€¦ · 2398 junichi aramaki where e and b are electric and magneticﬁeld, respectively. in the...

Applied Mathematical Sciences, Vol. 8, 2014, no. 49, 2397 - 2426HIKARI Ltd, www.m-hikari.com

http://dx.doi.org/10.12988/ams.2014.43219

On the Classical Solutions of the Extended

Magnetostatic Born-Infeld System

Junichi Aramaki

Division of Science, Faculty of Science and Engineering,Tokyo Denki University

Hatoyama-machi, Saitama 350-0394, Japan

Copyright c© 2014 Junichi Aramaki. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

Abstract

We consider the Born-Infeld system in a bounded domain. In orderto get nontrivial solution, we extend the original Born-Infeld system.For extended Born-Infeld system under the Dirichlet condition or theNeumann condition, we shall prove the existence of weak solution andits regularity.

Mathematics Subject Classification: 35Q60, 35J50, 78A30, 78A25

Keywords: The Born-Infeld system, Variational problem, Regularity ofsolutions

1 Introduction

The Born-Infeld theory is an extension of the Maxwell theory. In order toovercome the infinity problem associated with a point charge source in theoriginal Maxwell theory, the introduction of the Born-Infeld electromagneticfield theory is well recognized. See Born [2, 3], Born-Infeld [4, 5] and Yang[16]. In the Maxwell theory the action function is given by

L =1

2(|E|2 − |B|2)

2398 Junichi Aramaki

where E and B are electric and magnetic field, respectively. In the Born-Infeldtheory, the action density is replaced by

LB = b2

(1 −

√1 − 1

b2(|E|2 − |B|2)

)

where b > 0 is a scale parameter. See [2, 3]. Moreover taking the invarianceprinciple into consideration, The author of [2, 3] introduced the other actiondensity

LBI = b2

(1 −

√1 − 1

b2(|E|2 − |B|2) +

1

b4(E · B)2

).

In the magnetostatic case where B(x, t) = curlA(x) and E(x, t) = 0, we seethat

L = LB = LBI = −S(|curl A|2)where

S(t) = b2

(√1 +

1

b2t− 1

), t ≥ 0. (1.1)

We consider the functional

SBI [A] =

∫�3

−Ldx =

∫�3

S(|curlA|2)dx.

The Euler-Lagrange equation of this functional is

curl (S ′(|curl A|2)curl A) = 0 in R3. (1.2)

The author of [16] showed that the solution A of (1.2) with finite energysatisfies curl A = 0. If the solution A satisfies curl A ≡ 0, we say that thesolution A is trivial and if not so, we say that the solution is nontrivial. Thefact that the solution of (1.2) is trivial means that in the vacuum space R

3 themagnetic monopole without an external effect does not exist. For reason ofthis triviality, we change the equation (1.2) into a new equation with a lowerorder term in a bounded domain Ω in R

3, and with some boundary condition.Therefore our energy functional is of the form:∫

Ω

(S(|curl A|2) + F (x,A))dx. (1.3)

The Euler-Lagrange equation of this functional is

curl (S ′(|curl A|2)curl A) +1

2∇�F (x,A) = 0 in Ω. (1.4)

The extended Born-Infeld system 2399

The boundary condition is to prescribe the tangential component of A:

AT = A0T on ∂Ω (1.5)

or a natural condition

S ′(|curl A|2)(curl A)T = D0T on ∂Ω. (1.6)

The problem of such setting was considered by Chen and Pan [6]. They provedthat if F = 0, the solutions A of (1.4) with the boundary condition (1.6)are trivial, and if F = 0 and Ω is simply connected, without holes and ifν · curl A0

T = 0 on ∂Ω where ν is the outer normal unit vector to ∂Ω, thenthe solutions A of (1.4) with the boundary condition (1.5) are trivial. Thuswe modify (1.2) to get nontrivial solutions. In order to do so, we consider theextended Born-Infeld model in the magnetostatic case by adding a lower orderterm F (x,A) as in (1.3).

In this paper we consider the case where the lower order term F (x,A) isof the form

F (x,A) = 〈M(x)A,A〉 + 2b · A + c(x) (1.7)

where M(x) = (Mij(x)) is a given positively definite symmetric 3 × 3 matrix,b(x) is a given vector field and c(x) is a given function. The authors of [6]considered the case where F is of the form F (x,A) = a(x)|A|2 where a(x) isa positive scalar function, and got some interesting results. Our purpose is toextend their results to the case where a lower order term is of the form (1.7).When F (x,A) = a(x)|A|2, the authors of [6] observed that if the boundarydata is small, there exists a classical solution of (1.4) with (1.5) or (1.6).However, in the case (1.7), it will be seen that we can not ignore the effect ofthe vector field b.

Throughout this paper, we impose that the following assumptions hold.(A1) Ω is simply connected bounded domain in R

3 without holes and witha C4 boundary ∂Ω.

(A2) M(x) = (Mij(x))i,j=1,2,3 is a symmetric and positively definite matrixwith Mij ∈ C1,1(Ω) and, that is, Mij(x) = Mji(x) and there exists a constantm0 > 0 such that

3∑i,j=1

Mij(x)ξiξj ≥ m0|ξ|2 for all ξ = (ξ1, ξ2, ξ3) ∈ R3 and x ∈ Ω,

b(x) = (b1(x), b2(x), b3(x)) with bi ∈ C1,1(Ω) and c(x) ∈ C0(Ω).(A3) A0

T ∈ C2,α(∂Ω,R3) where AT is a tangent vector field to ∂Ω, andν · curl A0

T = 0 on ∂Ω where ν is the outer unit vector field to ∂Ω.(A4) DT ∈ C2,α(∂Ω,R3) where DT is a tangent vector field to ∂Ω.(A2’) (A2) holds with Mij ∈ C2,α(Ω) and bi ∈ C2,α(Ω).


