# On the Harmonic Problem with Nonlinear Boundary Integral Conditions

Authors: Saker Hacene
Journal: International Journal of Analysis
 (2014)

---

## Abstract

In the present work, we deal with the harmonic problems in a bounded domain ofℝ
2 with the nonlinear boundary integral conditions. After applying the Boundary integral method, a nonlinear boundary integral equation is obtained; the existence and uniqueness of the solution will be a consequence of applying theory of monotone operators.


---

## Body

## 1. Introduction

For the harmonic problem, the simplest boundary condition we can impose specifiesu at all points on the boundary Γ and is known as the Dirichlet boundary condition. The Dirichlet problem for the Laplace equation can easily be solved using the boundary integral equation [1]. If the normal derivative of u, that is, ∂
u
/
∂
n, where n is the outward normal to the boundary Γ, is specified at all points on the boundary Γ, that is, the Neumann boundary condition, with ∫
Γ
‍
(
∂
u
/
∂
n
)
d
s
=
0, then given the value of u at one point on Γ enables a unique solution to be obtained [1].In this work, we impose more general boundary conditions, namely, the nonlinear integral equation of Urysohn type [2, 3].Much attention has been paid to the resolution of boundary value problems for partial differential operators with nonlinear boundary conditions by the method of integral equations in many directions (see, e.g., Atkinson and Chandler [4, 5] and Ruotsalainen and Wendland [6]).Problems involving nonlinearities form a basis of mathematical models of various steady-state phenomena and processes in mechanics, physics, and many other areas of science. Among these is the steady-state heat transfer. Also some electromagnetic problems contain nonlinearities in the boundary conditions, for instance, problems where the electrical conductivity of the boundary is variable [7]. Further applications arise in heat radiation and heat transfer [7, 8].In the present paper, we look for the solution of the Laplacian equation with nonlinear data of the form(1)
Δ
u
(
x
)
=
0
,
x
∈
Ω
,
(2)
∂
u
∂
n
(
x
)
+
∫
Γ
‍
K
(
x
,
y
,
u
(
y
)
)
d
s
y
=
f
(
x
)
,
x
∈
Γ
.
We recall that the nonlinear boundary integral operator defined by
(3)
A
(
x
,
u
(
x
)
)
=
∫
Γ
‍
K
(
x
,
y
,
u
(
y
)
)
d
s
y
,
x
∈
Γ
,
is the nonlinear integral operator of Urysohn type.In (1), we assume Ω is an open bounded region in ℝ
2 with a smooth boundary Γ
=
∂
Ω, and
(4)
f
:
Γ
⟶
ℝ
,
K
:
Γ
×
Γ
×
ℝ
⟶
ℝ
are given real value functions.By the Green representation formula, we formulate a nonlinear integral equation on the boundaryΓ of the domain Ω. Under some assumptions on the Kernel of the nonlinear integral equation of Urysohn K
(
x
,
y
,
u
) we prove the existence and uniqueness of the solution.


### 1.1. Definitions and Notations

Definition 1 (see [1, 9]).
Letm
∈
ℕ; one denotes by H
m
(
Ω
) the Sobolev space:
(5)
H
m
(
Ω
)
=
{
u
∈
L
2
(
Ω
)
;
D
α
u
∈
L
2
(
Ω
)
,
|
α
|
≤
m
}
.Definition 2 (see [1, 9]).
Lets
∈
ℝ; one denotes by H
s
(
ℝ
n
) the Sobolev space:
(6)
H
s
(
ℝ
n
)
=
{
u
∈
L
2
(
ℝ
n
)
;
(
1
+
|
ξ
|
2
)
s
/
2
|
F
[
u
]
|
∈
L
2
(
ℝ
n
)
}
;
and the associated norm:
(7)
∥
u
∥
H
s
=
(
∫
ℝ
n
‍
(
1
+
|
ξ
|
2
)
s
|
F
[
u
]
|
2
d
ξ
)
1
/
2
,
with F
[
·
] the Fourier transform.Definition 3 (see [1, 9]).
LetΩ
⊂
ℝ
n a bounded domain and Γ
∶
=
∂
Ω; one defined
(8)
H
s
(
Ω
)
=
{
u
∣
Ω
:
u
∈
H
s
(
ℝ
n
)
}
,
s
∈
ℝ
H
s
(
Γ
)
=
{
{
u
∣
Γ
:
u
∈
H
s
+
(
1
/
2
)
(
ℝ
n
)
}
,
s
>
0
L
2
(
Γ
)
,
s
=
0
(
H
-
s
(
Γ
)
)
′
(
dual
space
)
,
s
<
0
.Definition 4 (see [1, 9]).
The Fichera trace spacesH
s
(
Γ
), for 0
<
s
<
1, is defined to be the completion of
(9)
C
s
0
(
Γ
)
∶
=
{
φ
∈
C
0
(
Γ
)
:
∥
φ
∥
H
s
(
Γ
)
<
∞
}
,
with respect to the norm
(10)
∥
u
∥
H
s
(
Γ
)
∶
=
{
∥
u
∥
L
2
(
Γ
)
2
+
∬
Γ
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
1
+
2
s
d
s
x
d
s
y
}
1
/
2
.


