a typical feature of thelogarithmic derivative of WKB solutions near turning points. This situationis also observed for (0.1) and we expect that the origin also plays the samerole as a turning point. This expectation has been validated in [K1] andWKB[K2], by analyzing the singularity structure of the Borel
solutions of (0.1) when $V_{1}=0$ and $V_{j}=0$ for $j\geq 3$ . In this paper we extend
the analysis of [K1] and [K2] to a more general potential as given in (0.2)$S_{\mathrm{y}}$$j$transformof

10\S 1. Connection formulas for simple pole type
operators of second order
We consider(1.1)$(- \frac{d^{2}}{dx^{2}}+\eta^{2}Q(x, \eta))\psi=0$

where$\eta$

denotes a large parameter, and$(1.2)\backslash$

with

,

$Q(x, \eta)=\frac{Q_{0}(x)}{x}+\eta^{-1}\frac{Q_{1}(x)}{x}+\sum_{j\geq 2}\eta^{-j}\frac{Q_{j}(x)}{x^{2}}$$\{Q_{j}\}$(A. 1) Eachsatisfying the conditions (A. ), (A.2) and (A.3) below:$\mathrm{I}$$Q_{j}$

are holomorphic in a neighborhood$Q_{0}(0)\neq 0$

(A.2)

.

$U\subset \mathbb{C}$of the origin, and.

is pre-Borel summable in , i.e., for any compact set $K$ in, $C_{K}>0$ for which
there exist constants$U$$\{Q_{j}\}$$\mathrm{U}$,$A_{K}$(A.3)$\sup_{x\in K}|Q_{j}(x)|\leq A_{K}C_{K^{j}}j!$holds for any$j\geq$Q.

(A.3) $Q_{j}(0)=0$ for $j\geq 3$ .

We study the analytic structure of Borel transform of WKB solutions of(1.1). We note that the equation (1.1) has a slightly different form from
(0.1); instead we assume (A.3). This is for a notational simplicity.
For this equation we can construct WKB solutions of the form
(1.4)$\psi_{\pm}=\frac{1}{\sqrt{S_{\mathrm{o}\mathrm{d}\mathrm{d}}(x,\eta)}}\exp(\pm\oint_{0}^{x}S_{\mathrm{o}\mathrm{d}\mathrm{d}}(x, \eta)dx)$

,

where
(1.5)

$S_{\mathrm{o}\mathrm{d}\mathrm{d}}(x, \eta)$$=$$(S^{+}(x, \eta)-S^{-}(x, \eta))/2$$=$$\eta S_{\mathrm{o}\mathrm{d}\mathrm{d},-1}(x)+S_{\mathrm{o}\mathrm{d}\mathrm{d},0}(x)+\eta^{-1}S_{\mathrm{o}\mathrm{d}\mathrm{d},1}(x)+\cdots$

11

with

, and

$S_{\mathrm{o}\mathrm{d}\mathrm{d},-1}(x)=\sqrt{Q_{0}(x)}/x$

(1.6)

with

$S^{\pm}(x_{f}\eta)=\eta S_{-1}^{\pm}(x)+S_{0}^{\pm}(x)+$$S_{-1}^{\pm}(x^{1},$$=\pm\sqrt{Q_{0}(x)}/x$(1.7)

op $-1S_{1}^{\pm}(x)+\cdots$

are two formal solutions of the Riccati equation$S^{2}+ \frac{dS}{dx}=\eta^{2}Q(x, \eta)$

associated to (1.1). We call WKB solutions of (1.1) normalized as (1.4) as
WKB solutions normalized at the origin.
WKB solutions (1.4) have the following expansion:
,

(1.8)

$\psi_{\pm}=e^{\pm\eta s(x)}\sum_{j=0}^{\infty}\psi_{\pm,j}(x)\eta^{-j-1/2}$

where
(1.9)

$s(x)= \oint_{0}^{x}S_{\mathrm{o}\mathrm{d}\mathrm{d},-1}(x)dx=\oint_{0}^{x}\sqrt{\frac{Q_{0}(x)}{x}}dx$.

Then the Borel transform of WKB solutions (1.4) are defined to be(1.10)$\psi_{\pm,B}(x, y)=\sum_{j=0}^{\infty}\frac{\psi_{\pm,j}(x)}{\Gamma(j+1/2)}(y\pm s(x))^{j-1/2}$

Concerning the analyticity of the Borel transform
we obtain$\emptyset\pm,B$.

of the WKB solutions

Theorem 1. We assume conditions (A.1); (A.2) and (A.3). Then for the
Borel transform (1.10) of $WKB$ solutions of (1.1), we can find a positive
constant for which the following hold:$r_{0}$(i) $(y+s(x))^{1/2}\psi_{+,B}$ and $(y-s(x))^{1/2}\psi_{-,B}$ converge and define holomorphic
respectively, where
andfunctions in$W_{+}(r_{0})$(1.11)(ii)$W_{-}(r_{0})$$W_{\pm}(r_{0})=\{(x, y)\in \mathbb{C};0<|x|<r_{0}, |y\pm s(x)|<2|s(x)|\}$.

can be analytically continued and define multi-valued
andanalytic functions in $W_{-}(r_{0})\backslash \{y=s(x)\}$ and $W_{+}(r_{0})\backslash \{y=-s(x)\}$ ,respectivel ,$\psi_{+,B}$$\psi_{-,B}$

$y$

12

(iii) The discontinuity of $\psi_{+,B}(x, y)$ (resp.

(1.12)$\psi_{-,B}$) along the cut$\{(x, y)\in \mathbb{C}^{2}; {\rm Im} y={\rm Im} s(x), {\rm Re} y\geq{\rm Re} s(x)\}${( ,(1.13) (resp.

