Variational Problem Involving Operator Curl in a Multiconnected Domain

Abstract

We shall study the problem of minimizing a functional involving the curl of vector fields in a three-dimensional, bounded multiconnected domain with prescribed tangential component on the boundary. The paper is an extension of minimization problem of the curl of vector fields. We shall prove the existence and the estimate of minimizers of more general functional which contains norm of the curl of vector fields.

1. Introduction

In this paper, we consider the following problem which was proposed by Pan [1, p. 9].

Problem A. Minimize the norm of the curl of vector fields in a given space with tangential trace on the boundary being prescribed.

The problem is related to the mathematical theory of liquid crystal, of superconductivity, and of electromagnetic field. When and is a simply connected domain without holes, Bates and Pan [2, 3] showed the existence of minimizer. For the multiconnected domain, the author of [1] obtained the existence of a minimizer of the Problem A in the case .

In the present paper we shall extend the results to more general functional containing Problem A.

More precisely, let be a Carathéodory function on and is a convex function with respect to ; moreover assume that for a.e. , , and there exist and such that for a.e. and all : Without loss of generality, we may assume that . We furthermore assume the following structure condition: Under (1) with , we have

For example, the function where is a measurable function satisfying for a.e. satisfies (1)-(2).

Let be a bounded domain in with boundary . Let be a given tangential vector field on . Let be the standard Sobolev space of vector fields. From now, we denote the tangential component of a vector field by ; that is, , where is the outer normal unit vector to the boundary . For any given tangential vector field on define a space of vector fieldsThen it is clear that is a closed convex set in . We consider the minimization problem

When , , and is a simply connected domain without holes, the authors of [2, 3] showed that (6) is achieved, and then in the case where , , and is bounded multiconnected domain, the author of [1] succeeded to show the existence of a minimizer of (6).

Since we allow to be a multiconnected domain in , throughout this paper, we assume that the domain satisfies the following (O1) and (O2) (cf. Dautray and Lions [4] and Amrouche and Seloula [5]).

(O1) is a bounded domain in with boundary . is locally situated on one side of ; has a finite number of connected components and denoting the boundary of the infinite connected component of .

(O2) There exist manifolds of dimension and of class denoted by such that and they are nontangential to and such that is simply connected and pseudo .

The number is called the first Betti number and the second Betti number of . We say that is simply connected if , and has no holes if . If we define the spacesthen it is well known that and . We note that and are contained in ; moreover, and are closed subspaces of . Also it will be shown in Lemma 4 that and are closed subspaces of . Thus since is a finite-dimensional closed subspace of , has a complement in ; that is, is a closed subspace of , , and (the direct sum). Therefore, for any , there exist uniquely and such that . We denote the projection by .

Define Note that if and , then the tangent trace is well defined as an element of (cf. [5, p. 45]), andMoreover, we note that if , then(cf. Amrouche and Seloula [6, Theorem ]). We will see, in Lemma 2 of Section 2, that We are in a position to state the main theorem.

Theorem 1. Let be a bounded domain satisfying (O1) and (O2), and let be a tangential vector field on . Then is achieved, and the minimizer of in satisfies the following estimate. There exists a constant independent of such that

2. Preliminaries

In this section, we shall give some lemmas as preliminaries.

Lemma 2. Let be a tangential vector field on . Then one has

Proof. Put Since , it is trivial that . For any , the problemhas a unique solution (cf. Girault and Raviart [7, Theorem ]). If we define , then in and . Thus . So we haveThus we have .

By Lemma 2, the minimization problem (1) reduces to the following problem.

Problem B. Find the minimizer such that

In the later, we frequently use the following lemma.

Lemma 3. (i) If , , , and , then , and there exists a constant such thatHere we note that if furthermore is simply connected, we can delete the first term in the right-hand side of (18).
(ii) If , , , and , then , and there exists a constant such thatWe note that if furthermore has no holes, we can delete the first term in the right-hand side of (19).

For the proof of (18) and (19), see [5, Theorem and Corollary ]. If is simply connected or has no holes, see Aramaki [8, Lemma ].

Lemma 4. The space is a closed subspace of .

Proof. Let in . Then from (19) we haveTherefore is a Cauchy sequence in . Hence there exists such that in , so we have and in as . It is clear that in , and on . This implies that .

3. Proof of the Main Theorem 1

In this section, we give a proof of Theorem 1. The proof consists of some lemmas and propositions. Throughout this section, we assume that is a given tangential vector field on .

Lemma 5. Let . Then the minimization problem has a unique minimizer.

Proof. From Lemma 4, we know that is a closed subspace of . Thus it is well known that (21) has a minimizer. For the uniqueness of the minimizer, it suffices to show that the unit sphere does not contain any line segment for and . (cf. Fujita et al. [9, p. 306 and the remark]). However, this is clear because the functionalis strictly convex.

For , let be a unique minimizer of (21) and define . Then since for any and , , we haveIf we define a spacethen we see that . Then we have the following.

