数理解析研究所講究録 1211 巻 2001 年 185-194

185

The Borel Sum of Divergent Barnes
Hypergeometric Series and its Application
to a Partial Differential Equation
Kunio Ichinobe (市延 邦夫)
Graduate School of Mathematics, Nagoya University(多元数理科学研究科, 名古屋大学)

1Introduction and Results

We shall treat of the divergent Barnes hypergeomtric series $qp-F1(p<q)$ which is defined
by
(1.1)

$qp-F1(\alpha;\gamma;z)=Fqp-1$

whereabbreviations$\alpha=(\alpha_{1}, \cdots, \alpha_{q})\in \mathbb{C}^{q}$,$(\alpha\gamma z)$$:= \sum_{n=0}^{\infty}\frac{(\alpha)_{n}}{(\gamma)_{n}}\frac{z^{n}}{n!}$,$( \gamma)_{n}=\prod_{m=1}^{p-1}(\gamma_{m})_{n}$$z\in \mathbb{C}$,and we use the following$\gamma=(\gamma_{1}, \cdots,\gamma_{p-1})\in \mathbb{C}^{p-1}$$( \alpha)_{n}=\prod_{\ell=1}^{q}(\alpha_{\ell})_{n}$,,with $(c)_{n}=\Gamma(c+n)/\Gamma(c)(c\in \mathbb{C})$ and denotes the Gamma function.
Throughout this paper, we assume
for all to make sense of this series andwe also assume
for all to avoid the trivial case where $\mathbb{Z}_{\leq 0}=\{0, -1, -2, \cdots\}$ .
We are concerned with the Borel summability of this divergent series (1.1). In theprevious papers [Ich] and [MI], we gave an explicit form of the Borel sum of this divergent
series (1.1) and its analytic continuation around the origin, which were given by alinear
conbination of $pq-F1$ . The explicit formula of the Borel sum means the rediscovery ofBarnes original one obained in [Bar], from the view point of Borel summability. In the
proof of previous papers we employed the Barnes type integral representation for thegeneralized hypergeometric function which is Borel transform of the divergent series (1.1).
In this paper we shall give another proof by employing the Euler integral representation
for the same function.
Before stating our results, we shall prepare some notations and definitions (cf. [Bal]).
$\Gamma$$\gamma_{j}\not\in \mathbb{Z}_{\leq 0}$$\alpha_{j}\not\in \mathbb{Z}_{\leq 0}$

For$d\in \mathbb{R}$

$j$

$j$

, $\beta>0$ and $\rho(0<\rho\leq\infty)$ , we define asector$S=\mathrm{S}(\mathrm{d},\beta, \rho)$$S(d, \beta, \rho):=\{z \in \mathbb{C};|d-\arg z|<\frac{\beta}{2},0<|z|<\rho\}$where , andrespectively$d$$\beta$$\rho$

by,are called the direction, the opening angle and the radius of$S(d,\beta, \rho)$,

186, which is called thebelongs to
For $>0$ , we define thatformal power series of Gevrey order $1/k$ , if there exist some positive constants and $K$
such that for any we have$\mathbb{C}[[z]]_{1/k}$$\hat{u}(z)=\Sigma_{n=0}^{\infty}u_{n}z^{n}$

$k$

$C$

$n$

$|u_{n}| \leq CK^{n}\Gamma(1+\frac{n}{k})$.

and $u(z)\in O(S)$ . Here $O(S)$ denotes the setLetof holomorphic functions on asector S. Then we define that$>0, \hat{u}(z)=\sum_{n=0}^{\infty}u_{n}z^{n}\in \mathbb{C}[[z]]_{1/k}$$k$$u(z)\cong_{k}$\^u (z)

in

$S$

,

if for any closed subsector $S’$ of , there exist some positive constants
for any $N$ we have$S$$|u(z)- \sum_{n=0}^{N-1}u_{n}z^{n}|\leq CK^{N}|z|^{N}\Gamma(1+\frac{N}{k})$,$z\in S’$$C$

and $K$ such that.

