1 Introduction
In this paper we consider the Dirichlet and the traction problems for the linearized n-dimensional elastostatics. The classical indirect methods for solving them consist in looking for the solution in the form of a double layer potential and a simple layer potential respectively. It is well-known that, if the boundary is sufficiently smooth, in both cases we are led to a singular integral system which can be reduced to a Fredholm one (see, e.g., 1).
Recently this approach was considered in multiply connected domains for several partial differential equations (see, e.g., 234567).
However these are not the only integral representations that are of importance. Another one consists in looking for the solution of the Dirichlet problem in the form of a simple layer potential. This approach leads to an integral equation of the first kind on the boundary which can be treated in different ways. For n = 2 and Ω simply connected see 8. A method hinging on the theory of reducible operators (see 910) and the theory of differential forms (see, e.g., 1112) was introduced in 13 for the n-dimensional Laplace equation and later extended to the three-dimensional elasticity in 14. This method can be considered as an extension of the one given by Muskhelishvili 15 in the complex plane. The double layer potential ansatz for the traction problem can be treated in a similar way, as shown in 16.
In the present paper we are going to consider these two last approaches in a multiply connected bounded domain of ℝn (n ≥ 2). Similar results for Laplace equation have been recently obtained in 17. We remark that we do not require the use of pseudo-differential operators nor the use of hypersingular integrals, differently from other methods (see, e.g., [18, Chapter 4] for the study of the Neumann problem for Laplace equation by means of a double layer potential).
After giving some notations and definitions in Section 2, we prove some preliminary results in Section 3. They concern the study of the first derivatives of a double layer potential. This leads to the construction of a reducing operator, which will be useful in the study of the integral system of the first kind arising in the Dirichlet problem.
Section 4 is devoted to the case n = 2, where there exist some exceptional boundaries in which we need to add a constant vector to the simple layer potential. In particular, after giving an explicit example of such boundaries, we prove that in a multiply connected domain the boundary is exceptional if, and only if, the external boundary is exceptional.
In Section 5 we find the solution of the Dirichlet problem in a multiply connected domain by means of a simple layer potential. We show how to reduce the problem to an equivalent Fredholm equation (see Remark 5.5).
Section 6 is devoted to the traction problem. It turns out that the solution of this problem does exist in the form of a double layer potential if, and only if, the given forces are balanced on each connected component of the boundary. While in a simply connected domain the solution of the traction problem can be always represented by means of a double layer potential (provided that, of course, the given forces are balanced on the boundary), this is not true in a multiply connected domain. Therefore the presence or absence of "holes" makes a difference.
We mention that lately we have applied the same method to the study of the Stokes system 19. Moreover the results obtained for other integral representations for several partial differential equations on domains with lower regularity (see, e.g., the references of 20 for C1 or Lipschitz boundaries and 21 for "worse" domains) lead one to hope that our approach could be extended to more general domains.
2 Notations and definitions
Throughout this paper we consider a domain (open connected set) Ω ⊂ ℝn, n ≥ 2, of the form Ω
=
Ω
0
\
⋃
j
=
1
m
Ω
̄
j
, where Ωj (j = 0, ..., m) are m + 1 bounded domains of ℝn with connected boundaries Σj ∈ C1, λ (λ ∈ (0, 1]) and such that
Ω
̄
j
⊂
Ω
0
and
Ω
̄
j
∩
Ω
̄
k
=
∅
,
j
,
k
=
1
,
.
.
.
,
m
,
j
≠
k
. For brevity, we shall call such a domain an (m + 1)-connected domain. We denote by ν the outwards unit normal on Σ = ∂Ω.
Let E be the partial differential operator
E
u
=
Δ
u
+
k
∇
div
u
,
where u = (u1, ..., un) is a vector-valued function and k > (n - 2)/n is a real constant. A fundamental solution of the operator - E is given by Kelvin's matrix whose entries are
Γ
i
j
(
x
,
y
)
=
1
2
π
-
k
+
2
2
(
k
+
1
)
δ
i
j
log
|
x
-
y
|
+
k
2
(
k
+
1
)
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
2
,
if
n
=
2
,
1
ω
n
-
k
+
2
2
(
k
+
1
)
δ
i
j
|
x
-
y
|
2
-
n
2
-
n
+
k
2
(
k
+
1
)
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
n
,
if
n
≥
3
,
i, j = 1, ..., n, ωn being the hypersurface measure of the unit sphere in ℝn.
As usual, we denote by E
(
u
,
v
)
the bilinear form defined as
E
(
y
,
v
)
=
2
σ
i
h
(
u
)
ε
i
h
(
v
)
=
2
σ
i
h
(
u
)
ε
i
h
(
u
)
,
where εih (u) and σih (u) are the linearized strain components and the stress components respectively, i.e.
ε
i
h
(
u
)
=
1
2
(
∂
i
u
h
+
∂
h
u
i
)
,
σ
i
h
(
u
)
=
ε
i
h
(
u
)
+
k
-
1
2
δ
i
h
ε
j
j
(
u
)
.
Let us consider the boundary operator Lξ whose components are
L
i
ξ
u
=
(
k
-
ξ
)
(
div
u
)
ν
i
+
ν
j
∂
j
u
i
+
ξ
ν
j
∂
i
u
j
,
i
=
1
,
.
.
.
,
n
,
ξ being a real parameter. We remark that the operator L1 is just the stress operator 2σihνh, which we shall simply denote by L, while Lk/(k+2) is the so-called pseudo-stress operator.
By the symbol
S
n
we denote the space of all constant skew-symmetric matrices of order n. It is well-known that the dimension of this space is n(n - 1)/2. From now on a + Bx stands for a rigid displacement, i.e. a is a constant vector and B
∈
S
n
. We denote by R
the space of all rigid displacements whose dimension is n(n + 1)/2. As usual {e1, ..., en} is the canonical basis for ℝn.
For any 1 < p < +∞ we denote by [Lp(Σ)]n the space of all measurable vector-valued functions u = (u1, ..., un) such that |uj|p is integrable over Σ (j = 1, ..., n). If h is any non-negative integer,
L
h
p
(
Σ
)
is the vector space of all differential forms of degree h defined on Σ such that their components are integrable functions belonging to Lp(Σ) in a coordinate system of class C1 and consequently in every coordinate system of class C1. The space
[
L
h
p
(
Σ
)
]
n
is constituted by the vectors (v1, ..., vn) such that vj is a differential form of
L
h
p
(
Σ
)
(
j
=
1
,
.
.
.
,
n
)
. [W1, p(Σ)]n is the vector space of all measurable vector-valued functions u = (u1, ..., un) such that uj belongs to the Sobolev space W1,p(Σ) (j = 1, ..., n).
If B and B' are two Banach spaces and S : B → B' is a continuous linear operator, we say that S can be reduced on the left if there exists a continuous linear operator S' : B' → B such that S'S = I + T, where I stands for the identity operator of B and T : B → B is compact. Analogously, one can define an operator S reducible on the right. One of the main properties of such operators is that the equation Sα = β has a solution if, and only if, 〈γ, β〉 = 0 for any γ such that S*γ = 0, S* being the adjoint of S (for more details see, e.g., 910).
We end this section by defining the spaces in which we look for the solutions of the BVPs we are going to consider.
Definition 2.1. The vector-valued function u belongs to
S
p
if, and only if, there exists φ ∈ [Lp(Σ)]n such that u can be represented by a simple layer potential
u
(
x
)
=
∫
Σ
Γ
(
x
,
y
)
φ
(
y
)
d
σ
y
,
x
∈
Ω
.
Definition 2.2. The vector-valued function w belongs to
D
p
if, and only if, there exists ψ ∈ [W1,p(Σ)]n such that w can be represented by a double layer potential
w
(
x
)
=
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
ψ
(
y
)
d
σ
y
,
x
∈
Ω
,
where [LyΓ(x, y)]' denotes the transposed matrix of Ly[Γ(x, y)].
3 Preliminary results
3.1 On the first derivatives of a double layer potential
Let us consider the boundary operator Lξ defined by (2). Denoting by Γj(x, y) the vector whose components are Γij(x, y), we have
L
i
,
y
ξ
[
Γ
j
(
x
,
y
)
]
=
-
1
ω
n
2
+
(
1
-
ξ
)
k
2
(
1
+
k
)
δ
i
j
+
n
k
(
ξ
+
1
)
2
(
k
+
1
)
(
y
i
-
x
i
)
(
y
j
-
x
j
)
|
y
-
x
|
2
(
y
p
-
x
p
)
ν
p
(
y
)
|
y
-
x
|
n
+
k
-
(
2
+
k
)
ξ
2
(
k
+
1
)
(
y
j
-
x
j
)
ν
i
(
y
)
-
(
y
i
-
x
i
)
ν
j
(
y
)
|
y
-
x
|
n
.
We recall that an immediate consequence of (5) is that, when ξ = k/(2 + k) we have
L
i
,
y
k
∕
(
2
+
k
)
[
Γ
j
(
x
,
y
)
]
=
O
(
|
x
-
y
|
1
-
n
+
λ
)
,
while for ξ ≠ k/(2 + k) the kernels
L
i
,
y
ξ
[
Γ
j
(
x
,
y
)
]
have a strong singularity on Σ.
Let us denote by wξ the double layer potential
w
j
ξ
(
x
)
=
∫
Σ
u
i
(
y
)
L
i
,
y
ξ
[
Γ
j
(
x
,
y
)
]
d
σ
y
,
j
=
1
,
.
.
.
,
n
.
