Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 765020, 12 pages
http://dx.doi.org/10.1155/2013/765020
Research Article
On the Dirichlet Problem for the Stokes System in
Multiply Connected Domains
Alberto Cialdea, Vita Leonessa, and Angelica Malaspina
Department of Mathematics, Computer Science and Economics, University of Basilicata, Viale dell’Ateneo Lucano 10,
85100 Potenza, Italy
Correspondence should be addressed to Alberto Cialdea; cialdea@email.it
Received 24 April 2012; Accepted 28 November 2012
Academic Editor: Chun-Lei Tang
Copyright © 2013 Alberto Cialdea et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The Dirichlet problem for the Stokes system in a multiply connected domain of R𝑛 (𝑛 ≥ 2) is considered in the present paper.
We give the necessary and sufficient conditions for the representability of the solution by means of a simple layer hydrodynamic
potential, instead of the classical double layer hydrodynamic potential.
1. Introduction
Potential theory methods have been employed for a long time
in the study of boundary value problems. In particular they
were widely used in BVPs for the Stokes system, starting from
[1, 2].
Recently some papers have used the integral representations of solutions for studying some BVPs for the Stokes
system also in multiply connected domains [3–8]. All these
papers concern the double layer hydrodynamic potential
approach for the Dirichlet problem and the simple layer
hydrodynamic potential approach for the traction problem.
The aim of the present paper is to investigate a different
integral representation for the Dirichlet problem for the
Stokes system in a multiply connected bounded domain of
R𝑛 (𝑛 ≥ 2). Namely, we consider the simple layer potential
approach for the Dirichlet problem in a domain
𝑚
Ω = Ω0 \ ⋃Ω𝑗 ,
𝑗=1
(1)
where Ω𝑗 (𝑗 = 0, . . . , 𝑚) are suitable domains with connected
boundaries in 𝐶1,𝜆 , 𝜆 ∈ (0, 1].
We use a new method which hinges on a singular integral
system in which the unknown is a usual vector valued
function, while the data is a vector whose components are
differential forms.
The paper is organized as follows. In Section 2 we give
an outlook of the method with a brief description of some
previous results.
After the preliminary Section 3, in Section 4 we study
in detail the case 𝑛 = 2, where some particular phenomena
appear.
Section 5 is devoted to determine the eigenspace of a
certain singular integral system in which the unknowns are
differential forms of degree 𝑛 − 2 on 𝜕Ω. In the same section,
we recall some known results concerning the eigenspaces of
some classical integral systems.
In Section 6 we construct a left reduction for the singular
integral system under study. Such a singular integral system is
equivalent in a precise sense to the Fredholm system obtained
through the reduction.
Finally, in the last section, we find the solution of
the Dirichlet problem for the Stokes system in a multiply
connected domain by means of a simple layer hydrodynamic
potential.
The main result is that, given 𝑓 ∈ [𝑊1,𝑝 (𝜕Ω)]𝑛 , we can
represent the solution of the Dirichlet problem
𝜇Δ𝑣 = ∇𝑟 in Ω,
div 𝑣 = 0
𝑣=𝑓
in Ω,
on 𝜕Ω,
(2)
2
Abstract and Applied Analysis
by means of a simple layer hydrodynamic potential if, and
only if, the conditions
∫
𝜕Ω𝑗
𝑓 ⋅ 𝜈𝑑𝜎 = 0,
𝑗 = 0, 1, . . . , 𝑚
(3)
are satisfied (𝜈 being the outwards unit normal on 𝜕Ω).
Moreover, if the data 𝑓 satisfies only the condition
∫ 𝑓 ⋅ 𝜈𝑑𝜎 = 0
(4)
𝜕Ω
(which is necessary for the existence of a solution of the
Dirichlet problem (2)) we show how to modify the integral
representation of the solution (see Theorem 23).
∫ 𝑑𝑔 ∧ ℎ = 0
Σ
(9)
𝑞
2. Sketch of the Method
The aim of this section is to give a better understanding of the
method we are going to use in the present paper.
We will do that by considering the Dirichlet problem for
Laplace equation in a bounded simply connected domain Ω ⊂
R𝑛 , whose boundary we denote by Σ as follows:
Δ𝑢 = 0 in Ω,
𝑢=𝑔
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 on 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.,
[10, 11]).
Let us denote by 𝑆𝜑 the left hand side of (8). In [9] a reducing operator 𝑆󸀠 was explicitly constructed. This implies that
there exists a solution of (8) if, and only if, the compatibility
conditions
(5)
on Σ.
Suppose that 𝑔 ∈ 𝑊1,𝑝 (Σ), 1 < 𝑝 < ∞. If we want to
find the solution in the form of a simple layer potential whose
density belongs to 𝐿𝑝 (Σ), we have to solve an integral equation
of the first kind on Σ as follows:
∫ 𝜑 (𝑦) 𝑠 (𝑥, 𝑦) 𝑑𝜎𝑦 = 𝑔 (𝑥) ,
Σ
𝑥 ∈ Σ,
(6)
where 𝑠(𝑥, 𝑦) is the fundamental solution of Laplace equation
1
1
{
if 𝑛 = 2,
− log 󵄨󵄨
󵄨,
{
{ 2𝜋
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
𝑠 (𝑥, 𝑦) = {
1
1
{
{−
󵄨𝑛−2 , if 𝑛 ≥ 3.
󵄨
󵄨
{ 𝜔𝑛 (𝑛 − 2) 󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
(7)
In [9] a new method for discussing such an equation was
proposed. Namely, the first step is to consider the differential
(in the sense of the theory of differential forms) of both sides
in (6). In this way we obtain the equation
∫ 𝜑 (𝑦) 𝑑𝑥 [𝑠 (𝑥, 𝑦)] 𝑑𝜎𝑦 = 𝑑𝑔 (𝑥) ,
Σ
𝑝
𝑥 ∈ Σ,
(8)
in which we look for a solution 𝜑 ∈ 𝐿 (Σ).
The integral on the left hand side is a singular integral
and it can be considered as a linear and continuous operator
𝑝
𝑝
from 𝐿𝑝 (Σ) to 𝐿 1 (Σ) (we denote by 𝐿 ℎ (Σ) the space of the
differential forms of degree ℎ whose coefficients belong to
𝐿𝑝 (Σ) in every local coordinate system).
It must be remarked that, if 𝑛 ≥ 3, the space in which we
look for the solution of (8) and the space in which the data is
given are different.
We recall that, if 𝐵 and 𝐵󸀠 are two Banach spaces and
𝑆 : 𝐵 → 𝐵󸀠 is a continuous linear operator, 𝑆 can be reduced
are satisfied for any ℎ ∈ 𝐿 𝑛−2 (Σ) (𝑞 = 𝑝/(𝑝 − 1)) such
that 𝑆∗ ℎ = 0. Moreover one can show that 𝑆∗ ℎ = 0 if, and
only if, ℎ is a weakly closed form. Therefore the compatibility
conditions (9) are satisfied, and there exists a solution 𝜑 ∈
𝐿𝑝 (Σ) of (8).
A left reduction is said to be equivalent if N(𝑆󸀠 ) = {0},
where N(𝑆󸀠 ) denotes the kernel of 𝑆󸀠 (see, e.g., [11, page
19-20]). Obviously this means that 𝑆𝑥 = 𝑦 if, and only if,
𝑆󸀠 𝑆𝑥 = 𝑆󸀠 𝑦. In [12] it was remarked that if N(𝑆󸀠 𝑆) = N(𝑆),
we still have a kind of equivalence. Indeed the coincidence of
these two kernels implies the following fact: if 𝑦 is such that
the equation 𝑆𝑥 = 𝑦 is solvable, then this equation is satisfied
if, and only if, 𝑆󸀠 𝑆𝑥 = 𝑆󸀠 𝑦.
Since N(𝑆󸀠 𝑆) = N(𝑆), then we have (8) equivalent to the
Fredholm equation 𝑆󸀠 𝑆𝜑 = 𝑆󸀠 (𝑑𝑔). These results lead to a
simple layer potential theory for the Dirichlet problem (5).
As a consequence one can obtain also a double layer representation for the Neumann problem for Laplace equation
[12].
A characteristic of this method is that it uses neither
the theory of pseudodifferential operators nor the concept of
hypersingular integrals.
This method has been used also for studying other BVPs.
In particular in [13] it was used to study the Dirichlet and the
Neumann problems in multiply connected domains. Among
other things, an interesting by-product of these results was
obtained as follows (see [13, Theorem 6.1]).
Let 𝑢 be a harmonic function of class 𝐶1 (Ω), where Ω
is the multiple connected domain (1). There exists a 2-form 𝑣
conjugate to 𝑢 in Ω if, and only if,
∫
𝜕Ω𝑗
𝜕𝑢
𝑑𝜎 = 0,
𝜕𝜈
𝑗 = 0, 1, . . . , 𝑚.
(10)
An explicit integral expression for 𝑣 was also given. We
recall that the 2-form 𝑣 is conjugate to 𝑢 if 𝑑𝑢 = 𝛿𝑣, 𝑑𝑣 = 0.
The method has been applied to different BVPs for several
PDEs (see [12–19]).
3. Preliminaries
In this paper Ω denotes an (𝑚 + 1)-connected domain of R𝑛
(𝑛 ≥ 2), that is an open-connected set of the form (1), where
each Ω𝑗 (𝑗 = 0, . . . , 𝑚) is a bounded domain of R𝑛 with
Abstract and Applied Analysis
3
connected boundaries Σ𝑗 ∈ 𝐶1,𝜆 (𝜆 ∈ (0, 1]), and such that
Ω𝑗 ⊂ Ω0 and Ω𝑗 ∩ Ω𝑘 = 0, 𝑗, 𝑘 = 1, . . . , 𝑚, 𝑗 ≠ 𝑘. Let 𝜈 be
the outwards unit normal on the boundary Σ = 𝜕Ω.