$y)\in \mathbb{C}^{2}$$x$;${\rm Im} y={\rm Im}$$(-s(x))$ , ${\rm Re} y\geq{\rm Re}$ $(-s(x))\}$ )coincides with
(1.14)$2\mathrm{i}\cos(\pi\sqrt{1+4Q_{2}(0)})\psi_{-,B}(x, y)$(1.15)(resp.

$2\mathrm{i}\cos(\pi\sqrt{1+4Q_{2}(0)})\psi_{-,B}(x,$ $y)$).

Here we note that $\sqrt{1+4Q_{2}(0)}$ is a difference value of two characteristic
exponents of (1.1) at the regular singular point $x=0$ .
If we assume endless continuability and some appropriate growth con(See [DP]. See also [V] and [DDP].), we obtain the
ditions onfollowing connection formulas of Borel sum of WKB solutions from Theorem: first, in view of singularity structure of
we define the Stokes$\psi_{\pm,B}(x, y)$$\mathrm{t}\mathrm{h}\mathrm{m}:\mathrm{m}\mathrm{a}\mathrm{i}\mathrm{n}$curve(1.16)$\emptyset\pm,B_{1}$$\gamma$${\rm Im}(s(x))=$Then when we cross
we obtain(1.17)(1.18)$\gamma$Jm $(-s(x))$$\Leftrightarrow$${\rm Im} \int_{0}^{x}\sqrt{\frac{Q_{0}(x)}{x}}dx=0$.

in a counterclockwise manner with a center the origin,$\psi_{+}$$-\neq$$\psi_{+}+2\mathrm{i}\cos(\pi\sqrt{1+4Q_{2}(0)})\psi_{-}$$\psi_{-}$$\vdasharrow$$\psi_{-}$,,

if ${\rm Re} \oint_{0}^{x}\sqrt{Q_{0}(x)/x}dx>0$ holds along , or$\gamma$(1.19)(1.20),$\psi_{+}$$\vdasharrow\psi_{+}$$\psi_{-}$$-\succ\psi_{-}+2\mathrm{i}\cos(\pi\sqrt{1+4Q_{2}(0)})\psi_{+}$,

if ${\rm Re} \int_{0}^{x}\sqrt{Q_{0}(x)/x}dx<0$ holds along .
$\gamma$

Remark. The Stokes multiplier $2\mathrm{i}\cos(\pi\sqrt{1+4Q_{2}(0)})$ depends on the subleading terms $Q_{2}(x)$ of the potential $Q(x, \eta)$ . This fact was essential in theargument discussed in [AKKT2]

13\S 2. Sketch of the proof of the connection formulas
To prove Theorem 1, we first transform (1.1) to a canonical equation(2.1)$(- \frac{d^{2}}{dx^{2}}+\eta^{2}(\frac{1}{x}+\eta^{-2}\frac{\lambda}{x^{2}}))\psi=0$,

C. Here, to distinguish the independent
with
where
variable and the unknown functions of (1.1) from (2.1), we use and as
the independent variable and the unknown functions of (1.1) respectively, In
fact we can prove the following:$\lambda=\lambda_{0}+\eta^{-1}\lambda_{1}+\cdots$$\lambda_{j}\in$$\tilde{\psi}$$\tilde{x}$Proposition 2. Assume (A.I) and (A . 2). Then we can find a neighborhoodandV of$\tilde{x}=0$(2.2)$x=x(\tilde{x}, \eta)=x_{0}(\tilde{x})+\eta^{-1}x_{1}(\tilde{x})+\cdots$

are holomorphic functions in some open neighborhood
where, and satisfy the following:$V\subseteq$$\{x_{j}(\tilde{x})\}_{g\geq 0}$$U$(i) $x_{0}(0)=0$ , $(dx_{0}/d\tilde{x})(0)\neq 0$ .
(n) $x_{j}(0)=0$ for $j\geq 1$ .

(iii) there exist positive constants(2.3), $C$ for which the following holds in V.

$\cdot$$A$$|x_{j}(\tilde{x})|\leq AC^{j-1}j!$.

(iv) The following relation holds degree by degree with respect to :$\eta$(2.4)$Q( \tilde{x}, \eta)=(\frac{\partial x}{\partial\tilde{x}})^{2}(\frac{1}{x(\tilde{x},\eta)}+\eta^{-2}\frac{\lambda}{x(\tilde{x},\eta)^{2}})-\frac{1}{2}\eta^{-2}\{x(\tilde{x}, \eta);\tilde{x}\}$,foris given by $\lambda_{j}=Q_{j+2}(0)$ . (Hence
derivative,
denotes the Schwarzian$j\geq 1$
if (A.3) holds.) Here,
where A$\lambda_{j}=0$$=\lambda_{0}+\eta^{-1}\lambda_{1}+\cdots$$\{x,\overline{x}\}$$\mathrm{i}.e.$(2.5)$\{x;\tilde{x}\}=\frac{x’}{x’}-\frac{3}{2}(\frac{x’}{x’})^{2}$,

where ’ denotes the differentiation with respect to .
$\tilde{x}$14(v) The follornirtg relation holds among the $WKB$ solutions
normalized at the origin, artdof (2.1):$\tilde{\emptyset}\pm$

of (1.1)$\psi\pm$(2.6)$\tilde{\psi}_{\pm}(\tilde{x}, \eta)=(\frac{\partial x}{\partial\tilde{x}})^{-1/2}\psi_{\pm}(x(\tilde{x}, \eta),$ $\eta)$.

