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Boundary Value Problems This Provisional PDF corresponds to the article as it appeared upon acceptance. Fully formatted PDF and full text (HTML) versions will be made available soon. Integral representations for solutions of some BVPs for the Lame' system in multiply connected domains Boundary Value Problems 2011, 2011:53 doi:10.1186/1687-2770-2011-53 Alberto Cialdea (cialdea@email.it) Vita Leonessa (vita.leonessa@unibas.it) Angelica Malaspina (angelica.malaspina@unibas.it) ISSN 1687-2770 Article type Research Submission date 21 May 2011 Acceptance date 12 December 2011 Publication date 12 December 2011 Article URL http://www.boundaryvalueproblems.com/content/2011/1/53 This peer-reviewed article was published immediately upon acceptance. It can be downloaded, printed and distributed freely for any purposes (see copyright notice below). For information about publishing your research in Boundary Value Problems go to http://www.boundaryvalueproblems.com/authors/instructions/ For information about other SpringerOpen publications go to http://www.springeropen.com © 2011 Cialdea et al. ; licensee Springer. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Integral representations for solutions of some BVPs for the Lam´ system in multiply connected domains e Alberto Cialdea∗ 1 , Vita Leonessa1 and Angelica Malaspina1 1 Department of Mathematics and Computer Science, University of Basilicata, V.le dell’Ateneo Lucano, 10, Campus of Macchia Romana, 85100 Potenza, Italy Email: Alberto Cialdea∗ - cialdea@email.it; Vita Leonessa - vita.leonessa@unibas.it; Angelica Malaspina - angelica.malaspina@unibas.it; ∗ Corresponding author Abstract The present paper is concerned with an indirect method to solve the Dirichlet and the traction problems for Lam´ system in a multiply connected bounded domain of Rn , n ≥ 2. It hinges on the theory of reducible operators e 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 and phrases. Lam´ system; boundary integral equations; potential theory; diﬀerential forms; e multiply connected domains. 1 Introduction In this paper we consider the Dirichlet and the traction problems for the linearized n-dimensional elasto- statics. 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., [2–7]). 1
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 Rn (n ≥ 2). Similar results for Laplace equation have been recently obtained in [17]. We remark that we do not require the use of pseudo-diﬀerential operators nor the use of hypersingular integrals, diﬀerently from other methods (see, e.g., [18, Chapter 4] for the study of the Neumann problem for Laplace equation by means of a double layer potential). After giving some notations and deﬁnitions in Section 2, we prove some preliminary results in Section 3. They concern the study of the ﬁrst derivatives of a double layer potential. This leads to the construction of a reducing operator, which will be useful in the study of the integral system of the ﬁrst kind arising in the Dirichlet problem. Section 4 is devoted to the case n = 2, where there exist some exceptional boundaries in which we need to add a constant vector to the simple layer potential. In particular, after giving an explicit example of such boundaries, we prove that in a multiply connected domain the boundary is exceptional if, and only if, the external boundary is exceptional. In Section 5 we ﬁnd the solution of the Dirichlet problem in a multiply connected domain by means of a simple layer potential. We show how to reduce the problem to an equivalent