# W1,p Regularity of Weak Solutions to Maxwell’s Equations

Authors: Zhihong Chen
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we study the steady-state Maxwell’s equations. The weak solution defined in weak formulation is considered, and the global existence is obtained in general bounded open domain. The interiorW1,p2≤p<∞ estimates of the weak solution are obtained, where the coefficient matrix is assumed to be BMO with small seminorm. The main analytical tools are the Vitali covering lemma, the maximal function technique, and the compactness method. We also consider the time-harmonic Maxwell’s equations and obtain the interior W1,p estimates.


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## Body

## 1. Introduction

It is well known that the classical Maxwell’s equations can be written in differential form as follows:(1)εEt+σE=∇×H,μHt+∇×E=F,where ε is the permittivity of the electric field, μ the permeability of the magnetic field, and σ the conductivity of the material.When the material is conductive, the current displacementεEt can be ignored since it is very small compared with the eddy current σE. Then, we have the following evolution system:(2)μHt+∇×σ−1∇×H=F,∇⋅H=0.IfH is assumed to be time independent, we can obtain(3)∇×Ax∇×H=F,∇⋅H=0,where Ax=σ−1.System (3) is an important mathematical model for the study of the penetration of magnetic field in materials. Yin [1] pointed out that this system is degenerated by the classical definition (see [2]). Thus, it has a different structure with general elliptic equations, and the regularity should be restudied. The existence of a unique weak solution can be found in [3, 4]. By using Campanato theory, this system has been studied in [1, 3, 5]. They showed that the weak solution is Hölder continuous with the assumption that A is a positive bounded scalar function. In [1, 5], they got the local Hölder continuity. Afterwards, Kang and Kim [3] obtained the global Hölder continuity on the Lipschitz domain. For the higher regularity, the interior C1,α estimate has been given in [5]. The W1,p estimate can be found in [6], in which A−1 is assumed to be in the VMO space and the domain is assumed to be C1.In this paper, we establish the existence theorem of weak solution of (3) in general bounded domain and study the W1,p2≤p<∞ regularity with the assumptions that Ax is defined on ℝ3 and has the small BMO seminorm (see Definition 2).Another goal of this paper is to establish theW1,p2≤p<∞ regularity of the following system:(4)∇×Ax∇×E^+E^=∇×F^,which can describe time-harmonic electromagnetic field. We prove that if the matrix A is uniformly positive definite and has the small BMO seminorm, then the weak solution of system (4) belongs to W1,p. We weaken the assumption in [7] that A is Lipschitz continuous and also generalize the assumption in [8] that A is a bounded scalar function and the real part of A has a positive lower bound.The remaining sections are organized in the following way. InSection 2, we introduce the relevant concepts and lemmas. In Section 3, we state our main theorems and give some remarks concerning them. In Section 4, the proofs of our main results are given.


## 2. Preliminaries

We introduce some notations and lemmas here.(1)
Br=y∈ℝ3:x<r is an open ball centered at origin with radius r and Ωr=Ω∩Br(2)
For two vector fieldsa∈ℝ3 and b∈ℝ3, a⋅b and a×b define the scalar product and the cross product, respectively(3)
∇g is the gradient of g; ∇⋅u is the divergence of u; and ∇×u is the curl of u.(4)
For a locally integrable functionf,(5)f¯Brx=1Br∫Brxfydy,is the average off over Brx.(6)ℳfx=supr>01Brx∫Brxfydy,is called the Hardy–Littlewood maximal function off. We also use ℳΩf, if f is not defined outside of Ω.(5)
A=∑i,j=13aij2,A∞=supyAy.(6)
H0div,Ω=u∈H01Ω;ℝ3:∇⋅u=0inΩ.(7)
Hcurl,Ω=u∈L2Ω;ℝ3:∇×u∈L2Ω;ℝ3.Definition 1.
We say that the matrixAx=aijx3×3 is uniformly positive definite if there exists λ>0:(7)λ−1ξ2≤Aξ⋅ξ≤λξ2,∀ξ∈ℝ3.Definition 2.
The matrixAx is called δ,R-vanishing if(8)sup0<r≤Rsupx∈ℝ31Br∫BrxAy−A¯Br2dy1/2≤δ,where A¯Br is the average of A over Brx.Lemma 1 (see [9]).
Letf be a measurable function in Ω, θ>0 and N>1 be constants. Then, for any 0<p<∞,(9)f∈LpΩiffS=∑k=1∞Nkpx∈Ω:f>θNk<∞,C−1S≤∫Ωfpdx≤CΩ+S,where C>0 is a constant depending only on θ,N, and p.Lemma 2 (see [10]).
(a)
Iff∈L1ℝn, then for every α>0,(10)x∈ℝn:ℳfx>α≤Cα∫fdx,whereC is a constant which depends only on the dimension n.(b)
Iff∈Lpℝn, with 1<p≤∞, then ℳf∈Lpℝn and(11)ℳfLp≤CfLp,where C depends only on p and the dimension n.Lemma 3 (see [11]).
Assume thatC and D are measurable sets of ℝn, C⊂D⊂B1, and that there exists an ε>0 such that C<εB1, and for all x∈B1 and for all r∈0,1 with C∩Brx≥εBrx implying that Brx∩B1⊂D. Then(12)C≤10nεD.Lemma 4 (see [7]).
LetΩ⊂ℝ3 be a bounded C1 domain with connected boundary. Let f∈LpΩ;ℝ3 with ∇⋅f=0 for some 1<p<∞. Then, there exists h∈W1,pΩ;ℝ3 such that ∇×h=f. Moreover, ∇⋅h=0 in Ω and hW1,pΩ≤CfLpΩ, where C depends only on p and Ω.