, we define that \^u(z) is summable inand
For $>0$ ,direction or Borel summable for short if there exist asector $S=S(d, \beta, \rho)$ with $\beta>\pi/k$
and $u(z)\in O(S)$ such that\^u(z) holds in .
$d\in \mathrm{R}$

$k$

$d$

$k$

$\hat{u}(z)\in \mathbb{C}[[z]]_{1/k}$$S$$u(z)\cong_{k}$Remark 1
satisfying $u(z)\cong_{k}$ \^u(z) in, then there are infinitely many(i) Iffor any and some $\rho>0$ .
(ii) If $\beta>\pi/k$ , then afunction $u(z)$ as mention above does not exist in general, but
it is unique if it does exist. In this sense such afunction is called the Borel sum of \^u indirection, or the Borel sum for short.
$S(d, \beta, \rho)$$u’ \mathrm{s}$$/\mathit{3}\leq\pi/k$

$d$

$u$

$d$

In what follows, we use the following abbreviations,,$\alpha+s=(\alpha_{1}+s, \alpha_{2}+s, \cdots, \alpha_{q}+s)\in \mathbb{C}^{q}$$\Gamma(\alpha)=\prod_{\ell=1}^{q}\Gamma(\alpha_{\ell})$,$\overline{\alpha_{j}}=(\alpha_{1}, \cdots, \alpha_{j-1}, \alpha_{j+1}, \cdots, \alpha_{q})\in \mathbb{C}^{q-1}$$\Gamma(\overline{\alpha_{j}})=\prod_{\ell=1,\ell\neq j}^{q}\Gamma(\alpha_{\ell})$,.

Now we put $\hat{f}(z)=F-1(qp\alpha;\gamma;z)\in \mathbb{C}[[z]]_{q-p}$ . Then our first result is stated as follows.
Theorem 1.1 (Borel sum)is l/(q-p)-summable in any direction. ThenAssume thatand its Borel sum $f(z)$ is given bysuch that$\alpha:-\alpha_{j}\not\in \mathbb{Z}(i\neq j)$

$d$

(1.2)

$\hat{f}(z)$$d\neq 0(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$$f(z)$$=$$\frac{C_{\alpha\gamma}}{2\pi i}\int_{I}\frac{\Gamma(\alpha+s)\Gamma(-s)}{\Gamma(\gamma+s)}(-z)^{s}ds$$=$$1+\alpha_{j}-\gamma 1+\alpha_{j}-\overline{\alpha_{j}}$$C_{\alpha\gamma} \sum_{j=1}^{q}C_{\alpha\gamma}(j)\cross(-z)_{p}^{-\alpha_{j}}F_{q-1}(\alpha_{j},$;$\frac{(-1)^{p-q}}{z})$

,

187

where $z\in S(\pi, (q-p+2)\pi$ ,(1.3)$\infty)$and$C_{\alpha\gamma}= \frac{\Gamma(\gamma)}{\Gamma(\alpha)}$,$C_{\alpha\gamma}(j)= \frac{\Gamma(\alpha_{j})\Gamma(\overline{\alpha_{j}}-\alpha_{j})}{\Gamma(\gamma-\alpha_{j})}$.

Here the path of integration I runs ffom $-i\infty to+i\infty$ on the imaginary axis in suchmanner that the poles ofare on the left side of I and the poles ofare onthe right side of .
$a$$\Gamma(\alpha+s)$$\Gamma(-s)$$I$

Next, our result for the analytic continuation of the Borel sum $f$ is stated as follows.

Theorem 1.2 (Analytic Continuation of Borel sum)
Under the same assumptions as in Theorem 1.1, we have
(1.4)$\frac{1}{2\pi i}\{f(z)-f(ze^{2\pi i})\}$$=$$\frac{C_{\alpha\gamma}}{2\pi i}\int_{I}\frac{\Gamma(\alpha+s)}{\Gamma(\gamma+s)\Gamma(1+s)}z^{s}ds$$=$$C_{\alpha\gamma} \sum_{j=1}^{q}\frac{C_{\alpha\gamma}(j)}{\Gamma(\alpha_{j})\Gamma(1-\alpha_{j})}z_{p}^{-\alpha_{j}}F_{q-1}$,
where $z\in S(0, (q-p)\pi$ ,andones as in Theorem 1.1, respectively.
$\infty)$$C_{\alpha\gamma}$$C_{\alpha\gamma}(j)$($\alpha_{j},$$1+\alpha_{j}-\gamma 1+\alpha_{j}-\overline{\alpha_{j}}$;$\frac{(-1)^{p-q}}{z}$),and the path of integration I are the same