Fredholm equation (see Remark 5.5). Section 6 is devoted to the traction problem. It turns out that the solution of this problem does exist in the form of a double layer potential if, and only if, the given forces are balanced on each connected component of the boundary. While in a simply connected domain the solution of the traction problem can be always represented by means of a double layer potential (provided that, of course, the given forces are balanced on the boundary), this is not true in a multiply connected domain. Therefore the presence or absence of “holes” makes a diﬀerence. We mention that lately we have applied the same method to the study of the Stokes system [19]. Moreover 2
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 C 1 or Lipschitz boundaries and [21] for ”worse” domains) lead one to hope that our approach could be extended to more general domains. 2 Notations and deﬁnitions Throughout this paper we consider a domain (open connected set) Ω ⊂ Rn , n ≥ 2, of the form Ω = m Ωj , where Ωj (j = 0, . . . , m) are m + 1 bounded domains of Rn with connected boundaries Ω0 \ j =1 Σj ∈ C 1,λ (λ ∈ (0, 1]) and such that Ωj ⊂ Ω0 and Ωj ∩ Ωk = ∅, j, k = 1, . . . , m, j = k . For brevity, we shall call such a domain an (m + 1)-connected domain. We denote by ν the outwards unit normal on Σ = ∂ Ω. Let E be the partial diﬀerential operator Eu = ∆u + k divu, where u = (u1 , . . . , un ) is a vector-valued function and k > (n − 2)/n is a real constant. A fundamental −E is given by Kelvin’s matrix whose entries are solution of the operator  (xi − yi )(xj − yj ) 1 k+2 k − δij log |x − y | + , if n = 2,   |x − y |2 2π 2(k + 1) 2(k + 1)  Γij (x, y ) = (1) |x − y |2−n (xi − yi )(xj − yj ) 1 k+2 k − , if n ≥ 3, δij +   |x − y |n 2−n  ωn 2(k + 1) 2(k + 1) i, j = 1, . . . , n, ωn being the hypersurface measure of the unit sphere in Rn . As usual, we denote by E (u, v ) the bilinear form deﬁned as E (u, v ) = 2σih (u) εih (v ) = 2σih (v ) εih (u), where εih (u) and σih (u) are the linearized strain components and the stress components respectively, i.e. k−1 1 εih (u) = (∂i uh + ∂h ui ), σih (u) = εih (u) + δih εjj (u) . 2 2 Let us consider the boundary operator Lξ whose components are Lξ u = (k − ξ )(div u) νi + νj ∂j ui + ξνj ∂i uj , i = 1, . . . , n, (2) i ξ being a real parameter. We remark that the operator L1 is just the stress operator 2σih νh , which we shall simply denote by L, while Lk/(k+2) is the so-called pseudo-stress operator. By the symbol Sn 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, 3
i.e. a is a constant vector and B ∈ Sn . We denote by R the space of all rigid displacements whose dimension is n(n + 1)/2. As usual {e1 , . . . , en } is the canonical basis for Rn . For any 1 < p < +∞ we denote by [Lp (Σ)]n the space of all measurable vector-valued functions u = (u1 , . . . , un ) such that |uj |p is integrable over Σ (j = 1, . . . , n). If h is any non-negative integer, Lp (Σ) is h the vector space of all diﬀerential forms of degree h deﬁned on Σ such that their components are integrable functions belonging to Lp (Σ) in a coordinate system of class C 1 and consequently in every coordinate system of class C 1 . The space [Lp (Σ)]n is constituted by the vectors (v1 , . . . , vn ) such that vj is a diﬀerential h form of Lp (Σ) (j = 1, . . . , n). [W 1,p (Σ)]n is the vector space of all measurable vector-valued functions h u = (u1 , . . . , un ) such that uj belongs to the Sobolev space W 1,p (Σ) (j = 1, . . . , n). If B and B are two Banach spaces and S : B → B is a continuous linear operator, we say