## 3. Main Theorems

In the following, we assume thatB6⊂BR⊂Ω and δ is a small positive constant. Our first theorem is the well-posedness in H1Ω;ℝ3. Considering the weak solution defined in weak formulation (see Definition 3), we have the following theorem.Theorem 1.
LetΩ be a bounded open domain and Ax be uniformly positive definite. Then, for f∈L2Ω;ℝ3, the Dirichlet problem(13)∇×Ax∇×u=∇×f,∇⋅u=0,inΩ,u=0,on∂Ω.has a unique weak solution u∈H01Ω;ℝ3 with(14)∇uL2Ω≤CfL2Ω,where C depends on λ.Remark 1.
We point out that this theorem holds in general bounded open domain, which may not be Lipschitz, like the Reifenberg flat domain (see [11, 12]). When ∂Ω∈C0,1, this theorem has been proved in many papers (see [1, 3, 4, 13]) because for the Lipschitz domain, the following identity (19) can be easily obtained by performing integration by parts. Here, we will give a proof of (19) in general bounded open domain by density.Theorem 2.
Let2≤p<∞ and B6⊂BR⊂Ω. There is a small δ=δλ,p,R>0 such that for all A with A uniformly positive definite and δ,R-vanishing, and for all f with f∈LpB6;ℝ3, if u is a weak solution of (13) in B6, then u belongs to W1,pB1;ℝ3 with the estimate(15)∇uLpB1≤CuLpB6+fLpB6,where the constant C is independent of u and f.Remark 2.
We remark that our assumption thatA is δ,R-vanishing weakens the assumption in [6] that A−1 belongs to VMO. Since the linear system (13) is degenerate (see [1]), the regularity theory of elliptic systems cannot be applied directly. We established some useful lemmas to handle the difficulty. Our basic tools are the Vitali covering lemma, the Hardy–Littlewood maximal function, and the compactness method, which have been used in [11] to deal with elliptic equations.Theorem 3.
Let2≤p<∞ and B6⊂BR⊂Ω. There is a small δ=δλ,p,R>0 such that for all A with A uniformly positive definite and δ,R-vanishing, and for all f with f∈LpB6;ℝ3, if u∈Hcurl,Ω is a weak solution of(16)∇×Ax∇×u+u=∇×f,inΩ,then(17)∇uLpB1≤CuLpB6+fLpB6,where the constant C is independent of u and f.Remark 3.
Here,Hcurl,Ω=u∈L2Ω;ℝ3:∇×u∈L2Ω;ℝ3. The existence of weak solution of (16) in Hcurl,Ω can be found in [8]. We point out that our assumption that A is δ,R-vanishing weakens the assumption in [7] that A is Lipschitz continuous.Remark 4.
Ifp>3, we will have the following interior Hölder estimate:(18)uC0,αB¯1≤CuLpB6+fLpB6,where α=1−3/p. We should remark that the Hölder estimate has been established in [8], but they did not give the concrete value of the Hölder exponent.


