Volume 2013, Article ID 765020,12pages 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; [email protected] 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 ofR𝑛(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 represen- tations 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
Ω = Ω0\⋃𝑚
𝑗=1
Ω𝑗, (1)
whereΩ𝑗(𝑗 = 0, . . . , 𝑚) are suitable domains with connected boundaries in𝐶1,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 2we 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 case𝑛 = 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 degree𝑛 − 2on𝜕Ω. In the same section, we recall some known results concerning the eigenspaces of some classical integral systems.
In Section6we 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, given𝑓 ∈ [𝑊1,𝑝(𝜕Ω)]𝑛, we can represent the solution of the Dirichlet problem
𝜇Δ𝑣 = ∇𝑟 inΩ, div𝑣 = 0 inΩ,
𝑣 = 𝑓 on𝜕Ω,
(2)
by means of a simple layer hydrodynamic potential if, and only if, the conditions
∫𝜕Ω𝑗𝑓 ⋅ 𝜈𝑑𝜎 = 0, 𝑗 = 0, 1, . . . , 𝑚 (3) are satisfied (𝜈 being the outwards unit normal on 𝜕Ω).
Moreover, if the data𝑓satisfies only the condition
∫𝜕Ω𝑓 ⋅ 𝜈𝑑𝜎 = 0 (4)
(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 Theorem23).
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 domainΩ ⊂ R𝑛, whose boundary we denote byΣas follows:
Δ𝑢 = 0 inΩ,
𝑢 = 𝑔 on Σ. (5)
Suppose that𝑔 ∈ 𝑊1,𝑝(Σ), 1 < 𝑝 < ∞. If we want to find the solution in the form of a simple layer potential whose density belongs to𝐿𝑝(Σ), we have to solve an integral equation of the first kind onΣas follows:
∫Σ𝜑 ( 𝑦) 𝑠 ( 𝑥, 𝑦) 𝑑𝜎𝑦= 𝑔 (𝑥) , 𝑥 ∈ Σ, (6) where𝑠(𝑥, 𝑦)is the fundamental solution of Laplace equation
𝑠 ( 𝑥, 𝑦) = {{ {{ {{ {
− 1
2𝜋log 1
𝑥 − 𝑦, if 𝑛 = 2,
− 1
𝜔𝑛(𝑛 − 2)
𝑥 − 𝑦1 𝑛−2, if 𝑛 ≥ 3. (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
∫Σ𝜑 (𝑦) 𝑑𝑥[𝑠 (𝑥, 𝑦)] 𝑑𝜎𝑦= 𝑑𝑔 (𝑥) , 𝑥 ∈ Σ, (8) in which we look for a solution𝜑 ∈ 𝐿𝑝(Σ).
The integral on the left hand side is a singular integral and it can be considered as a linear and continuous operator from𝐿𝑝(Σ)to𝐿𝑝1(Σ)(we denote by𝐿𝑝ℎ(Σ) the space of the differential forms of degreeℎwhose coefficients belong to 𝐿𝑝(Σ)in every local coordinate system).
It must be remarked that, if𝑛 ≥ 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, if𝐵 and 𝐵 are two Banach spaces and 𝑆 : 𝐵 → 𝐵is a continuous linear operator,𝑆can be reduced
on the left if there exists a continuous linear operator𝑆 : 𝐵 → 𝐵such that𝑆𝑆 = 𝐼 + 𝑇, where𝐼stands for the identity operator on𝐵, and𝑇 : 𝐵 → 𝐵is compact. Analogously, one can define an operator𝑆reducible on the right. One of the main properties of such operators is that the equation𝑆𝛼 = 𝛽 has a solution if, and only if,⟨𝛾, 𝛽⟩ = 0for any𝛾such that 𝑆∗𝛾 = 0,𝑆∗being the adjoint of𝑆(for more details see, e.g., [10,11]).
Let us denote by𝑆𝜑the left hand side of (8). In [9] a reduc- ing operator𝑆was explicitly constructed. This implies that there exists a solution of (8) if, and only if, the compatibility conditions
∫Σ𝑑𝑔 ∧ ℎ = 0 (9)
are satisfied for any ℎ ∈ 𝐿𝑞𝑛−2(Σ) (𝑞 = 𝑝/(𝑝 − 1)) such that𝑆∗ℎ = 0. Moreover one can show that𝑆∗ℎ = 0if, and only if,ℎis 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 equivalentif N(𝑆) = {0}, where N(𝑆) denotes the kernel of 𝑆 (see, e.g., [11, page 19-20]). Obviously this means that𝑆𝑥 = 𝑦if, and only if, 𝑆𝑆𝑥 = 𝑆𝑦. In [12] it was remarked that if N(𝑆𝑆) = N(𝑆), we still have a kind of equivalence. Indeed the coincidence of these two kernels implies the following fact: if𝑦is such that the equation𝑆𝑥 = 𝑦is solvable, then this equation is satisfied if, and only if,𝑆𝑆𝑥 = 𝑆𝑦.
Since N(𝑆𝑆) = N(𝑆), then we have (8) equivalent to the Fredholm equation𝑆𝑆𝜑 = 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 rep- resentation 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 class𝐶1(Ω), whereΩ is the multiple connected domain(1). There exists a2-form𝑣 conjugate to𝑢inΩif, and only if,
∫𝜕Ω𝑗
𝜕𝑢
𝜕𝜈𝑑𝜎 = 0, 𝑗 = 0, 1, . . . , 𝑚. (10) An explicit integral expression for𝑣was also given. We recall that the2-form𝑣is conjugate to𝑢if𝑑𝑢 = 𝛿𝑣, 𝑑𝑣 = 0.
The method has been applied to different BVPs for several PDEs (see [12–19]).
3. Preliminaries
In this paperΩdenotes an(𝑚 + 1)-connected domainofR𝑛 (𝑛 ≥ 2), that is an open-connected set of the form (1), where each Ω𝑗 (𝑗 = 0, . . . , 𝑚) is a bounded domain ofR𝑛 with
connected boundariesΣ𝑗 ∈ 𝐶1,𝜆(𝜆 ∈ (0, 1]), and such that Ω𝑗 ⊂ Ω0 andΩ𝑗 ∩ Ω𝑘 = 0,𝑗, 𝑘 = 1, . . . , 𝑚, 𝑗 ̸= 𝑘. Let𝜈be the outwards unit normal on the boundaryΣ = 𝜕Ω.