Remark 2In the casefor some and , we can prove the similar results to
Theorems, where the logarithmic terms appear (see [Ich]).
$i$$\alpha_{i}-\alpha_{j}\in \mathbb{Z}$$j$2Proof of Theorem 1.1
In order to prove Theorems we use the following lemma for the Borel summability.

Lemma 2.1 Let $k>0$ ,andthree propositions are equivalent:(1) \^u (z) is -summable in direction.
(2) Letbe the formal -Borel transformation of \^u (z)$d\in \mathbb{R}$$k$$g(\zeta)$(2.1)$\hat{u}(z)=\Sigma_{n=0}^{\infty}u_{n}z^{n}\in \mathbb{C}[[z]]_{1/k}$. Then the following

$d$

$k$

$g( \zeta)=(\hat{B}_{k}\hat{u})(\zeta):=\sum_{n=0}^{\infty}\frac{u_{n}}{\Gamma(1+n/k)}\zeta^{n}$,which is holomorphic in a nighbourhood of $\zeta=0$ . Then
can be continued analyticallyinpositivesome
constant
growthand satisfies a
condition of exponentia
for$g(\zeta)$$S(d, \epsilon, \infty)$$\epsilon$

$l$

188
order at most
have
(2.2)

$k$

there, that is, there exist some positive numbers $C$ and$|g(\zeta)|\leq C\exp\{\delta|\zeta|^{k}\}$,$\zeta\in S(d, \epsilon, \infty)$

such that we$\delta$.

In this case, the Borel sum $u(z)$ is obtained after an analytic continuation of the follow $ing$
Laplace integral(2.3)$u(z)=( \mathcal{L}_{k}g)(z):=\frac{1}{z^{k}}\int_{0}^{\infty(d)}e^{-(\zeta/z)^{k}}g(\zeta)d(\zeta^{k})$where $z\in S(d, \beta, \rho)$ with $\beta<\pi/k$ and $\rho>0$ and the path of integration is taken from 0
to oo along the half line of argument .
, $k_{j}>0$ satisfy $1/k=1/k_{1}+\cdots+1/k_{j}$ . Let(3) Let $j\geq 2$ and $k_{1}>0$ ,be the\^u(z)following iterated formal Borel transformations of$d$$h(\zeta)$$\cdots$(2.4)$h(\zeta)=(\hat{B}_{k_{1}}0\cdots 0\hat{B}_{k_{j}}\hat{u})(\zeta)$.

above. In this case, the Borel sum $u(z)$ is
Thenholds the same properties as
obtained after an analytic continuation of the following iterated Laplace integrals$g(\zeta)$$h(\zeta)$(2.5)$u(z)=(\mathcal{L}_{k_{j}}0\cdots 0\mathcal{L}_{k_{1}}h)(z)$.

The equivalence of (i) and (ii) is given in [Bal] and the equivalence of (iii) with others
is proved in [Miy].

Proof of Theorem 1.1. Letof$h(\zeta)$

be the $(q-p)$ times iterated formal 1-Borel transformations$\hat{f}(z)$(2.6)$h(\zeta)=(\hat{B}_{1}^{q-p}\hat{f})(\zeta)=Fqq-1(\gamma,$$1,\alpha\cdots$,1;$\zeta)$.

andThis series is convergent in $|\zeta|<1$ . Then we can see that,
equation
becausesatisfies aFuchsian
withhas at most polynomial growth asanyl/(q-p)-summable. Thereforeinisdirection suchsingular points$f(z)$
followingbyis giventheiterated Laplace
and the Borel sum
that
integrals and its analytic continuation$h(\zeta)\in O(\mathbb{C}\backslash [1, \infty))$$h(\zeta)$$\zetaarrow\infty$$\{0, 1, \infty\}$$h(\zeta)$$d$$\hat{f}(z)$$d\neq 0(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$(2.7) $f(z)$

...

