Poisson and heat semigroups for Bessel operator and on the hyperbolic space

TL;DR

This paper derives explicit formulas for Poisson and heat semigroups related to the Bessel operator and hyperbolic spaces, enhancing understanding of their mathematical properties and potential applications.

Contribution

It provides new explicit formulas for semigroups associated with the Bessel operator and hyperbolic spaces, which were previously not fully characterized.

Findings

Explicit formulas for Poisson semigroup on hyperbolic space

Explicit formulas for heat semigroup on hyperbolic space

Enhanced mathematical understanding of Bessel operator semigroups

Abstract

In this paper we find explicit formulas for the Poisson and heat semigroups associated to the modified Bessel operator and on the hyperbolic spaces $\mathbb{H}^n$.

Equations

$$\displaystyle\left\{\begin{array}[]{cc}L^{a}u(y,x)=-\frac{\partial^{2}}{\partial y^{2}}u(y,x),(y,x)\in I\!\!R^{+}\times I\!\!R\\ u(0,x)=u_{0}(x),u_{0}\in C^{\infty}_{0}(I\!\!R^{+})\end{array}\right.,$$

$$\displaystyle\left\{\begin{array}[]{cc}{\cal L}_{n}U(y,w)=-\frac{\partial^{2}}{\partial y^{2}}U(y,w),(y,w)\in I\!\!R^{+}\times I\!\!H^{n}\\ U(0,w)=U_{0}(w),U_{0}\in C^{\infty}_{0}(I\!\!H^{n})\end{array}\right.,$$

$$\displaystyle\left\{\begin{array}[]{cc}{L}^{a}v(t,x)=\frac{\partial}{\partial t}v(t,x),(t,x)\in I\!\!R^{+}\times I\!\!R\\ v(0,x)=v_{0}(x),v_{0}\in C^{\infty}_{0}(I\!\!R^{+})\end{array}\right.,$$

$$\displaystyle\left\{\begin{array}[]{cc}{\cal L}_{n}V(t,w)=\frac{\partial}{\partial t}V(t,w),(t,w)\in I\!\!R^{+}\times I\!\!H^{n}\\ V(0,w)=V_{0}(w),V_{0}\in C^{\infty}_{0}(I\!\!H^{n})\end{array}\right.,$$

$$L^a=x^2\frac{{\partial }^2}{\partial x^2}+x\frac{\partial }{\partial x}-a^2{x}^2,$$

$${\mathcal{L}}_n=x_n^2{\Delta }_{n-1}+x_n^2\frac{{\partial }^2}{\partial x_n^2}+(2-n)x_n\frac{\partial }{\partial x_n}+\frac{(n-1{)}^2}{4},$$

$$u(y,x)={\int }_0^{\infty }{p}_a(y,x,x^{\prime })u_0(x^{\prime })\frac{d{x}^{\prime }}{x^{\prime }},$$

$$p_a(y,x,x^{\prime })=\frac{|a|}{\pi }\frac{x{x}^{\prime }\sin y{K}_1\left(|a|\sqrt{x^2+x^{\prime 2}-2x{x}^{\prime }\cos y}\right)}{\sqrt{x^2+x^{\prime 2}-2x{x}^{\prime }\cos y}},$$

$$\frac{\partial \phi }{\partial x}=(2x-2x^{\prime }\cos y)\frac{\partial \varphi }{\partial z},\frac{{\partial }^2\phi }{\partial x^2}=(2x-2x^{\prime }\cos y{)}^2\frac{{\partial }^2\varphi }{\partial z^2}+2\frac{\partial \varphi }{\partial z},$$

$$\frac{\partial \phi }{\partial y}=2x{x}^{\prime }\sin y\frac{\partial \varphi }{\partial z},\frac{{\partial }^2\phi }{\partial y^2}=(2x{x}^{\prime }\sin y{)}^2\frac{{\partial }^2\varphi }{\partial z^2}+2x{x}^{\prime }\cos y\frac{\partial \varphi }{\partial z}.$$

$$\left(L^a+\frac{{\partial }^2}{\partial y^2}\right)\phi =4x^2\left(z\frac{{\partial }^2\varphi }{\partial z^2}+\frac{\partial \varphi }{\partial z}-\frac{a^2}{4}\varphi \right),$$

