# Sharp Estimates for Green’s Functions of Cone-Type Planar Domains

Authors: Mohamed Amine Ben Boubaker; Mohamed Selmi
Journal: Journal of Complex Analysis
 (2016)

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## Abstract

We establish sharp estimates for Green’s functions of cone-type planar domains. Our work generalizes all estimates given by Zhao in 1988 and Selmi in 2000. Our principal idea is to use conformal mappings.


---

## Body

## 1. Introduction

We work in the Euclidean spaceR
2. By G
Ω, we denote the Green function for the Laplacian in a domain Ω. We write δ
(
x
) for the distance from x
∈
Ω to the Euclidean boundary ∂
Ω of Ω. For two positive functions f and g on a set Ω, we say that f is comparable to g on Ω and we denote f
≃
g, if there exists A
≥
1, such that ∀
x
∈
Ω, 1
/
A
g
(
x
)
≤
f
(
x
)
≤
A
g
(
x
). The constant A will be called the constant of comparison. For a
,
b
∈
R, we denote (
a
∨
b
)
=
max
⁡
(
a
,
b
) and (
a
∧
b
)
=
min
⁡
(
a
,
b
). D
(
a
,
s
) denote the open disk of center a and radius s. For z
∈
C, we denote r
=
z and θ
=
arg
⁡
z. In [1], Zhao established estimates for the Green function on bounded C
2 domains Ω in R
2. More precisely, he proved that(1)
G
Ω
x
,
y
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
,
on
Ω
×
Ω
.This result was generalized later by Selmi in [2]. He established inequalities for the Green function of a class of Dini-smooth Jordan domains in R
2. He proved that if Ω is a bounded multiply connected Dini-smooth Jordan domain in R
2, then inequality (1) holds. In [3], under restricted conditions on Ω, Sweers established an implicit formula for the estimate of the Green function; the estimate is expressed as a function of the first eigenfunction. The aim of the present paper is to give estimates for the Green function in some particular bounded and unbounded nonsmooth domains in R
2. Our work generalizes all estimates given by Zhao in [1], Selmi in [2], and Sweers in [3]. Our principal idea is to use conformal mappings. Note that, by the Riemann Theorem (see [4], pages 133–136 and pages 229–233), if Ω is a nonempty open simply connected domain in R
2, distinct of R
2, then there exists a conformal mapping from Ω onto the unit disk D. Consequently, for all x
,
y
∈
Ω
(2)
G
Ω
x
,
y
=
G
D
ϕ
x
,
ϕ
y
≃
l
n
1
+
1
-
ϕ
x
2
1
-
ϕ
y
2
ϕ
x
-
ϕ
y
2
on
Ω
×
Ω
.Thus, to give estimates for the Green function, it is sufficient to give estimates for ϕ
(
x
)
-
ϕ
(
y
) and 1
-
ϕ
(
x
)
2 on Ω. The problem reduces to understanding distortion introduced by ϕ. The paper is organized as follows. In Section 2, we establish some preliminary results that will be necessary throughout this paper. In Section 3, we give estimates for the Green function in S
α
=
{
z
∈
C
:
θ
<
π
/
2
α
}, α
∈
]
1
/
2
,
+
∞
[
∖
{
1
}. We put Δ
1
=
{
z
∈
C
:
r
≥
0
,
θ
=
π
/
2
α
} and Δ
2
=
{
z
∈
C
:
r
≥
0
,
θ
=
-
π
/
2
α
}
. For z
∈
S
α, we denote d
1
(
z
)
=
d
(
z
,
Δ
1
) (resp., d
2
(
z
)
=
d
(
z
,
Δ
2
)), the distance from z to Δ
1 (resp., from z to Δ
2). If x
,
y
∈
S
α, we define the bracket [
x
,
y
]
α
,
0
=
x
-
y
2
+
4
d
1
(
x
)
d
1
(
y
)
+
4
d
2
(
x
)
d
2
(
y
). We prove the following results.Theorem 1.
Letα
>
1
/
2, α
≠
1. Then(3)
G
S
α
x
,
y
≃
ln
⁡
1
+
x
∧
y
x
∨
y
α
-
2
d
1
x
d
1
y
d
2
x
d
2
y
x
-
y
2
x
,
y
α
,
0
≃
ln
⁡
1
+
x
∧
y
x
∨
y
α
-
2
d
1
x
d
1
y
d
2
x
d
2
y
x
-
y
2
x
∨
y
2
,
o
n
S
α
×
S
α
.Corollary 2.
Letα
>
1
/
2, α
≠
1. Then(4)
G
S
α
x
,
y
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
x
∧
y
x
∨
y
α
-
1
,
o
n
S
α
×
S
α
.In Section4, we give estimates for the Green function in S
α
D
=
{
z
∈
C
:
0
<
r
<
1
,
θ
<
π
/
2
α
}, α
∈
]
1
/
2
,
+
∞
[
∖
1
. We put Δ
0,2
α
=
{
z
∈
C
:
0
<
r
<
1
,
θ
=
π
/
2
α
}, Δ
0
,
-
2
α
=
{
z
∈
C
:
0
<
r
<
1
,
θ
=
-
π
/
2
α
}, and C
1,2
α
=
{
z
∈
C
:
r
=
1
,
θ
≤
π
/
2
α
}. For z
∈
S
α
(
D
), we denote δ
1
(
z
)
=
d
(
z
,
Δ
0,2
α
) (resp., δ
2
(
z
)
=
d
(
z
,
Δ
0
,
-
2
α
); resp., δ
3
(
z
)
=
d
(
z
,
C
1,2
α
)), the distance from z to Δ
0,2
α (resp., from z to Δ
0
-
2
α; resp., from z to C
1,2
α). For x
