Poisson kernel of half spaces in real hyperbolic spaces

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Poisson kernel of half spaces in real hyperbolic spaces

Tomasz Byczkowski, Piotr Graczyk, Andrzej Stos

To cite this version:

Tomasz Byczkowski, Piotr Graczyk, Andrzej Stos. Poisson kernel of half spaces in real hyperbolic

spaces. Revista Matemática Iberoamericana, European Mathematical Society, 2007, 23 (2007), no. 1,

pp.85–126. �hal-00079091�

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ccsd-00079091, version 1 - 8 Jun 2006

Poisson kernels of half–spaces in real hyperbolic spaces

T. Byczkowski

Institute of Mathematics, Wroc law University of Technology, Poland and

P. Graczyk

Département de Mathématiques, Université d’Angers, France and

A. St´os

Institute of Mathematics, Wroc law University of Technology Laboratoire de Math´ematique, UMR 6620, Universit´e Blaise Pascal

Clermont-Ferrand, France

Abstract

We provide an integral formula for the Poisson kernel of half- spaces for Brownian motion in real hyperbolic space H

ⁿ

. This enables us to find asymptotic properties of the kernel. We also show convergence to the Poisson kernel of the whole space H

ⁿ

. For n = 3, 4 or 6 we compute explicit formulas for the Poisson kernel itself.

2000 MS Classification: Primary 60J45; Secondary 58J65. Key words and phrases:

hyperbolic spaces, Brownian motion, Poisson kernel. Research partially supported by

KBN grant 2 P03A 041 22 and RTN Harmonic Analysis and Related Problems contract

HPRN-CT-2001-00273-HARP

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1 Introduction

In recent years there is a growing interest in research of Brownian motion on hyperbolic spaces (cf. [Y3], [BJ]). More advanced theory requires a detailed knowledge of some basic potential-theoretic objects. The most fundamental turns out to be the Poisson kernel for a domain. If we know the Poisson kernel then we are able, for example, to solve the classical Dirichlet boundary problem for a domain. Another basic object in potential theory is the Green function of a domain. The ”sweeping out” method allows to recover the Green function from the Poisson kernel. The Green function of a domain is a very useful tool in solving various important problems in potential theory.

In particular, it enables to determine conditions under which a reasonable potential theory of the Schr¨odinger operator is feasible for a given domain and a particular potential ([ChZ]). In the classical situation of the Laplacian in R

ⁿ

, the exact formula for the Poisson kernel (or Green function) always simpliﬁes the arguments as well as indicates the future areas of investigation.

We believe that the same situation will occur in the case of hyperbolic spaces.

The aim of this paper is to give a representation formula for the Poisson kernel of a half–space in the real hyperbolic space H

ⁿ

, i.e. for the probability distribution of the hyperbolic Brownian motion stopped when leaving a half–

space, and to use it in order to prove exact asymptotics of the kernel. Note that the boundary of the considered half–space is a horocycle in H

ⁿ

.

The Poisson kernel of a half–space is closely related to stable laws and functionals of the Brownian motion ([BCF], [BCFY], [Y1], [Y2]). Another motivation comes from the risk theory in financial mathematics ([D]). Our kernel, up to a passage from the dimension 2 to the dimension n, was identified in terms of its Fourier transform in [BCF]. It turns out, however, that it is not sufficient for most applications. In particular, a formula for the kernel itself or its asymptotical behaviour were not identified (cf. [BCFY], p. 589).

From the technical point of view, the main diﬃculty is that the inverse Fourier transform (or the Hankel transform) leads to an integral containing Bessel functions which has oscillatory character, see (5) below. Moreover, for integrals like (5), the Lebesgue bounded convergence theorem is often not applicable (for example when | y | → ∞ ) and we are left with a nontrivial problem of obtaining the asymptotics of the kernel.

The paper is organized as follows. In Section 2, after some preliminaries, we provide, for convenience of the reader, proof of the formula for the Fourier transform of the Poisson kernel of a half–space in H

ⁿ

. We also state a ﬁrst integral formula (5) for the Poisson kernel of a half–space, based on the inverse Fourier transform.

In Section 3, in Theorem 3.2 we obtain a second integral formula for the

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Poisson kernel of a half–space. This is our main representation formula. It is much more suitable for further applications than (5). Section 3 ends with explicit integral formulas for the Poisson kernel of a half–space, that arise in lower dimensions.

In Section 4 we study the above mentioned asymptotics of the Poisson kernel of a half–space in H

ⁿ

. We use our main representation formula from Theorem 3.2 as well as the semigroup and homogeneity properties of the Poisson kernel.

2 Preliminaries

Consider the half-space model of the n-dimensional real hyperbolic space H

ⁿ

= { (x

1

, . . . , x

n−1

, x

n

) ∈ R

ⁿ⁻¹

× R : x

n

> 0 } .

The Riemannian metric, the volume element and the Laplace-Beltrami operator are given by

ds

²

= dx

²₁

+ ... + dx

²_n₋₁

+ dx

²_n

x

²_n

,

dV = dx

1

...dx

n−1

dx

n

x

ⁿ_n

,

∆ = x

²_n

(

n

X

i=1

∂

_i²

) − (n − 2)x

n

∂

n

, respectively (here ∂

i

=

_∂x^∂_i

, i = 1, ..., n).

Let (B

i

(t))

i=1...n

be a family of independent classical Brownian motions on R with the generator

_dx^d²2

(and not

¹₂_dx^d²2

) i.e. the variance E

⁰

B

_i²

(t) = 2t.

Then the Brownian motion on H

ⁿ

, X = (X

i

)

i=1...n

, can be described by the following system of stochastic diﬀerential equations



 



 



dX

1

(t) = X

n

(t)dB

1

(t) dX

2

(t) = X

n

(t)dB

2

(t)

. . .

dX

n

(t) = X

n

(t)dB

n

(t) − (n − 2)X

n

(t)dt.

By the Itô formula one verifies that the generator of the solution of this system is ∆. Moreover, it can be easily verified that the solution is given by



 



 



X

1

(t) = X

1

(0) + R

t

0

X

n

(0) exp(B

n

(s) − (n − 1)s)dB

1

(s) X

₂

(t) = X

₂

(0) + R

t

0

X

_n

(0) exp(B

_n

(s) − (n − 1)s)dB

₂

(s) . . .