$$z^2{\varphi }_{zz}+z{\varphi }_z-\frac{a^2}{4}z\varphi =0,$$

$$[z^2\frac{{\partial }^2\varphi }{\partial z^2}+(1-2\alpha )z\frac{\partial \varphi }{\partial z}-(\beta \gamma z^{\gamma }{)}^2\varphi +({\alpha }^2-{\nu }^2{\gamma }^2)\varphi =0,$$

$$L_x^a\phi (y,x,x^{\prime })=L_{x^{\prime }}^a\phi (y,x,x^{\prime })=-\frac{{\partial }^2}{\partial y^2}\phi (y,x,x^{\prime }),$$

$$p_a(y,x,x^{\prime })=-\frac{1}{\pi }\frac{\partial }{\partial y}\phi (y,x,x^{\prime })$$

$$u(y,x)=\tilde{u}(y,X)={\int }_0^{\infty }{P}_a(y,X,X^{\prime }){\tilde{u}}_0(X^{\prime })d{X}^{\prime },$$

$$P_a(y,X,X^{\prime })=\frac{|a|}{\pi }\frac{e^{(X+X^{\prime })/2}\sin y{K}_1\left(|a|\sqrt{{\sinh }^2\frac{(X-X^{\prime })}{2}+{\sin }^{2}y/2}\right)}{\sqrt{{\sinh }^2\frac{(X-X^{\prime })}{2}+{\sin }^{2}y/2}}.$$

$$\frac{K_1\left(2|a|e^{X+arg\sinh (s\sin y/2)}\sin (y/2)\sqrt{1+s^2}\right)}{\sqrt{1+s^2}}\frac{{\tilde{u}}_0(X+arg\sinh (s\sin y/2))2ds}{\sqrt{1+s^2{\sin }^2(y/2)}}.$$

$$\underset{y⟶0}{\lim }u(y,x)=\underset{y⟶0}{\lim }\tilde{u}(y,X)={\tilde{u}}_0(X)\frac{1}{\pi }{\int }_{-\infty }^{\infty }\frac{ds}{1+s^2}={\tilde{u}}_0(X)=u_0(x)$$

$$d{s}^2=\frac{d{x}_1^2+d{x}_2^2+...+d{x}_n^2}{x_n^2},$$

$$d\mu (w)=\frac{d{x}_{1}d{x}_2...d{x}_n}{x_n^n},$$

$${\cosh }^2(\rho (w,w^{\prime })/2)=\frac{|w-w^{\prime }{|}^2}{4x_n{x}_n^{\prime }}+1,$$

$${\mathcal{L}}_n=x_n^2{\Delta }_n+(2-n)\frac{\partial }{\partial x_n}+((n-1)/2{)}^2,$$

$${\mathcal{F}}^{-1}\left[p_{|\xi |}(y,x_n,x_n^{\prime })\right](x)=\frac{2^{(n-1)/2}\Gamma ((n+1)/2)}{\pi }\frac{x_n{x}_n^{\prime }\sin y}{(z^2+|x{|}^2{)}^{(n+1)/2}},$$

$${\mathcal{F}}^{-1}[f](|x|)=|x{|}^{1-n/2}{\int }_0^{+\infty }{J}_{\frac{n-2}{2}}(\rho |x|)f(\rho ){\rho }^{n/2}d\rho ,$$

$${\int }_0^{+\infty }{x}^{\alpha -1}{J}_{\mu }(bx)K_{\nu }(cx)dx=A_{\mu ,\nu }^{\alpha },$$

$$A_{\mu ,\nu }^{\alpha }=2^{\alpha -2}{b}^{\mu }{c}^{-(\alpha +\mu )}\frac{\Gamma ((\alpha +\mu +\nu )/2\Gamma ((\alpha +\mu -\nu )/2)}{\Gamma (\mu +1)}\times$$

$${}_2{F}_1((\alpha +\mu +\nu )/2,(\alpha +\mu -\nu )/2,\mu +1,-\frac{b^2}{c^2}),$$

$${\int }_0^{+\infty }{K}_1(r|z)J_{\frac{n-3}{2}}(r\left|x\right|)r^{\frac{n+1}{2}}dr=A_{1,(n-3)/2}^{(n+3)/2},$$