We consider the classical Stokes system for the incom- pressible viscous fluid
𝜇Δ𝑢 = ∇𝑝,
div𝑢 = 0, inΩ, (11)
where the unknowns 𝑢 = (𝑢1, . . . , 𝑢𝑛) and 𝑝 = 𝑝(𝑥) are the velocity and pressure of the fluid flow, respectively, and the constant𝜇 > 0is 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
𝛾𝑖𝑗(𝑥, 𝑦) = {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
− 1
4𝜋𝜇[𝛿𝑖𝑗log 1
𝑥 − 𝑦
+(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦2 ] , if𝑛 = 2,
− 1 2𝜔𝑛𝜇[ 𝛿𝑖𝑗
𝑛 − 2
𝑥 − 𝑦1 𝑛−2 +(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦𝑛 ] , if𝑛 ≥ 3, (12) 𝜀𝑗(𝑥, 𝑦) = − 1
𝜔𝑛 𝑥𝑗− 𝑦𝑗
𝑥 − 𝑦𝑛, (13) (𝑖, 𝑗 = 1, . . . , 𝑛),𝜔𝑛 being the hypersurface measure of the unit sphere inR𝑛. For a solution(𝑢, 𝑝)of (11) we consider the following classical boundary operators:
𝑇𝑗𝑢 = [−𝛿𝑖𝑗𝑝 + 𝜇 (𝜕𝑗𝑢𝑖+ 𝜕𝑖𝑢𝑗)] 𝜈𝑖,
𝑇𝑗𝑢 = [𝛿𝑖𝑗𝑝 + 𝜇 (𝜕𝑗𝑢𝑖+ 𝜕𝑖𝑢𝑗)] 𝜈𝑖, 𝑗 = 1, . . . , 𝑛. (14) Through this paper,𝑝indicates 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 is any nonnegative integer, 𝐿𝑝ℎ(Σ) is the vector space of all differential forms of degreeℎ (brieflyℎ-forms) defined on Σ such that their components are integrable functions belonging to𝐿𝑝(Σ) in a coordinate system of class𝐶1 and consequently in every coordinate system of class 𝐶1. The space[𝐿𝑝ℎ(Σ)]𝑛is constituted by the vectors(𝑣1, . . . , 𝑣𝑛)such that𝑣𝑗is a differential form of𝐿𝑝ℎ(Σ)(𝑗 = 1, . . . , 𝑛).[𝑊1,𝑝(Σ)]𝑛 is the vector space of all measurable vector-valued functions 𝑢 = (𝑢1, . . . , 𝑢𝑛)such that𝑢𝑗 belongs to the Sobolev space 𝑊1,𝑝(Σ)(𝑗 = 1, . . . , 𝑛).
The pair(𝑣, 𝑟)with components 𝑣𝑖(𝑥) = − ∫
Σ𝛾𝑖𝑗(𝑥, 𝑦) 𝜑𝑗(𝑦) 𝑑𝜎𝑦, 𝑖 = 1, . . . , 𝑛, 𝑥 ∈R𝑛, (15) 𝑟 (𝑥) = − ∫
Σ𝜀𝑗(𝑥, 𝑦) 𝜑𝑗(𝑦) 𝑑𝜎𝑦, 𝑥 ∈R𝑛 (16)
is the simple layer hydrodynamic potential with density 𝜑.
The pair(𝑤, 𝑞)with components 𝑤𝑖(𝑥)=∫
Σ𝑇𝑗,𝑦 [𝛾𝑖(𝑥, 𝑦)] 𝜓𝑗(𝑦) 𝑑𝜎𝑦, 𝑖=1, . . . , 𝑛, 𝑥 ∈R𝑛, (17) 𝑞 (𝑥) = 2𝜇 ∫
Σ
𝜕
𝜕𝜈𝑦[𝜀𝑗(𝑥, 𝑦)] 𝜓𝑗(𝑦) 𝑑𝜎𝑦, 𝑥 ∈R𝑛 (18) is the double layer hydrodynamic potential with density𝜓.
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
∫|𝑦|=1log𝑥 − 𝑦𝑑𝑠𝑦= 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 domainΩis exceptionalif there exists some constant vector which cannot be represented inΩby a simple layer potential.
Denoting by Σ𝑅 the circle of radius 𝑅 centered at the origin, we have the following lemma.
Lemma 1. The circleΣ𝑅with𝑅 = exp(1/2)is exceptional for the Stokes system.
Proof. Keeping in mind that (see, e.g., [16, Section 4])
∫Σ𝑅
log𝑥 − 𝑦𝑑𝑠𝑦= 2𝜋𝑅log𝑅,
∫Σ𝑅
(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦2 𝑑𝑠𝑦= 𝛿𝑖𝑗𝜋𝑅, |𝑥| < 𝑅,
(20)
we find
∫Σ𝑅𝛾𝑖𝑗(𝑥, 𝑦) 𝑑𝑠𝑦= 𝑅
4𝜇𝛿𝑖𝑗(2log𝑅 − 1) , |𝑥| < 𝑅. (21) Taking𝑅 =exp(1/2)we obtain the result.
Let us consider now the exceptional boundaries of not simply connected domains.
Proposition 2. LetΩ ⊂R2be an(𝑚 + 1)-connected domain.
Denote byPthe eigenspace in[𝐿𝑝(Σ)]2of the singular integral system
∫Σ𝜑𝑗(𝑦) 𝜕
𝜕𝑠𝑥𝛾𝑖𝑗(𝑥, 𝑦) 𝑑𝑠𝑦= 0, 𝑎.𝑒. 𝑥 ∈ Σ, 𝑖 = 1, 2. (22) ThendimP= 2(𝑚 + 1).
Proof. As in the proof of [16, Lemma 12], one can show that
𝜕
𝜕𝑠𝑥𝛾𝑖𝑗(𝑥, 𝑦) = 1 4𝜋𝜇𝛿𝑖𝑗 𝜕
𝜕𝑠𝑥log𝑥 − 𝑦 +O(𝑦 − 𝑥ℎ−1) , (23) deduce that system (22) can be regularized to a Fredholm one, and see that its index is zero. Since the vectors𝑒𝑖𝜒Σ𝑗 (by𝜒𝑋we denote the characteristic function of the set𝑋) (𝑖 = 1, 2,𝑗 = 0, 1, . . . , 𝑚) are the only eigensolutions of the adjoint system
∫Σ𝜑𝑗(𝑦) 𝜕
𝜕𝑠𝑦𝛾𝑖𝑗(𝑥, 𝑦) 𝑑𝑠𝑦= 0, a.e.𝑥 ∈ Σ, 𝑖 = 1, 2, (24) we have dimP= 2(𝑚 + 1).