,
y
∈
S
α
(
D
), we define the brackets [
x
,
y
]
α
,
1
=
x
-
y
2
+
4
δ
1
(
x
)
δ
1
(
y
)
+
4
δ
2
(
x
)
δ
2
(
y
), [
x
,
y
]
α
,
2
=
x
-
y
2
+
4
δ
1
(
x
)
δ
1
(
y
)
+
(
1
-
x
2
)
(
1
-
y
2
), [
x
,
y
]
α
,
3
=
x
-
y
2
+
4
δ
2
(
x
)
δ
2
(
y
)
+
(
1
-
x
2
)
(
1
-
y
2
), and [
x
,
y
]
α
=
(
x
∨
y
)
2
(
x
-
e
i
π
/
2
α
∨
y
-
e
i
π
/
2
α
)
2
(
x
-
e
-
i
π
/
2
α
∨
y
-
e
-
i
π
/
2
α
)
2. We prove the following.Theorem 3.
Letα
>
1
/
2, α
≠
1. Then(5)
G
S
α
D
x
,
y
≃
ln
⁡
1
+
δ
1
x
δ
1
y
δ
2
x
δ
2
y
δ
3
x
δ
3
y
x
-
y
2
x
,
y
α
,
1
x
,
y
α
,
2
x
,
y
α
,
3
x
∧
y
x
∨
y
α
-
2
,
o
n
S
α
D
×
S
α
D
,
(6)
G
S
α
D
x
,
y
≃
ln
⁡
1
+
δ
1
x
δ
1
y
δ
2
x
δ
2
y
δ
3
x
δ
3
y
x
-
y
2
x
,
y
α
x
∧
y
x
∨
y
α
-
2
,
o
n
S
α
D
×
S
α
D
.Corollary 4.
Letα
>
1
/
2, α
≠
1. Then, for all x
,
y
∈
S
α
(
D
),(7)
G
S
α
D
x
,
y
≃
ln
⁡
1
+
x
-
e
i
π
/
2
α
∧
y
-
e
i
π
/
2
α
x
-
e
i
π
/
2
α
∨
y
-
e
i
π
/
2
α
x
-
e
-
i
π
/
2
α
∧
y
-
e
-
i
π
/
2
α
x
-
e
-
i
π
/
2
α
∨
y
-
e
-
i
π
/
2
α
x
∧
y
x
∨
y
α
-
1
δ
x
δ
y
x
-
y
2
.In Section5, we give estimates for the Green function in D
∩
D
(
a
,
s
). To study the intersection of two open disk, let us remark that using translations, rotations, and homotheties we reduce the problem to study the situation of D and D
(
a
,
s
), where s
∈
R
∗
+, 0
<
s
≤
1, and 1
-
s
<
a
<
1
+
s. The two boundary circles C
(
0,1
) and C
(
a
,
s
) intersect nontangentially at e
±
i
θ
0
=
a
+
e
±
i
θ
1 with cos
⁡
θ
0
=
1
+
a
2
-
s
2
/
2
a and sin
⁡
θ
0
=
[
s
2
-
(
a
-
1
)
2
]
[
(
1
+
a
)
2
-
s
2
]
/
2
a. The interior angle φ of the intersection is given by cos
⁡
φ
=
1
+
s
2
-
a
2
/
2
s and the boundary of D
∩
D
(
a
,
s
) is formed by the arcs of circles(8)
C
1
=
z
∈
C
:
z
=
1
,
arg
⁡
z
≤
θ
0
,
C
2
=
z
∈
C
:
z
-
a
=
s
,
arg
⁡
z
≥
θ
1
.We denote by δ
i
⋆
(
z
) the distance of z
∈
D
∩
D
(
a
,
r
0
) to the boundary C
i. For x
,
y
∈
D
∩
D
(
a
,
s
), we write (9)
Δ
x
,
y
=
δ
1
∗
x
δ
1
∗
y
δ
2
∗
x
δ
2
∗
y
x
-
y
2
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
2
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
2
.We prove the following.Theorem 5.
For allx
,
y
∈
D
∩
D
(
a
,
s
),(10)
G
D
∩
D
a
,
s
x
,
y
≃
ln
⁡
1
+
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
α
-
2
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
α
-
2
Δ
x
,
y
,where α
=
π
/
π
+
θ
0
-
θ
1
>
1.Corollary 6.
For allx
,
y
∈
D
∩
D
(
a
,
s
), (11)
G
D
∩
D
a
,
s
x
,
y
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
α
-
1
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
α
-
1
.In Section6, we give estimates for the Green function in D
∪
D
(
a
,
s
). We reduce the problem as in Section 5. We prove the following.Theorem 7.
For allx
,
y
∈
D
∪
D
(
a
,
s
), (12)
G
D
∪
D
a
,
s
x
,
y
≃
ln
⁡
1
+
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
β
-
1
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
β
-
1
δ
x
δ
y
x
-
y
2
,where β
=
π
/
π
-
θ
0
+
θ
1.In Section7, we give estimates for the Green function in D
∖
D
(
a
,
s
). We reduce the problem as in Section 5. We prove the following.Theorem 8.
For allx
,
y
∈
D
∖
D
(
a
,
s
), (13)
G
D
∖
D
a
,
s
x
,
y
≃
ln
⁡
1
+
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
γ
-
1
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
γ
-
1
δ
x
δ
y
x
-
y
2
,where γ
=
π
/
θ
1
-
θ
0.In Section8, we give estimates for the Green function in D
1
=
D
∩
{
z
∈
C
:
R
e
(
z
)
>
cos
⁡
θ
0
}, 0
<
θ
0
<
π
/
2. We prove the following.Theorem 9.
For allx
,
y
∈
D
1,(14)
G
D
1
x
,
y
≃
ln
⁡
1
+
δ
1
′
′
′
x
δ
1
′
′
′
y
δ
2
′
′
′
x
δ
2
′
′
′
y
x
-
y
2
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
2
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
2
×
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
ν
-
2
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
ν
-
2
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
x
-
e
i
θ
0
∧
y
-
e
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
ν
-
1
x
-
e
-
i
θ
0
∧
y
-
e
-
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
ν
-
1
,where ν
=
π
/
θ
0
≥
2, δ
1
′
′
′
(
x
)
=
1
-
x, and δ
2
′
′
′
(
x
)
=
R
e
(
x
)
-
cos
⁡
θ
0.