## 1.1. Definitions and Notations

Definition 1 (see [1, 9]).
Letm
∈
ℕ; one denotes by H
m
(
Ω
) the Sobolev space:
(5)
H
m
(
Ω
)
=
{
u
∈
L
2
(
Ω
)
;
D
α
u
∈
L
2
(
Ω
)
,
|
α
|
≤
m
}
.Definition 2 (see [1, 9]).
Lets
∈
ℝ; one denotes by H
s
(
ℝ
n
) the Sobolev space:
(6)
H
s
(
ℝ
n
)
=
{
u
∈
L
2
(
ℝ
n
)
;
(
1
+
|
ξ
|
2
)
s
/
2
|
F
[
u
]
|
∈
L
2
(
ℝ
n
)
}
;
and the associated norm:
(7)
∥
u
∥
H
s
=
(
∫
ℝ
n
‍
(
1
+
|
ξ
|
2
)
s
|
F
[
u
]
|
2
d
ξ
)
1
/
2
,
with F
[
·
] the Fourier transform.Definition 3 (see [1, 9]).
LetΩ
⊂
ℝ
n a bounded domain and Γ
∶
=
∂
Ω; one defined
(8)
H
s
(
Ω
)
=
{
u
∣
Ω
:
u
∈
H
s
(
ℝ
n
)
}
,
s
∈
ℝ
H
s
(
Γ
)
=
{
{
u
∣
Γ
:
u
∈
H
s
+
(
1
/
2
)
(
ℝ
n
)
}
,
s
>
0
L
2
(
Γ
)
,
s
=
0
(
H
-
s
(
Γ
)
)
′
(
dual
space
)
,
s
<
0
.Definition 4 (see [1, 9]).
The Fichera trace spacesH
s
(
Γ
), for 0
<
s
<
1, is defined to be the completion of
(9)
C
s
0
(
Γ
)
∶
=
{
φ
∈
C
0
(
Γ
)
:
∥
φ
∥
H
s
(
Γ
)
<
∞
}
,
with respect to the norm
(10)
∥
u
∥
H
s
(
Γ
)
∶
=
{
∥
u
∥
L
2
(
Γ
)
2
+
∬
Γ
|
u
(
x
)
-
u
(
y
)
|
2
|
x
-
y
|
1
+
2
s
d
s
x
d
s
y
}
1
/
2
.