Lemma 6. One can see that

Proof. For any , as the above we can writeWe show the uniqueness of the above decomposition. If we can writewhere and , then . Therefore we haveHenceHere we use the following inequality. There exists a constant such thatfor all . For the proof of this inequality, see DiBenedetto [10, Lemma ] for , and see Miranda et al. [11, (7C’)]. Applying (30) with to (29), we haveFrom these equalities, we have , so .

Now we state a refinement of Fatou’s lemma (cf. Evans [12, pp. 11-12]).

Lemma 7. Assume that . Let weakly in and a.e. in . Then one hasIf furthermorethenwhere denotes the conjugate exponent of ; that is, . In particular, if strongly in and a.e. in , then (34) holds.

Proof. We use an elementary estimate. Let . Then, for any fixed , there exists a constant such that for any (cf. [12, (1.13)]). Definewhere for . Then we have If we apply (35) with and , we haveWe note that the right-hand side is integrable. By the hypothesis, we can see that a.e. in . Therefore by the Lebesgue dominated theorem, we haveTherefore we have Since weakly in , is bounded. Since is arbitrary, we haveIf furthermorethen we have

Lemma 8. is a weakly closed set in .

Proof. Let weakly in . Then we have in , on , andPassing to a subsequence, we may assume that strongly in and a.e. in . Thus from Lemma 7, we have in . Therefore we haveThis implies that .

Lemma 9. There exists a constant such that for all satisfying in andone has

Proof. If the conclusion (47) is false, there exists a sequence satisfying in andsuch that as . After passing to a subsequence, we may assume that weakly in , strongly in , and a.e. in . Therefore we have in and on , so . From Lemma 7,Thus we have . Hence strongly in . From (19), we see thatas . This contradicts .

Proposition 10. Let . Then the minimization problemis achieved and

Proof. By Lemma 2, we can see thatSince , it is clear thatOn the other hand, for any , we can write , where , and . Hence we haveThus (52) holds. We show that the right-hand side of (52) has a minimizer. Let be a minimizing sequence. ThenBy (1), we haveThus, by Lemma 9, is bounded in . Passing to a subsequence, we may assume that weakly in , strongly in , and a.e. in . Therefore we have on . Sinceit follows from Lemma 7 thatTherefore . It suffices to prove thatIn fact, we can choose a subsequence of so thatSince weakly in , it follows from the Mazur theorem that there exist such that convex hull of and strongly in . Hence we can choose a subsequence of so that strongly in and a.e. in . By the Fatou lemma, we haveSince is a convex function with respect to , we haveTherefore we have This completes the proof.

Lemma 11. Let be a minimizer of . Then is a weak solution of the following system:

Proof. If is a minimizer of , then we can see that, for any , we haveThus we have for all . We claim that In fact, since it is clear that , we haveConversely let . Choose to be a solution ofBy the elliptic regularity theorem, we see that . Define . Then , in , and on . Therefore and .
Hence (67) holds for any . Since , it follows from (67) that is a weak solution of (65).

Remark 12. The system (65) is so called the -curl system. When is a bounded, simply connected domain in without holes and with boundary for some . If , then [8] showed that the weak solution of system (65) satisfies the fact that for some and there exists a constant depending only on such that .

Lemma 13. Let be a minimizer of (52). Then any minimizer of (17) must have the form where . In particular, the minimizer of (52) is unique.

Proof. Since for any , we see thatThus is a minimizer of (17). On the other hand, for any minimizer of (17), define . Then . From (67), we haveTherefore,By the structure condition (2), we have in , so .
If is a minimizer of (52), we can write , where . If follows from Lemma 6 that we see that . Thus the minimizer of (52) in is unique.

For , let be a minimizer of (17). Then there exist uniquely which is a minimizer of (52) and such that We note that .

In order to show the estimate in Theorem 1, it suffices to prove the following proposition.

Proposition 14. There exists a constant independent of such that

Proof. Assume that the conclusion is false. Then there exists a sequence such that andFor brevity of notation, we write . Passing to a subsequence, we may assume that weakly in , strongly in , and a.e. in . Thus , in , and on . Since satisfiesand strongly in and a.e. in , it follows from Lemma 7 that Hence we have . On the other hand, is a weak solution of Since and , we see that . Since , it follows from the Green formula that where denotes the duality bracket of the spaces and . Here we have Since weakly in , we see that is bounded. Since , we haveas . Since is equivalent to , using (80), we have Since , we see that , so . From (78) with , we haveTherefore in , so weakly in and strongly in . From (80), we can see that . By (19),as . This contradicts .

Proof of Theorem 1. The proof of Theorem 1 follows from Lemma 2 and Propositions 10 and 14.

Remark 15. Instead of minimizing , it is also interesting to minimize . This problem is related to the mathematical theory of liquid crystals. For and , see Aramaki [13].

Competing Interests

The author declares that there is no conflict of interests regarding the publication of this paper.