We consider the classical Stokes system for the incompressible viscous fluid
𝜇Δ𝑢 = ∇𝑝,
div 𝑢 = 0,
in Ω,
1
1
{
{
[𝛿𝑖𝑗 log 󵄨󵄨
−
󵄨
{
{
4𝜋𝜇
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
{
{
{
{
{
{
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
{
{
{
+
],
{
󵄨2
󵄨󵄨
{
{
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
𝛾𝑖𝑗 (𝑥, 𝑦) = {
𝛿𝑖𝑗
{
1
1
{
{
−
[
{
{
󵄨𝑛−2
󵄨
{
󵄨
2𝜔
𝜇
𝑛
−
2
{
𝑛
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
{
{
{
{
{
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
{
{
{
+
],
󵄨𝑛
󵄨󵄨
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
{
if 𝑛 = 2,
if 𝑛 ≥ 3,
(12)
(13)
𝑗 = 1, . . . , 𝑛.
(14)
Σ
𝑖 = 1, . . . , 𝑛, 𝑥 ∈ R𝑛 ,
(15)
𝑟 (𝑥) = − ∫ 𝜀𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦 ,
Σ
𝑞 (𝑥) = 2𝜇 ∫
Σ
𝜕
[𝜀 (𝑥, 𝑦)] 𝜓𝑗 (𝑦) 𝑑𝜎𝑦 ,
𝜕𝜈𝑦 𝑗
𝑥 ∈ R𝑛
(18)
is the double layer hydrodynamic potential with density 𝜓.
4. On the Bidimensional Case
It is wellknown that there are some exceptional plane
domains in which no every harmonic function can be
represented by a simple layer potential. The simplest example
of this kind is given by the unit disk, for which one has
󵄨
󵄨
log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 𝑑𝑠𝑦 = 0,
|𝑥| < 1.
(19)
It is also known that such domains do not occur in higher
dimensions. For similar questions for the Laplace equation
and the elasticity system, see [13, Section 3] and [16, Section
4], respectively.
In this section we show that also for the Stokes system
there are similar domains. We say that the boundary of the
domain Ω is exceptional if there exists some constant vector
which cannot be represented in Ω by a simple layer potential.
Denoting by Σ𝑅 the circle of radius 𝑅 centered at the
origin, we have the following lemma.
Lemma 1. The circle Σ𝑅 with 𝑅 = exp(1/2) is exceptional for
the Stokes system.
Proof. Keeping in mind that (see, e.g., [16, Section 4])
Through this paper, 𝑝 indicates a real number such that
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
𝑝
is any nonnegative integer, 𝐿 ℎ (Σ) is the vector space of
all differential forms of degree ℎ (briefly ℎ-forms) 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
𝑝
space [𝐿 ℎ (Σ)]𝑛 is constituted by the vectors (𝑣1 , . . . , 𝑣𝑛 ) such
𝑝
that 𝑣𝑗 is a differential form of 𝐿 ℎ (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).
The pair (𝑣, 𝑟) with components
𝑣𝑖 (𝑥) = − ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦 ,
𝑖 = 1, . . . , 𝑛, 𝑥 ∈ R𝑛 ,
(17)
|𝑦|=1
(𝑖, 𝑗 = 1, . . . , 𝑛), 𝜔𝑛 being the hypersurface measure of the
unit sphere in R𝑛 . For a solution (𝑢, 𝑝) of (11) we consider the
following classical boundary operators:
𝑇𝑗󸀠 𝑢 = [𝛿𝑖𝑗 𝑝 + 𝜇 (𝜕𝑗 𝑢𝑖 + 𝜕𝑖 𝑢𝑗 )] 𝜈𝑖 ,
Σ
∫
1 𝑥𝑗 − 𝑦𝑗
,
𝜔𝑛 󵄨󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨󵄨𝑛
𝑇𝑗 𝑢 = [−𝛿𝑖𝑗 𝑝 + 𝜇 (𝜕𝑗 𝑢𝑖 + 𝜕𝑖 𝑢𝑗 )] 𝜈𝑖 ,
󸀠
𝑤𝑖 (𝑥) = ∫ 𝑇𝑗,𝑦
[𝛾𝑖 (𝑥, 𝑦)] 𝜓𝑗 (𝑦) 𝑑𝜎𝑦 ,
(11)
where the unknowns 𝑢 = (𝑢1 , . . . , 𝑢𝑛 ) and 𝑝 = 𝑝(𝑥) are
the velocity and pressure of the fluid flow, respectively, and
the constant 𝜇 > 0 is the kinematic viscosity of the fluid. A
fundamental solution for this system is given by the pair of
fundamental velocity tensor and its associated pressure vector
𝜀𝑗 (𝑥, 𝑦) = −
is the simple layer hydrodynamic potential with density
𝜑.
The pair (𝑤, 𝑞) with components
𝑥∈R
𝑛
(16)
󵄨
󵄨
∫ log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 𝑑𝑠𝑦 = 2𝜋𝑅 log 𝑅,
Σ
𝑅
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
𝑑𝑠𝑦 = 𝛿𝑖𝑗 𝜋𝑅,
∫
󵄨2
󵄨󵄨
Σ𝑅
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
(20)
|𝑥| < 𝑅,
we find
∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝑑𝑠𝑦 =
Σ𝑅
𝑅
𝛿 (2 log 𝑅 − 1) ,
4𝜇 𝑖𝑗
|𝑥| < 𝑅.
(21)
Taking 𝑅 = exp(1/2) we obtain the result.
Let us consider now the exceptional boundaries of not
simply connected domains.
Proposition 2. Let Ω ⊂ R2 be an (𝑚 + 1)-connected domain.
Denote by P the eigenspace in [𝐿𝑝 (Σ)]2 of the singular integral
system
∫ 𝜑𝑗 (𝑦)
Σ
𝜕
𝛾 (𝑥, 𝑦) 𝑑𝑠𝑦 = 0,
𝜕𝑠𝑥 𝑖𝑗
Then dim P = 2(𝑚 + 1).
𝑎.𝑒. 𝑥 ∈ Σ, 𝑖 = 1, 2. (22)
4
Abstract and Applied Analysis
Proof. As in the proof of [16, Lemma 12], one can show that
𝜕
𝜕
1
󵄨ℎ−1
󵄨
󵄨
󵄨
𝛾 (𝑥, 𝑦) =
log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 + O (󵄨󵄨󵄨𝑦 − 𝑥󵄨󵄨󵄨 ) ,
𝛿
𝜕𝑠𝑥 𝑖𝑗
4𝜋𝜇 𝑖𝑗 𝜕𝑠𝑥
(23)
deduce that system (22) can be regularized to a Fredholm one,
and see that its index is zero. Since the vectors 𝑒𝑖 𝜒Σ𝑗 (by 𝜒𝑋 we
denote the characteristic function of the set 𝑋) (𝑖 = 1, 2, 𝑗 =
0, 1, . . . , 𝑚) are the only eigensolutions of the adjoint system
𝜕
𝛾 (𝑥, 𝑦) 𝑑𝑠𝑦 = 0,
∫ 𝜑𝑗 (𝑦)
𝜕𝑠𝑦 𝑖𝑗
Σ
a.e. 𝑥 ∈ Σ, 𝑖 = 1, 2, (24)
Theorem 3. Let Ω ⊂ R2 be an (𝑚 + 1)-connected domain. The
following conditions are equivalent
(1) There exists a Hölder continuous vector function 𝜑 ≢ 0
such that
∫ 𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝑠𝑦 = 0,
𝑥 ∈ Σ.
(3) Σ0 is exceptional.
(4) Let 𝜑1 , . . . , 𝜑2𝑚+2 be linearly independent vectors of P
(see Proposition 2), and let 𝑐𝑗𝑘 = (𝛼𝑗𝑘 , 𝛽𝑗𝑘 ) ∈ R2 be
given by
∫ 𝛾 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝑠𝑦 = 𝑐𝑗𝑘 ,
Σ
(26)
(27)
5. Some Eigenspaces
We determine the structure of the kernel of a particular
singular integral system. Namely, let us denote by N𝑝 the
𝑝
space of 𝜓 ∈ [𝐿 𝑛−2 (Σ)]𝑛 such that
Σ
a.e. on Σ, 𝑖 = 1, . . . , 𝑛.
We begin by proving the following result.
𝑠 (𝑥, 𝑦) 𝑑𝑦,
(29)
where 𝛾(𝑥, 𝑦) and 𝑠(𝑥, 𝑦) are given by (12) and (7), respectively.
Proof. By the well-known Stokes identity we have
𝑢𝑖 (𝑥) = ∫ Δ𝑢𝑖 (𝑦) 𝑠 (𝑥, 𝑦) 𝑑𝑦 = ∫ Δ𝑢𝑗 (𝑦) 𝛿𝑖𝑗 𝑠 (𝑥, 𝑦) 𝑑𝑦.
R𝑛
(30)
Since, for every 𝑛 ≠ 2, 4,
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
𝜕2 󵄨󵄨
1
󵄨4−𝑛
=
󵄨𝑥 − 𝑦󵄨󵄨󵄨
󵄨󵄨𝑛
󵄨󵄨
(4 − 𝑛) (2 − 𝑛) 𝜕𝑦𝑖 𝜕𝑦𝑗 󵄨
󵄨󵄨𝑥 − 𝑦󵄨󵄨
− 𝛿𝑖𝑗 𝜔𝑛 𝑠 (𝑥, 𝑦) ,
𝛾𝑖𝑗 (𝑥, 𝑦) =
𝛿𝑖𝑗
2𝜇
𝑠 (𝑥, 𝑦) −
1 (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
,
󵄨𝑛
󵄨󵄨
2𝜇𝜔𝑛
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
∀𝑛 ≥ 2,
we can rewrite
𝛿𝑖𝑗
1
𝜕2
1
𝛾𝑖𝑗 (𝑥, 𝑦) =
𝑠 (𝑥, 𝑦) −
𝜇
2𝜇𝜔𝑛 (4 − 𝑛) (2 − 𝑛) 𝜕𝑦𝑖 𝜕𝑦𝑗
(31)
(32)
× 󵄨󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨󵄨4−𝑛 .