## 2. Preliminaries Results

In this section, the following notations will be adopted. Forα
>
1
/
2 and 0
<
ε
<
1, we denote (15)
D
α
=
z
∈
C
:
0
<
r
<
1
,
0
<
θ
<
π
α
,
Ω
α
,
ε
=
z
∈
C
:
ε
<
r
<
1
,
θ
<
π
α
,
Δ
ε
,
α
=
z
∈
C
:
ε
<
r
<
1
,
θ
=
π
α
,
Δ
ε
,
-
α
=
z
∈
C
:
ε
<
r
<
1
,
θ
=
π
-
α
,
C
ε
,
α
=
z
∈
C
:
r
=
ε
,
θ
≤
π
α
.Lemma 10.
Letα
>
1
/
2, α
≠
1. Then, (16)
sin
⁡
α
θ
≃
sin
⁡
θ
,
o
n
0
,
π
2
α
.Proof.
It suffices to see that(17)
φ
α
θ
=
sin
⁡
α
θ
sin
⁡
θ
if
θ
∈
0
,
π
2
α
α
if
θ
=
0is continuous and does not vanish on [
0
,
π
/
2
α
].Lemma 11.
Letα
>
1. Then (18)
1
-
z
α
≃
1
-
z
,
o
n
S
α
/
2
D
.Proof.
Letε
<
1
/
4. The function(19)
f
α
z
=
1
-
z
α
1
-
z
if
z
≠
1
α
if
z
=
1is holomorphic on Ω
α
,
ε and continuous and does not vanish on Ω
α
,
ε
¯. Then, by maximum and minimum principles, it is sufficient to control f
α on ∂
Ω
α
,
ε. We have to discuss three cases.
Case 1. If z
∈
C
1
,
α, then (20)
1
-
z
α
1
-
z
2
=
sin
⁡
α
θ
/
2
sin
⁡
θ
/
2
2
.The result follows from Lemma 10.
Case 2. If z
∈
C
ε
,
α, then (21)
3
8
≤
1
-
z
α
1
-
z
≤
8
3
.
Case 3. If z
∈
Δ
ε
,
α
∪
Δ
ε
,
-
α, then(22)
1
-
z
α
1
-
z
2
=
1
+
r
α
2
1
+
r
2
-
2
r
cos
⁡
π
/
α
≤
1
+
r
α
2
1
-
r
2
+
4
r
sin
2
⁡
π
/
2
α
.So, we get (23)
1
4
≤
1
-
z
α
1
-
z
2
≤
20
sin
2
⁡
π
/
2
α
.Since the constants are independent of ε, we obtain, for all z
∈
S
α
/
2
(
D
), (24)
1
-
z
α
≃
1
-
z
.Lemma 12.
Letα
∈
]
1
/
2
,
1
[. Then(25)
1
-
z
α
≃
1
-
z
,
o
n
D
α
.Proof.
The proof is similar to that of Lemma11. Let ε
<
1
/
4. The function (26)
g
α
z
=
z
α
+
i
z
-
e
-
i
π
/
2
α
if
z
≠
e
-
i
π
/
2
α
,
α
e
-
i
π
α
-
1
/
2
α
if
z
=
e
-
i
π
/
2
αis holomorphic on Ω
2
α
,
ε and continuous and does not vanish on Ω
2
α
,
ε
¯. By maximum and minimum principles, it is sufficient to control g
α on ∂
Ω
2
α
,
ε. We have to discuss four cases.
Case 1. If z
∈
C
1,2
α, then(27)
g
α
z
2
=
1
-
sin
⁡
α
θ
1
-
cos
⁡
θ
+
π
/
2
α
=
sin
⁡
α
ν
sin
⁡
ν
2
,where ν
=
θ
+
π
/
2
α
/
2
∈
[
0
,
π
/
2
α
]. The result follows from Lemma 10.
Case 2. If z
∈
C
ε
,
2
α, then(28)
1
4
<
1
-
ε
α
1
+
ε
≤
g
α
z
≤
1
+
ε
α
1
-
ε
<
8
3
.
Case 3. If z
∈
Δ
ε
,
2
α, then (29)
1
4
≤
g
α
z
2
=
1
+
r
α
2
r
2
+
1
-
2
r
cos
⁡
π
/
α
≤
20
sin
2
⁡
π
/
2
α
.
Case 4. If z
∈
Δ
ε
,
-
2
α, then (30)
1
2
≤
g
α
z
=
1
-
r
α
1
-
r
=
∫
r
1
α
t
α
-
1
d
t
∫
r
1
d
t
≤
1
+
α
.As the constants are independent of ε, we obtain for all z
∈
S
α
(
D
)
(31)
z
α
+
i
≃
z
-
e
-
i
π
/
2
α
.Now let z
∈
D
α; then e
-
i
π
/
2
α
z
∈
S
α
(
D
). By using the previous relation, we get, for all z
∈
D
α,(32)
z
α
-
1
≃
z
-
1
.By using the conjugate expression in the last lemma, we obtain the following.Lemma 13.
Letα
∈
]
1
/
2
,
1
[. Then (33)
1
-
z
α
≃
1
-
z
,
o
n
D
∖
D
α
/
2
α
-
1
.Lemma 14.
Letα
>
1
/
2, α
≠
1. Then, (34)
x
α
-
y
α
≃
x
-
y
x
∨
y
α
-
1
,
o
n
S
α
×
S
α
.Proof.
We have to discuss two cases.
Case 1. If α
>
1, assume that x
>
y and put z
=
y
/
x. Then, z
<
1 and a
r
g
z
≤
π
/
α. Since(35)
x
α
-
y
α
x
-
y
=
x
α
-
1
1
-
z
α
1
-
z
,then the result follows by Lemma 11.
Case 2. If α
∈
]
1
/
2
,
1
[, assume that x
>
y and put z
=
y
/
x. Then, z
<
1 and arg
⁡
z
∈
[
0
,
π
/
α
], or arg
⁡
z
∈
[
π
/
α
/
2
α
-
1
,
2
π
]. Since (36)
x
α
-
y
α
x
-
y
=
x
α
-
1
1
-
z
α
1
-
z
,then the result follows by Lemmas 12 and 13.Lemma 15.
Letα
>
1
/
2, α
≠
1. Then (37)
cos
⁡
α
θ
≃
sin
2
⁡
π
2
α
-
θ
+
0
∧
cos
⁡
π
2
α
-
θ
2
1
/
2
sin
2
⁡
π
2
α
+
θ
+
0
∧
cos
⁡
π
2
α
+
θ
2
1
/
2
,
o
n
-
π
2
α
,
π
2
α
.Proof.
The function(38)
f
α
θ
=
cos
⁡
α
θ
sin
2
⁡
π
/
2
α
-
θ
+
0
∧
cos
⁡
π
/
2
α
-
θ
2
1
/
2
sin
2
⁡
π
/
2
α
+
θ
+
0
∧
cos
⁡
π
/
2
α
+
θ
2
1
/
2is continuous and even and does not vanish on ]
-
π
/
2
α
,
π
/
2
α
[. To study f
α on a neighborhood of π
/
2
α, we put ν
=
π
/
2
α
-
θ
∈
[
0
,
ε
[
⊂
[
0
,
π
/
2
]. Hence (39)
f
α
θ
=
f
α
π
2
α
-
ν
=
sin
⁡
α
ν
sin
⁡
ν
1
sin
2
⁡
π
/
α
-
ν
+
0
∧
cos
⁡
π
/
α
-
ν
2
,
lim
ν
→
0
⁡
f
α
π
2
α
-
ν
=
α
if
cos
⁡
π
α
<
0
α
sin
⁡
π
/
α
if
cos
⁡
π
α
≥
0
.As lim
θ
→
π
/
2
⁡
f
α
(
θ
)
≠
0, the function f
α can be extended by continuity on [
-
π
/
2
α
,
π
/
2
α
] and the compacity argument finishes the proof.