Proof of this Proposition 2 will be given in the subsequent sections.
Once this proposition is proved, then we can prove Theorem 1 in the samemanner as in [K2]. In fact, by considering the Borel transform of (2.6), vxe
obtain(2.7)$\tilde{\psi}_{\pm,B}(\tilde{x}, y)=(P_{\mp 2\sqrt{x}}\psi_{\pm,B}(x, y))_{x=x_{0}(\tilde{x})}$

,

where
(2.8)
$P_{y0}(x; \frac{\partial}{\partial x}, \frac{\partial}{\partial y})$.
$=J(x) \sum_{N=0\mu+\nu_{0}}^{\infty}\mathrm{I}_{n=N}\mu+\nu_{1}^{1}+_{n}=\nu\mu_{f}0\mathrm{I}_{n=\mu}(-1)^{n_{\frac{\Gamma(n+1/2)}{\Gamma(1/2)m!n!}}}$$\mathrm{x}\tilde{x}_{\mu_{1}+1}(x)\cdots$$\tilde{x}_{\mu_{m}+1}(x)\tilde{x}_{\nu_{1}+1}’(x)\cdots$ $\tilde{x}_{\nu_{n}+1}’(x)(\frac{\partial}{\partial x})^{m}(\frac{\partial}{\partial y})_{y0}^{-N}$

,

and
(2.9)

$(x_{0}^{l}(\tilde{x}))^{-1/2}=J(x_{0}(\tilde{x}))$,$x_{j}(\tilde{x})=\tilde{x}_{j}(x_{0}(\tilde{x}))$.

(See [K2] for the definition of) In [K2] we study the analyticity,of the right-hand side of (2.7) by using the explicit description ofwhich are expressed by Gauss hypergeometric functions (Here we note thatsince we assume (A.3) in Theorem 1.):$(\partial/\partial y)_{y0}^{-N}.$$\psi_{\pm,B}(x, y)$$\lambda=\lambda_{0}$$\psi_{+,B}(x, y)=\frac{1}{\sqrt{4\pi}}s^{-1/2}F(\alpha-\frac{1}{2},$$\beta-\frac{1}{2}$,$\frac{1}{2};s)|_{s=_{\overline{4}\overline{\sqrt{x}}}^{p}+\frac{1}{2}}$$\psi_{-,B}(x, y)=\frac{1}{\sqrt{-4\pi}}(1-s)^{-1/2}F(\alpha-\frac{1}{2},$ $\beta-\frac{1}{2}$,,$\frac{1}{2};1-s)|_{s=\frac{y}{4\sqrt{x}}+\frac{1}{2}}$,where and are constants satisfying $\alpha+\beta=2$ and
4Ao- There wehave only used the properties in Proposition 2. Hence once Proposition 2 isproven, then we can use the same argument to obtain Theorem 1.
$\alpha$$\beta$$\alpha\beta=$

15\S 3. Formal coordinate transformation to a canonical equation
so thatIn this section we construct the transformation function
satisfies the properties (i), (ii), (iv) andit satisfies (2.4), and then this
2 (iii) will be given in \S 4.
Proposition
proofof(v) in Proposition 2. The
By comparing (2.4) degree by degree with respect to , we obtain$\{x_{j}(\tilde{x})\}$$\{x_{j}\}$$\eta$$( \frac{dx_{0}}{d\tilde{x}})^{2}\frac{1}{x_{0}}=\frac{Q_{0}(\tilde{x})}{\tilde{x}}$(3.1)(3.1.0),$\{$$(2 \frac{x_{0}’}{x_{0}}\frac{d}{d\tilde{x}}-(\frac{x_{0}’}{x_{0}})^{2})x_{n}(\tilde{x})=F_{n}(\tilde{x})-(\frac{x_{0}’}{x_{0}})^{2}\lambda_{n-2}$

for $n\geq 1$ (we set(3.2)$F_{1}(\tilde{x})$

$=$

(3.3)

$F_{2}(\tilde{x})$$=$(3.1.n)for the convenience). Here$\mathrm{A}_{-}\mathrm{i}=0$$\frac{Q_{1}(\tilde{x})}{\tilde{x}}$,$\frac{Q_{2}(\tilde{x})}{\tilde{x}^{2}}-\frac{x_{1}^{\prime 2}}{x_{0}}-(\frac{x_{0}’x_{1}}{x_{0}})^{2}+\frac{2x_{0}’x_{1}’x_{1}}{x_{0}^{2}}+\frac{1}{2}\{x_{0};\tilde{x}\}$

,

and
(3.4)
$F_{n}(\tilde{x})$$=$$\sum$$\sum$$\frac{Q_{n}(\tilde{x})}{\tilde{x}^{2}}+$$\mu+\nu+l=n\mu_{1}\dagger$.

$(-1)^{l+1}x_{\nu_{1}}’x_{\nu_{2}}’ \frac{x_{\mu_{1}+1}\cdots x_{\mu_{l}+1}}{x_{0^{l+1}}}$$+\mu_{l}=\mu$$\nu_{1}+\nu_{2}=\nu$$0\leq\mu_{\mathrm{j}}\leq n-2$$0\leq\nu_{\mathrm{j}}\leq n-1$$+$$\sum$$\sum$$($-1$)^{l+1}(l +1) \lambda_{k}x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}\cdots x_{\mu\iota+1}}{x_{0^{l+2}}}$$\mu+\nu+l+k=n-2\mu_{1}+\cdot+\mu_{l}=\mu$$k\neq n-2$$\nu_{1}+\nu_{2}=\nu$$+ \frac{1}{2}\mathrm{I}_{n-2}\sum_{\mu_{1}\mu+l+\cdots+\mu_{l}=\mu}($$($$+ \frac{3}{4}\mathrm{I}\sum_{n\mu+l-2\mu_{1}+\cdot+\mu_{\mathrm{t}}=\mu}$

-1

-1

,

$)^{l}x_{k}’(x)$ $\frac{x_{\mu_{1}+1}’\cdots x_{\mu \mathfrak{x}+1}’}{x_{0}^{l+1}}$$)^{l+1}(l +1)x_{\nu_{1}}’x_{\nu+2}’ \frac{x_{\mu_{1}+1}’\cdots x_{\mu_{l}+1}’}{x_{0^{l+1}}’}$$\nu_{1}+\nu_{2}=\nu$

We will now solve (3.1.n) for n(3.5)$\geq 0$step by step. First we obtain$x_{0}( \tilde{x})=(\frac{1}{2}\int_{0}^{\tilde{x}}\sqrt{\frac{Q_{0}(\tilde{x})}{\tilde{x}}}d\tilde{x})^{1/2}$.

