Open Access

Integral representations for solutions of some BVPs for the Lamé system in multiply connected domains

Boundary Value Problems20112011:53

DOI: 10.1186/1687-2770-2011-53

Received: 21 May 2011

Accepted: 12 December 2011

Published: 12 December 2011

Abstract

The present paper is concerned with an indirect method to solve the Dirichlet and the traction problems for Lamé system in a multiply connected bounded domain of n , n ≥ 2. It hinges on the theory of reducible operators and on the theory of differential forms. Differently from the more usual approach, the solutions are sought in the form of a simple layer potential for the Dirichlet problem and a double layer potential for the traction problem.

2000 Mathematics Subject Classification. 74B05; 35C15; 31A10; 31B10; 35J57.

Keywords

Lamé system boundary integral equations potential theory differential forms multiply connected domains

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., [27]).

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 [9, 10]) and the theory of differential forms (see, e.g., [11, 12]) 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 n (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 n , n ≥ 2, of the form Ω = Ω 0 \ j = 1 m Ω ̄ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq1_HTML.gif, where Ω j (j = 0, ..., m) are m + 1 bounded domains of n with connected boundaries Σ j C1, λ(λ (0, 1]) and such that Ω ̄ j Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq2_HTML.gif and Ω ̄ j Ω ̄ k = , j , k = 1 , . . . , m , j k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq3_HTML.gif. 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equa_HTML.gif
where u = (u1, ..., u n ) 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ1_HTML.gif
(1)

i, j = 1, ..., n, ω n being the hypersurface measure of the unit sphere in n .

As usual, we denote by E ( u , v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq4_HTML.gif 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equb_HTML.gif
where ε ih (u) and σ ih (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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equc_HTML.gif
Let us consider the boundary operator L ξ 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ2_HTML.gif
(2)

ξ being a real parameter. We remark that the operator L1 is just the stress operator 2σ ih ν h , which we shall simply denote by L, while Lk/(k+2)is the so-called pseudo-stress operator.

By the symbol S n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq5_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq6_HTML.gif. We denote by R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq7_HTML.gif the space of all rigid displacements whose dimension is n(n + 1)/2. As usual {e1, ..., e n } is the canonical basis for n .

For any 1 < p < +∞ we denote by [L p (Σ)] n the space of all measurable vector-valued functions u = (u1, ..., u n ) such that |u j | p is integrable over Σ (j = 1, ..., n). If h is any non-negative integer, L h p ( Σ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq8_HTML.gif is the vector space of all differential forms of degree h defined on Σ such that their components are integrable functions belonging to L p (Σ) in a coordinate system of class C1 and consequently in every coordinate system of class C1. The space [ L h p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq9_HTML.gif is constituted by the vectors (v1, ..., v n ) such that v j is a differential form of L h p ( Σ ) ( j = 1 , . . . , n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq10_HTML.gif. [W1, p(Σ)] n is the vector space of all measurable vector-valued functions u = (u1, ..., u n ) such that u j belongs to the Sobolev space W1,p(Σ) (j = 1, ..., n).

If B and B' are two Banach spaces and S : BB' 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 : BB 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 = 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., [9, 10]).

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq11_HTML.gif if, and only if, there exists φ [L p (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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ3_HTML.gif
(3)
Definition 2.2. The vector-valued function w belongs to D p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq12_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ4_HTML.gif
(4)

where [L y Γ(x, y)]' denotes the transposed matrix of L y [Γ(x, y)].

3 Preliminary results

3.1 On the first derivatives of a double layer potential

Let us consider the boundary operator L ξ defined by (2). Denoting by Γ j (x, y) the vector whose components are Γ ij (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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ5_HTML.gif
(5)
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ6_HTML.gif
(6)

while for ξk/(2 + k) the kernels L i , y ξ [ Γ j ( x , y ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq13_HTML.gif have a strong singularity on Σ.

