# Variational Problem Involving Operator Curl in a Multiconnected Domain

Authors: Junichi Aramaki
Journal: Chinese Journal of Mathematics
 (2016)

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## 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 ofL2 minimization problem of the curl of vector fields. We shall prove the existence and the estimate of minimizers of more general functional which contains Lp norm of the curl of vector fields.


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## Body

## 1. Introduction

In this paper, we consider the following problem which was proposed by Pan [1, p. 9].Problem A.
Minimize theLp 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. Whenp=2 and Ω 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 p=2.In the present paper we shall extend the results to more general functional containing Problem A.More precisely, letS(x,t) be a Carathéodory function on Ω×[0,∞) and S(x,t2) is a convex function with respect to t; moreover assume that for a.e. x∈Ω, S(x,t)∈C1((0,∞)), and there exist 1<p<∞ and λ,Λ>0 such that for a.e. x∈Ω and all t>0: (1)λtp-2/2≤Stx,t≔∂∂tSx,t≤Λtp-2/2.Without loss of generality, we may assume that S(x,0)=0. We furthermore assume the following structure condition: (2)Stx,a2a-Stx,b2b·a-b>0for  anya,b∈R3witha≠b.Under (1) with S(x,0)=0, we have (3)2pλtp/2≤Sx,t≤2pΛtp/2.For example, the functionS(x,t)=ν(x)tp/2 where ν(x) is a measurable function satisfying 0<ν∗≤ν(x)≤ν∗<∞ for a.e. x∈Ω satisfies (1)-(2).LetΩ be a bounded domain in R3 with C2 boundary ∂Ω. Let HT be a given tangential vector field on ∂Ω. Let W1,p(Ω,R3) be the standard Sobolev space of vector fields. From now, we denote the tangential component of a vector field u by uT; that is, uT=u-(u·ν)ν, where ν is the outer normal unit vector to the boundary ∂Ω. For any given tangential vector field on ∂Ω(4)HT∈W1-1/p,p∂Ω,R3,define a space of vector fields(5)Wt1,pΩ,R3,HT=u∈W1,pΩ,R3;uT=HTon∂Ω.Then it is clear that Wt1,p(Ω,R3,HT) is a closed convex set in W1,p(Ω,R3). We consider the minimization problem (6)RtpHT=infu∈Wt1,pΩ,R3,HT⁡∫ΩSx,curl⁡u2dx.Whenp=2, S(x,t)=t, and Ω is a simply connected domain without holes, the authors of [2, 3] showed that (6) is achieved, and then in the case where p=2, S(x,t)=T, and Ω is bounded multiconnected domain, the author of [1] succeeded to show the existence of a minimizer of (6).Since we allowΩ to be a multiconnected domain in R3, throughout this paper, we assume that the domain Ω satisfies the following (O1) and (O2) (cf. Dautray and Lions [4] and Amrouche and Seloula [5]).(O1)Ω is a bounded domain in R3 with C2 boundary ∂Ω. Ω is locally situated on one side of ∂Ω; ∂Ω has a finite number of connected components Γ1,…,Γm+1(m≥0) and Γm+1 denoting the boundary of the infinite connected component of R3∖Ω¯.