Then
𝜕2 󵄨󵄨
1
󵄨4−𝑛
󵄨𝑥−𝑦󵄨󵄨󵄨 ,
2𝜔𝑛 (4−𝑛) (2−𝑛) 𝜕𝑦𝑖 𝜕𝑦𝑗 󵄨
𝑢𝑖 (𝑥) = 𝜇 ∫ Δ𝑢𝑗 (𝑦) 𝛾𝑖𝑗 (𝑥, 𝑦) 𝑑𝑦
+
Proof. The proof runs as in [16, Theorem 1] with obvious
modifications. We omit the details.
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0,
𝜕𝑦𝑖 𝜕𝑦𝑗
R𝑛
Then det C = 0, where
𝛼2𝑚+2,0
⋅⋅⋅
𝛼2𝑚+2,𝑚 )
.
𝛽2𝑚+2,0
⋅⋅⋅
𝛽2𝑚+2,𝑚 )
𝜕2 𝑢𝑗 (𝑦)
R𝑛
𝛿𝑖𝑗 𝑠 (𝑥, 𝑦) = 𝜇𝛾𝑖𝑗 (𝑥, 𝑦)+
𝑥 ∈ Σ𝑘 , 𝑗 = 1, . . . , 2𝑚 + 2, 𝑘 = 0, 1, . . . , 𝑚.
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
⋅⋅⋅
R𝑛
(25)
(2) There exists a constant vector which cannot be represented in Ω by a simple layer potential;
𝛼1,0
⋅⋅⋅
𝛼
C = ( 1,𝑚
𝛽1,0
⋅⋅⋅
𝛽
( 1,𝑚
𝑢𝑖 (𝑥) = 𝜇 ∫ Δ𝑢𝑗 (𝑦) 𝛾𝑖𝑗 (𝑥, 𝑦) 𝑑𝑦 + ∫
R𝑛
we have dim P = 2(𝑚 + 1).
Σ
Lemma 4. Let 𝑢 ∈ [𝐶0∞ (R𝑛 )]𝑛 . Then, for any 𝑥 ∈ R𝑛 ,
(28)
1
𝜕2 󵄨󵄨
󵄨4−𝑛
∫ Δ𝑢𝑖 (𝑦)
󵄨󵄨𝑥−𝑦󵄨󵄨󵄨 𝑑𝑦.
𝑛
2𝜔𝑛 (4−𝑛) (2−𝑛) R
𝜕𝑦𝑖 𝜕𝑦𝑗
(33)
Integrating by parts, it follows that the last integral is equal to
𝜕2 𝑢𝑗 (𝑦) 󵄨
1
󵄨4−𝑛
Δ 󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
𝑑𝑦
∫
2𝜔𝑛 (4 − 𝑛) (2 − 𝑛) R𝑛 𝜕𝑦𝑖 𝜕𝑦𝑗 𝑦 󵄨
=∫
R𝑛
𝜕2 𝑢𝑗 (𝑦)
𝜕𝑦𝑖 𝜕𝑦𝑗
(34)
𝑠 (𝑥, 𝑦) 𝑑𝑦,
since Δ 𝑦 |𝑥 − 𝑦|4−𝑛 = 2(4 − 𝑛)|𝑥 − 𝑦|2−𝑛 . Then the claim holds
for 𝑛 ≠ 2, 4.
In the same manner it is possible to show formula (29) for
𝑛 = 2 and 𝑛 = 4 after observing that, if 𝑛 = 2, we have
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 ) 1 𝜕2 󵄨
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨2 log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
=
󵄨
󵄨
󵄨
󵄨
󵄨󵄨2
󵄨󵄨
2
𝜕𝑦
𝜕𝑦
𝑖 𝑗
󵄨󵄨𝑥 − 𝑦󵄨󵄨
󵄨 1
󵄨
− 𝛿𝑖𝑗 log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 − 𝛿𝑖𝑗 ,
2
(35)
Abstract and Applied Analysis
5
Δ 𝑦 |𝑥 − 𝑦|2 log |𝑥 − 𝑦| = 4(log |𝑥 − 𝑦| + 1), while, for 𝑛 = 4,
(𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
𝛿𝑖𝑗
1 𝜕2
󵄨
󵄨
=
+
log 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 ,
4
2
󵄨󵄨
󵄨
󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
󵄨
2
𝜕𝑦
𝜕𝑦
2󵄨󵄨𝑥 − 𝑦󵄨󵄨
𝑖 𝑗
󵄨
󵄨
(36)
Taking the exterior angular boundary value (for the
definition of internal (external) angular boundary values see,
e.g., [20, page 53] or [21, page 293]), we have
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0
Σ𝑘
2
Δ 𝑦 log |𝑥 − 𝑦| = 2/|𝑥 − 𝑦| .
𝑝
Lemma 5. Let 𝜁1 , . . . , 𝜁𝑛 be differential forms in 𝐿 𝑛−2 (Σ) such
that 𝑑𝜁𝑗 = (−1)𝑛−1 𝜈𝑗 𝑑𝜎 on Σ. One has 𝜓 ∈ N𝑝 if, and only if,
𝑚
𝜓𝑗 = ∑ 𝑐ℎ 𝜒Σℎ 𝜁𝑗 + 𝜂𝑗 ,
where 𝑐0 , . . . , 𝑐𝑚 ∈ R and 𝜂1 . . . , 𝜂𝑛 are weakly closed forms
𝑝
belonging to 𝐿 𝑛−2 (Σ).
Proof. It is easy to construct the differential forms 𝜁1 , . . . ,
𝜁𝑛 . For example, one can take the restriction on Σ of
the following forms: 𝜁1 = (−1)𝑛−1 𝑥2 𝑑𝑥3 ⋅ ⋅ ⋅ 𝑑𝑥𝑛 , 𝜁𝑗 =
(−1)𝑛−𝑗 𝑥1 𝑑𝑥2 ⋅ ⋅ ⋅ 𝑗̂ ⋅ ⋅ ⋅ 𝑑𝑥𝑛 (𝑗 = 2, . . . , 𝑛). We remark that (37)
holds if, and only if, the weak differentials 𝑑𝜓𝑗 exist and
𝑗 = 1, . . . , 𝑛,
also in Ω𝑘 . Summing over 𝑘 we find
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0,
Σ
(38)
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0,
Σ
that is,
𝑚
Σ
Σℎ
ℎ=0
(46)
for every 𝑥 ∈ R𝑛 \ Σ and a.e. on Σ. In particular 𝜓 is the
solution of the singular integral system (28).
Conversely, suppose (28) holds. Arguing again as in [9,
pages 189-190], from (28) it follows that
ℎ=0
∫ 𝜓𝑗 ∧ 𝑑𝑢𝑗 = ∑ 𝑐ℎ ∫ 𝑢 ⋅ 𝜈𝑑𝜎,
(45)
Σ𝑘
(37)
ℎ=0
𝑚
a.e. on Σ𝑘 . Arguing as in [9, pages 189-190], this implies that
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0
𝑗 = 1, . . . , 𝑛,
𝑑𝜓𝑗 = (−1)𝑛−1 ∑ 𝑐ℎ 𝜒Σℎ 𝜈𝑗 𝑑𝜎,
(44)
𝑛
∀𝑢 ∈ [𝐶0∞ (R𝑛 )] . (39)
Since 𝜀𝑗 (𝑥, 𝑦)
=
𝑥 ∉ Σ.
(47)
−𝜕𝑥𝑗 𝑠(𝑥, 𝑦), system (11) implies that
Δ 𝑥 [𝛾𝑖𝑗 (𝑥, 𝑦)] = −(1/𝜇)(𝜕2 /𝜕𝑥𝑖 𝜕𝑥𝑗 )𝑠(𝑥, 𝑦). Hence,
Let us prove that (39) holds if, and only if,
∫ 𝜓𝑗 ∧ 𝑑𝑢𝑗 = 𝑐𝑘 ∫ 𝑢 ⋅ 𝜈𝑑𝜎,
Σ𝑘
Σ𝑘
𝑛
∀𝑢 ∈ [𝐶0∞ (R𝑛 )] ,
𝜕2
∫ 𝜓 (𝑦) ∧ 𝑑𝑦 [𝑠 (𝑥, 𝑦)] = 0,
𝜕𝑥𝑖 𝜕𝑥𝑗 Σ 𝑗
(40)
𝑘 = 0, . . . , 𝑚.
𝑥 ∉ Σ.
(48)
Therefore, there exist some constants 𝑎0 , 𝑎1 , . . . , 𝑎𝑚 such that
It is obvious that (40) implies (39).
Conversely, suppose that (39) is true. Define 𝑈𝑘𝜀 = {𝑥 ∈
R𝑛 | dist(𝑥, Σ𝑘 ) < 𝜀}, where 0 < 𝜀 < min0≤ℎ<𝑘≤𝑚 dist(Σℎ , Σ𝑘 ).