Let us denote by w ξ 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ7_HTML.gif
(7)
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ8_HTML.gif
(8)
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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equd_HTML.gif
and s h (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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Eque_HTML.gif
Since, for a scalar function f and for a fixed h, we have *df dx h = (- 1)n- 1 h f 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ9_HTML.gif
(9)
where, for every ψ L 1 p ( Σ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq14_HTML.gif,
Θ h ( ψ ) ( x ) = * Σ d x [ s n - 2 ( x , y ) ] ψ ( y ) d x h , x Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ10_HTML.gif
(10)

The following lemma can be considered as an extension of formula (9) to elasticity. Here du denotes the vector (du1, ..., du n ) and ψ = (ψ1, ..., ψ n ) is an element of [ L 1 p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq15_HTML.gif.

Lemma 3.1. Let w ξ 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ11_HTML.gif
(11)
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ12_HTML.gif
(12)
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ13_HTML.gif
(13)

and Θ h is given by (10), h = 1, ..., n.

Proof. Let n ≥ 3. Denote by M hi the tangential operators M hi = ν h i - ν i 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equf_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equg_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equh_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ14_HTML.gif
(14)
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equi_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equj_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equk_HTML.gif

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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ15_HTML.gif
(15)

where the limit has to be understood as an internal angular boundary value1.

Proof. Let h pj (x) = x p x j |x|-n. Since h C( n \{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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ16_HTML.gif
(16)
where γ p j ( η ) = - 2 π 2 F ( h p j ) ( ν η ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq16_HTML.gif, F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq17_HTML.gif being the Fourier transform
F ( h ) ( x ) = n h ( y ) e - 2 π i x y d y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equl_HTML.gif
(see also [24] and note that in [23, 24] 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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equm_HTML.gif
and, since
F ( | x | 2 - n ) = π n / 2 - 2 Γ ( n / 2 - 1 ) | x | - 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equn_HTML.gif
(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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equo_HTML.gif
Finally, keeping in mind that ω n = 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 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equp_HTML.gif

Combining this formula with (16) we get (15).   □

Lemma 3.3. Let ψ L 1 p ( Σ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq18_HTML.gif. Let us write ψ as ψ = ψ h dx h with
ν h ψ h = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ17_HTML.gif
(17)
Then, for almost every η Σ,
lim x η Θ s ( ψ ) ( x ) = - 1 2 ψ s ( η ) + Θ s ( ψ ) ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ18_HTML.gif
(18)

where Θ s 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 ψ h defined on Σ such that ψ = ψ h dx h 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equq_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equr_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq19_HTML.gif and (18) is proved.   □

Lemma 3.4. Let ψ L 1 p ( Σ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq20_HTML.gif. Let us write ψ as ψ = ψ h dx h 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ19_HTML.gif
(19)

where K ξ 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equs_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equt_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equu_HTML.gif

and the result follows.   □

Lemma 3.5. Let ψ = ( ψ 1 , , ψ n ) [ L 1 p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq21_HTML.gif. 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ20_HTML.gif
(20)

K ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq22_HTML.gif being as in (12) and the limit has to be understood as an internal angular boundary value.

Proof. Let us write ψ i as ψ i = ψ ih dx h with
ν h ψ i h = 0 , i = 1 , , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ21_HTML.gif
(21)
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equv_HTML.gif
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equw_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equx_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equy_HTML.gif

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 w ξ 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ22_HTML.gif
(22)

a.e. on Σ, where L + ξ ( w ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq23_HTML.gif and L - ξ ( w ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq24_HTML.gif denote the internal and the external angular boundary limit of L ξ (w ξ ) respectively and K ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq25_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq26_HTML.gif be the following singular integral operator
R φ ( x ) = Σ d x [ Γ ( x , y ) ] φ ( y ) d σ y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ23_HTML.gif
(23)
Let us define R ξ : [ L 1 p ( Σ ) ] n [ L p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq27_HTML.gif 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ24_HTML.gif
(24)
Then
R ξ R φ = - 1 4 φ + ( T ξ ) 2 φ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ25_HTML.gif
(25)
where
T ξ φ ( x ) = Σ L x ξ [ Γ ( x , y ) ] φ ( y ) d σ y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ26_HTML.gif
(26)
Proof. Let u be the simple layer potential with density φ [L p (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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equz_HTML.gif
where w ξ 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaa_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equab_HTML.gif