X

n

(t) = X

n

(0) exp(B

n

(t) − (n − 1)t)

(5)

Convention: by c (or C) we always denote a general constant that depends on n and other constant parameters only. The value of these constants may change in the same string of estimates.

Below we identify the Poisson kernel (the function y → P

a

(x, y)) in terms of its Fourier transform. Theorems 2.1 and 2.2 cover facts that are essentially known. Similar results can be found in both [BCF] and [BCFY], in a slightly diﬀerent setting of more general generators on the space H

²

. In order to make our paper self-contained and for the convenience of the reader, we include these facts here, in the present setting of H

ⁿ

.

Deﬁne the projection ˜ : R

ⁿ

∋ u = (u

1

, ..., u

n

) → u ˜ = (u

1

, ..., u

n−1

) ∈ R

ⁿ−1

. In particular, ˜ X (t) = (X

1

(t), ..., X

n−1

(t)).

Consider a half-space D = { x ∈ H

ⁿ

: x

n

> a } for some ﬁxed a > 0.

Deﬁne

τ = inf { t > 0 : X(t) ∈ / D } = inf { t > 0 : X

n

(t) = a } .

By P

a

(x, dy), x = (x

1

, x

2

, ..., x

n

) ∈ D, y = (y

1

, y

2

, ..., y

n−1

, a) ∈ ∂D we denote the Poisson kernel of D, ie. the distribution of X(τ) starting at x (since X

n

(τ ) = a it is enough to consider the distribution of ˜ X(τ)).

Theorem 2.1.

F [P

_a

(x, · )](u) = E

^x

e

ⁱ^h^u,^X^˜^τⁱ

= e

ⁱ^h^{x, u}^˜ ⁱ

x

_n

a

ⁿ⁻¹₂

K

ⁿ⁻¹

2

( | u | x

_n

) K

ⁿ⁻¹

2

( | u | a) , x ∈ D, u ∈ R

ⁿ⁻¹

where K

ν

, ν > 0, is the modified Bessel function of the third kind, called also Macdonald function.

Proof. Since B

_i

(t) are independent and τ depends only on X

_n

(i.e. B

_n

) we obtain for x = (˜ x, x

n

) = (x

1

, x

2

, . . . , x

n

)

E

^x

exp(i h u, X(τ) ˜ i ) = E

^xⁿ

n−1

Y

j=1

E

^x^j

exp(iu

j

X

j

(τ )). (1) We adopt here a useful notation that E

^x^j

Y for the expectation with re- spect to j’th component B

_j

of our basic (n − 1)-dimensional Brownian motion B = (B

1

, . . . , B

n

), starting from x

n

. We compute the integral R

τ

0

X

n

(s) dB

j

(s) by approximation of 1

_{s<τ}

(s)X

n

(s) by simple processes of the form P

k

f

_k

1

_[t_k_,t_k+1₎

, where f

_k

∈ F (B

_n

(t

_k

)). Using independence of

the increments of B

j

and the fact that the function under the integral below

(6)

is independent of B

j

, we get

E

^x^j

exp(iu

j

X

j

(τ)) = E

^x^j

exp

iu

j

(X

j

(0) + Z

τ

0

X

n

dB

j

)

= e

^iu^j^x^j

exp

− u

²_j

Z

τ

0

X

_n²

(s) ds

. This and (1) imply that

E

^x

exp(i h u, X(τ ˜ ) i ) = e

ⁱ^h^{x, u}^˜ ⁱ

E

^xⁿ

e

^−|^u^|²^R⁰^τ^Xⁿ²^(s)ds

= e

ⁱ^h^{x, u}^˜ ⁱ

E

^xⁿ

e

_q

(τ ), where e

q

(τ) = exp( R

τ

0

q(X

n

(s)))ds with q(y) = − ( | u | y)

²

. Observe that the function ϕ(y) = E

^y

e

q

(τ) is by deﬁnition the gauge for the Schr¨odinger operator L + q based on the generator L of X

_n

and the potential q. By general theory (see e.g. [ChZ], Prop. 4.13, p. 119) it is a solution for the Schr¨odinger equation. Since dX

n

(t) = X

n

(t)dB

n

(t) − (n − 2)X

n

(t)dt, by a standard argument based on the Itˆo formula, we get the generator of X

_n

L = x

²_n

d

²

dx

²_n

− (n − 2)x

n

d dx

n

. Consequently, ϕ satisﬁes the following equation

y

²

ϕ

^′′

(y) − (n − 2)yϕ

^′

(y) − | u |

²

y

²

ϕ(y) = 0 (2) on the positive half-line. Let ϕ(y) = y

ⁿ⁻¹²

g(y). Then ϕ

^′

(y) =

ⁿ⁻₂¹

y

ⁿ⁻³²

g(y) + y

ⁿ⁻¹²

g

^′

(y), ϕ

^′′

(y) =

⁽ⁿ⁻¹⁾⁽ⁿ₄ ⁻³⁾

y

ⁿ⁻⁵²

g(y) + (n − 1)y

ⁿ⁻³²

g

^′

(y) + y

ⁿ⁻¹²

g

^′′

(y) and consequently (2) reads as

y

²

g

^′′

(y) + yg

^′

(y) − | u |

²

y

²

+ ((n − 1)/2)

²

g(y) = 0.

Substituting | u | y = z and g(y) = h(z) we get

z

²

h

^′′

(z) + zh

^′

(z) − z

²

+ ((n − 1)/2)

²

h(z) = 0. (3)

This is the modiﬁed Bessel equation of order (n − 1)/2. Taking into account the form of the general solution of (3) we infer that

ϕ(y) = y

ⁿ⁻¹²

(c

1

I

ⁿ⁻¹

2

( | u | y) + c

2

K

ⁿ⁻¹

2

( | u | y)), for an appropriate choice of c

1

and c

2

, where I

ⁿ⁻¹

2

( · ) and K

ⁿ⁻¹

2

( · ) are the modified Bessel function of the first and third kind, respectively. Observe that by definition ϕ(y) is bounded in y and ϕ(a) = 1. Since I

ⁿ⁻¹

2

( | u | y) is

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unbounded and K

ⁿ⁻¹

2

( | u | y) is bounded when y → ∞ , it follows that c

1

= 0.