that S can be reduced on the left if there exists a continuous linear operator S : B → B such that S S = I + T , where I stands for the identity operator of B and T : B → B is compact. Analogously, one can deﬁne an operator S reducible on the right. One of the main properties of such operators is that the equation Sα = β has a solution if, and only if, γ , β = 0 for any γ such that S ∗ γ = 0, S ∗ being the adjoint of S (for more details see, e.g., [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 if, and only if, there exists ϕ ∈ [Lp (Σ)]n such that u can be represented by a simple layer potential x ∈ Ω. u(x) = Γ(x, y ) ϕ(y ) dσy , (3) Σ Deﬁnition 2.2. The vector-valued function w belongs to Dp if, and only if, there exists ψ ∈ [W 1,p (Σ)]n such that w can be represented by a double layer potential x ∈ Ω, w ( x) = [Ly Γ(x, y )] ψ (y ) dσy , (4) Σ where [Ly Γ(x, y )] denotes the transposed matrix of Ly [Γ(x, y )]. 4
3 Preliminary results 3.1 On the ﬁrst derivatives of a double layer potential Let us consider the boundary operator Lξ deﬁned by (2). Denoting by Γj (x, y ) the vector whose components are Γij (x, y ), we have 2 + (1 − ξ )k nk (ξ + 1) (yi − xi )(yj − xj ) (yp − xp )νp (y ) 1 Lξ [Γj (x, y )] = − δij + i,y |y − x|2 |y − x|n ωn 2(1 + k ) 2(k + 1) k − (2 + k )ξ (yj − xj )νi (y ) − (yi − xi )νj (y ) + . (5) |y − x|n 2(k + 1) We recall that an immediate consequence of (5) is that, when ξ = k/(2 + k ) we have k/(2+k) [Γj (x, y )] = O(|x − y |1−n+λ ) , Li,y (6) while for ξ = k/(2 + k ) the kernels Lξ [Γj (x, y )] have a strong singularity on Σ. i,y Let us denote by wξ the double layer potential ξ ui (y )Lξ [Γj (x, y )] dσy , wj (x) = j = 1, . . . , n. (7) i,y Σ It is known that the ﬁrst derivatives of a harmonic double layer potential with density ϕ belonging to W 1,p (Σ) can be written by means of the formula proved in [13, p. 187] ∂s(x, y ) ∗d dϕ(y ) ∧ sn−2 (x, y ), x ∈ Ω. ϕ(y ) dσy = dx (8) ∂νy Σ Σ Here ∗ and d denote the Hodge star operator and the exterior derivative respectively, s(x, y ) is the funda- mental solution of Laplace equation  1 log |x − y | ,  if n = 2,  2π s(x, y ) = 1 |x − y |2−n , if n ≥ 3  (2 − n)ωn  and sh (x, y ) is the double h-form introduced by Hodge in [22] s(x, y )dxj1 . . . dxjh dy j1 . . . dy jh . sh (x, y ) = j1
where, for every ψ ∈ Lp (Σ), 1 dx [sn−2 (x, y )] ∧ ψ (y ) ∧ dxh , Θh (ψ )(x) = ∗ x ∈ Ω. (10) Σ The following lemma can be considered as an extension of formula (9) to elasticity. Here du denotes the vector (du1 , . . . , dun ) and ψ = (ψ1 , . . . , ψn ) is an element of [Lp (Σ)]n . 1 Lemma 3.1. Let wξ be the double layer potential (7) with density u ∈ [W 1,p (Σ)]n . Then ξ ξ ∂s wj (x) = Kjs (du)(x), x ∈ Ω, j, s = 1, . . . , n, (11) where 1 ξ ξ δ 123...n ∂xs Khj (x, y ) ∧ ψi (y ) ∧ dy j3 . . . dy jn , Kjs (ψ )(x) = Θs (ψj )(x) − (12) (n − 2)! hij3 ...jn Σ 1 k (ξ + 1) (yl − xl )(yj − xj ) k − (2 + k )ξ ξ Khj (x, y ) = + δlj s(x, y ) , (13) |y − x|n ωn 2(k + 1) 2(k + 1) and Θh is given by (10), h = 1, . . . , n. Proof. Let n ≥ 3. Denote by M hi the tangential operators M hi = νh ∂i − νi ∂h , h, i = 1, . . . , n. By observing that xh xj δij xh νh xi xj xh νh M hi −n = , |x|n |x|n |x|n+2 we ﬁnd in Ω (yh − xh )νh (y ) k (ξ + 1) hi (yh − xh )(yj − xj ) 1 ξ wj (x) = − − ui (y ) δij M 2(k + 1) y |y − x|n |y − x|n ωn Σ k − (2 + k )ξ ij |y − x|2−n + My dσy = 2−n 2(k + 1) k (ξ + 1) hi (yh − xh )(yj − xj ) ∂s(x, y ) − (2 − n)s(x, y ) uj (y ) dσy + ui (y ) M 2(k + 1) y |y − x|2 ∂νy Σ Σ k − (2 + k )ξ ij − My [s(x, y )] dσy . 2(k + 1) An integration by parts on Σ leads to k (ξ + 1)(2 − n) (yh − xh )(yj − xj ) ∂s(x, y ) ξ M hi [ui (y )] w j ( x) = − dσy − uj (y ) |y − x|2 ∂νy 2(k + 1) Σ Σ k − (2 + k )ξ ∂s(x, y ) ξ M hi [ui (y )] Khj (x, y ) dσy . s(x, y ) dσy = − dσy − + δhj uj (y ) 2(k + 1) ∂νy Σ Σ 6