Theorem 3. LetΩ ⊂R2be an(𝑚+1)-connected domain. The following conditions are equivalent
(1)There exists a H¨older continuous vector function𝜑 ̸≡ 0 such that
∫Σ𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝑠𝑦= 0, 𝑥 ∈ Σ. (25) (2)There exists a constant vector which cannot be repre-
sented inΩby a simple layer potential;
(3)Σ0is exceptional.
(4)Let𝜑1, . . . , 𝜑2𝑚+2be linearly independent vectors ofP (see Proposition2), and let𝑐𝑗𝑘 = (𝛼𝑗𝑘, 𝛽𝑗𝑘) ∈ R2 be given by
∫Σ𝛾 ( 𝑥, 𝑦) 𝜑𝑗(𝑦) 𝑑𝑠𝑦= 𝑐𝑗𝑘, 𝑥 ∈ Σ𝑘, 𝑗 = 1, . . . , 2𝑚 + 2, 𝑘 = 0, 1, . . . , 𝑚.
(26)
ThendetC= 0, where
C= ( (
𝛼1,0 ⋅ ⋅ ⋅ 𝛼2𝑚+2,0
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 𝛼1,𝑚 ⋅ ⋅ ⋅ 𝛼2𝑚+2,𝑚
𝛽1,0 ⋅ ⋅ ⋅ 𝛽2𝑚+2,0
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 𝛽1,𝑚 ⋅ ⋅ ⋅ 𝛽2𝑚+2,𝑚
)
)
. (27)
Proof. The proof runs as in [16, Theorem 1] with obvious modifications. We omit the details.
5. Some Eigenspaces
We determine the structure of the kernel of a particular singular integral system. Namely, let us denote byN𝑝 the space of𝜓 ∈ [𝐿𝑝𝑛−2(Σ)]𝑛such that
∫Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0, a.e. on Σ, 𝑖 = 1, . . . , 𝑛.
(28) We begin by proving the following result.
Lemma 4. Let𝑢 ∈ [𝐶∞0 (R𝑛)]𝑛. Then, for any𝑥 ∈R𝑛, 𝑢𝑖(𝑥) = 𝜇 ∫
R𝑛Δ𝑢𝑗(𝑦) 𝛾𝑖𝑗(𝑥, 𝑦) 𝑑𝑦 + ∫
R𝑛
𝜕2𝑢𝑗(𝑦)
𝜕𝑦𝑖𝜕𝑦𝑗 𝑠 ( 𝑥, 𝑦) 𝑑𝑦, (29) where𝛾(𝑥, 𝑦)and𝑠(𝑥, 𝑦)are given by(12)and(7), respectively.
Proof. By the well-known Stokes identity we have 𝑢𝑖(𝑥) = ∫
R𝑛Δ𝑢𝑖(𝑦) 𝑠 (𝑥, 𝑦) 𝑑𝑦 = ∫
R𝑛Δ𝑢𝑗(𝑦) 𝛿𝑖𝑗𝑠 (𝑥, 𝑦) 𝑑𝑦.
(30) Since, for every𝑛 ̸= 2, 4,
(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦𝑛 = 1 (4 − 𝑛) (2 − 𝑛)
𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗𝑥 − 𝑦4−𝑛
− 𝛿𝑖𝑗𝜔𝑛𝑠 ( 𝑥, 𝑦) , 𝛾𝑖𝑗(𝑥, 𝑦) = 𝛿𝑖𝑗
2𝜇𝑠 ( 𝑥, 𝑦) − 1 2𝜇𝜔𝑛
(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦𝑛 , ∀𝑛 ≥ 2, (31) we can rewrite
𝛾𝑖𝑗(𝑥, 𝑦) = 𝛿𝑖𝑗
𝜇 𝑠 ( 𝑥, 𝑦) − 1 2𝜇𝜔𝑛
1 (4 − 𝑛) (2 − 𝑛)
𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗
× 𝑥 − 𝑦4−𝑛.
(32)
Then
𝛿𝑖𝑗𝑠 ( 𝑥, 𝑦)=𝜇𝛾𝑖𝑗(𝑥, 𝑦)+ 1 2𝜔𝑛(4−𝑛) (2−𝑛)
𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗𝑥−𝑦4−𝑛, 𝑢𝑖(𝑥)=𝜇 ∫
R𝑛Δ𝑢𝑗(𝑦) 𝛾𝑖𝑗(𝑥, 𝑦) 𝑑𝑦
+ 1
2𝜔𝑛(4−𝑛) (2−𝑛)∫
R𝑛Δ𝑢𝑖(𝑦) 𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗𝑥−𝑦4−𝑛𝑑𝑦.
(33) Integrating by parts, it follows that the last integral is equal to
1
2𝜔𝑛(4 − 𝑛) (2 − 𝑛)∫
R𝑛
𝜕2𝑢𝑗(𝑦)
𝜕𝑦𝑖𝜕𝑦𝑗 Δ𝑦𝑥 − 𝑦4−𝑛𝑑𝑦
= ∫R𝑛
𝜕2𝑢𝑗(𝑦)
𝜕𝑦𝑖𝜕𝑦𝑗 𝑠 ( 𝑥, 𝑦) 𝑑𝑦,
(34)
sinceΔ𝑦|𝑥 − 𝑦|4−𝑛= 2(4 − 𝑛)|𝑥 − 𝑦|2−𝑛. Then the claim holds for𝑛 ̸= 2, 4.
In the same manner it is possible to show formula (29) for 𝑛 = 2and𝑛 = 4after observing that, if𝑛 = 2, we have
(𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦2 = 1 2
𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗𝑥 − 𝑦2log𝑥 − 𝑦
− 𝛿𝑖𝑗log𝑥 − 𝑦 − 12𝛿𝑖𝑗,
(35)
Δ𝑦|𝑥 − 𝑦|2log|𝑥 − 𝑦| = 4(log|𝑥 − 𝑦| + 1), while, for𝑛 = 4, (𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦4 = 𝛿𝑖𝑗 2𝑥 − 𝑦2 +1
2
𝜕2
𝜕𝑦𝑖𝜕𝑦𝑗log𝑥 − 𝑦, (36) Δ𝑦log|𝑥 − 𝑦| = 2/|𝑥 − 𝑦|2.