16from (3.1.0). Then we easily confirm that (i) is satisfied since
expanded as(3.6)$x_{0}(\tilde{x})$can be$x_{0}(\tilde{x})=\sqrt{Q_{0}(\mathrm{O})}\tilde{x}+O(\tilde{x}^{2})$near the origin, and $Q_{0}(0)\neq 0$ by our assumption (A. ). We take theandis holomorphic inof the origin so thatneighborhood
does not vanish in . The holomorphic solution (3.1.1) is obtained by$\mathrm{I}$$U_{0}\subset U$$U_{0}$$x_{0}’$

(3.7)

since
in

$U_{0}$$x_{0}$$U_{0}$(3.8)$\mathrm{x}\mathrm{o}(\mathrm{x})=\sqrt{x_{0}(\tilde{x})}\int_{0}^{\overline{x}}\frac{\sqrt{x_{0}(\tilde{x})}}{2x_{0}(\tilde{x})},F_{1}(\tilde{x})d\tilde{x}$has a simple pole at the origin (See (3.2).). This is holomorphic
and vanish at the origin. Now we solve (3.1.2). We first observe that

$F_{1}(\overline{x})$$x_{1}$$x_{2}( \tilde{x})=\sqrt{x_{0}(\tilde{x})}\oint_{0}^{\tilde{x}}\frac{\sqrt{x_{0}(\tilde{x})}}{2x_{0}(\tilde{x})},(F_{2}(\tilde{x})-(\frac{x_{0}’}{x_{0}})^{2}\lambda_{0})d\tilde{x}$give a holomorphic solution of (3.1.2). Then we choose(3.9)$\lambda_{0}=(\frac{x_{0}}{x_{0}},$ $)^{2}F_{2}(\tilde{x})|_{\tilde{x}=0}$

to ensure that this solution vanish at the origin.
Remark. At the level $n=2$ , we can obtain a holomorphic solution of (3.1.n)even if we do not assume the condition (3.9). However, the resulting solution
does not vanish at the origin. Hence
would have higher order$(\geq 2)$ poles at the origin by its definition, in general.
Thus we can not expect
a holomorphic solution of (3.1.n) near the origin. We can obtain holomorphic
solution of (3.1.2) without the condition (3.9).
$F_{n}(\tilde{x})$$x_{2}(\tilde{x})$We now inductively determine
havefor $n\geq 3$ . We first note thata double pole at the origin by our induction hypothesis. Then we choose$x_{n}(\tilde{x})$$F_{n}(\tilde{x})$$\lambda_{n-2}$as(3.10)$\lambda_{n-2}=(\frac{x_{0}}{x_{0}},$ $)^{2}F_{n}(\tilde{x})|_{\tilde{x}=0}$

17

By this choice of the constant(3.11)$\lambda_{n-2}$, we obtain a holomorphic solution$x_{n}( \tilde{x})=\sqrt{x_{0}(\tilde{x})}\int_{0}^{\tilde{x}}\frac{\sqrt{x_{0}(\tilde{x})}}{2x_{0}(\tilde{x})},(F_{n}(\tilde{x})-(\frac{x_{0}’}{x_{0}})^{2}\lambda_{0})d\tilde{x}$is holomorphic inof (3.1.n) which vanishes at the origin. Thishas been comof
runs
and
the
construction
induction
Hence ourpleted.
We then prove $\lambda_{j}=Q_{j+2}(0)$ for $j\geq 0$ . Multiplying (2.4) by , andtaking the limit tends to zero, we obtain$U_{0}$$x_{n}(\tilde{x})$$\}$$\{x_{j}(\tilde{x})\}$$x\sim 2$$\tilde{x}$(3.12)$\lim_{\tilde{x}arrow 0}\tilde{x}^{2}Q(\tilde{x}, \eta)=\eta^{-2}\lambda\lim_{\tilde{x}arrow 0}(\frac{\partial x}{\partial\tilde{x}})^{2}\frac{\tilde{x}^{2}}{x(\tilde{x}_{7}\eta)^{2}}$.

The right-hand side of (3.12) becomes(3.13)$\eta^{-2}(Q_{2}(0)+\eta^{-1}Q_{3}(0)+\cdots )$while the lefthand side of (3.12) becomes
Hence we obtain(3.14)$\eta^{-2}\lambda$

because $x_{j}(0)=0$ for a1I$j$.

$\lambda=Q_{2}(0)+\eta^{-1}Q_{3}(0)+\cdots$

Since (v) is a direct consequence of (2.4), the remaining part of the proofof Proposition 2 is (iii), the pre-Borel summability of the transformation. We prove this pre-Borel summability in the next section.
function$x(\tilde{x}, \eta)$\S 4. Pre-Borel summability of the transformation function.
In this section we prove Proposition 2 (iii). We can assume that there
exist positive constants $B$ , $D$ and so that following hold:
$R$

(a)

$x_{0}(\tilde{x})$(b)$x_{0}(\tilde{x})$is holomorphic in $\{x;|x|\leq R\}$ .

is holomorphic in $\{x;|x|\leq R\}$ .

(c) For every ,
$n$

(4.1)

$|^{\sup_{x|\leq R}}|( \frac{x_{0}}{x_{0}},$$)^{2} \frac{Q_{n}(\tilde{x})}{\tilde{x}^{2}}|\leq n!BD^{n}$.

