Journal
of Applied Mathematics
(1997), 191-196.
and Stochastic Analysis, 10:2
NEW GENERALIZATIONS OF THE POISSON
KERNEL
University
HIROSHI HARUKI
of Waterloo, Department of Pure
Mathematics
Waterloo, Ontario, Canada N2L 3G1
THEMISTOCLES M. RASSIAS
of La Verne, Department of Mathematics
P.O. Box 51105, Kifissia, Athens 14510, Greece
University
(Received December, 1995; Revised April, 1996)
The purpose of this paper is to give new generalizations of the Poisson
Kernel in two dimensions and discuss integral formulas for them. This
paper concludes with an open problem.
Key words: Poisson Kernel, Integral Formula, Functional Equation,
Residue Theorem.
AMS subject classifications: 31A05, 31A10, 39B10.
1. Introduction
The Poisson Kernel in two dimensions is defined by
1 r2
1 2r cos0 + r 2
Then, as is well-known, the integral formula
1
P(O, r)de__f
(1 reiO)(1
r2
re
ie)"
(1)
2r
1
2
/ P(O,r)dO-
(2)
1
0
holds. Here r is a real parameter satisfying rl < 1.
In [3] which is a motive of our present paper, a proof of
functional equation
F(r )
(2)
is given by using the
F(),
where
2
F(r)
defl=--/
P(O;r)dO
0
in
Irl < 1.
In this paper
we shall treat generalizations of
Printed in the U.S.A.
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(1)
and
(2):
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191
HIROSHI HAIUKI and THEMISTOCLES M. RASSIAS
192
First, if we set
Q(O;a,b)de_f
1 -ab
be -io)’
where a,b are complex parameters satisfying a < 1 and bl < 1.
By taking a-r and b-r in (3) we find that (3) is a generalization of
Section 2 we shall prove the integral formula for Q(O; a, b)
(1 aeiO)(1
(3)
(1). In
27t"
Q(0; a, b)de
2--
(4)
1,
0
where a,b are complex parameters satisfying a < 1 and bl < 1.
By taking a r and b r in (4) we find that (4)is a generalization of (2). The
method of proof of (4) in this paper is similar to the pooe given for (2) in [3], i.e.,
the method is by applying a functional equation.
Second, if we set
R(0; a, b, c, d)
L(a, b, c, d)
(1 aeie)(1
be -i0)(1
where a,b,c,d are complex parameters satisfying
Idl < 1 and
cei)(1
a
< 1,
(5)
de
bl < 1,
< 1 and
c
ab)(1 ad)(1 bc)(1 cd)
L(a,b,c, d) d=ef(1
1 -abcd
By taking c 0 and d 0 in (5) we find that (5)is a generalization of (3).
In Section 3 we shall prove the integral formula for R(O; a, b, c, d)
(6)
2r
1
27r
/
R(0; a b, c, d)dO
(7)
1,
0
where a,b,c,d are complex parameters satisfying a < 1, b] < 1, c < 1 and
dl < 1. The method of proof of (7) is the calculus of residues (cf. [1, pp. 147-151]).
Remark 1: The purpose of this paper is to prove (4) and (7).
2. Proof of the Integral Formula
(4)
Theorem 1:
2rr
1/
2:r
Q(O; a ,b)dO
1,
0
where a, b are complex parameters satisfying
Proof: If we set
J
0
1 and
bl
<1.
2r
2-
G(a,b)de=f2
a] <
Q(O;a,b)dO
2J
0
1-ab
(1-aei)(l-be -O)
then G(a,b) is a continuous function of a,b when
clear that
G(0, 0) 1.
a
<
1 and
.dO (by
(3)),
bl < 1.
Also, it is
(9)
New Generalizations of the Poisson Kernel
By (8) let
193
us write
27r
"
2 / (1-aei)(1-beiO) dO+2J
G(a,b)
1-ab
r
0
1-ab
(1- aei)(1
be
i)
dO
In
for all complex a,b satisfying
< 1 and bl < 1.
Maki.ng the substitution 0- + r in the second integral and using the formulas
e in’- e- *"
1, one obtains
0
ab
1
2---
(1 aeiO)(1
0
io) + (1 + aeiO)(1 + be io)
be
for all complex values of a,b satisfying
From the identity
< 1 and b] <
a
(1 + aeiO)(1 + be io) + (1 aei)(1
we
(1
0
dO
1.
io) 2(1 + ab),
be
get
2
G(a,b)
0
2(1 a2b 2)
dO.
(1 -a2e2i)(1 -b2e 2io)
1/2 in the above integral yields
Making the substitution 0
2-
2-
G(a,b)
2
1
0
a2b 2
a2ei)(1-b2e -i) de
(1
for all complex numbers a,b satisfying
In view of (8), (10) we obtain
a
2
0
< 1 and b <
for all complex numbers a,b satisfying a < 1 and
By repeated applications of (11) we have
for all complex values of a, b satisfying
rt
b2n)(n
a
for all a,b satisfying
a
(11)
b] <
1.
1, 2, 3,...)
< 1 and bl <
Letting n+ + oo in the above inequality, using
follow from the hypothesis that
G(a, b) at (0, 1) yields
d0(10)
1.
G(a 2, b 2)
G(a, b)
G(a, b) + G(a 2
(1- a 2 eiO)(1-b2e iO)
a] < 1 and b] <
a(a,b) G(0, 0)
< 1 and ]b < 1.
By (9), (12)we obtain
G(a,b)-i
1.
lira a 2n
lira b 2n
0 which
1 and applying the continuity of
(12)
HIROSHI HARUKI and THEMISTOCLES M. RASSIAS
194
for all complex numbers a, b satisfying
Ibl < 1. Hence,
and
(4).
by
(8)
we get
Q.E.D.
