- About this Journal ·
- Abstracting and Indexing ·
- Advance Access ·
- Aims and Scope ·
- Annual Issues ·
- Article Processing Charges ·
- Articles in Press ·
- Author Guidelines ·
- Bibliographic Information ·
- Citations to this Journal ·
- Contact Information ·
- Editorial Board ·
- Editorial Workflow ·
- Free eTOC Alerts ·
- Publication Ethics ·
- Reviewers Acknowledgment ·
- Submit a Manuscript ·
- Subscription Information ·
- Table of Contents
Abstract and Applied Analysis
Volume 2013 (2013), Article ID 765020, 12 pages
On the Dirichlet Problem for the Stokes System in Multiply Connected Domains
Department of Mathematics, Computer Science and Economics, University of Basilicata, Viale dell'Ateneo Lucano 10, 85100 Potenza, Italy
Received 24 April 2012; Accepted 28 November 2012
Academic Editor: Chun-Lei Tang
Copyright © 2013 Alberto Cialdea et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The Dirichlet problem for the Stokes system in a multiply connected domain of is considered in the present paper. We give the necessary and sufficient conditions for the representability of the solution by means of a simple layer hydrodynamic potential, instead of the classical double layer hydrodynamic potential.
Recently some papers have used the integral representations of solutions for studying some BVPs for the Stokes system also in multiply connected domains [3–8]. All these papers concern the double layer hydrodynamic potential approach for the Dirichlet problem and the simple layer hydrodynamic potential approach for the traction problem.
The aim of the present paper is to investigate a different integral representation for the Dirichlet problem for the Stokes system in a multiply connected bounded domain of (). Namely, we consider the simple layer potential approach for the Dirichlet problem in a domain where () are suitable domains with connected boundaries in , .
We use a new method which hinges on a singular integral system in which the unknown is a usual vector valued function, while the data is a vector whose components are differential forms.
The paper is organized as follows. In Section 2 we give an outlook of the method with a brief description of some previous results.
Section 5 is devoted to determine the eigenspace of a certain singular integral system in which the unknowns are differential forms of degree on . In the same section, we recall some known results concerning the eigenspaces of some classical integral systems.
In Section 6 we construct a left reduction for the singular integral system under study. Such a singular integral system is equivalent in a precise sense to the Fredholm system obtained through the reduction.
Finally, in the last section, we find the solution of the Dirichlet problem for the Stokes system in a multiply connected domain by means of a simple layer hydrodynamic potential.
The main result is that, given , we can represent the solution of the Dirichlet problem by means of a simple layer hydrodynamic potential if, and only if, the conditions are satisfied ( being the outwards unit normal on ). Moreover, if the data satisfies only the condition (which is necessary for the existence of a solution of the Dirichlet problem (2)) we show how to modify the integral representation of the solution (see Theorem 23).
2. Sketch of the Method
The aim of this section is to give a better understanding of the method we are going to use in the present paper.
We will do that by considering the Dirichlet problem for Laplace equation in a bounded simply connected domain , whose boundary we denote by as follows:
Suppose that , . If we want to find the solution in the form of a simple layer potential whose density belongs to , we have to solve an integral equation of the first kind on as follows: where is the fundamental solution of Laplace equation
In  a new method for discussing such an equation was proposed. Namely, the first step is to consider the differential (in the sense of the theory of differential forms) of both sides in (6). In this way we obtain the equation in which we look for a solution .
The integral on the left hand side is a singular integral and it can be considered as a linear and continuous operator from to (we denote by the space of the differential forms of degree whose coefficients belong to in every local coordinate system).
It must be remarked that, if , the space in which we look for the solution of (8) and the space in which the data is given are different.
We recall that, if and are two Banach spaces and is a continuous linear operator, can be reduced on the left if there exists a continuous linear operator such that , where stands for the identity operator on , and is compact. Analogously, one can define an operator reducible on the right. One of the main properties of such operators is that the equation has a solution if, and only if, for any such that , being the adjoint of (for more details see, e.g., [10, 11]).