18

Then we can find a positive constant(4.2)$|x|\leq R\mathrm{s}\mathrm{u}\mathrm{p}|x_{0}(x)|\leq C_{1}$and$C_{1}$

so that$\frac{1}{C_{1}}\leq|x|\leq R\mathrm{s}\mathrm{u}\mathrm{p}|x_{0}’(x)|\leq C_{1}$.

for $j\geq 1$ isFurthermore, as we have shown in the previous section,
holomorphic in $\{x; |x|\leq R\}$ .
The pre-Borel summability ofnear the origin is a consequence ofthe following:$x_{j}(\tilde{x})$$\{x_{j}(\tilde{x})\}$

Lemma 3. We can find positive constants A and C so that the following
inequalities hold for any sufficiently small
and n:$\epsilon>0$$|x_{n}(\tilde{x})|\leq n!AC^{n-1}\epsilon^{-n}$,$| \tilde{x}|\leq R-\epsilon|\overline{x}|\leq R-\epsilon\sup|x_{n}’(\tilde{x})|\leq n!AC^{n-1}\epsilon^{-n}$,$\sup$(4.3)$\geq 1$$\{$.

$| \tilde{x}|\leq R-\epsilon\sup|\frac{x_{n}(\tilde{x})}{x_{0}(\tilde{x})}|\leq n!\mathrm{A}C^{n-1}\epsilon^{-n}$To prove this lemma we prepare the following:

Lemma 4. ([Kl, Lemma 2.33) Let $R’$ be a positive number, $v(t)$ a holomorphic function on
satisfying $v(0)=0$ . Then the differential$\{t\in \mathbb{C};|t|<\mathrm{R}\mathrm{f}\}$equation(4.4)$(t \frac{d}{dt}-\frac{1}{2})u(t)=v(t)$has a unique holomorphic solution on $\{t\in \mathbb{C};|t|<R’\}_{7}$ which satisfies thefollowing inequalities for any positive $R’<R’$ :
(4.5)

(4.6)
(4.7)

$|t| \leq R’’|t|\leq R’\mathrm{s}\mathrm{u}\mathrm{p}|u(t)|\leq 2\sup,|v(t)|$,$\sup_{|t|\leq R^{\mathit{1}\prime}}|u’(t)|\leq\frac{2}{R’},\sup_{t||\leq R’},|v(t)|_{7}$.

$|t| \leq R’\mathrm{s}\mathrm{u}\mathrm{p},|\frac{u(t)}{t}|\leq\frac{2}{R’},\sup_{t||\leq R’}|v(t)|$See [Kl, pp.43-44] for the proof of Lemma 4.

Ia

By changing a local coordinate through(4.8)$(t \frac{d}{dt}-\frac{1}{2})x_{n}=\frac{1}{2}\{(\frac{x_{0}}{x_{0}},$$t=x_{0}(\tilde{x})$in (3.1), we obtain.

$)^{2}F_{n}-\lambda_{n-2}\}$

as (3.10), the right-hand side of (4.8) have a zero at the
Since we choosesuch that for any
origin. Hence by Lemma 4, we find a positive constantsufficiently sm all positive ,$\lambda_{n}$$C_{2}$$\epsilon$(4.9)$| \tilde{x}|\leq R-\epsilon\sup|x_{n}(\tilde{x})|_{7}\sup|\overline{x}|\leq R-\epsilon|x_{n}’(\tilde{x})|$and$| \tilde{x}|\leq R-\epsilon\sup|\frac{x_{n}(\tilde{x})}{x_{0}(\tilde{x})}|$are dominated by(4.10)$C_{2} \sup|\tilde{x}|\leq R-\epsilon|(\frac{x_{0}}{x_{0}},)^{2}F_{n}-\lambda_{n-2}|$

To give the estimation of (4.10) we decompose(4.11)$F_{n}$.

as$F_{n}( \overline{x})=\frac{Q_{n}(\tilde{x})}{\tilde{x}^{2}}+F_{n,\mathrm{I}}+F_{n,\mathrm{I}\mathrm{I}}+F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}+F_{n_{\gamma}\mathrm{I}\mathrm{V}}$

,

where
,

$F_{n,\mathrm{I}}$$= \sum_{\mu+\nu+l=n\mu}$$\mathfrak{a}^{\nu_{1}+,=\nu}0\leq\nu_{j}^{J}\cdot\leq n-11\leq\mu\leq n-2++\mu p=\mu\sum_{\nu_{2}}(-1)^{l+1}x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}\cdots x_{\mu\iota+1}}{x_{0^{l+1}}}$$F_{n,\mathrm{I}\mathrm{I}}$,$=$$2 \nu_{1}+\nu_{2}=n-2\mu_{1}++\iota_{y}=\mu\sum_{\mu},(-1)^{l+1}(l+1)\lambda_{k}x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}\cdots x_{\mu’+1}}{x_{0^{l+2}}}$$\mu+\nu_{k}\mathrm{I}_{=}$$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}$$F_{n,\mathrm{I}\mathrm{V}}$