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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ27_HTML.gif
(27)

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 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ28_HTML.gif
(28)
W ± = φ [ L p ( Σ ) ] n : ± 1 2 φ + T φ = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ29_HTML.gif
(29)
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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ30_HTML.gif
(30)

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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq28_HTML.gif and W - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq29_HTML.gif have dimension n(n + 1)m/2. Moreover
V + = { v h X Σ j : h = 1 , , n ( n + 1 ) / 2 , j = 1 , , m } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equac_HTML.gif

where {v h : h = 1 ..., n(n + 1)/2} is an orthonormal basis of the space R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq30_HTML.gif and X Σ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq31_HTML.gif is the characteristic function of Σ j .

Proof. We define the vector-valued functions α j , j = 1, ..., m as α j ( x ) = ( a + B x ) X Σ j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq32_HTML.gif, x Σ. For a fixed j = 1, ..., m, the function α j (x) belongs to V + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq33_HTML.gif; 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equad_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ31_HTML.gif
(31)
Now we prove that the following n(n + 1)m/2 eigensolutions of V + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq34_HTML.gif
w h j ( x ) = v h ( x ) X Σ j ( x ) , h = 1 , , n ( n + 1 ) / 2 , j = 1 , , m , x Σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equae_HTML.gif
are linearly independent. Indeed, if h = 1 n ( n + 1 ) 2 j = 1 m c h j w h j = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq35_HTML.gif, we have
h = 1 n ( n + 1 ) 2 c h j v h ( x ) = 0 , x Σ j , j = 1 , , m . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaf_HTML.gif
Then, by applying a classical uniqueness theorem to the domain Ω j ,
h = 1 n ( n + 1 ) 2 c h j v h ( x ) = 0 , x Ω j , j = 1 , , m , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equag_HTML.gif
from which it easily follows that
c h j = 0 , h = 1 , , n ( n + 1 ) / 2 , j = 1 , , m . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equah_HTML.gif
Thus, dim V + n ( n + 1 ) m / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq36_HTML.gif. On the other hand, suppose φ W - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq37_HTML.gif and let u be the simple layer potential with density φ. Since E u = 0 in Ω j and L - u = 0 on Σ j , u = a j + B j x on each connected component Ω j , j = 1, ..., m, and u = 0 in n \ Ω ¯ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq38_HTML.gif. Note that this is true also for n = 2, because φ W - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq39_HTML.gif implies Σ φ d σ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq40_HTML.gif. We can define a linear map τ as follows
τ : W - ( n × S n ) m φ ( a 1 , B 1 , , a m , B m ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equai_HTML.gif

If τ(φ) = 0, from a classical uniqueness theorem, we have that φ ≡ 0 in n . Thus, τ is an injective map and dim W - n ( n + 1 ) m / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq41_HTML.gif. The assertion follows from (30).   □

Lemma 3.12. The spaces V - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq42_HTML.gif and W + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq43_HTML.gif have dimension n(n + 1)/2. Moreover V - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq44_HTML.gif is constituted by the restrictions to Σ of the rigid displacements.

Proof. Let α R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq45_HTML.gif. 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaj_HTML.gif
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 ¯ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equak_HTML.gif
This shows that the restriction to Σ of α belongs to V - https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq46_HTML.gif and then dim V - dim R = n ( n + 1 ) / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq47_HTML.gif. On the other hand, suppose ϕ W + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq48_HTML.gif 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 + n × l n ϕ ( a , B ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equal_HTML.gif

If n ≥ 3, we have that σ(ϕ) = 0 implies u ≡ 0 in n 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq49_HTML.gif. We have σ | W + 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq50_HTML.gif is injective and its range does not contain the vectors ((1, 0), 0) and ((0, 1), 0)2. Therefore dim W + 0 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq51_HTML.gif. On the other hand, dim W + - 2 dim W + 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq52_HTML.gifand then dim W + 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq53_HTML.gif. In any case, dim W + n ( n + 1 ) / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq54_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equam_HTML.gif

Similar domains occur also in elasticity. In order to give explicitly such an example, let us prove the following lemma.