(O2) There existn manifolds of dimension 2 and of class C2 denoted by Σ1,…,Σn(n≥0) such that Σi∩Σj=∅(i≠j) and they are nontangential to ∂Ω and such that Ω∖(⋃i=1nΣi) is simply connected and pseudo C1,1.The numbern is called the first Betti number and m the second Betti number of Ω. We say that Ω is simply connected if n=0, and Ω has no holes if m=0. If we define the spaces(7)KNpΩ=u∈W1,pΩ,R3;curl⁡u=0,div⁡u=0inΩ,ν·u=0on∂Ω,KTpΩ=u∈W1,pΩ,R3;curl⁡u=0,div⁡u=0inΩ,uT=0on∂Ω,then it is well known that dim⁡KNp(Ω)=n and dim⁡KTp(Ω)=m. We note that KNp(Ω) and KTp(Ω) are contained in W1,p(Ω,R3); moreover, KNp(Ω) and KTp(Ω) are closed subspaces of W1,p(Ω,R3). Also it will be shown in Lemma 4 that KNp(Ω) and KTp(Ω) are closed subspaces of Lp(Ω,R3). Thus since KTp(Ω) is a finite-dimensional closed subspace of Lp(Ω,R3), KTp(Ω) has a complement Lp in Lp(Ω,R3); that is, Lp is a closed subspace of Lp(Ω,R3), Lp∩KTp(Ω)={0}, and Lp(Ω,R3)=Lp⊕KTp(Ω) (the direct sum). Therefore, for any w∈Lp(Ω,R3), there exist uniquely v∈Lp and u∈KTp(Ω) such that w=v+u. We denote the projection P:Lp(Ω,R3)→Lp by Pw=v.Define(8)HpΩ,curl,div⁡0=u∈LpΩ,R3;curl⁡u∈LpΩ,R3,div⁡u=0inΩ,HtpΩ,curl,div⁡0,HT=u∈HpΩ,curl,div⁡0;uT=HTon∂Ω.Note that if u∈Lp(Ω,R3) and curl⁡u∈Lp(Ω,R3), then the tangent trace uT is well defined as an element of W-1/p,p(∂Ω,R3) (cf. [5, p. 45]), and(9)HpΩ,curl,div⁡0∩W1,pΩ,R3=u∈W1,pΩ,R3;div⁡u=0inΩ.Moreover, we note that if HT∈W1-1/p,p(∂Ω,R3), then(10)HtpΩ,curl,div⁡0,HT⊂Wt1,pΩ,R3,HT.(cf. Amrouche and Seloula [6, Theorem 2.3]). We will see, in Lemma 2 of Section 2, that (11)RtpHT=infv∈HtpΩ,curl,div0,HT⁡∫ΩSx,curl⁡v2dx.We are in a position to state the main theorem.Theorem 1.
LetΩ⊂R3 be a bounded domain satisfying (O1) and (O2), and let HT∈W1-1/p,p(∂Ω,R3) be a tangential vector field on ∂Ω. Then Rtp(HT) is achieved, and the minimizer A of Rtp(HT) in Htp(Ω,curl,div⁡0,HT) satisfies the following estimate. There exists a constant C=C(Ω)>0 independent of HT such that(12)PAW1,pΩ≤CHTW1-1/p,p∂Ω.


## 2. Preliminaries

In this section, we shall give some lemmas as preliminaries.Lemma 2.
LetHT∈W1-1/p,p(∂Ω,R3) be a tangential vector field on ∂Ω. Then one has (13)RtpHT=infv∈HtpΩ,curl,div⁡0,HT⁡∫ΩSx,curl⁡v2dx.Proof.
Put(14)α=infu∈Wt1,pΩ,R3,HT⁡∫ΩSx,curl⁡u2dx,β=infv∈HtpΩ,curl,div⁡0,HT⁡∫ΩSx,curl⁡v2dx.Since Htp(Ω,curl,div⁡0,HT)⊂Wt1,p(Ω,R3,HT), it is trivial that α≤β. For any u∈W1,p(Ω,R3,HT), the problem(15)Δφ=div⁡uinΩ,φ=0on∂Ωhas a unique solution φ∈W2,p(Ω) (cf. Girault and Raviart [7, Theorem 1.8]). If we define v=u-∇φ∈W1,p(Ω,R3), then curl⁡v=curl⁡u,div⁡v=div⁡u-Δφ=0 in Ω and vT=uT-(∇φ)T=uT=HT. Thus v∈Htp(Ω,curl,div⁡0,HT). So we have(16)∫ΩSx,curl⁡u2dx=∫ΩSx,curl⁡v2dx≥β.Thus we have α≥β.By Lemma2, the minimization problem (1) reduces to the following problem.Problem B.
Find the minimizeru∈Htp(Ω,curl,div⁡0,HT) such that (17)RtpHT=infv∈HtpΩ,curl,div⁡0,HT⁡∫ΩSx,curl⁡v2dx.In the later, we frequently use the following lemma.Lemma 3.
(i) Ifu∈Lp(Ω,R3), curl⁡u∈Lp(Ω,R3), div⁡u∈Lp(Ω), and u·ν∈W1-1/p,p(∂Ω), then u∈W1,p(Ω,R3), and there exists a constant c1(Ω)>0 such that(18)uW1,pΩ≤c1ΩuLpΩ+curl⁡uLpΩ+div⁡uLpΩ+u·νW1-1/p,p∂Ω.Here we note that if furthermore Ω is simply connected, we can delete the first term uLp(Ω) in the right-hand side of (18).