## 4. Proofs of Main Theorems

### 4.1. Proof of Theorem1

Let us first prove the following important equality.Lemma 5.
LetΩ⊂ℝ3 be bounded and u∈H01Ω;ℝ3. We then have the following identity:(19)∇uL2Ω2=∇×uL2Ω2+∇⋅uL2Ω2.Proof.
Letϕ∈C0∞Ω;ℝ3, then we have(20)−Δϕ=∇×∇×ϕ−∇∇⋅ϕ,inΩ.
We multiply this identity byu and integrate(21)−∫ΩΔϕ⋅udx=∫Ω∇×∇×ϕ⋅udx−∫Ω∇∇⋅ϕ⋅udx.
Sinceu∈H01Ω;ℝ3, by using the definition of weak derivative, we have(22)−∫ΩΔϕ⋅udx=∫Ω∇ϕ:∇udx.
Similarly,(23)∫Ω∇×∇×ϕ⋅udx=∫Ω∇×ϕ⋅∇×udx,−∫Ω∇∇⋅ϕ⋅udx=∫Ω∇⋅ϕ∇⋅udx.
Hence, we obtain(24)∫Ω∇ϕ:∇udx=∫Ω∇×ϕ⋅∇×udx+∫Ω∇⋅ϕ∇⋅udx.
Then, if we takeϕ⟶u in H01Ω;ℝ3, (19) can be proved by the density.Remark 5.
In the proof of (19), we do not use integration by parts. Thus, we do not need the domain to be Lipschitz.Remark 6.
The identity (19) implies that the norm of H01Ω;ℝ3 is equivalent to the right hand of it raised to the power 1/2. This also means we can consider the weak solution of (13) in H0div,Ω with the norm ∇×uL2Ω instead of H01Ω;ℝ3, where H0div,Ω=u∈H01Ω;ℝ3:∇⋅u=0inΩ.Thus, we can define the weak solution of (13) as follows.Definition 3.
A vector fieldu∈H0div,Ω is said to be a weak solution of (13), if the following identity holds:(25)∫ΩAx∇×u⋅∇×vdx=∫Ωf⋅∇×vdx,for any v∈H0div,Ω.Now, we give the proof of Theorem1.Proof of Theorem 1.
In order to prove the existence and uniqueness of weak solution, we define the bilinear form as follows:(26)Bu,v=∫ΩAx∇×u⋅∇×vdx,for any u,v∈H0div,Ω. Since Ax is uniformly positive definite with λ, we have(27)Bu,u≥λ−1∇×uL2Ω2.
This means that the bilinear formBu,v is coercive on H0div,Ω. Moreover,(28)Bu,v≤λ∇×uL2Ω∇×vL2Ω.
Now fixf∈L2Ω;ℝ3 and set(29)f,v≔∫Ωf⋅∇×vdx.
This is a bounded linear functional onH0div,Ω.
Thus, we can apply Lax–Milgram theorem (see [14]) to find a unique function u∈H0div,Ω satisfying(30)Bu,v=f,v,for all v∈H0div,Ω; u is consequently the weak solution of (13).
Moreover, we can chooseu as a test function to get(31)∇uL2Ω=∇×uL2Ω≤CfL2Ω,where C depends on λ and the theorem is proved.