## 2. The Boundary Integral Method

### 2.1. Representative Formula and Boundary Operator

We introduce the fundamental solution of the Laplacian operator in the plane defined by:(11)
E
(
x
,
y
)
=
1
2
π
log
⁡
|
x
-
y
|
.
We first consider some standard boundary integral operators.Forx
∈
Ω, the single layer potential is
(12)
S
Ω
u
(
x
)
∶
=
-
∫
Γ
E
(
x
,
y
)
u
(
y
)
d
s
y
and the double layer potential is
(13)
D
Ω
u
(
x
)
∶
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
.Using Green’s identity for harmonic functions, we get(14)
u
(
x
)
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
-
∫
Γ
∂
u
(
y
)
∂
n
y
E
(
x
,
y
)
d
s
y
for x
∈
Ω, which can be written as
(15)
u
(
x
)
=
D
Ω
u
(
x
)
+
S
Ω
∂
u
(
x
)
∂
n
,
for
x
∈
Ω
.
Sending in (15) x
→
Γ. The continuity of the simple layer potential S
Ω and the jump relation of the double layer potential D
Ω. we can write the integral equation on the boundary as follows:
(16)
u
(
x
)
-
D
u
(
x
)
=
S
∂
u
(
x
)
∂
n
,
x
∈
Γ
,
where
(17)
S
∂
u
(
x
)
∂
n
:
=
-
2
∫
Γ
E
(
x
,
y
)
∂
u
(
y
)
∂
n
d
s
y
,
x
∈
Γ
,
D
u
(
x
)
:
=
2
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
,
x
∈
Γ
.
Clearly, if u
∈
H
1
(
Ω
) is the solution of (1), then the Cauchy data u
|
Γ and ∂
u
/
∂
n
|
Γ satisfies the integral equation (16).Then the boundary conditions(18)
∂
u
∂
n
(
x
)
=
-
A
(
x
,
u
(
x
)
)
+
f
(
x
)
yield
(19)
u
(
x
)
-
D
u
(
x
)
=
-
S
(
A
(
x
,
u
(
x
)
)
)
+
S
f
(
x
)
,
x
∈
Γ
.
Equation (19) can be written as
(20)
(
I
-
D
)
u
(
x
)
+
S
(
A
(
x
,
u
(
x
)
)
)
=
S
f
(
x
)
,
x
∈
Γ
.
Conversely, if u
|
Γ solves (20), then the solution of (1) can be given by the representation formula (15) and will satisfy
(21)
∂
u
∂
n
(
x
)
=
-
A
(
x
,
u
(
x
)
)
+
f
(
x
)
due to (20). For studying the solvability of the nonlinear equation (20), we give some assumptions to be made here.(H1)
We assume adiam
⁡
(
Ω
)
<
1.
(H2)
The KernelK
(
·
,
·
,
·
) of the Urysohn operator is a Caratheodory function [3].
(H3)
We assume that∂
K
(
x
,
y
,
u
)
/
∂
u is measurable satisfying
(22)
0
<
a
≤
∂
K
(
x
,
y
,
u
)
∂
u
≤
b
<
+
∞
,for some constants a and b.Remark 5.
(
1
) The operator S may have eigenfunctions [1]; then (
H
1
) ensures that the integral operator
(23)
S
:
H
s
(
Γ
)
⟶
H
s
+
1
(
Γ
)
is an isomorphism for every s
∈
ℝ and
(24)
(
S
μ
,
μ
)
≥
c
∥
μ
∥
H
-
1
/
2
2
for all μ
∈
H
-
1
/
2 with some positive constant c
>
0 [1]. By (
·
,
·
) we denote the L
2
(
Γ
) scalar product.
(
2
) The Kernel K
(
·
,
·
,
·
) is a Caratheodory function (
H
2
); that is, K
(
·
,
·
,
u
) is measurable for all u
∈
ℝ and K
(
x
,
y
,
·
) is continuous for almost all x
,
y
∈
Γ.
(
3
) The assumption (
H
3
) implies that the Nemytskii operator
(25)
A
:
L
2
(
Γ
)
⟶
L
2
(
Γ
)
is Lipschitz continuous and strongly monotonous such that
(26)
(
A
u
-
A
v
,
u
-
v
)
≤
b
mes
(
Γ
)
∥
u
-
v
∥
0
2
,
(
A
u
-
A
v
,
u
-
v
)
≥
a
mes
(
Γ
)
∥
u
-
v
∥
0
2
,