It is known that the first derivatives of a harmonic double layer potential with density φ belonging to W1,p(Σ) can be written by means of the formula proved in [13, p. 187]
*
d
∫
Σ
φ
(
y
)
∂
s
(
x
,
y
)
∂
ν
y
d
σ
y
=
d
x
∫
Σ
d
φ
(
y
)
∧
s
n
-
2
(
x
,
y
)
,
x
∈
Ω
.
Here * and d denote the Hodge star operator and the exterior derivative respectively, s(x, y) is the fundamental solution of Laplace equation
s
(
x
,
y
)
=
1
2
π
log
|
x
-
y
|
,
if
n
=
2
,
1
(
2
-
n
)
ω
n
|
x
-
y
|
2
-
n
,
if
n
≥
3
and sh(x, y) is the double h-form introduced by Hodge in 22
s
h
(
x
,
y
)
=
∑
j
1
<
…
<
j
h
s
(
x
,
y
)
d
x
j
1
…
d
x
j
h
d
y
j
1
…
d
y
j
h
.
Since, for a scalar function f and for a fixed h, we have *df ∧ dxh = (-1)n-1∂hf dx, denoting by w the harmonic double layer potential with density φ ∈ W1,p(Σ), (8) implies
∂
h
w
(
x
)
=
-
Θ
h
(
d
φ
)
(
x
)
,
x
∈
Ω
where, for every ψ
∈
L
1
p
(
Σ
)
,
Θ
h
(
ψ
)
(
x
)
=
*
∫
Σ
d
x
[
s
n
-
2
(
x
,
y
)
]
∧
ψ
(
y
)
∧
d
x
h
,
x
∈
Ω
.
The following lemma can be considered as an extension of formula (9) to elasticity. Here du denotes the vector (du1, ..., dun) and ψ = (ψ1, ..., ψn) is an element of
[
L
1
p
(
Σ
)
]
n
.
Lemma 3.1. Let wξ be the double layer potential (7) with density u ∈ [W1,p(Σ)]n. Then
∂
s
w
j
ξ
(
x
)
=
K
j
s
ξ
(
d
u
)
(
x
)
,
x
∈
Ω
,
j
,
s
=
1
,
…
,
n
,
where
K
j
s
ξ
(
ψ
)
(
x
)
=
Θ
s
(
ψ
j
)
(
x
)
-
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
.
.
.
n
∫
Σ
∂
x
s
K
h
j
ξ
(
x
,
y
)
∧
ψ
i
(
y
)
∧
d
y
j
3
…
d
y
j
n
,
K
h
j
ξ
(
x
,
y
)
=
1
ω
n
k
(
ξ
+
1
)
2
(
k
+
1
)
(
y
l
-
x
l
)
(
y
j
-
x
j
)
|
y
-
x
|
n
+
k
-
(
2
+
k
)
ξ
2
(
k
+
1
)
δ
l
j
s
(
x
,
y
)
,
and Θh is given by (10), h = 1, ..., n.
Proof. Let n ≥ 3. Denote by Mhi the tangential operators Mhi = νh∂i - νi∂h, h, i = 1, ..., n. By observing that
M
h
i
x
h
x
j
|
x
|
n
=
δ
i
j
x
h
ν
h
|
x
|
n
-
n
x
i
x
j
x
h
ν
h
|
x
|
n
+
2
,
we find in Ω
w
j
ξ
(
x
)
=
-
1
ω
n
∫
Σ
u
i
(
y
)
δ
i
j
(
y
h
-
x
h
)
ν
h
(
y
)
|
y
-
x
|
n
-
k
(
ξ
+
1
)
2
(
k
+
1
)
M
y
h
i
(
y
h
-
x
h
)
(
y
j
-
x
j
)
|
y
-
x
|
n
+
k
-
(
2
+
k
)
ξ
2
(
k
+
1
)
M
y
i
j
|
y
-
x
|
2
-
n
2
-
n
d
σ
y
=
-
∫
Σ
u
j
(
y
)
∂
s
(
x
,
y
)
∂
ν
y
d
σ
y
+
∫
Σ
u
i
(
y
)
k
(
ξ
+
1
)
2
(
k
+
1
)
M
y
h
i
(
y
h
-
x
h
)
(
y
j
-
x
j
)
|
y
-
x
|
2
(
2
-
n
)
s
(
x
,
y
)
-
k
-
(
2
+
k
)
ξ
2
(
k
+
1
)
M
y
i
j
s
(
x
,
y
)
d
σ
y
.
An integration by parts on Σ leads to
w
j
ξ
(
x
)
=
-
∫
Σ
u
j
(
y
)
∂
s
(
x
,
y
)
∂
ν
y
d
σ
y
-
∫
Σ
M
h
i
[
u
i
(
y
)
]
k
(
ξ
+
1
)
(
2
-
n
)
2
(
k
+
1
)
(
y
h
-
x
h
)
(
y
j
-
x
j
)
|
y
-
x
|
2
+
k
-
(
2
+
k
)
ξ
2
(
k
+
1
)
δ
h
j
s
(
x
,
y
)
d
σ
y
=
-
∫
Σ
u
j
(
y
)
∂
s
(
x
,
y
)
∂
ν
y
d
σ
y
-
∫
Σ
M
h
i
[
u
i
(
y
)
]
K
h
j
ξ
(
x
,
y
)
d
σ
y
.
Therefore, by recalling (9),
∂
s
w
j
ξ
(
x
)
=
Θ
s
(
d
u
j
)
(
x
)
-
∫
Σ
M
h
i
[
u
i
(
y
)
]
∂
x
s
[
K
h
j
ξ
(
x
,
y
)
]
d
σ
y
.
If f is a scalar function, we may write
M
h
i
(
f
)
d
σ
=
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
…
n
d
f
∧
d
x
j
3
…
d
x
j
n
.
This identity is established by observing that on Σ we have
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
…
n
d
f
∧
d
x
j
3
…
d
x
j
n
=
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
…
n
∂
j
2
f
d
x
j
2
∧
…
d
x
j
n
=
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
…
n
δ
j
1
…
j
n
1
…
n
ν
j
1
∂
j
2
f
d
σ
=
δ
j
1
j
2
h
i
ν
j
1
∂
j
2
f
d
σ
=
(
ν
h
∂
i
f
-
ν
i
∂
h
f
)
d
σ
.
Then we can rewrite (14) as
∂
s
w
j
ξ
(
x
)
=
Θ
s
(
d
u
j
)
(
x
)
-
1
(
n
-
2
)
!
δ
h
i
j
3
…
j
n
123
…
n
∫
Σ
∂
x
s
[
K
h
j
ξ
(
x
,
y
)
]
∧
d
u
i
(
y
)
∧
d
y
j
3
…
d
y
j
n
.
Similar arguments prove the result if n = 2. We omit the details. □
3.2 Some jump formulas
Lemma 3.2. Let f ∈ L1(Σ). If η ∈ Σ is a Lebesgue point for f, we have
lim
x
→
η
∫
Σ
f
(
y
)
∂
x
s
(
y
p
-
x
p
)
(
y
j
-
x
j
)
|
y
-
x
|
n
d
σ
y
=
ω
n
2
(
δ
p
j
-
2
ν
j
(
η
)
ν
p
(
η
)
)
ν
s
(
η
)
f
(
η
)
+
∫
Σ
f
(
y
)
∂
x
s
(
y
p
-
η
p
)
(
y
j
-
η
j
)
|
y
-
η
|
n
d
σ
y
,
where the limit has to be understood as an internal angular boundary value1.
Proof. Let hpj(x) = xpxj|x|-n. Since h ∈ C∞(ℝn\{0}) is even and homogeneous of degree 2 - n, due to the results proved in 23, we have
lim
x
→
η
∫
Σ
f
(
y
)
∂
x
s
(
y
p
-
x
p
)
(
y
j
-
x
j
)
|
y
-
x
|
n
d
σ
y
=
-
ν
s
(
η
)
γ
p
j
(
η
)
f
(
η
)
+
∫
Σ
f
(
y
)
∂
x
s
(
y
p
-
η
p
)
(
y
j
-
η
j
)
|
y
-
η
|
n
d
σ
y
,
where
γ
p
j
(
η
)
=
-
2
π
2
F
(
h
p
j
)
(
ν
η
)
, F
being the Fourier transform
F
(
h
)
(
x
)
=
∫
ℝ
n
h
(
y
)
e
-
2
π
i
x
⋅
y
d
y
(see also 24 and note that in 2324 ν is the inner normal). On the other hand
F
(
h
p
j
)
(
x
)
=
1
2
-
n
F
(
x
p
∂
j
(
|
x
|
2
-
n
)
)
=
-
1
(
2
-
n
)
2
π
i
∂
p
F
(
∂
j
(
|
x
|
2
-
n
)
)
=
-
1
2
-
n
∂
p
(
x
j
F
(
|
x
|
2
-
n
)
)
and, since
F
(
|
x
|
2
-
n
)
=
π
n
/
2
-
2
Γ
(
n
/
2
-
1
)
|
x
|
-
2
(see, e.g., [25, p. 156]), we find
F
(
h
p
j
)
(
x
)
=
π
n
∕
2
-
2
(
n
-
2
)
Γ
(
n
∕
2
-
1
)
∂
p
(
x
j
|
x
|
-
2
)
=
π
n
∕
2
-
2
(
n
-
2
)
Γ
(
n
∕
2
-
1
)
(
δ
p
j
|
x
|
-
2
-
2
x
j
x
p
|
x
|
-
4
)
.