### 4.2. Proof of Theorem2

For simplicity, we takeR>6 and assume B6⊂BR⊂Ω. We will locally approximate solution (13) by a function satisfying a suitable homogeneous problem. We need some lemmas here. The first one is the following energy estimate.Lemma 6.
Assume thatu is a weak solution of (13) in B1. Then(32)∫B1ϕ2∇u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,for any ϕ∈C0∞B1, where C depends on λ.Proof.
First note thatϕ2u∈H01B1;ℝ3, so we have(33)∫B1Ax∇×u⋅∇×ϕ2udx=∫B1f⋅∇×ϕ2udx.
Using the identity∇×ϕ2u=ϕ2∇×u+2ϕ∇ϕ×u, we obtain(34)∫B1ϕ2∇×u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,where we used the Hölder inequality, and C depends on λ.
Moreover, we know thatϕu∈H01B1;ℝ3. By Lemma 5, we have(35)∫B1∇ϕu2dx=∫B1∇×ϕu2dx+∫B1∇⋅ϕu2dx.
Therefore,(36)∫B1ϕ2∇u2dx≤C∫B1∇ϕu2dx+∫B1∇ϕ2u2dx=C∫B1∇×ϕu2dx+∫B1∇⋅ϕu2dx+∫B1∇ϕ2u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,where we have used the fact that ∇⋅u=0 in B1.Lemma 7.
For anyε>0, there is a small δ=δε>0 such that for any weak solution u of (13) in B4 with(37)1B4∫B4∇u2dx≤1,1B4∫B4A−A¯B42+f2dx≤δ2,there exists a weak solution v of(38)∇×A¯B4∇×v=0,inB4,∇⋅v=0,inB4,such that(39)1B2∫B2∇u−v2dx≤ε2.Proof.
Firstly, we claim that, for anyη>0, there is a small δ=δη>0 and a weak solution v of (38), such that(40)1B4∫B4u−v2dx≤η2.
Suppose it is false. Then, we can findη0>0 and sequences Akk=1∞, ukk=1∞ and fkk=1∞, such that uk is a weak solution of(41)∇×Ak∇×uk=∇×fk,inB4,∇⋅uk=0,inB4,with(42)1B4∫B4∇uk2dx≤1,1B4∫B4Ak−A¯kB42+fk2dx≤1k2.
But for any weak solutionvk of(43)∇×A¯kB4∇×vk=0,inB4,∇⋅vk=0,inB4,we have(44)1B4∫B4uk−vk2dx>η02.
By (42) and Poincaré inequality, uk−u¯kB4k=1∞ is bounded in H1B4;ℝ3. So there exist u0∈H1B4;ℝ3 and a subsequence, still denoted as uk−u¯k, such that uk−u¯k⟶u0 strongly in L2B4;ℝ3 and ∇×uk⟶∇×u0 weakly in L2B4;ℝ3. Since A¯kB4k=1∞ is a bounded sequence of constant matrices, there exist a constant matrix A0 and a subsequence, still denoted as A¯k, such that A¯k⟶A0. Combining (42), we know that Akk=1∞ has a subsequence, denoted also as Ak, such that Ak⟶A0 strongly in L2B4. Thus, u0 satisfies the following system:(45)∇×A0∇×u0=0,inB4,∇⋅u0=0,inB4.
Takevk=u¯k+u0−hk, where hk∈H01B4;ℝ3 satisfies the following system:(46)∇×A¯k∇×hk=∇×A¯k−A0∇×u0,inB4,∇⋅hk=0,inB4,hk=0,on∂B4.
Using Theorem1, we have(47)hkL2B4≤C∇hkL2B4≤CA¯k−A0∇×u0L2B4≤CA¯k−A0.
Moreover,vk satisfies system (43), and(48)uk−vkL2B4≤uk−u¯k−u0L2B4+hkL2B4≤uk−u¯k−u0L2B4+CA¯k−A0.
This means(49)uk−vkL2B4⟶0ask⟶∞.
But this is a contradiction to (44), and thus, (40) holds and the claim is proved.
Now, we give the proof of (39). It is easy to see that u−v satisfies the following system:(50)∇×Ax∇×u−v=∇×f−Ax−A¯B4∇×v,inB4,∇⋅u−v=0,inB4.
By Lemma6, we have(51)1B2∫B2∇u−v2dx≤C1B3∫B3f−A−A¯B4∇×v2dx+1B4∫B4u−v2dx≤C1B4∫B4f2dx+1B4∫B4A−A¯B42dx+1B4∫B4u−v2dx.
Here, we used the interiorW1,∞ regularity of v (see Theorem 2.2 of [5]). Combining (37) and (40), we conclude(52)1B2∫B2∇u−v2dx≤Cδ2+η2=ε2,by taking η and δ satisfying the last identity. This finishes the proof of this lemma.Lemma 8.
Letu be a weak solution of (13) in Ω⊃B6. There exists a constant M such that for any ε>0, there exists a small δ=δε>0, if(53)B1∩x∈Ω:ℳ∇u2≤1∩x∈Ω:ℳf2≤δ2≠∅,then(54)x∈Ω:ℳ∇u2>M2∩B1<εB1.Proof.