$=$

$\frac{1}{z}\int_{0}^{\infty(d)}\exp(-\frac{s_{1}}{z})ds_{1^{\frac{1}{s_{1}}}}\int_{0}^{\infty(d)}\exp(-\frac{s_{2}}{s_{1}})ds_{2}\cdots$$\cross\frac{1}{s_{q-p-2}}\int_{0}^{\infty(d)}\exp(-\frac{s_{q-p-1}}{s_{q-p-2}})ds_{q-p-1}\frac{1}{s_{q-p-1}}\int_{0}^{\infty(d)}\exp(-\frac{\zeta}{s_{q-p-1}})h(\zeta)d\zeta$

where $d=\arg\langle=\arg$(2.8)$\frac{s_{1}}{z}=u_{1}$,$s_{j}\neq 0(\mathrm{m}\mathrm{o}\mathrm{d} 2\pi)$$\frac{s_{2}}{s_{1}}=u_{2}$,$\cdot$.. ,and$|d-\mathrm{a}r\mathrm{g}z|<\pi/2$$\frac{s_{q-p-1}}{s_{q-p-2}}=u_{q-p-1}$,. By achange of variables$\frac{\zeta}{s_{q-p-1}}=u_{q-p}$

,

,

189

(2.9)

$f(z)$$\int_{0}^{\infty(a)}e^{-u_{1}}du_{1}\int_{0}^{\infty(0)}e^{-u_{2}}du_{2}\cdots\int_{0}^{\infty(0)}e^{-u_{q-p}}h(uz)du_{q-p}$$=$,where $a=d-\arg z(|a|<\pi/2)$ and $u=u_{1}\cdot u_{2}\cdots u_{q-p}$ .
To caluclate these integrals, we employ the Euler integral representation ofis given by(2.11)$\mathrm{A}(\mathrm{C})$$=$$t=t_{1}\cdot t_{2}\cdots t_{q-1}$(2.11)

which$C_{0} \prod_{j=1}^{p-1}\int_{0}^{1}t_{j}^{\alpha_{\mathrm{j}}-1}(1-t_{j})^{\gamma_{j}-\alpha_{\mathrm{j}}-1}dt_{j}\prod_{j=p}^{q-2}\int_{0}^{1}t_{j}^{\alpha_{j}-1}(1-t_{j})^{-\alpha_{\mathrm{j}}}dt_{j}$$\cross\int_{0}^{1}t_{q-1^{\alpha_{q-1}-1}}(1-t_{p-1})^{-\alpha_{q-1}}(1-t\zeta)^{-\alpha_{q}}dt_{q-1}$

where$h(\zeta)$,

and$C_{0}= \frac{\Gamma(\gamma)}{\Pi_{j=1}^{p-1}\Gamma(\alpha_{j})\Gamma(\gamma_{j}-\alpha_{j})}$.

$\frac{\Gamma(1)}{\prod_{j=p}^{q-1}\Gamma(\alpha_{j})\Gamma(1-\alpha_{j})}$.

Here in order to make sense of these integrals in (2.10), we assume the following integrability conditions(2.12)${\rm Re}\gamma_{j}>{\rm Re}\alpha_{j}>0(j=1, \cdots,p-1)$

Moreover we assume(2.13),$0<{\rm Re}\alpha_{j}<1(j=p, \cdots, q-1)$$0<{\rm Re}\alpha_{q}<1$.

.

We remark that we can remove such restrictions at the end of proof.
Then we obtain the following fundamental formula for the Borel sum(2.14)$f(z)$$=C_{0} \int_{0}^{\infty(0)}e^{-u_{2}}du_{2}\int_{0}^{\infty(0)}e^{-u_{3}}du_{3}\cdots\int_{0}^{\infty(0)}e^{-u_{q-\mathrm{p}}}du_{q-p}$$\cross\prod_{j=1}^{p-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1}(1-t_{j})^{\gamma_{j}-\alpha_{j}-1}dt_{j}\prod_{j=p}^{q-1}\mathit{1}^{1}t_{j}^{\alpha_{j}-1}(1-t_{j})^{1-\alpha_{\mathrm{j}}-1}dt_{j}$$\cross\int_{0}^{\infty(a)}e^{-u_{1}}(1-tuz)^{-\alpha_{q}}du_{1}$.