From the other condition we get the normalizing constant

c

₂

= 1

a

ⁿ⁻¹²

K

ⁿ⁻¹

2

( | u | a) . This completes the proof.

Remark. As K

ⁿ⁻¹

2

(x) ∼ √ π(2x)

⁻^1/2

e

⁻^x

when x → ∞ , the Fourier transform of our kernel is in L

¹

. Thus there exists the corresponding density which we denote by P

a

(x, y).

From Theorem 2.1 it follows that nonzero ˜ x gives rise just to a translation of P

a

(x, y ) as a function of y. Therefore, in what follows we may and do assume ˜ x = 0. Consequently, we may simplify the notation by identifying P

a

(x, y) = P

a

((0, ..., 0, x), y), x > 0.

For z > 0 we deﬁne ([GR], 8.432.8) m

ⁿ₂₋1

(z) =

Z

_∞

0

e

⁻^u

u

ⁿ²⁻¹

(u + 2z)

ⁿ²⁻¹

du = d

n

e

^z

z

ⁿ⁻¹²

K

ⁿ⁻¹

2

(z), (4)

where d

n

= π

⁻¹²

2

ⁿ⁺¹²

Γ

ⁿ₂

. Observe that for n ∈ 2 N the function m

ⁿ₂₋1

is just a polynomial of the degree

ⁿ₂

− 1. In this case we regard m

ⁿ₂₋1

(z) as deﬁned for all complex numbers. By [GR] (8.468, p. 915), for n ∈ 2 N we get

m

ⁿ₂₋1

(z) = d

n

r π 2

n 2−1

X

j=0

(n − 2 − j)!2

^j+ⁿ²⁺¹

j!(

ⁿ₂

− 1 − j)! z

^j

.

In particular, m

0

(z) = 1, m

1

(z) = 2(1 + z) and m

2

(z) = 8(z

²

+ 3z + 3).

Theorem 2.2 (Poisson kernel formula). Let a > 0, x > a and y ∈ R

ⁿ⁻¹

. If | y | > 0 then

P

a

(x, y) = x 2πa

ⁿ⁻¹₂

| y |

⁻ⁿ⁻³²

Z

_∞

0

K

ⁿ⁻¹

2

(rx) K

ⁿ⁻¹

2

(ra) J

ⁿ⁻³

2

(r | y | )r

ⁿ⁻¹²

dr, (5) and when | y | = 0 it is understood in the limiting sense, i.e.

P

a

(x, 0) = 2

²⁻ⁿ

Γ

ⁿ⁻₂¹

x πa

ⁿ⁻¹₂

Z

_∞

0

K

ⁿ⁻¹

2

(rx) K

ⁿ⁻¹

2

(ra) r

ⁿ⁻²

dr. (6)

Equivalently, we can write

P

a

(x, y) = (2π)

¹⁻ⁿ²

| y |

⁻ⁿ⁻³²

Z

_∞

0

e

⁻^r(x⁻^a)

m

ⁿ₂₋1

(rx)

m

ⁿ₂₋1

(ra) r

ⁿ⁻¹²

J

ⁿ⁻³

2

(r | y | )dr, | y | > 0.

(7)

(8)

The special case | y | = 0 reads as P

a

(x, 0) = 2

²⁻ⁿ

π

¹⁻ⁿ²

Γ

ⁿ⁻₂¹

Z

_∞

0

e

⁻^r(x⁻^a)

m

ⁿ₂₋1

(rx)

m

ⁿ₂₋1

(ra) r

ⁿ⁻²

dr.

Proof. Recall that if f is a radial function, f (y) = f

o

( | y | ), then so is F f and the Fourier inversion formula in R

ⁿ⁻¹

reads, up to a factor (2π)

⁻⁽ⁿ⁻¹⁾

, as the Hankel transform of order (n − 3)/2 ([F], (7.38), p. 247):

f

o

( | y | ) = (2π)

⁻ⁿ⁻¹²

Z

_∞

0

( F f )

o

(r)(r | y | )

¹⁻ⁿ⁻¹²

J

ⁿ⁻¹

2 −1

(r | y | )r

ⁿ⁻²

dr.

This gives (5). Now, (7) is immediate and the special cases y = 0 follow from the asymptotics of the Bessel function (see e.g. [GR], 8.440 or [F], (5.10), p. 130)

J

ν

(z) ∼ 1

2

^ν

Γ(1 + ν) z

^ν

, z → 0. (8) The proof is complete.

Corollary 2.3. The Poisson kernel P

a

(x, y), as a function of three variables (a, x, y), is a homogeneous function of order − n + 1:

P

ta

(tx, ty) = t

⁻ⁿ⁺¹

P

a

(x, y ), t > 0.

Proof. The homogeneity property can be derived easily from the scaling properties of the process (X

1

(t), . . . , X

n

(t)). This is also obvious by change of variables ˜ r = tr in the formula (5) written for P

ta

(tx, ty).

Remark. Certainly, when n ∈ 2 N then also J

ⁿ⁻³

2

(r | y | ) simpliﬁes to an elementary function. This fact, however, is not very useful in what follows and we will not pursue this further.

3 Poisson kernel of half-space

In this section we give a representation formula for the Poisson kernel. For n = 2 the resulting formula coincides with the one of R

²

, so that below we shall always tacitly assume n ≥ 3 (note, however, that a great part of our argument remains valid also for n = 2).

From now on we use the following notation, partially introduced in the preceding section: λ = x − a.

The following technical lemma is essential in what follows.