Let 𝑣𝑘 ∈ 𝐶0∞ (𝑈𝑘𝜀 ) be such that 𝑣𝑘 = 1 in 𝑈𝑘𝜀/2 . Since 𝑣𝑘 𝑢 ∈
[𝐶0∞ (R𝑛 )]𝑛 , we may write
−𝑎
{
{ ℎ
𝜕𝑗 Ψ𝑗 (𝑥) = {−𝑎0
{
{0
𝑥 ∈ Ωℎ , ℎ = 1, . . . , 𝑚,
𝑥 ∈ Ω,
𝑥 ∈ R𝑛 \ Ω,
(49)
𝑚
∫ 𝜓𝑗 ∧ 𝑑 (𝑣𝑘 𝑢𝑗 ) = ∑ 𝑐ℎ ∫ 𝑣𝑘 𝑢 ⋅ 𝜈𝑑𝜎,
Σ
ℎ=0
Σℎ
(41)
where
and (40) follows immediately.
Suppose now that (39) is true. From (40) it follows that
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 𝑐𝑘 ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜈𝑗 (𝑦) 𝑑𝜎𝑦 ,
Σ𝑘
Σ𝑘
∀𝑥 ∉ Σ𝑘 .
(42)
An integration by parts shows that
∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] = 0,
Σ𝑘
Ψ𝑗 (𝑥) = ∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝑠 (𝑥, 𝑦)] .
Σ
Then, on account of Lemma 4, for every 𝑢 ∈ [𝐶0∞ (R𝑛 )]𝑛 ,
∫ 𝜓𝑗 ∧ 𝑑𝑢𝑗 = 𝜇 ∫ Δ𝑢𝑗 (𝑥) 𝑑𝑥 ∫ 𝜓𝑗 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)]
Σ
R𝑛
+∫
∀𝑥 ∉ Ω𝑘 .
(43)
(50)
R𝑛
Σ
2
𝜕
𝑢 (𝑥) 𝑑𝑥∫ 𝜓𝑗 (𝑦)∧𝑑𝑦 [𝑠 (𝑥, 𝑦)] .
𝜕𝑥𝑖 𝜕𝑥𝑗 𝑖
Σ
(51)
6
Abstract and Applied Analysis
The first term of the right hand side vanishes because of
(47). As far as the second one is concerned, integrating by
parts we get
∫
R𝑛
𝜕2
𝑢 (𝑥) Ψ𝑗 (𝑥) 𝑑𝑥
𝜕𝑥𝑖 𝜕𝑥𝑗 𝑖
𝑚
= ∑∫
ℎ=1 Ωℎ
Σ
1
W± = {𝜑𝑘 ∈ 𝐿𝑝 (Σ) : ∓ 𝜑𝑘 (𝑥)
2
+ ∫ 𝐹𝑖𝑘 (𝑦, 𝑥) 𝜑𝑖 (𝑦) 𝑑𝜎𝑦 = 0, 𝑘 = 1, . . . , 𝑛} ,
𝜕2
𝑢𝑖 (𝑥) Ψ𝑗 (𝑥) 𝑑𝑥
+∫
Ω 𝜕𝑥𝑖 𝜕𝑥𝑗
Σ
𝜕
𝑢 (𝑥) Ψ𝑗 (𝑥) 𝑑𝑥
𝜕𝑥𝑖 𝜕𝑥𝑗 𝑖
𝑚
(52)
𝑚
ℎ=1 Σℎ
ℎ=1 Ωℎ
(55)
𝑛 (𝑥𝑘 − 𝑦𝑘 ) (𝑥𝑖 − 𝑦𝑖 ) (𝑥𝑗 − 𝑦𝑗 )
𝜈
(𝑦)
.
𝑗
𝑛+2
󵄨
󵄨󵄨
𝜔𝑛
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
+ ∫ 𝜕𝑖 𝑢𝑖 Ψ𝑗 𝜈𝑗 𝑑𝜎 − ∫ 𝜕𝑖 𝑢𝑖 𝜕𝑗 Ψ𝑗 𝑑𝑥
For the proofs of the following two results see [7, Lemma 3.3]
and [8, Theorem 3.2], respectively.
− ∫ 𝜕𝑖 𝑢𝑖 Ψ𝑗 𝜈𝑗 𝑑𝜎 + ∫
Proposition 7. The sets V+ and W− are linear subspaces of
𝐿1 (Σ) and
Σ
Ω
R𝑛 \Ω0
Σ0
𝜕𝑖 𝑢𝑖 𝜕𝑗 Ψ𝑗 𝑑𝑥
dim (V+ ) = dim (W− ) = 1 +
𝑚
= ∑ ∫ 𝜕𝑖 𝑢𝑖 𝜕𝑗 Ψ𝑗 𝑑𝑥 − ∫ 𝜕𝑖 𝑢𝑖 𝜕𝑗 Ψ𝑗 𝑑𝑥.
ℎ=1 Ωℎ
Ω
𝜕2
𝑢 (𝑥) Ψ𝑗 (𝑥) 𝑑𝑥
𝜕𝑥𝑖 𝜕𝑥𝑗 𝑖
𝑚
= − ∑ 𝑎ℎ ∫ 𝜕𝑖 𝑢𝑖 𝑑𝑥 + 𝑎0 ∫ 𝜕𝑖 𝑢𝑖 𝑑𝑥
ℎ=1
Ωℎ
Ω
(53)
𝑚
= ∑ 𝑎ℎ ∫ 𝑢 ⋅ 𝜈𝑑𝜎 + 𝑎0 ∫ 𝑢 ⋅ 𝜈𝑑𝜎
Σℎ
ℎ=1
𝑛 (𝑛 + 1) 𝑚
.
2
(56)
A basis of W− is expressed by the fields {𝜓𝑖ℎ , 𝜈 : 𝑖 =
1, . . . , 𝑛(𝑛 + 1)/2, ℎ = 1, . . . , 𝑚}. The simple layer potentials
𝑣𝑖ℎ whose densities are 𝜓𝑖𝑘 such that: 𝑣𝑖ℎ |Ω𝑘 = 𝛿ℎ𝑘 𝜌𝑖 ,
𝑖 = 1, . . . , 𝑛(𝑛 + 1)/2, ℎ, 𝑘 = 1, . . . , 𝑚, where 𝜌𝑖 are rigid
displacement in R𝑛 , specifically 𝜌𝑖 (𝑥) = 𝑒𝑖 , 𝑖 = 1, . . . , 𝑛, and,
for 𝑖 = 𝑛 + 1, . . . , 𝑛(𝑛 + 1)/2, 𝜌𝑖 (𝑥) = (𝑒ℎ ∧ 𝑒𝑘 )𝑥, ℎ =
1, . . . , 𝑛 − 1, 𝑘 = ℎ + 1, . . . , 𝑛, ℎ[𝑛 − (ℎ + 1)/2] + 𝑘 = 𝑖.
In addition, every 𝜓 ∈ W− has the property that 𝑣|Σ0 = 0,
where 𝑣 is the simple layer potential with density 𝜓.
Hence, by (49),
R𝑛
󸀠
𝐹𝑘𝑖 (𝑥, 𝑦) : = 𝑇𝑖,𝑦
[𝛾𝑘 (𝑥, 𝑦)]
=−
= − ∑ ∫ 𝜕𝑖 𝑢𝑖 Ψ𝑗 𝜈𝑗 𝑑𝜎 + ∑ ∫ 𝜕𝑖 𝑢𝑖 𝜕𝑗 Ψ𝑗 𝑑𝑥
∫
(54)
where (see, e.g., [22])
2
R𝑛 \Ω0
1
V± = {𝜑𝑘 ∈ 𝐿𝑝 (Σ) : ± 𝜑𝑘 (𝑥)
2
+ ∫ 𝐹𝑘𝑖 (𝑥, 𝑦) 𝜑𝑖 (𝑦) 𝑑𝜎𝑦 = 0, 𝑘 = 1, . . . , 𝑛} ,
𝜕2
𝑢 (𝑥) Ψ𝑗 (𝑥) 𝑑𝑥
𝜕𝑥𝑖 𝜕𝑥𝑗 𝑖
+∫
We conclude this section by recalling some properties
concerning the following eigenspaces:
Σ
Proposition 8. The sets V− and W+ are linear subspaces of
𝐿1 (Σ) and
dim (V− ) = dim (W+ ) =
𝑛 (𝑛 + 1)
+ 𝑚.
2
(57)
By setting 𝑐0 = 𝑎0 and 𝑐ℎ = 𝑎0 + 𝑎ℎ (ℎ = 1, . . . , 𝑚) we get the
claim.
A basis for W+ is expressed by the fields {𝜓𝑖 , 𝜈𝜒Σℎ : 𝑖 =
1, . . . , 𝑛(𝑛 + 1)/2, ℎ = 1, . . . , 𝑚}, where 𝜓𝑖 , 𝑖 = 1, . . . , 𝑛(𝑛 + 1)/2
are zero on Σ \ Σ0 , and such that the simple layer potentials
with density 𝜓𝑖 are 𝑛(𝑛+1)/2 rigid displacement in Ω0 (linearly
independent for 𝑛 ≥ 3).
Finally, every function 𝜑 which is the restriction to Σ of a
rigid displacement belongs to V− .
Remark 6. Lemma 5 shows that the dimension of the kernel
N𝑝 is infinite. However, if we consider the quotient space
N𝑝 /Ξ𝑝 , Ξ𝑝 being the space of weakly closed differential forms
𝑝
in 𝐿 𝑛−2 (Σ), we have dim(N𝑝 /Ξ𝑝 ) = 𝑚 + 1.
One recalls that if 𝜑 ∈ [𝐿1 (Σ)]𝑛 belongs to one of the
eigenspaces V± , W± , then 𝜑 ∈ [𝐶𝜆 (Σ)]𝑛 . This follows from
general results about integral equations (see [8, Lemma 31]
and [7, page 81]).