In the following for any vector field A, we denote the tangential componentof A by AT , that is, A = A − (ν · A)ν.

We call the following system the Dirichlet problem.{curl (S ′(|curl A|2)curl A) +M(x)A + b = 0 in Ω,AT = A0

T on ∂Ω.(1.8)

Then we get the following result.

Theorem 1.1. Assume that (A1), (A2’) and (A3) with 0 < α < 1 hold.Then there exists R1 > 0 such that if

‖A0T‖C2,α(∂Ω) + ‖b‖C1,α(Ω) ≤ R1,

then the system (1.8) has a classical solution A ∈ C2,α(Ω,R3).

Remark 1.2. If A0T �≡ 0, the above solution is nontrivial.

Next, we call the following system the Neumann problem.{curl (S ′(|curl A|2)curl A) +M(x)A + b = 0 in Ω,S ′(|curl A|2)(curl A)T = D0

T on ∂Ω.(1.9)

Then we get the following result.

Theorem 1.3. Assume that (A1), (A2) and (A4) with 0 < α < 1 hold.Then there exists R2 > 0 such that if

‖D0T‖C2,α(∂Ω) + ‖b‖C1,α(Ω) ≤ R2,

then the system (1.9) has a solution A, and A can be written in the form:

A = v + ∇φ

where v ∈ C2,α(Ω,R3) satisfies ν · v = 0 on ∂Ω and div v = 0 in Ω, andφ ∈ C2,α(Ω). In addition to the above hypotheses, if Mij ∈ C2,α(Ω) andν · curl DT ∈ C2,α(∂Ω), then we get v ∈ C3,α(Ω,R3) and φ ∈ C3,α(Ω).

This paper consists of the following sections. In section 2, we give somepreliminaries. We attempt to modify the given function S(t) in (1.1) for t > Kwith 0 < K < b2 so that the modified function SK has quadratic growth of|curl A|, and we see some properties of SK as in [6]. In section 3, we considerthe Dirichlet problem (1.8) and give the proof of Theorem 1.1. Section 4 isdevoted to the Neumann problem (1.9) and to the proof of Theorem 1.3.


2 Preliminaries

First, we consider the Dirichlet problem (1.8). Since the corresponding func-tional

S+[A] =

∫Ω

(S(|curl A|2) + 〈M(x)A,A〉 + 2b · A + c(x))dx (2.1)

does not have weak compactness in the admissible space which we treat, wemodify S(t) for t > K with 0 < K < b2 in the functional (2.1) to get a strictlyincreasing function SK(t) which has quadratic growth in |curl A| at infinity.Then we consider the modified functional

S+K [A] =

∫Ω

(SK(|curlA|2) + 〈M(x)A,A〉 + 2b · A + c(x))dx. (2.2)

If we take a minimizer AK of S+K in some admissible space, and if ‖curl AK‖L∞(Ω) ≤√

K, then we will be able to see that AK is a critical point of the original func-tional (2.1).

Next, for the Neumann problem (1.9) we consider the modified functional

H+K [A] =

∫Ω

(SK(|curlA|2) + 〈M(x)A,A〉 + 2b · A + c(x))dx

+ 2

∫∂Ω

(DT × AT ) · νdS (2.3)

where dS denotes the surface element. The Euler-Lagrange equations of (2.2)and (2.3) are following, respectively.{

curl (S ′K(|curlA|2)curl A) +M(x)A + b = 0 in Ω,

AT = A0T on ∂Ω.

(2.4)

and {curl (S ′

K(|curlA|2)curl A) +M(x)A + b = 0 in Ω,S ′(|curl A|2)(curl A)T = D0

T on ∂Ω.(2.5)

Now we introduce some function spaces. First we define subspaces ofL2(Ω,R3)

H1(Ω) = {v ∈ L2(Ω,R3); curlv = 0, div v = 0 in Ω,ν · v = 0 on ∂Ω},H2(Ω) = {v ∈ L2(Ω,R3); curlv = 0, div v = 0 in Ω,ν × v = 0 on ∂Ω}.

We say that Ω is simply connected if dim H1(Ω) = 0, and Ω has no holes ifdim H2(Ω) = 0. Next we define

H2(Ω, curl ) = {u ∈ L2(Ω,R3); curlu ∈ L2(Ω,R3)},H2(Ω, div ) = {u ∈ L2(Ω,R3); div u ∈ L2(Ω)},

H2(Ω, curl , div ) = H(Ω, curl ) ∩H(Ω, div ).


It is easy to show that these spaces are Banach spaces with respect to thenorm

‖u‖H2(Ω,curl ) = ‖u‖L2(Ω) + ‖curl u‖L2(Ω),

‖u‖H2(Ω,div ) = ‖u‖L2(Ω) + ‖div u‖L2(Ω),

‖u‖H2(Ω,curl ,div ) = ‖u‖L2(Ω) + ‖curl u‖L2(Ω) + ‖div u‖L2(Ω),

respectively.The next lemma was shown by Dautray and Lions [8, p. 204].

Lemma 2.1. Let Ω be a bounded domain in R3 with a C2 boundary. Then

the following holds.(i) The normal trace map u �→ ν ·u is a continuous map from H2(Ω, div )

to H−1/2(Ω).(ii) The tangential trace map u → uT = u − (u · ν)ν = ν × (u × ν) is a

continuous map from H2(Ω, curl ) to H−1/2(Ω,R3).

Next lemma is followed from [8, Proposition 6] and Pan [15].

Lemma 2.2. Let Ω be a bounded domain in R3 with C2 boundary. Then

the following holds.(i) If Ω has no holes, u ∈ H2(Ω, curl , div ) and ν × u ∈ H1/2(∂Ω), then

u ∈ H1(Ω,R3) and there exists a constant C(Ω) > 0 such that

‖u‖H1(Ω) ≤ C(Ω)(‖curlu‖L2(Ω) + ‖div u‖L2(Ω) + ‖ν × u‖H1/2(∂Ω)).