By assumption (53), there is a point x0∈B1 such that for all r>0,(55)1Br∫Ωrx0∇u2dx≤1,1Br∫Ωrx0f2dx≤δ2.
SinceB4⊂B5x0, we conclude that(56)1B4∫B4∇u2dx≤543,1B4∫B4f2dx≤543δ2.
Applying Lemma7 to 4/53/2u and 4/53/2f, we have(57)1B2∫B2∇u−v2dx≤η2.
Note thatv satisfies system (38). We can find a constant N such that(58)∇vL∞B3≤N.
TakeM2:=max33,4N2. Now suppose that(59)x1∈B1∩x∈Ω:ℳB2∇u−v2≤N2.
Whenr≤1, then Brx1⊂B2. Hence, we have(60)1Br∫Brx1∇u2dx≤2Br∫Brx1∇u−v2+∇v2dx≤4N2.
Whenr>1, then Brx1⊂B3rx0. Hence, by (55), we have(61)1Br∫Ωrx1∇u2dx≤33B3r∫Ω3rx0∇u2dx≤33.
The above two inequalities show that(62)x1∈B1∩x∈Ω:ℳ∇u2≤M2.
Combining (59) and (62), we have(63)B1∩x∈Ω:ℳ∇u2>M2⊂B1∩x∈Ω:ℳB2∇u−v2>N2.
Consequently,(64)x∈Ω:ℳ∇u2>M2∩B1≤x∈Ω:ℳB2∇u−v2>N2∩B1≤CN2∫B2∇u−v2dx≤CN2η2<εB1,by taking η satisfying the last inequality above. This finishes the proof.Lemma 9.
Letu be a weak solution of (13) in Ω⊃B6. There exists a constant M such that for any ε>0 and r∈0,1, there exists a small δ=δε>0, if(65)x∈Ω:ℳ∇u2>M2∩Br≥εBr,then(66)Br⊂x∈Ω:ℳ∇u2>1∪x∈Ω:ℳf2>δ2.Proof.
We argue by contradiction. If conclusion (66) is false, then(67)Br∩x∈Ω:ℳ∇u2≤1∩x∈Ω:ℳf2≤δ2≠∅.
Let us consider the functions(68)u˜x=urxr,A˜x=Arx,f˜x=frx,with x∈B1. Then, it is easy to check that u˜,A˜,f˜ satisfy the conditions of Lemma 8, and(69)x∈Ω:ℳ∇u˜2>M2∩B1<εB1.
Scaling back in the above estimate yields(70)x∈Ω:ℳ∇u2>M2∩Br<εBr,which is contradiction to (65).Now takeM,ε, and the corresponding δ>0 given by Lemma 8.Lemma 10.
Assume thatA is uniformly positive definite and δ,R-vanishing. Suppose that u is a weak solution of (13) in Ω⊃BR⊃B6 and(71)x∈B1:ℳ∇u2>M2<εB1.
Letk be a positive integer and set ε1=103ε. Then, we have(72)x∈B1:ℳ∇u2>M2k≤∑i=1kε1ix∈B1:ℳf2>δ2M2k−i+ε1kx∈B1:ℳ∇u2>1.Proof.
We intend to prove by induction onk. For the case k=1, let(73)C=x∈B1:ℳ∇u2>M2,D=x∈B1:ℳf2>δ2∪x∈B1:ℳ∇u2>1.
Then, in view of (71), Lemmas 9 and 3, we see that C≤ε1D, and our conclusion is valid for k=1.
Assume that the conclusion is valid for some positive integerk≥2. Let u1=u/M and corresponding f1=f/M. Then u1 is the weak solution of(74)∇×A∇×u1=∇×f1,inB6,∇⋅u1=0,inB6,and the following inequality holds:(75)x∈B1:ℳ∇u12>M2<εB1.
Thus, we have(76)x∈B1:ℳ∇u2>M2k+1=x∈B1:ℳ∇u12>M2k≤∑i=1kε1ix∈B1:ℳf12>δ2M2k−i+ε1kx∈B1:ℳ∇u12>1≤∑i=1k+1ε1ix∈B1:ℳf2>δ2M2k+1−i+ε1k+1x∈B1:ℳ∇u2>1.
This estimate in turn completes the induction onk.Finally, we give the proof of Theorem2.Proof of Theorem 2.
We will consider the casep>2 only. The case p=2 is Theorem 1. Since f∈LpB6;ℝ3, then ℳf2∈Lp/2B6 by Lemma 2. In view of Lemma 1, there is a constant C depending only on δ,p,M such that(77)∑k=1∞Mpkx∈B1:ℳf2>δ2M2k≤Cℳf2Lp/2B6p/2.
Let us suppose thatfLpB6 and ∇uL2B6 are small enough such that(78)∑k=1∞Mpkx∈B1:ℳf2>δ2M2k≤CMf2Lp/2B6p/2≤1,x∈B6:ℳ∇u2>M2∩B1<εB1.
These assumptions are reasonable, since we can multiply system (13) by a small constant depending on fLpB6 and ∇uL2B6.
Let us compute(79)∑k=0∞Mpkx∈B1:ℳ∇u2>M2k≤∑k=1∞Mpk∑i=1kε1ix∈B1:ℳf2>δ2M2k−i+ε1kx∈B1:ℳ∇u2>1=∑i=1∞Mpε1i∑k=i∞Mpk−ix∈B1:ℳf2>δ2M2k−i+∑k=1∞Mpε1kx∈B1:ℳ∇u2>1≤C∑k=1∞Mpε1k<∞,where we used Lemma 10 and selected ε1 such that Mpε1<1. Then by Lemma 1, we get ℳ∇u2∈Lp/2B1, and this gives u∈W1,pB1;ℝ3. The proof is now completed.