$=$

$=$

,

$\frac{1}{2}\mathrm{I}_{n-2}\sum_{\mu_{1}\mu+l+\cdots+\mu\iota=\mu}(-1)^{l}x_{k}’(x)\frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0}^{l+1}},$$\frac{3}{4}$$\sum$$\mu+\nu+l=n-2\mu 1+\cdot\cdot+\mu\iota=\mu\sum_{\nu_{1\tau^{1}}\nu_{2}=\nu}(-1)^{l+1}(l+1)x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0}^{l+1}},\cdot$,,
In the following we give the estimation of
tively. Without loss of generality, we can assume that
$F_{n,\mathrm{I}}$(4.12)$D<C$

and$F_{n,\mathrm{I}\mathrm{I}}$$\epsilon<1$

respecandis so large that

$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}$$C$$F_{n,\mathrm{I}\mathrm{V}}$20holds.
1’) To give the estimation of(4.13)$F_{n,\mathrm{I}}$, we write$F_{n,\mathrm{I}}=F_{n,\mathrm{I}}^{(1)}+2F_{n,\mathrm{I}}^{(2)}+F_{n,\mathrm{I}}^{(3)}$,$=$$F_{n,\mathrm{I}}^{(1)}$$\mu+l=n\mu+\sum_{0^{1}\leq}..\sum_{\mu_{j}\leq n^{l}-2}+\mu=\mu(-1)^{l+1}x_{0}’x_{0}’\frac{x_{\mu_{1}+1}\cdots x_{\mu_{l}+1}}{x_{0^{l+1}}}$(4.14),$=$$F_{n,\mathrm{I}}^{(2)}$$1 \leq k\leq n-1\sum_{\mu+k+l=n}\mu++\mu=\mu\sum_{0^{1}\leq\mu_{j}\leq n^{l}-2}(-1)^{l+1}x_{0}’x_{k}’\frac{x_{\mu_{1}+1}\cdots x_{\mu\iota+1}}{x_{0^{l+1}}}$(4.15)$=$$F_{n,\mathrm{I}}^{(3)}$.

$\mu+\nu+l=n\sum_{\mu_{1}\dagger}\mathrm{I}_{\iota=\mu}(-1)^{l+1}x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}\cdots x_{\mu\iota+1}}{x_{0^{l+1}}}$$0\leq\mu_{j}\leq n-21\leq\nu_{j}\leq n-1\nu_{1}+\nu_{2}=\nu$

We fist estimate(4.16)$F_{n,\mathrm{I}}^{(1)}$$( \frac{x_{0}}{x_{0}’})^{2}F_{n,\mathrm{I}}^{(1)}$. We obtain$\leq$$\frac{|x_{0}|^{2}}{|x_{0}|^{2}},\sum_{\mu+l=n}\sum_{0^{1}\leq\mu_{j}\leq n-2}\frac{|x_{0}’||x_{0}’|}{|x_{0}|}|\frac{x_{\mu+1}}{x_{0}}|\cdots|\frac{x_{\mu+1}}{x_{0}}|\mu++\mu_{l}=\mu$$=$.

$| \frac{x_{\mu+1}}{x_{0}}|\cdots|\frac{x_{\mu+1}}{x_{0}}|$$|x_{0}| \sum_{\mu+l=n\mu}$$++ \mu_{l}=\mu\sum_{0^{1}\leq\mu_{j}\leq n-2},$

By using (4.2) and (4.3) we find that

(4.17)$|(x_{0}/x_{0}’)^{2}F_{n,\mathrm{I}}^{(1)}|$is dominated by$A^{l}C^{\mu}\epsilon^{-}"-l(\mu_{1}+1)!\cdots(\mu_{l}+1)!$$C_{1} \sum_{\mu+l=n\mu_{1}+,0\leq}$$\mu_{j}\cdot\cdot\leq n-2\sum_{+\mu_{l}=\mu},$$\sum$$=$$(\mu_{\mathrm{I}}+1)!\cdots(\mu_{l}+1)!$$C_{1}C^{n} \epsilon^{-n}\sum_{l=2}^{n}(\frac{A}{C})^{l}\mu \mathrm{x}+\cdot\cdot+\mu_{l}=n-p0\leq\mu_{j}\leq n-2$

Then we use the formula(4.18)!

$\cdots n_{l}!\leq n!$$n_{1}+n_{2}++n_{l}=n \sum_{n_{\mathrm{j}}\geq 1}n_{1}$to obtain(4.19)$( \frac{x_{0}}{x_{0}},$$)^{2}F_{n,\mathrm{I}}^{(1)}$$\leq$$n!C_{1}C^{n} \epsilon^{-n}\sum_{l=2}^{n}(\frac{A}{C})^{l}$$\leq$$n!AC^{n-1} \epsilon^{-n}\cdot\frac{AC_{1}}{C(1-A/C)}$

.

with

21

In a similar manner we can give the estimation ofand$F_{n,\mathrm{I}}^{(2)}$$F_{n,\mathrm{I}}^{(3)}$as follows:(4.20)$( \frac{x_{0}}{x_{0}’})^{2}F_{n,\mathrm{I}}^{(2)}$$\leq$$C_{1}^{2}$$A^{l+1}C^{\mu+\nu-1}\epsilon^{-\mu-\nu-l}$$\sum$$\sum$$1\leq k\leq n-1\mu+k+\iota=n\mu_{1}+\cdot\cdot+\mu_{l}=\mu 0\leq\mu_{j}\leq n-2$$\mathrm{X}$$=$$\nu!(\mu_{1}+1)$! $\cdots(\mu_{l}+1)$ !

$AC_{1}^{2}C^{n-1} \epsilon^{-n}\sum_{l=1}^{n}(\frac{A}{C})l0\leq\mu_{j}\leq n-2,1\leq k\leq n-1\sum_{\mu_{1}++\mu_{l}+k=n-l}k$$\leq$$n!AC_{1}^{2}C^{n-1} \epsilon^{-n}\sum_{l=1}^{n}(\frac{A}{C})l$$\leq$$n$ $AC^{n-1}\epsilon_{7}^{-n_{\frac{C_{1}^{2}}{C(1-A/C)}}}$! $(\mu_{1}+1)$ ! $\cdots(\mu_{t}+1)$ !

!