(ii) If Ω has simply connected, u ∈ H2(Ω, curl , div ) and ν ·u ∈ H1/2(∂Ω),then u ∈ H1(Ω,R3) and there exists a constant C(Ω) > 0 such that

‖u‖H1(Ω) ≤ C(Ω)(‖curlu‖L2(Ω) + ‖div u‖L2(Ω) + ‖ν · u‖H1/2(∂Ω)).

Here we introduce the notion of weak solution of the modified system.For the Dirichlet problem (2.4), we call A ∈ H1(Ω,R3,A0

T ) := {A ∈H1(Ω,R3); AT = A0

T} is a weak solution of (2.4) if∫Ω

(S ′K(|curl A|2)curl A · curl H + (MA) · H + b · H)dx = 0 (2.6)

for any H ∈ H1t0(Ω,R

3) := {H ∈ H1(Ω,R3),HT = 0 on ∂Ω}.For the Neumann problem (2.5), we call A ∈ H2(Ω, curl ) is a weak solution

of (2.5) if∫Ω

(S ′K(|curlA|2)curl A · curl H + (MA) · H + b · H)dx

+

∫∂Ω

(DT × HT ) · νdS = 0 (2.7)


for any H ∈ C1(Ω,R3).Finally following [6], we construct a modified function SK(t) from S(t) in

(1.1). First we note that S(t) ∈ C∞([0,∞)) is a positive, strictly increasing,and strictly concave function. If we define

Φ(t) = t(S ′(t))2, t ≥ 0,

then Φ(t) ∈ C∞([0,∞)) is a positive, strictly increasing, and strictly concavefunction on (0,∞). Moreover,

2Φ′(t) − Φ(t)

t> 0 for all t ∈ (0, b2).

We can construct the modified function SK(t) as follows (cf. [6]).

Lemma 2.3. For any K > 0, we can construct a function SK(t) ∈ C3([0,∞))such that for some small δ > 0,

(i)

0 < SK(t) =

⎧⎨⎩S(t) t ∈ [0, K],C3-connected t ∈ [K,K + δ],aKt+ b1 t ∈ [K + δ,∞)

(2.8)

where aK is a positive constant and b1 is also a positive constant satisfying

b1 > b2

(√1 +

1

b2K − 1

)− aK(K + δ).

0 < S ′K(t) =

⎧⎨⎩S ′(t) t ∈ [0, K],C2-connected t ∈ [K,K + δ],aK t ∈ [K + δ,∞).

(2.9)

If we define ΦK(t) = t(S ′K(t))2, then ΦK(t) ∈ C2([0,∞)) is strictly increasing

and

ΦK(t) =

⎧⎨⎩Φ(t) t ∈ [0, K],concave t ∈ [K,K + δ],a2Kt t ∈ [K + δ,∞).

(2.10)

(ii) Since ρ = ΦK(t) is strictly increasing, the inverse function t = Φ−1K (ρ)

is defined. Define a function fK as

fK(ρ) =1

S ′K(Φ−1

K (ρ)). (2.11)

Then fK ∈ C2([0,∞)) and

0 < fK(ρ) =

⎧⎪⎨⎪⎩1

S′(Φ−1K (ρ))

ρ ∈ [0, K4(1+K/b2)

],

C2-connected ρ ∈ [ K4(1+K/b2)

, a2K(K + δ)

],

1aK

ρ ∈ [a2K(K + δ),∞).

(2.12)


(iii) If 0 < K < b2, we can choose the small δ > 0 such that

2Φ′K(t) − ΦK(t)

t≥ l(K) > 0 for all t ∈ [K,K + δ], (2.13)

andfK(ρ) − 2f ′

K(ρ)ρ ≥ λ(K, δ) > 0 for all ρ ∈ [0,∞). (2.14)

Moreover, we can show that

S ′K(t) + 2tS ′′

K(t) > 0 for all t ∈ [0,∞). (2.15)

From the definition of SK , we can easily show that the functional∫Ω

SK(|curl v|2)dx

is a strictly convex functional on H2(Ω, curl ). Moreover, we can see that

0 < inf0≤t<∞

S ′K(t) < sup

0≤t<∞S ′K(t) <∞.

3 The Dirichlet problem

In this section we shall prove the existence of the minimizer of the functionalS+K and its regularity of the minimizer for the Dirichlet problem (1.8). SinceS+K has not the term div A, the functional has not compactness in the natural

space H1t (Ω,R

3,A0T ). To overcome this, we decompose the minimizer problem

into two steps. First we define

X 2 = H1t (Ω, div 0,A0

T ) ⊕ gradH10 (Ω)

where

H1t (Ω, div 0,A0

T ) = {v ∈ H1(Ω,R3); div v = 0 in Ω,vT = A0T on ∂Ω}.

Define

at(A0T ,X 2) = inf

�+∇φ∈X 2S+K [v + ∇φ].

Then we can prove the existence of the unique minimizer.

Lemma 3.1. Let Ω be a bounded domain in R3 with C4 boundary and A0

T ∈H1/2(∂Ω,R3), and let 0 < K < ∞. Then S+

K has a unique minimizer v∗K +

∇φ∗K ∈ X 2, that is

at(A0T ,X 2) = S+

K [v∗K + ∇φ∗

K ].