## 3. Estimates for the Green Function on the SectorS
α

It is clear that(40)
d
1
z
=
r
sin
2
⁡
π
2
α
-
θ
+
0
∧
cos
⁡
π
2
α
-
θ
2
,
d
2
z
=
r
sin
2
⁡
π
2
α
+
θ
+
0
∧
cos
⁡
π
2
α
+
θ
2
.Lemma 16.
Letα
>
1
/
2, α
≠
1, Then(41)
x
,
y
α
,
0
≃
x
∨
y
2
,
o
n
S
α
.Proof.
First, we remark that(42)
x
,
y
α
,
0
≤
x
-
y
2
+
8
x
y
≤
x
2
+
y
2
+
10
x
y
≤
6
x
2
+
y
2
≤
12
x
2
∨
y
2
.On the other hand, the bracket [
x
,
y
]
α
,
0 satisfies(43)
x
,
y
α
,
0
≥
x
2
+
y
2
+
2
x
y
-
cos
2
⁡
π
2
α
cos
⁡
arg
⁡
x
cos
⁡
arg
⁡
y
-
sin
2
⁡
π
2
α
sin
⁡
arg
⁡
x
sin
⁡
arg
⁡
y
.Let(44)
K
α
:
-
π
2
α
,
π
2
α
×
-
π
2
α
,
π
2
α
⟶
R
θ
,
φ
⟼
-
cos
2
⁡
π
2
α
cos
⁡
θ
cos
⁡
φ
-
sin
2
⁡
π
2
α
sin
⁡
θ
sin
⁡
φ
.Obviously, K
α is continuous and K
α
≤
1. Assume that K
α
(
θ
,
φ
)
=
1; we see that cos
⁡
π
/
2
α
≠
0 and sin
⁡
π
/
2
α
≠
0. In consequence, cos
2
⁡
π
/
2
α
(
1
+
cos
⁡
θ
cos
⁡
φ
)
+
sin
2
⁡
π
/
2
α
(
1
+
sin
⁡
θ
sin
⁡
φ
)
=
0. This implies the contradiction (1
+
cos
⁡
θ
cos
⁡
φ
=
0 and 1
+
sin
⁡
θ
sin
⁡
φ
=
0). Also, K
α
(
θ
,
φ
)
=
-
1 implies the contradiction (1
-
cos
⁡
θ
cos
⁡
φ
=
0 and 1
-
sin
⁡
θ
sin
⁡
φ
=
0). By compacity argument, min
⁡
(
K
α
(
θ
,
φ
)
)
=
k
α with -
1
<
k
α
<
1. This leads us to(45)
x
2
+
y
2
+
2
x
y
K
α
arg
⁡
x
,
arg
⁡
y
≥
x
2
+
y
2
+
2
x
y
k
α
≥
1
-
k
α
2
x
2
∨
y
2
2
,
with
1
-
k
α
2
>
0
.Lemma 17.
Letα
>
1
/
2, α
≠
1. Then(46)
d
1
z
∨
d
2
z
≃
z
o
n
S
α
.Proof.
It is clear thatd
i
(
z
)
≤
z for all z
∈
S
α, i
=
1,2. Moreover(47)
d
1
z
+
d
2
z
≥
sin
⁡
π
2
α
-
θ
+
sin
⁡
π
2
α
+
θ
z
.The function(48)
h
:
-
π
2
α
,
π
2
α
⟶
R
σ
⟼
h
σ
=
sin
⁡
π
2
α
-
σ
+
sin
⁡
π
2
α
+
σis continuous and even and does not vanish. In fact sin
⁡
(
π
/
2
α
-
σ
)
=
sin
⁡
(
π
/
2
α
+
σ
)
=
0 allows us to conclude that π
/
α
=
n
π with n
∈
N, where the contradiction is clear enough. Then inf
⁡
h
>
0.Forα
>
1
/
2, α
≠
1, z
∈
S
α, we denote by δ
(
z
)
=
(
d
1
(
z
)
∧
d
2
(
z
)
) the distance of z to the boundary of S
α.Proof of Theorem1.
Ifα
=
1, then ∀
x
,
y
∈
S
1: (49)
G
S
1
x
,
y
=
ln
⁡
x
+
y
¯
x
-
y
=
1
2
ln
⁡
1
+
4
R
e
x
R
e
y
x
-
y
2
.As, for α
≠
1, the function z
↦
z
α is a conformal mapping from S
α onto S
1, then we have (50)
G
S
α
x
,
y
=
1
2
ln
⁡
1
+
x
α
cos
⁡
α
arg
⁡
x
y
α
cos
⁡
α
arg
⁡
y
x
α
-
y
α
,
∀
x
,
y
∈
S
α
.By Lemma 15, we remark that x
2
cos
⁡
(
α
arg
⁡
x
)
≃
d
1
(
x
)
d
2
(
x
) and y
2
cos
⁡
(
α
arg
⁡
y
)
≃
d
1
(
y
)
d
2
(
y
). Then, (51)
x
α
cos
⁡
α
arg
⁡
x
y
α
cos
⁡
α
arg
⁡
y
≃
d
1
x
d
1
y
d
2
x
d
2
y
x
α
-
2
y
α
-
2
.On the other hand, by Lemmas 11, 14, and 16, we get(52)
x
α
-
y
α
2
≃
x
-
y
2
x
∨
y
2
α
-
2
≃
x
-
y
2
x
,
y
α
,
0
x
∨
y
2
α
-
4
.So, we obtain(53)
G
S
α
x
,
y
≃
ln
⁡
1
+
d
1
x
d
1
y
d
2
x
d
2
y
x
-
y
2
x
,
y
α
,
0
x
∧
y
x
∨
y
α
-
2
≃
ln
⁡
1
+
d
1
x
d
1
y
d
2
x
d
2
y
x
-
y
2
x
2
∨
y
2
x
∧
y
x
∨
y
α
-
2
.Now we present two proofs of Corollary2.First Proof of Corollary2. By using Lemma 17, we get, for all z
∈
S
α,(54)
δ
z
=
d
1
z
∧
d
2
z
≃
d
1
z
d
2
z
d
1
z
+
d
2
z
≃
d
1
z
d
2
z
z
.By substituting the previous expression in relation (3), we get(55)
G
S
α
x
,
y
≃
ln
⁡
1
+
d
1
x
d
1
y
d
2
x
d
2
y
x
-
y
2
x
∨
y
2
x
∧
y
x
∨
y
α
-
2
≃
ln
⁡
1
+
d
1
x
d
2
x
/
x
d
1
y
d
2
y
/
y
x
y
x
-
y
2
x
∨
y
2
x
∧
y
x
∨
y
α
-
2
≃
ln
⁡
1
+
δ
x
δ
y
x
y
x
-
y
2
x
∨
y
2
x
∧
y
x
∨
y
α
-
2
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
x
∧
y
x
∨
y
α
-
1
,
on
S
α
.Second Proof of Corollary2. As, for α
≠
1, the conformal mapping ϕ
α
(
z
)
=
z
α
-
1
/
z
α
+
1, from S
α onto D, 0 is the only singular point of order α
-
1 and ϕ
α
(
z
)
≃
ϕ
α
(
0
)
+
2
z
α in a neighborhood of 0. Moreover for x
,
y
∈
S
α
(56)
G
S
α
x
,
y
=
G
D
ϕ
α
x
,
ϕ
α
y
≃
ln
⁡
1
+
1
-
ϕ
α
x
2
1
-
ϕ
α
y
2
ϕ
α
x
-
ϕ
α
y
2
.We observe that(57)
ϕ
α
x
-
ϕ
α
y
2
=
4
x
α
-
y
α
2
x
α
+
1
y
α
+
1
2
≃
4
x
-
y
2
x
∨
y
2
α
-
2
x
α
+
1
2
y
α
+
1
2and for all z
∈
S
α
(58)
1
-
ϕ
α
z
2
=
4
z
α
-
1
z
cos
⁡
α
arg
⁡
z
z
α
+
1
2
≃
δ
z
ϕ
α
′
z
.Consequently,(59)
ϕ
α
x
-
ϕ
α
y
2
≃
4
x
-
y
2
ϕ
α
′
x
ϕ
α
′
y
x
∨
y
2
α
-
2
x
α
-
1
y
α
-
1
.This implies that for all x
,
y
∈
S
α
(60)
G
S
α
x
,
y
≃
ln
⁡
1
+
δ
x
δ
y
x
-
y
2
x
∧
y
x
∨
y
α
-
1
.