Therefore, by recalling (9), ξ ξ M hi [ui (y )] ∂xs [Khj (x, y )] dσy . ∂s wj (x) = Θs (duj )(x) − (14) Σ If f is a scalar function, we may write 1 M hi (f ) dσ = δ 123...n df ∧ dxj3 . . . dxjn . (n − 2)! hij3 ...jn This identity is established by observing that on Σ we have 1 1 δ 123...n df ∧ dxj3 . . . dxjn = δ 123...n ∂j f dxj2 ∧ . . . dxjn = (n − 2)! hij3 ...jn (n − 2)! hij3 ...jn 2 1 δ 123...n δ 1...n νj ∂j f dσ = δj1 j2 νj1 ∂j2 f dσ = (νh ∂i f − νi ∂h f ) dσ . hi (n − 2)! hij3 ...jn j1 ...jn 1 2 Then we can rewrite (14) as 1 ξ ξ δ 123...n ∂xs [Khj (x, y )] ∧ dui (y ) ∧ dy j3 . . . dy jn . ∂s wj (x) = Θs (duj )(x) − (n − 2)! hij3 ...jn Σ 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 (yp − xp )(yj − xj ) lim f (y )∂xs dσy = |y − x|n x→ η Σ (15) (yp − ηp )(yj − ηj ) ωn (δpj − 2νj (η )νp (η )) νs (η ) f (η ) + f (y )∂xs dσy , |y − η |n 2 Σ where the limit has to be understood as an internal angular boundary value1 . Proof. Let hpj (x) = xp xj |x|−n . Since h ∈ C ∞ (Rn \ {0}) is even and homogeneous of degree 2 − n, due to the results proved in [23], we have (yp − xp )(yj − xj ) (yp − ηp )(yj − ηj ) dσy = −νs (η )γpj (η ) f (η ) + lim f (y )∂xs f (y )∂xs dσy , (16) |y − x|n |y − η |n x→η Σ Σ where γpj (η ) = −2 π 2 F (hpj )(νη ), F being the Fourier transform h(y ) e−2π i x·y dy F (h)(x) = Rn (see also [24] and note that in [23, 24] ν is the inner normal). On the other hand 1 1 1 F (xp ∂j (|x|2−n )) = − ∂p F (∂j (|x|2−n )) = − ∂p (xj F (|x|2−n )) F (hpj )(x) = 2−n (2 − n) 2πi 2−n 7
and, since π n/2−2 F (|x|2−n ) = |x|−2 Γ(n/2 − 1) (see, e.g., [25, p. 156]), we ﬁnd π n/2−2 π n/2−2 ∂p (xj |x|−2 ) = (δpj |x|−2 − 2xj xp |x|−4 ). F (hpj )(x) = (n − 2) Γ(n/2 − 1) (n − 2) Γ(n/2 − 1) Finally, keeping in mind that ωn = n π n/2 /Γ(n/2 + 1) and Γ(n/2 + 1) = n(n − 2) Γ(n/2 − 1)/4, we obtain π n/2 ωn γpj (η ) = −2 (δpj − 2νj (η )νp (η )) = − (δpj − 2νj (η )νp (η )). (n − 2) Γ(n/2 − 1) 2 Combining this formula with (16) we get (15). Lemma 3.3. Let ψ ∈ Lp (Σ). Let us write ψ as ψ = ψh dxh with 1 νh ψh = 0. (17) Then, for almost every η ∈ Σ, 1 lim Θs (ψ )(x) = − ψs (η ) + Θs (ψ )(η ), (18) 2 x→η where Θs is given by (10) and the limit has to be understood as an internal angular boundary value. Proof. First we note that the assumption (17) is not restrictive, because, given the 1-form ψ on Σ, there exist scalar functions ψh deﬁned on Σ such that ψ = ψh dxh and (17) holds (see [26, p. 41]). We have ∂xi [s(x, y )]ψh (y ) dy j1 . . . dy jn−2 dy h dxi dxj1 . . . dxjn−2 dxs ∗ Θs (ψ )(x) = = Σ j1
Proof. We have 1 ∂x K ξ (x, y ) ∧ ψ (y ) ∧ dy j3 . . . dy jn = δ 123...n (n − 2)! lij3 ...jn Σ s lj 1 ∂x K ξ (x, y )ψh (y )νr (y ) dσy = δ 123...n δ 123...n (n − 2)! lij3 ...jn rhj3 ...jn Σ s lj ξ li δrh ∂xs Klj (x, y )ψh (y )νr (y ) dσy . Σ Keeping in mind (13), formula (15) leads to 1 ξ δ 123...n ∂xs Klj (x, y ) ∧ ψ (y ) ∧ dy j3 . . . dy jn = lim (n − 2)! lij3 ...jn x→η Σ k − (2 + k )ξ k (ξ + 1) li (δlj − 2νj (η )νl (η ))νs (η ) − δrh δlj νs (η ) νr (η ) ψh (η ) 4(k + 1) 4(k + 1) 1 ξ δ 123...n ∂xs Klj (η, y ) ∧ ψ (y ) ∧ dy j3 . . . dy jn . + (n − 2)! lij3 ...jn Σ On the other hand k − (2 + k )ξ k (ξ + 1) ξ k (ξ + 1) li li (δlj − 2νj νl )νs − δlj νs − δrh δlj νs νr ψh = δrh νj νl νs νr ψh = 4(k + 1) 4(k + 1) 2 2(k + 1) k−ξ ξ k (ξ + 1) ξ δlj νs − νj νl νs (νl ψi − νi ψl ) = − νj νs ψi − νi νs ψj , 2 2(k + 1) 2(k + 1) 2 and the result follows. Lemma 3.5. Let ψ = (ψ1 , . . . , ψn ) ∈ [Lp (Σ)]n . Then, for almost every η ∈ Σ, 1 ξ ξ ξ lim [(k − ξ )Kjj (ψ )(x)νi (η ) + νj (η )Kij (ψ )(x) + ξνj (η )Kji (ψ )(x)] = x→η ξ ξ ξ (k − ξ )Kjj (ψ )(η )νi (η ) + νj (η )Kij (ψ )(η ) + ξνj (η )Kji (ψ )(η ) , (20) Kξ being as in (12) 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, i = 1, . . . , n. (21) In view of Lemmas 3.3 and 3.4 we have k−ξ 1 ξ ξ ξ lim Kjs (ψ )(x) = − ψjs (η ) + νj (η )ψhh (η ) + νh (η ) ψhj (η ) νs (η ) + Kjs (ψ )(η ). 2 2(k + 1) 2 x→η 9