Proof. Since we can construct the divergence-free lifting v ∈ H1(Ω,R3) of A0T

(for example see Aramaki [1]), we can see X 2 �= ∅. Let {vn + ∇φn} ⊂ X 2 bea minimizing sequence of S+

K . Let

Ωn = {x ∈ Ω; |curl vn(x)| ≥ K + δ}.We will use an elementary inequality

ab ≤ εa2 +1

4εb2 for any ε > 0 and a, b ≥ 0. (3.1)

Then using (A2) and (3.1), we have

S+K [vn + ∇φn] =

∫Ω

(SK(|curl vn|2) + 〈M(vn + ∇φn), vn + ∇φn〉+b · (vn + ∇φn) + c)dx

≥∫

Ωn

SK(|curl vn|2)dx+

∫Ω

(〈M(vn + ∇φn), vn + ∇φn〉+b · (vn + ∇φn) + c)dx

≥∫

Ωn

(aK |curl vn|2 + b1)dx +

∫Ω

(m0|vn + ∇φn|2

−|b||vn + ∇φn|)dx− ‖c‖C0(Ω)|Ω|≥ aK

∫Ω

|curl vn|2dx−∫

Ω\Ωn

(aK |curl vn|2 + b1)dx

+

∫Ω

{(m0 − ε)|vn + ∇φn|2 − 1

4ε|b|2}dx− ‖c‖C0(Ω)|Ω|

≥ aK‖curl vn‖L2(Ω) + (m0 − ε)‖vn + ∇φn‖2L2(Ω)

−(aK(K + δ)2 + b1 + ‖c‖C0(Ω))|Ω|where |Ω| denotes the volume of Ω. Since vn and ∇φn are orthogonal inL2(Ω,R3) to each other, we see that ‖vn+∇φn‖2

L2(Ω) = ‖vn‖2L2(Ω)+‖∇φn‖2

L2(Ω).

If we put ε = m0/2, we see that {curl vn}, {vn} and {∇φn} are bounded inL2(Ω,R3). From the Poincare inequality, {φn} is bounded in H1

0 (Ω). Sincediv vn = 0 in Ω and vn,T = A0

T on ∂Ω, it follows from Lemma 2.2 that {vn} isbounded inH1(Ω,R3). Passing to subsequences, we may assume that vn → v∗

K

weakly in H1(Ω,R3) and φn → φ∗K weakly in H1

0 (Ω). Then we can see thatdiv v∗

K = 0 in Ω and v∗K,T = A0

T on ∂Ω. Thus v∗K ∈ H1

t (Ω, div 0,A0T ). Since

S+K is convex functional and lower semi continuous on X 2, it is weakly lower

semi-continuous in X 2. Thus v∗K + ∇φ∗

K is a minimizer of S+K . Since S+

K isstrictly convex, the uniqueness follows.

Secondly we define

at(A0T , H

1) = inf�∈H1

t (Ω,�3,�0T )S+K [A].


Then we have at(A0T , H

1) = at(A0T ,X 2). In fact, let A ∈ H1

t (Ω,R3,A0

T ). Thenthe following Dirichlet problem{

Δφ = div A in Ω,φ = 0 on ∂Ω

has a unique solution φ ∈ H2(Ω) ∩ H10 (Ω). If we define v = A − ∇φ ∈

H1(Ω,R3), we have div v = 0 in Ω and vT = AT − (∇φ)T = AT = A0T . Hence

v ∈ H1t (Ω, div 0,A0

T ), so A = v+∇φ ∈ X 2. That is to say, H1t (Ω, div 0,A0

T ) ⊂X 2. This implies that at(A

0T ,X 2) ≤ at(A

0T , H

1). On the other hand, sinceA∗K = v∗

K + ∇φ∗K ∈ H1

t (Ω,R3,A0

T ), we see that

at(A0T ,X 2) = S+

K [A∗K ] ≥ at(A

0T , H

1).

This fact is a reason of decomposition of the above minimizing problem intotwo steps.

Summing up, since A∗K = v∗

K + ∇φ∗K ∈ H1

t (Ω,R3,A0

T ), we obtain thefollowing.

Proposition 3.2. Let Ω be a bounded domain in R3 with a C4 boundary,

A0T ∈ H1/2(∂Ω,R3) and 0 < K < ∞. Then S+

K has a unique minimizerA∗K ∈ H1

t (Ω,R3,A0

T ).

In the following we shall give the regularity of the minimizer A∗K = v∗

K +∇φ∗

K . First we examine the H2 regularity of φ∗K .

Lemma 3.3. Let Ω be a bounded domain in R3 with a C2 boundary and

0 < K <∞. Then we see that φ∗K ∈ H2(Ω) ∩H1

0 (Ω) and

‖φ∗K‖H2(Ω) ≤ C(‖φ∗

K‖L2(Ω) + ‖v∗K‖H1(Ω) + ‖b‖H1(Ω)) (3.2)

where C depends on Ω,m0 and ‖M‖C0(Ω).

Proof. Since φ∗K is the minimizer of the functional S+

K [v∗K + ∇φ] on H1

0 (Ω),the Euler-Lagrange equation becomes{

div (M(x)(v∗K + ∇φ) + b) = 0 in Ω,

φ = 0 on ∂Ω.(3.3)

We rewrite (3.3) into the Dirichlet system for the second order elliptic linearequation with respect to φ:{ −∑3

i,j=1 ∂i(Mij(x)∂jφ) =∑3

i,j=1 ∂i(Mijv∗K,j) + div b in Ω,

φ = 0 on ∂Ω(3.4)

where v∗K = (v∗K,1, v

∗K,2, v

∗K,3). Since Mij ∈W 1,∞(Ω) and the right hand side of

the first equation of (3.4) belongs to L2(Ω), it follows from Chen and Wu [7,Chapter 1, Theorem 5.2] that φ∗

K ∈ H2(Ω) and the estimate (3.2) holds.