Let us denote by the left hand side of (8). In  a reducing operator was explicitly constructed. This implies that there exists a solution of (8) if, and only if, the compatibility conditions are satisfied for any () such that . Moreover one can show that if, and only if, is a weakly closed form. Therefore the compatibility conditions (9) are satisfied, and there exists a solution of (8).
A left reduction is said to be equivalent if , where denotes the kernel of (see, e.g., [11, page ]). Obviously this means that if, and only if, . In  it was remarked that if , we still have a kind of equivalence. Indeed the coincidence of these two kernels implies the following fact: if is such that the equation is solvable, then this equation is satisfied if, and only if, .
As a consequence one can obtain also a double layer representation for the Neumann problem for Laplace equation .
A characteristic of this method is that it uses neither the theory of pseudodifferential operators nor the concept of hypersingular integrals.
This method has been used also for studying other BVPs. In particular in  it was used to study the Dirichlet and the Neumann problems in multiply connected domains. Among other things, an interesting by-product of these results was obtained as follows (see [13, Theorem 6.1]).
Let be a harmonic function of class , where is the multiple connected domain (1). There exists a 2-form conjugate to in if, and only if,
An explicit integral expression for was also given. We recall that the -form is conjugate to if .
In this paper denotes an -connected domain of (), that is an open-connected set of the form (1), where each () is a bounded domain of with connected boundaries (), and such that and , . Let be the outwards unit normal on the boundary .
We consider the classical Stokes system for the incompressible viscous fluid where the unknowns and are the velocity and pressure of the fluid flow, respectively, and the constant is the kinematic viscosity of the fluid. A fundamental solution for this system is given by the pair of fundamental velocity tensor and its associated pressure vector (), being the hypersurface measure of the unit sphere in . For a solution of (11) we consider the following classical boundary operators:
Through this paper, indicates a real number such that . We denote by the space of all measurable vector-valued functions such that is integrable over (). If is any nonnegative integer, is the vector space of all differential forms of degree (briefly -forms) defined on such that their components are integrable functions belonging to in a coordinate system of class and consequently in every coordinate system of class . The space is constituted by the vectors such that is a differential form of (). is the vector space of all measurable vector-valued functions such that belongs to the Sobolev space ().
The pair with components is the simple layer hydrodynamic potential with density .
The pair with components is the double layer hydrodynamic potential with density .
4. On the Bidimensional Case
It is wellknown that there are some exceptional plane domains in which no every harmonic function can be represented by a simple layer potential. The simplest example of this kind is given by the unit disk, for which one has
In this section we show that also for the Stokes system there are similar domains. We say that the boundary of the domain is exceptional if there exists some constant vector which cannot be represented in by a simple layer potential.
Denoting by the circle of radius centered at the origin, we have the following lemma.
Lemma 1. The circle with is exceptional for the Stokes system.
Proof. Keeping in mind that (see, e.g., [16, Section 4])
Taking we obtain the result.
Let us consider now the exceptional boundaries of not simply connected domains.
Proposition 2. Let be an -connected domain. Denote by the eigenspace in of the singular integral system Then .
Proof. As in the proof of [16, Lemma 12], one can show that deduce that system (22) can be regularized to a Fredholm one, and see that its index is zero. Since the vectors (by we denote the characteristic function of the set ) (, ) are the only eigensolutions of the adjoint system we have .
Theorem 3. Let be an -connected domain. The following conditions are equivalent(1)There exists a Hölder continuous vector function such that (2)There exists a constant vector which cannot be represented in by a simple layer potential;(3) is exceptional.(4)Let be linearly independent vectors of (see Proposition 2), and let be given by Then , where
Proof. The proof runs as in [16, Theorem 1] with obvious modifications. We omit the details.
5. Some Eigenspaces
We determine the structure of the kernel of a particular singular integral system. Namely, let us denote by the space of such that
We begin by proving the following result.