Lemma 5. Let𝜁1, . . . , 𝜁𝑛be differential forms in𝐿𝑝𝑛−2(Σ)such that𝑑𝜁𝑗 = (−1)𝑛−1𝜈𝑗𝑑𝜎onΣ. One has𝜓 ∈N𝑝if, and only if,
𝜓𝑗 =∑𝑚
ℎ=0
𝑐ℎ𝜒Σℎ𝜁𝑗+ 𝜂𝑗, 𝑗 = 1, . . . , 𝑛, (37) where𝑐0, . . . , 𝑐𝑚 ∈ Rand𝜂1. . . , 𝜂𝑛 are weakly closed forms belonging to𝐿𝑝𝑛−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−1𝑥2𝑑𝑥3⋅ ⋅ ⋅ 𝑑𝑥𝑛, 𝜁𝑗 = (−1)𝑛−𝑗𝑥1𝑑𝑥2⋅ ⋅ ⋅ ̂𝑗⋅ ⋅ ⋅ 𝑑𝑥𝑛(𝑗 = 2, . . . , 𝑛). We remark that (37) holds if, and only if, the weak differentials𝑑𝜓𝑗exist and
𝑑𝜓𝑗 = (−1)𝑛−1∑𝑚
ℎ=0
𝑐ℎ𝜒Σℎ𝜈𝑗𝑑𝜎, 𝑗 = 1, . . . , 𝑛, (38) that is,
∫Σ𝜓𝑗∧ 𝑑𝑢𝑗 =∑𝑚
ℎ=0
𝑐ℎ∫
Σℎ
𝑢 ⋅ 𝜈𝑑𝜎, ∀𝑢 ∈ [𝐶∞0 (R𝑛)]𝑛. (39) Let us prove that (39) holds if, and only if,
∫Σ𝑘
𝜓𝑗∧ 𝑑𝑢𝑗 = 𝑐𝑘∫
Σ𝑘
𝑢 ⋅ 𝜈𝑑𝜎,
∀𝑢 ∈ [𝐶∞0 (R𝑛)]𝑛, 𝑘 = 0, . . . , 𝑚.
(40)
It is obvious that (40) implies (39).
Conversely, suppose that (39) is true. Define𝑈𝑘𝜀 = {𝑥 ∈ R𝑛 |dist(𝑥, Σ𝑘) < 𝜀}, where0 < 𝜀 <min0≤ℎ<𝑘≤𝑚dist(Σℎ, Σ𝑘).
Let𝑣𝑘 ∈ 𝐶∞0 (𝑈𝑘𝜀)be such that𝑣𝑘 = 1in𝑈𝑘𝜀/2. Since𝑣𝑘𝑢 ∈ [𝐶∞0 (R𝑛)]𝑛, we may write
∫Σ𝜓𝑗∧ 𝑑 (𝑣𝑘𝑢𝑗) =∑𝑚
ℎ=0
𝑐ℎ∫
Σℎ𝑣𝑘𝑢 ⋅ 𝜈𝑑𝜎, (41) and (40) follows immediately.
Suppose now that (39) is true. From (40) it follows that
∫Σ𝑘
𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 𝑐𝑘∫
Σ𝑘
𝛾𝑖𝑗(𝑥, 𝑦) 𝜈𝑗(𝑦) 𝑑𝜎𝑦,
∀𝑥 ∉ Σ𝑘. (42) An integration by parts shows that
∫Σ𝑘𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0, ∀𝑥 ∉ Ω𝑘. (43)
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
∫Σ𝑘
𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0 (44)
a.e. onΣ𝑘. Arguing as in [9, pages 189-190], this implies that
∫Σ𝑘
𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0 (45)
also inΩ𝑘. Summing over𝑘we find
∫Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0, (46)
for every𝑥 ∈ R𝑛 \ Σand a.e. onΣ. In particular 𝜓is 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
∫Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] = 0, 𝑥 ∉ Σ. (47)
Since 𝜀𝑗(𝑥, 𝑦) = −𝜕𝑥𝑗𝑠(𝑥, 𝑦), system (11) implies that Δ𝑥[𝛾𝑖𝑗(𝑥, 𝑦)] = −(1/𝜇)(𝜕2/𝜕𝑥𝑖𝜕𝑥𝑗)𝑠(𝑥, 𝑦). Hence,
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗 ∫
Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝑠 (𝑥, 𝑦)] = 0, 𝑥 ∉ Σ. (48) Therefore, there exist some constants𝑎0, 𝑎1, . . . , 𝑎𝑚such that
𝜕𝑗Ψ𝑗(𝑥) ={{ {{ {
−𝑎ℎ 𝑥 ∈ Ωℎ, ℎ = 1, . . . , 𝑚,
−𝑎0 𝑥 ∈ Ω, 0 𝑥 ∈R𝑛\ Ω,
(49)
where
Ψ𝑗(𝑥) = ∫
Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝑠 (𝑥, 𝑦)] . (50) Then, on account of Lemma4, for every𝑢 ∈ [𝐶∞0 (R𝑛)]𝑛,
∫Σ𝜓𝑗∧ 𝑑𝑢𝑗= 𝜇 ∫
R𝑛Δ𝑢𝑗(𝑥) 𝑑𝑥 ∫
Σ𝜓𝑗(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)]
+∫R𝑛
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) 𝑑𝑥∫
Σ𝜓𝑗(𝑦)∧𝑑𝑦[𝑠 (𝑥, 𝑦)] . (51)
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
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) Ψ𝑗(𝑥) 𝑑𝑥
=∑𝑚
ℎ=1
∫Ωℎ
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) Ψ𝑗(𝑥) 𝑑𝑥
+ ∫Ω
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) Ψ𝑗(𝑥) 𝑑𝑥
+ ∫R𝑛\Ω0
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) Ψ𝑗(𝑥) 𝑑𝑥
= −∑𝑚
ℎ=1
∫Σℎ𝜕𝑖𝑢𝑖Ψ𝑗𝜈𝑗𝑑𝜎 +∑𝑚
ℎ=1
∫Ωℎ𝜕𝑖𝑢𝑖𝜕𝑗Ψ𝑗𝑑𝑥
+ ∫Σ𝜕𝑖𝑢𝑖Ψ𝑗𝜈𝑗𝑑𝜎 − ∫
Ω𝜕𝑖𝑢𝑖𝜕𝑗Ψ𝑗𝑑𝑥
− ∫Σ0
𝜕𝑖𝑢𝑖Ψ𝑗𝜈𝑗𝑑𝜎 + ∫
R𝑛\Ω0
𝜕𝑖𝑢𝑖𝜕𝑗Ψ𝑗𝑑𝑥
=∑𝑚
ℎ=1
∫Ωℎ
𝜕𝑖𝑢𝑖𝜕𝑗Ψ𝑗𝑑𝑥 − ∫
Ω𝜕𝑖𝑢𝑖𝜕𝑗Ψ𝑗𝑑𝑥.