for all u
,
v
∈
L
2
(
Γ
).Theorem 6.
Let assumptions (H1), (H2), and (H3) hold. Then, for everyf
∈
H
-
1
/
2 the nonlinear boundary integral equation (20) has a unique solution in H
1
/
2
(
Γ
).Proof.
The proof follows from the well-known theorem by Browder and Minty on monotone operators [6, 10].
Since the simple layer potential operator onΓ
(27)
S
:
H
-
1
/
2
(
Γ
)
⟶
H
1
/
2
(
Γ
)
is an isomorphism, it is sufficient to consider the unique solvability of the following equation:
(28)
B
u
(
x
)
:
=
S
-
1
(
I
-
D
)
u
(
x
)
+
A
(
x
,
u
(
x
)
)
=
f
(
x
)
,
x
∈
Γ
.
We will prove that the operator
(29)
B
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous and strongly monotonous.(i)
In the first we show thatB is continuous.
It is clear from the continuity of the mapping properties of the simple and double layer operators that(30)
S
-
1
(
I
-
D
)
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous. And from (H3),
(31)
A
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous. Hence the boundary integral operator
(32)
B
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous.(ii)
In the second we show thatB is strongly monotone operator.
The functionμ
∈
H
-
1
/
2
(
Γ
) defined by
(33)
μ
(
x
)
∶
=
S
-
1
(
I
-
D
)
u
(
x
)
for u
(
x
)
∈
H
1
/
2
(
Γ
) is the normal derivative of the harmonic function
(34)
w
(
x
)
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
-
∫
Γ
μ
(
y
)
E
(
x
,
y
)
d
s
y
for x
∈
Ω; this means that w satisfies the problem
(35)
Δ
w
(
x
)
=
0
,
x
∈
Ω
,
w
(
x
)
=
u
(
x
)
,
x
∈
Γ
.
Then Green’s theorem yields(36)
(
S
-
1
(
I
-
D
)
u
,
u
)
=
∫
Γ
μ
u
d
s
=
∫
Γ
∂
w
∂
n
u
d
s
=
∫
Γ
∂
w
∂
n
w
d
s
=
∫
Ω
(
∇
w
)
2
d
x
.
Hence, for all u
,
v
∈
H
1
/
2
(
Γ
),
(37)
(
S
-
1
(
I
-
D
)
(
u
-
v
)
,
u
-
v
)
=
∫
‍
Ω
(
∇
(
w
1
-
w
2
)
)
2
d
x
=
|
w
1
-
w
2
|
H
1
(
Ω
)
2
,
where (
w
1
-
w
2
) denotes the harmonic function corresponding to the Cauchy data u
-
v and S
-
1
(
I
-
D
)
(
u
-
v
).
On the other hand, we note that there exists(
ν
1
-
ν
2
)
∈
H
-
1
/
2
(
Γ
), such that
(38)
S
(
ν
1
-
ν
2
)
=
u
-
v
on Γ [1]. Hence for all x
∈
Ω, we have
(39)
S
Ω
(
ν
1
-
ν
2
)
=
w
1
-
w
2
.
The simple layer potential
(40)
S
Ω
:
H
s
(
Γ
)
⟶
H
s
+
(
3
/
2
)
(
Ω
)
is continuous, for all s
∈
ℝ [1]. Hence for s
=
-
3
/
2, we find
(41)
∥
w
1
-
w
2
∥
L
2
(
Ω
)
≤
c
1
∥
ν
1
-
ν
2
∥
H
-
3
/
2
(
Γ
)
≤
c
2
∥
u
-
v
∥
H
-
1
/
2
(
Γ
)
≤
c
3
∥
u
-
v
∥
0
,
for some positive constants c
1, c
2, and c
3.
Hence we have(42)
∥
u
-
v
∥
0
≥
1
c
3
∥
w
1
-
w
2
∥
L
2
(
Ω
)
.
Then with (28) and (37) we get
(43)
(
B
u
-
B
v
,
u
-
v
)
=
(
S
-
1
(
I
-
D
)
(
u
-
v
)
,
u
-
v
)
+
(
A
u
-
A
v
,
u
-
v
)
=
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
(
A
u
-
A
v
,
u
-
v
)
,
and with (26) we get the inequality
(44)
(
B
u
-
B
v
,
u
-
v
)
≥
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
a
mes
(
Γ
)
∥
u
-
v
∥
0
2
;
hence with (42) we have
(45)
(
B
u
-
B
v
,
u