Remark 3.4. In the right hand side of the estimate (3.2), we can removethe term ‖φ∗

K‖L2(Ω). In fact, if we multiply the first equation of (3.3) by φ∗K,

and then integrate by parts, we see that∫Ω

M(x)∇φ∗K · ∇φ∗

Kdx =

∫Ω

div (M(x)v∗K + b)φ∗

Kdx

≤ C(‖v∗K‖H1(Ω) + ‖b‖H1(Ω))‖φ∗

K‖L2(Ω)

where C depends on ‖M‖C1(Ω). If we use the Poincare inequality:

‖φ∗K‖L2(Ω) ≤ C(Ω)‖∇φ∗

K‖L2(Ω)

and positivity of the matrix M(x), we easily see that

‖φ∗K‖L2(Ω) ≤ C(Ω)‖∇φ∗

K‖L2(Ω) ≤ C(‖v∗K‖H1(Ω) + ‖b‖H1(Ω))

where C depends on Ω,m0 and ‖M‖C1(Ω). Thus we can remove ‖φ∗K‖L2(Ω) from

(3.2).

We shall give the C2,α estimate of A∗K = v∗

K + ∇φ∗K and prove that if the

boundary data A0T and b are small, then ‖curl A∗

K‖C0(Ω) is small. Then wewill see that A∗

K is a classical solution of the extended magnetostatic Born-Infeld system (1.8). First the minimizer A∗

K of at(A0T , H

1) can be writteninto the form A∗

K = v∗K + ∇φ∗

K where v∗K ∈ H1

t (Ω, div 0,A0T ) and φ∗

K ∈H2(Ω) ∩ H1

0 (Ω). Moreover, v∗K and φ∗

K are weak solutions of the followingequations, respectively.{

curl (S ′K(|curl v|2)curl v) +M(x)(v + ∇φ∗

K) + b = 0 in Ω,vT = A0

T on ∂Ω(3.5)

and {div (M(x)(v∗

K + ∇φ) + b) = 0 in Ω,φ = 0 on ∂Ω.

(3.6)

Lemma 3.5. Let Ω be a bounded domain in R3 with a C4 boundary, A0

T ∈H1/2(∂Ω) and 0 < K <∞. Then we get the estimate

‖v∗K‖H1(Ω) + ‖φ∗

K‖H1(Ω) ≤ C(‖A0T‖H1/2(∂Ω) + ‖b‖L2(Ω)) (3.7)

where C depends on Ω,m0, ‖M‖C0(Ω) and S ′K.

Proof. For brevity of notations, we write v∗K and φ∗

K by v and φ, respectively.It is well known that there exists a divergence-free extension Ae ∈ H1(Ω,R3)of A0

T such that ‖Ae‖H1(Ω) ≤ C(Ω)‖A0T‖H1/2(∂Ω). For example, see [1]. Define


u = v − Ae ∈ H1(Ω,R3). Then div u = 0 in Ω and uT = 0 on ∂Ω. If wechoose u as a test field of (2.4), we see that

0 =

∫Ω

(S ′K(|curl v|2)curl v · curl u +M(x)(v + ∇φ) · u + b · u)dx

=

∫Ω

(S ′K(|curl v|2)curl v · (curl v − curl Ae)

+M(x)(v + ∇φ) · (v + ∇φ−∇φ− Ae)

+b · (v + ∇φ−∇φ− Ae)dx.

From (3.6), we see that∫Ω

M(x)(v + ∇φ) + b) · ∇φdx = 0.

For any ε > 0, using (3.1) we have

minS ′K‖curl v‖2

L2(Ω) +m0‖v + ∇φ‖2L2(Ω)

≤∫

Ω

S ′K(|curl v|2)|curl v|2dx+

∫Ω

M(x)(v + ∇φ) · Aedx

+

∫Ω

S ′K(|curlv|2)curl v · curl Aedx

+

∫Ω

b · (v + ∇φ)dx+

∫Ω

b · Aedx

≤ ε‖curl v‖2L2(Ω) + C(ε)‖curlAe‖2

L2(Ω) + ε‖v + ∇φ‖2L2(Ω)

+C(ε)‖Ae‖2L2(Ω) + C(ε)‖b‖2

L2(Ω).

Since minS ′K > 0 and m0 > 0, if we choose ε > 0 small enough, then we have

‖curl v‖2L2(Ω) + ‖v + ∇φ‖2

L2(Ω) ≤ C(‖Ae‖2H1(Ω) + ‖b‖2

L2(Ω))

≤ C1(‖A0T‖2

H1/2(∂Ω) + ‖b‖2L2(Ω)).

Since v and ∇φ are orthogonal to each other in L2(Ω,R3), we have ‖v +∇φ‖2

L2(Ω) = ‖v‖2L2(Ω) + ‖∇φ‖2

L2(Ω). Since div v = 0 in Ω and vT = A0T ∈

H1/2(∂Ω), it follows from [8, p. 212, Corollary 1] that

‖v‖H1(Ω) ≤ C(Ω)(‖v‖L2(Ω) + ‖curl v‖L2(Ω) + ‖A0T‖H1/2(∂Ω)).

Thus we have

‖v‖H1(Ω) + ‖∇φ‖L2(Ω) ≤ C(‖A0T‖H1/2(∂Ω) + ‖b‖L2(Ω))

where C depends on Ω,m0, ‖M‖C0(Ω) and S ′K . Taking the Poincare inequality

into the consideration, we get the estimate (3.7).


Next we give the regularity of v∗K and φ∗

K .

Proposition 3.6. Assume that Ω,M(x) and b(x) satisfy (A1), (A2’) and(A3) with 0 < α < 1, respectively, and 0 < K < b2. Let A∗

K = v∗K + ∇φ∗

K bea minimizer of S+

K. Then v∗K ∈ C2,α(Ω,R3), φ∗

K ∈ C3,α(Ω) and

‖v∗K‖C2,α(Ω) + ‖∇φ∗

K‖C2,α(Ω) ≤ C

where C depends on Ω, α,m0, ‖M‖C1,1(Ω), ‖b‖C1,1(Ω), K, δ and ‖A0T‖C2,α(∂Ω).