## 4. Estimates of the Green Function onS
α
(
D
)

It is clear that(61)
δ
1
z
=
r
sin
2
⁡
π
2
α
-
θ
+
0
∧
cos
⁡
π
2
α
-
θ
2
,
δ
2
z
=
r
sin
2
⁡
π
2
α
+
θ
+
0
∧
cos
⁡
π
2
α
+
θ
2and δ
3
(
z
)
=
1
-
r
≃
1
-
r
2. We see that [
x
,
y
]
α
,
2
>
0 on S
α
(
D
)
×
S
α
(
D
)
¯
∖
(
e
i
π
/
2
α
,
e
i
π
/
2
α
) and [
x
,
y
]
α
,
3
>
0 on S
α
(
D
)
×
S
α
(
D
)
¯
∖
(
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
)
. In the same way in Lemmas 11, 12, and 13, we prove the following lemma.Lemma 18.
Consider(62)
1
+
z
α
≃
1
-
e
i
π
/
α
z
1
-
e
-
i
π
/
α
z
o
n
S
α
/
2
D
,
i
f
α
>
1
1
-
e
i
π
/
α
z
1
-
e
-
i
π
/
α
z
o
n
D
α
,
o
r
D
∖
D
α
/
2
α
-
1
i
f
α
∈
1
2
,
1
.By using the last Lemma, we get the following.Lemma 19.
Letα
>
1
/
2, α
≠
1. Then(63)
1
+
x
α
y
α
≃
1
-
e
i
π
/
α
x
y
1
-
e
-
i
π
/
α
x
y
,
o
n
S
α
D
×
S
α
D
.Lemma 20.
Letα
>
1
/
2, α
≠
1. Then(64)
1
-
e
-
i
π
/
α
x
y
2
≃
x
,
y
α
,
2
,
1
-
e
i
π
/
α
x
y
2
≃
x
,
y
α
,
3
o
n
S
α
D
×
S
α
D
.Proof.
By symmetry, it is sufficient to prove the result for[
x
,
y
]
α
,
2. We see that, for all α
>
1
/
2 and x
,
y
∈
S
α
(
D
), (65)
1
-
e
-
i
π
/
α
x
y
2
≤
x
,
y
α
,
2
.Moreover, if α
≥
2,(66)
x
,
y
α
,
2
=
1
-
e
-
i
π
/
α
x
y
2
.If 1
<
α
<
2, then for z
∈
S
α
(
D
)
(67)
δ
1
z
≤
r
sin
⁡
π
/
2
α
-
θ
cos
⁡
π
/
2
α
.Since sin
⁡
(
π
/
2
α
-
θ
) is positive and cos
⁡
(
π
/
2
α
-
θ
)
<
0 for π
/
2
<
π
/
2
α
-
θ
<
π
/
2
+
π
/
2
α
<
π, then sin
⁡
(
π
/
2
α
-
θ
)
≥
sin
⁡
π
/
2
+
π
/
2
α
=
cos
⁡
π
/
2
α
>
0. This implies for x
,
y
∈
S
α
(
D
) that (68)
x
,
y
α
,
2
≤
1
cos
2
⁡
π
/
2
α
1
-
e
-
i
π
/
α
x
y
2
.Finally, for 1
/
2
<
α
<
1, we consider the function (69)
f
α
x
,
y
=
x
,
y
α
,
2
1
-
e
-
i
π
/
α
x
y
2
,
x
,
y
∈
S
α
D
×
S
α
D
¯
.
f
α is continuous and does not vanish in S
α
(
D
)
×
S
α
(
D
)
¯
∖
(
e
i
π
/
2
α
,
e
i
π
/
2
α
). We will prove that f
α is continuous at (
e
i
π
/
2
α
,
e
i
π
/
2
α
) and does not vanish. Let ε
>
0. Every z
∈
D
(
e
i
π
/
2
α
,
ε
)
∩
S
α
(
D
)
¯ can be written as z
=
e
i
π
/
2
α
+
t
e
i
θ with 0
≤
t
≤
ε. Besides, z
∈
D
(
e
i
π
/
2
α
,
ε
)
∩
S
α
(
D
)
¯ if and only if cos
⁡
(
π
/
2
α
-
θ
)
<
-
t
/
2
∈
]
-
ε
/
2
,
0
]. This implies that (
π
/
2
α
-
θ
)
∈
[
π
/
2
,
arccos
⁡
(
-
ε
/
2
)
[, and so sin
⁡
(
π
/
2
α
-
θ
)
≥
0 and δ
1
(
z
)
=
I
m
(
z
e
-
i
π
/
α
)
=
t
sin
⁡
(
π
/
2
α
-
t
)
=
±
t
sin
⁡
(
π
/
2
α
-
θ
). This proves that 1
-
e
-
i
π
/
α
x
y
2
=
[
x
,
y
]
α
,
2 for x
,
y
∈
D
(
e
i
π
/
2
α
,
ε
)
∩
S
α
(
D
)
¯ and f
α
(
x
,
y
)
=
1 in a neighborhood of (
e
i
π
/
2
α
)
,
e
i
π
/
2
α
). This enables us to extend f
α continuously on (
e
i
π
/
2
α
,
e
i
π
/
2
α
) by 1. The compacity argument ensures the proof.Lemma 21.
Letα
>
1
/
2, α
≠
1. Then (70)
x
,
y
α
,
2
x
,
y
α
,
3
≃
1
+
x
α
y
α
2
,
o
n
S
α
D
.Proof.
The proof follows from Lemmas19 and 20.Proposition 22.
Letα
>
1
/
2. Then (71)
1
+
x
α
y
α
≃
x
-
e
i
π
/
2
α
∨
y
-
e
i
π
/
2
α
x
-
e
-
i
π
/
2
α
∨
y
-
e
-
i
π
/
2
α
,
o
n
S
α
D
×
S
α
D
.Proof.
We equipC
×
C with the norm (
x
,
y
)
=
x
2
+
y
2, (
x
,
y
)
∈
C
×
C. We have (72)
x
∨
y
≤
x
,
y
≤
2
x
∨
y
.Let α
>
1
/
2. The function(73)
f
α
:
C
∖
-
∞
,
0
2
⟶
C
x
,
y
⟼
x
α
y
αis continuous and differentiable in two points (
e
i
π
/
2
α
,
e
i
π
/
2
α
), (
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
) and(74)
lim
x
,
y
→
e
i
π
/
2
α
,
e
i
π
/
2
α
⁡
f
α
x
,
y
-
f
α
e
i
π
/
2
α
,
e
i
π
/
2
α
x
,
y
-
e
i
π
/
2
α
,
e
i
π
/
2
α
=
lim
x
,
y
→
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
⁡
f
α
x
,
y
-
f
α
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
x
,
y
-
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
=
α
2
.It follows that(75)
lim
x
,
y
→
e
i
π
/
2
α
,
e
i
π
/
2
α
⁡
1
+
x
α
y
α
x
,
y
-
e
i
π
/
2
α
,
e
i
π
/
2
α
=
lim
x
,
y
→
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
⁡
1
+
x
α
y
α
x
,
y
-
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