Finally, keeping in mind that ωn = n πn/2/Γ(n/2 + 1) and Γ(n/2 + 1) = n(n - 2)Γ(n/2 - 1)/4, we obtain
γ
p
j
(
η
)
=
-
2
π
n
∕
2
(
n
-
2
)
Γ
(
n
∕
2
-
1
)
(
δ
p
j
-
2
ν
j
(
η
)
ν
p
(
η
)
)
=
-
ω
n
2
(
δ
p
j
-
2
ν
j
(
η
)
ν
p
(
η
)
)
.
Combining this formula with (16) we get (15). □
Lemma 3.3. Let ψ
∈
L
1
p
(
Σ
)
. Let us write ψ as ψ = ψhdxh with
ν
h
ψ
h
=
0
.
Then, for almost every η ∈ Σ,
lim
x
→
η
Θ
s
(
ψ
)
(
x
)
=
-
1
2
ψ
s
(
η
)
+
Θ
s
(
ψ
)
(
η
)
,
where Θs is given by (10) and the limit has to be understood as an internal angular boundary value.
Proof. First we note that the assumption (17) is not restrictive, because, given the 1-form ψ on Σ, there exist scalar functions ψh defined on Σ such that ψ = ψhdxh and (17) holds (see [26, p. 41]). We have
Θ
s
(
ψ
)
(
x
)
=
∑
j
1
<
…
<
j
n
-
2
*
∫
Σ
∂
x
i
[
s
(
x
,
y
)
]
ψ
h
(
y
)
d
y
j
1
…
d
y
j
n
-
2
d
y
h
d
x
i
d
x
j
1
…
d
x
j
n
-
2
d
x
s
=
∑
j
1
<
…
<
j
n
-
2
δ
k
j
1
.
.
.
j
n
-
2
h
12
.
.
.
.
.
.
.
.
.
.
.
n
δ
i
j
1
.
.
.
j
n
-
2
s
12
.
.
.
.
.
.
.
.
.
.
.
n
∫
Σ
∂
x
i
[
s
(
x
,
y
)
]
ν
k
(
y
)
ψ
h
(
y
)
d
σ
y
=
δ
k
h
i
s
∫
Σ
∂
x
i
[
s
(
x
,
y
)
]
ν
k
(
y
)
ψ
h
(
y
)
d
σ
y
and then
lim
x
→
η
Θ
s
(
ψ
)
(
x
)
=
-
1
2
δ
k
h
i
s
ν
i
(
η
)
ν
k
(
η
)
ψ
h
(
η
)
+
Θ
s
(
ψ
)
(
η
)
a.e. on Σ. From (17) it follows that
δ
k
h
i
s
ν
i
ν
k
ψ
h
=
ν
i
ν
i
ψ
s
-
ν
i
ν
s
ψ
i
=
ψ
s
and (18) is proved. □
Lemma 3.4. Let ψ
∈
L
1
p
(
Σ
)
. Let us write ψ as ψ = ψhdxh and suppose that (17) holds. Then, for almost every η ∈ Σ,
lim
x
→
η
1
(
n
-
2
)
!
δ
l
i
j
3
.
.
.
j
n
123
.
.
.
n
∫
Σ
∂
x
s
K
l
j
ξ
(
x
,
y
)
∧
ψ
(
y
)
∧
d
y
j
3
…
d
y
j
n
=
-
k
-
ξ
2
(
k
+
1
)
ν
j
(
η
)
ψ
i
(
η
)
+
ξ
2
ν
i
(
η
)
ψ
j
(
η
)
ν
s
(
η
)
+
1
(
n
-
2
)
!
δ
l
i
j
3
.
.
.
j
n
123
.
.
.
.
n
∫
Σ
∂
x
s
K
l
j
ξ
(
η
,
y
)
∧
ψ
(
y
)
∧
d
y
j
3
…
d
y
j
n
,
where Kξ is defined by (13) and the limit has to be understood as an internal angular boundary value.
Proof. We have
1
(
n
-
2
)
!
δ
l
i
j
3
…
j
n
123
…
n
∫
Σ
∂
x
s
K
l
j
ξ
(
x
,
y
)
∧
ψ
(
y
)
∧
d
y
j
3
…
d
y
j
n
=
1
(
n
-
2
)
!
δ
l
i
j
3
…
j
n
123
…
n
δ
r
h
j
3
…
j
n
123
…
n
∫
Σ
∂
x
s
K
l
j
ξ
(
x
,
y
)
ψ
h
(
y
)
ν
r
(
y
)
d
σ
y
=
δ
r
h
l
i
∫
Σ
∂
x
s
K
l
j
ξ
(
x
,
y
)
ψ
h
(
y
)
ν
r
(
y
)
d
σ
y
.
Keeping in mind (13), formula (15) leads to
lim
x
→
η
1
(
n
-
2
)
!
δ
l
i
j
3
…
j
n
123
…
n
∫
Σ
∂
x
s
K
l
j
ξ
(
x
,
y
)
∧
ψ
(
y
)
∧
d
y
j
3
…
d
y
j
n
=
δ
r
h
l
i
k
(
ξ
+
1
)
4
(
k
+
1
)
(
δ
l
j
-
2
ν
j
(
η
)
ν
l
(
η
)
)
ν
s
(
η
)
-
k
-
(
2
+
k
)
ξ
4
(
k
+
1
)
δ
l
j
ν
s
(
η
)
ν
r
(
η
)
ψ
h
(
η
)
+
1
(
n
-
2
)
!
δ
l
i
j
3
…
j
n
123
…
n
∫
Σ
∂
x
s
K
l
j
ξ
(
η
,
y
)
∧
ψ
(
y
)
∧
d
y
j
3
…
d
y
j
n
.
On the other hand
δ
r
h
l
i
k
(
ξ
+
1
)
4
(
k
+
1
)
(
δ
l
j
-
2
ν
j
ν
l
)
ν
s
-
k
-
(
2
+
k
)
ξ
4
(
k
+
1
)
δ
l
j
ν
s
ν
r
ψ
h
=
δ
r
h
l
i
ξ
2
δ
l
j
ν
s
-
k
(
ξ
+
1
)
2
(
k
+
1
)
ν
j
ν
l
ν
s
ν
r
ψ
h
=
ξ
2
δ
l
j
ν
s
-
k
(
ξ
+
1
)
2
(
k
+
1
)
ν
j
ν
l
ν
s
(
ν
l
ψ
i
-
ν
i
ψ
l
)
=
-
k
-
ξ
2
(
k
+
1
)
ν
j
ν
s
ψ
i
-
ξ
2
ν
i
ν
s
ψ
j
,
and the result follows. □
Lemma 3.5. Let ψ
=
(
ψ
1
,
…
,
ψ
n
)
∈
[
L
1
p
(
Σ
)
]
n
. Then, for almost every η ∈ Σ,
lim
x
→
η
[
(
k
-
ξ
)
K
j
j
ξ
(
ψ
)
(
x
)
ν
i
(
η
)
+
ν
j
(
η
)
K
i
j
ξ
(
ψ
)
(
x
)
+
ξ
ν
j
(
η
)
K
j
i
ξ
(
ψ
)
(
x
)
]
=
(
k
-
ξ
)
K
j
j
ξ
(
ψ
)
(
η
)
ν
i
(
η
)
+
ν
j
(
η
)
K
i
j
ξ
(
ψ
)
(
η
)
+
ξ
ν
j
(
η
)
K
j
i
ξ
(
ψ
)
(
η
)
,
K
ξ
being as in (12) and the limit has to be understood as an internal angular boundary value.
Proof. Let us write ψi as ψi = ψihdxh with
ν
h
ψ
i
h
=
0
,
i
=
1
,
…
,
n
.
In view of Lemmas 3.3 and 3.4 we have
lim
x
→
η
K
j
s
ξ
(
ψ
)
(
x
)
=
-
1
2
ψ
j
s
(
η
)
+
k
-
ξ
2
(
k
+
1
)
ν
j
(
η
)
ψ
h
h
(
η
)
+
ξ
2
ν
h
(
η
)
ψ
h
j
(
η
)
ν
s
(
η
)
+
K
j
s
ξ
(
ψ
)
(
η
)
.
Therefore
lim
x
→
η
[
(
k
-
ξ
)
K
j
j
ξ
(
ψ
)
(
x
)
ν
i
(
η
)
+
ν
j
(
η
)
K
i
j
ξ
(
ψ
)
(
x
)
+
ξ
ν
j
(
η
)
K
j
i
ξ
(
ψ
)
(
x
)
]
=
Φ
(
ψ
)
(
η
)
+
(
k
-
ξ
)
K
j
j
ξ
(
ψ
)
(
η
)
ν
i
(
η
)
+
ν
j
(
η
)
K
i
j
ξ
(
ψ
)
(
η
)
+
ξ
ν
j
(
η
)
K
j
i
ξ
(
ψ
)
(
η
)
,
where
Φ
(
ψ
)
=
(
k
-
ξ
)
-
1
2
ψ
j
j
+
k
-
ξ
2
(
k
+
1
)
ν
j
ψ
h
h
+
ξ
2
ν
h
ψ
h
j
ν
j
ν
i
+
ν
j
-
1
2
ψ
i
j
+
k
-
ξ
2
(
k
+
1
)
ν
i
ψ
h
h
+
ξ
2
ν
h
ψ
h
i
ν
j
+
ξ
ν
j
-
1
2
ψ
j
i
+
k
-
ξ
2
(
k
+
1
)
ν
j
ψ
h
h
+
ξ
2
ν
h
ψ
h
j
ν
i
.