Therefore ξ ξ ξ lim [(k − ξ )Kjj (ψ )(x)νi (η ) + νj (η )Kij (ψ )(x) + ξνj (η )Kji (ψ )(x)] = x→η ξ ξ ξ Φ(ψ )(η ) + (k − ξ )Kjj (ψ )(η )νi (η ) + νj (η )Kij (ψ )(η ) + ξνj (η )Kji (ψ )(η ), where k−ξ 1 ξ Φ(ψ ) = (k − ξ ) − ψjj + νj ψhh + νh ψhj νj νi 2 2(k + 1) 2 k−ξ k−ξ 1 ξ 1 ξ +νj − ψij + νi ψhh + νh ψhi νj + ξ νj − ψji + νj ψhh + νh ψhj νi . 2 2(k + 1) 2 2 2(k + 1) 2 Conditions (21) lead to k−ξ k−ξ k−ξ 1 Φ(ψ ) = − (k − ξ ) 1 − − −ξ νi ψhh . 2 k+1 k+1 k+1 The bracketed expression vanishing, Φ = 0 and the result is proved. Remark 3.6. In Lemmas 3.2, 3.3, 3.4 and 3.5 we have considered internal angular boundary values. It is clear that similar formulas hold for external angular boundary values. We have just to change the sign in the ﬁrst term on the right hand sides in (15), (18) and (19), while (20) remains unchanged. 3.3 Reduction of a certain singular integral operator The results of the previous subsection imply the following lemmas. Lemma 3.7. Let wξ be the double layer potential (7) with density u ∈ [W 1,p (Σ)]n . Then Lξ ,i (wξ ) = Lξ ,i (wξ ) = (k − ξ )Kjj (du)νi + νj Kij (du) + ξνj Kji (du) ξ ξ ξ (22) − + a.e. on Σ, where Lξ (wξ ) and Lξ (wξ ) denote the internal and the external angular boundary limit of Lξ (wξ ) − + respectively and Kξ 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 : [Lp (Σ)]n → [Lp (Σ)]n be the following singular integral operator 1 Rϕ(x) = dx [Γ(x, y )] ϕ(y ) dσy . (23) Σ 10
Let us deﬁne R ξ : [Lp (Σ)]n → [Lp (Σ)]n to be the singular integral operator 1 Riξ (ψ )(x) = (k − ξ )Kjj (ψ )(x)νi (x) + νj (x)Kij (ψ )(x) + ξνj (x)Kji (ψ )(x). ξ ξ ξ (24) Then 1 R ξ Rϕ = − ϕ + (T ξ )2 ϕ, (25) 4 where T ξ ϕ(x) = Lξ [Γ(x, y )] ϕ(y ) dσy . (26) x Σ Proof. Let u be the simple layer potential with density ϕ ∈ [Lp (Σ)]n . In view of Lemma 3.7, we have a.e. on Σ Riξ (Rϕ) = (k − ξ )Kjj (du)νi + νj Kij (du) + ξνj Kji (du) = Lξ (wξ ), ξ ξ ξ i where wξ is the double layer potential (7) with density u. Moreover, if x ∈ Ω, ξ ui (y ) Lξ [Γj (x, y )] dσy = −uj (x) + Lξ [u(y )] Γij (x, y ) dσy wj (x) = i,y i Σ Σ and then, on account of (26), 1 1 1 1 1 Lξ wξ = − Lξ u + T ξ (Lξ u) = − ϕ + T ξϕ + T ξ ϕ + T ξϕ = − ϕ + (T ξ )2 ϕ. 2 2 2 2 4 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. x ∈ Σ, T ϕ(x) = Lx [Γ(x, y )] ϕ(y ) dσy , (27) Σ and denote by T ∗ its adjoint. In this subsection we determine the dimension of the following eigenspaces 1 ϕ + T ∗ϕ = 0 ; ϕ ∈ [Lp (Σ)]n : V± = (28) 2 11
1 ϕ ∈ [Lp (Σ)]n : ± ϕ + T ϕ = 0 . W± = (29) 2 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 h¨lder-continuous and then these eigenspaces do not depend on p. This implies that o dim V+ = dim W− , dim V− = dim W+ . (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+ and W− have dimension n(n + 1)m/2. Moreover V+ = {vh χΣj : h = 1, . . . , n(n + 1)/2, j = 1, . . . , m} , where {vh : h = 1 . . . , n(n + 1)/2} is an orthonormal basis of the space R and χΣj is the characteristic function of Σj . Proof. We deﬁne the vector-valued functions αj , j = 1, . . . , m as αj (x) = (a + Bx)χΣj (x), x ∈ Σ. For a ﬁxed j = 1, . . . , m, the function αj (x) belongs to V+ ; indeed 1 1 − (a + Bx)χΣj (x)+ [Ly Γ(x, y )] (a + By ) χΣj (y ) dσy = − (a + Bx)χΣj (x)+ [Ly Γ(x, y )] (a + By ) dσy = 2 2 Σ Σj 1 1 − (a + Bx) + (a + Bx) = 0, x ∈ Σj , 2 2 because of  x ∈ Ωj , αj (x)  x ∈ Σj , [Ly Γ(x, y )] αj (y ) dσy = αj (x)/2 (31) Σ  x ∈ Ωj . 