(4.21)( ,)$F_{n,\mathrm{I}}^{(3)}$$\leq$$\frac{x_{0}}{x_{0}}$$C_{1}^{3} \mathrm{I}_{=n\mu 1}\sum_{+++\mu_{l}\mu=\mu}A^{l+2}C^{\mu+\nu}\epsilon^{-\mu-\nu-l}$$0\leq\mu_{j}\leq n-21\leq\nu_{f}\leq n-1\nu_{1}+\nu_{2}=\nu$$\mathrm{X}$$y_{1}$! ! $(\mu_{1}+1)$ ! $\cdots(\mu_{l}+1)$ !
$\nu_{2}$$\leq$$A^{2}C_{1}^{3} \sum_{l=0}^{n}(\frac{A}{C})l\mu_{1}+\cdot\cdot+\mu_{l}+\nu_{1}+\nu_{2}=n-l\sum_{\nu_{1},\nu_{2}\geq 1}\nu_{1}$$\leq$$n!AC^{n-1} \epsilon^{-n}\frac{AC_{1}^{3}}{C(1-A/C)}$.

!

$\nu_{2}$! $(\mu_{1}+1)$ ! $\cdots(\mu_{l}+1)’$.

222’) We give the estimation ofwith$F_{n,\mathrm{I}\mathrm{I}}$. We first write$F_{n,\mathrm{I}\mathrm{I}}=F_{n,\mathrm{I}\mathrm{I}}^{(1)}+F_{n,\mathrm{I}\mathrm{I}}^{(2)}+F_{n,\mathrm{I}\mathrm{I}}^{(3)}$(4.22),$F_{n,\mathrm{I}\mathrm{I}}^{(1)}=$$\mu+l+k=n-2\sum_{k\neq n-2}\sum_{\mu_{1}+\cdots+\mu_{l}=\mu}(-1)^{l+1}(l+1)\lambda_{k}x_{0^{X}0^{\frac{x_{\mu_{1}+1}\cdots x_{\mu_{l}+1}}{x_{0^{l+2}}}}}’$$F_{n,\mathrm{I}\mathrm{I}}^{(2)}=$$k \neq n-2,\nu\neq 0\sum_{\mu+\nu+l+k=n-2\mu_{1}+}\mathrm{I}_{\iota=\mu}(-1)^{l+1}(l+1)\lambda_{k}x_{0}’x_{\nu}’\frac{x_{\mu_{1}+1}\cdots x_{\mu \mathrm{r}+1}}{x_{0^{l+2}}}$$F_{n,\mathrm{I}\mathrm{I}}(3)$$=$$\sum$$\mathrm{I}$$\mu+\nu+l+k=n-2$$\mu_{1}+\cdot\cdot+\mu_{t}=\mu$$k\neq n-2$$\nu_{1}+\nu_{2}=\nu$$(-1$ $)^{l+1}(l$ $+1$,$) \lambda_{k}x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}\cdots x_{\mu_{l}+1}}{x_{0}^{l+2}}$.

$\nu 1_{?^{\mathcal{U}}2}\neq 0$

We first note that we obtain(4.23)$|\lambda_{k}|\leq k!B’D^{k}$$(k\geq 0)$

for some positive constant $B’$ since $\lambda_{k}=Q_{k}(0)$ and (4.1). By using thesimilar argument as 1’), we find(4.24)$|( \frac{x_{0}}{x_{0}},$$)F_{n,\mathrm{I}\mathrm{I}}^{(1)}| \leq B’C^{n-2}\epsilon^{-n+2}\sum_{l=0}^{n-2}(\frac{A}{C})^{l}\sum_{k=0}^{n-l-2}(\frac{D\epsilon}{C})^{k}k!(n-2-k)$

Since we assume (4.12), we obtain(4.25)$\sum_{k=0}^{n-l-2}(\frac{D\epsilon}{C})^{k}k!(n-2-k)!\leq\sum_{k=0}^{n-l-2}k!(n-2-k)!\leq 3(n-2)!$

Hence we conclude th at(4 . 26)$|( \frac{x_{0}}{x_{0}’})^{2}F_{n,\mathrm{I}\mathrm{I}}^{(1)}|\leq(n-2)!AC^{n-1}\epsilon^{-n}\frac{3B’}{AC(1-A/C)}$.

In a similar manner we obtain(4.27)(4.28)$|( \frac{x_{0}}{x_{0}’})F_{n,\mathrm{I}\mathrm{I}}^{(2)}|$$\leq$$(n-2)!AC^{n-1} \epsilon^{-n}\frac{3B’C_{1}\epsilon^{2}}{C^{2}(1-A/C)}$,$|( \frac{x_{0}}{x_{0}’})F_{n,\mathrm{I}\mathrm{I}}^{(3)}|$$\leq$$(n-2)!AC^{n-1} \epsilon^{-n}\frac{3AB’C_{1}^{2}\epsilon^{2}}{C^{3}(1-A/C)}$

.

.

!.

23

3’) We give the estimation ofwith$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}$. As in the previous estimation we write$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}=F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}^{(1)}+F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}^{(2)}$(4.29)$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}^{(1)}$(4.30)$F_{n,\mathrm{I}\mathrm{I}\mathrm{I}}^{(2)}$

$=$

$=$

$\frac{1}{2}\sum_{\mu+l=n-2}\sum_{\mu_{1}+\cdots+\mu\iota=\mu}(-1)^{l}x_{0}’(x)\frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0^{l+1}}’}$,$\frac{1}{2}\sum_{k\neq 0^{n-2}}\sum_{\mu_{1}\mu+\iota+k=+\cdots+\mu\iota=\mu}(-1)^{l}x_{k}’(x)\frac{x_{\mu_{1}+1+1}’x_{\mu f}’}{x_{0^{l+1}}’}\ldots$.