$$A_{1,(n-3)/2}^{(n+3)/2}=\frac{|x{|}^{(3-n)/2}}{\pi z}{2}^{(n-1)/2}\frac{|x{|}^{(n-3)/2}}{z^n}\Gamma ((n+1)/2)\times$$

$$F\left(((n+1)/2,(n-1)/2(n-1)/2-\frac{|x{|}^2}{z^2}\right).$$

Full text

1 Introduction

The differential operators of Bessel type and the Laplace-Beltrami operator on the hyperbolic space are known as very important operators in analysis and its applications. This paper deals with the Poisson and heat semigroups associated to these second order differential operators. In the last decades the Poisson and heat semigroups associated to many second differential operators have been studied and computed explicitly and there is many interesting papers published in this area of reaserch(see for example Betancor et al.[2], Isolda Cardoso[7], Keles and Bayrakci[15], Stein [16] and the references theirin). The main objective of this paper is to solve explicitly the following Poisson and heat problems

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

where

[TABLE]

and

[TABLE]

are respectively the Bessel operator on $I\!\!R^{+}$ and the Laplace-Beltrami operator on the half space model of the hyperbolic space $I\!\!H^{n}$.

2 Poisson semigroup associated to Bessel operator

In this section we give explicit formulas for the Poisson semigroup associated to the Bessel operator $L^{a}$, that is we prove the following theorem.

Theorem 2.1**.**

For $a\in I\!\!R^{*}$ the Poisson problem (1.3) has the solution given by

[TABLE]

with

[TABLE]

and $K_{1}$ is the modified Bessel functions of second kind.

Proof.

To see that the function $u(y,x)$ satisfes the Poisson equation in (1.3), set $\varphi(y,x,x^{\prime})=\phi(z)$, with $z=x^{2}+x^{\prime 2}-2xx^{\prime}\cos y$, then we have

[TABLE]

and

[TABLE]

Using the above formulas we have

[TABLE]

and we see that the first equation in the problem (1.3) is equivalent to

[TABLE]

which is a particular case of Lommel differential equation for modified Bessel functions

[TABLE]

with $\alpha=0$, $\nu=0$, $\beta=1$ and $\gamma=1/2$, an approprite solution is $\phi(z)=cK_{0}(z^{1/2}),$ where $K_{0}$ is the modified Bessel function of second kind.

This means that the function $\varphi(y,x,x^{\prime})=cK_{0}\left(|a|\sqrt{x^{2}+x^{\prime 2}-2xx^{\prime}\cos y}\right)$ satisfies the equations

[TABLE]

and in consequence it is a solution of the first equation in (1.3).

Using the formula $K_{0}^{\prime}(z)=-K_{1}(z)$ we see that

[TABLE]

and $p_{a}(y,x,x^{\prime})$ satisfies the same equation in (1.3).

To finish the proof of Theorem 2.1 it remains to show the limit condition. For this set $z=x^{2}+x^{\prime 2}-2xx^{\prime}\cos y=2xx^{\prime}(\frac{x^{2}+x^{\prime 2}}{2xx^{\prime}}-\cos y)$ and $x=e^{X}$ and $x^{\prime}=e^{X^{\prime}}$ to obtain $z=4e^{X+X^{\prime}}\{\sinh^{2}\frac{(X-X^{\prime})}{2}+\sin^{2}(y/2)\}$.

Replacing in (2.1) we obtain

[TABLE]

with

[TABLE]

Setting $\sinh\frac{(X^{\prime}-X)}{2}=s\sin y/2$ or $X^{\prime}=X+2arg\sinh(s\sin y/2)$ we can write

$u(y,x)=\tilde{u}(y,X)=\frac{|a|}{\pi}\int_{-\infty}^{\infty}e^{X+arg\sinh(s\sin y/2)}\sin y\times$

[TABLE]

Now we use the asymptotic formula for the modified Bessel function of second kind (Lebedev [9] p.136) $K_{\nu}(z)\sim\frac{2^{\nu-1}\Gamma(\nu)}{z^{\nu}},z\longrightarrow 0$ we obtain

[TABLE]

and this finishes the proof of Theorem 2.1. ∎

3 Poisson equation on the hyperbolic space

In this section we consider the Poisson equation on the hyperbolic upper half space.