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Lemma 3.1. Let

Q(z) = z − n(n − 2) λ

8ax , z ∈ C . Define F

λ

(z) by the following formula:

λF

λ

(z) = (z/a)e

^λz/a

(x/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz/a) − (x/a)

ⁿ²⁻¹

Q(z/a)K

ⁿ⁻¹

2

(z) K

ⁿ⁻¹

2

(z) . (9)

Then

F

λ

(z) = O(z

⁻¹

), z → ∞ . (10) Proof. Using the asymptotic expansions for the modiﬁed Bessel function K

ⁿ⁻¹

2

(z) ([GR], 8.451.6, p. 910) we get e

^z

z

ⁿ⁻¹²

K

ⁿ⁻¹

2

(z) = z

ⁿ²⁻¹

(c

0

+ c

1

2z + R

2

), (11)

where

c

_k

= c

⁽ⁿ⁾_k

= r π

2 Γ(n/2 + k)

k!Γ(n/2 − k) , k = 0, 1,

R

2

= O(z

⁻²

) and | z | is large enough. Hence, it is enough to show (z/a)e

^xz/a

(xz/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz/a) − (x/a)

ⁿ²⁻¹

Q(z/a)e

^z

z

ⁿ⁻¹²

K

ⁿ⁻¹

2

(z)

= O(z

ⁿ²⁻²

), z → ∞ . (12) From (11) it follows that on one hand we have

(z/a)e

^xz/a

(xz/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz/a)

= c

0

(x/a)

ⁿ²⁻¹

(z

ⁿ²

/a) + (c

1

/(2x)) (x/a)

ⁿ²⁻¹

z

ⁿ²⁻¹

+ O(z

ⁿ²⁻²

).

On the other hand, using

ⁿ⁽ⁿ₄⁻²⁾

c

0

= c

1

, we get x

a

ⁿ₂₋1

Q z a

e

^z

z

ⁿ⁻¹²

K

ⁿ⁻¹

2

(z)

= x a

ⁿ₂₋1

z

a − n(n − 2) 8

λ ax

(c

0

z

ⁿ²⁻¹

+ 1

2 c

1

z

ⁿ²⁻²

+ O(z

ⁿ²⁻³

))

= c

o

a x

a

ⁿ₂₋1

z

ⁿ²

+ x a

ⁿ₂₋1

c

1

2a − n(n − 2) 4 c

0

λ 2ax

z

ⁿ²⁻¹

+ O(z

ⁿ²⁻²

)

= c

_o

a

x a

ⁿ₂₋1

z

ⁿ²

+ x a

ⁿ₂₋1

c

₁

2x z

ⁿ²⁻¹

+ O(z

ⁿ²⁻²

).

Then (12) is obviously satisﬁed and the assertion follows.

(10)

Remark. The advantage of this lemma is due to the fact that we may and do use it for z ∈ C . This fact is exploited below.

Observe that the function K

ⁿ₂₋1

(z) has no zeros in {ℜ (z) ≥ 0 } (cf. [E1], p. 62) and hence F

λ

(z) is analytic in this half-plane. Moreover, by the inverse Laplace transform theorem ([F], Theorem 8.5) together with (10) we get that F

λ

is the Laplace transform of some function w

λ

, i.e.

F

_λ

(z) = Z

_∞

0

e

⁻^zv

w

_λ

(v)dv, ℜ (z) > 0, (13) under the additional condition that for some b > 0 the following limit

r

lim

→∞

1 2πi

Z

b+ir b−ir

F

λ

(z)e

^vz

dz (14)

exists for all v > 0 and it is a piecewise continuous function of v admitting the Laplace transform. Then the limit is equal to w

_λ

(v).

The existence of the above limit is shown in Theorem 3.3, together with an explicit formula for the function w

λ

itself.

We are ready to state our representation formula. Recall that λ = x − a.

Theorem 3.2.

P

a

(x, y) = Γ(

ⁿ₂

− 1) 2π

^n/2

λ (λ

²

+ | y |

²

)

^n/2

Z

_∞

0

w

λ

(v)L(λ, y, v)

((λ + av)

²

+ | y |

²

)

ⁿ²⁻¹

dv, (15) with L(λ, y, v) defined by

L(λ, y, v)

(λ

²

+ | y |

²

)((λ + av)

²

+ | y |

²

)

ⁿ²⁻¹

=

λ

²

+ | y |

²

(λ + av)

²

+ | y |

²

ⁿ₂−1

− 1 +

n 2

− 1

av(2λ + av)

λ

²

+ | y |

²

. (16) Proof. Recall that for ν > 0 we have ([E], vol I, (7) and (8) p. 182 or [GR], 17.13.43, 17.13.44)

Z

_∞

0

e

⁻^λr

r

^ν

J

ν−1

(r | y | )dr = 2

^ν

π

⁻¹²

Γ(ν + 1

2 ) | y |

^ν⁻¹

λ (λ

²

+ | y |

²

)

^ν+¹²

and

Z

_∞

0

e

⁻^λr

r

^ν⁻¹

J

_ν₋₁

(r | y | )dr = 2

^ν⁻¹

π

⁻¹²

Γ(ν − 1

2 ) | y |

^ν⁻¹

1 (λ

²

+ | y |

²

)

^ν⁻¹²

.

(11)

For z = ra we have

λF

λ

(ra) = rm

ⁿ₂₋1

(rx) − (x/a)

ⁿ²⁻¹

Q(r)m

ⁿ₂₋1

(ra)

m

ⁿ₂₋1

(ra) , (17)

so rm

ⁿ₂₋1

(rx)

m

ⁿ₂₋1

(ra) = x a

ⁿ₂₋1

Q(r) + λF

λ

(ra) and hence by (4)

x a

ⁿ⁻¹₂

rK

ⁿ⁻¹

2

(rx) K

ⁿ⁻¹

2

(ra) = e

⁻^λr

[ x a

ⁿ₂₋1

Q(r) + λF

λ

(ra)].

Putting this into the Hankel transform formula (5) and using (13) we get 2π

^n/2

Γ(

ⁿ₂

− 1) P

a

(x, y) = λ(n − 2) 2

x a

ⁿ₂₋1

2 (λ

²

+ | y |

²

)

ⁿ²

− 1 4xa

n (λ

²

+ | y |

²

)

ⁿ²⁻¹

+ λ Z

_∞

0

w

λ

(v)

((λ + av)

²

+ | y |

²

)

ⁿ²⁻¹

dv.