Proof. By the well-known Stokes identity we have
Since, for every ,
we can rewrite
Then Integrating by parts, it follows that the last integral is equal to since . Then the claim holds for .
In the same manner it is possible to show formula (29) for and after observing that, if , we have , while, for , .
Lemma 5. Let be differential forms in such that on . One has if, and only if, where and are weakly closed forms belonging to .
Proof. It is easy to construct the differential forms . For example, one can take the restriction on of the following forms: , (). We remark that (37) holds if, and only if, the weak differentials exist and
Let us prove that (39) holds if, and only if,
It is obvious that (40) implies (39).
Conversely, suppose that (39) is true. Define , where . Let be such that in . Since , we may write and (40) follows immediately.
Suppose now that (39) is true. From (40) it follows that
An integration by parts shows that
Taking the exterior angular boundary value (for the definition of internal (external) angular boundary values see, e.g., [20, page 53] or [21, page 293]), we have a.e. on . Arguing as in [9, pages 189-190], this implies that also in . Summing over we find for every and a.e. on . In particular is the solution of the singular integral system (28).
Conversely, suppose (28) holds. Arguing again as in [9, pages 189-190], from (28) it follows that Since , system (11) implies that . Hence, Therefore, there exist some constants such that where Then, on account of Lemma 4, for every ,
The first term of the right hand side vanishes because of (47). As far as the second one is concerned, integrating by parts we get Hence, by (49), By setting and () we get the claim.
Remark 6. Lemma 5 shows that the dimension of the kernel is infinite. However, if we consider the quotient space , being the space of weakly closed differential forms in , we have .
We conclude this section by recalling some properties concerning the following eigenspaces: where (see, e.g., ) For the proofs of the following two results see [7, Lemma 3.3] and [8, Theorem 3.2], respectively.
Proposition 7. The sets and are linear subspaces of and
A basis of is expressed by the fields . The simple layer potentials whose densities are such that: , , where are rigid displacement in , specifically , and, for , , .
In addition, every has the property that , where is the simple layer potential with density .
Proposition 8. The sets and are linear subspaces of and
A basis for is expressed by the fields , where , are zero on , and such that the simple layer potentials with density are rigid displacement in (linearly independent for ).
Finally, every function which is the restriction to of a rigid displacement belongs to .
Remark 9. We can make the statement of Proposition 8 slightly more precise, saying that the simple layer potentials with density are rigid displacement in linear independent for any , unless and is exceptional. Indeed, let us show that if and is not exceptional, such rigid displacements are linearly independent. Let be such that We have also Let . In view of the equivalence between and of Theorem 3, has to vanish. Therefore () because of the linearly independence of . On the other hand, if and is exceptional, Theorem 3 shows that the potentials with densities are linearly dependent.
6. Reduction of a Certain Singular Integral Operator
For every , let be the operator defined by where and denote the Hodge star operator and the exterior derivative, respectively, and is the double -form introduced by Hodge in  as follows: Note that the operator satisfies the equation for each , since (see [9, page 187])
Moreover we introduce the operators defined as for every , where
In the sequel denotes the vector whose elements are -forms, and .
Proof. Note that, even if one could prove (66)-(67) directly, it seems easier to deduce them from the similar results we have already obtained for the elasticity system (see [16, Section 3]). For , let be the double layer elastic potential with density , that is,
where and are the stress operator and the Kelvin's matrix associated to the Lamé system , respectively.
Thanks to [16, Lemma 1], we know that where and is given by (60).
From [16, formula (5)] (where we set ), letting , we get , from which as . Therefore we obtain formula (66) by letting in (69). Formula (67) is an immediate consequence of (62) because .
For the next lemma it is convenient to recall here two jump formulas proved in [16, Lemmas 2 and 3].
Let . If is a Lebesgue point for , we get where the limit has to be understood as an internal angular boundary value, and the integral in the right hand side is a singular integral.