(ii) Ifu∈Lp(Ω,R3), curl⁡u∈Lp(Ω,R3), div⁡u∈Lp(Ω), and uT∈W1-1/p,p(∂Ω,R3), then u∈W1,p(Ω,R3), and there exists a constant c2(Ω)>0 such that(19)uW1,pΩ≤c2ΩuLpΩ+curl⁡uLpΩ+div⁡uLpΩ+uTW1-1/p,p∂Ω.We note that if furthermore Ω has no holes, we can delete the first term uLp(Ω) in the right-hand side of (19).For the proof of (18) and (19), see [5, Theorem 3.4 and Corollary 5.2]. If Ω is simply connected or has no holes, see Aramaki [8, Lemma 2.2].Lemma 4.
The spaceKTp(Ω) is a closed subspace of Lp(Ω,R3).Proof.
LetKTp(Ω)∋uj→u in Lp(Ω,R3). Then from (19) we have(20)uj-ukW1,pΩ≤c2Ωuj-ukLpΩ.Therefore {uj} is a Cauchy sequence in W1,p(Ω,R3). Hence there exists u0∈W1,p(Ω,R3) such that uj→u0 in W1,p(Ω,R3), so we have u=u0 and uj→u in W1,p(Ω,R3) as j→∞. It is clear that curl⁡u=0,div⁡u=0 in Ω, and uT=0 on ∂Ω. This implies that u∈KTp(Ω).


## 3. Proof of the Main Theorem1

In this section, we give a proof of Theorem1. The proof consists of some lemmas and propositions. Throughout this section, we assume that HT is a given tangential vector field on ∂Ω.Lemma 5.
LetA∈Htp(Ω,curl,div⁡0,HT). Then the minimization problem (21)γ=infu∈KTpΩ⁡A-uLpΩhas a unique minimizer.Proof.
From Lemma4, we know that KTp(Ω) is a closed subspace of Lp(Ω,R3). Thus it is well known that (21) has a minimizer. For the uniqueness of the minimizer, it suffices to show that the unit sphere B={u∈Lp(Ω,R3);uLp(Ω)=1} does not contain any line segment [u,v]={λu+(1-λ)v;0≤λ≤1} for u,v∈B and u≠v. (cf. Fujita et al. [9, p. 306 and the remark]). However, this is clear because the functional(22)fu=∫Ωupdxis strictly convex.ForA∈Htp(Ω,curl,div⁡0,HT), let u∈KTp(Ω) be a unique minimizer of (21) and define B=A-u. Then since for any z∈KTp(Ω) and t∈R, BLp(Ω)p≤B+tzLp(Ω)p, we have(23)0=ddtt=0∫ΩB+tzpdx=p∫ΩBp-2B·zdx.If we define a space(24)BΩ,HT=B∈LpΩ,R3;curl⁡B∈LpΩ,R3,div⁡B=0inΩ,BT=HTon∂Ω,∫ΩBp-2B·zdx=0∀z∈KTpΩ,then we see that B∈B(Ω,HT). Then we have the following.Lemma 6.
One can see that(25)HtpΩ,curl,div⁡0,HT=BΩ,HT⊕KTpΩthe  direct  sum.Proof.
For anyA∈Htp(Ω,curl,div⁡0,HT), as the above we can write(26)A=B+u,whereB∈BΩ,HT,u∈KTpΩ.We show the uniqueness of the above decomposition. If we can write(27)A=B1+u1=B2+u2,where B1,B2∈BΩ,HT,u1 and u2∈KTp(Ω), then B1-B2=u2-u1∈KTp(Ω). Therefore we have(28)∫ΩB1p-2B1·B1-B2dx=0,∫ΩB2p-2B2·B1-B2dx=0.Hence(29)∫ΩB1p-2B1-B2p-2B2·B1-B2dx=0.Here we use the following inequality. There exists a constant c>0 such that(30)ap-2a-bp-2b·a-b≥ca-bpifp≥2,ca+bp-2a-b2if1<p<2for all a,b∈R3. For the proof of this inequality, see DiBenedetto [10, Lemma 4.4] for p≥2, and see Miranda et al. [11, (7C’)]. Applying (30) with a=B1,b=B2 to (29), we have(31)∫ΩB1-B2pdx=0forp≥2,∫ΩB1+B2p-2B1-B22dx=0for1<p<2.From these equalities, we have B1=B2, so u1=u2.Now we state a refinement of Fatou’s lemma (cf. Evans [12, pp. 11-12]).Lemma 7.