Putting r = 0 in (17) we get λF

λ

(0) = λ

Z

_∞

0

w

λ

(v)dv

= − (x/a)

ⁿ²⁻¹

Q(0) = n(n − 2) (x/a)

ⁿ²⁻¹

λ

8xa (18)

so that

λ(F

_λ

(ra) − F

_λ

(0))m

ⁿ

2−1

(ra) = rm

ⁿ

2−1

(rx) − (x/a)

ⁿ²⁻¹

rm

ⁿ

2−1

(ra).

Dividing both sides by r and taking the limit r → 0 we obtain λaF

_λ^′

(0) = − λa

Z

_∞

0

vw

λ

(v)dv = 1 − (x/a)

ⁿ²⁻¹

.

We used the fact that vw

_λ

(v) allows the Laplace transform which is evident from Theorem 3.3. Hence

(x/a)

ⁿ²⁻¹

= 1 + λa Z

_∞

0

vw

λ

(v )dv Moreover,

λ[(F

λ

(ra) − F

λ

(0)) − raF

_λ^′

(0)]m

ⁿ₂₋1

(ra) = rm

ⁿ₂₋1

(rx) − rm

ⁿ₂₋1

(ra). (19)

(12)

Again, dividing both sides by (ra)

²

, letting r → 0 and using m

ⁿ₂₋1

(0) = m

^′ⁿ

2−1

(0) we get

λF

_λ^′′

(0)/2 = λ/a

²

so that

1 = a

²

2 Z

_∞

0

v

²

w

λ

(v)dv. (20)

The facts that F

_λ^′′

(0) exists and that the function v

²

w

λ

(v) admits the Laplace transform follow from Theorem 3.3. Consequently, we have

x a

ⁿ₂₋1

= 1 + λa Z

_∞

0

vw

λ

(v)dv

= a

²

2 Z

_∞

0

v

²

w

λ

(v)dv + λa Z

_∞

0

vw

λ

(v)dv

= 1

2 Z

_∞

0

av(2λ + av)w

λ

(v)dv.

Finally, 2π

^n/2

Γ(

ⁿ₂

− 1)

⁻¹

P

a

(x, y) is equal to λ(n − 2)

2 Z

_∞

0

av(2λ + av)w

λ

(v) (λ

²

+ | y |

²

)

ⁿ²

dv − λ

Z

_∞

0

w

λ

(v)

(λ

²

+ | y |

²

)

ⁿ²⁻¹

dv +λ

Z

_∞

0

w

λ

(v )

((λ + av)

²

+ | y |

²

)

ⁿ²⁻¹

dv

= λ

(λ

²

+ | y |

²

)

ⁿ²

Z

_∞

0

w

λ

(v)L(λ, y, v) ((λ + av)

²

+ | y |

²

)

ⁿ²⁻¹

dv and the assertion follows.

Below we give a description of the function w

λ

. The formula depends on the zeros of the function K

ⁿ⁻¹

2

(z). Even if in general the values of these zeros are not given explicitly, we are able to prove some important properties (as boundedeness or asymptotics) of w

_λ

, which are essential in applications.

Moreover, in lower dimensions we provide explicit formulas as well (see Sec- tion 3).

The function K

ⁿ⁻¹

2

(z) extends to an entire function when n is even and has a holomorphic extension to C \ ( −∞ , 0] when n is odd. Denote the set of zeros of the function K

ⁿ⁻¹

2

(z) by Z = { z

₁

, ..., z

_k_n

} . We now provide some information about these zeros, needed in the sequel (cf. [E1], p. 62). Recall that in the case of even dimensions the functions m

ⁿ₂₋₁

(z) are polynomials of degree

ⁿ₂

− 1. They always have the same zeros as K

ⁿ⁻¹

2

, so k

n

= (n/2) − 1

when n ∈ 2 N . For n = 2k + 1, k

n

is the even number closest to (n/2) − 1.

(13)

In particular, we have k

3

= 0, and for n = 5 and 7 we have k

n

= 2. The functions K

ⁿ⁻¹

2

and K

ⁿ⁻³

2

have no common zeros.

In order to describe the function w

λ

we introduce additional notation.

Let us deﬁne

w

1,λ

(v) = − (x/a)

ⁿ⁻¹²

λa

kn

X

i=1

z

i

e

^λzⁱ^/a

K

ⁿ⁻¹

2

(xz

i

/a) K

ⁿ⁻³

2

(z

_i

) e

^zⁱ^v

, (21)

and w

₁^∗

(v ) = sup

_0<λ6a

| w

_1,λ

(v) | .

Using the functions m

ⁿ₂₋1

, the formula (21) reads as follows:

w

1,λ

(v) = − 1 (n − 2)λa

kn

X

i=1

m

ⁿ₂₋1

(xz

i

/a) m

ⁿ₂₋2

(z

i

) e

^zⁱ^v

. We deﬁne additionally in the case of n odd

w

2,λ

(v) = ( − 1)

ⁿ⁺¹²

(x/a)

ⁿ⁻¹²

λa (22)

× Z

_∞

0

I

ⁿ⁻¹

2

(xu/a) K

ⁿ⁻¹

2

(u) − I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a) K

²n−1

2

(u) + π

²

I

²n−1 2

(u) e

⁻^λu/a

e

⁻^vu

udu, and, as before w

₂^∗

(v ) = sup

_0<λ6a

| w

2,λ

(v) | .

We also need the following asymptotic formulas for the modiﬁed Bessel functions I

ν

, K

ν

: For u > 1 we have ([F], [GR])

I

ν

(u) = (2πu)

⁻¹²

e

^u

[1 + E

1

(u)], K

ν

(u) = π

¹²

(2u)

⁻¹²

e

⁻^u

[1 + E

2

(u)], (23) where E

1

(u), E

2

(u) = O(u

⁻¹

), u → ∞ .

When u → 0 we have for ν > 0:

I

ν

(u) ∼ c

ν

u

^ν

, K

ν

(u) ∼ c

^′_ν

u

⁻^ν

, (24) with c

ν

= 2

⁻^ν

/Γ(ν + 1) and c

^′_ν

= 2

^ν⁻¹

Γ(ν). Whenever ν = 0 one has I

0

(u) ∼ 1, K

0

(u) ∼ log(2/u). We now formulate and prove our representation theorem for the function w

λ

.