(52)
Hence, by (49),
∫R𝑛
𝜕2
𝜕𝑥𝑖𝜕𝑥𝑗𝑢𝑖(𝑥) Ψ𝑗(𝑥) 𝑑𝑥
= −∑𝑚
ℎ=1
𝑎ℎ∫
Ωℎ𝜕𝑖𝑢𝑖𝑑𝑥 + 𝑎0∫
Ω𝜕𝑖𝑢𝑖𝑑𝑥
=∑𝑚
ℎ=1
𝑎ℎ∫
Σℎ
𝑢 ⋅ 𝜈𝑑𝜎 + 𝑎0∫
Σ𝑢 ⋅ 𝜈𝑑𝜎
= 𝑎0∫
Σ0
𝑢 ⋅ 𝜈𝑑𝜎 +∑𝑚
ℎ=1
(𝑎0+ 𝑎ℎ) ∫
Σℎ
𝑢 ⋅ 𝜈𝑑𝜎.
(53)
By setting𝑐0 = 𝑎0and𝑐ℎ= 𝑎0+ 𝑎ℎ(ℎ = 1, . . . , 𝑚) we get the claim.
Remark 6. Lemma5shows that the dimension of the kernel N𝑝 is infinite. However, if we consider the quotient space N𝑝/Ξ𝑝,Ξ𝑝being the space of weakly closed differential forms in𝐿𝑝𝑛−2(Σ), we have dim(N𝑝/Ξ𝑝) = 𝑚 + 1.
We conclude this section by recalling some properties concerning the following eigenspaces:
V±= {𝜑𝑘∈ 𝐿𝑝(Σ) : ±1 2𝜑𝑘(𝑥)
+ ∫Σ𝐹𝑘𝑖(𝑥, 𝑦) 𝜑𝑖(𝑦) 𝑑𝜎𝑦 = 0, 𝑘 = 1, . . . , 𝑛} , W±= {𝜑𝑘∈ 𝐿𝑝(Σ) : ∓1
2𝜑𝑘(𝑥)
+ ∫Σ𝐹𝑖𝑘(𝑦, 𝑥) 𝜑𝑖(𝑦) 𝑑𝜎𝑦= 0, 𝑘 = 1, . . . , 𝑛} , (54) where (see, e.g., [22])
𝐹𝑘𝑖(𝑥, 𝑦) := 𝑇𝑖,𝑦 [𝛾𝑘(𝑥, 𝑦)]
= −𝑛 𝜔𝑛
(𝑥𝑘− 𝑦𝑘) ( 𝑥𝑖− 𝑦𝑖) (𝑥𝑗− 𝑦𝑗)
𝑥 − 𝑦𝑛+2 𝜈𝑗(𝑦) . (55) For the proofs of the following two results see [7, Lemma 3.3]
and [8, Theorem 3.2], respectively.
Proposition 7. The setsV+ andW− are linear subspaces of 𝐿1(Σ)and
dim(V+) =dim(W−) = 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 = 𝛿ℎ𝑘𝜌𝑖, 𝑖 = 1, . . . , 𝑛(𝑛 + 1)/2, ℎ, 𝑘 = 1, . . . , 𝑚, where 𝜌𝑖 are rigid displacement inR𝑛, specifically𝜌𝑖(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, every𝜓 ∈W−has the property that𝑣|Σ0 = 0, where𝑣is the simple layer potential with density𝜓.
Proposition 8. The setsV−andW+ are linear subspaces of 𝐿1(Σ)and
dim(V−) =dim(W+) = 𝑛 (𝑛 + 1)
2 + 𝑚. (57)
A basis for W+ is expressed by the fields {𝜓𝑖, 𝜈𝜒Σℎ : 𝑖 = 1, . . . , 𝑛(𝑛+1)/2, ℎ = 1, . . . , 𝑚}, where𝜓𝑖,𝑖 = 1, . . . , 𝑛(𝑛+1)/2 are zero onΣ \ Σ0, and such that the simple layer potentials with density𝜓𝑖are𝑛(𝑛+1)/2rigid displacement inΩ0(linearly independent for𝑛 ≥ 3).
Finally, every function𝜑which is the restriction toΣof a rigid displacement belongs toV−.
One recalls that if 𝜑 ∈ [𝐿1(Σ)]𝑛 belongs to one of the eigenspacesV±,W±, then𝜑 ∈ [𝐶𝜆(Σ)]𝑛. This follows from general results about integral equations (see [8, Lemma 31]
and [7, page 81]).
Remark 9. We can make the statement of Proposition 8 slightly more precise, saying thatthe simple layer potentials with density𝜓𝑖are𝑛(𝑛 + 1)/2rigid displacement inΩ0linear independent for any𝑛 ≥ 2, unless𝑛 = 2andΣ0is exceptional.
Indeed, let us show that if𝑛 = 2andΣ0is not exceptional, such rigid displacements are linearly independent. Let𝑐𝑖be such that
∑3 𝑖=1
𝑐𝑖∫
Σ0
𝜓𝑖(𝑦) 𝛾 (𝑥, 𝑦) 𝑑𝜎𝑦= 0, in Ω0. (58) We have also
∫Σ0
∑3 𝑖=1
𝑐𝑖𝜓𝑖(𝑦) 𝛾 (𝑥, 𝑦) 𝑑𝜎𝑦= 0, a.e.𝑥 ∈ Σ0. (59) Let𝜑 = ∑3𝑖=1𝑐𝑖𝜓𝑖. In view of the equivalence between (1) and (3) of Theorem3, 𝜑 has to vanish. Therefore 𝑐𝑖 = 0 (𝑖 = 1, 2, 3) because of the linearly independence of𝜓𝑖. On the other hand, if𝑛 = 2andΣ0 is exceptional, Theorem3 shows that the potentials with densities{𝜓𝑖}𝑖=1,2,3are linearly dependent.
6. Reduction of a Certain Singular Integral Operator
For every𝜓 ∈ 𝐿𝑝1(Σ), letΘℎbe the operator defined by Θℎ(𝜓) (𝑥) = ∗ (∫
Σ𝑑𝑥[𝑠𝑛−2( 𝑥, 𝑦)] ∧ 𝜓 ( 𝑦) ∧ 𝑑𝑥ℎ) , 𝑥 ∈ Ω,
(60)
where ∗ and 𝑑 denote the Hodge star operator and the exterior derivative, respectively, and𝑠ℎ(𝑥, 𝑦)is the doubleℎ- form introduced by Hodge in [23] as follows:
𝑠ℎ(𝑥, 𝑦) = ∑
𝑗1<⋅⋅⋅<𝑗ℎ
𝑠 ( 𝑥, 𝑦) 𝑑𝑥𝑗1⋅ ⋅ ⋅ 𝑑𝑥𝑗ℎ𝑑𝑦𝑗1⋅ ⋅ ⋅ 𝑑𝑦𝑗ℎ. (61) Note that the operatorΘℎsatisfies the equation
𝜕ℎ∫
Σ𝑢 ( 𝑦) 𝜕
𝜕𝜈𝑦𝑠 ( 𝑥, 𝑦) 𝑑𝜎𝑦= −Θℎ(𝑑𝑢) , 𝑥 ∈ Ω, (62) for each𝑢 ∈ 𝑊1,𝑝(Σ), since (see [9, page 187])
∗ 𝑑 ∫
Σ𝑢 ( 𝑦) 𝜕
𝜕𝜈𝑦𝑠 ( 𝑥, 𝑦) 𝑑𝜎𝑦
= 𝑑𝑥∫
Σ𝑑𝑢 (𝑦) ∧ 𝑠𝑛−2(𝑥, 𝑦) , 𝑥 ∈ Ω.