The proof of Proposition 3.6 consists of following several lemmas.First we consider the following div-curl system.⎧⎨⎩

curl Q = −M(x)(v∗K + ∇φ∗

K) − b(x) in Ω,div Q = 0 in Ω,ν · Q = 0 on ∂Ω.

(3.8)

We define a space Hm(Ω, div 0,R3) = {u ∈ Hm(Ω,R3); div u = 0 in Ω} form = 1, 2. First we have the following lemma.

Lemma 3.7. The above system (3.8) has a unique solution QK ∈ H2(Ω, div 0,R3)and

‖Q‖H2(Ω) ≤ C(‖A0T‖H1/2(∂Ω) + ‖b‖H1(Ω))

where C depends on Ω,m0, ‖M‖C1(Ω) and S ′K.

Proof. SinceM(v∗K+∇φ∗

K)+b ∈ H1(Ω, div 0,R3) and Ω has no holes, it followsfrom [15, Lemma 5.7] or [1] that (3.8) has a solution QK ∈ H2(Ω, div 0,R3).Since Ω is simply connected, dim H1(Ω) = {0}, so the uniqueness follows.Moreover, we have

‖Q‖H2(Ω) ≤ C(Ω)‖M(v∗K + ∇φ∗

K) + b‖H1(Ω)

≤ C(‖v∗K‖H1(Ω) + ‖∇φ∗

K‖H1(Ω) + ‖b‖H1(Ω))

where C depends on Ω and ‖M‖C1(Ω). From Lemma 3.3, Remark 3.4 andLemma 3.5, we get the conclusion.

From (2.4) and (3.8), we have

curl (S ′K(|curl v∗

K |2)curl v∗K) = curlQK

in Ω. Since Ω is simply connected, there exists ψK ∈ H1(Ω) such that

S ′K(|curl v∗

K |2)curl v∗K = QK + ∇ψK . (3.9)


From (2.11), we can write

curl A∗K = curlv∗

K =QK + ∇ψK

S ′K(Φ−1(|QK + ∇ψK |2))

= fK(|QK + ∇ψK |2)(QK + ∇ψK).

Since ν · curl v∗K = ν · curl A0

T = 0 by the hypothesis (A3), and ν · QK = 0on ∂Ω by (3.8), it follows from (3.9) that ∂ψK

∂�= 0 on ∂Ω. Here we used the

fact ν · curl v = ν · curl vT according to Monneau [12]. Thus ψK ∈ H1(Ω) is aweak solution of the following system.{

div (fK(|QK + ∇ψ|2)(QK + ∇ψ)) = 0 in Ω,∂ψ∂�

= 0 on ∂Ω.(3.10)

Lemma 3.8. Let ψK ∈ H1(Ω) be a weak solution of (3.10). Then for any1 < q <∞, ψK ∈W 1,q(Ω) and

‖ψK‖W 1,q(Ω) ≤ C(aK + ‖A0T‖H1/2(∂Ω) + ‖b‖H1(Ω))

where C depends on Ω, q,K, δ and fK.

Proof. We rewrite the system (3.10) as a linear equation of ψK .{a−1K ΔψK = div f in Ω,∂ψK

∂�= 0 on ∂Ω

where f = a−1K ∇ψK − fK(|QK + ∇ψK |2)(QK + ∇ψK). Note that∫

Ω

div fdx =

∫∂Ω

f · νdS = 0

because of (3.8) and (3.10), and by the definition of fK we see that if |QK(x)+∇ψK(x)|2 ≥ a2

K(K + δ), then f(x) = −fK(|QK + ∇ψK |2)QK . From Lemma3.7 and the Sobolev embedding theorem, we see that QK ∈ C0,1/2(Ω,R3).Since |∇ψK | ≤ |QK + ∇ψK | + |QK |, we see that

|f(x)| ≤ C(1 + a−1K |QK(x)|)

where C depends on K, δ and fK . In particular, f ∈ L∞(Ω,R3). By theclassical Lq Schauder estimate (cf. Morrey [13, Theorem 5.5.5’ and the ramarksin p.157], Lions and Magenes [11]), we see that ψK ∈ W 1,q(Ω) for any 1 < q <∞ and

‖ψK‖W 1,q(Ω) ≤ C(Ω, q)‖f‖Lq(Ω) ≤ C(aK + ‖QK‖Lq(Ω))

where C depends on Ω, q,K, δ and fK . Since H2(Ω) ↪→ C0,1/2(Ω) ↪→ Lq(Ω),using Lemma 3.7

‖QK‖Lq(Ω) ≤ C(Ω)‖QK‖H2(Ω) ≤ C(‖A0T‖H1/2(∂Ω) + ‖b‖H1(Ω)).


Since W 1,q(Ω) ↪→ Cτ (Ω) for τ = 1 − 3/q > 0 by the Sobolev embeddingtheorem, we see that ψK ∈ C0,τ (Ω) for such τ . Since 1 < q < ∞ is arbitrary,it follows that ψK ∈ C0,τ (Ω) for any τ ∈ (0, 1).

Next we examine the regularity of v∗K .

Lemma 3.9. For any 1 < q <∞, v∗K ∈W 1,q(Ω,R3) and

‖v∗K‖W 1,q(Ω) ≤ C(Ω, q)‖curl AK‖Lq(Ω)

≤ C(Ω, q)(do‖QK + ∇ψK‖Lq(Ω) + ‖A0T‖W 1−1/q,q(∂Ω))

where d0 = ‖fK‖C0([0,∞)). In particular, v∗K ∈ C0,τ (Ω,R3) for any τ ∈ (0, 1).