Let $I\!\!H^{n}=\{w=(x_{1},x_{2},...x_{n})\in I\!\!R^{n},x_{n}>0\}$ be the hyperbolic half space endowed with the usual hyperbolic metric

[TABLE]

the metric $ds$ is invariant with respect to the motion group $G=SO(n,1),$ the hyperbolic volume form $d\mu(w)$ is

[TABLE]

and the hyperbolic distance $\rho(w,w^{\prime})$ given as

[TABLE]

with the Laplace Beltrami operator

[TABLE]

where $\Delta_{n}=\sum_{j=1}^{n}\frac{\partial^{2}}{\partial x_{j}^{2}}$ is the Euclidean Laplacian on $I\!\!R^{n}$. Before giving the main result of this section we start by the following lemma in which we compute the Fourier transform of the Poisson semigroup $p_{|\xi|}$ for the Bessel operator with respect to the parameter $|\xi|.$

Lemma 3.1**.**

Set $\xi=(\xi_{1},\xi_{2},...,\xi_{n-1})$ and $x=(x_{1},x_{2},...,x_{n-1})$, let $p_{|\xi|}(y,x_{n},x_{n}^{\prime})$ be the kernel of the Poisson semigroup for Bessel operator given in (2.2), then the following formula holds.

[TABLE]

with $z=\sqrt{x^{2}+x^{\prime 2}-2xx^{\prime}\cos y}.$

Proof.

From the formula giving the Fourier transform of a radial function

[TABLE]

we obtain

${\cal F}^{-1}\left[P_{|\xi|}(y,x_{n},x_{n}^{\prime})\right](x)=\frac{x_{n}x^{\prime}_{n}\sin y}{\pi z}|x|^{3-n)/2}\int^{+\infty}_{0}K_{1}(\rho|z)J_{\frac{n-3}{2}}(\rho|x|)\rho^{\frac{n+1}{2}}d\rho.$

Using formula Prudnikov([14] $p.365$)

[TABLE]

where

[TABLE]

with $\alpha=(n+3)/2,\mu=(n-3)/2,\nu=1,b=|x|,c=z$ we have

[TABLE]

[TABLE]

Now from the formula $F\left(a,b,b,z\right)=\left(1-z\right)^{-a}$ we obtain the result of Lemma 3.1. ∎

Theorem 3.1**.**

The Poisson problem (1.6) in hyperbolic space $I\!\!H^{n}$ has the solution given by

[TABLE]

with

[TABLE]

Proof.

Using the following formula intertwining the Laplace Beltrami operator ${\cal L}_{n}$ on the hyperbolic space $I\!\!H^{n}$ and the Bessel operator $L^{|\xi|}$

[TABLE]

The Poisson problem on the hyperbolic space (1.6) is transformed into the Bessel Poisson problem (1.3), with $u(y,x_{n})={\cal F}\left[x^{(1-n)/2}U(y,x,x_{n})\right](\xi)$ and $u_{0}(x_{n})=x_{n}^{(1-n)/2}{\cal F}[U_{0}(x,x_{n})](\xi)$

${\cal F}\left[x^{(1-n)/2}U(y,x,x_{n})\right](\xi)=$

[TABLE]

[TABLE]

$U(y,x,x_{n})=(2\pi)^{-(n-1)/2}\times$

[TABLE]

$U(y,x,x_{n})=(2\pi)^{-(n-1)/2}\int_{0}^{\infty}\int_{I\!\!R^{n-1}}{\cal F}^{-1}\left[P_{|\xi|}(y,x_{n},x_{n}^{\prime})\right](x-x^{\prime})\times$

[TABLE]

[TABLE]

with

[TABLE]

and the proof of Theorem 3.1 is finished. ∎

Proposition 3.1**.**

*Let $P^{I\!\!H}_{n}\left(y,w,w^{\prime}\right)$ be the Poisson kernel on the hyperbolic space $I\!\!H^{n}$ then we have

i)$\left(-\frac{\partial}{2\pi\sinh\rho\partial\rho}\right)P^{I\!\!H}_{n}\left(y,\rho(w,w^{\prime})\right)=P^{I\!\!H}_{n+2}\left(y,\rho(w,w^{\prime})\right),$

ii) $\int_{r}^{\infty}\frac{P^{I\!\!H}_{n+1}\left(y,\rho\right)}{\sqrt{\cosh^{2}\rho/2-\cosh^{2}r/2}}\sinh\rho d\rho=P^{I\!\!H}_{n}\left(y,r\right)$.