Further, let and write as with Assumption (73) is not restrictive, because, given the 1-form on , there exist scalar functions defined on such that and (73) holds (see [24, page 41]). Then, for almost every , where is given by (60), and the limit has to be understood again as an internal angular boundary value.
Lemma 14. Let be the double layer potential (17) with density . Then a.e. on , where and denote the internal and the external angular boundary limits of , respectively, and is given by (60) and by (64).
Proposition 15. Let be the following singular integral operator Let one define to be the singular integral operator Then where
7. The Dirichlet Problem
Let us consider the Dirichlet problem for the Stokes system where the given data satisfies the compatibility condition (4).
By the symbol we mean the class of the simple layer hydrodynamic potentials (15)-(16) with density in . Whenever and is exceptional (see Section 4), we say that belongs to if, and only if, where and .
We will see that condition (4) is not sufficient to prove the existence of the solution in the class , but it must be satisfied on each , .
We begin by proving the following result.
Theorem 16. Given , there exists a solution of the singular integral system if, and only if, for every such that the weak differentials exist and (38) holds for some real constants .
Proof. Consider the adjoint of (see (83)), , that is, the operator whose components are given by Proposition 15 implies that the integral system (92) has a solution if, and only if, for each such that . The result follows from Lemma 5.
Proposition 17. Given , there exists a solution of the BVP if, and only if, conditions (3) are satisfied. The density of the pair (see (15)-(16)) solves the singular integral system , where is given by (83).
Proof. Clearly, there exists a solution of this BVP if, and only if, there exists a solution of the singular integral system
In view of Theorem 16, there exists a solution of this system if, and only if, for every satisfying , that is, such that the weak differentials exist and (39) holds for some real constants . Equation (39) being true for any , we can write because of a density argument. In view of the arbitrariness of , (98) is satisfied if, and only if, (3) holds.
Proposition 18. Let . Let , , , be the elements of the basis of given by Proposition 7. The pair is the solution of the BVP
Theorem 19. Given , the Dirichlet problem is solvable if, and only if, conditions (3) are satisfied. Moreover the solution is unique ( is unique up to an additive constant).
Proof. Suppose conditions (3) are satisfied. Let be a solution of the problem (96). Since on , on () for some . The pair , where and are given by (100), solves the problem (103).
Conversely, if there exists a solution of (103), the compatibility condition (4) has to be satisfied. Moreover, for any , is the solution of the Stokes system also in . Therefore conditions (3) are satisfied for . These, together with (4), imply (3) also for . The uniqueness is known [7, Theorem 5.5].
Remark 20. The density of can be written as , where solves the singular integral system (97), and is the density of a simple layer potential which is constant on every connected component of .
Remark 21. If or and is not exceptional, denoting by the density of the simple layer potential (15)-(16) obtained in Theorem 19, we have that solves the integral system of the first kind
on . Therefore, Theorem 19 can be seen as an existence theorem for the integral system of the first kind (104) in .
If and is exceptional, we have the existence of a solution of the integral equation
Remark 22. Observe that the solvability of the Dirichlet problem (90) by means of a simple layer potential hinges on the singular integral system (97). Thanks to Proposition 15, the operator provides a left reduction for such a system. This reduction is not an equivalent one, but, as in [25, pages 253-254], one can show that is a weakly equivalent reduction (see definition in Section 3). Since the system is solvable, we have if, and only if, is solution of the Fredholm system . In this sense, such Fredholm system is equivalent to the problem (103).
In order to obtain a similar integral representation for the solution of the Dirichlet problem (90) when satisfies the only condition (4), we need to modify the representation of the solution by adding an extra term.
By we denote the space of all pairs written as where and belong to .
Theorem 23. Given satisfying (4), the Dirichlet problem has one, and only one, solution given by where is solution of the integral system of the first kind
Proof. Let be given by (108); imposing the boundary condition, we get (the symbol () stands for the interior (exterior) value of the double layer potential (17) on ) In view of Remark 21 such a system is solvable if, and only if, On the other hand, because of the jump formulas, we have