In order to calculate the last integral, we give the following lemma.

Lemma 2.2 Let $0<{\rm Re}\beta<1$ . Then we have
(2.15)$\int_{0}^{\infty(a)}e^{-u}(1-zu)^{-\beta}du=\frac{\{\Gamma(\beta)\}^{-1}}{2\pi i}\int_{\tilde{I}}\Gamma(\beta+\mathrm{s})\mathrm{T}(\mathrm{s}+1)\Gamma(-s)(-z)^{s}ds$,

where $a=(d-\arg z)$ $(0<d<2\pi)$ with $|a|<\pi/2$ and the path of integrationfrom $-i\infty$ to $+i\infty with-{\rm Re}\beta<\kappa$ $<0$ .
$\kappa$$\kappa$$\tilde{I}mns$

190
Proof of Lemma 2.2. First, we notice(2.15)$(1-zu)^{-\beta}$$=$.

$\frac{\{\Gamma(\beta)\}^{-1}}{2\pi i}\int_{\tilde{I}}\Gamma(\beta+s)\Gamma(-s)(-zu)^{s}ds$

Therefore by substituting this formula into the left side of (2.15) and exchanging the orderof integration, we obtain the conclusion.
By using Lemma 2.2 and exchanging the order of integrations in (2.14), we have$\square$(2.17)$f(z)$$=$$\frac{C_{0}}{2\pi i}\int_{\tilde{I_{1}}}\Gamma(1+s)\Gamma(\alpha_{q}+s)\Gamma(-s)(-z)^{s}ds$$\cross\int_{0}^{\infty(0)}e^{-u_{2}}u_{2^{S}}du_{2}$$\int_{0}^{\infty(0)}e^{-u_{3}}u_{\mathit{3}^{S}}du_{\mathit{3}}$$\cdots$ $\int_{0}^{\infty(0)}e^{-u_{q-\mathrm{p}}}u_{q-p^{S}}du_{q-p}$$\cross\prod_{j=1}^{p-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1+s}(1-t_{j})^{\gamma_{j}-\alpha_{j}-1}dt_{j}\prod_{j=p}^{q-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1+s}(1-t_{j})^{-\alpha_{j}}dt_{j}$$=$$\frac{C_{\alpha\gamma}}{2\pi i}\int_{\tilde{I_{1}}}\frac{\Gamma(\alpha+s)\Gamma(-s)}{\Gamma(\gamma+s)}(-z)^{s}ds$,

runs fromis the constant given by (1.3). Here the path of integrationand it is possible to take such . Finally,to $by changing the path of integration into I in Theorem 1.1, we can remove the restriction(2.12) and (2.13) for the parameters and 7, so we obtain the desired first formula (1.2)
of integral representation for the Borel sum. In addition, by residue theorem, we obtain
the desired the second formula (cf. [Ich, Theorem 2.1]).

where$\tilde{I_{1}}$$C_{\alpha\gamma}$$\kappa_{1}-i\infty$$\kappa_{1}$$i \infty \mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}-\min\{\alpha_{j}\}<\kappa_{1}<0$$\kappa_{1}$$\tilde{I_{1}}$$\alpha$$\square$

3Proof of Theorem 1.2
Proof of Theorem 1.2. Prom (2.14), we get the following formula(3.1):$f(z)-f(ze^{2\pi})$$=$$C_{0} \int_{0}^{\infty(0)}e^{-u_{2}}du_{2}\int_{0}^{\infty(0)}e^{-us}du_{3}\cdots\int_{0}^{\infty(0)}e^{-u_{q-\mathrm{p}}}du_{q-p}$$\cross\prod_{j=1}^{p-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1}(1-t_{j})^{\gamma_{j}-\alpha_{j}-1}dt_{j}\prod_{j=p}^{q-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1}(1-t_{j})^{-\alpha_{j}}dt_{j}$$\cross\int_{c_{+}(t\hat{u_{1}}z)}e^{-\mathrm{u}_{1}}(1-tuz)^{-\alpha_{q}}du_{1}$.