Proof.

The part i) is simple, to prove ii) set

[TABLE]

[TABLE]

with $C_{n+1}=\frac{\Gamma(n+2)/2)}{(2\pi)^{(n+2)/2}}$. Set $\cosh\rho=\sigma$ and set $\sigma-\cosh r=\mu$, we see that

[TABLE]

[TABLE]

$I=\sqrt{2}C_{n+1}\sin y\left(\cosh r-\cos y\right)^{-(n+2)/2}\times$

[TABLE]

Using the formula (Magnus et al.[11] p.13)

[TABLE]

where $B$ is the beta function, with $x=1/2$ and $y=(n+1)/2$ we obtain

[TABLE]

[TABLE]

thus we obtain ii) and the proof of Theorem 3.1is finished. ∎

4 Heat semigroup on hyperbolic space

In this section we give a new explicit formula for the heat kernel on the hyperbolic space $I\!\!H^{n}$.

Proposition 4.1**.**

*Let $e^{-y\sqrt{-{\cal L}_{n}}}$ and $e^{t{\cal L}_{n}}$ be the Poisson and heat semigroups on the hyperbolic space $I\!\!H^{n}$ then we have

i) $e^{t{\cal L}_{n}}=(4t)^{-1/2}L_{y^{2}}^{-1}\left[\frac{\sqrt{\pi}e^{-y\sqrt{-{\cal L}_{n}}}}{y}\right](1/4t)$, where $L_{y^{2}}^{-1}$ is the Laplace inverse transform with respect to $y^{2}$.

ii) $\left(-\frac{\partial}{2\pi\sinh\rho\partial\rho}\right)K^{I\!\!H}_{n}\left(t,\rho(w,w^{\prime})\right)=K^{I\!\!H}_{n+2}\left(t,\rho(w,w^{\prime})\right),$

iii) $\int_{r}^{\infty}\frac{K^{I\!\!H}_{n+1}\left(t,\rho\right)}{\sqrt{\cosh^{2}\rho/2-\cosh^{2}r/2}}\sinh\rho d\rho=K^{I\!\!H}_{n}\left(t,r\right).$

Proof.

To prove i) use the subordination formula (Strichartz [17] $p.50$).

$\frac{e^{-y\lambda}}{y}=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-uy^{2}}u^{-1/2}e^{-\lambda^{2}/4u}du$ or $\frac{\sqrt{\pi}e^{-y\lambda}}{y}=L\left(u^{-1/2}e^{-\lambda^{2}/4u}\right)(y^{2}),$

where $(Lf)(p)$ is the Laplace transform and $e^{\frac{-\lambda^{2}}{4u}}=u^{1/2}L^{-1}_{y^{2}}\left(\sqrt{\pi}\frac{e^{-y\lambda}}{y}\right)(u)$.

Set $\lambda=\sqrt{{\cal L}_{n}}$ and $\frac{1}{4u}=t$ in the last formula we can write

[TABLE]

where $L^{-1}$ is the inverse Laplace transform.

The parts ii) and iii) are consequence of i) and Proposition 3.1 ∎

Theorem 4.1**.**

The heat Cauchy problem on hyperbolic space (1.12) has the unique solution given by

[TABLE]

with

[TABLE]

Proof.

using (3.13) we see the formula (4.2) and the proof of Theorem4.1 is finished.

∎

Corollary 4.1**.**

*(Davies-Mandouvalos[4] and Lohoue and Rychener[10]) Let $K_{n}(t,w,w^{\prime})$ be the heat kernel on the hyperbolic space $I\!\!H^{n}$ then we have

i) For $n$ odd   $K_{n}(t,w,w^{\prime})=\left(-\frac{\partial}{2\pi\sinh\rho\partial\rho}\right)^{\frac{n-1}{2}}\frac{e^{\frac{-\rho^{2}}{4t}}}{(4\pi t)^{1/2}}$,

ii) for $n$ even   $K_{n}(t,w,w^{\prime})=\left(-\frac{\partial}{2\pi\sinh\rho\partial\rho}\right)^{\frac{n-2}{2}}\int_{\rho}^{\infty}\left(\cosh^{2}s/2-\cosh^{2}\rho/2\right)^{-1/2}\frac{e^{\frac{-s^{2}}{4t}}}{(4\pi t)^{3/2}}sds.$

Proof.