Conditions (21) lead to
Φ
(
ψ
)
=
-
1
2
(
k
-
ξ
)
1
-
k
-
ξ
k
+
1
-
k
-
ξ
k
+
1
-
ξ
k
-
ξ
k
+
1
ν
i
ψ
h
h
.
The bracketed expression vanishing, Φ = 0 and the result is proved. □
Remark 3.6. In Lemmas 3.2, 3.3, 3.4 and 3.5 we have considered internal angular boundary values. It is clear that similar formulas hold for external angular boundary values. We have just to change the sign in the first term on the right hand sides in (15), (18) and (19), while (20) remains unchanged.
3.3 Reduction of a certain singular integral operator
The results of the previous subsection imply the following lemmas.
Lemma 3.7. Let wξ be the double layer potential (7) with density u ∈ [W1,p(Σ)]n. Then
L
+
,
i
ξ
(
w
ξ
)
=
L
-
,
i
ξ
(
w
ξ
)
=
(
k
-
ξ
)
K
j
j
ξ
(
d
u
)
ν
i
+
ν
j
K
i
j
ξ
(
d
u
)
+
ξ
ν
j
K
j
i
ξ
(
d
u
)
a.e. on Σ, where
L
+
ξ
(
w
ξ
)
and
L
-
ξ
(
w
ξ
)
denote the internal and the external angular boundary limit of Lξ(wξ) respectively and
K
ξ
is given by (12).
Proof. It is an immediate consequence of (11), (20) and Remark 3.6. □
Remark 3.8. The previous result is connected to [1, Theorem 8.4, p. 320].
Lemma 3.9. Let R :
[
L
p
(
Σ
)
]
n
→
[
L
1
p
(
Σ
)
]
n
be the following singular integral operator
R
φ
(
x
)
=
∫
Σ
d
x
[
Γ
(
x
,
y
)
]
φ
(
y
)
d
σ
y
.
Let us define
R
′
ξ
:
[
L
1
p
(
Σ
)
]
n
→
[
L
p
(
Σ
)
]
n
to be the singular integral operator
R
′
i
ξ
(
ψ
)
(
x
)
=
(
k
-
ξ
)
K
j
j
ξ
(
ψ
)
(
x
)
ν
i
(
x
)
+
ν
j
(
x
)
K
i
j
ξ
(
ψ
)
(
x
)
+
ξ
ν
j
(
x
)
K
j
i
ξ
(
ψ
)
(
x
)
.
Then
R
′
ξ
R
φ
=
-
1
4
φ
+
(
T
ξ
)
2
φ
,
where
T
ξ
φ
(
x
)
=
∫
Σ
L
x
ξ
[
Γ
(
x
,
y
)
]
φ
(
y
)
d
σ
y
.
Proof. Let u be the simple layer potential with density φ ∈ [Lp(Σ)]n. In view of Lemma 3.7, we have a.e. on Σ
R
′
i
ξ
(
R
φ
)
=
(
k
-
ξ
)
K
j
j
ξ
(
d
u
)
ν
i
+
ν
j
K
i
j
ξ
(
d
u
)
+
ξ
ν
j
K
j
i
ξ
(
d
u
)
=
L
i
ξ
(
w
ξ
)
,
where wξ is the double layer potential (7) with density u. Moreover, if x ∈ Ω,
w
j
ξ
(
x
)
=
∫
Σ
u
i
(
y
)
L
i
,
y
ξ
[
Γ
j
(
x
,
y
)
]
d
σ
y
=
-
u
j
(
x
)
+
∫
Σ
L
i
ξ
[
u
(
y
)
]
Γ
i
j
(
x
,
y
)
d
σ
y
and then, on account of (26),
L
ξ
w
ξ
=
-
1
2
L
ξ
u
+
T
ξ
(
L
ξ
u
)
=
-
1
2
1
2
φ
+
T
ξ
φ
+
T
ξ
1
2
φ
+
T
ξ
φ
=
-
1
4
φ
+
(
T
ξ
)
2
φ
.
□
Corollary 3.10. The operator R defined by (23) can be reduced on the left. A reducing operator is given by R'ξ with ξ = k/(2 + k).
Proof. This follows immediately from (25), because of the weak singularity of the kernel in (26) when ξ = k/(2 + k) (see (6)). □
3.4 The dimension of some eigenspaces
Let T be the operator defined by (26) with ξ = 1, i.e.
T
φ
(
x
)
=
∫
Σ
L
x
[
Γ
(
x
,
y
)
]
φ
(
y
)
d
σ
y
,
x
∈
Σ
,
and denote by T* its adjoint.
In this subsection we determine the dimension of the following eigenspaces
V
±
=
φ
∈
[
L
p
(
Σ
)
]
n
:
∓
1
2
φ
+
T
*
φ
=
0
;
W
±
=
φ
∈
[
L
p
(
Σ
)
]
n
:
±
1
2
φ
+
T
φ
=
0
.
We first observe that the (total) indices of singular integral systems in (28)-(29) vanish. This can be proved as in [1, pp. 235-238]. Moreover, by standard techniques, one can prove that all the eigenfunctions are hölder-continuous and then these eigenspaces do not depend on p. This implies that
dim
V
+
=
dim
W
-
,
dim
V
-
=
dim
W
+
⋅
The next two lemmas determine such dimensions. Similar results for Laplace equation can be found in [27, Chapter 3].
Lemma 3.11. The spaces
V
+
and
W
-
have dimension n(n + 1)m/2. Moreover
V
+
=
{
v
h
X
Σ
j
:
h
=
1
,
…
,
n
(
n
+
1
)
/
2
,
j
=
1
,
…
,
m
}
,
where {vh : h = 1 ..., n(n + 1)/2} is an orthonormal basis of the space R
and
X
Σ
j
is the characteristic function of Σj.
Proof. We define the vector-valued functions αj, j = 1, ..., m as
α
j
(
x
)
=
(
a
+
B
x
)
X
Σ
j
(
x
)
, x ∈ Σ. For a fixed j = 1, ..., m, the function αj(x) belongs to
V
+
; indeed
-
1
2
(
a
+
B
x
)
X
Σ
j
(
x
)
+
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
(
a
+
B
y
)
X
Σ
j
(
y
)
d
σ
y
=
-
1
2
(
a
+
B
x
)
X
Σ
j
(
x
)
+
∫
Σ
j
[
L
y
Γ
(
x
,
y
)
]
′
(
a
+
B
y
)
d
σ
y
=
-
1
2
(
a
+
B
x
)
+
1
2
(
a
+
B
x
)
=
0
,
x
∈
Σ
j
,
because of
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
α
j
(
y
)
d
σ
y
=
α
j
(
x
)
x
∈
Ω
j
,
α
j
(
x
)
/
2
x
∈
Σ
j
,
0
x
∉
Ω
¯
j
.
Now we prove that the following n(n + 1)m/2 eigensolutions of
V
+
w
h
j
(
x
)
=
v
h
(
x
)
X
Σ
j
(
x
)
,
h
=
1
,
…
,
n
(
n
+
1
)
/
2
,
j
=
1
,
…
,
m
,
x
∈
Σ
are linearly independent. Indeed, if
∑
h
=
1
n
(
n
+
1
)
∕
2
∑
j
=
1
m
c
h
j
w
h
j
=
0
, we have
∑
h
=
1
n
(
n
+
1
)
∕
2
c
h
j
v
h
(
x
)
=
0
,
x
∈
Σ
j
,
j
=
1
,
…
,
m
.
Then, by applying a classical uniqueness theorem to the domain Ωj,
∑
h
=
1
n
(
n
+
1
)
∕
2
c
h
j
v
h
(
x
)
=
0
,
x
∈
Ω
j
,
j
=
1
,
…
,
m
,
from which it easily follows that
c
h
j
=
0
,
h
=
1
,
…
,
n
(
n
+
1
)
/
2
,
j
=
1
,
…
,
m
.
Thus,
dim
V
+
≥
n
(
n
+
1
)
m
/
2
. On the other hand, suppose
φ
∈
W
-
and let u be the simple layer potential with density φ. Since Eu = 0 in Ωj and L-u = 0 on Σj, u = aj + Bjx on each connected component Ωj, j = 1, ..., m, and u = 0 in
ℝ
n
\
Ω
¯
0
. Note that this is true also for n = 2, because
φ
∈
W
-
implies
∫
Σ
φ
d
σ
=
0
. We can define a linear map τ as follows
τ
:
W
-
→
(
ℝ
n
×
S
n
)
m
φ
→
(
a
1
,
B
1
,
…
,
a
m
,
B
m
)
.
If τ(φ) = 0, from a classical uniqueness theorem, we have that φ ≡ 0 in ℝn. Thus, τ is an injective map and
dim
W
-
≤
n
(
n
+
1
)
m
/
2
. The assertion follows from (30). □
Lemma 3.12. The spaces
V
-
and
W
+
have dimension n(n + 1)/2. Moreover
V
-
is constituted by the restrictions to Σ of the rigid displacements.
Proof. Let
α
∈
R
. If x ∈ Σ, we have
1
2
α
(
x
)
+
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
α
(
y
)
d
σ
y
=
1
2
α
(
x
)
-
1
2
α
(
x
)
=
0
,
thanks to
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
α
(
y
)
d
σ
y
=
-
α
(
x
)
x
∈
Ω
,
-
α
(
x
)
/
2
x
∈
Σ
,
0
x
∉
Ω
¯
.