Theorem 3.3. In the even dimensions w

λ

(v) = w

1,λ

(v);

while, in the odd dimensions

w

_λ

(v) = w

_1,λ

(v) + w

_2,λ

(v).

(14)

Moreover, we have | w

^∗_i

(v) | 6 C

i

(n, a), i = 1, 2 and w

1

(v) = lim

x→a+

w

1,λ

(v ) = 1 a

²

kn

X

i=1

z

_i²

e

^zⁱ^v

;

( − 1)

ⁿ⁺¹²

w

_2,λ

(v ) > 0, v > 0, (n odd);

w

2

(v) = lim

x→a+

w

2,λ

(v) = ( − 1)

ⁿ⁺¹²

a

²

Z

_∞

0

ue

⁻^vu

du K

²n−1

2

(u) + π

²

I

²n−1 2

(u) , (n odd);

Z

_∞

0

v

^k

w

^∗₁

(v )dv < ∞ , k = 1, 2, . . . ;

Z

_∞

0

v

ⁿ⁻¹

w

₂^∗

(v)dv < ∞ ;

v

lim

→∞

v

^k

w

1,λ

(v) = 0, k = 1, 2, . . . ;

v

lim

→∞

v

ⁿ⁺¹

w

2,λ

(v) = ( − 1)

ⁿ⁺¹²

n!

2

ⁿ⁻²

Γ(

ⁿ⁻₂¹

)Γ(

ⁿ⁺¹₂

)

(x/a)

ⁿ⁻¹

− 1

λa , (n odd).

Proof. We recall the basic formula (9) λF

λ

(z) = (z/a)e

^λz/a

(x/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz/a) − (x/a)

ⁿ²⁻¹

Q(z/a)K

ⁿ⁻¹

2

(z) K

ⁿ⁻¹

2

(z) .

By standard rules for computing residues of meromorphic functions and using the following formula for derivatives of Bessel functions (cf. [E1], 7.11(22) p. 79)

d

dz (z

^ν

K

ν

(z)) = − z

^ν

K

ν−1

(z), we obtain

Res

_z_i

F

_λ

= − (x/a)

ⁿ⁻¹²

λa

z

i

e

^λzⁱ^/a

K

ⁿ⁻¹

2

(xz

i

/a) K

ⁿ⁻³

2

(z

i

) . (25)

Using the functions m

ⁿ₂₋1

, we obtain Res

_z_i

F

_λ

= − 1

(n − 2)λa

m

ⁿ₂₋1

(xz

i

/a)

m

ⁿ₂₋2

(z

i

) . (26) As mentioned before (see (13) and (14)), by the inversion theorem for the Laplace transform we have

w

_λ

(v) = 1 2πi lim

r→∞

Z

b+ir b−ir

F

_λ

(z)e

^zv

dz

(15)

for some b > 0. We show the existence of the above limit together with computing formula for the function w

_λ

.

The technique of integration is diﬀerent in even and odd dimensions.

This is due to the fact that in the ﬁrst case the function under the integral extends to a meromorphic one while in the odd dimension we have to deal with a branch cut.

For n ∈ 2 N we choose any b > 0. All the zeros of m

ⁿ₂₋1

(z) satisfy ℜ (z

i

) < b (actually, we have in general ℜ (z

i

) < 0, i = 1, ..., s, cf. [E1], p.

62). To calculate w

λ

we integrate over the rectangular contour with corners at b − ir, b + ir, − r − ir, − r + ir. By (10) we infer that integrals over the upper, left, and bottom side of the rectangle tend to 0 as r → ∞ . Hence, by the residue theorem, the limit in (14) exists and is equal to the sum of all residues of the function F

λ

(z)e

^zλ

. Thus, we have w

λ

= w

1,λ

and the assertion follows.

In the odd dimensions, however, the function under integral is no longer meromorphic. We make the branch cut along the negative real axis ( −∞ , 0]

and change the contour of integration to wrap around this line (see the picture).

- 6

_b

−ir ir

−r

γ1

6

γ2

-

q q

First, we examine behaviour of our function near the negative axis ( −∞ , 0).

For z = − y (y > 0) we have (see [E1], (45), p. 80)

ǫ

lim

→0+

K

ⁿ⁻¹

2

( − y + iǫ) = e

^iπ(1−n)²

K

ⁿ⁻¹

2

(y) − iπI

ⁿ⁻¹

2

(y),

(16)

ǫ

lim

→0+

K

ⁿ⁻¹

2

( − y − iǫ) = e

^iπ(n−1)²

K

ⁿ⁻¹

2

(y) + iπI

ⁿ⁻¹

2

(y).

Now, observe that, similarly as before, the integrals over the left, upper and bottom side of our rectangular contour vanish as r → ∞ by (10). The same holds true for the half-circle with radius ǫ → 0 around the origin. Note that the branch cut and the residues for F

λ

(z) are due to the term

F ˜

λ

(z) = z(x/a)

ⁿ⁻¹²

e

^λz/a

K

ⁿ⁻¹

2

(xz/a) λaK

ⁿ⁻¹

2

(z) ,

the rest of the function F

λ

(z) being holomorphic in C . Therefore 1

2πi Z

b+ir

b−ir

F

λ

(z)e

^zv

dz = w

1,λ

(v) − 1 2πi

Z

γ1

+ Z

γ2

F ˜

λ

(z)e

^zv

dz.

After taking the limits r → ∞ and ǫ → 0, we get Z

γ1

+ Z

γ2

F ˜

λ

(z)e

^zv

dz

= (x/a)

ⁿ⁻¹²

λa



 − Z

_∞

0

u(e

^iπ(1−n)²

K

ⁿ⁻¹

2

(xu/a) − iπI

ⁿ⁻¹

2

(xu/a)) e

^iπ(1−n)²

K

ⁿ⁻¹

2

(u) − iπI

ⁿ⁻¹

2

(u) e

⁻^λu/a

e

⁻^vu

du +

Z

_∞

0

u(e

^iπ(n−1)²

K

ⁿ⁻¹

2

(xu/a) + iπI

ⁿ⁻¹

2

(xu/a)) e

^iπ(n−1)²

K

ⁿ⁻¹

2

(u) + iπI

ⁿ⁻¹

2

(u) e

⁻^λu/a

e

⁻^vu

du





= ( − 1)

ⁿ⁺¹²

(x/a)

ⁿ⁻¹²

λa 2πi

× Z

_∞

0

u[I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a) − I

ⁿ⁻¹

2

(xu/a)K

ⁿ⁻¹

2

u)]

K

²n−1 2

(u) + π

²

I

²n−1 2

(u) e

⁻^λu/a

e

⁻^vu

du.