Assume that1<p<∞. Let Bj→B weakly in Lp(Ω,R3) and a.e. in Ω. Then one has(32)limj→∞⁡∫ΩBjp-Bjp-2Bj-Bp-2Bp′dx=∫ΩBpdx.If furthermore(33)limj→∞∫ΩBjpdx=∫ΩBpdx,then(34)Bjp-2Bj⟶Bp-2Bstrongly  inLp′Ω,R3,where p′ denotes the conjugate exponent of p; that is, (1/p)+(1/p′)=1. In particular, if Bj→B strongly in Lp(Ω,R3) and a.e. in Ω, then (34) holds.Proof.
We use an elementary estimate. Let1≤q<∞. Then, for any fixed ε>0, there exists a constant C=C(ε,q)>0 such that (35)a+bq-aq≤εaq+Cbqfor any a,b∈R3 (cf. [12, (1.13)]). Define(36)gjε=Bjp-2Bjp′-Bjp-2Bj-Bp-2Bp′-Bp-2Bp′-εBjp-2Bj-Bp-2Bp′+,where [a]+=max⁡{a,0} for a∈R. Then we have (37)gjε≤Bjp-2Bjp′-Bjp-2Bj-Bp-2Bp′+Bp-2Bp′-εBjp-2Bj-Bp-2Bp′+=Bjp-2Bj-Bp-2B+Bp-2Bp′-Bjp-2Bj-Bp-2Bp′+Bp-2Bp′-εBjp-2Bj-Bp-2Bp′+.If we apply (35) with a=Bjp-2Bj-|B|p-2B,b=Bp-2B and q=p′, we have(38)gjε≤C+1Bp-2Bp′=C+1Bp.We note that the right-hand side is integrable. By the hypothesis, we can see that gjε→0 a.e. in Ω. Therefore by the Lebesgue dominated theorem, we have(39)limj→∞∫Ωgjεdx=0.Therefore we have (40)limsupj→∞∫ΩBjp-2Bjp′-Bjp-2Bj-Bp-2Bp′-Bp-2Bp′dx≤εlimsupj→∞∫ΩBjp-2Bj-Bp-2Bp′dx≤ε2p′limsupj→∞∫ΩBjp-2Bjp′+Bp-2Bp′dx=ε2p′limsupj→∞∫ΩBjp+Bpdx.Since Bj→B weakly in Lp(Ω,R3), BjLp(Ω) is bounded. Since ε is arbitrary, we have(41)limj→∞∫ΩBjp-Bjp-2Bj-Bp-2Bp′dx=∫ΩBpdx.If furthermore(42)limj→∞∫ΩBjpdx=∫ΩBpdx,then we have(43)limj→∞∫ΩBjp-2Bj-Bp-2Bp′dx=0.Lemma 8.
B ( Ω , H T ) is a weakly closed set in W1,p(Ω,R3).Proof.
LetBj∈B(Ω,HT),Bj→B weakly in W1,p(Ω,R3). Then we have curl⁡B∈Lp(Ω,R3),div⁡B=0 in Ω, BT=HT on ∂Ω, and(44)∫ΩBjp-2Bj·zdx=0∀z∈KTpΩ.Passing to a subsequence, we may assume that Bj→B strongly in Lp(Ω,R3) and a.e. in Ω. Thus from Lemma 7, we have |Bj|p-2Bj→|B|p-2B in Lp′(Ω,R3). Therefore we have(45)∫ΩBp-2B·zdx=0∀z∈KTpΩ.This implies that B∈B(Ω,HT).Lemma 9.
There exists a constantc(Ω)>0 such that for all B∈W1,p(Ω,R3) satisfying div⁡B=0 in Ω and(46)∫ΩBp-2B·zdx=0∀z∈KTpΩ,one has (47)BW1,pΩ≤cΩcurl⁡BLpΩ+BTW1-1/p,p∂Ω.Proof.
