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 ℝ^{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 domains1 Introduction
In this paper we consider the Dirichlet and the traction problems for the linearized ndimensional 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 wellknown 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 ndimensional Laplace equation and later extended to the threedimensional 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 ℝ^{n }(n ≥ 2). Similar results for Laplace equation have been recently obtained in [17]. We remark that we do not require the use of pseudodifferential 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 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 ⊂ ℝ^{n}, n ≥ 2, of the form , where Ω_{j }(j = 0, ..., m) are m + 1 bounded domains of ℝ^{n }with connected boundaries Σ_{j }∈ C^{1, λ }(λ ∈ (0, 1]) and such that and . 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
where u = (u_{1}, ..., u_{n}) is a vectorvalued 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
i, j = 1, ..., n, ω_{n }being the hypersurface measure of the unit sphere in ℝ^{n}.
As usual, we denote by the bilinear form defined as
where ε_{ih }(u) and σ_{ih }(u) are the linearized strain components and the stress components respectively, i.e.
Let us consider the boundary operator L^{ξ }whose components are
ξ being a real parameter. We remark that the operator L^{1 }is just the stress operator 2σ_{ih}ν_{h}, which we shall simply denote by L, while L^{k/(k+2) }is the socalled pseudostress operator.
By the symbol we denote the space of all constant skewsymmetric matrices of order n. It is wellknown 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 . We denote by the space of all rigid displacements whose dimension is n(n + 1)/2. As usual {e_{1}, ..., e_{n}} is the canonical basis for ℝ^{n}.
For any 1 < p < +∞ we denote by [L^{p}(Σ)]^{n }the space of all measurable vectorvalued functions u = (u_{1}, ..., u_{n}) such that u_{j}^{p }is integrable over Σ (j = 1, ..., n). If h is any nonnegative integer, 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 C^{1 }and consequently in every coordinate system of class C^{1}. The space is constituted by the vectors (v_{1}, ..., v_{n}) such that v_{j }is a differential form of . [W^{1, p}(Σ)]^{n }is the vector space of all measurable vectorvalued functions u = (u_{1}, ..., u_{n}) such that u_{j }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 vectorvalued function u belongs to if, and only if, there exists φ ∈ [L^{p}(Σ)]^{n }such that u can be represented by a simple layer potential
Definition 2.2. The vectorvalued function w belongs to if, and only if, there exists ψ ∈ [W^{1,p}(Σ)]^{n }such that w can be represented by a double layer potential
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
We recall that an immediate consequence of (5) is that, when ξ = k/(2 + k) we have
while for ξ ≠ k/(2 + k) the kernels have a strong singularity on Σ.
Let us denote by w^{ξ }the double layer potential
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]
Here * and d denote the Hodge star operator and the exterior derivative respectively, s(x, y) is the fundamental solution of Laplace equation
and s_{h}(x, y) is the double hform introduced by Hodge in [22]
Since, for a scalar function f and for a fixed h, we have *df ∧ dx^{h }= (1)^{n1}∂_{h}f dx, denoting by w the harmonic double layer potential with density φ ∈ W^{1,p}(Σ), (8) implies
The following lemma can be considered as an extension of formula (9) to elasticity. Here du denotes the vector (du_{1}, ..., du_{n}) and ψ = (ψ_{1}, ..., ψ_{n}) is an element of .
Lemma 3.1. Let w^{ξ }be the double layer potential (7) with density u ∈ [W^{1,p}(Σ)]^{n}. Then
where
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
we find in Ω
An integration by parts on Σ leads to
Therefore, by recalling (9),
If f is a scalar function, we may write
This identity is established by observing that on Σ we have
Then we can rewrite (14) as
Similar arguments prove the result if n = 2. We omit the details. □
3.2 Some jump formulas
Lemma 3.2. Let f ∈ L^{1}(Σ). If η ∈ Σ is a Lebesgue point for f, we have
where the limit has to be understood as an internal angular boundary value^{1}.
Proof. Let h_{pj}(x) = x_{p}x_{j}x^{n}. Since h ∈ C^{∞}(ℝ^{n}\{0}) is even and homogeneous of degree 2  n, due to the results proved in [23], we have
where , being the Fourier transform
(see also [24] and note that in [23,24]ν is the inner normal). On the other hand
and, since
(see, e.g., [[25], p. 156]), we find
Finally, keeping in mind that ω_{n }= n π^{n/2}/Γ(n/2 + 1) and Γ(n/2 + 1) = n(n  2)Γ(n/2  1)/4, we obtain
Combining this formula with (16) we get (15). □
Lemma 3.3. Let . Let us write ψ as ψ = ψ_{h}dx^{h }with
Then, for almost every η ∈ Σ,
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 1form 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
and then
a.e. on Σ. From (17) it follows that and (18) is proved. □
Lemma 3.4. Let . Let us write ψ as ψ = ψ_{h}dx^{h }and suppose that (17) holds. Then, for almost every η ∈ Σ,
where K^{ξ }is defined by (13) and the limit has to be understood as an internal angular boundary value.