This shows that the restriction to Σ of α belongs to
V
-
and then
dim
V
-
≥
dim
R
=
n
(
n
+
1
)
/
2
. On the other hand, suppose
ϕ
∈
W
+
and let u be the simple layer potential with density ϕ. Since Eu = 0 in Ω and L+u = 0 on Σ, u = a + Bx in Ω. Let σ be the linear map
σ
:
W
+
→
ℝ
n
×
l
n
ϕ
→
(
a
,
B
)
.
If n ≥ 3, we have that σ(ϕ) = 0 implies u ≡ 0 in ℝn and then ϕ ≡ 0 on Σ, in view of classical uniqueness theorems.
If n = 2, define
W
+
0
=
ϕ
∈
W
+
/
∫
Σ
ϕ
d
σ
=
0
. We have
σ
|
W
+
0
is injective and its range does not contain the vectors ((1, 0), 0) and ((0, 1), 0)2. Therefore
dim
W
+
0
≤
1
. On the other hand,
dim
W
+
-
2
≤
dim
W
+
0
and then
dim
W
+
≤
3
. In any case,
dim
W
+
≤
n
(
n
+
1
)
/
2
and the result follows from (30). □
4 The bidimensional case
The case n = 2 requires some additional considerations. It is well-known that there are some domains in which no every harmonic function can be represented by means of a harmonic simple layer potential. For instance, on the unit disk we have
∫
|
y
|
=
1
log
|
x
-
y
|
d
s
y
=
0
,
|
x
|
<
1
.
Similar domains occur also in elasticity. In order to give explicitly such an example, let us prove the following lemma.
Lemma 4.1. Let ΣR be the circle of radius R centered at the origin. We have
∫
Σ
R
|
x
-
y
|
2
log
|
x
-
y
|
d
s
y
=
2
π
R
(
R
2
log
R
+
(
1
+
log
R
)
|
x
|
2
)
,
|
x
|
<
R
.
Proof. Denote by u(x) the function on the left hand side of (32) and by ΩR the ball of radius R centered at the origin. Let us fix x0 ∈ ΣR. For any x ∈ ΣR we have
∫
Σ
R
|
x
-
y
|
2
log
|
x
-
y
|
d
s
y
=
∫
Σ
R
|
x
0
-
y
|
2
log
|
x
0
-
y
|
d
s
y
and then u is constant on ΣR. Moreover
Δ
u
(
x
)
=
4
∫
Σ
R
(
1
+
log
|
x
-
y
|
)
d
s
y
and then also Δu is constant on ΣR. Since Δu is harmonic in ΩR and continuous on
Ω
¯
R
, it is constant in ΩR and then
Δ
u
(
x
)
=
Δ
u
(
0
)
=
4
∫
Σ
R
(
1
+
log
|
y
|
)
d
s
y
=
8
π
R
(
1
+
log
R
)
,
x
∈
Ω
R
.
The function u(x) - 2πR(1 + log R) |x|2 is continuous on
Ω
¯
R
, harmonic in ΩR and constant on ΣR. Then it is constant in ΩR and
u
(
x
)
-
2
π
R
(
1
+
log
R
)
|
x
|
2
=
u
(
0
)
=
∫
Σ
R
|
y
|
2
log
|
y
|
d
σ
y
=
2
π
R
3
log
R
.
□
Corollary 4.2. Let ΣR be the circle of radius R centered at the origin. We have
∫
Σ
R
Γ
i
j
(
x
,
y
)
d
s
y
=
δ
i
j
R
4
(
k
+
1
)
(
k
-
2
(
k
+
2
)
log
R
)
,
|
x
|
<
R
.
Proof. Since
∂
11
∫
Σ
R
|
x
-
y
|
2
log
|
x
-
y
|
d
s
y
=
2
∫
Σ
R
log
|
x
-
y
|
d
s
y
+
2
∫
Σ
R
(
x
1
-
y
1
)
2
|
x
-
y
|
2
d
s
y
+
2
π
R
,
formula (32) implies
∫
Σ
R
(
x
1
-
y
1
)
2
|
x
-
y
|
2
d
s
y
=
π
R
,
|
x
|
<
R
.
In a similar way
∫
Σ
R
(
x
2
-
y
2
)
2
|
x
-
y
|
2
d
s
y
=
π
R
,
|
x
|
<
R
.
From (32) we have also
∂
12
∫
Σ
R
|
x
-
y
|
2
log
|
x
-
y
|
d
s
y
=
2
∫
Σ
R
(
x
1
-
y
1
)
(
x
2
-
y
2
)
|
x
-
y
|
2
d
s
y
=
0
,
|
x
|
<
R
.
Keeping in mind the expression (1), (33) follows. □
This corollary shows that, if R = exp[k/(2(k + 2))], we have
∫
Σ
R
Γ
(
x
,
y
)
e
1
d
s
y
=
∫
Σ
R
Γ
(
x
,
y
)
e
2
d
s
y
=
0
,
|
x
|
<
R
.
This implies that in ΩR, for such a value of R, we cannot represent any smooth solution of the system Eu = 0 by means of a simple layer potential.
If there exists some constant vector which cannot be represented in the simply connected domain Ω by a simple layer potential, we say that the boundary of Ω is exceptional. We have proved that
Lemma 4.3. The circle ΣR with R = exp[k/(2(k + 2))] is exceptional for the operator Δ + k∇div.
Due to the results in 28, one can scale the domain in such a way that its boundary is not exceptional.
Here we show that also in some (m + 1)-connected domains one cannot represent any constant vectors by a simple layer potential and that this happens if, and only if, the exterior boundary Σ0 (considered as the boundary of the simply connected domain Ω0) is exceptional.
We note that, if any constant vector c can be represented by a simple layer potential, then any sufficiently smooth solution of the system Eu = 0 can be represented by a simple layer potential as well (see Section 5 below).
We first prove a property of the singular integral system
∫
Σ
φ
j
(
y
)
∂
∂
s
x
Γ
i
j
(
x
,
y
)
d
s
y
=
0
,
x
∈
Σ
,
i
=
1
,
2
.
Lemma 4.4. Let Ω ⊂ ℝ2 be an (m + 1)-connected domain. Denote by P
the eigenspace in [Lp(Σ)]2 of the system (34). Then
dim
P
=
2
(
m
+
1
)
.
Proof. We have
∂
Γ
i
j
∂
s
x
(
x
,
y
)
=
1
2
π
∂
∂
s
x
-
(
k
+
2
)
δ
i
j
2
(
k
+
1
)
log
|
x
-
y
|
+
k
2
(
k
+
1
)
(
y
i
-
x
i
)
(
y
j
-
x
j
)
|
x
-
y
|
2
d
s
y
and, since
∂
∂
s
x
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
2
=
ẋ
i
x
j
-
y
j
|
x
-
y
|
2
+
ẋ
j
x
i
-
y
i
|
x
-
y
|
2
-
2
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
3
∂
∂
s
x
|
x
-
y
|
=
ẋ
i
∂
∂
x
j
log
|
x
-
y
|
+
ẋ
j
∂
∂
x
i
log
|
x
-
y
|
-
2
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
2
∂
∂
s
x
log
|
x
-
y
|
=
2
ẋ
i
ẋ
j
-
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
2
∂
∂
s
x
log
|
x
-
y
|
+
O
(
|
y
-
x
|
h
-
1
)
(the dot denotes the derivative with respect to the arc length on Σ), we find3
∂
∂
s
x
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
2
=
O
(
|
y
-
x
|
h
-
1
)
.
We have proved that4
∂
∂
s
x
Γ
i
j
(
x
,
y
)
=
-
1
2
π
k
+
2
2
(
k
+
1
)
δ
i
j
∂
∂
s
x
log
|
x
-
y
|
+
O
(
|
y
-
x
|
h
-
1
)
and then the system (34) is of regular type (see 1529). From the general theory we know that such a system can be regularized to a Fredholm one. Let us consider now the adjoint system
∫
Σ
φ
j
(
y
)
∂
∂
s
y
Γ
i
j
(
x
,
y
)
d
s
y
=
0
,
x
∈
Σ
,
i
=
1
,
2
.
It is not difficult to see that the index is zero and then systems (34) and (35) have the same number of eigensolutions.
The vectors
e
i
X
Σ
j
(
i
=
1
,
2
,
j
=
0
,
1
,
…
,
m
)
are the only linearly independent eigensolutions of (35). Indeed it is obvious that such vectors satisfy the system (35). On the other hand, if ψ satisfies the system (35) then
∫
Σ
ψ
∂
f
∂
s
d
s
=
0
for any f ∈ [C∞ (ℝ2)]2. This can be siproved by the same method in [13, pp. 189-190]. Therefore ψ has to be constant on each curve Σj (j = 0, ..., m), i.e. ψ is a linear combination of
e
i
X
Σ
j
(
i
=
1
,
2
,
j
=
0
,
1
,
…
,
m
)
. □
Theorem 4.5. Let Ω ⊂ ℝ2 be an (m + 1)-connected domain. The following conditions are equivalent:
I. there exists a Hölder continuous vector function
φ
≢
0
such that
∫
Σ
Γ
(
x
,
y
)
φ
(
y
)
d
s
y
=
0
,
x
∈
Σ
;
II. there exists a constant vector which cannot be represented in Ω by a simple layer potential (i.e., there exists c ∈ ℝ2 such that
c
∉
S
p
);
III. Σ0 is exceptional;
IV. let φ1, ..., φ2m+2 be linearly independent functions of P
and let cjk = (αjk, βjk) ∈ ℝ2 be given by
∫
Σ
Γ
(
x
,
y
)
φ
j
(
y
)
d
s
y
=
c
j
k
,
x
∈
Σ
k
,
j
=
1
,
…
,
2
m
+
2
,
k
=
0
,
1
,
…
,
m
.