### 4.3. Proof of Theorem3

Now we give the proof of Theorem3Proof of Theorem 3.
Since∇⋅u=0 in B6, then, by Lemma 4, there exists w∈W1,pB6;ℝ3 satisfying ∇×w=u and wW1,pB6≤CuLpB6. Thus, we can rewrite system (16) as(80)∇×Ax∇×u=∇×f−w,∇⋅u=0,inΩ.
Thus, we can use Theorem2 and get(81)∇uLpB1≤CuLpB6+f−wLpB6≤CuLpB6+fLpB6+wLpB6≤CuLpB6+fLpB6.
This completes the proof.


## 4.1. Proof of Theorem1

Let us first prove the following important equality.Lemma 5.
LetΩ⊂ℝ3 be bounded and u∈H01Ω;ℝ3. We then have the following identity:(19)∇uL2Ω2=∇×uL2Ω2+∇⋅uL2Ω2.Proof.
Letϕ∈C0∞Ω;ℝ3, then we have(20)−Δϕ=∇×∇×ϕ−∇∇⋅ϕ,inΩ.
We multiply this identity byu and integrate(21)−∫ΩΔϕ⋅udx=∫Ω∇×∇×ϕ⋅udx−∫Ω∇∇⋅ϕ⋅udx.
Sinceu∈H01Ω;ℝ3, by using the definition of weak derivative, we have(22)−∫ΩΔϕ⋅udx=∫Ω∇ϕ:∇udx.
Similarly,(23)∫Ω∇×∇×ϕ⋅udx=∫Ω∇×ϕ⋅∇×udx,−∫Ω∇∇⋅ϕ⋅udx=∫Ω∇⋅ϕ∇⋅udx.
Hence, we obtain(24)∫Ω∇ϕ:∇udx=∫Ω∇×ϕ⋅∇×udx+∫Ω∇⋅ϕ∇⋅udx.
Then, if we takeϕ⟶u in H01Ω;ℝ3, (19) can be proved by the density.Remark 5.
In the proof of (19), we do not use integration by parts. Thus, we do not need the domain to be Lipschitz.Remark 6.
The identity (19) implies that the norm of H01Ω;ℝ3 is equivalent to the right hand of it raised to the power 1/2. This also means we can consider the weak solution of (13) in H0div,Ω with the norm ∇×uL2Ω instead of H01Ω;ℝ3, where H0div,Ω=u∈H01Ω;ℝ3:∇⋅u=0inΩ.Thus, we can define the weak solution of (13) as follows.Definition 3.
A vector fieldu∈H0div,Ω is said to be a weak solution of (13), if the following identity holds:(25)∫ΩAx∇×u⋅∇×vdx=∫Ωf⋅∇×vdx,for any v∈H0div,Ω.Now, we give the proof of Theorem1.Proof of Theorem 1.
In order to prove the existence and uniqueness of weak solution, we define the bilinear form as follows:(26)Bu,v=∫ΩAx∇×u⋅∇×vdx,for any u,v∈H0div,Ω. Since Ax is uniformly positive definite with λ, we have(27)Bu,u≥λ−1∇×uL2Ω2.
This means that the bilinear formBu,v is coercive on H0div,Ω. Moreover,(28)Bu,v≤λ∇×uL2Ω∇×vL2Ω.
Now fixf∈L2Ω;ℝ3 and set(29)f,v≔∫Ωf⋅∇×vdx.
This is a bounded linear functional onH0div,Ω.
Thus, we can apply Lax–Milgram theorem (see [14]) to find a unique function u∈H0div,Ω satisfying(30)Bu,v=f,v,for all v∈H0div,Ω; u is consequently the weak solution of (13).
Moreover, we can chooseu as a test function to get(31)∇uL2Ω=∇×uL2Ω≤CfL2Ω,where C depends on λ and the theorem is proved.