Proof. We have
Keeping in mind (13), formula (15) leads to
On the other hand
and the result follows. □
Lemma 3.5. Let . Then, for almost every η ∈ Σ,
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
In view of Lemmas 3.3 and 3.4 we have
Therefore
where
Conditions (21) lead to
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 ∈ [W^{1,p}(Σ)]^{n}. Then
a.e. on Σ, where and denote the internal and the external angular boundary limit of L^{ξ}(w^{ξ}) respectively and 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 : be the following singular integral operator
Let us define to be the singular integral operator
Then
where
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
where w^{ξ }is the double layer potential (7) with density u. Moreover, if x ∈ Ω,
and then, on account of (26),
□
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.
and denote by T* its adjoint.
In this subsection we determine the dimension of the following eigenspaces
We first observe that the (total) indices of singular integral systems in (28)(29) vanish. This can be proved as in [[1], pp. 235238]. Moreover, by standard techniques, one can prove that all the eigenfunctions are höldercontinuous and then these eigenspaces do not depend on p. This implies that
The next two lemmas determine such dimensions. Similar results for Laplace equation can be found in [[27], Chapter 3].
Lemma 3.11. The spaces and have dimension n(n + 1)m/2. Moreover
where {v_{h }: h = 1 ..., n(n + 1)/2} is an orthonormal basis of the space and is the characteristic function of Σ_{j}.
Proof. We define the vectorvalued functions α_{j}, j = 1, ..., m as , x ∈ Σ. For a fixed j = 1, ..., m, the function α_{j}(x) belongs to ; indeed
because of
Now we prove that the following n(n + 1)m/2 eigensolutions of
are linearly independent. Indeed, if , we have
Then, by applying a classical uniqueness theorem to the domain Ω_{j},
from which it easily follows that
Thus, . On the other hand, suppose 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 . Note that this is true also for n = 2, because implies . We can define a linear map τ as follows
If τ(φ) = 0, from a classical uniqueness theorem, we have that φ ≡ 0 in ℝ^{n}. Thus, τ is an injective map and . The assertion follows from (30). □
Lemma 3.12. The spaces and have dimension n(n + 1)/2. Moreover is constituted by the restrictions to Σ of the rigid displacements.
Proof. Let . If x ∈ Σ, we have
thanks to
This shows that the restriction to Σ of α belongs to and then . On the other hand, suppose 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
If n ≥ 3, we have that σ(ϕ) = 0 implies u ≡ 0 in ℝ^{n }and then ϕ ≡ 0 on Σ, in view of classical uniqueness theorems.
If n = 2, define . We have is injective and its range does not contain the vectors ((1, 0), 0) and ((0, 1), 0)^{2}. Therefore . On the other hand, and then . In any case, and the result follows from (30). □
4 The bidimensional case
The case n = 2 requires some additional considerations. It is wellknown 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
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
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 x_{0 }∈ Σ_{R}. For any x ∈ Σ_{R }we have
and then u is constant on Σ_{R}. Moreover
and then also Δu is constant on Σ_{R}. Since Δu is harmonic in Ω_{R }and continuous on , it is constant in Ω_{R }and then
The function u(x)  2πR(1 + log R) x^{2 }is continuous on , harmonic in Ω_{R }and constant on Σ_{R}. Then it is constant in Ω_{R }and
□
Corollary 4.2. Let Σ_{R }be the circle of radius R centered at the origin. We have
Proof. Since
formula (32) implies
In a similar way
From (32) we have also
Keeping in mind the expression (1), (33) follows. □
This corollary shows that, if R = exp[k/(2(k + 2))], we have
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 + 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 Σ_{0 }(considered as the boundary of the simply connected domain Ω_{0}) 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
Lemma 4.4. Let Ω ⊂ ℝ^{2 }be an (m + 1)connected domain. Denote by the eigenspace in [L^{p}(Σ)]^{2 }of the system (34). Then .