This ends the proof of the ﬁrst part of the theorem.

All what remains is to show the corresponding properties of the functions

w

i,λ

, i = 1, 2. We begin with w

1,λ

, which is easier to analyze. First of all,

observe that ℜ (z

i

) < 0, so for ﬁxed λ > 0 the function w

1,λ

is bounded

and lim

v→∞

v

^k

w

1,λ

(v ) = 0, for all k = 1, 2, . . .. To see what happens when

λ → 0 we use the formula for the residue of F

λ

(see (25)), together with the

(17)

Lagrange formula. Since K

ⁿ⁻¹

2

(z

i

) = 0 we get Res

z_i

F

λ

= 1

λ

(z

i

/a)e

^λzⁱ^/a

(xz

i

/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz

i

/a)

− z

_iⁿ⁻¹²

K

ⁿ⁻³

2

(z

i

)

= − e

^λzⁱ^/a

a

²

z

_i

z

_iⁿ⁻³²

K

ⁿ⁻³

2

(z

i

)

(xz

i

/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xz

i

/a) − z

_iⁿ⁻¹²

K

ⁿ⁻¹

2

(z

i

) (xz

i

/a) − z

i

= e

^λzⁱ^/a

a

²

z

i

z

_iⁿ⁻³²

K

ⁿ⁻³

2

(z

i

)

(ξz

i

)

ⁿ⁻¹²

K

ⁿ⁻³

2

(ξz

i

)

→ z

i

z

n−1 2

i

K

ⁿ⁻³

2

(z

i

) a

²

z

_iⁿ⁻³²

K

ⁿ⁻³

2

(z

i

)

= z

i

a

2

,

because 1 < ξ < x/a and ξ → 1 as λ → 0.

Furthermore, for 0 < λ 6 a we have

| Res

zi

F

λ

| 6 | z

i

|

a

2

ⁿ⁻¹²

sup

16ξ62

K

ⁿ⁻³

2

(ξz

i

) K

ⁿ⁻³

2

(z

i

) .

Since ℜ (z

i

) < 0, we have obtained that | w

^∗₁

(v) | is bounded by a constant C

1

(n, a) and that w

^∗₁

integrates all powers of v .

We now prove the corresponding statements for w

2,λ

. Observe that the numerator in (22) is equal to

K

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a) I

ⁿ⁻¹

2

(xu/a) K

ⁿ⁻¹

2

(xu/a) − I

ⁿ⁻¹

2

(u) K

ⁿ⁻¹

2

(u)

!

and hence is positive, because the function I

ν

(u)/K

ν

(u), u > 0, is obviously increasing.

Using the Lagrange formula once again and taking into account (z

^ν

I

ν

(z))

^′

= z

^ν

I

ν−1

(z)

(see [E1] 7.11(19) p. 79) we obtain for 0 < λ 6 a

(18)

1 λ

[I

ⁿ⁻¹

2

(xu/a) K

ⁿ⁻¹

2

(u) − I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a)]

K

²n−1 2

(u) + π

²

I

²n−1 2

(u)

= λu a

1 λ

[(xu/a)

ⁿ⁻¹²

I

ⁿ⁻¹

2

(xu/a) − u

ⁿ⁻¹²

I

ⁿ⁻¹

2

(u)]

(xu/a) − u

× K

ⁿ⁻¹

2

(u) (xu/a)

ⁿ⁻¹²

[K

²n−1

2

(u) + π

²

I

²n−1 2

(u)]

− λu a

1 λ

[(xu/a)

ⁿ⁻¹²

K

ⁿ⁻¹

2

(xu/a) − u

ⁿ⁻¹²

K

ⁿ⁻¹

2

(u)]

(xu/a) − u

× I

ⁿ⁻¹

2

(u) (xu/a)

ⁿ⁻¹²

[K

²n−1

2

(u) + π

²

I

²n−1 2

(u)]

= u

a

(ξ

1

u)

ⁿ⁻¹²

I

ⁿ⁻³

2

(ξ

1

u) K

ⁿ⁻¹

2

(u) + (ξ

2

u)

ⁿ⁻¹²

I

ⁿ⁻¹

2

(u)K

ⁿ⁻³

2

(ξ

2

u) (xu/a)

ⁿ⁻¹²

[K

²n−1

2

(u) + π

²

I

²n−1 2

(u)]

→ u a

[I

ⁿ⁻³

2

(u) K

ⁿ⁻¹

2

(u) + I

ⁿ⁻¹

2

(u)K

ⁿ⁻³

2

(u)]

K

²n−1 2

(u) + π

²

I

²n−1 2

(u)

= 1

a

1 K

²n−1

2

(u) + π

²

I

²n−1 2

(u) ,

where for the last equality we used [E1] 7.11(39) p. 80 . Here the convergence takes place when λ → 0 and 0 < ξ

1

, ξ

2

6 x/a 6 2 and ξ

1

, ξ

2

→ 1 as λ → 0. Thus, we have obtained

λ

lim

→0

w

_2,λ

(v ) = ( − 1)

ⁿ⁺¹²

a

²

Z

_∞

0

ue

⁻^vu

du K

²n−1

2

(u) + π

²

I

²n−1 2

(u) ,

since the passage to the limit under the integral sign is justiﬁed by (28) below.