Set $\cos y=z$

[TABLE]

To prove the first statment i) we have

[TABLE]

that is

[TABLE]

and

[TABLE]

using ii) of 4.1 we have i).

To prove iii) we can write

[TABLE]

and

[TABLE]

this gives

[TABLE]

and finally

[TABLE]

using iii) of Theorem 4.1 we have

[TABLE]

Combining (4.5) and the part i) of Proposition 4.1 we obtain ii) and the proof of Corollary 4.2 is finished. ∎

Note that the wave equation on hyperbolic space is studied in intissar-Ould Moustapha[6] Bunk et al.[3] Lax-Phillips [8]

5 Heat kernel for the Bessel operator

Proposition 5.1**.**

i) The modified Laplace-Beltrami operator ${\cal L}_{2}$ on the hyperbolic space and Bessel operator $L^{a}$ on $I\!\!R^{+}$ are connected via the formulas

[TABLE]

where the Fourier transform is given by

[TABLE]

ii) The heat kernels for Bessel operator $H^{a}(t,x_{2},x_{2}^{\prime})$ is connected to the heat kernel on the hyperbolic half plane $H_{2}(t,z,z^{\prime})$ via the formula

[TABLE]

Proof.

The proof of this proposition is simple and in consequence is left to the reader. ∎

Theorem 5.1**.**

The heat Cauchy problem for the Bessel operator $L^{a}$ has the unique solution given by

[TABLE]

*with

$K_{a}(t,x,x^{\prime})=\frac{1}{4\sqrt{\pi}t^{3/2}}\times$*

[TABLE]

Proof.

The proof of this theorem follows from the proposition 5.3, the Fubini theorem and the formula (Lebedev[9] p.114)

[TABLE]

∎

6 Applications

In this section we give some applications of our results. As an application of Theorem 2.1 and 5.1 we give the following corollary giving explicit solution to the Poisson and heat problems with Morse potential. For recent work on Morse potential the reader can consult ( Abdelhaye et al. [1], Ikeda-Matsumoto[5], Morse [12] and Ould Moustapha [13]).

Corollary 6.1**.**

For $a\in I\!\!R$ the problem

[TABLE]

has the solution given by

[TABLE]

with

[TABLE]

where $K_{1}$ is the modified Bessel functions of second kind.

Proof.

Set $X=\ln x$ the problem (6.3) is transformed into the problem (1.3) and it is not hard to see the result of theorem from 2.1. ∎

Theorem 6.1**.**

The heat Cauchy problem with Morse Potential

[TABLE]

has the unique solution given by

[TABLE]

*with

$K_{a}(t,X,X^{\prime})=\frac{1}{4\sqrt{\pi}t^{3/2}}\times$*

[TABLE]

Proof.

Set $X=\ln x$ the problem (6.8) is transformed into the problem (1.9) and it is not hard to see the result of theorem from Theorem 5.1. ∎

Remark 6.1**.**

The Poisson and heat semigroups of the operator of bessel type

[TABLE]

*are considered in Betancor et al. [2].

It is not hard to see that*

[TABLE]

Corollary 6.2**.**

If $a\in iI\!\!R^{*}$ and $a=ib$ the problem (2.1) has the solution given by

[TABLE]

with

[TABLE]

[TABLE]

wih $H^{(1)}_{1}$, $H^{(1)}_{0}$ are the Bessel function of the third kind.

Proof.

Using formula $K_{\nu}(ze^{\frac{-i\pi\nu}{2}})=1/2i\pi e^{\frac{i\pi\nu}{2}}H^{(n)}_{\nu}(z)$ (Magnus et al.[11]$p67.$) ∎

The last application of our result is the explicit formula of the Poisson semigroup on the sphere $S^{n}$.

Corollary 6.3**.**

The Poisson equation in the sphere $S^{n}$ has the unique solution given by

[TABLE]

with

[TABLE]

Proof.

By comparing the radial parts of the Laplace Beltrami operators on the spaces $I\!\!H^{n}$ and $S^{n}$ given respectively by

[TABLE]

and

[TABLE]

the corollary 6.3 can be seen from the theorem 3.1 by an argument of analytic continuation. ∎

Note that the result of the corollary agrees with the formula $(4.9)$ in (Taylor [18] p. 114).