Proof. We have
and, since
(the dot denotes the derivative with respect to the arc length on Σ), we find^{3}
We have proved that^{4}
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
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 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
for any f ∈ [C^{∞ }(ℝ^{2})]^{2}. This can be siproved by the same method in [[13], pp. 189190]. Therefore ψ has to be constant on each curve Σ_{j }(j = 0, ..., m), i.e. ψ is a linear combination of . □
Theorem 4.5. Let Ω ⊂ ℝ^{2 }be an (m + 1)connected domain. The following conditions are equivalent:
I. there exists a Hölder continuous vector function such that
II. there exists a constant vector which cannot be represented in Ω by a simple layer potential (i.e., there exists c ∈ ℝ^{2 }such that );
III. Σ_{0 }is exceptional;
IV. let φ_{1}, ..., φ_{2m+2 }be linearly independent functions of and let c_{jk }= (α_{jk}, β_{jk}) ∈ ℝ^{2 }be given by
Then
where
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
If II is not true, we can find two linear independent vector functions ψ_{1 }and ψ_{2 }such that
Arguing as before, we find ψ_{j }= 0 on Σ_{k}, k = 1, ..., m, j = 1, 2, and then
Since φ, ψ_{1}, ψ_{2 }belong to the kernel of the system
Lemma 4.4 shows that they are linearly dependent. Let λ, μ_{1}, μ_{2 }∈ ℝ such that (λ, μ_{1}, μ_{2}) ≠ (0, 0, 0) and
This implies
i.e. μ_{1}e_{1 }+ μ_{2}e_{2 }= 0, and then μ_{1 }= μ_{2 }= 0. Now (38) leads to λφ = 0 and thus λ = 0, which is absurd.
II ⇒ III. If Σ_{0 }is not exceptional, for any c ∈ ℝ^{2 }there exists ϱ ∈ [C^{λ}(Σ_{0})]^{2 }such that
Setting
we can write
and this contradicts II.
III ⇒ IV. Let us suppose . For any c = (α, β) ∈ ℝ^{2 }there exists λ = (λ_{1}, ..., λ_{2m+2}) solution of the system
i.e.
Therefore
Arguing as before, this leads to on Σ_{k }for k = 1, ..., m. Then Σ_{0 }is not exceptional.
IV ⇒ I. From (37) it follows that there exists an eigensolution λ = (λ_{1}, ..., λ_{2m+2}) of the homogeneous system
Set
In view of the linear independence of φ_{1}, ..., φ_{2m+2}, the vector function φ does not identically vanish and it is such that (36) holds. □
Definition 4.6. Whenever n = 2 and Σ_{0 }is exceptional, we say that u belongs to if, and only if,
where φ ∈ [L^{p}(Σ)]^{2 }and c ∈ ℝ^{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
where f ∈ [W^{1,p}(Σ)]^{n}.
We establish some preliminary results.
Theorem 5.1. Given , there exists a solution of the singular integral system
if, and only if,
for every such that γ is a weakly closed (n  2)form.
Proof. Denote by the adjoint of R (see (23)), i.e. the operator whose components are given by
Thanks to Corollary 3.10, the integral system (41) admits a solution φ ∈ [L^{p}(Σ)]^{n }if, and only if,
for any 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 ∈ [W^{1,p}(Σ)]^{n }there exists a solution of the BVP
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:
in which the unknown is φ ∈ [L^{p}(Σ)]^{n }and the datum is . 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 of the Fredholm integral system
The dimension of is nm. This can be proved as in [[30], p. 63], where the case n = 3 is considered.
Theorem 5.3. Given c_{0}, c_{1}, ..., c_{m }∈ ℝ^{n}, there exists a solution of the BVP
It is given by
where satisfy the following conditions
Proof. Let ψ_{1}, ..., ψ_{nm }be nm linearly independent eigensolutions of the space . For a fixed j = 1, ..., nm we set
Then on Σ. As in [[30], Theorem III, p. 45], this implies that V_{j }is constant on each connected component of . Then V_{j }= 0 in and in Ω _{k }(k = 1, ..., m). For every k = 1, ..., m, consider the n × nm matrix defined as follows
The nm × nm matrix has a not vanishing determinant. Indeed, if , the linear system admits an eigensolution λ = (λ_{1}, ..., λ_{nm}) ∈ ℝ^{nm}. Hence the potential
vanishes not only on , but also on Ω_{k }(k = 1, ..., m). Since this implies W = 0 on Σ, we find W = 0 in Ω, thanks to the classical uniqueness theorem for the Dirichlet problem. Accordingly, W = 0 all over ℝ^{n}, from which and this is absurd.