𝑚
= 𝑎0 ∫ 𝑢 ⋅ 𝜈𝑑𝜎 + ∑ (𝑎0 + 𝑎ℎ ) ∫ 𝑢 ⋅ 𝜈𝑑𝜎.
Σ0
ℎ=1
Σℎ
Abstract and Applied Analysis
7
Remark 9. We can make the statement of Proposition 8
slightly more precise, saying that the simple layer potentials
with density 𝜓𝑖 are 𝑛(𝑛 + 1)/2 rigid displacement in Ω0 linear
independent for any 𝑛 ≥ 2, unless 𝑛 = 2 and Σ0 is exceptional.
Indeed, let us show that if 𝑛 = 2 and Σ0 is not exceptional,
such rigid displacements are linearly independent. Let 𝑐𝑖 be
such that
3
∑𝑐𝑖 ∫ 𝜓𝑖 (𝑦) 𝛾 (𝑥, 𝑦) 𝑑𝜎𝑦 = 0,
Σ0
𝑖=1
in Ω0 .
(58)
𝑝
for every 𝜓 ∈ [𝐿 1 (Σ)]𝑛 , where
𝐻𝑙𝑗 (𝑥, 𝑦) =
∫ ∑𝑐𝑖 𝜓𝑖 (𝑦) 𝛾 (𝑥, 𝑦) 𝑑𝜎𝑦 = 0,
Σ0 𝑖=1
a.e. 𝑥 ∈ Σ0 .
Lemma 10. Let (𝑤, 𝑞) be the double layer hydrodynamic
potential of (17)-(18) with density 𝑢 ∈ [𝑊1,𝑝 (Σ)]𝑛 . Then, for
𝑥 ∉ Σ,
(59)
Let 𝜑 = ∑3𝑖=1 𝑐𝑖 𝜓𝑖 . In view of the equivalence between (1)
and (3) of Theorem 3, 𝜑 has to vanish. Therefore 𝑐𝑖 = 0
(𝑖 = 1, 2, 3) because of the linearly independence of 𝜓𝑖 . On
the other hand, if 𝑛 = 2 and Σ0 is exceptional, Theorem 3
shows that the potentials with densities {𝜓𝑖 }𝑖=1,2,3 are linearly
dependent.
Θℎ (𝜓) (𝑥) = ∗ (∫ 𝑑𝑥 [𝑠𝑛−2 (𝑥, 𝑦)] ∧ 𝜓 (𝑦) ∧ 𝑑𝑥ℎ ) ,
Σ
(𝑘)
∑ 𝑠 (𝑥, 𝑦) 𝑑𝑥𝑗1 ⋅ ⋅ ⋅ 𝑑𝑥𝑗ℎ 𝑑𝑦𝑗1 ⋅ ⋅ ⋅ 𝑑𝑦𝑗ℎ .
Σ
(𝑘)
where
(𝑘)
H𝑗ℎ (𝜓) (𝑥) = Θℎ (𝜓𝑗 ) (𝑥) −
𝑥 ∈ Ω,
𝜕
𝑠 (𝑥, 𝑦) 𝑑𝜎𝑦
𝜕𝜈𝑦
= 𝑑𝑥 ∫ 𝑑𝑢 (𝑦) ∧ 𝑠𝑛−2 (𝑥, 𝑦) ,
Σ
(𝑛 − 2)!
Σ
(62)
(𝑘)
𝐻𝑙𝑗 (𝑥, 𝑦) =
H𝑗ℎ (𝜓) (𝑥) = Θℎ (𝜓𝑗 ) (𝑥) −
(𝑦𝑙 − 𝑥𝑙 ) (𝑦𝑗 − 𝑥𝑗 )
𝑘
󵄨𝑛
󵄨󵄨
𝜔𝑛 (𝑘 + 1)
󵄨󵄨𝑦 − 𝑥󵄨󵄨󵄨
−
(63)
𝑥 ∈ Ω.
Moreover we introduce the operators H𝑗ℎ defined as
123⋅⋅⋅𝑛
𝛿𝑙𝑖𝑗
3 ⋅⋅⋅𝑗𝑛
(𝑛 − 2)!
× ∫ 𝜕𝑥ℎ 𝐻𝑙𝑗 (𝑥, 𝑦) ∧ 𝜓𝑖 (𝑦) ∧ 𝑑𝑦𝑗3 ⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛 ,
Σ
123⋅⋅⋅𝑛
𝛿𝑙𝑖𝑗
3 ⋅⋅⋅𝑗𝑛
× ∫ 𝜕𝑥ℎ 𝐻𝑙𝑗 (𝑥, 𝑦) ∧ 𝜓𝑖 (𝑦) ∧ 𝑑𝑦𝑗3 ⋅ ⋅ ⋅ 𝑑𝑦𝑗𝑛 ,
for each 𝑢 ∈ 𝑊1,𝑝 (Σ), since (see [9, page 187])
Σ
(69)
(𝑘)
𝜕
𝑠 (𝑥, 𝑦) 𝑑𝜎𝑦 = −Θℎ (𝑑𝑢) ,
𝜕𝜈𝑦
∗ 𝑑 ∫ 𝑢 (𝑦)
(𝑘)
𝜕ℎ 𝑤 𝑗 (𝑥) = H𝑗ℎ (𝑑𝑢) (𝑥) ,
Note that the operator Θℎ satisfies the equation
Σ
(68)
where 𝐿 and Γ are the stress operator and the Kelvin’s
matrix associated to the Lamé system −Δ𝑢 − 𝑘∇ div 𝑢 = 0,
respectively.
Thanks to [16, Lemma 1], we know that
(61)
𝑗1 <⋅⋅⋅<𝑗ℎ
(𝑘)
(𝑥) = ∫ 𝑢𝑖 (𝑦) 𝐿 𝑖,𝑦 [Γ𝑗 (𝑥, 𝑦)] 𝑑𝜎𝑦 ,
(𝑘)
where ∗ and 𝑑 denote the Hodge star operator and the
exterior derivative, respectively, and 𝑠ℎ (𝑥, 𝑦) is the double ℎform introduced by Hodge in [23] as follows:
𝜕ℎ ∫ 𝑢 (𝑦)
(67)
(𝑘)
𝑥 ∈ Ω,
𝑠ℎ (𝑥, 𝑦) =
𝑞 (𝑥) = 2𝜇Θℎ (𝑑𝑢ℎ ) (𝑥) ,
For 𝑘 > (𝑛 − 2)/𝑛, let 𝑤 be the double layer elastic potential
with density 𝑢, that is,
(𝑘)
(60)
(66)
Proof. Note that, even if one could prove (66)-(67) directly, it
seems easier to deduce them from the similar results we have
already obtained for the elasticity system (see [16, Section 3]).
(𝑘)
𝑤𝑗
𝑝
𝜕ℎ 𝑤𝑗 (𝑥) = H𝑗ℎ (𝑑𝑢) (𝑥) ,
where H𝑗ℎ and Θℎ are given by (60) and (64), respectively.
6. Reduction of a Certain Singular
Integral Operator
For every 𝜓 ∈ 𝐿 1 (Σ), let Θℎ be the operator defined by
(65)
In the sequel 𝑑𝑢 denotes the vector (𝑑𝑢1 , . . . , 𝑑𝑢𝑛 ) whose
𝑝
elements are 1-forms, and 𝜓 = (𝜓1 , . . . , 𝜓𝑛 ) ∈ [𝐿 1 (Σ)]𝑛 .
We have also
3
1 (𝑦𝑙 − 𝑥𝑙 ) (𝑦𝑗 − 𝑥𝑗 )
.
󵄨𝑛
󵄨󵄨
𝜔𝑛
󵄨󵄨𝑦 − 𝑥󵄨󵄨󵄨
(64)
1
𝛿 𝑠 (𝑥, 𝑦) ,
𝑘 + 1 𝑙𝑗
(70)
and Θℎ is given by (60).
From [16, formula (5)] (where we set 𝜉 = 1), letting 𝑘 →
+∞, we get
(𝑘)
(𝑘)
𝜕𝑥ℎ {𝐿 𝑖,𝑦 [Γ𝑗 (𝑥, 𝑦)]}
󳨀→ −
(𝑦𝑖 − 𝑥𝑖 ) (𝑦𝑗 − 𝑥𝑗 ) (𝑦𝑘 − 𝑥𝑘 )
(71)
𝑛
𝜕𝑥ℎ {
𝜈𝑘 (𝑦)}
𝑛+2
󵄨
󵄨
󵄨󵄨𝑥 − 𝑦󵄨󵄨
𝜔𝑛
󵄨
󵄨
󸀠
= 𝜕𝑥ℎ 𝑇𝑖,𝑦
[𝛾𝑗 (𝑥, 𝑦)] ,
8
Abstract and Applied Analysis
(𝑘)
𝑥 ∉ Σ, from which 𝜕ℎ𝑤 → 𝜕ℎ 𝑤 as 𝑘 → +∞. Therefore we
obtain formula (66) by letting 𝑘 → +∞ in (69). Formula
(67) is an immediate consequence of (62) because 𝜀𝑗 (𝑥, 𝑦) =
−𝜕𝑥𝑗 𝑠(𝑥, 𝑦).
For the next lemma it is convenient to recall here two
jump formulas proved in [16, Lemmas 2 and 3].