By a straightforward computation we obtain(4.31)$|( \frac{x_{0}}{x_{0}},$ $)^{2}F_{n,\mathrm{I}\mathrm{I}\mathrm{I}- 1}|$$\leq$$(n-2)!AC^{n-1}\epsilon^{-n_{\frac{C_{1}^{6}}{2AC(1-AC_{1}/C)}}}$(4.32)$|( \frac{x_{0}}{x_{0}},)^{2}F_{n,\mathrm{I}\mathrm{I}\mathrm{I}\sim 2}|$$\leq$$n!AC^{n-1}\epsilon^{-n_{\frac{e^{3}AC_{[perp]}^{5}\}{2C^{2}(1-AC_{1}/C)}}}$.

Here we have used the inequality(4.33)$| \tilde{x}|\leq R-\epsilon\sup|\frac{d^{k}x_{0}}{d\tilde{x}^{k}}|\leq C_{1}\epsilon^{n-1}$

and(4.34)$(n\geq 1)$$| \tilde{x}|\leq R-\epsilon\sup|\frac{d^{k}x_{n}}{d\tilde{x}^{k}}|\leq(n+k)!AC^{n-1}\epsilon^{-n-k}e^{k}$

In fact, (4.33) follows from(4.35),$\frac{d^{k}x_{0}}{d\tilde{x}^{k}}=\frac{1}{2\pi \mathrm{i}}\oint_{|\zeta-x|=\epsilon}\frac{x_{0}’(\zeta)}{(\zeta-\tilde{x})^{n}}d\zeta$and (4.34) can be obtained inductively by using(4.36)$\frac{d^{k}x_{n}}{d\tilde{x}^{k}}$$=$$\frac{1}{2\pi \mathrm{i}}\oint_{|\zeta-x|=\frac{\epsilon}{n+1}}\frac{x_{n}^{(k-1)}(\zeta)}{(\zeta-\tilde{x})^{2}}d\zeta$

.

.

,

24

4’) We give the estimation ofwith$F_{n,\mathrm{I}\mathrm{V}}$. We write$F_{n,\mathrm{I}\mathrm{V}}=F_{n,\mathrm{I}\mathrm{V}}^{(1)}+2F_{n,\mathrm{I}\mathrm{V}}^{(2)}+F_{n,\mathrm{I}\mathrm{V}}^{(3)}$(4.37)$F_{n,1\mathrm{V}}^{(1)}$$=$$\frac{3}{4}$$\sum$$\sum$$($-1$)^{1+1}$ $(l+1)x_{0}’x_{0}’ \frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0^{l+1}}’}$,$\mu+l=n-2\mu_{1}+\cdots+\mu_{l}=\mu$(4.38)$F_{n,\mathrm{I}\mathrm{V}}^{(2)}$$=$$\frac{3}{4}$$\sum$$\sum$$\mu+l+\nu=n-2$ ,$\mu_{1}+\cdots+\mu_{l}=\mu$$(-1)^{l+1}(l+1)x_{0}’x_{\nu}’ \frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0}^{l+1}},$,$\nu\neq 1$(4.39)$F_{n,\mathrm{I}\mathrm{V}}^{(3)}$.

$=$$\frac{3}{4}\sum_{\mu+l+\nu=n-2}\sum_{\mu_{l}\mu_{1}+\cdot\cdot+=\mu}(-1)^{l+1}(l\nu_{f},\nu_{2}\neq 0^{y} +1)x_{\nu_{1}}’x_{\nu_{2}}’\frac{x_{\mu_{1}+1}’\cdots x_{\mu\iota+1}’}{x_{0}^{\prime l+1}}$

We then obtain(4.40)(4.41)(4.42)$\leq$$(n-2)!AC^{n-1} \epsilon^{-n}\frac{3C_{1}^{7}\epsilon}{4AC(1-A/C)^{2}}$$|( \frac{x_{0}}{x_{0}},$ $)^{2}F_{n,\mathrm{I}\mathrm{V}}^{(2)}|$$\leq$$(n-1)!AC^{n-1} \epsilon^{-n}\frac{3eC_{1}^{6}}{4C^{2}(1-AC_{1}/C)^{2}}$$|( \frac{x_{0}}{x_{0}},$ $)^{2}F_{n,\mathrm{I}\mathrm{V}}^{(3)}|$$\leq$$n!AC^{n-1}\epsilon^{-n_{\frac{e^{2}AC_{1}^{4}}{4C^{3}(1-AC_{1}/C)}}}$

Summing up, we conclude that (4.10) is dominated by(4.43),$|( \frac{x_{0}}{x_{0}},$ $)^{2}F_{n,\mathrm{I}\mathrm{V}}^{(1)}|$$n!AC^{n-1}\epsilon^{n-1}\mathrm{x}$$\mathrm{C}2\mathrm{C}3$

.

,

25

where
(4.44)$C_{3}$$=$$\frac{BD}{A}+(\frac{AC_{1}}{C(1-A/C)}+\frac{2C_{1}^{2}}{C(1-A/C)}+\frac{AC_{1}^{3}}{C(1-A/C)})$$+( \frac{3B’}{AC(1-A/C)}+\frac{6B’C_{1}}{C^{2}(1-A/C)}+\frac{3AB’C_{1}^{2}}{C^{3}(1-A/C)})$$+\{$ $\frac{C_{1}^{6}}{2AC(1-AC_{1}/C)}+\frac{e^{3}AC_{1}^{5}}{2C^{2}(1-AC_{1}/C)})$$+( \frac{3C_{1}^{7}}{4AC(1-A/C)^{2}}+\frac{6C_{1}^{6}}{4C^{2}(1-AC_{1}/C)^{2}}+\frac{e^{2}AC_{1}^{4}}{4C^{3}(1-AC_{1}/C)})$.

Hence we first choose so that $BD<A$ and (4.3) holds for $n=1$ . Then
we chose so large that $D<C$ and $C_{2}C_{3}<1$ . Then then our induction
proceeds. This prove Lemma 4.3.
$A$

$C$

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[DP]

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[K1]

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[K2]

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