-
v
)
≥
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
a
mes
(
Γ
)
c
3
2
∥
w
1
-
w
2
∥
L
2
(
Ω
)
2
≥
min
⁡
{
1
,
a
mes
(
Γ
)
c
3
2
}
×
(
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
∥
w
1
-
w
2
∥
L
2
(
Ω
)
2
)
≥
min
⁡
{
1
,
a
mes
(
Γ
)
c
3
2
}
∥
w
1
-
w
2
∥
H
1
(
Ω
)
2
≥
c
4
∥
u
-
v
∥
H
1
/
2
(
Γ
)
2
by the trace theorem [1, 9], which completes the proof.Now we prove the regularity of the solution of the nonlinear boundary integral equation (20).Theorem 7.
For allS
f
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2, the unique solution of the nonlinear boundary integral equation (20) belongs to the space H
s
(
Γ
).In the proof of this theorem we will need the following lemma.Lemma 8.
For everyu
∈
H
s
(
Γ
), 0
≤
s
≤
1, one has A
u
∈
H
s
(
Γ
) and the mapping A
:
H
s
(
Γ
)
→
H
s
(
Γ
) is bounded.Proof.
Fors
=
0, u
∈
H
0
(
Γ
)
:
=
L
2
(
Γ
) has already been proved.
Fors
=
1, u
∈
H
1
(
Γ
), u is an absolutely continuous function. By assumption (H3) the function A
u
(
x
) is Lipschitz continuous. Hence A
u
(
x
) is also absolutely continuous function.
It remains to prove the case0
<
s
<
1; by the assumption (H3) and due to the definition of the Sobolev space in Definition 4, we have
(46)
∬
Γ
|
A
u
(
x
)
-
A
u
(
y
)
|
2
|
x
-
y
|
1
+
2
s
d
s
x
d
s
y
≤
b
2
(
mes
(
Γ
)
)
2
∥
u
∥
H
s
(
Γ
)
2
,
which completes the proof of Lemma 8.Proof of Theorem7.
LetS
f
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2, be given. By Theorem 6, there exists a unique solution u
∈
H
1
/
2
(
Γ
) of the nonlinear boundary integral equation:
(47)
(
I
-
D
)
u
+
S
A
u
=
S
f
.
Lemma 8 ensure that
(48)
S
f
-
S
A
u
∈
H
s
(
Γ
)
therefore
(49)
(
I
-
D
)
u
∈
H
s
(
Γ
)
.
This implies together with the Fredholm property of the double layer potential operator thatu
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2.
Example
9. Here we give an example to illustrate the theoretical results. We consider the harmonic problems:
(50)
Δ
u
(
x
)
=
0
,
x
∈
Ω
∂
u
∂
n
(
x
)
+
∫
Γ
‍
(
2
u
(
y
)
+
sin
u
(
y
)
)
d
s
y
=
f
(
x
)
,
x
∈
Γ
,
where the nonlinear boundary integral equation of Urysohn type is defined by
(51)
A
u
(
x
)
=
∫
Γ
‍
(
2
u
(
y
)
+
sin
u
(
y
)
)
d
s
y
,
x
∈
Γ
and the domain is
(52)
Ω
=
{
x
=
(
x
1
,
x
2
)
x
1
2
+
x
2
2
<
r
2
<
1
4
}
.
Clearly, the nonlinearity satisfies our assumptions (
H
1
), (
H
2
), and (
H
3
) such that
(53)
diam
⁡
(
Ω
)
=
2
r
<
1
.
The Kernel(
2
u
(
y
)
+
sin
u
(
y
)
) of the nonlinear boundary integral equation of Urysohn type is a Caratheodory function. And
(54)
∂
K
(
x
,
y
,
u
)
∂
u
=
2
+
cos
⁡
u
(
y
)
is measurable satisfying
(55)
1
≤
∂
(
2
u
(
y
)
+
sin
u
(
y
)
)
∂
u
≤
3
<
+
∞
,
implying that the Nemytskii operator
(56)
A
:
L
2
(
Γ
)
⟶
L
2
(
Γ
)
is Lipschitz continuous and strongly monotonous such that
(57)
2
π
r
∥
u
-
v
∥
0
2
≤
(
A
u
-
A
v
,
u
-
v
)
≤
6
π
r
∥
u
-
v
∥
0
2
for all u
,
v
∈
L
2
(
Γ
).