Then
det
C
=
0
,
where
C
=
α
1
,
0
⋯
α
2
m
+
2
,
0
⋯
⋯
⋯
α
1
,
m
⋯
α
2
m
+
2
,
m
β
1
,
0
⋯
β
2
m
+
2
,
0
⋯
⋯
⋯
β
1
,
m
⋯
β
2
m
+
2
,
m
.
Proof. I ⇒ II. Let u be the simple layer potential (3) with density φ.
Since u = 0 in Ω, and then on Σk, we find that u = 0 also in Ωk (k = 1, ..., m) in view of a known uniqueness theorem.
On the other hand L+u - L-u = φ on Σ and φ = 0 on Σk, k = 1, ..., m. This means that
∫
Σ
0
Γ
(
x
,
y
)
φ
(
y
)
d
s
y
=
0
,
x
∈
Ω
0
.
If II is not true, we can find two linear independent vector functions ψ1 and ψ2 such that
∫
Σ
Γ
(
x
,
y
)
ψ
j
(
y
)
d
s
y
=
e
j
,
x
∈
Ω
,
j
=
1
,
2
.
Arguing as before, we find ψj = 0 on Σk, k = 1, ..., m, j = 1, 2, and then
∫
Σ
0
Γ
(
x
,
y
)
ψ
j
(
y
)
d
s
y
=
e
j
,
x
∈
Ω
0
,
j
=
1
,
2
.
Since φ, ψ1, ψ2 belong to the kernel of the system
∫
Σ
0
∂
∂
s
x
Γ
(
x
,
y
)
ψ
(
y
)
d
s
y
=
0
,
x
∈
Σ
0
,
Lemma 4.4 shows that they are linearly dependent. Let λ, μ1, μ2 ∈ ℝ such that (λ, μ1, μ2) ≠ (0, 0, 0) and
λ
φ
+
μ
1
ψ
1
+
μ
2
ψ
2
=
0
on
Σ
0
.
This implies
∫
Σ
0
Γ
(
x
,
y
)
(
λ
φ
(
y
)
+
μ
1
ψ
1
(
y
)
+
μ
2
ψ
2
(
y
)
)
d
s
y
=
0
,
x
∈
Ω
0
,
i.e. μ1e1 + μ2e2 = 0, and then μ1 = μ2 = 0. Now (38) leads to λφ = 0 and thus λ = 0, which is absurd.
II ⇒ III. If Σ0 is not exceptional, for any c ∈ ℝ2 there exists ϱ ∈ [Cλ(Σ0)]2 such that
∫
Σ
0
Γ
(
x
,
y
)
ϱ
(
y
)
d
s
y
=
c
,
x
∈
Ω
0
.
Setting
φ
(
y
)
=
ϱ
(
y
)
y
∈
Σ
0
,
0
y
∈
Σ
\
Σ
0
,
we can write
∫
Σ
Γ
(
x
,
y
)
φ
(
y
)
d
s
y
=
c
,
x
∈
Ω
,
and this contradicts II.
III ⇒ IV. Let us suppose
det
C
≠
0
. For any c = (α, β) ∈ ℝ2 there exists λ = (λ1, ..., λ2m+2) solution of the system
∑
j
=
1
2
m
+
2
λ
j
α
j
k
=
α
,
∑
j
=
1
2
m
+
2
λ
j
β
j
k
=
β
,
k
=
0
,
…
,
m
,
i.e.
∑
j
=
1
2
m
+
2
λ
j
c
j
k
=
c
,
k
=
0
,
…
,
m
.
Therefore
∫
Σ
Γ
(
x
,
y
)
∑
j
=
1
2
m
+
2
λ
j
φ
j
(
y
)
d
s
y
=
c
,
x
∈
Σ
.
Arguing as before, this leads to
∑
j
=
1
2
m
+
2
λ
j
φ
j
=
0
on Σk for k = 1, ..., m. Then Σ0 is not exceptional.
IV ⇒ I. From (37) it follows that there exists an eigensolution λ = (λ1, ..., λ2m+2) of the homogeneous system
∑
j
=
1
2
m
+
2
λ
j
c
j
k
=
0
,
k
=
0
,
…
,
m
.
Set
φ
(
x
)
=
∑
j
=
1
2
m
+
2
λ
j
φ
j
(
x
)
.
In view of the linear independence of φ1, ..., φ2m+2, the vector function φ does not identically vanish and it is such that (36) holds. □
Definition 4.6. Whenever n = 2 and Σ0 is exceptional, we say that u belongs to
S
p
if, and only if,
u
(
x
)
=
∫
Σ
Γ
(
x
,
y
)
φ
(
y
)
d
s
y
+
c
,
x
∈
Ω
,
where φ ∈ [Lp(Σ)]2 and c ∈ ℝ2.
5 The Dirichlet problem
The purpose of this section is to represent the solution of the Dirichlet problem in an (m + 1)-connected domain by means of a simple layer potential. Precisely we give an existence and uniqueness theorem for the problem
u
∈
S
p
,
E
u
=
0
in
Ω
,
u
=
f
on
Σ
,
where f ∈ [W1,p(Σ)]n.
We establish some preliminary results.
Theorem 5.1. Given
ω
∈
[
L
1
p
(
Σ
)
]
n
, there exists a solution of the singular integral system
∫
Σ
d
x
[
Γ
(
x
,
y
)
]
φ
(
y
)
d
σ
y
=
ω
(
x
)
,
φ
∈
[
L
p
(
Σ
)
]
n
,
x
∈
Σ
if, and only if,
∫
Σ
γ
∧
ω
i
=
0
,
i
=
1
,
…
,
n
for every
γ
∈
L
n
-
2
q
(
Σ
)
(
q
=
p
/
(
p
-
1
)
)
such that γ is a weakly closed (n - 2)-form.
Proof. Denote by
R
*
:
[
L
n
-
2
q
(
Σ
)
]
n
→
[
L
q
(
Σ
)
]
n
the adjoint of R (see (23)), i.e. the operator whose components are given by
R
j
*
ψ
(
x
)
=
∫
Σ
ψ
i
(
y
)
∧
d
y
[
Γ
i
j
(
x
,
y
)
]
,
x
∈
Σ
.
Thanks to Corollary 3.10, the integral system (41) admits a solution φ ∈ [Lp(Σ)]n if, and only if,
∫
Σ
ψ
i
∧
ω
i
=
0
for any
ψ
=
(
ψ
1
,
…
,
ψ
n
)
∈
[
L
n
-
2
q
(
Σ
)
]
n
such that R*ψ = 0. Arguing as in 13, R*ψ = 0 if, and only if, all the components of ψ are weakly closed (n - 2)-forms. It is clear that (43) is equivalent to conditions (42). □
Lemma 5.2. For any f ∈ [W1,p(Σ)]n there exists a solution of the BVP
w
∈
S
p
,
E
w
=
0
i
n
Ω
,
d
w
=
d
f
o
n
Σ
.
It is given by (3), where the density φ ∈ [Lp(Σ)]n solves the singular integral system Rφ = df with R as in (23).
Proof. Consider the following singular integral system:
∫
Σ
d
x
[
Γ
(
x
,
y
)
]
φ
(
y
)
d
σ
y
=
d
f
(
x
)
,
x
∈
Σ
,
in which the unknown is φ ∈ [Lp(Σ)]n and the datum is
d
f
∈
[
L
1
p
(
Σ
)
]
n
. In view of Theorem 5.1, there exists a solution φ of system (45) because conditions (42) are satisfied. □
In the next result we consider the eigenspace F
of the Fredholm integral system
-
1
2
ψ
(
x
)
+
∫
Σ
L
x
k
∕
(
k
+
2
)
[
Γ
(
x
,
y
)
]
ψ
(
y
)
d
σ
y
=
0
,
x
∈
Σ
.
The dimension of F
is nm. This can be proved as in [30, p. 63], where the case n = 3 is considered.
Theorem 5.3. Given c0, c1, ..., cm ∈ ℝn, there exists a solution of the BVP
v
∈
S
p
,
E
v
=
0
i
n
Ω
,
v
=
c
k
o
n
Σ
k
,
k
=
0
,
…
,
m
.
It is given by
v
(
x
)
=
∑
h
=
1
m
∑
i
=
1
n
(
c
h
i
-
c
0
i
)
∫
Σ
Γ
(
x
,
y
)
Ψ
h
,
i
(
y
)
d
σ
y
+
c
0
,
x
∈
Ω
,
where
Ψ
h
,
i
∈
F
(
h
=
1
,
…
,
m
,
i
=
1
,
…
,
n
)
satisfy the following conditions
∫
Σ
Γ
(
x
,
y
)
Ψ
h
,
i
(
y
)
d
σ
y
=
δ
h
k
e
i
,
x
∈
Ω
¯
k
,
k
=
1
,
…
,
m
.
Proof. Let ψ1, ..., ψnm be nm linearly independent eigensolutions of the space F
. For a fixed j = 1, ..., nm we set
V
j
(
x
)
=
∫
Σ
Γ
(
x
,
y
)
ψ
j
(
y
)
d
σ
y
,
x
∈
Ω
.