0 /  Now we prove that the following n(n + 1)m/2 eigensolutions of V+ h = 1, . . . , n(n + 1)/2, j = 1, . . . , m, x ∈ Σ whj (x) = vh (x)χΣj (x), n(n+1)/2 m are linearly independent. Indeed, if chj whj = 0, we have j =1 h=1 n(n+1)/2 x ∈ Σj , j = 1, . . . , m. chj vh (x) = 0, h=1 Then, by applying a classical uniqueness theorem to the domain Ωj , n(n+1)/2 x ∈ Ωj , j = 1, . . . , m, chj vh (x) = 0, h=1 12
from which it easily follows that chj = 0, h = 1, . . . , n(n + 1)/2, j = 1, . . . , m. Thus, dim V+ ≥ n(n + 1)m/2. On the other hand, suppose ϕ ∈ W− and let u be the simple layer potential with density ϕ. Since Eu = 0 in Ωj and L− u = 0 on Σj , u = aj + B j x on each connected component Ωj , j = 1, . . . , m, and u = 0 in Rn \ Ω0 . Note that this is true also for n = 2, because ϕ ∈ W− implies ϕ dσ = 0. We can deﬁne a linear map τ as follows Σ τ : W− −→ (Rn × Sn )m ϕ −→ (a1 , B 1 , . . . , am ,B m ). If τ (ϕ) = 0, from a classical uniqueness theorem, we have that ϕ ≡ 0 in Rn . Thus, τ is an injective map and dim W− ≤ n(n + 1)m/2. The assertion follows from (30). Lemma 3.12. The spaces V− and W+ have dimension n(n + 1)/2. Moreover V− is constituted by the restrictions to Σ of the rigid displacements. Proof. Let α ∈ R. If x ∈ Σ, we have 1 1 1 α(x) − α(x) = 0, α(x) + [Ly Γ(x, y )] α(y ) dσy = 2 2 2 Σ thanks to  −α(x) x ∈ Ω,  [Ly Γ(x, y )] α(y ) dσy = −α(x)/2 x ∈ Σ, Σ  x ∈ Ω. 0 /  This shows that the restriction to Σ of α belongs to V− and then dim V− ≥ dim R = n(n + 1)/2. On the other hand, suppose φ ∈ W+ and let u be the simple layer potential with density φ. Since Eu = 0 in Ω and L+ u = 0 on Σ, u = a + Bx in Ω. Let σ be the linear map σ : W+ −→ Rn × Sn φ −→ (a, B ). If n ≥ 3, we have that σ (φ) = 0 implies u ≡ 0 in Rn and then φ ≡ 0 on Σ, in view of classical uniqueness theorems. 0 If n = 2, deﬁne W+ = {φ ∈ W+ / φ dσ = 0}. We have σ |W+ is injective and its range does not contain 0 Σ 2 0 0 . Therefore dim W+ ≤ 1. On the other hand, dim W+ − 2 ≤ dim W+ the vectors (1, 0), 0 and (0, 1), 0 and then dim W+ ≤ 3. In any case, dim W+ ≤ n(n + 1)/2 and the result follows from (30). 13
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 log |x − y | dsy = 0, |x| < 1. |y |=1 Similar domains occur also in elasticity. In order to give explicitly such an example, let us prove the following lemma. Lemma 4.1. Let ΣR be the circle of radius R centered at the origin. We have |x − y |2 log |x − y | dsy = 2πR(R2 log R + (1 + log R)|x|2 ), |x| < R. (32) ΣR Proof. Denote by u(x) the function on the left hand side of (32) and by ΩR the ball of radius R centered at the origin. Let us ﬁx x0 ∈ ΣR . For any x ∈ ΣR we have |x − y |2 log |x − y | dsy = |x0 − y |2 log |x0 − y | dsy ΣR ΣR and then u is constant on ΣR . Moreover (1 + log |x − y |) dsy ∆u(x) = 4 ΣR and then also ∆u is constant on ΣR . Since ∆u is harmonic in ΩR and continuous on ΩR , it is constant in ΩR and then (1 + log |y |) dsy = 8πR(1 + log R), x ∈ ΩR . ∆u(x) = ∆u(0) = 4 ΣR The function u(x) − 2πR(1 + log R)|x|2 is continuous on ΩR , harmonic in ΩR and constant on ΣR . Then it is constant in ΩR and u(x) − 2πR(1 + log R)|x|2 = u(0) = |y |2 log |y | dσy = 2πR3 log R. ΣR Corollary 4.2. Let ΣR be the circle of radius R centered at the origin. We have R (k − 2(k + 2) log R), |x| < R. Γij (x, y ) dsy = δij (33) 4(k + 1) ΣR 14