## 4.2. Proof of Theorem2

For simplicity, we takeR>6 and assume B6⊂BR⊂Ω. We will locally approximate solution (13) by a function satisfying a suitable homogeneous problem. We need some lemmas here. The first one is the following energy estimate.Lemma 6.
Assume thatu is a weak solution of (13) in B1. Then(32)∫B1ϕ2∇u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,for any ϕ∈C0∞B1, where C depends on λ.Proof.
First note thatϕ2u∈H01B1;ℝ3, so we have(33)∫B1Ax∇×u⋅∇×ϕ2udx=∫B1f⋅∇×ϕ2udx.
Using the identity∇×ϕ2u=ϕ2∇×u+2ϕ∇ϕ×u, we obtain(34)∫B1ϕ2∇×u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,where we used the Hölder inequality, and C depends on λ.
Moreover, we know thatϕu∈H01B1;ℝ3. By Lemma 5, we have(35)∫B1∇ϕu2dx=∫B1∇×ϕu2dx+∫B1∇⋅ϕu2dx.
Therefore,(36)∫B1ϕ2∇u2dx≤C∫B1∇ϕu2dx+∫B1∇ϕ2u2dx=C∫B1∇×ϕu2dx+∫B1∇⋅ϕu2dx+∫B1∇ϕ2u2dx≤C∫B1ϕ2f2dx+∫B1∇ϕ2u2dx,where we have used the fact that ∇⋅u=0 in B1.Lemma 7.
For anyε>0, there is a small δ=δε>0 such that for any weak solution u of (13) in B4 with(37)1B4∫B4∇u2dx≤1,1B4∫B4A−A¯B42+f2dx≤δ2,there exists a weak solution v of(38)∇×A¯B4∇×v=0,inB4,∇⋅v=0,inB4,such that(39)1B2∫B2∇u−v2dx≤ε2.Proof.
Firstly, we claim that, for anyη>0, there is a small δ=δη>0 and a weak solution v of (38), such that(40)1B4∫B4u−v2dx≤η2.
Suppose it is false. Then, we can findη0>0 and sequences Akk=1∞, ukk=1∞ and fkk=1∞, such that uk is a weak solution of(41)∇×Ak∇×uk=∇×fk,inB4,∇⋅uk=0,inB4,with(42)1B4∫B4∇uk2dx≤1,1B4∫B4Ak−A¯kB42+fk2dx≤1k2.
But for any weak solutionvk of(43)∇×A¯kB4∇×vk=0,inB4,∇⋅vk=0,inB4,we have(44)1B4∫B4uk−vk2dx>η02.
By (42) and Poincaré inequality, uk−u¯kB4k=1∞ is bounded in H1B4;ℝ3. So there exist u0∈H1B4;ℝ3 and a subsequence, still denoted as uk−u¯k, such that uk−u¯k⟶u0 strongly in L2B4;ℝ3 and ∇×uk⟶∇×u0 weakly in L2B4;ℝ3. Since A¯kB4k=1∞ is a bounded sequence of constant matrices, there exist a constant matrix A0 and a subsequence, still denoted as A¯k, such that A¯k⟶A0. Combining (42), we know that Akk=1∞ has a subsequence, denoted also as Ak, such that Ak⟶A0 strongly in L2B4. Thus, u0 satisfies the following system:(45)∇×A0∇×u0=0,inB4,∇⋅u0=0,inB4.
Takevk=u¯k+u0−hk, where hk∈H01B4;ℝ3 satisfies the following system:(46)∇×A¯k∇×hk=∇×A¯k−A0∇×u0,inB4,∇⋅hk=0,inB4,hk=0,on∂B4.
Using Theorem1, we have(47)hkL2B4≤C∇hkL2B4≤CA¯k−A0∇×u0L2B4≤CA¯k−A0.
Moreover,vk satisfies system (43), and(48)uk−vkL2B4≤uk−u¯k−u0L2B4+hkL2B4≤uk−u¯k−u0L2B4+CA¯k−A0.
This means(49)uk−vkL2B4⟶0ask⟶∞.
But this is a contradiction to (44), and thus, (40) holds and the claim is proved.
Now, we give the proof of (39). It is easy to see that u−v satisfies the following system:(50)∇×Ax∇×u−v=∇×f−Ax−A¯B4∇×v,inB4,∇⋅u−v=0,inB4.
By Lemma6, we have(51)1B2∫B2∇u−v2dx≤C1B3∫B3f−A−A¯B4∇×v2dx+1B4∫B4u−v2dx≤C1B4∫B4f2dx+1B4∫B4A−A¯B42dx+1B4∫B4u−v2dx.
Here, we used the interiorW1,∞ regularity of v (see Theorem 2.2 of [5]). Combining (37) and (40), we conclude(52)1B2∫B2∇u−v2dx≤Cδ2+η2=ε2,by taking η and δ satisfying the last identity. This finishes the proof of this lemma.Lemma 8.
Letu be a weak solution of (13) in Ω⊃B6. There exists a constant M such that for any ε>0, there exists a small δ=δε>0, if(53)B1∩x∈Ω:ℳ∇u2≤1∩x∈Ω:ℳf2≤δ2≠∅,then(54)x∈Ω:ℳ∇u2>M2∩B1<εB1.Proof.
By assumption (53), there is a point x0∈B1 such that for all r>0,(55)1Br∫Ωrx0∇u2dx≤1,1Br∫Ωrx0f2dx≤δ2.
SinceB4⊂B5x0, we conclude that(56)1B4∫B4∇u2dx≤543,1B4∫B4f2dx≤543δ2.
Applying Lemma7 to 4/53/2u and 4/53/2f, we have(57)1B2∫B2∇u−v2dx≤η2.