For each h = 1, ..., m and i = 1, ..., n, let be the solution of the system
Setting
Put
The potential v belongs to , thanks to the isomorphism σ introduced in the proof of Lemma 3.12 (for n = 2 see Definition 4.6). Moreover
i.e. v = ck on Σ_{k }(k = 0, 1, ..., m). This shows that v is solution of (46). □
We are now in a position to establish the main result of this section.
Theorem 5.4. The Dirichlet problem (40) has a unique solution u for every f ∈ [W^{1,p}(Σ)]^{n}. If n ≥ 3 or n = 2 with Σ_{0 }is not exceptional, u is given by (3). If n = 2 and Σ_{0 }is exceptional, it is given by (39). In any case, the density φ solves the singular system (45).
Proof. Let w be a solution of the problem (44). Since dw = df on Σ, w = f + c_{h }on Σ_{h }(h = 0, ..., m) for some c_{h }∈ ℝ^{n}. The function u = w  v, where v is given by (47), solves the problem (40).
In order to show the uniqueness, suppose that (3) is solution of (40) with f = 0. From Corollary 3.10 it follows that the condition u = 0 on Σ implies that
where T^{k/(k+2) }is the compact operator given by (26). By bootstrap techniques, (48) implies that φ is a Hölder function on Σ. Then u belongs to and we get that
from which
The solution of (49) is u(x) = a + Bx, where a ∈ ℝ^{n }and are arbitrary. Finally, u = 0 in by virtue of the classical uniqueness theorem for the Dirichlet problem. □
Remark 5.5. In order to solve the Dirichlet problem (40), we need to solve the singular integral system (45). We know that this system can be reduced to a Fredholm one by means of the operator R'^{k/(k+2)}. This reduction is not an equivalent reduction in the usual sense (for this definition see, e.g., [[10], p. 19]), because , being the kernel of the operator R'^{k/(k+2)}.
However R'^{k/(k+2) }still provides a kind of equivalence. In fact, as in [[31], pp. 253254], one can prove that . This implies that if ψ is such that there exists at least a solution of the equation Rφ = ψ, then Rφ = ψ if, and only if, R'^{k/(k+2) }Rφ = R'^{k/(k+2)}ψ.
Since we know that the system Rφ = df is solvable, we have that Rφ = df if, and only if, φ is solution of the Fredholm system R'^{k/(k+2)}Rφ = R'^{k/(k+2)}df.
Therefore, even if we do not have an equivalent reduction in the usual sense, such Fredholm system is equivalent to the Dirichlet problem (40).
6 The traction problem
The aim of this section is to study the possibility of representing the solution of the traction problem by means of a double layer potential. As we shall see, in an (m + 1)connected domain this is possible if, and only if, the given forces are balanced on each connected component Σ_{j }of the boundary.
More precisely, we consider the problem
where f ∈ [L^{p}(Σ)]^{n }is such that
We shall prove that, in order to have the existence of a solution of such a problem, condition (51) is not sufficient, but it must be satisfied on each Σ_{j}, j = 0, 1, ..., m (see Theorem 6.2 below).
If f satisfies the only condition (51), we need to modify the representation of the solution by adding some extra terms (see Theorem 6.4 below).
Lemma 6.1. Let be a double layer potential with density ψ ∈ [W^{1,2}(Σ)]^{n}. Then
Proof. Let (ψ_{k})_{k≥1 }be a sequence of functions in [C^{1,h}(Σ)]^{n }(0 < h < λ) such that ψ_{k }→ ψ in [W^{1,2}(Σ)]^{n}.
Setting
we have that , Ew_{k }= 0 and then
From ψ_{k }→ ψ in [L^{2}(Σ)]^{n}, it follows that w_{k }→ w in [L^{2}(Σ)]^{n }because of wellknown properties of singular integral operators.
On the other hand we have that in L^{2}(Ω). By applying formula (11), we see that ∇w_{k }→ ∇w in [L^{2}(Ω)]^{n}. Moreover, since also in L^{2}(Σ), (22) shows that Lw_{k }→ Lw in [L^{2}(Σ)]^{n}. We get the claim by letting k → +∞ in (53). □
Theorem 6.2. Given f ∈ [L^{p}(Σ)]^{n}, the traction problem (50) admits a solution if, and only if,
for every j = 0, 1, ..., m, a ∈ ℝ^{n }and . The solution is determined up to an additive rigid displacement.
Moreover, (4) is a solution of (50) if, and only if, its density ψ is given by
ϕ being a solution of the singular integral system
where T is given by (27).