Lemma 4.1. Let Σ R 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ32_HTML.gif
(32)
Proof. Denote by u(x) the function on the left hand side of (32) and by Ω R the ball of radius R centered at the origin. Let us fix x0 Σ R . For any x Σ R 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equan_HTML.gif
and then u is constant on Σ R . Moreover
Δ u ( x ) = 4 Σ R ( 1 + log | x - y | ) d s y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equao_HTML.gif
and then also Δu is constant on Σ R . Since Δu is harmonic in Ω R and continuous on Ω ¯ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq55_HTML.gif, it is constant in Ω R 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equap_HTML.gif
The function u(x) - 2πR(1 + log R) |x|2 is continuous on Ω ¯ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq56_HTML.gif, harmonic in Ω R and constant on Σ R . Then it is constant in Ω R 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaq_HTML.gif

Corollary 4.2. Let Σ R 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ33_HTML.gif
(33)
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equar_HTML.gif
formula (32) implies
Σ R ( x 1 - y 1 ) 2 | x - y | 2 d s y = π R , | x | < R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equas_HTML.gif
In a similar way
Σ R ( x 2 - y 2 ) 2 | x - y | 2 d s y = π R , | x | < R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equat_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equau_HTML.gif

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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equav_HTML.gif

This implies that in Ω R , for such a value of R, we cannot represent any smooth solution of the system E u = 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 Σ R with R = exp[k/(2(k + 2))] is exceptional for the operator Δ + kdiv.

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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ34_HTML.gif
(34)

Lemma 4.4. Let Ω 2 be an (m + 1)-connected domain. Denote by P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq57_HTML.gif the eigenspace in [L p (Σ)]2 of the system (34). Then dim P = 2 ( m + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq58_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaw_HTML.gif
and, since
s x ( x i - y i ) ( x j - y j ) | x - y | 2 = i x j - y j | x - y | 2 + j x i - y i | x - y | 2 - 2 ( x i - y i ) ( x j - y j ) | x - y | 3 s x | x - y | = i x j log | x - y | + j x i log | x - y | - 2 ( x i - y i ) ( x j - y j ) | x - y | 2 s x log | x - y | = 2 i j - ( x i - y i ) ( x j - y j ) | x - y | 2 s x log | x - y | + O ( | y - x | h - 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equax_HTML.gif
(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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equay_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equaz_HTML.gif
and then the system (34) is of regular type (see [15, 29]). 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ35_HTML.gif
(35)

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq59_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equba_HTML.gif

for any f [C (2)]2. This can be siproved by the same method in [[13], pp. 189-190]. Therefore ψ has to be constant on each curve Σ j (j = 0, ..., m), i.e. ψ is a linear combination of e i X Σ j ( i = 1 , 2 , j = 0 , 1 , , m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq60_HTML.gif.   □

Theorem 4.5. Let Ω 2 be an (m + 1)-connected domain. The following conditions are equivalent:

I. there exists a Hölder continuous vector function φ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq61_HTML.gif such that
Σ Γ ( x , y ) φ ( y ) d s y = 0 , x Σ ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ36_HTML.gif
(36)

II. there exists a constant vector which cannot be represented in Ω by a simple layer potential (i.e., there exists c 2 such that c S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq62_HTML.gif);

III. Σ0 is exceptional;

IV. let φ1, ..., φ2m+2be linearly independent functions of P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq63_HTML.gif and let c jk = (α jk , β jk ) 2 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbb_HTML.gif
Then
det C = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ37_HTML.gif
(37)
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbc_HTML.gif

Proof. I II. Let u be the simple layer potential (3) with density φ.