starts at
on $\arg u_{1}=-\arg X$ ,where the path of integration $C_{+}(X)$ withencircles the point $u_{1}=X^{-1}$ in the positive direction and returns to the starting point.
In order to calculate the last integral, we give the following lemma$X=t\overline{\mathrm{u}_{1}}z$$\infty$

191
Lemma 3.1 Let $0<{\rm Re}\beta<1$ . Then we have
(3.2)$\int_{0}^{\infty(a)}e^{-u}\{(1-zu)^{-\beta}-(1-zue^{2\pi:})^{-\beta}\}$$\int_{c_{+}(z)}e^{-u}(1-zu)^{-\beta}du=$

,

$=$

where

$a$

and the path of integration

du$\frac{1}{\Gamma(\beta)}\int_{\tilde{I}}\Gamma(\beta+s)z^{s}ds$$\overline{I}$are the same ones as in Lemma 2.2, respectively.

Proof of Lemma 3.1. Since we notice that $(1-zu)^{-\beta}$ is univalent in$u$

on $|u|<|z|$ , wecan prove (3.2) in the same manner as in Lemma 2.2.
By using Lemma 3.1 and exchanging the order of integrations in (3.1), we have(3.3) $f(z)-f(ze^{2\pi i})$$=$$\square$$\frac{C_{0}}{\Gamma(\alpha_{q})}\int_{\overline{I_{1}}}\Gamma(\alpha_{q}+s)z^{s}ds\prod_{j=2}^{q-p}\int_{0}^{\infty(0)}e^{-u_{j}}u_{j^{S}}du_{j}$$\cross\prod_{j=1}^{p-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1+s}(1-t_{j})^{\gamma_{j}-\alpha_{j}-1}dt_{j}\prod_{j=p}^{q-1}\int_{0}^{1}t_{j}^{\alpha_{j}-1+s}(1-t_{j})^{-\alpha_{j}}dt_{j}$$=$$C_{\alpha\gamma} \int_{\overline{I_{1}}}\frac{\Gamma(\alpha+s)}{\Gamma(\gamma+s)\Gamma(1+s)}z^{s}ds$,

where the path of integration is the same one as in (2.17). Therefore, by changing thepath of integration, we can remove the restriction (2.12) and (2.13) for the parameters.
This is the desired first formula (1.4) of integral representation and by using residue
theorem we obtain the desired second formula (cf. [Ich, Theorem 2.2]).
$\overline{I_{1}}$$\square$4Appliction to aPartial Differential Equation
In this section, we shall give an application to apartial differential equation of Theorem
1.2. Let us consider the following Cauchy problem.
$\partial_{t}^{p}u(t, x)=\partial_{x}^{q}u(t, x)$

(4.1)

$\{$

$u(0, x)=\varphi(x)$,,$\dot{\theta}_{t}u(0, x)=0(1\leq j\leq p-1)$,

is holomorphic in aneighbourhood at, $p<q$ and the Cauchy data
where ,
the origin.
This Cauchy problem has aunique formal power series solution in t-variable$t$$x\in \mathbb{C}$$\varphi(x)$

\^u

(4.2)

$(t, x)= \sum_{n=0}^{\infty}\varphi^{(qn)}(x)\frac{t^{pn}}{(pn)!}$.

Miyake [Miy] proved that the formal solution (4.2) of (4.1) is Borel summable insatisfies the following two
direction in -plane if and only if the Cauchy data
conditions
$d$

$t$

$\varphi(x)$192(i) the Cauchy data$\varphi(x)$(4.3)

can be continued analytically in$\Omega(p, q;d, \epsilon)=\cup Sj=0q-1(\frac{pd+2\pi ij}{q},$.
for some(ii) the Cauchy data$q/(q-p)$ in.