Proof. Since (x1 − y1 )2 |x − y |2 log |x − y | dsy = 2 log |x − y | dsy + 2 ∂11 dsy + 2πR, |x − y |2 ΣR ΣR ΣR formula (32) implies (x1 − y1 )2 |x| < R. dsy = πR, |x − y |2 ΣR In a similar way ( x2 − y2 ) 2 |x| < R. dsy = πR, |x − y |2 ΣR From (32) we have also (x1 − y1 )(x2 − y2 ) |x − y |2 log |x − y | dsy = 2 |x| < R. ∂12 dsy = 0, |x − y |2 ΣR ΣR Keeping in mind the expression (1), (33) follows. This corollary shows that, if R = exp[k/(2(k + 2))], we have |x| < R. Γ(x, y ) e1 dsy = Γ(x, y ) e2 dsy = 0, ΣR ΣR This implies that in ΩR , for such a value of R, we cannot represent any smooth solution of the system Eu = 0 by means of a simple layer potential. If there exists some constant vector which cannot be represented in the simply connected domain Ω by a simple layer potential, we say that the boundary of Ω is exceptional. We have proved that Lemma 4.3. The circle ΣR with R = exp[k/(2(k + 2))] is exceptional for the operator ∆ + k div. Due to the results in [28], one can scale the domain in such a way that its boundary is not exceptional. Here we show that also in some (m + 1)-connected domains one cannot represent any constant vectors by a simple layer potential and that this happens if, and only if, the exterior boundary Σ0 (considered as the boundary of the simply connected domain Ω0 ) is exceptional. We note that, if any constant vector c can be represented by a simple layer potential, then any suﬃciently smooth solution of the system Eu = 0 can be represented by a simple layer potential as well (see Section 5 below). We ﬁrst prove a property of the singular integral system ∂ x ∈ Σ, i = 1, 2. ϕj (y ) Γij (x, y ) dsy = 0, (34) ∂sx Σ 15
Lemma 4.4. Let Ω ⊂ R2 be an (m + 1)-connected domain. Denote by P the eigenspace in [Lp (Σ)]2 of the system (34). Then dim P = 2(m + 1). Proof. We have (yi − xi )(yj − xj ) ∂ Γij 1∂ (k + 2)δij k − log |x − y | + (x, y ) = dsy |x − y |2 ∂sx 2π ∂sx 2(k + 1) 2(k + 1) and, since ∂ (xi − yi )(xj − yj ) xj − yj xi − yi (xi − yi )(xj − yj ) ∂ −2 |x − y | = = xi ˙ + xj ˙ 2 2 2 |x − y |3 |x − y | |x − y | |x − y | ∂sx ∂sx (xi − yi )(xj − yj ) ∂ ∂ ∂ log |x − y | + xj log |x − y | − 2 log |x − y | = xi ˙ ˙ |x − y |2 ∂xj ∂xi ∂sx (xi − yi )(xj − yj ) ∂ log |x − y | + O(|y − x|h−1 ) 2 xi xj − ˙˙ |x − y |2 ∂sx (the dot denotes the derivative with respect to the arc length on Σ), we ﬁnd3 ∂ (xi − yi )(xj − yj ) = O(|y − x|h−1 ). |x − y |2 ∂sx We have proved that4 ∂ 1 k+2 ∂ log |x − y | + O(|y − x|h−1 ) Γij (x, y ) = − δij ∂sx 2π 2(k + 1) ∂sx 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 ∂ x ∈ Σ, i = 1, 2. ϕj (y ) Γij (x, y ) dsy = 0, (35) ∂sy Σ 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 ei χΣj (i = 1, 2, j = 0, 1, . . . , m) are the only linearly independent eigensolutions of (35). Indeed it is obvious that such vectors satisfy the system (35). On the other hand, if ψ satisﬁes the system (35) then ∂f ψ ds = 0 ∂s Σ for any f ∈ [C ∞ (R2 )]2 . This can be siproved by the same method in [13, pp. 189–190]. Therefore ψ has to be constant on each curve Σj (j = 0, . . . , m), i.e. ψ is a linear combination of ei χΣj (i = 1, 2, j = 0, 1, . . . , m). Theorem 4.5. Let Ω ⊂ R2 be an (m + 1)-connected domain. The following conditions are equivalent: 16