Then
L
-
k
∕
(
k
+
2
)
V
j
=
0
on Σ. As in [30, Theorem III, p. 45], this implies that Vj is constant on each connected component of
ℝ
n
\
Ω
¯
. Then Vj = 0 in
ℝ
n
\
Ω
¯
0
5
and
V
j
(
x
)
=
a
j
k
in Ω k (k = 1, ..., m). For every k = 1, ..., m, consider the n × nm matrix
D
k
defined as follows
D
k
=
a
1
,
1
k
a
1
,
2
k
⋯
a
1
,
n
m
k
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
⋯
a
n
,
1
k
a
n
,
2
k
⋯
a
n
,
n
m
k
.
The nm × nm matrix
D
=
(
D
1
…
D
m
)
′
has a not vanishing determinant. Indeed, if
det
D
=
0
, the linear system
D
λ
=
0
admits an eigensolution λ = (λ1, ..., λnm) ∈ ℝnm. Hence the potential
W
(
x
)
=
∑
j
=
1
n
m
λ
j
V
j
(
x
)
vanishes not only on
ℝ
n
\
Ω
¯
0
, but also on Ωk (k = 1, ..., m). Since this implies W = 0 on Σ, we find W = 0 in Ω, thanks to the classical uniqueness theorem for the Dirichlet problem. Accordingly, W = 0 all over ℝn, from which
∑
j
=
1
n
m
λ
j
ψ
j
≡
0
and this is absurd.
For each h = 1, ..., m and i = 1, ..., n, let
(
λ
i
,
1
h
,
…
,
λ
i
,
n
m
h
)
∈
ℝ
n
m
be the solution of the system
∑
j
=
1
n
m
λ
i
,
j
h
a
j
k
=
δ
h
k
e
i
,
k
=
1
,
…
,
m
.
Setting
V
¯
h
,
i
(
x
)
=
∑
j
=
1
n
m
λ
i
,
j
h
V
j
(
x
)
,
x
∈
Ω
,
we get
E
V
¯
h
,
i
=
0
,
V
¯
h
,
i
|
Σ
0
=
0
and
V
¯
h
,
i
|
Σ
k
=
∑
j
=
1
n
m
λ
i
,
j
h
a
j
k
=
δ
h
k
e
i
,
k
=
1
,
…
,
m
.
Put
v
(
x
)
=
∑
h
=
1
m
∑
i
=
1
n
(
c
h
i
-
c
0
i
)
V
¯
h
,
i
(
x
)
+
c
0
.
The potential v belongs to
S
p
, thanks to the isomorphism σ introduced in the proof of Lemma 3.12 (for n = 2 see Definition 4.6). Moreover
v
(
x
)
|
Σ
k
=
∑
h
=
1
m
∑
i
=
1
n
(
c
h
i
-
c
0
i
)
δ
h
k
e
i
+
c
0
,
i.e. v = ck on Σk (k = 0, 1, ..., m). This shows that v is solution of (46). □
We are now in a position to establish the main result of this section.
Theorem 5.4. The Dirichlet problem (40) has a unique solution u for every f ∈ [W1,p(Σ)]n. If n ≥ 3 or n = 2 with Σ0 is not exceptional, u is given by (3). If n = 2 and Σ0 is exceptional, it is given by (39). In any case, the density φ solves the singular system (45).
Proof. Let w be a solution of the problem (44). Since dw = df on Σ, w = f + ch on Σh (h = 0, ..., m) for some ch ∈ ℝn. The function u = w - v, where v is given by (47), solves the problem (40).
In order to show the uniqueness, suppose that (3) is solution of (40) with f = 0. From Corollary 3.10 it follows that the condition u = 0 on Σ implies that
-
1
4
φ
+
T
k
∕
(
k
+
2
)
2
φ
=
0
,
where Tk/(k+2) is the compact operator given by (26). By bootstrap techniques, (48) implies that φ is a Hölder function on Σ. Then u belongs to
[
C
1
,
λ
(
Ω
¯
)
∩
C
2
(
Ω
)
]
n
and we get that
∫
Ω
E
(
u
,
u
)
d
x
=
0
,
from which
E
(
u
,
u
)
=
0
in
Ω
.
The solution of (49) is u(x) = a + Bx, where a ∈ ℝn and B ∈ S n
are arbitrary. Finally, u = 0 in
Ω
̄
by virtue of the classical uniqueness theorem for the Dirichlet problem. □
Remark 5.5. In order to solve the Dirichlet problem (40), we need to solve the singular integral system (45). We know that this system can be reduced to a Fredholm one by means of the operator R'k/(k+2). This reduction is not an equivalent reduction in the usual sense (for this definition see, e.g., [10, p. 19]), because
N
(
R
′
k
∕
(
k
+
2
)
)
≠
{
0
}
,
N
(
R
′
k
∕
(
k
+
2
)
)
being the kernel of the operator R'k/(k+2).
However R'k/(k+2) still provides a kind of equivalence. In fact, as in [31, pp. 253-254], one can prove that
N
(
R
′
k
∕
(
k
+
2
)
R
)
=
N
(
R
)
. This implies that if ψ is such that there exists at least a solution of the equation Rφ = ψ, then Rφ = ψ if, and only if, R'k/(k+2) Rφ = R'k/(k+2)ψ.
Since we know that the system Rφ = df is solvable, we have that Rφ = df if, and only if, φ is solution of the Fredholm system R'k/(k+2)Rφ = R'k/(k+2)df.
Therefore, even if we do not have an equivalent reduction in the usual sense, such Fredholm system is equivalent to the Dirichlet problem (40).
6 The traction problem
The aim of this section is to study the possibility of representing the solution of the traction problem by means of a double layer potential. As we shall see, in an (m + 1)-connected domain this is possible if, and only if, the given forces are balanced on each connected component Σj of the boundary.
More precisely, we consider the problem
w
∈
D
p
,
E
u
=
0
in
Ω
,
L
w
=
f
on
Σ
,
where f ∈ [Lp(Σ)]n is such that
∫
Σ
f
(
x
)
(
a
+
B
x
)
d
σ
x
=
0
,
a
∈
ℝ
n
,
B
∈
S
n
.
We shall prove that, in order to have the existence of a solution of such a problem, condition (51) is not sufficient, but it must be satisfied on each Σj, j = 0, 1, ..., m (see Theorem 6.2 below).
If f satisfies the only condition (51), we need to modify the representation of the solution by adding some extra terms (see Theorem 6.4 below).
Lemma 6.1. Let
w
∈
D
2
be a double layer potential with density ψ ∈ [W1,2(Σ)]n. Then
∫
Ω
E
(
w
,
w
)
d
x
=
∫
Σ
w
L
w
d
σ
.
Proof. Let (ψk)k≥1 be a sequence of functions in [C1,h(Σ)]n (0 < h < λ) such that ψk → ψ in [W1,2(Σ)]n.
Setting
w
k
(
x
)
=
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
ψ
k
(
y
)
d
σ
y
,
we have that
w
k
∈
[
C
1
,
h
(
Ω
¯
)
]
n
, Ewk = 0 and then
∫
Ω
E
(
w
k
,
w
k
)
d
x
=
∫
Σ
w
k
L
w
k
d
σ
.
From ψk → ψ in [L2(Σ)]n, it follows that wk → w in [L2(Σ)]n because of well-known properties of singular integral operators.
On the other hand we have that
K
s
j
(
d
ψ
k
)
→
K
s
j
(
d
ψ
)
in L2(Ω). By applying formula (11), we see that ∇wk → ∇w in [L2(Ω)]n. Moreover, since
K
s
j
(
d
ψ
k
)
→
K
s
j
(
d
ψ
)
also in L2(Σ), (22) shows that Lwk → Lw in [L2(Σ)]n. We get the claim by letting k → +∞ in (53). □
Theorem 6.2. Given f ∈ [Lp(Σ)]n, the traction problem (50) admits a solution if, and only if,
∫
Σ
j
f
(
x
)
(
a
+
B
x
)
d
σ
x
=
0
for every j = 0, 1, ..., m, a ∈ ℝn and B ∈ S n
. The solution is determined up to an additive rigid displacement.
Moreover, (4) is a solution of (50) if, and only if, its density ψ is given by
ψ
(
x
)
=
∫
Σ
Γ
(
x
,
y
)
ϕ
(
y
)
d
σ
y
,
x
∈
Σ
,
ϕ being a solution of the singular integral system
-
1
4
ϕ
+
T
2
ϕ
=
f
,
where T is given by (27).
Proof. Assume that conditions (54) hold. If u is the double layer potential with density ψ ∈ [W1,p(Σ)]n, in view of (22) the boundary condition Lu = f turns into the equation
R
′
(
d
ψ
)
=
f
,
where R' is given by (24) with ξ = 1.
On account of Theorem 5.4, if n = 2 and Σ0 is exceptional, any ψ ∈ [W1,p(Σ)]2 can be written as
∫
Σ
Γ
(
x
,
y
)
ϕ
(
y
)
d
σ
y
+
c
,
with ϕ ∈ [Lp(Σ)]2, c ∈ ℝ2. In all other cases, ψ can be written as (55) with ϕ ∈ [Lp(Σ)]n. In any case, since dψ = Rϕ (R being defined by (23)), we infer R'(dψ) = R'Rϕ. Keeping in mind Lemma 3.9, we find that equation (57) is equivalent to (56), with ψ given by (55).
Therefore there exists a solution of the traction problem (50) if, and only if, the singular integral system (56) is solvable.
On the other hand, there exists a solution γ ∈ [Lp(Σ)]n of the singular integral system
1
2
γ
+
T
γ
=
f
if, and only if, f is orthogonal to
V
-
. In view of Lemma 3.12, this occurs if, and only if, (51) is satisfied. Then conditions (54) imply the existence of a solution of (58).
Consider now the singular integral system
-
1
2
ϕ
+
T
ϕ
=
γ
.