,$\epsilon$$q$sectors$\infty)$$\epsilon>0$$\varphi(x)$has the growth condition of exponential order at most$\Omega(p, q;d, \epsilon)$

Under the above conditions, we get the integral representation of the Borel sum of(4.2) by using the kernel function as follows.
Theorem 4.1 (Integral representation of Borel sum)
, the Borel sum $u(t, x)$ of
Under the conditions (i) and (ii) for the Cauchy data
the $fo$ rmal solution (4.2) of the Cauchy problem (4.1) is given by$\varphi(x)$(4.4)$u(t, x)= \int_{0}^{\infty(pd/q)}\Phi(x, \zeta)k(t, \zeta)d\zeta$

,

where
(4.5)

$\Phi(x, \zeta)=\sum_{j=0}^{q-1}\varphi(x+\zeta\omega^{j})$and the kemel function(4.6)$k(t, \zeta)$$=$$k(t, \zeta)$,$\omega$$=\exp(2\pi i/q)$,

is given by$\frac{D_{pq}}{\zeta}\sum_{j=0}^{q-1}D_{pq}(j)X_{p}^{-j/q}F_{q-1}(1+j/q-(q^{\frac{p/p}{/q}})_{j}1+j/q-$;$\frac{(-1)^{p-q}}{X})$

,

where
(4.7)

$X= \frac{q^{q}}{p^{p}}\frac{t^{p}}{\zeta^{q}}$and $p=$ $(1,2, \cdots,p)$ ,$q=(1,2, \cdots, q)$,,$D_{pq}= \frac{\Gamma(p/p)}{\Gamma(q/q)}$,$D_{pq}(j)= \frac{\Gamma((q\overline{/q})_{j}-j/q)}{\Gamma(p/p-j/q)}$

Theorem 4.1. Let $v(s, x)$ be the $(q-p)$ times iterated -Borel transforms in-variable of the formal solution (4.2)$Pro\mathrm{o}/of$
$t$

.

(4.8)

$p$

$v(s, x)=(( \hat{B}_{p})^{q-p}\hat{u}(\cdot, x))(s)=\sum_{n=0}^{\infty}\frac{\varphi^{(qn)}(x)}{(pn)!}\frac{s^{pn}}{(n!)^{q-p}}$

By Cauchy’s integral formula, for sufficiently small(4.9)$v(s, x)$

$|s|$

and

$|x|$

.

we have$=$$\frac{1}{2\pi i}\oint_{|\zeta|=r}\frac{\varphi(x+\zeta)}{\zeta}\sum_{n=0}^{\infty}\frac{(qn)!}{(pn)!(n!)^{q-p}}(\frac{s^{p}}{\zeta^{q}})^{n}d\zeta$$=$$\frac{1}{2\pi i}\oint_{|\zeta|=\mathrm{r}}\frac{\varphi(x+\zeta)}{\zeta}h(s, \zeta)d\zeta$,

193, 1; Y) with.
where $r>\sqrt{q(q^{q}/p^{p})|s^{p}|}$ and $h(s, \zeta)=Fqq-1$ ( $q/q;p/p$ , 1,for
Here we notice that $h(s$ , (;) has singular points in -plane at roots of(the root with argument $dp/q$ ),afixed $s\neq 0$ with $\arg s=d$ . We put
and we denote by $[0, a]$ the segment joining the origin and . Since we notice that $h(s$ , (;)(outside of segments), we can deform the contour ofis univalent in
integration for (4.9) as follows.
$\mathrm{Y}=q^{q}/p^{\mathrm{p}}\cdot s^{p}/\zeta^{q}$$\cdots$$\zeta$$q$$\zeta^{q}=(q^{q}/p^{\mathrm{p}})s^{p}$$q$$a=(q^{q}/p^{\mathrm{p}})^{1/q}s^{p/q}$$a$$\mathbb{C}_{\zeta}\backslash \bigcup_{j=0}^{q-1}[0, a\omega^{j}]$$q$.