Proof. We note that QK + ∇ψK ∈ Lq(Ω,R3) for any 1 < q <∞, and so

curl A∗K = fK(|QK + ∇ψK |2)(QK + ∇ψK) ∈ Lq(Ω,R3).

Here we remember that v∗K is a solution of the following system⎧⎨⎩

curl v = curlA∗K = fK(|QK + ∇ψK |2)(QK + ∇ψK) in Ω,

div v = 0 in Ω,vT = A0

T on ∂Ω.(3.11)

By the hypothesis (A3), ν ·curl A0T = 0 on ∂Ω. Since Ω has no holes, it follows

from [1] that (3.11) has a unique solution v∗K ∈W 1,q(Ω,R3) satisfying

‖v∗K‖W 1,q(Ω) ≤ C(Ω, q)(‖fK(|QK + ∇ψK |2)(QK + ∇ψK)‖Lq(Ω)

+‖A0T‖W 1−1/q,q(∂Ω))

≤ C(Ω, q)(do‖QK + ∇ψK‖Lq(Ω) + ‖A0T‖W 1−1/q,q(∂Ω)).

Since q is arbitrary, it follows from the Sobolev embedding theorem that v∗K ∈

Cτ (Ω,R3) for any τ ∈ (0, 1).

We examine the C1,τ regularity of φ∗K .

Lemma 3.10. For any τ ∈ (0, 1), φ∗K ∈ C1,τ (Ω) and

‖φ∗K‖C1,τ (Ω) ≤ C(‖v∗

K‖W 1,q(Ω) + ‖b‖W 1,q(Ω))

where C depends on Ω, q, τ and ‖M‖C1(Ω).

Proof. Since φ∗K is a solution of (3.3), we can rewrite (3.3) into the form⎧⎨⎩

−∑3i,j=1Mij(x)∂i∂jφ

∗K −∑3

j=1(∑3

i=1(∂iMij(x)))∂jφ∗K

=∑3

i,j=1 ∂i(Mij(x)v∗K,j) + div b in Ω,

φ∗K = 0 on ∂Ω

(3.12)


where we denote v∗K = (v∗K,1, v

∗K,2, v

∗K,3). Since Mij ∈ C0(Ω) satisfies (A2) and

the right hand side of the first equation in (3.12) belongs to Lq(Ω) for any1 < q < ∞, it follows from Gilbarg and Trudinger [9, Theorem 9.15] or [7,Chapter 3, Theorem 6.3] that φ∗

K ∈W 2,q(Ω) and

‖φ∗K‖W 2,q(Ω) ≤ C(Ω, q)(‖∇(Mv∗

K)‖Lq(Ω) + ‖div b‖Lq(Ω)

≤ C(‖v∗K‖W 1,q(Ω) + ‖b‖W 1,q(Ω))

where C depends on Ω, q and ‖M‖C1(Ω). By the Sobolev embedding theorem:

W 2,q(Ω) ↪→ C1,1−3/q(Ω) for any 1 < q <∞. Thus the conclusion holds.

We examine C1,τ regularity of QK .

Lemma 3.11. QK ∈ C1,τ (Ω,R3) for any τ ∈ (0, 1) and

‖QK‖C1,τ (Ω) ≤ C(‖v∗K‖C0,τ (Ω) + ‖∇φ∗

K‖C0,τ (Ω) + ‖b‖C0,τ (Ω))

where C depends on Ω, τ and ‖M‖C0,τ (Ω).

Proof. From Lemma 3.9, 3.10 and (A2), the right hand side of the first equa-tion in (3.8) belongs to C0,τ (Ω). Since Ω has no holes, it follows from theregularity of div-curl system (cf. [15, Lemma 5.7 and Corollary 5.4] thatQK ∈ C1,τ (Ω,R3) and

‖QK‖C1,τ (Ω) ≤ C(Ω, q)‖M(v∗K + ∇φ∗

K) + b‖C0,τ (Ω).

We examine C1,θ estimate of ψK for some θ ∈ (0, 1).

Lemma 3.12. ψK ∈ C1,θ(Ω) for some θ ∈ (0, 1) and ‖ψK‖C1,θ(Ω) ≤ Cwhere C depends on Ω, ‖QK‖C1(Ω), ‖ψK‖C0(Ω), m0 and ‖M‖C0(Ω).

Proof. Define

A(x,z) = (A1(x,z), A2(x,z), A3(x,z)) = fK(|QK(x) + z|2)(QK + z).

Then by Lemma 3.11, we know QK ∈ C1,τ (Ω,R3), so A(x,z) ∈ C1,τ (Ω ×R

3,R3). From (2.14), we have

3∑i,j=1

∂Ai∂zj

(x,z)ξiξj ≥ λ(K, δ)|ξ|2 for all ξ = (ξ1, ξ2, ξ3) ∈ R3.

Moreover, we have

(1 + |z|2)∣∣∣∣∂Ai∂zj

(x,z)

∣∣∣∣ + (1 + |z|)(∣∣∣∣∂Ai∂xj

(x,z)

∣∣∣∣ + |Ai(x,z)|)

≤ Λ(1 + |z|2)


for some constant Λ > 0. Therefore ψK is a solution of the Neumann problem{div A(x,∇ψ) = 0 in Ω,∂ψ∂�

= 0 on ∂Ω.

Applying Ladyzhenskaya and Ural’tzeva [10, Chapter 10, Theorem 2.1], thereexists θ ∈ (0, 1) and C > 0 depending on Ω, ‖Q‖C1(Ω), ‖ψK‖C0(Ω), λ(K, δ) and

Λ such that ψK ∈ C1,θ(Ω) and ‖ψK‖C1,θ(Ω) ≤ C.

We examine C1,θ estimate of A∗K .