## 2.1. Representative Formula and Boundary Operator

We introduce the fundamental solution of the Laplacian operator in the plane defined by:(11)
E
(
x
,
y
)
=
1
2
π
log
⁡
|
x
-
y
|
.
We first consider some standard boundary integral operators.Forx
∈
Ω, the single layer potential is
(12)
S
Ω
u
(
x
)
∶
=
-
∫
Γ
E
(
x
,
y
)
u
(
y
)
d
s
y
and the double layer potential is
(13)
D
Ω
u
(
x
)
∶
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
.Using Green’s identity for harmonic functions, we get(14)
u
(
x
)
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
-
∫
Γ
∂
u
(
y
)
∂
n
y
E
(
x
,
y
)
d
s
y
for x
∈
Ω, which can be written as
(15)
u
(
x
)
=
D
Ω
u
(
x
)
+
S
Ω
∂
u
(
x
)
∂
n
,
for
x
∈
Ω
.
Sending in (15) x
→
Γ. The continuity of the simple layer potential S
Ω and the jump relation of the double layer potential D
Ω. we can write the integral equation on the boundary as follows:
(16)
u
(
x
)
-
D
u
(
x
)
=
S
∂
u
(
x
)
∂
n
,
x
∈
Γ
,
where
(17)
S
∂
u
(
x
)
∂
n
:
=
-
2
∫
Γ
E
(
x
,
y
)
∂
u
(
y
)
∂
n
d
s
y
,
x
∈
Γ
,
D
u
(
x
)
:
=
2
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
,
x
∈
Γ
.
Clearly, if u
∈
H
1
(
Ω
) is the solution of (1), then the Cauchy data u
|
Γ and ∂
u
/
∂
n
|
Γ satisfies the integral equation (16).Then the boundary conditions(18)
∂
u
∂
n
(
x
)
=
-
A
(
x
,
u
(
x
)
)
+
f
(
x
)
yield
(19)
u
(
x
)
-
D
u
(
x
)
=
-
S
(
A
(
x
,
u
(
x
)
)
)
+
S
f
(
x
)
,
x
∈
Γ
.
Equation (19) can be written as
(20)
(
I
-
D
)
u
(
x
)
+
S
(
A
(
x
,
u
(
x
)
)
)
=
S
f
(
x
)
,
x
∈
Γ
.
Conversely, if u
|
Γ solves (20), then the solution of (1) can be given by the representation formula (15) and will satisfy
(21)
∂
u
∂
n
(
x
)
=
-
A
(
x
,
u
(
x
)
)
+
f
(
x
)
due to (20). For studying the solvability of the nonlinear equation (20), we give some assumptions to be made here.(H1)
We assume adiam
⁡
(
Ω
)
<
1.
(H2)
The KernelK
(
·
,
·
,
·
) of the Urysohn operator is a Caratheodory function [3].
(H3)
We assume that∂
K
(
x
,
y
,
u
)
/
∂
u is measurable satisfying
(22)
0
<
a
≤
∂
K
(
x
,
y
,
u
)
∂
u
≤
b
<
+
∞
,for some constants a and b.Remark 5.
(
1
) The operator S may have eigenfunctions [1]; then (
H
1
) ensures that the integral operator
(23)
S
:
H
s
(
Γ
)
⟶
H
s
+
1
(
Γ
)
is an isomorphism for every s
∈
ℝ and
(24)
(
S
μ
,
μ
)
≥
c
∥
μ
∥
H
-
1
/
2
2
for all μ
∈
H
-
1
/
2 with some positive constant c
>
0 [1]. By (
·
,
·
) we denote the L
2
(
Γ
) scalar product.
(
2
) The Kernel K
(
·
,
·
,
·
) is a Caratheodory function (
H
2
); that is, K
(
·
,
·
,
u
) is measurable for all u
∈
ℝ and K
(
x
,
y
,
·
) is continuous for almost all x
,
y
∈
Γ.
(
3
) The assumption (
H
3
) implies that the Nemytskii operator
(25)
A
:
L
2
(
Γ
)
⟶
L
2
(
Γ
)
is Lipschitz continuous and strongly monotonous such that
(26)
(
A
u
-
A
v
,
u
-
v
)
≤
b
mes
(
Γ
)
∥
u
-
v
∥
0
2
,
(
A
u
-
A
v
,
u
-
v
)
≥
a
mes
(
Γ
)
∥
u
-
v
∥
0
2
,
for all u
,
v
∈