From Lemma 3.11 the dimension of the kernel
N
(
-
I
∕
2
+
T
*
)
=
V
+
is n(n + 1)m/2 and
{
v
h
X
Σ
j
:
j
=
1
,
…
,
m
,
h
=
1
,
…
,
n
(
n
+
1
)
∕
2
}
is a basis of it. The equation (59) has a solution if, and only if,
∫
Σ
j
γ
v
h
d
σ
=
0
,
j
=
1
,
…
,
m
,
h
=
1
,
…
,
n
(
n
+
1
)
∕
2
.
Since γ is solution of (58), conditions (60) are fulfilled. Indeed, picking j = 1, ..., m and h = 1, ..., n(n+1)/2, by integrating (58) on Σj we find (see (31))
∫
Σ
j
f
v
h
d
σ
=
1
2
∫
Σ
j
γ
v
h
d
σ
+
∫
Σ
j
v
h
(
x
)
d
σ
x
∫
Σ
L
x
[
Γ
(
x
,
y
)
]
γ
(
y
)
d
σ
y
=
1
2
∫
Σ
j
γ
v
h
d
σ
+
∫
Σ
γ
(
y
)
d
σ
y
∫
Σ
j
[
L
x
Γ
(
x
,
y
)
]
′
v
h
(
x
)
d
σ
x
=
∫
Σ
j
γ
v
h
d
σ
.
Conditions (60) follow from (54) since the last ones are equivalent to
∫
Σ
j
f
v
h
d
σ
=
0
,
j
=
1
,
…
,
m
,
h
=
1
,
…
,
n
(
n
+
1
)
∕
2
.
Let ϕ be a solution of (59); taking (58) into account, we have that ϕ solves (56) and then the traction problem (50) admits a solution.
Conversely, if u is a solution of (50), from Lemma 3.7, we have that
∫
Σ
j
f
(
x
)
(
a
+
B
x
)
d
σ
x
=
∫
Σ
j
L
+
u
(
x
)
(
a
+
B
x
)
d
σ
x
=
∫
Σ
j
L
-
u
(
x
)
(
a
+
B
x
)
d
σ
x
.
By Lemma 6.1, for any fixed j = 1, ..., m we have
∫
Σ
j
f
(
x
)
(
a
+
B
x
)
d
σ
x
=
∫
Ω
j
E
(
u
,
a
+
B
x
)
d
x
=
0
since E
(
u
,
a
+
B
x
)
=
0
.
Now we pass to discuss the uniqueness. Let u be a solution of (50) with datum f = 0. As we know, the condition L+u = 0 is equivalent to the singular integral system
-
1
4
ϕ
+
T
2
ϕ
=
0
, ϕ being as in (55), which can be written as
-
I
2
+
T
ϕ
2
+
T
ϕ
=
0
.
Set
Ξ
=
ϕ
2
+
T
ϕ
.
Since -Ξ/2+TΞ = 0 and the operator -I/2+T can be reduced to Fredholm one, as shown by Kupradze [1, Chapter IV, §7], Ξ has to be Hölder continuous. By a similar argument, the vector-valued function ϕ, being solution of the singular integral system (61), is Hölder continuous. Therefore the relevant simple layer potential ψ belongs to W1,2(Σ), i.e.
u
∈
D
2
. By applying formula (52), we get that u is a rigid displacement in Ω. □
We remark that, by Theorem 6.2, a solution of the traction problem (50) can be written as a double layer potential if, and only if, conditions (54) are satisfied.
In order to consider the problem (50) under the only condition (51), we introduce the following space.
Definition 6.3. We define
D
̃
p
as the space of all the functions w written as
w
(
x
)
=
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
ψ
(
y
)
d
σ
y
+
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
c
j
h
∫
Σ
j
Γ
(
x
,
y
)
v
h
(
y
)
d
σ
y
,
x
∈
Ω
,
where ψ ∈ [W1,p(Σ)]n, {vh : h = 1, ..., n(n + 1)/2} is an orthonormal basis for R
and cjh ∈ ℝ.
Theorem 6.4. Given f ∈ [Lp(Σ)]n satisfying (51), the traction problem
w
∈
D
̃
p
,
E
w
=
0
i
n
Ω
,
L
w
=
f
o
n
∑
admits a solution given by
w
(
x
)
=
∫
Σ
[
L
y
Γ
(
x
,
y
)
]
′
ψ
(
y
)
d
σ
y
+
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
j
Γ
(
x
,
y
)
v
h
(
y
)
d
σ
y
,
x
∈
Ω
,
where ψ ∈ [W1,p(Σ)]n is solution of the system
R
′
(
d
ψ
)
(
x
)
=
f
(
x
)
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
1
2
v
h
(
x
)
X
Σ
j
(
x
)
+
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
o
n
Σ
.
The solution is uniquely determined up to an additive rigid displacement.
Proof. First observe that
L
x
∫
Σ
j
Γ
(
x
,
y
)
v
h
(
y
)
d
σ
y
=
1
2
v
h
(
x
)
+
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
x
∈
Σ
j
,
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
x
∈
Σ
\
Σ
j
,
for h = 1, ..., n(n + 1)/2 and j = 1, ..., m. If w is given by (63), taking into account (57), we find that Lw = f if, and only if, is (64) satisfied.
Denote by g the right hand side of (64). In view of Theorem 6.2, R'(dψ) = g has a solution if, and only if,
∫
Σ
k
g
v
l
d
σ
=
0
for any k = 0, 1, ..., m, l = 1, ..., n(n + 1)/2. By integrating on Σk (k = 1, ..., m), for every l we get
∫
Σ
k
g
(
x
)
v
l
(
x
)
d
σ
x
=
∫
Σ
k
f
(
x
)
v
l
(
x
)
d
σ
x
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
k
1
2
v
h
(
x
)
X
Σ
j
(
x
)
+
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
v
l
(
x
)
d
σ
x
=
∫
Σ
k
f
(
x
)
v
l
(
x
)
d
σ
x
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
k
1
2
v
h
(
x
)
X
Σ
j
(
x
)
v
l
(
x
)
d
σ
x
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
k
v
l
(
x
)
d
σ
x
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
=
∫
Σ
k
f
(
x
)
v
l
(
x
)
d
σ
x
-
|
Σ
k
|
2
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
δ
j
k
δ
h
l
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
j
v
h
(
y
)
d
σ
y
∫
Σ
k
[
L
x
Γ
(
x
,
y
)
]
′
v
l
(
x
)
d
σ
x
=
1
2
∫
Σ
k
f
(
x
)
v
l
(
x
)
d
σ
x
-
|
Σ
k
|
2
∑
j
=
1
m
δ
j
k
δ
h
l
|
Σ
j
|
∫
Σ
j
f
(
x
)
v
h
(
x
)
d
σ
x
=
0
.
On the other hand
∫
Σ
0
g
(
x
)
v
l
(
x
)
d
σ
x
=
∫
Σ
0
f
(
x
)
v
l
(
x
)
d
σ
x
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
0
1
2
v
h
(
x
)
X
Σ
j
(
x
)
+
∫
Σ
j
L
x
[
Γ
(
x
,
y
)
]
v
h
(
y
)
d
σ
y
v
l
(
x
)
d
σ
x
=
∫
Σ
0
f
(
x
)
v
l
(
x
)
d
σ
x
-
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
j
v
h
(
y
)
d
σ
y
∫
Σ
0
[
L
x
Γ
(
x
,
y
)
]
′
v
l
(
x
)
d
σ
x
=
∫
Σ
0
f
(
x
)
v
l
(
x
)
d
σ
x
+
∑
j
=
1
m
∑
h
=
1
n
(
n
+
1
)
∕
2
1
|
Σ
j
|
∫
Σ
j
f
(
t
)
v
h
(
t
)
d
σ
t
∫
Σ
j
v
l
(
y
)
v
h
(
y
)
d
σ
y
=
∫
Σ
0
f
(
x
)
v
l
(
x
)
d
σ
x
+
∑
j
=
1
m
∫
Σ
j
f
(
x
)
v
l
(
x
)
d
σ
x
=
∫
Σ
f
(
x
)
v
l
(
x
)
d
σ
x
=
0
.
Finally, assume that w is solution of (62) with f = 0. From (63) it follows that
w
∈
D
p
and then w is a rigid displacement in Ω by virtue of the uniqueness proved in Theorem 6.2. □
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
The authors declare that the work was realized in collaboration with same responsibility. All authors read and approved the final manuscript.
Endnotes
1For the definition of internal (external) angular boundary values see, e.g., [23, p. 53].
2If a simple layer potential u, whose density belongs to
W
+
0
, is such that u(x) = c in Ω, then u(x) = c in Ω0. Since u(∞) = 0, we find u(x) = 0 in
ℝ
2
\
Ω
¯
0
and this leads to u = 0 in ℝ2, c = 0.
3It is not difficult to see that
ẋ
i
ẋ
j
-
(
x
i
-
y
i
)
(
x
j
-
y
j
)
|
x
-
y
|
-
2
=
O
(
|
y
-
x
|
h
)
,
x
,
y
∈
Σ
.
4We remark that for n ≥ 3 the formula
d
x
[
Γ
i
j
(
x
,
y
)
]
=
-
1
ω
n
k
+
2
2
(
k
+
1
)
δ
i
j
2
-
n
d
x
[
|
x
-
y
|
2
-
n
]
+
O
(
|
y
-
x
|
h
-
n
+
1
)
is false.
5This is true also for n = 2 because
∫
Σ
ψ
j
d
σ
=
0
.