Let 𝑓 ∈ 𝐿1 (Σ). If 𝜂 ∈ Σ is a Lebesgue point for 𝑓, we get
lim ∫ 𝑓 (𝑦) 𝜕𝑥𝑠
𝑥→𝜂 Σ
(𝑦𝑙 − 𝑥𝑙 ) (𝑦𝑗 − 𝑥𝑗 )
𝑑𝜎𝑦
󵄨𝑛
󵄨󵄨
󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨
𝜔
= 𝑛 (𝛿𝑙𝑗 − 2𝜈𝑗 (𝜂) 𝜈𝑙 (𝜂)) 𝜈𝑠 (𝜂) 𝑓 (𝜂)
2
Hence, by (65) and (72),
lim
𝑥 → 𝜂 (𝑛
1
𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3 ⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
− 2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
𝑙𝑖
𝛿𝑟ℎ
(𝛿𝑙𝑗 − 2𝜈𝑗 (𝜂) 𝜈𝑙 (𝜂)) 𝜈𝑠 (𝜂) 𝜈𝑟 (𝜂) 𝜓ℎ (𝜂)
2
=
+
1
𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝜂, 𝑦)∧𝜓 (𝑦)∧𝑑𝑦𝑗3 ⋅ ⋅ ⋅∧𝑑𝑦𝑗𝑛 .
(𝑛−2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
(77)
Keeping in mind (73), we find
(72)
(𝑦𝑙 − 𝜂𝑙 ) (𝑦𝑗 − 𝜂𝑗 )
𝑑𝜎𝑦 ,
󵄨󵄨𝑛
󵄨󵄨
Σ
󵄨󵄨𝑥 − 𝑦󵄨󵄨
where the limit has to be understood as an internal angular
boundary value, and the integral in the right hand side is a
singular integral.
𝑝
Further, let 𝜓 ∈ 𝐿 1 (Σ) and write 𝜓 as 𝜓 = 𝜓ℎ 𝑑𝑥ℎ with
𝑙𝑖
𝛿𝑟ℎ
1
(𝛿𝑙𝑗 − 2𝜈𝑗 𝜈𝑙 ) 𝜈𝑠 𝜈𝑟 𝜓ℎ = ( 𝛿𝑙𝑗 𝜈𝑠 − 𝜈𝑗 𝜈𝑙 𝜈𝑠 ) (𝜈𝑙 𝜓𝑖 − 𝜈𝑖 𝜓𝑙 )
2
2
1
1
= − 𝜈𝑠 𝜈𝑗 𝜓𝑖 − 𝜈𝑠 𝜈𝑖 𝜓𝑗 ,
2
2
+ ∫ 𝑓 (𝑦) 𝜕𝑥𝑠
𝜈ℎ 𝜓ℎ = 0.
(73)
Assumption (73) is not restrictive, because, given the 1-form
𝜓 on Σ, there exist scalar functions 𝜓ℎ defined on Σ such that
𝜓 = 𝜓ℎ 𝑑𝑥ℎ and (73) holds (see [24, page 41]). Then, for almost
every 𝜂 ∈ Σ,
1
lim Θℎ (𝜓) (𝑥) = − 𝜓ℎ (𝜂) + Θℎ (𝜓) (𝜂) ,
2
𝑥→𝜂
(74)
where Θℎ is given by (60), and the limit has to be understood
again as an internal angular boundary value.
and the result follows.
𝑝
Lemma 12. Let 𝜓 = (𝜓1 , . . . , 𝜓𝑛 ) ∈ [𝐿 1 (Σ)]𝑛 . Then, for almost
every 𝜂 ∈ Σ,
lim 𝜇 [2𝛿𝑖𝑗 Θℎ (𝜓ℎ ) (𝑥) + H𝑖𝑗 (𝜓) (𝑥) + H𝑗𝑖 (𝜓) (𝑥)] 𝜈𝑖 (𝑥)
𝑥→𝜂
= 𝜇 [2𝛿𝑖𝑗 Θℎ (𝜓ℎ ) (𝜂) + H𝑖𝑗 (𝜓) (𝜂) + H𝑗𝑖 (𝜓) (𝜂)] 𝜈𝑖 (𝜂) ,
(79)
Θℎ and H being as in (60) and (64), respectively, and the limit
has to be understood as an internal angular boundary value.
Proof. Let us write 𝜓𝑖 as 𝜓𝑖 = 𝜓𝑖ℎ 𝑑𝑥ℎ with
𝜈ℎ 𝜓𝑖ℎ = 0,
𝑝
Lemma 11. Let 𝜓 ∈ 𝐿 1 (Σ). Let one write 𝜓 as 𝜓 = 𝜓ℎ 𝑑𝑥ℎ and
suppose that (73) holds. Then, for almost every 𝜂 ∈ Σ,
lim
𝑥 → 𝜂 (𝑛
1
𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3 ⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
− 2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
1
= − [𝜈𝑗 (𝜂) 𝜓𝑖 (𝜂) + 𝜈𝑖 (𝜂) 𝜓𝑗 (𝜂)] 𝜈𝑠 (𝜂)
2
Proof. We have
1
𝛿123⋅⋅⋅𝑛 𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝑥, 𝑦) 𝜓ℎ (𝑦) 𝜈𝑟 (𝑦) 𝑑𝜎𝑦
=
(𝑛 − 2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 𝑟ℎ𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
=
∫ 𝜕𝑥𝑠 𝐻𝑙𝑗 (𝑥, 𝑦) 𝜓ℎ (𝑦) 𝜈𝑟 (𝑦) 𝑑𝜎𝑦 .
Σ
(76)
(80)
lim 𝜇 [2𝛿𝑖𝑗 Θℎ (𝜓ℎ ) (𝑥) + H𝑖𝑗 (𝜓) (𝑥) + H𝑗𝑖 (𝜓) (𝑥)] 𝜈𝑖 (𝑥)
𝑥→𝜂
= 𝜇Ψ𝑖𝑗 (𝜓) (𝜂) 𝜈𝑖 (𝜂)
+ 𝜇 [2𝛿𝑖𝑗 Θℎ (𝜓ℎ ) (𝜂)
+H𝑖𝑗 (𝜓) (𝜂) + H𝑗𝑖 (𝜓) (𝜂)] 𝜈𝑖 (𝜂) ,
(81)
where
1
1
Ψ𝑖𝑗 (𝜓) = − 𝛿𝑖𝑗 𝜓ℎℎ − 𝜓𝑖𝑗 + (𝜈𝑖 𝜓𝑠𝑠 + 𝜈𝑠 𝜓𝑠𝑖 ) 𝜈𝑗
2
2
1
1
− 𝜓𝑗𝑖 + (𝜈𝑗 𝜓𝑠𝑠 + 𝜈𝑠 𝜓𝑠𝑗 ) 𝜈𝑖 .
2
2
1
𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3 ⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
(𝑛 − 2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
𝑙𝑖
𝛿𝑟ℎ
𝑖 = 1, . . . , 𝑛.
On account of (72) and (74), we infer
1
+
𝛿123⋅⋅⋅𝑛 ∫ 𝜕 𝐻 (𝜂, 𝑦)∧𝜓 (𝑦)∧𝑑𝑦𝑗3 ⋅ ⋅ ⋅∧𝑑𝑦𝑗𝑛 ,
(𝑛−2)! 𝑙𝑖𝑗3 ⋅⋅⋅𝑗𝑛 Σ 𝑥𝑠 𝑙𝑗
(75)
where 𝐻𝑙𝑗 is defined by (65), and the limit has to be understood
as an internal angular boundary value.
(78)
(82)
By (80) we get Ψ𝑖𝑗 (𝜓)𝜈𝑖 = −𝜓ℎℎ 𝜈𝑗 − 𝜓𝑖𝑗 𝜈𝑖 /2 + 𝜓𝑠𝑠 𝜈𝑗 + 𝜈𝑠 𝜓𝑠𝑗 /2 =
0.
Remark 13. Whenever we consider external boundary values,
we have just to change the sign in the first term on the
right hand sides in (72), (74), and (75), while (79) remains
unchanged.
Abstract and Applied Analysis
9
Lemma 14. Let 𝑤 be the double layer potential (17) with density 𝑢 ∈ [𝑊1,𝑝 (Σ)]𝑛 . Then 𝑇+,𝑗 𝑤 = 𝑇−,𝑗 𝑤 = 𝜇[2𝛿𝑖𝑗 Θℎ (𝑑𝑢ℎ ) +
H𝑖𝑗 (𝑑𝑢) + H𝑗𝑖 (𝑑𝑢)]𝜈𝑖 a.e. on Σ, where 𝑇+ 𝑤 and 𝑇− 𝑤 denote
the internal and the external angular boundary limits of 𝑇𝑤,
respectively, and Θℎ is given by (60) and H by (64).
Proof. It is an immediate consequence of (66), (67), (79), and
Remark 13.
𝑝
Proposition 15. Let 𝑅 : [𝐿𝑝 (Σ)]𝑛 → [𝐿 1 (Σ)]𝑛 be the following singular integral operator
where the given data 𝑓 ∈ [𝑊1,𝑝 (Σ)]𝑛 satisfies the compatibility condition (4).
The aim of the present section is to study the representability of the solution of this problem by means of a
simple layer hydrodynamic potential (15)-(16).
By the symbol S𝑝 we mean the class of the simple layer
hydrodynamic potentials (15)-(16) with density in [𝐿𝑝 (Σ)]𝑛 .
Whenever 𝑛 = 2 and Σ0 is exceptional (see Section 4), we say
that (𝑣, 𝑟) belongs to S𝑝 if, and only if,
𝑣 (𝑥) = − ∫ 𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝜎𝑦 + 𝑐,
Σ
𝑅𝜑 (𝑥) = − ∫ 𝑑𝑥 [𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦 .
Σ
󸀠
Let one define 𝑅 :
integral operator
𝑝
[𝐿 1 (Σ)]𝑛
𝑝
(83)
𝑛
→ [𝐿 (Σ)] to be the singular
𝑅𝑗󸀠 (𝜓) (𝑥) = 𝜇 [2𝛿𝑖𝑗 Θℎ (𝜓ℎ ) (𝑥) + H𝑖𝑗 (𝜓) (𝑥)
+H𝑗𝑖 (𝜓) (𝑥)] 𝜈𝑖 (𝑥) .