L
2
(
Γ
).Theorem 6.
Let assumptions (H1), (H2), and (H3) hold. Then, for everyf
∈
H
-
1
/
2 the nonlinear boundary integral equation (20) has a unique solution in H
1
/
2
(
Γ
).Proof.
The proof follows from the well-known theorem by Browder and Minty on monotone operators [6, 10].
Since the simple layer potential operator onΓ
(27)
S
:
H
-
1
/
2
(
Γ
)
⟶
H
1
/
2
(
Γ
)
is an isomorphism, it is sufficient to consider the unique solvability of the following equation:
(28)
B
u
(
x
)
:
=
S
-
1
(
I
-
D
)
u
(
x
)
+
A
(
x
,
u
(
x
)
)
=
f
(
x
)
,
x
∈
Γ
.
We will prove that the operator
(29)
B
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous and strongly monotonous.(i)
In the first we show thatB is continuous.
It is clear from the continuity of the mapping properties of the simple and double layer operators that(30)
S
-
1
(
I
-
D
)
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous. And from (H3),
(31)
A
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous. Hence the boundary integral operator
(32)
B
:
H
1
/
2
(
Γ
)
⟶
H
-
1
/
2
(
Γ
)
is continuous.(ii)
In the second we show thatB is strongly monotone operator.
The functionμ
∈
H
-
1
/
2
(
Γ
) defined by
(33)
μ
(
x
)
∶
=
S
-
1
(
I
-
D
)
u
(
x
)
for u
(
x
)
∈
H
1
/
2
(
Γ
) is the normal derivative of the harmonic function
(34)
w
(
x
)
=
∫
Γ
u
(
y
)
∂
∂
n
y
E
(
x
,
y
)
d
s
y
-
∫
Γ
μ
(
y
)
E
(
x
,
y
)
d
s
y
for x
∈
Ω; this means that w satisfies the problem
(35)
Δ
w
(
x
)
=
0
,
x
∈
Ω
,
w
(
x
)
=
u
(
x
)
,
x
∈
Γ
.
Then Green’s theorem yields(36)
(
S
-
1
(
I
-
D
)
u
,
u
)
=
∫
Γ
μ
u
d
s
=
∫
Γ
∂
w
∂
n
u
d
s
=
∫
Γ
∂
w
∂
n
w
d
s
=
∫
Ω
(
∇
w
)
2
d
x
.
Hence, for all u
,
v
∈
H
1
/
2
(
Γ
),
(37)
(
S
-
1
(
I
-
D
)
(
u
-
v
)
,
u
-
v
)
=
∫
‍
Ω
(
∇
(
w
1
-
w
2
)
)
2
d
x
=
|
w
1
-
w
2
|
H
1
(
Ω
)
2
,
where (
w
1
-
w
2
) denotes the harmonic function corresponding to the Cauchy data u
-
v and S
-
1
(
I
-
D
)
(
u
-
v
).
On the other hand, we note that there exists(
ν
1
-
ν
2
)
∈
H
-
1
/
2
(
Γ
), such that
(38)
S
(
ν
1
-
ν
2
)
=
u
-
v
on Γ [1]. Hence for all x
∈
Ω, we have
(39)
S
Ω
(
ν
1
-
ν
2
)
=
w
1
-
w
2
.
The simple layer potential
(40)
S
Ω
:
H
s
(
Γ
)
⟶
H
s
+
(
3
/
2
)
(
Ω
)
is continuous, for all s
∈
ℝ [1]. Hence for s
=
-
3
/
2, we find
(41)
∥
w
1
-
w
2
∥
L
2
(
Ω
)
≤
c
1
∥
ν
1
-
ν
2
∥
H
-
3
/
2
(
Γ
)
≤
c
2
∥
u
-
v
∥
H
-
1
/
2
(
Γ
)
≤
c
3
∥
u
-
v
∥
0
,
for some positive constants c
1, c
2, and c
3.
Hence we have(42)
∥
u
-
v
∥
0
≥
1
c
3
∥
w
1
-
w
2
∥
L
2
(
Ω
)
.
Then with (28) and (37) we get
(43)
(
B
u
-
B
v
,
u
-
v
)
=
(
S
-
1
(
I
-
D
)
(
u
-
v
)
,
u
-
v
)
+
(
A
u
-
A
v
,
u
-
v
)
=
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
(
A
u
-
A
v
,
u
-
v
)
,
and with (26) we get the inequality
(44)
(
B
u
-
B
v
,
u
-
v
)
≥
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
a
mes
(
Γ
)
∥
u
-
v
∥
0
2
;
hence with (42) we have
(45)
(
B
u
-
B
v
,
u
-
v
)
≥