Note thatv satisfies system (38). We can find a constant N such that(58)∇vL∞B3≤N.
TakeM2:=max33,4N2. Now suppose that(59)x1∈B1∩x∈Ω:ℳB2∇u−v2≤N2.
Whenr≤1, then Brx1⊂B2. Hence, we have(60)1Br∫Brx1∇u2dx≤2Br∫Brx1∇u−v2+∇v2dx≤4N2.
Whenr>1, then Brx1⊂B3rx0. Hence, by (55), we have(61)1Br∫Ωrx1∇u2dx≤33B3r∫Ω3rx0∇u2dx≤33.
The above two inequalities show that(62)x1∈B1∩x∈Ω:ℳ∇u2≤M2.
Combining (59) and (62), we have(63)B1∩x∈Ω:ℳ∇u2>M2⊂B1∩x∈Ω:ℳB2∇u−v2>N2.
Consequently,(64)x∈Ω:ℳ∇u2>M2∩B1≤x∈Ω:ℳB2∇u−v2>N2∩B1≤CN2∫B2∇u−v2dx≤CN2η2<εB1,by taking η satisfying the last inequality above. This finishes the proof.Lemma 9.
Letu be a weak solution of (13) in Ω⊃B6. There exists a constant M such that for any ε>0 and r∈0,1, there exists a small δ=δε>0, if(65)x∈Ω:ℳ∇u2>M2∩Br≥εBr,then(66)Br⊂x∈Ω:ℳ∇u2>1∪x∈Ω:ℳf2>δ2.Proof.
We argue by contradiction. If conclusion (66) is false, then(67)Br∩x∈Ω:ℳ∇u2≤1∩x∈Ω:ℳf2≤δ2≠∅.
Let us consider the functions(68)u˜x=urxr,A˜x=Arx,f˜x=frx,with x∈B1. Then, it is easy to check that u˜,A˜,f˜ satisfy the conditions of Lemma 8, and(69)x∈Ω:ℳ∇u˜2>M2∩B1<εB1.
Scaling back in the above estimate yields(70)x∈Ω:ℳ∇u2>M2∩Br<εBr,which is contradiction to (65).Now takeM,ε, and the corresponding δ>0 given by Lemma 8.Lemma 10.
Assume thatA is uniformly positive definite and δ,R-vanishing. Suppose that u is a weak solution of (13) in Ω⊃BR⊃B6 and(71)x∈B1:ℳ∇u2>M2<εB1.
Letk be a positive integer and set ε1=103ε. Then, we have(72)x∈B1:ℳ∇u2>M2k≤∑i=1kε1ix∈B1:ℳf2>δ2M2k−i+ε1kx∈B1:ℳ∇u2>1.Proof.
We intend to prove by induction onk. For the case k=1, let(73)C=x∈B1:ℳ∇u2>M2,D=x∈B1:ℳf2>δ2∪x∈B1:ℳ∇u2>1.
Then, in view of (71), Lemmas 9 and 3, we see that C≤ε1D, and our conclusion is valid for k=1.
Assume that the conclusion is valid for some positive integerk≥2. Let u1=u/M and corresponding f1=f/M. Then u1 is the weak solution of(74)∇×A∇×u1=∇×f1,inB6,∇⋅u1=0,inB6,and the following inequality holds:(75)x∈B1:ℳ∇u12>M2<εB1.
Thus, we have(76)x∈B1:ℳ∇u2>M2k+1=x∈B1:ℳ∇u12>M2k≤∑i=1kε1ix∈B1:ℳf12>δ2M2k−i+ε1kx∈B1:ℳ∇u12>1≤∑i=1k+1ε1ix∈B1:ℳf2>δ2M2k+1−i+ε1k+1x∈B1:ℳ∇u2>1.
This estimate in turn completes the induction onk.Finally, we give the proof of Theorem2.Proof of Theorem 2.
We will consider the casep>2 only. The case p=2 is Theorem 1. Since f∈LpB6;ℝ3, then ℳf2∈Lp/2B6 by Lemma 2. In view of Lemma 1, there is a constant C depending only on δ,p,M such that(77)∑k=1∞Mpkx∈B1:ℳf2>δ2M2k≤Cℳf2Lp/2B6p/2.
Let us suppose thatfLpB6 and ∇uL2B6 are small enough such that(78)∑k=1∞Mpkx∈B1:ℳf2>δ2M2k≤CMf2Lp/2B6p/2≤1,x∈B6:ℳ∇u2>M2∩B1<εB1.
These assumptions are reasonable, since we can multiply system (13) by a small constant depending on fLpB6 and ∇uL2B6.
Let us compute(79)∑k=0∞Mpkx∈B1:ℳ∇u2>M2k≤∑k=1∞Mpk∑i=1kε1ix∈B1:ℳf2>δ2M2k−i+ε1kx∈B1:ℳ∇u2>1=∑i=1∞Mpε1i∑k=i∞Mpk−ix∈B1:ℳf2>δ2M2k−i+∑k=1∞Mpε1kx∈B1:ℳ∇u2>1≤C∑k=1∞Mpε1k<∞,where we used Lemma 10 and selected ε1 such that Mpε1<1. Then by Lemma 1, we get ℳ∇u2∈Lp/2B1, and this gives u∈W1,pB1;ℝ3. The proof is now completed.


## 4.3. Proof of Theorem3

Now we give the proof of Theorem3Proof of Theorem 3.
Since∇⋅u=0 in B6, then, by Lemma 4, there exists w∈W1,pB6;ℝ3 satisfying ∇×w=u and wW1,pB6≤CuLpB6. Thus, we can rewrite system (16) as(80)∇×Ax∇×u=∇×f−w,∇⋅u=0,inΩ.
Thus, we can use Theorem2 and get(81)∇uLpB1≤CuLpB6+f−wLpB6≤CuLpB6+fLpB6+wLpB6≤CuLpB6+fLpB6.
This completes the proof.



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*Source: 1080594-2020-06-23.xml*