(63)
Moreover we introduce the operatorsH𝑗ℎdefined as H𝑗ℎ(𝜓) (𝑥) = Θℎ(𝜓𝑗) (𝑥) − 𝛿123⋅⋅⋅𝑛𝑙𝑖𝑗3⋅⋅⋅𝑗𝑛
(𝑛 − 2)!
× ∫Σ𝜕𝑥ℎ𝐻𝑙𝑗(𝑥, 𝑦) ∧ 𝜓𝑖(𝑦) ∧ 𝑑𝑦𝑗3⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛, (64)
for every𝜓 ∈ [𝐿𝑝1(Σ)]𝑛, where 𝐻𝑙𝑗(𝑥, 𝑦) = 1
𝜔𝑛
(𝑦𝑙− 𝑥𝑙) (𝑦𝑗− 𝑥𝑗)
𝑦 − 𝑥𝑛 . (65) In the sequel𝑑𝑢denotes the vector(𝑑𝑢1, . . . , 𝑑𝑢𝑛)whose elements are1-forms, and𝜓 = (𝜓1, . . . , 𝜓𝑛) ∈ [𝐿𝑝1(Σ)]𝑛. Lemma 10. Let (𝑤, 𝑞) be the double layer hydrodynamic potential of (17)-(18)with density𝑢 ∈ [𝑊1,𝑝(Σ)]𝑛. Then, for 𝑥 ∉ Σ,
𝜕ℎ𝑤𝑗(𝑥) =H𝑗ℎ(𝑑𝑢) (𝑥) , (66) 𝑞 (𝑥) = 2𝜇Θℎ(𝑑𝑢ℎ) (𝑥) , (67) whereH𝑗ℎandΘℎare given by(60)and(64), respectively.
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]).
For𝑘 > (𝑛 − 2)/𝑛, let(𝑘)𝑤be the double layer elastic potential with density𝑢, that is,
(𝑘)𝑤𝑗(𝑥) = ∫
Σ𝑢𝑖(𝑦)𝐿(𝑘)𝑖,𝑦[(𝑘)Γ𝑗 (𝑥, 𝑦)] 𝑑𝜎𝑦, (68) where (𝑘)𝐿 and (𝑘)Γ are the stress operator and the Kelvin’s matrix associated to the Lam´e system−Δ𝑢 − 𝑘∇div𝑢 = 0, respectively.
Thanks to [16, Lemma 1], we know that
𝜕ℎ(𝑘)𝑤𝑗(𝑥) =H(𝑘)𝑗ℎ(𝑑𝑢) (𝑥) , (69) where
H(𝑘)𝑗ℎ(𝜓) (𝑥) = Θℎ(𝜓𝑗) (𝑥) − 𝛿123⋅⋅⋅𝑛𝑙𝑖𝑗3⋅⋅⋅𝑗𝑛 (𝑛 − 2)!
× ∫Σ𝜕𝑥ℎ(𝑘)𝐻𝑙𝑗(𝑥, 𝑦) ∧ 𝜓𝑖(𝑦) ∧ 𝑑𝑦𝑗3⋅ ⋅ ⋅ 𝑑𝑦𝑗𝑛,
(𝑘)𝐻𝑙𝑗(𝑥, 𝑦) = 𝑘 𝜔𝑛(𝑘 + 1)
(𝑦𝑙− 𝑥𝑙) (𝑦𝑗− 𝑥𝑗)
𝑦 − 𝑥𝑛
− 1
𝑘 + 1𝛿𝑙𝑗𝑠 ( 𝑥, 𝑦) ,
(70) andΘℎis given by (60).
From [16, formula (5)] (where we set𝜉 = 1), letting𝑘 → +∞, we get
𝜕𝑥ℎ{𝐿(𝑘)𝑖,𝑦[(𝑘)Γ𝑗(𝑥, 𝑦)]}
→ − 𝑛
𝜔𝑛𝜕𝑥ℎ{(𝑦𝑖− 𝑥𝑖) (𝑦𝑗− 𝑥𝑗) ( 𝑦𝑘− 𝑥𝑘)
𝑥 − 𝑦𝑛+2 𝜈𝑘(𝑦)}
= 𝜕𝑥ℎ𝑇𝑖,𝑦 [𝛾𝑗(𝑥, 𝑦)] ,
(71)
𝑥 ∉ Σ, from which𝜕ℎ(𝑘)𝑤 → 𝜕ℎ𝑤as𝑘 → +∞. Therefore we obtain formula (66) by letting𝑘 → +∞in (69). Formula (67) is an immediate consequence of (62) because𝜀𝑗(𝑥, 𝑦) =
−𝜕𝑥𝑗𝑠(𝑥, 𝑦).
For the next lemma it is convenient to recall here two jump formulas proved in [16, Lemmas 2 and 3].
Let𝑓 ∈ 𝐿1(Σ). If𝜂 ∈ Σis a Lebesgue point for𝑓, we get
𝑥 → 𝜂lim∫
Σ𝑓 ( 𝑦) 𝜕𝑥𝑠(𝑦𝑙− 𝑥𝑙) (𝑦𝑗− 𝑥𝑗)
𝑥 − 𝑦𝑛 𝑑𝜎𝑦
= 𝜔𝑛
2 (𝛿𝑙𝑗− 2𝜈𝑗(𝜂) 𝜈𝑙(𝜂)) 𝜈𝑠(𝜂) 𝑓 (𝜂) + ∫Σ𝑓 ( 𝑦) 𝜕𝑥𝑠(𝑦𝑙− 𝜂𝑙) (𝑦𝑗− 𝜂𝑗)
𝑥 − 𝑦𝑛 𝑑𝜎𝑦,
(72)
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.