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
a
mes
(
Γ
)
c
3
2
∥
w
1
-
w
2
∥
L
2
(
Ω
)
2
≥
min
⁡
{
1
,
a
mes
(
Γ
)
c
3
2
}
×
(
|
w
1
-
w
2
|
H
1
(
Ω
)
2
+
∥
w
1
-
w
2
∥
L
2
(
Ω
)
2
)
≥
min
⁡
{
1
,
a
mes
(
Γ
)
c
3
2
}
∥
w
1
-
w
2
∥
H
1
(
Ω
)
2
≥
c
4
∥
u
-
v
∥
H
1
/
2
(
Γ
)
2
by the trace theorem [1, 9], which completes the proof.Now we prove the regularity of the solution of the nonlinear boundary integral equation (20).Theorem 7.
For allS
f
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2, the unique solution of the nonlinear boundary integral equation (20) belongs to the space H
s
(
Γ
).In the proof of this theorem we will need the following lemma.Lemma 8.
For everyu
∈
H
s
(
Γ
), 0
≤
s
≤
1, one has A
u
∈
H
s
(
Γ
) and the mapping A
:
H
s
(
Γ
)
→
H
s
(
Γ
) is bounded.Proof.
Fors
=
0, u
∈
H
0
(
Γ
)
:
=
L
2
(
Γ
) has already been proved.
Fors
=
1, u
∈
H
1
(
Γ
), u is an absolutely continuous function. By assumption (H3) the function A
u
(
x
) is Lipschitz continuous. Hence A
u
(
x
) is also absolutely continuous function.
It remains to prove the case0
<
s
<
1; by the assumption (H3) and due to the definition of the Sobolev space in Definition 4, we have
(46)
∬
Γ
|
A
u
(
x
)
-
A
u
(
y
)
|
2
|
x
-
y
|
1
+
2
s
d
s
x
d
s
y
≤
b
2
(
mes
(
Γ
)
)
2
∥
u
∥
H
s
(
Γ
)
2
,
which completes the proof of Lemma 8.Proof of Theorem7.
LetS
f
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2, be given. By Theorem 6, there exists a unique solution u
∈
H
1
/
2
(
Γ
) of the nonlinear boundary integral equation:
(47)
(
I
-
D
)
u
+
S
A
u
=
S
f
.
Lemma 8 ensure that
(48)
S
f
-
S
A
u
∈
H
s
(
Γ
)
therefore
(49)
(
I
-
D
)
u
∈
H
s
(
Γ
)
.
This implies together with the Fredholm property of the double layer potential operator thatu
∈
H
s
(
Γ
), 1
/
2
≤
s
≤
3
/
2.
Example
9. Here we give an example to illustrate the theoretical results. We consider the harmonic problems:
(50)
Δ
u
(
x
)
=
0
,
x
∈
Ω
∂
u
∂
n
(
x
)
+
∫
Γ
‍
(
2
u
(
y
)
+
sin
u
(
y
)
)
d
s
y
=
f
(
x
)
,
x
∈
Γ
,
where the nonlinear boundary integral equation of Urysohn type is defined by
(51)
A
u
(
x
)
=
∫
Γ
‍
(
2
u
(
y
)
+
sin
u
(
y
)
)
d
s
y
,
x
∈
Γ
and the domain is
(52)
Ω
=
{
x
=
(
x
1
,
x
2
)
x
1
2
+
x
2
2
<
r
2
<
1
4
}
.
Clearly, the nonlinearity satisfies our assumptions (
H
1
), (
H
2
), and (
H
3
) such that
(53)
diam
⁡
(
Ω
)
=
2
r
<
1
.
The Kernel(
2
u
(
y
)
+
sin
u
(
y
)
) of the nonlinear boundary integral equation of Urysohn type is a Caratheodory function. And
(54)
∂
K
(
x
,
y
,
u
)
∂
u
=
2
+
cos
⁡
u
(
y
)
is measurable satisfying
(55)
1
≤
∂
(
2
u
(
y
)
+
sin
u
(
y
)
)
∂
u
≤
3
<
+
∞
,
implying that the Nemytskii operator
(56)
A
:
L
2
(
Γ
)
⟶
L
2
(
Γ
)
is Lipschitz continuous and strongly monotonous such that
(57)
2
π
r
∥
u
-
v
∥
0
2
≤
(
A
u
-
A
v
,
u
-
v
)
≤
6
π
r
∥
u
-
v
∥
0
2
for all u
,
v
∈
L
2
(
Γ
).



---

*Source: 976520-2014-02-23.xml*