I. there exists a H¨lder continuous vector function ϕ ≡ 0 such that o x ∈ Σ; Γ(x, y ) ϕ(y ) dsy = 0, (36) Σ II. there exists a constant vector which cannot be represented in Ω by a simple layer potential (i.e., there exists c ∈ R2 such that c ∈ S p ); / III. Σ0 is exceptional; IV. let ϕ1 , . . . , ϕ2m+2 be linearly independent functions of P and let cjk = (αjk , βjk ) ∈ R2 be given by x ∈ Σk , j = 1, . . . , 2m + 2, k = 0, 1, . . . , m. Γ(x, y ) ϕj (y ) dsy = cjk , Σ Then det C = 0 , (37) where   α1,0 . . . α2m+2,0  ... ... ...    α . . . α2m+2,m  C =  1,m .  β1,0 . . . β2m+2,0     ... ... ...  β1,m . . . β2m+2,m Proof. I ⇒ II. Let u be the simple layer potential (3) with density ϕ. Since u = 0 in Ω, and then on Σk , we ﬁnd that u = 0 also in Ωk (k = 1, . . . , m) in view of a known uniqueness theorem. On the other hand L+ u − L− u = ϕ on Σ and ϕ = 0 on Σk , k = 1, . . . , m. This means that x ∈ Ω0 . Γ(x, y ) ϕ(y ) dsy = 0, Σ0 If II is not true, we can ﬁnd two linear independent vector functions ψ1 and ψ2 such that x ∈ Ω, j = 1, 2. Γ(x, y ) ψj (y ) dsy = ej , Σ Arguing as before, we ﬁnd ψj = 0 on Σk , k = 1, . . . , m, j = 1, 2, and then x ∈ Ω0 , j = 1, 2. Γ(x, y ) ψj (y ) dsy = ej , Σ0 Since ϕ, ψ1 , ψ2 belong to the kernel of the system ∂ x ∈ Σ0 , Γ(x, y ) ψ (y ) dsy = 0, ∂sx Σ0 17
Lemma 4.4 shows that they are linearly dependent. Let λ, µ1 , µ2 ∈ R such that (λ, µ1 , µ2 ) = (0, 0, 0) and λϕ + µ1 ψ1 + µ2 ψ2 = 0 on Σ0 . (38) This implies x ∈ Ω0 , Γ(x, y ) (λϕ(y ) + µ1 ψ1 (y ) + µ2 ψ2 (y )) dsy = 0, Σ0 i.e. µ1 e1 + µ2 e2 = 0, and then µ1 = µ2 = 0. Now (38) leads to λϕ = 0 and thus λ = 0, which is absurd. II ⇒ III. If Σ0 is not exceptional, for any c ∈ R2 there exists ∈ [C λ (Σ0 )]2 such that x ∈ Ω0 . Γ(x, y ) (y ) dsy = c, Σ0 Setting ( y ) y ∈ Σ0 , ϕ(y ) = y ∈ Σ \ Σ0 , 0 we can write x ∈ Ω, Γ(x, y ) ϕ(y ) dsy = c, Σ and this contradicts II. III ⇒ IV. Let us suppose det C = 0. For any c = (α, β ) ∈ R2 there exists λ = (λ1 , . . . , λ2m+2 ) solution of the system 2m+2 2m+2 λj αjk = α, λj βjk = β, k = 0, . . . , m, j =1 j =1 i.e. 2m+2 λj cjk = c, k = 0, . . . , m. j =1 Therefore 2m+2 x ∈ Σ. Γ(x, y ) λj ϕj (y ) dsy = c, Σ j =1 2m+2 Arguing as before, this leads to λj ϕj = 0 on Σk for k = 1, . . . , m. Then Σ0 is not exceptional. j =1 IV ⇒ I. From (37) it follows that there exists an eigensolution λ = (λ1 , . . . , λ2m+2 ) of the homogeneous system 2m+2 λj cjk = 0, k = 0, . . . , m. j =1 Set 2m+2 ϕ(x) = λj ϕj (x) . j =1 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. 18
Deﬁnition 4.6. Whenever n = 2 and Σ0 is exceptional, we say that u belongs to S p if, and only if, x ∈ Ω, u(x) = Γ(x, y ) ϕ(y ) dsy + c, (39) Σ where ϕ ∈ [Lp (Σ)]2 and c ∈ R2 . 5 The Dirichlet problem The purpose of this section is to represent the solution of the Dirichlet problem in an (m + 1)-connected domain by means of a simple layer potential. Precisely we give an existence and uniqueness theorem for the problem   u ∈ S p, E u = 0 in Ω, (40) u=f on Σ,  where f ∈ [W 1,p (Σ)]n . We establish some preliminary results. Theorem 5.1. Given ω ∈ [Lp (Σ)]n , there exists a solution of the singular integral system 1 ϕ ∈ [Lp (Σ)]n , x ∈ Σ dx [Γ(x, y )] ϕ(y ) dσy = ω (x), (41) Σ if, and only if, γ ∧ ωi = 0, i = 1, . . . , n (42) Σ for every γ ∈ Lq −2 (Σ) (q = p/(p − 1)) such that γ is a weakly closed (n − 2)-form. n Proof. Denote by R∗ : [Lq −2 (Σ)]n −→ [Lq (Σ)]n the adjoint of R (see (23)), i.e. the operator whose compo- n nents are given by ∗ ψi (y ) ∧ dy [Γij (x, y )], x ∈ Σ. Rj ψ (x) = Σ Thanks to Corollary 3.10, the integral system (41) admits a solution ϕ ∈ [Lp (Σ)]n if, and only if, ψi ∧ ωi = 0 (43) Σ for any ψ = (ψ1 , . . . , ψn ) ∈ [Lq −2 (Σ)]n such that R∗ ψ = 0. Arguing as in [13], R∗ ψ = 0 if, and only if, all the n 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 ∈ [W 1,p (Σ)]n there exists a solution of the BVP   w ∈ S p, Ew = 0 in Ω, (44) dw = df on Σ.  19