=
α
2
.Hence, the function (76)
g
α
x
,
y
=
1
+
x
α
y
α
x
,
y
-
e
i
π
/
2
α
,
e
i
π
/
2
α
x
,
y
-
e
-
i
π
/
2
α
,
e
-
i
π
/
2
αdefined on S
α
(
D
)
×
S
α
(
D
)
¯ is continuous, with g
α
(
e
i
π
/
2
α
,
e
i
π
/
2
α
)
=
g
α
(
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
)
=
α
/
2
s
i
n
π
/
2
α
≠
0
. Moreover, the numerator and the denominator of g
α do not vanish on S
α
(
D
)
×
S
α
(
D
)
¯
∖
(
e
i
π
/
2
α
,
e
i
π
/
2
α
), (
e
-
i
π
/
2
α
,
e
-
i
π
/
2
α
). In fact, 1
+
x
α
y
α
=
0 implies x
α
y
α
=
-
1 and then x
=
y
=
1 and α
(
arg
⁡
x
+
arg
⁡
y
)
=
±
π. In consequence,(77)
arg
⁡
x
=
arg
⁡
y
=
π
2
α
or
arg
⁡
x
=
arg
⁡
y
=
-
π
2
α
.For the denominator the argument is clear. Finally, g
α is continuous on the compact subset S
α
(
D
)
×
S
α
(
D
)
¯ and does not vanish. In consequence, g
α
≃
1, on S
α
(
D
)
×
S
α
(
D
)
¯.Remark 23.
Letα
>
1, α
≠
1. Then(78)
δ
1
z
δ
2
z
δ
3
z
z
z
-
e
i
π
/
2
α
z
-
e
-
i
π
/
2
α
≃
δ
z
,
on
S
α
D
.Proof.
We have to discuss three cases.
Case 1. If δ
1
(
z
)
≤
δ
2
(
z
), then sin
⁡
π
/
2
α
≤
z
-
e
-
i
π
/
2
α
≤
2, which implies that(79)
δ
z
=
δ
1
z
∧
δ
3
z
≃
δ
1
z
δ
3
z
δ
1
z
+
δ
3
z
≃
δ
1
z
δ
3
z
z
-
e
i
π
/
2
α
≃
δ
1
z
δ
2
z
δ
3
z
z
-
e
i
π
/
2
α
z
≃
δ
1
z
δ
2
z
δ
3
z
z
-
e
i
π
/
2
α
z
-
e
-
i
π
/
2
α
z
.
Using the same argument, we obtain the result ifδ
1
(
z
)
≥
δ
2
(
z
).
Case 2. If δ
3
(
z
)
≤
δ
1
(
z
), then δ
1
(
z
)
≃
z
-
e
i
π
/
2
α and z
≥
1
/
2, which implies (80)
δ
z
=
δ
2
z
∧
δ
3
z
≃
δ
2
z
δ
3
z
δ
2
z
+
δ
3
z
≃
δ
2
z
δ
3
z
z
-
e
-
i
π
/
2
α
≃
δ
1
z
δ
2
z
δ
3
z
z
-
e
i
π
/
2
α
z
-
e
-
i
π
/
2
α
z
.
In the same way, we prove the result forδ
1
(
z
)
≤
δ
3
(
z
).
Case 3. If δ
3
(
z
)
≤
δ
2
(
z
), then z
≥
1
/
2, which implies (81)
δ
z
=
δ
1
z
∧
δ
3
z
≃
δ
1
z
δ
3
z
z
-
e
i
π
/
2
α
≃
δ
1
z
δ
3
z
δ
2
z
z
-
e
i
π
/
2
α
z
-
e
-
i
π
/
2
α
≃
δ
1
z
δ
2
z
δ
3
z
z
-
e
i
π
/
2
α
z
-
e
-
i
π
/
2
α
z
.Remark 24.
Ifα
=
1, then for all z
∈
S
1
(
D
)
(82)
δ
1
z
δ
2
z
δ
3
z
=
z
R
e
z
1
-
z
.Moreover, since 1
+
z
2
=
4
(
R
e
(
z
)
)
2
+
(
1
-
z
2
)
2, then(83)
z
-
i
z
+
i
≃
R
e
z
∨
1
-
z
,
on
S
1
D
.This implies that (84)
δ
1
z
δ
2
z
δ
3
z
=
z
R
e
z
1
-
z
≃
z
R
e
z
∧
1
-
z
R
e
z
∨
1
-
z
≃
z
δ
z
z
-
i
z
+
i
,
on
S
1
D
.Proof of Theorem3.
The Green functionG
D of D is given, for all x
,
y
∈
D, by (85)
G
D
x
,
y
=
ln
⁡
1
-
x
y
¯
x
-
y
=
1
2
ln
⁡
1
+
1
-
x
2
1
-
y
2
x
-
y
2
.Hence, the Green function of S
1
(
D
) is(86)
G
S
1
D
x
,
y
=
ln
⁡
1
-
x
y
¯
x
-
y
-
ln
⁡
1
+
x
y
x
+
y
¯
=
ln
⁡
1
-
x
y
¯
x
+
y
¯
x
-
y
1
+
x
y
=
1
2
ln
⁡
1
+
4
x
1
y
1
1
-
x
2
1
-
y
2
x
-
y
2
1
+
x
y
2
.Since, the function z
↦
z
α is a conformal mapping from S
α
(
D
) onto S
1
(
D
), then for all x
,
y
∈
S
α
(
D
)
(87)
G
S
α
D
x
,
y
=
G
S
1
D
x
α
,
y
α
=
1
2
ln
⁡
1
+
4
R
e
x
α
·
R
e
y
α
1
-
x
2
α
1
-
y
2
α
x
α
-
y
α
2
1
+
x
α
y
α
2
.On the other hand, if t
∈
[
0,1
], then 1
≤
1
-
t
2
α
/
1
-
t
≤
2
α. It follows that (88)
1
-
z
2
α
≃
1
-
z
=
δ
3
z
,
∀
z
∈
S
α
D
.By Lemma 15, for all z
∈
S
α
(
D
), we have R
e
(
z
α
)
=
z
α
cos
⁡
(
α
arg
⁡
z
)
≃
z
α
-
2
δ
1
(
z
)
δ
2
(
z
). By Lemma 14, x
α
-
y
α
2
≃
(
x
∨
y
)
2
α
-
2
x
-
y
2. By Lemma 21, 1
+
x
α
y
α
2
≃
[
x
,
y
]
α
,
2
[
x
,
y
]
α
,
3. By Lemma 16, [
x
,
y
]
α
,
1
≃
(
x
∨
y
)
2. Hence, we obtain(89)
G
S
α
D
x
,
y
≃
l
n
1
+
δ
1
x
δ
1
y
δ
2
x
δ
2
y
δ
3
x
δ
3
y
x
α
-
2
y
α
-
2
x
-
y
2
x
∨
y
2
α
-
2
x
,
y
α
,
2
x
,
y
α
,
3
≃
l
n
1
+
δ
1
x
δ
1
y
δ
2
x
δ
2
y
δ
3
x
δ
3
y
x
-
y
2
x
,
y
α
,
1
x
,
y
α
,
2
x
,
y
α
,
3
x
∧
y
x
∨
y
α
-
2
.Relation (6) follows by using (87), Lemmas 14 and 15, and Proposition 22.Proof of Corollary4.
The results follows by using (78) and (6) and the fact that x
y
=
(
x
∨
y
)
(
x
∧
y
).Remark 25.
By using relations (83) and (86) of Proposition 22, we get (90)
G
S
1
D
x
,
y
≃
ln
⁡
1
+
x
-
i
∧
y
-
i
x
-
i
∨
y
-
i
x
+
i
∧
y
+
i
x
+
i
∨
y
+
i
δ
x
δ
y
x
-
y
2
on
S
1
D
×
S
1
D
.