If the conclusion (47) is false, there exists a sequence {Bj}⊂W1,p(Ω,R3) satisfying div⁡Bj=0 in Ω and(48)∫ΩBjp-2Bj·zdx=0∀z∈KTpΩ,such that BjW1,p(Ω)=1,curl⁡BjLp(Ω)→0,Bj,TW1-1/p,p(∂Ω)→0 as j→∞. After passing to a subsequence, we may assume that Bj→B0 weakly in W1,p(Ω,R3), strongly in Lp(Ω,R3), and a.e. in Ω. Therefore we have div⁡B0=0,curl⁡B0=0 in Ω and B0,T=0 on ∂Ω, so B0∈KTp(Ω). From Lemma 7,(49)∫ΩB0pdx=∫ΩB0p-2B0·B0dx=limj→∞∫ΩBjp-2Bj·B0dx=0.Thus we have B0=0. Hence Bj→0 strongly in Lp(Ω,R3). From (19), we see that(50)BjW1,pΩ≤c2ΩBjLpΩ+curl⁡BjLpΩ+Bj,TW1-1/p,p∂Ω⟶0as j→∞. This contradicts BjW1,p(Ω)=1.Proposition 10.
LetHT∈W1-1/p,p(∂Ω,R3). Then the minimization problem(51)infB∈BΩ,HT⁡∫ΩSx,curl⁡B2dxis achieved and (52)RtpHT=infB∈BΩ,HT⁡∫ΩSx,curl⁡B2dxProof.
By Lemma2, we can see that(53)RtpHT=infA∈HtpΩ,curl,div⁡0,HT⁡∫ΩSx,curl⁡A2dx.Since B(Ω,HT)⊂Htp(Ω,curl,div⁡0,HT), it is clear that(54)RtpHT≤infB∈BΩ,HT⁡∫ΩSx,curl⁡B2dx.On the other hand, for any A∈Htp(Ω,curl,div⁡0,HT), we can write A=B+u, where B∈BΩ,HT, and u∈KTp(Ω). Hence we have(55)∫ΩSx,curl⁡A2dx=∫ΩSx,curl⁡B2dx≥infB∈BΩ,HT⁡∫ΩSx,curl⁡B2dx.Thus (52) holds. We show that the right-hand side of (52) has a minimizer. Let {Bj}⊂B(Ω,HT) be a minimizing sequence. Then(56)∫ΩSx,curl⁡Bj2dx=RtpHT+o1asj⟶∞.By (1), we have(57)2pλ∫Ωcurl⁡Bjpdx≤∫ΩSx,curl⁡Bj2dx=RtpHT+o1.Thus, by Lemma 9, {Bj} is bounded in W1,p(Ω,R3). Passing to a subsequence, we may assume that Bj→B0 weakly in W1,p(Ω,R3), strongly in Lp(Ω,R3), and a.e. in Ω. Therefore we have div⁡B0=0,B0,T=HT on ∂Ω. Since(58)∫ΩBjp-2Bj·zdx=0∀z∈KTpΩ,it follows from Lemma 7 that(59)∫ΩB0p-2B0·zdx=0∀z∈KTpΩ.Therefore B0∈B(Ω,HT). It suffices to prove that(60)∫ΩSx,curl⁡B02dx≤liminfj→∞⁡∫ΩSx,curl⁡Bj2dx.In fact, we can choose a subsequence {curl⁡Bjk} of {curl⁡Bj} so that(61)limk→∞∫ΩSx,curl⁡Bjk2dx=liminfj→∞⁡∫ΩSx,curl⁡Bj2dx.Since curl⁡Bjk→curl⁡B0 weakly in Lp(Ω,R3), it follows from the Mazur theorem that there exist gl∈Lp(Ω,R3) such that gl∈ convex hull of {curl⁡Bjk;k≥l} and gl→curl⁡B0 strongly in Lp(Ω,R3). Hence we can choose a subsequence {glm} of {gl} so that glm→curl⁡B0 strongly in Lp(Ω,R3) and a.e. in Ω. By the Fatou lemma, we have(62)∫ΩSx,curl⁡B02dx≤liminfm→∞⁡∫ΩSx,glm2dx.Since S(x,t2) is a convex function with respect to t, we have(63)∫ΩSx,glm2dx≤sup⁡∫ΩSx,curl⁡Bjk2dx;k≥lm.Therefore we have (64)∫ΩSx,curl⁡B02dx≤liminfm→∞⁡∫ΩSx,glm2dx≤limm→∞sup⁡∫ΩSx,curl⁡Bjk2dx;k≥lm=limk→∞∫ΩSx,curl⁡Bjk2dx=liminfj→∞⁡∫ΩSx,curl⁡Bj2dx.This completes the proof.Lemma 11.