Moreover, using the above equations and the asymptotic behavior (23) of I

ⁿ⁻¹

2

and K

ⁿ⁻¹

2

, we obtain for u > 1 and 0 < λ ≤ a 1

λ [I

ⁿ⁻¹

2

(xu/a)K

ⁿ⁻¹

2

(u) − I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a)]

K

²n−1 2

(u) + π

²

I

²n−1 2

(u) e

⁻^xu/a

e

^u

(19)

6 u a

2

ⁿ⁻¹²

[e

⁻^xu/a

I

ⁿ⁻³

2

(xu/a)e

^u

K

ⁿ⁻¹

2

(u) + e

⁻^u

I

ⁿ⁻¹

2

(u)e

^u

K

ⁿ⁻³

2

(u)]

(x/a)

ⁿ⁻¹²

(K

²n−1 2

(u) + π

²

I

²n−1 2

(u))

6 cu

²

[1 + E

1

(xu/a)][1 + E

2

(u)] + [1 + E

1

(u)][1 + E

2

(u)]

a cosh(2u)

6 Cu

²

cosh(2u) . (27)

For u 6 1 we have 1

λ [I

ⁿ⁻¹

2

(xu/a)K

ⁿ⁻¹

2

(u) − I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a)]

K

ⁿ⁻₂¹²

(u) + π

²

I

²n−1 2

(u) e

⁻^xu/a

e

^u

6 2

ⁿ⁻¹²

u a

I

ⁿ⁻³

2

(2u)K

ⁿ⁻¹

2

(u) + I

ⁿ⁻¹

2

(u)K

ⁿ⁻³

2

(u) K

²n−1

2

(u) + π

²

I

²n−1 2

(u) e

^u

. Now, if

ⁿ⁻₂¹

− 1 > 0 (i.e. n > 3), using the asymptotics (24) we obtain that the above expression is bounded from above by

Cu

ⁿ⁻¹

1 + u

²

1 + π

²

u

²ⁿ⁻²

≤ Cu ˜

ⁿ⁻¹

, u ∈ (0, 1).

For

ⁿ⁻₂¹

= 1 (i.e. n = 3) one obtains in fact the same bound Cu

²

1 + u

²

log(2/u)

1 + u

⁴

6 Cu

²

, u ∈ (0, 1).

Thus, we ﬁnally get for

ⁿ⁻₂¹

> 1 w

^∗₂

(v) = sup

0<λ6a

| w

2,λ

(v ) | 6 C

1

Z

1 0

u

ⁿ

e

⁻^vu

du + C

2

Z

_∞

1

u

³

e

⁻^vu

du cosh(2u) . Now, one easily obtains

w

^∗₂

(v) 6 C(n, a) and w

₂^∗

(v) 6 C

1

/v

ⁿ⁺¹

+ C

2

e

⁻^v

, (28) and the conclusions concerning the function w

^∗₂

(v ) follow.

To ﬁnish the proof we show the existence and compute the limit

v

lim

→∞

v

ⁿ⁺¹

w

2,λ

(v).

As before, we take into account the expression under the integral sign in

(22) multiplied by v

ⁿ⁺¹

and, after changing variables t = vu we obtain

(20)

v

ⁿ⁺¹

u[I

ⁿ⁻¹

2

(xu/a) K

ⁿ⁻¹

2

(u) − I

ⁿ⁻¹

2

(u)K

ⁿ⁻¹

2

(xu/a)]

K

²n−1 2

(u) + π

²

I

²n−1 2

(u) e

⁻^λu/a

e

⁻^vu

du

= v

ⁿ⁻¹

t[I

ⁿ⁻¹

2

(xt/av) K

ⁿ⁻¹

2

(t/v) − I

ⁿ⁻¹

2

(t/v)K

ⁿ⁻¹

2

(xt/av)]

K

²n−1 2

(t/v) + π

²

I

²n−1 2

(t/v) e

⁻^λt/av

e

⁻^t

dt.

(29) Using the same formulas (24) as before, we obtain that for any ﬁxed t > 0 the expression above has the following asymptotics when v → ∞

v

ⁿ⁻¹

tc

_n

c

^′_n

[(xt/av)

ⁿ⁻¹²

(t/v)

⁻ⁿ⁻¹²

− (xt/av)

⁻ⁿ⁻¹²

(t/v)

ⁿ⁻¹²

]

(c

^′_n

)

²

(t/v)

¹⁻ⁿ

+ π

²

c

²_n

(t/v)

ⁿ⁻¹

e

⁻^λt/av

e

⁻^t

= (x/a)

ⁿ⁻¹

− 1 (x/a)

ⁿ⁻¹²

c

n

c

^′_n

t

ⁿ

(c

^′_n

)

²

+ π

²

c

²_n

(t/v)

²ⁿ⁻²

e

⁻^λt/av

e

⁻^t

→ (x/a)

ⁿ⁻¹

− 1 (x/a)

ⁿ⁻¹²

c

n

c

^′_n

t

ⁿ

e

⁻^t

.

Observe that, for the sake of simplicity, we wrote here c

_n

and c

^′_n

instead of c

ⁿ⁻¹

2

and c

^′n−1 2

. Note that for any ﬁxed t > 0 and v such that t < v we get that (29) is bounded by c(x, a, n)t

ⁿ

e

⁻^t

.

Now, we write

v

ⁿ⁺¹

w

2,λ

(v) = ( − 1)

ⁿ⁺¹²

(x/a)

ⁿ⁻¹²

λa

Z

v 0

. . . + Z

_∞

v

. . .

.

We use (27) with t/v = u > 1 and we observe that the expression in the second integral is bounded from above by

t

²

(x/a)

ⁿ⁻¹²

v

ⁿ⁻²

exp(xt/av − t/v)

cosh(2t/v) e

⁻^λt/av

e

⁻^t

6 v

ⁿ⁻²

(x/a)

⁻ⁿ⁻¹²

t

²

e

⁻^t

. Since v → ∞ , the second integral tends to 0, while the ﬁrst one converges to the following limit

v

lim

→∞

v

ⁿ⁺¹

w

2,λ

(v) = ( − 1)

ⁿ⁺¹²

(x/a)

ⁿ⁻¹²

λa

(x/a)

ⁿ⁻¹

− 1 (x/a)

ⁿ⁻¹²

c

n

c

^′_n

Z

_∞

0

t

ⁿ

e

⁻^t

dt

= ( − 1)

ⁿ⁺¹²

n! (x/a)

ⁿ⁻¹

− 1 λa

c

_n

c

^′_n

.

This ends the proof of the theorem.