Further, let𝜓 ∈ 𝐿𝑝1(Σ)and write𝜓as𝜓 = 𝜓ℎ𝑑𝑥ℎwith
𝜈ℎ𝜓ℎ= 0. (73)
Assumption (73) is not restrictive, because, given the 1-form 𝜓onΣ, there exist scalar functions𝜓ℎdefined onΣsuch that 𝜓 = 𝜓ℎ𝑑𝑥ℎand (73) holds (see [24, page 41]). Then, for almost every𝜂 ∈ Σ,
𝑥 → 𝜂limΘℎ(𝜓) (𝑥) = −1
2𝜓ℎ(𝜂) + Θℎ(𝜓) (𝜂) , (74) whereΘℎis given by (60), and the limit has to be understood again as an internal angular boundary value.
Lemma 11. Let𝜓 ∈ 𝐿𝑝1(Σ). Let one write𝜓as𝜓 = 𝜓ℎ𝑑𝑥ℎand suppose that(73)holds. Then, for almost every𝜂 ∈ Σ,
𝑥 → 𝜂lim 1
(𝑛 − 2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
= −1
2[𝜈𝑗(𝜂) 𝜓𝑖(𝜂) + 𝜈𝑖(𝜂) 𝜓𝑗(𝜂)] 𝜈𝑠(𝜂)
+ 1
(𝑛−2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝜂, 𝑦)∧𝜓 (𝑦)∧𝑑𝑦𝑗3⋅ ⋅ ⋅∧𝑑𝑦𝑗𝑛, (75) where𝐻𝑙𝑗is defined by(65), and the limit has to be understood as an internal angular boundary value.
Proof. We have 1
(𝑛 − 2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
= 1
(𝑛 − 2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛𝛿123⋅⋅⋅𝑛𝑟ℎ𝑗3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝑥, 𝑦) 𝜓ℎ(𝑦) 𝜈𝑟(𝑦) 𝑑𝜎𝑦
= 𝛿𝑟ℎ𝑙𝑖 ∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝑥, 𝑦) 𝜓ℎ(𝑦) 𝜈𝑟(𝑦) 𝑑𝜎𝑦.
(76)
Hence, by (65) and (72),
𝑥 → 𝜂lim 1
(𝑛 − 2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝑥, 𝑦) ∧ 𝜓 (𝑦) ∧ 𝑑𝑦𝑗3⋅ ⋅ ⋅ ∧ 𝑑𝑦𝑗𝑛
=𝛿𝑙𝑖𝑟ℎ
2 (𝛿𝑙𝑗− 2𝜈𝑗(𝜂) 𝜈𝑙(𝜂)) 𝜈𝑠(𝜂) 𝜈𝑟(𝜂) 𝜓ℎ(𝜂)
+ 1
(𝑛−2)!𝛿𝑙𝑖𝑗123⋅⋅⋅𝑛3⋅⋅⋅𝑗𝑛∫
Σ𝜕𝑥𝑠𝐻𝑙𝑗(𝜂, 𝑦)∧𝜓 (𝑦)∧𝑑𝑦𝑗3⋅ ⋅ ⋅∧𝑑𝑦𝑗𝑛. (77) Keeping in mind (73), we find
𝛿𝑙𝑖𝑟ℎ
2 (𝛿𝑙𝑗− 2𝜈𝑗𝜈𝑙) 𝜈𝑠𝜈𝑟𝜓ℎ= (1
2𝛿𝑙𝑗𝜈𝑠− 𝜈𝑗𝜈𝑙𝜈𝑠) ( 𝜈𝑙𝜓𝑖− 𝜈𝑖𝜓𝑙)
= −1
2𝜈𝑠𝜈𝑗𝜓𝑖−1 2𝜈𝑠𝜈𝑖𝜓𝑗,
(78) and the result follows.
Lemma 12. Let𝜓 = (𝜓1, . . . , 𝜓𝑛) ∈ [𝐿𝑝1(Σ)]𝑛. Then, for almost every𝜂 ∈ Σ,
𝑥 → 𝜂lim𝜇 [2𝛿𝑖𝑗Θℎ(𝜓ℎ) (𝑥) +H𝑖𝑗(𝜓) (𝑥) +H𝑗𝑖(𝜓) (𝑥)] 𝜈𝑖(𝑥)
= 𝜇 [2𝛿𝑖𝑗Θℎ(𝜓ℎ) ( 𝜂) +H𝑖𝑗(𝜓) (𝜂) +H𝑗𝑖(𝜓) (𝜂)] 𝜈𝑖(𝜂) , (79) ΘℎandHbeing as in(60)and(64), respectively, and the limit has to be understood as an internal angular boundary value.
Proof. Let us write𝜓𝑖as𝜓𝑖= 𝜓𝑖ℎ𝑑𝑥ℎwith
𝜈ℎ𝜓𝑖ℎ= 0, 𝑖 = 1, . . . , 𝑛. (80) On account of (72) and (74), we infer
𝑥 → 𝜂lim𝜇 [2𝛿𝑖𝑗Θℎ(𝜓ℎ) (𝑥) +H𝑖𝑗(𝜓) (𝑥) +H𝑗𝑖(𝜓) (𝑥)] 𝜈𝑖(𝑥)
= 𝜇Ψ𝑖𝑗(𝜓) (𝜂) 𝜈𝑖(𝜂) + 𝜇 [2𝛿𝑖𝑗Θℎ(𝜓ℎ) ( 𝜂)
+H𝑖𝑗(𝜓) (𝜂) +H𝑗𝑖(𝜓) (𝜂)] 𝜈𝑖(𝜂) ,
(81) where
Ψ𝑖𝑗(𝜓) = − 𝛿𝑖𝑗𝜓ℎℎ−1 2𝜓𝑖𝑗+1
2(𝜈𝑖𝜓𝑠𝑠+ 𝜈𝑠𝜓𝑠𝑖) 𝜈𝑗
−1 2𝜓𝑗𝑖+1
2(𝜈𝑗𝜓𝑠𝑠+ 𝜈𝑠𝜓𝑠𝑗) 𝜈𝑖.
(82)
By (80) we getΨ𝑖𝑗(𝜓)𝜈𝑖= −𝜓ℎℎ𝜈𝑗− 𝜓𝑖𝑗𝜈𝑖/2 + 𝜓𝑠𝑠𝜈𝑗+ 𝜈𝑠𝜓𝑠𝑗/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.
Lemma 14. Let𝑤be the double layer potential(17)with den- sity𝑢 ∈ [𝑊1,𝑝(Σ)]𝑛. Then𝑇+,𝑗𝑤 = 𝑇−,𝑗𝑤 = 𝜇[2𝛿𝑖𝑗Θℎ(𝑑𝑢ℎ) + H𝑖𝑗(𝑑𝑢) +H𝑗𝑖(𝑑𝑢)]𝜈𝑖a.e. onΣ, where𝑇+𝑤and𝑇−𝑤denote the internal and the external angular boundary limits of𝑇𝑤, respectively, andΘℎis given by(60)andHby(64).