LetA∈Htp(Ω,curl,div⁡0,HT) be a minimizer of Rtp(HT). Then A is a weak solution of the following system:(65)curl⁡Stx,curl⁡A2curl⁡A=0,div⁡A=0inΩ,AT=HTon∂Ω.Proof.
IfA∈Htp(Ω,curl,div⁡0,HT) is a minimizer of Rtp(HT), then we can see that, for any w∈Htp(Ω,curl,div⁡0,0), we have(66)ddtt=0∫ΩSx,curl⁡A+tcurl⁡w2dx=0.Thus we have (67)∫ΩStx,curl⁡A2curl⁡A·curl⁡wdx=0for all w∈Htp(Ω,curl,div⁡0,0). We claim that (68)curl⁡HtpΩ,curl,div⁡0,0=curl⁡Wt1,pΩ,R3,0.
In fact, since it is clear thatHtp(Ω,curl,div⁡0,0)⊂Wt1,p(Ω,R3,0), we have(69)curl⁡HtpΩ,curl,div⁡0,0⊂curl⁡Wt1,pΩ,R3,0.Conversely let u∈Wt1,p(Ω,R3,0). Choose ϕ to be a solution of(70)Δϕ=div⁡uinΩ,ϕ=0on∂Ω.By the elliptic regularity theorem, we see that ϕ∈W2,p(Ω). Define v=u-∇ϕ. Then curl⁡v=curl⁡u∈Lp(Ω,R3), div⁡v=div⁡u-Δϕ=0 in Ω, and vT=uT-(∇ϕ)T=uT=0 on ∂Ω. Therefore v∈Htp(Ω,curl,div⁡0,0) and curl⁡u=curl⁡v∈curl⁡[Htp(Ω,curl,div⁡0,0)].
Hence (67) holds for any w∈Wt1,p(Ω,R3,0). Since D(Ω,R3)⊂Wt1,p(Ω,R3,0), it follows from (67) that A is a weak solution of (65).Remark 12.
The system (65) is so called the p-curl system. When Ω is a bounded, simply connected domain in R3 without holes and with C2+α boundary for some α∈(0,1). If HT=0, then [8] showed that the weak solution A of system (65) satisfies the fact that A∈C1+β(Ω¯,R3) for some β∈(0,1) and there exists a constant C depending only on p,Ω such that AC1+β(Ω¯)≤C.Lemma 13.
LetB0∈B(Ω,HT) be a minimizer of (52). Then any minimizer A∈Htp(Ω,curl,div⁡0,HT) of (17) must have the form A=B0+u where u∈KTp(Ω). In particular, the minimizer of (52) is unique.Proof.
Since for anyu∈KTp(Ω), we see that(71)B0+u∈HtpΩ,curl,div⁡0,HT,∫ΩcurlB0+updx=∫Ωcurl⁡B0pdx=RtpHT.Thus B0+u is a minimizer of (17). On the other hand, for any minimizer A∈Htp(Ω,curl,div⁡0,HT) of (17), define w=A-B0. Then w∈Htp(Ω,curl,div⁡0,0). From (67), we have(72)∫ΩStx,curl⁡A2curl⁡A·curl⁡wdx=∫ΩStx,curl⁡B02curl⁡B0·curl⁡wdx=0.Therefore,(73)∫ΩStx,curl⁡A2curl⁡A-Stx,curl⁡B02curl⁡B0·curl⁡A-curl⁡B0dx=0.By the structure condition (2), we have curl⁡(A-B0)=0 in Ω, so A-B0∈KTp(Ω).
IfB∈B(Ω,HT)⊂Htp(Ω,curl,div⁡0,HT) is a minimizer of (52), we can write B=B0+u, where u∈KTp(Ω). If follows from Lemma 6 that we see that u=0. Thus the minimizer of (52) in B(Ω,HT) is unique.ForHT∈W1-1/p,p(∂Ω,R3), let A=A(HT)∈Htp(Ω,curl,div⁡0,HT) be a minimizer of (17). Then there exist uniquely B0=B0(HT)∈B(Ω,HT) which is a minimizer of (52) and u=u(HT)∈KTp(Ω) such that (74)AHT=B0HT+uHT.We note that PA(HT)=B0(HT).In order to show the estimate in Theorem1, it suffices to prove the following proposition.Proposition 14.
There exists a constantc=c(Ω) independent of HT such that(75)B0HTW1,pΩ≤cHTW1-1/p,p∂Ω.Proof.
