Chinese Journal of Mathematics
Volume 2016 (2016), Article ID 2459694, 8 pages
http://dx.doi.org/10.1155/2016/2459694

Research Article

Variational Problem Involving Operator Curl in a Multiconnected Domain

Junichi Aramaki

Division of Science, Faculty of Science and Engineering, Tokyo Denki University, Hatoyama-machi, Saitama 350-0394, Japan

Received 26 August 2016; Accepted 12 October 2016

Academic Editor: Zui-Cha Deng

Copyright © 2016 Junichi Aramaki. 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.

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 fields

Then 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 spaces

then 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]), and

Moreover, 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 problem

has a unique solution (cf. Girault and Raviart [7, Theorem ]). If we define , then in and . Thus . So we have

Thus 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 that

Here 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 that

We 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 have

Therefore 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 functional

is strictly convex.

For , let be a unique minimizer of (21) and define . Then since for any and , , we have

If we define a space

then we see that . Then we have the following.

Lemma 6. One can see that

Proof. For any , as the above we can write

We show the uniqueness of the above decomposition. If we can write

where and , then . Therefore we have

Hence

Here we use the following inequality. There exists a constant such that

for 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 have

From 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 has

If furthermore

then

where 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)]). Define

where for . Then we have

If we apply (35) with and , we have

We 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 have

Therefore we have

Since weakly in , is bounded. Since is arbitrary, we have

If furthermore

then we have

Lemma 8. is a weakly closed set in .

Proof. Let weakly in . Then we have in , on , and

Passing to a subsequence, we may assume that strongly in and a.e. in . Thus from Lemma 7, we have in . Therefore we have

This implies that .

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

one has

Proof. If the conclusion (47) is false, there exists a sequence satisfying in and

such 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 that

as . This contradicts .

Proposition 10. Let . Then the minimization problem

is achieved and

Proof. By Lemma 2, we can see that

Since , it is clear that

On the other hand, for any , we can write , where , and . Hence we have

Thus (52) holds. We show that the right-hand side of (52) has a minimizer. Let be a minimizing sequence. Then

By (1), we have

Thus, 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 . Since

it follows from Lemma 7 that

Therefore . It suffices to prove that

In fact, we can choose a subsequence of so that

Since 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 have

Since is a convex function with respect to , we have

Therefore 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 have

Thus we have

for all . We claim that

In fact, since it is clear that , we have

Conversely let . Choose to be a solution of

By 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 that

Thus is a minimizer of (17). On the other hand, for any minimizer of (17), define . Then . From (67), we have

Therefore,

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 and

For 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 satisfies

and 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 have

as . Since is equivalent to , using (80), we have

Since , we see that , so . From (78) with , we have

Therefore 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.

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