Remark: Another proof of Theorem 1 is given as Corollary 1 to Theorem 2.
(7)
3. Proof of the Integral Formula
Theorem 2:
2rl /
R(O; a, b, c, d)dO
1,
0
where a,b,c,d are complex parameters satisfying
bl < 1,
[a < 1,
and
Proof: We have
27r
if (1
if"
2rr
dO
be -i)(1
aeiO)(1
0
cei)(1
de -io)
(13)
27I"
1
2i
0
e iO
(1 aeiO)(e
iO
b)(1 ceiO)(e iO d)
ieiOdo.
e iO then we obtain
If we set z
Furthermore,
ieidO- dz.
(14)
z
f(z) de__f
--(1-az)(z-b)(1-cz)(z-d)"
(15)
we set
Hence, by (13), (14) and (15)
we obtain
27r
2r
o
(1 aeiO)(1
be
i)(1 cei)(1
de
io)
27ri
I1
-1
where the right-hand side means the complex integral of the function f(z) along the
unit circle zl = 1 on the z-plane in the positive direction.
By (15) we note that f(z)is an analytic function in z < 1 except at z b and
z d each of which is a simple pole of f.
We consider two cases.
Case 1: Let b
-
d.
Suppose that R 1 and R 2 denote the residues of f(z) at z-b and z-d,
respectively. By the Residue Theorem (cf. [1, pp. 147-151]) we get
Next, by
a standard method
1 f
2ri
/
f(z)dz-R I+R 2.
(cf. [2, p. 242]),
(17)
we shall calculate
R 1 and R 2. By (15)
we have
/1
--lim ((z- b)f(z))- lim,
z--+b
z-+b[l
z
az)(1 cz)(z d)
b
(1 ab)(1 bc)(b d) (18)
New Generalizations of the Poisson Kernel
195
and
By (17), (18)and (19)we
[
1
2ri j
I1
d
z
az)(z b)(1 cz)-(1 ad)(d b)(1 cd) "(19)
R2--1im((z-d)f(z))-lzmd(lz--.d
can write
f(z)dz
d
b
(1 ab)(1 bc)(b d) + (1 ad)(d b)(1 cd)
=1
abcd
1
(20)
(1 ab)(1 ad)(1 bc)(1 cd)
1
L(a, b, c, d)
(by (6)).
By (16), (20)we obtain
2r
2r
(1 aeiO)(1
0
cei)(1
be -io)(1
L(a,b,c,d)
de
and therefore
2r
2r
L(a,b,c,d)
(1 aei)(1
0
)(1 cei)(1
be
for all complex values of a,b,c,d satisfying
a
de
io)
< 1, bl < 1,
dO- 1
c
(21)
< 1 and dl <
1.
By (5), (21)we get (7).
Case 2: Letb-d.
In this case, by (15)
we have
z
f(z)
(1 az)(1 -cz)(z-b) 2"
f(z) is an analytic function in z _< 1 except
By (22) we see that
double pole of the function.
In this case, let R denote the residue of f(z) at z
By the Residue Theorem we get
1/
Izl
27ri
-
()
at z
b.
f(z)dz- R.
I(z)dz
(23)
=1
In the following, we shall calculate R.
By Cauchy’s Integral Formula for the derivative (cf. [2, pp. 178-179])
1]
b which is a
1J
we obtain
z
b)2dz
(1 az)(1 cz) / (z
))
(1-az)(1-cz) z=b
1 ab2c
(1 -ab)2(1 -bc) 2
(24)
HIROSHI HARUKI and THEMISTOCLES M. RASSIAS
196
By (16), (24)
we obtain
1
L(a, b, c, b) (by (6)).
(note that b- d)
27v
1
2
J
dO
(1 aei)(1
0
i)(1 cei)(1
be
L(a,b,c,b)
be
and thus
27v
2- 1 ]
L (a, b, c, b
io)(1 cei)(1
for all complex values ofa, b, csatisfying al <1, b
(1 aeie)(1
0
be
be
i)
dO
1
<land
cl
<1.
for all complex numbersa, b, csatisfying al <1, bl <1 and
From Case 1 and Case 2 we get the desired result (7).
el
<1.
By (5), (25)
we
get
R(O; a, b, c, b)
1
Q.E.D.
Corollary 1: (to Theorem 2) If we set c 0 and d 0 in Theorem 2, we obtain
Theorem 1. Therefore, Theorem 2 gives another proof of Theorem 1.
Corollary2:
(to Theorem 2) If we
l_j_
2
where
/
1 + ab
1 -ab’
Q(0; a, b)2dO
0
Q(0; a, b)
1- ab
(1 aeie)(1
where a,b are complex parameters satisfying
b in Theorem2 we obtain
a and d
set c
be
ie)
]a <
(se e (3))
1 and
b[ <
l-ab
-be -0
)n+l
1.
4. Open Problem
Let
2r
2r
I n defl
/
Q(O;a, b) + l dO
o
n
2-%/ ( (1 -aeiO)(1
o
where a,b are complex parameters satisfying
By Theorems 1 and 2 we obtain
I o- 1 and
Open Problem: Compute I n for
n
a
11
dO
< 1 and bl <
(n
O, 1,...),
(26)
1.
1 +ab
1-ab"
2, 3, 4,
References
[1]
[2]
[a]
Ahlfors, LV., Complex Analysis, McGraw-Hill (2nd Edition) 1966.
Hille, E., Analytic Function Theory, Volume I, Ginn and Company 1959.
Taylor, A.E., A note on the Poisson kernel, Amer. Math. Monthly, y 57 (1950),
478-479.