Assume that the conclusion is false. Then there exists a sequence{Hj,T}⊂W1-1/p,p(∂Ω,R3) such that B0(Hj,T)W1,p(Ω)=1 and(76)Hj,TW1-1/p,p∂Ω⟶0asj⟶∞.For brevity of notation, we write Bj=B0(Hj,T). Passing to a subsequence, we may assume that Bj→B weakly in W1,p(Ω,R3), strongly in Lp(Ω,R3), and a.e. in Ω. Thus curl⁡B∈Lp(Ω,R3), div⁡B=0 in Ω, and BT=0 on ∂Ω. Since Bj satisfies(77)∫ΩBjp-2Bj·zdx=0∀z∈KTpΩand Bj→B strongly in Lp(Ω,R3) and a.e. in Ω, it follows from Lemma 7 that (78)∫ΩBp-2B·zdx=0∀z∈KTpΩ.Hence we have B∈B(Ω,0). On the other hand, Bj is a weak solution of (79)curl⁡Stx,curl⁡Bj2curl⁡Bj=0inΩ,Bj,T=Hj,Ton∂Ω.Since Stx,curlBj2curl⁡Bj∈Lp′(Ω,R3) and curl⁡Stx,curlBj2curl⁡Bj=0, we see that Stx,curlBj2curl⁡Bj∂Ω∈W-1/p′,p′(∂Ω,R3). Since ν×Hj,T∈W1-1/p,p(∂Ω,R3)=W1/p′,p(∂Ω,R3), it follows from the Green formula that (80)0=∫Ωcurl⁡Stx,curl⁡Bj2curl⁡Bj·Bjdx=∫ΩStx,curl⁡Bj2curl⁡Bj·curl⁡Bjdx+∫∂ΩBj,T,ν×Stx,curl⁡Bj2curl⁡BjdS,where 〈·,·〉 denotes the duality bracket of the spaces W1/p′,p(∂Ω,R3) and W-1/p′,p′(∂Ω,R3). Here we have (81)∫∂ΩHj,T,ν×Stx,curl⁡Bj2curl⁡BjdS≤Hj,TW1-1/p,p∂ΩStx,curl⁡Bj2curl⁡BjLp′Ω≤Hj,TW1-1/p,p∂Ω∫ΩΛcurl⁡Bjp-1p′dx1/p′≤ΛHj,TW1-1/p,p∂Ωcurl⁡BjLpΩp/p′.Since curl⁡Bj→curl⁡B weakly in Lp(Ω,R3), we see that curlBjLp(Ω) is bounded. Since Hj,TW1-1/p,p(∂Ω)→0, we have(82)∫∂Ων×Hj,T,Stx,curl⁡Bj2curl⁡BjdS⟶0as j→∞. Since S(x,t2)t2 is equivalent to S(x,t), using (80), we have (83)∫ΩStx,curl⁡B2curl⁡B2dx≤liminfj→∞⁡∫ΩStx,curl⁡Bj2curl⁡Bj2dx=liminfj→∞⁡∫ΩStx,curl⁡Bj2curl⁡Bj2dx+∫∂Ων×Hj,T,Stx,curl⁡Bj2curl⁡BjdS=limsupj→∞∫ΩStx,curl⁡Bj2curl⁡Bj2dx+∫∂Ων×Hj,T,Stx,curl⁡Bj2curl⁡BjdS=0.Since St(x,curlB2)curlB2≥λ|curlB|p, we see that curl⁡B=0, so B∈KTp(Ω). From (78) with z=B, we have(84)0=∫ΩBp-2B·Bdx=∫ΩBpdx.Therefore B=0 in Ω, so Bj→0 weakly in W1,p(Ω,R3) and strongly in Lp(Ω,R3). From (80), we can see that curlBjLp(Ω)→0. By (19),(85)BjW1,pΩ≤c2ΩBjLpΩ+curl⁡BjLpΩ+Hj,TW1-1/p,p∂Ω⟶0as j→∞. This contradicts BjW1,p(Ω)=1.Proof of Theorem1.
The proof of Theorem1 follows from Lemma 2 and Propositions 10 and 14.Remark 15.
Instead of minimizingS(t,curlu2), it is also interesting to minimize S(x,div⁡u2). This problem is related to the mathematical theory of liquid crystals. For p=2 and S(x,t)=t, see Aramaki [13].



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*Source: 2459694-2016-11-27.xml*