## 5. Estimates of the Green Function onD
∩
D
(
a
,
s
)

We can see that(91)
δ
1
⋆
z
=
1
-
z
if
arg
⁡
z
≤
θ
0
z
-
e
i
θ
0
∧
z
-
e
-
i
θ
0
if
arg
⁡
z
≥
θ
0
,
δ
2
⋆
z
=
s
-
z
-
a
if
arg
⁡
z
-
a
≥
θ
1
z
-
e
i
θ
1
∧
z
-
e
-
i
θ
1
if
arg
⁡
z
-
a
≤
θ
1
.Lemma 26.
Leta
1
,
a
2
,
b
1
,
b
2
,
k
1
,
k
2
>
0 such that k
1
≤
a
1
+
a
2
≤
k
2 and k
1
≤
b
1
+
b
2
≤
k
2. Then, (92)
1
a
1
b
2
∨
a
2
b
1
≃
a
1
∨
b
1
a
2
∨
b
2
,
2
a
1
b
2
∧
a
2
b
1
≃
a
1
∧
b
1
a
2
∧
b
2
.Proof.
(1) It is clear that (
a
1
b
2
∨
a
2
b
1
)
≤
(
a
1
∨
b
1
)
(
a
2
∨
b
2
). Inversely, we must prove that a
1
b
2
+
a
2
b
1
/
(
a
1
+
b
1
)
(
a
2
+
b
2
) is lower bounded.(a)
Ifa
1
≤
a
2, b
1
≤
b
2 (or a
2
≤
a
1, b
2
≤
b
1
), then a
2
≥
k
1
/
2, b
2
≥
k
1
/
2. This implies (93)
a
1
b
2
+
a
2
b
1
a
1
+
b
1
a
2
+
b
2
≥
k
1
2
k
2
.
(b)
Ifa
1
≤
a
2, b
2
≤
b
1(or a
2
≤
a
1, b
1
≤
b
2), then a
2
≥
k
1
/
2, b
1
≥
k
2
/
2. This implies k
1
/
2
≤
a
1
+
b
1
≤
2
k
2, k
1
/
2
≤
a
2
+
b
2
≤
2
k
2, k
1
2
/
4
≤
a
2
b
1 and so (94)
a
1
b
2
+
a
2
b
1
a
1
+
b
1
a
2
+
b
2
≥
k
1
2
16
k
2
2
.
For (2), we use the fact that(
a
∧
b
)
=
a
b
/
(
a
∨
b
).Proof of Theorem5.
The conformal mappingϕ
(
z
)
=
e
i
π
-
θ
0
-
θ
1
/
2
e
i
θ
0
-
z
/
z
-
e
-
i
θ
0 is a one to one application from D
∩
D
(
a
,
r
) onto the sector S
α
=
{
z
∈
C
:
a
r
g
z
<
π
+
θ
0
-
θ
1
/
2
=
π
/
2
α
}, where α
=
π
/
π
+
θ
0
-
θ
1
>
1. More precisely,(95)
ϕ
C
1
=
t
e
i
π
/
2
α
,
t
≥
0
,
ϕ
C
2
=
t
e
-
i
π
/
2
α
,
t
≥
0
.The Green function of D
′
=
D
∩
D
(
a
,
r
) is given for x
,
y
∈
D
′ by (96)
G
D
′
x
,
y
=
ln
⁡
ϕ
α
x
+
ϕ
α
y
¯
ϕ
α
x
-
ϕ
α
y
.From (4), the Green function estimates on S
α, we have (97)
G
D
′
x
,
y
≃
ln
⁡
1
+
δ
1
ϕ
x
δ
1
ϕ
y
δ
2
ϕ
x
δ
2
ϕ
y
ϕ
x
-
ϕ
y
2
ϕ
x
∨
ϕ
y
2
ϕ
x
∧
ϕ
y
ϕ
x
∨
ϕ
y
α
-
2
.But, (98)
ϕ
x
-
ϕ
y
2
=
x
-
y
2
x
-
e
-
i
θ
0
2
y
-
e
-
i
θ
0
2
.Besides, by Lemma 26,(99)
ϕ
x
∨
ϕ
y
=
x
-
e
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
i
θ
0
y
-
e
-
i
θ
0
=
1
x
-
e
-
i
θ
0
y
-
e
-
i
θ
0
x
-
e
i
θ
0
y
-
e
-
i
θ
0
∨
y
-
e
i
θ
0
x
-
e
-
i
θ
0
≃
1
x
-
e
-
i
θ
0
y
-
e
-
i
θ
0
x
-
e
i
θ
0
∨
y
-
e
i
θ
0
x
-
e
-
i
θ
0
∨
y
-
e
-
i
θ
0
.Moreover, we can verify that, for i
=
1, i
=
2
:
δ
i
(
ϕ
(
z
)
)
≃
δ
i
∗
(
z
)
/
z
-
e
-
i
θ
0
2. Thus the results follow.Proof of Corollary6.
From the property of Lemma17, we deduce for all z
∈
D
∩
D
(
a
,
r
)
(100)
δ
1
∗
z
∨
δ
2
∗
z
≃
z
-
e
i
θ
0
z
-
e
-
i
θ
0
.For z
∈
D
∩
D
(
a
,
r
), we have (101)
δ
1
ϕ
z
∨
δ
2
ϕ
z
≃
ϕ
z
,
δ
i
ϕ
z
≃
δ
i
∗
z
z
-
e
-
i
θ
0
.Now, we see that, for z
∈
D
∩
D
(
a
,
r
),(102)
δ
∗
z
=
δ
1
∗
z
∧
δ
2
∗
z
≃
δ
1
∗
z
δ
2
∗
z
δ
1
∗
z
+
δ
2
∗
z
≃
δ
1
∗
z
δ
2
∗
z
z
-
e
i
θ
0
z
-
e
-
i
θ
0
.Thus the results follow.