Proof. Assume that conditions (54) hold. If u is the double layer potential with density ψ ∈ [W^{1,p}(Σ)]^{n}, in view of (22) the boundary condition Lu = f turns into the equation
where R' is given by (24) with ξ = 1.
On account of Theorem 5.4, if n = 2 and Σ_{0 }is exceptional, any ψ ∈ [W^{1,p}(Σ)]^{2 }can be written as
with ϕ ∈ [L^{p}(Σ)]^{2}, c ∈ ℝ^{2}. In all other cases, ψ can be written as (55) with ϕ ∈ [L^{p}(Σ)]^{n}. In any case, since dψ = Rϕ (R being defined by (23)), we infer R'(dψ) = R'Rϕ. Keeping in mind Lemma 3.9, we find that equation (57) is equivalent to (56), with ψ given by (55).
Therefore there exists a solution of the traction problem (50) if, and only if, the singular integral system (56) is solvable.
On the other hand, there exists a solution γ ∈ [L^{p}(Σ)]^{n }of the singular integral system
if, and only if, f is orthogonal to . In view of Lemma 3.12, this occurs if, and only if, (51) is satisfied. Then conditions (54) imply the existence of a solution of (58).
Consider now the singular integral system
From Lemma 3.11 the dimension of the kernel is n(n + 1)m/2 and is a basis of it. The equation (59) has a solution if, and only if,
Since γ is solution of (58), conditions (60) are fulfilled. Indeed, picking j = 1, ..., m and h = 1, ..., n(n+1)/2, by integrating (58) on Σ_{j }we find (see (31))
Conditions (60) follow from (54) since the last ones are equivalent to
Let ϕ be a solution of (59); taking (58) into account, we have that ϕ solves (56) and then the traction problem (50) admits a solution.
Conversely, if u is a solution of (50), from Lemma 3.7, we have that
By Lemma 6.1, for any fixed j = 1, ..., m we have
Now we pass to discuss the uniqueness. Let u be a solution of (50) with datum f = 0. As we know, the condition L_{+}u = 0 is equivalent to the singular integral system , ϕ being as in (55), which can be written as
Set
Since Ξ/2+TΞ = 0 and the operator I/2+T can be reduced to Fredholm one, as shown by Kupradze [[1], Chapter IV, §7], Ξ has to be Hölder continuous. By a similar argument, the vectorvalued function ϕ, being solution of the singular integral system (61), is Hölder continuous. Therefore the relevant simple layer potential ψ belongs to W^{1,2}(Σ), i.e. . By applying formula (52), we get that u is a rigid displacement in Ω. □
We remark that, by Theorem 6.2, a solution of the traction problem (50) can be written as a double layer potential if, and only if, conditions (54) are satisfied.
In order to consider the problem (50) under the only condition (51), we introduce the following space.
Definition 6.3. We define as the space of all the functions w written as
where ψ ∈ [W^{1,p}(Σ)]^{n}, {v_{h }: h = 1, ..., n(n + 1)/2} is an orthonormal basis for and c_{jh }∈ ℝ.
Theorem 6.4. Given f ∈ [L^{p}(Σ)]^{n }satisfying (51), the traction problem
admits a solution given by
where ψ ∈ [W^{1,p}(Σ)]^{n }is solution of the system
The solution is uniquely determined up to an additive rigid displacement.
Proof. First observe that
for h = 1, ..., n(n + 1)/2 and j = 1, ..., m. If w is given by (63), taking into account (57), we find that Lw = f if, and only if, is (64) satisfied.
Denote by g the right hand side of (64). In view of Theorem 6.2, R'(dψ) = g has a solution if, and only if, for any k = 0, 1, ..., m, l = 1, ..., n(n + 1)/2. By integrating on Σ_{k }(k = 1, ..., m), for every l we get
On the other hand
Finally, assume that w is solution of (62) with f = 0. From (63) it follows that and then w is a rigid displacement in Ω by virtue of the uniqueness proved in Theorem 6.2. □
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
The authors declare that the work was realized in collaboration with same responsibility. All authors read and approved the final manuscript.
Endnotes
^{1}For the definition of internal (external) angular boundary values see, e.g., [[23], p. 53].
^{2}If a simple layer potential u, whose density belongs to , is such that u(x) = c in Ω, then u(x) = c in Ω_{0}. Since u(∞) = 0, we find u(x) = 0 in and this leads to u = 0 in ℝ^{2}, c = 0.
^{3}It is not difficult to see that
^{4}We remark that for n ≥ 3 the formula
is false.
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