(84)
𝑥 ∈ Ω,
(91)
𝑟 (𝑥) = − ∫ 𝜀𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦 ,
Σ
𝑥 ∈ Ω,
where 𝜑 ∈ [𝐿𝑝 (Σ)]2 and 𝑐 ∈ R2 .
We will see that condition (4) is not sufficient to prove
the existence of the solution in the class S𝑝 , but it must be
satisfied on each Σ𝑗 , 𝑗 = 0, 1, . . . , 𝑚.
We begin by proving the following result.
𝑝
Then
1
𝑅󸀠 𝑅𝜑 = 𝜑 − 𝐾2 𝜑,
4
(85)
Theorem 16. Given 𝜔 ∈ [𝐿 1 (Σ)]𝑛 , there exists a solution 𝜑 ∈
[𝐿𝑝 (Σ)]𝑛 of the singular integral system
− ∫ 𝑑𝑥 [𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦 = 𝜔 (𝑥) ,
where
Σ
𝐾𝜑 (𝑥) = − ∫ 𝑇𝑥 [𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦 .
Σ
(86)
Proof. Let 𝑣 be the simple layer potential (15) with density 𝜑 ∈
[𝐿𝑝 (Σ)]𝑛 . In view of Lemma 14, we have a.e. on Σ
𝑎.𝑒. 𝑥 ∈ Σ,
(92)
if, and only if,
∫ 𝜓𝑖 ∧ 𝜔𝑖 = 0,
(93)
Σ
𝑞
𝑅𝑗󸀠 (𝑅𝜑) = 𝜇 [2𝛿𝑖𝑗 Θℎ (𝑑𝑣ℎ ) + H𝑖𝑗 (𝑑𝑣) + H𝑗𝑖 (𝑑𝑣)] 𝜈𝑖 = 𝑇𝑗 𝑤,
(87)
for every 𝜓 = (𝜓1 , . . . , 𝜓𝑛 ) ∈ [𝐿 𝑛−2 (Σ)]𝑛 (𝑞 = 𝑝/(𝑝 − 1)) such
that the weak differentials 𝑑𝜓𝑗 exist and (38) holds for some
real constants 𝑐0 , . . . , 𝑐𝑚 .
where 𝑤 is the double layer potential (17) with density 𝑣.
Moreover, if 𝑥 ∈ Ω,
Proof. Consider the adjoint of 𝑅 (see (83)), 𝑅∗ : [𝐿 𝑛−2 (Σ)]𝑛 →
[𝐿𝑞 (Σ)]𝑛 , that is, the operator whose components are given by
𝑅𝑖∗ 𝜓 (𝑥) = − ∫ 𝜓𝑖 (𝑦) ∧ 𝑑𝑦 [𝛾𝑖𝑗 (𝑥, 𝑦)] .
󸀠
[𝛾𝑘 (𝑥, 𝑦)] 𝑑𝜎𝑦
𝑤𝑘 (𝑥) = ∫ 𝑣𝑖 (𝑦) 𝑇𝑖,𝑦
Σ
(88)
= 𝑣𝑘 (𝑥) + ∫ 𝛾𝑖𝑘 (𝑥, 𝑦) 𝑇𝑖 [𝑣 (𝑦)] 𝑑𝜎𝑦 ,
Σ
and then, on account of (86),
1 1
1
1
𝑇𝑤 = 𝑇𝑣 − 𝐾 (𝑇𝑣) = ( 𝜑 + 𝐾𝜑) − 𝐾 ( 𝜑 + 𝐾𝜑)
2
2 2
2
1
= 𝜑 − 𝐾2 𝜑.
4
𝑞
(89)
Σ
(94)
Proposition 15 implies that the integral system (92) has a
solution 𝜑 ∈ [𝐿𝑝 (Σ)]𝑛 if, and only if,
∫ 𝜓𝑖 ∧ 𝜔𝑖 = 0,
Σ
(95)
𝑞
for each 𝜓 = (𝜓1 , . . . , 𝜓𝑛 ) ∈ [𝐿 𝑛−2 (Σ)]𝑛 such that 𝑅∗ 𝜓 = 0.
The result follows from Lemma 5.
Proposition 17. Given 𝑓 ∈ [𝑊1,𝑝 (Σ)]𝑛 , there exists a solution
of the BVP
(̃𝑣, 𝑟̃) ∈ S𝑝 ,
7. The Dirichlet Problem
𝜇Δ̃𝑣 = ∇̃𝑟 𝑖𝑛 Ω,
Let us consider the Dirichlet problem for the Stokes system
div 𝑣̃ = 0 𝑖𝑛 Ω,
𝜇Δ𝑣 = ∇𝑟 in Ω,
div 𝑣 = 0 in Ω,
𝑣=𝑓
on Σ,
𝑑̃𝑣 = 𝑑𝑓
(90)
(96)
𝑜𝑛 Σ,
if, and only if, conditions (3) are satisfied. The density 𝜑 of
the pair (̃𝑣, 𝑟̃) (see (15)-(16)) solves the singular integral system
𝑅𝜑 = 𝑑𝑓, where 𝑅 is given by (83).
10
Abstract and Applied Analysis
Proof. Clearly, there exists a solution of this BVP if, and only
if, there exists a solution 𝜑 ∈ [𝐿𝑝 (Σ)]𝑛 of the singular integral
system
− ∫ 𝑑𝑥 [𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦 = 𝑑𝑓 (𝑥) ,
Σ
a.e. 𝑥 ∈ Σ.
Theorem 19. Given 𝑓 ∈ [𝑊1,𝑝 (Σ)]𝑛 , the Dirichlet problem
(𝑣, 𝑟) ∈ S𝑝 ,
𝜇Δ𝑣 = ∇𝑟 𝑖𝑛 Ω,
(97)
In view of Theorem 16, there exists a solution 𝜑 of this
system if, and only if,
∫ 𝜓𝑖 ∧ 𝑑𝑓𝑖 = 0,
Σ
(98)
𝑞
for every 𝜓 = (𝜓1 , . . . , 𝜓𝑛 ) ∈ [𝐿 𝑛−2 (Σ)]𝑛 satisfying 𝑅∗ 𝜓 =
0, that is, such that the weak differentials 𝑑𝜓𝑗 exist and (39)
holds for some real constants 𝑐0 , . . . , 𝑐𝑚 . Equation (39) being
true for any 𝑢 ∈ [𝑊1,𝑝 (Σ)]𝑛 , we can write
𝑚
∫ 𝜓𝑖 ∧ 𝑑𝑓𝑖 = ∑ 𝑐ℎ ∫ 𝑓 ⋅ 𝜈𝑑𝜎,
Σ
ℎ=0
Σℎ
(99)
because of a density argument. In view of the arbitrariness of
𝑐0 , . . . , 𝑐𝑚 , (98) is satisfied if, and only if, (3) holds.
Proposition 18. Let 𝑎ℎ ∈ R𝑛 (ℎ = 0, . . . , 𝑚). Let 𝜓𝑖𝑘 , 𝑖 =
1, . . . , 𝑛, 𝑘 = 1, . . . , 𝑚, be the elements of the basis of W− given
by Proposition 7. The pair
𝑚 𝑛
𝑣0 (𝑥) = ∑ ∑ (𝑎𝑘𝑖 −𝑎0𝑖 ) ∫ 𝛾 (𝑥, 𝑦) 𝜓𝑖𝑘 (𝑦) 𝑑𝜎𝑦 +𝑎0 ,
Σ
𝑘=1 𝑖=1
𝑚 𝑛
𝑟0 (𝑥) = ∑ ∑ (𝑎𝑘𝑖 − 𝑎0𝑖 ) ∫ 𝜀 (𝑥, 𝑦) 𝜓𝑖𝑘 (𝑦) 𝑑𝜎𝑦 ,
Σ
𝑘=1 𝑖=1
𝑥 ∈ Ω,
𝑥 ∈ Ω,
(100)
𝑣=𝑓
Proof. Suppose conditions (3) are satisfied. Let (̃𝑣, 𝑟̃) be a
solution of the problem (96). Since 𝑑̃𝑣 = 𝑑𝑓 on Σ, 𝑣̃ = 𝑓 + 𝑎ℎ
on Σℎ (ℎ = 0, . . . , 𝑚) for some 𝑎ℎ ∈ R𝑛 . The pair (𝑣, 𝑟) =
(̃𝑣, 𝑟̃) − (𝑣0 , 𝑟0 ), where 𝑣0 and 𝑟0 are given by (100), solves the
problem (103).
Conversely, if there exists a solution (𝑣, 𝑟) of (103), the
compatibility condition (4) has to be satisfied. Moreover, for
any 𝑗 = 1, . . . , 𝑚, (𝑣, 𝑟) is the solution of the Stokes system
also in Ω𝑗 . Therefore conditions (3) are satisfied for 𝑗 =
1, . . . , 𝑚. These, together with (4), imply (3) also for 𝑗 = 0.
The uniqueness is known [7, Theorem 5.5].
Remark 20. The density (𝜑, 𝜀) of (𝑣, 𝑟) can be written as
(𝜑, 𝜀) = (𝜑0 + 𝜆 0 , 𝜀), where 𝜑0 solves the singular integral
system (97), and (𝜆 0 , 𝜀) is the density of a simple layer
potential which is constant on every connected component
of Σ.