Since u = 0 in Ω, and then on Σ k , we find that u = 0 also in Ω k (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 Σ k , k = 1, ..., m. This means that
Σ 0 Γ ( x , y ) φ ( y ) d s y = 0 , x Ω 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbd_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Eqube_HTML.gif
Arguing as before, we find ψ j = 0 on Σ k , 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbf_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbg_HTML.gif
Lemma 4.4 shows that they are linearly dependent. Let λ, μ1, μ2 such that (λ, μ1, μ2) ≠ (0, 0, 0) and
λ φ + μ 1 ψ 1 + μ 2 ψ 2 = 0 on  Σ 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ38_HTML.gif
(38)
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbh_HTML.gif

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 2 there exists ϱ [C λ 0)]2 such that
Σ 0 Γ ( x , y ) ϱ ( y ) d s y = c , x Ω 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbi_HTML.gif
Setting
φ ( y ) = ϱ ( y ) y Σ 0 , 0 y Σ \ Σ 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbj_HTML.gif
we can write
Σ Γ ( x , y ) φ ( y ) d s y = c , x Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbk_HTML.gif

and this contradicts II.

III IV. Let us suppose det C 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq64_HTML.gif. For any c = (α, β) 2 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbl_HTML.gif
i.e.
j = 1 2 m + 2 λ j c j k = c , k = 0 , , m . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbm_HTML.gif
Therefore
Σ Γ ( x , y ) j = 1 2 m + 2 λ j φ j ( y ) d s y = c , x Σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbn_HTML.gif

Arguing as before, this leads to j = 1 2 m + 2 λ j φ j = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq65_HTML.gif on Σ k 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbo_HTML.gif
Set
φ ( x ) = j = 1 2 m + 2 λ j φ j ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbp_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq66_HTML.gif if, and only if,
u ( x ) = Σ Γ ( x , y ) φ ( y ) d s y + c , x Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ39_HTML.gif
(39)

where φ [L p (Σ)]2 and c 2.

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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ40_HTML.gif
(40)

where f [W1,p(Σ)] n .

We establish some preliminary results.

Theorem 5.1. Given ω [ L 1 p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq67_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ41_HTML.gif
(41)
if, and only if,
Σ γ ω i = 0 , i = 1 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ42_HTML.gif
(42)

for every γ L n - 2 q ( Σ ) ( q = p / ( p - 1 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq68_HTML.gif such that γ is a weakly closed (n - 2)-form.

Proof. Denote by R * : [ L n - 2 q ( Σ ) ] n [ L q ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq69_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbq_HTML.gif
Thanks to Corollary 3.10, the integral system (41) admits a solution φ [L p (Σ)] n if, and only if,
Σ ψ i ω i = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ43_HTML.gif
(43)

for any ψ = ( ψ 1 , , ψ n ) [ L n - 2 q ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq70_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ44_HTML.gif
(44)

It is given by (3), where the density φ [L p (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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ45_HTML.gif
(45)

in which the unknown is φ [L p (Σ)] n and the datum is d f [ L 1 p ( Σ ) ] n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq71_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq72_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbr_HTML.gif

The dimension of F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq73_HTML.gif is nm. This can be proved as in [[30], p. 63], where the case n = 3 is considered.

Theorem 5.3. Given c0, c1, ..., c m n , 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ46_HTML.gif
(46)
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equ47_HTML.gif
(47)
where Ψ h , i F ( h = 1 , , m , i = 1 , , n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq74_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbs_HTML.gif
Proof. Let ψ1, ..., ψ nm be nm linearly independent eigensolutions of the space F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_IEq75_HTML.gif. For a fixed j = 1, ..., nm we set
V j ( x ) = Σ Γ ( x , y ) ψ j ( y ) d σ y , x Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-53/MediaObjects/13661_2011_Article_98_Equbt_HTML.gif
Then