## 6. Estimates of the Green Function onD
∪
D
(
a
,
s
)

The conformal mapping(103)
ϕ
z
=
e
i
π
-
θ
0
-
θ
1
/
2
e
i
θ
0
-
z
z
-
e
-
i
θ
0is a one to one application from D
∪
D
(
a
,
s
) onto S
β
=
{
z
∈
C
,
a
r
g
z
≤
π
/
2
β
}, where β
=
α
/
2
α
-
1. The Green function is given for x
,
y
∈
D
′
′
=
D
∪
D
(
a
,
s
) by (104)
G
D
′
′
x
,
y
=
ln
⁡
ϕ
β
x
+
ϕ
β
y
¯
ϕ
β
x
-
ϕ
β
y
.The same analysis leads to analogous estimates. The boundary of D
∪
D
(
a
,
s
) is formed by the two arcs of circles (105)
C
1
′
=
z
∈
C
,
z
=
1
,
arg
⁡
z
≥
θ
0
,
C
2
′
=
z
∈
C
,
z
-
a
=
s
,
arg
⁡
z
≤
θ
1
.We denote by δ
i
′
(
z
) the distance of z
∈
D
∪
D
(
a
,
s
) to the boundary C
i
′, i
∈
1,2. For z
∈
D
∪
D
(
a
,
s
)
(106)
δ
1
′
z
=
1
-
z
if
arg
⁡
z
≥
θ
0
z
-
e
i
θ
0
∧
z
-
e
-
i
θ
0
if
arg
⁡
z
≤
θ
0
δ
2
′
z
=
s
-
z
-
a
if
arg
⁡
z
-
a
≤
θ
1
z
-
e
i
θ
1
∧
z
-
e
-
i
θ
1
if
arg
⁡
z
-
a
≥
θ
1
δ
z
=
δ
1
′
z
∧
δ
2
′
z
≃
δ
1
′
z
δ
2
′
z
δ
1
′
z
+
δ
2
′
z
≃
δ
1
′
z
δ
2
′
z
z
-
e
i
θ
0
z
-
e
-
i
θ
0
.Thus, we obtain Theorem 7.


## 7. Estimates of the Green Function onD
∖
D
(
a
,
s
)

The conformal mapping(107)
i
ϕ
z
=
i
e
i
π
-
θ
0
-
θ
1
/
2
e
i
θ
0
-
z
z
-
e
-
i
θ
0is a one to one application from D
0
=
D
∖
D
(
a
,
r
) onto the sector S
γ
=
{
z
∈
C
,
a
r
g
z
<
α
-
1
/
α
π
=
π
/
2
γ
}, where γ
=
α
/
α
-
1
>
1. The Green function is given for x
,
y
∈
D
0
=
D
∖
D
(
a
,
r
) by (108)
G
D
0
x
,
y
=
ln
⁡
i
ϕ
x
γ
+
i
ϕ
y
γ
¯
i
ϕ
x
γ
-
i
ϕ
y
γ
.The boundary of D
∖
D
(
a
,
r
) is formed by two arcs of circles (109)
C
1
′
′
=
z
∈
C
,
z
=
1
,
arg
⁡
z
≥
θ
0
,
C
2
′
′
=
z
∈
C
,
z
-
a
=
s
,
arg
⁡
z
-
a
≥
θ
1
.We denote by δ
i
′
′
(
z
) the distance of z
∈
D
∖
D
(
a
,
s
) to the boundary C
i
′
′, i
=
1,2. Consider(110)
δ
1
′
′
z
=
1
-
z
if
arg
⁡
z
≥
θ
0
z
-
e
i
θ
0
∧
z
-
e
-
i
θ
0
if
arg
⁡
z
≤
θ
0
δ
2
′
′
z
=
z
-
a
-
s
if
arg
⁡
z
-
a
≥
θ
1
z
-
e
i
θ
1
∧
z
-
e
-
i
θ
1
if
arg
⁡
z
-
a
≤
θ
1
.We find using the same analysis Theorem 8.


## 8. Estimates of the Green Function onD
∩
{
z
∈
C
,
R
e
(
z
)
>
cos
⁡
θ
0
}
,
0
<
θ
0
<
π
/
2

The conformal mapping(111)
ψ
z
=
e
-
i
θ
0
/
2
e
i
θ
0
-
z
z
-
e
-
i
θ
0is a one to one application from this domain onto the sector S
ν
=
{
z
∈
C
,
a
r
g
z
<
θ
0
/
2
=
π
/
2
ν
}, where ν
=
π
/
θ
0
≥
2. The Green function is given for x
,
y
∈
D
1
=
D
∩
{
z
∈
C
,
R
e
(
z
)
>
cos
⁡
θ
0
} by (112)
G
D
1
x
,
y
=
l
n
ψ
ν
x
+
ψ
ν
y
¯
ψ
ν
x
-
ψ
ν
y
.The boundary of this domain is formed by the arc of the unit circle {
z
∈
C
,
z
=
1
,
a
r
g
z
<
θ
0
} and the segment [
e
-
i
θ
0
,
e
i
θ
0
]. For z
∈
D
∩
{
z
∈
C
,
R
e
(
z
)
>
cos
⁡
θ
0
}, let us put(113)
δ
1
′
′
′
z
=
1
-
z
,
δ
2
′
′
′
z
=
R
e
z
-
cos
⁡
θ
0
.We find using the same analysis Theorem 9.



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*Source: 7208285-2016-02-10.xml*