Proof. It is an immediate consequence of (66), (67), (79), and Remark13.
Proposition 15. Let𝑅 : [𝐿𝑝(Σ)]𝑛 → [𝐿𝑝1(Σ)]𝑛be the follow- ing singular integral operator
𝑅𝜑 (𝑥) = − ∫
Σ𝑑𝑥[𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦. (83) Let one define𝑅 : [𝐿𝑝1(Σ)]𝑛 → [𝐿𝑝(Σ)]𝑛to be the singular integral operator
𝑅𝑗(𝜓) (𝑥) = 𝜇 [2𝛿𝑖𝑗Θℎ(𝜓ℎ) (𝑥) +H𝑖𝑗(𝜓) (𝑥)
+H𝑗𝑖(𝜓) (𝑥)] 𝜈𝑖(𝑥) . (84) Then
𝑅𝑅𝜑 =1
4𝜑 − 𝐾2𝜑, (85)
where
𝐾𝜑 (𝑥) = − ∫
Σ𝑇𝑥[𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦. (86) Proof. Let𝑣be the simple layer potential (15) with density𝜑 ∈ [𝐿𝑝(Σ)]𝑛. In view of Lemma14, we have a.e. onΣ
𝑅𝑗(𝑅𝜑) = 𝜇 [2𝛿𝑖𝑗Θℎ(𝑑𝑣ℎ) +H𝑖𝑗(𝑑𝑣) +H𝑗𝑖(𝑑𝑣)] 𝜈𝑖= 𝑇𝑗𝑤, (87) where 𝑤 is the double layer potential (17) with density𝑣.
Moreover, if𝑥 ∈ Ω, 𝑤𝑘(𝑥) = ∫
Σ𝑣𝑖(𝑦) 𝑇𝑖,𝑦 [𝛾𝑘(𝑥, 𝑦)] 𝑑𝜎𝑦
= 𝑣𝑘(𝑥) + ∫
Σ𝛾𝑖𝑘(𝑥, 𝑦) 𝑇𝑖[𝑣 (𝑦)] 𝑑𝜎𝑦, (88)
and then, on account of (86), 𝑇𝑤 = 1
2𝑇𝑣 − 𝐾 (𝑇𝑣) = 1 2(1
2𝜑 + 𝐾𝜑) − 𝐾 (1
2𝜑 + 𝐾𝜑)
=1
4𝜑 − 𝐾2𝜑.
(89)
7. The Dirichlet Problem
Let us consider the Dirichlet problem for the Stokes system 𝜇Δ𝑣 = ∇𝑟 inΩ,
div𝑣 = 0 inΩ, 𝑣 = 𝑓 on Σ,
(90)
where the given data𝑓 ∈ [𝑊1,𝑝(Σ)]𝑛satisfies the compatibil- ity condition (4).
The aim of the present section is to study the repre- sentability of the solution of this problem by means of a simple layer hydrodynamic potential (15)-(16).
By the symbolS𝑝we mean the class of the simple layer hydrodynamic potentials (15)-(16) with density in[𝐿𝑝(Σ)]𝑛. Whenever𝑛 = 2andΣ0is exceptional (see Section4), we say that(𝑣, 𝑟)belongs toS𝑝if, and only if,
𝑣 (𝑥) = − ∫
Σ𝛾 (𝑥, 𝑦) 𝜑 (𝑦) 𝑑𝜎𝑦+ 𝑐, 𝑥 ∈ Ω, 𝑟 (𝑥) = − ∫
Σ𝜀𝑗(𝑥, 𝑦) 𝜑𝑗(𝑦) 𝑑𝜎𝑦, 𝑥 ∈ Ω,
(91)
where𝜑 ∈ [𝐿𝑝(Σ)]2and𝑐 ∈R2.
We will see that condition (4) is not sufficient to prove the existence of the solution in the classS𝑝, but it must be satisfied on eachΣ𝑗,𝑗 = 0, 1, . . . , 𝑚.
We begin by proving the following result.
Theorem 16. Given𝜔 ∈ [𝐿𝑝1(Σ)]𝑛, there exists a solution𝜑 ∈ [𝐿𝑝(Σ)]𝑛of the singular integral system
− ∫Σ𝑑𝑥[𝛾 (𝑥, 𝑦)] 𝜑 (𝑦) 𝑑𝜎𝑦= 𝜔 (𝑥) , 𝑎.𝑒. 𝑥 ∈ Σ, (92) if, and only if,
∫Σ𝜓𝑖∧ 𝜔𝑖= 0, (93) for every𝜓 = (𝜓1, . . . , 𝜓𝑛) ∈ [𝐿𝑞𝑛−2(Σ)]𝑛 (𝑞 = 𝑝/(𝑝 − 1))such that the weak differentials𝑑𝜓𝑗 exist and(38)holds for some real constants𝑐0, . . . , 𝑐𝑚.
Proof. Consider the adjoint of𝑅(see (83)),𝑅∗: [𝐿𝑞𝑛−2(Σ)]𝑛→ [𝐿𝑞(Σ)]𝑛, that is, the operator whose components are given by
𝑅∗𝑖𝜓 (𝑥) = − ∫
Σ𝜓𝑖(𝑦) ∧ 𝑑𝑦[𝛾𝑖𝑗(𝑥, 𝑦)] . (94) Proposition 15 implies that the integral system (92) has a solution𝜑 ∈ [𝐿𝑝(Σ)]𝑛if, and only if,
∫Σ𝜓𝑖∧ 𝜔𝑖= 0, (95) for each𝜓 = (𝜓1, . . . , 𝜓𝑛) ∈ [𝐿𝑞𝑛−2(Σ)]𝑛such that𝑅∗𝜓 = 0.
The result follows from Lemma5.
Proposition 17. Given𝑓 ∈ [𝑊1,𝑝(Σ)]𝑛, there exists a solution of the BVP
(̃𝑣, ̃𝑟) ∈S𝑝, 𝜇Δ̃𝑣 = ∇̃𝑟 𝑖𝑛 Ω,
diṽ𝑣 = 0 𝑖𝑛 Ω, 𝑑̃𝑣 = 𝑑𝑓 𝑜𝑛 Σ,
(96)
if, and only if, conditions (3) are satisfied. The density𝜑of the pair(̃𝑣, ̃𝑟)(see(15)-(16)) solves the singular integral system 𝑅𝜑 = 𝑑𝑓, where𝑅is given by(83).