Remark 21. If 𝑛 ≥ 3 or 𝑛 = 2 and Σ0 is not exceptional,
denoting by 𝜑 the density of the simple layer potential (15)(16) obtained in Theorem 19, we have 𝜑 that solves the integral
system of the first kind
− ∫ 𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝜎𝑦 = 𝑓 (𝑥)
Σ
𝑝
(𝑣0 , 𝑟0 ) ∈ S ,
𝜇Δ𝑣0 = ∇𝑟0
𝑖𝑛 Ω,
div 𝑣0 = 0 𝑖𝑛 Ω,
(101)
𝑜𝑛 Σℎ , ℎ = 0, . . . , 𝑚.
Proof. The pair (𝑣0 , 𝑟0 ) belongs to S𝑝 (for 𝑛 = 2, see
Remark 9). Obviously it satisfies the Stokes system, and it satisfies the boundary conditions since, thanks to Proposition 7,
𝑚 𝑛
𝑣0 |Σ0 = ∑ ∑ (𝑎𝑘𝑖 − 𝑎0𝑖 ) 𝑣𝑖𝑘 |Σ0 + 𝑎0 = 𝑎0 ,
𝑘=1 𝑖=1
𝑚 𝑛
𝑣0 |Σℎ = ∑ ∑ (𝑎𝑘𝑖 − 𝑎0𝑖 ) 𝑣𝑖𝑘 |Σℎ + 𝑎0
𝑘=1 𝑖=1
𝑚 𝑛
= ∑ ∑ (𝑎𝑘𝑖 − 𝑎0𝑖 ) 𝛿ℎ𝑘 𝑒𝑖 + 𝑎0 = 𝑎ℎ ,
𝑘=1 𝑖=1
for any ℎ = 1, . . . , 𝑚.
𝑜𝑛 Σ,
is solvable if, and only if, conditions (3) are satisfied. Moreover
the solution (𝑣, 𝑟) is unique (𝑟 is unique up to an additive
constant).
is the solution of the BVP
𝑣0 = 𝑎ℎ
(103)
div 𝑣 = 0 𝑖𝑛 Ω,
(102)
(104)
on Σ. Therefore, Theorem 19 can be seen as an existence
theorem for the integral system of the first kind (104) in 𝐿𝑝 (Σ).
If 𝑛 = 2 and Σ0 is exceptional, we have the existence of a
solution (𝜑, 𝑐) ∈ [𝐿𝑝 (Σ)]2 × R2 of the integral equation
− ∫ 𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝜎𝑦 = 𝑓 (𝑥) + 𝑐
Σ
on Σ.
(105)
Remark 22. Observe that the solvability of the Dirichlet
problem (90) by means of a simple layer potential hinges on
the singular integral system (97). Thanks to Proposition 15,
the operator 𝑅󸀠 provides a left reduction for such a system.
This reduction is not an equivalent one, but, as in [25,
pages 253-254], one can show that 𝑅󸀠 is a weakly equivalent
reduction (see definition in Section 3). Since the system 𝑅𝜑 =
𝑑𝑓 is solvable, we have 𝑅𝜑 = 𝑑𝑓 if, and only if, 𝜑 is solution
of the Fredholm system 𝑅󸀠 𝑅𝜑 = 𝑅󸀠 𝑑𝑓. In this sense, such
Fredholm system is equivalent to the problem (103).
In order to obtain a similar integral representation for the
solution of the Dirichlet problem (90) when 𝑓 satisfies the
only condition (4), we need to modify the representation of
the solution by adding an extra term.
Abstract and Applied Analysis
11
̃ 𝑝 we denote the space of all pairs (𝑣, 𝑟) written as
By S
𝑓𝑖 (𝑥) − 𝑤+,𝑖 (𝑥)
𝑣𝑖 (𝑥) = − ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦
Σ
+∫
Σ
󸀠
𝑇𝑗,𝑦
𝑖
[𝛾 (𝑥, 𝑦)] 𝜓𝑗 (𝑦) 𝑑𝜎𝑦 , 𝑖 = 1, . . . , 𝑛, 𝑥 ∈ Ω,
Σ
Σ
𝜕
[𝜀 (𝑥, 𝑦)] 𝜓𝑗 (𝑦) 𝑑𝜎𝑦 ,
𝜕𝜈𝑦 𝑗
𝑥 ∈ Ω,
(106)
where 𝜑 and 𝜓 belong to [𝐿𝑝 (Σ)]𝑛 .
Theorem 23. Given 𝑓 ∈ [𝑊1,𝑝 (Σ)]𝑛 satisfying (4), the
Dirichlet problem
̃𝑝,
(𝑣, 𝑟) ∈ S
𝜇Δ𝑣 = ∇𝑟 𝑖𝑛 Ω,
(107)
div 𝑣 = 0 𝑖𝑛 Ω,
𝑣=𝑓
Therefore, conditions (112) become − ∫Σ 𝑤− ⋅ 𝜈𝑑𝜎 = 0, ℎ =
ℎ
0, . . . , 𝑚. Since 𝑤− can be considered as the datum of the
interior Dirichlet problem in Ωℎ , for ℎ = 1, . . . , 𝑚, we have
∫ 𝑤− ⋅ 𝜈𝑑𝜎 = 0,
Σℎ
(116)
𝑚
Σ
𝜕
[𝜀 (𝑥, 𝑦)] 𝑓𝑗 (𝑦) 𝑑𝜎𝑦 ,
𝜕𝜈𝑦 𝑗
(109)
= ∑ ∫ 𝑤− ⋅ 𝜈𝑑𝜎 = 0.
𝑗=1 Σ𝑗
𝑥 ∈ Ω,
where 𝜑 ∈ [𝐿𝑝 (Σ)]𝑛 is solution of the integral system of the first
kind
− ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦
Finally, assume that (𝑣, 𝑟) is the solution of (107) with the data
𝑓 = 0. The integral representation (108) shows that (𝑣, 𝑟) ∈
S𝑝 , and then the uniqueness follows from Theorem 19.
References
Σ
1
󸀠
= 𝑓𝑖 (𝑥) − ∫ 𝑇𝑗,𝑦
[𝛾𝑖 (𝑥, 𝑦)] 𝑓𝑗 (𝑦) 𝑑𝜎𝑦 ,
2
Σ
𝑎.𝑒. on Σ.
(110)
Proof. Let 𝑣 be given by (108); imposing the boundary
condition, we get (the symbol 𝑤+ (𝑤− ) stands for the interior
(exterior) value of the double layer potential (17) on Σ)
− ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦 = 𝑓𝑖 (𝑥) − 𝑤+,𝑖 (𝑥) ,
a.e. 𝑥 ∈ Σ.
(111)
In view of Remark 21 such a system is solvable if, and only if,
∫ (𝑓𝑖 − 𝑤+,𝑖 ) 𝜈𝑖 𝑑𝜎 = 0,
1
∫ 𝑓 ⋅ 𝜈𝑑𝜎 − ∫ 𝑤 ⋅ 𝜈𝑑𝜎
2 Σ0
Σ0
𝑚
1𝑚
= − ∑ ∫ 𝑓 ⋅ 𝜈𝑑𝜎 + ∑ ∫ 𝑤 ⋅ 𝜈𝑑𝜎
2 𝑗=1 Σ𝑗
𝑗=1 Σ𝑗
𝑥 ∈ Ω,
𝑟 (𝑥) = − ∫ 𝜀𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦
Σ
(114)
1
0 = ∫ 𝑤+ ⋅ 𝜈𝑑𝜎 = ∫ ( 𝑓 + 𝑤) ⋅ 𝜈𝑑𝜎 = ∫ 𝑤 ⋅ 𝜈𝑑𝜎. (115)
Σ
Σ 2
Σ
(108)
󸀠
+ ∫ 𝑇𝑗,𝑦
[𝛾𝑖 (𝑥, 𝑦)] 𝑓𝑗 (𝑦) 𝑑𝜎𝑦 ,
Σ
ℎ = 1, . . . , 𝑚.
As far as ℎ = 0 is concerned, first we remark that (4) implies
∫Σ 𝑤 ⋅ 𝜈𝑑𝜎 = 0, because
=
Σ
+ 2𝜇 ∫
(113)
1
∫ ( 𝑓 − 𝑤) ⋅ 𝜈𝑑𝜎
Σ0 2
𝑣𝑖 (𝑥) = − ∫ 𝛾𝑖𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦
Σ
a.e. 𝑥 ∈ Σ.
Keeping in mind (4) and (114), this leads to
𝑜𝑛 Σ,
has one, and only one, solution (𝑣, 𝑟) given by
Σℎ
1
󸀠
= 𝑓𝑖 (𝑥) − ( 𝑓𝑖 (𝑥) + ∫ 𝑇𝑗,𝑦
[𝛾𝑖 (𝑥, 𝑦)] 𝑓𝑗 (𝑦) 𝑑𝜎𝑦 )
2
Σ
1
:= 𝑓𝑖 (𝑥) − 𝑤𝑖 (𝑥) = −𝑤−,𝑖 (𝑥) ,
2
𝑟 (𝑥) = − ∫ 𝜀𝑗 (𝑥, 𝑦) 𝜑𝑗 (𝑦) 𝑑𝜎𝑦
+ 2𝜇 ∫
On the other hand, because of the jump formulas, we have
ℎ = 0, . . . , 𝑚.
(112)
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International Journal of
Advances in
Combinatorics
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Mathematical Physics
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Volume 2014
Journal of
Complex Analysis
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Volume 2014
International
Journal of
Mathematics and
Mathematical
Sciences
Journal of
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Stochastic Analysis
Abstract and
Applied Analysis
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International Journal of
Mathematics
Volume 2014
Volume 2014
Discrete Dynamics in
Nature and Society
Volume 2014
Volume 2014
Journal of
Journal of
Discrete Mathematics
Journal of
Volume 2014
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Applied Mathematics
Journal of
Function Spaces
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Volume 2014
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Volume 2014
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Volume 2014
Optimization
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Volume 2014
Hindawi Publishing Corporation
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Volume 2014