(4.10)$v(s, x)= \frac{1}{2\pi i}\int_{0}^{\infty(pd/q)}\frac{\Phi(x,\zeta)}{\zeta}\{h(s, \zeta)-h(s, \zeta\omega^{-1})\}d\zeta$

Hence the Borel sum $u(t, x)$ is given by the following iterated Laplace transforms of $v(s, x)$(4.11)$u(t, x)$$=$$((\mathcal{L}_{p})^{q-p}v(\cdot, x))(t)$$=$$\frac{1}{2\pi i}\int_{0}^{\infty(pd/q)}\frac{\Phi(x,\zeta)}{\zeta}\{(\mathcal{L}_{p})^{q-p}(h(\cdot, \zeta)-h(\cdot, \zeta\omega^{-1}))\}(t)d\zeta$

This observation shows that the kernel function(4.12)$k(t, \zeta)$.

is given by$k(t, \zeta)=\frac{1}{2\pi i}\frac{1}{\zeta}\{((\mathcal{L}_{p})^{q-p}h)(\cdot, \zeta)-((\mathcal{L}_{p})^{q-p}h)(\cdot, \zeta\omega^{-1})\}(t)$.

is an iterated formal Borel transforms
Now, we shall prove that the functionof the formal solution of the following Cauchy problem for the adjoint equation$h(s, \zeta)/\zeta$(4.13)$\partial_{t}^{p}u(t, \zeta)=(-\partial_{\zeta})^{q}u(t, \zeta)$$\{$$u(0, \zeta)=1/\zeta$,,$\partial_{t}^{j}u(0, \zeta)=0(1\leq j\leq p-1)$.

This Cauchy problem (4.13) has aunique formal solution\^e(t, )(4.14)$\zeta$.
$=$$\frac{1}{\zeta}\sum_{n=0}^{\infty}\frac{(qn)}{(pn)}!$$=$$\frac{1}{\zeta}qp-F1(\frac{q}{q};\overline{\frac{p_{p}}{p}};\frac{q^{q}}{p^{p}}\frac{t^{p}}{\zeta^{q}})\mathrm{p}\mathrm{u}\mathrm{t}=\frac{1}{\zeta}\hat{f}(X)$$( \frac{t^{p}}{\zeta^{q}})^{n}$where $X$ is given by (4.7). Let$g(\mathrm{Y})=((\hat{B}_{1})^{q-p}\hat{f})(\mathrm{Y})$(4.15),$g(\mathrm{Y})=h(s, \zeta)$(4.16). Then we can see that$((\mathcal{L}_{1})^{q-p}g)(X)=((\mathcal{L}_{p})^{q-p}h(\cdot, \zeta))(t)$Hence, by letting $f(X)$ be the Borel sum ofof \^e(t, ) is given by
the Borel sum$e(t, \langle)$,$\hat{f}(X)$.

, we have $f(X)=((\mathcal{L}_{1})^{q-p}g)(X)$ and$\zeta$$e(t, \zeta)=\frac{1}{r}f(X)$.

194Thus we can see that the kernel function $k(t$ , (;) is given by(4.17)$k(t, \zeta)$$=$$\frac{1}{2\pi i}\{e(t, \zeta)-e(t, \zeta e^{-^{2}i^{\pi}}q)\}$$=$$\frac{1}{2\pi i}\cross\frac{1}{\zeta}\{f(X)-f(Xe^{2\pi i})\}$,$X=q^{q}t^{p}/p^{\mathrm{p}}\zeta^{q}$

Therefore by using Theorem 1.2, we get the kernel function
This completes the proof of Theorem 4.1.

$k(t, \zeta)$.

which is given by (4.6).
$\square$

References[Bal] W. Balser, Prom Divergent Power Series to Analytic Functions, Springer Lecture
Notes, No. 1582, 1994.

[Bar] E. W. Barnes, The asymptotic expansion of integral functions defined by generalized
hypergeometric series, Proc. London Math. Soc, (2) 5(1907), 59-116.
[Ich] K. Ichinobe, The Borel sum of Divergent Barnes Hypergeometric Series and its. Inst. Math.
Application to a Partial Differential Equation, to appear in Publ.
Sci. 37 no. 1.
${\rm Res}$[Miy] M. Miyake, Borel summability of divergent solutions of the Cauchy problem to nonKowalevskian equations, Partial differential equations and their applications (Wuhan,1999), 225-239, World Sci. Publ. River Edge, NJ, 1999.
[MI] M. Miyake and K. Ichinobe, On the Borel summability of divergent solutions ofparabolic type equations and Barnes generalized hypergeometric functions, Surikaiseki
kenky\^usho K\^oky\^uroku, Kyoto Univ. No. 1158. (2000), 43-57