Lemma 3.13. We can see that A∗K = v∗

K+∇φ∗K ∈ C1,θ(Ω,R3) and ‖A∗

K‖C1,θ(Ω) ≤C where C depends on Ω, θ, ‖fK‖C1([0,∞)), ‖QK‖C0,θ(Ω), ‖∇ψ∗

K‖C0,θ(Ω), ‖A0T‖H1/2(∂Ω)

and ‖b‖C1,θ(Ω).

Proof. First we return to the system (3.11). From Lemma 3.11 we see thatQK ∈ C1,τ (Ω,R3). Thus the right hand side of the first equation of (3.11)belongs to C0,θ(Ω,R3) according to Lemma 3.12. We can easily see that

div (fK(|QK + ∇ψK |2)(QK + ∇ψK)) = div (curl A∗K) = 0 in Ω,

ν · A0T = 0 on ∂Ω, and using (A3)

ν · fK(|QK + ∇ψK |2)(QK + ∇ψK) = 0

= ν · curl ((ν × A0T ) × ν) = ν · curl A0

T on ∂Ω.

If we take the lifting H of −ν × A0T , then ν × H = A0

T . Since Ω is simplyconnected, it follows from [15, Corollary 5.6] that v∗

K ∈ C1,θ(Ω,R3) and

‖v∗K‖C1,θ(Ω) ≤ C(Ω, θ)(‖curl A∗

K‖C0,θ(Ω) + ‖A0T‖C1,θ(∂Ω)) ≤ C

where C depends on Ω, θ, ‖fK‖C1([0,∞)), ‖Q∗K‖C0,θ(Ω) and ‖∇ψK‖C0,θ(Ω). Here

we return to the system (3.4). Since the right hand side of the first equation of(3.4) belongs to C0,θ(Ω), by the Schauder theory we have φ∗

K ∈ C2,θ(Ω). ThusA∗K = v∗

K + ∇φ∗K ∈ C1,θ(Ω,R3) and we get the concluding estimate.

We examine C2,θ estimate of ψK .

Lemma 3.14. We can see that ψK ∈ C2,θ(Ω) and ‖ψK‖C2,θ(Ω) ≤ C whereC depends on Ω, θ, λ(Ω, δ), ‖fK‖C1,θ([0,∞)), ‖QK‖C1,θ(Ω), and ‖ψK‖C1,θ(Ω).

Proof. First we consider the system (3.8). From Lemma 3.13, the right handside of the first equation of (3.8) belongs to C1,θ(Ω,R3). Therefore it followsfrom the regularity of the div-curl system that QK ∈ C2,θ(Ω,R3) and

‖QK‖C2,θ(Ω) ≤ C(‖M(v∗K + ∇φ∗

K) + b‖C1,θ(Ω)

≤ C(‖v∗K‖C1,θ(Ω) + ‖∇φ∗

K‖C1,θ(Ω) + ‖b‖C1,θ(Ω))


where C depends on Ω, θ, and ‖M‖C1,θ(Ω).Next we rewrite (3.10) into the following linear equation{ ∑3

i,j=1 aij(x)∂2ψk

∂xi∂xj= h in Ω,

∂ψK

∂�= 0 on ∂Ω

(3.13)

where

aij = fK(|QK + ∇ψK |2)δij+2f ′

K(|QK + ∇ψK |2)(QK,i + ∂iψK)(QK,j + ∂jψK),

h = −2f ′K(|QK + ∇ψK |2)〈∇QK(QK + ∇ψK),QK + ∇ψK〉

−fK(|QK + ∇ψK |2)div QK

where QK = (QK,1, QK,2, QK,3). We note that aij ∈ C0,θ(Ω) and the system isuniformly elliptic according to (2.14). Since h ∈ C0,θ(Ω) and fK ∈ C2([0,∞)),we see that ψK ∈ C2,θ(Ω) and ‖ψK‖C2,θ(Ω) ≤ C‖h‖C1,θ(Ω) ≤ C1 where C1

depends on Ω, θ, λ(K, δ), ‖fK‖C1,θ([0,∞)), ‖QK‖C1,θ(Ω) and ‖ψK‖C1,θ(Ω).

Finally we examine C2,θ estimate of v∗K .

Lemma 3.15. It follows that v∗K ∈ C2,θ(Ω,R3) and

‖v∗K‖C2,θ(Ω) ≤ C(Ω, θ)‖curl A∗

K‖C1,θ(Ω).

Proof. Since QK + ∇ψK ∈ C1,θ(Ω,R3) by Lemma 3.11 and 3.14, it followsfrom the regularity of the div-curl system (3.11) that v∗

K ∈ C2,θ(Ω,R3),

curl A∗K = fK(|QK + ∇ψK |2)(QK + ∇ψK) ∈ C1,θ(Ω,R3)

and ‖v∗K‖C2,θ(Ω) ≤ C‖curl A∗

K‖C1,θ(Ω).

Since the right hand side of the first equation of (3.12) belongs to C1,θ(Ω) ⊂C0,α(Ω), we see that φ∗

K ∈ C2,α(Ω) ∩ C3,θ(Ω), and so A∗K = v∗

K + ∇φ∗K ∈

C2,θ(Ω,R3) + gradC2,α(Ω) ⊂ C1,α(Ω,R3).End of the proof of Proposition 3.6.When {θ, α} := min{θ, α} = α, since v∗

K ∈ C2,α(Ω,R3) and φ∗K ∈ C3,α(Ω),

the proof is done.When {θ, α} = θ, since A∗

K = v∗K + ∇φ∗

K ∈ C1,α(Ω,R3), it follows from(3.8) that QK ∈ C2,α(Ω,R3) and

‖QK‖C2,α(Ω) ≤ C(‖v∗K‖C1,α(Ω) + ‖∇φ∗

K‖C1,α(Ω) + ‖b‖C1,α(Ω)).

By Lemma 3.14, ψK ∈ C2,θ(Ω) ⊂ C1,α(Ω). We consider the system (3.13). Wenote that aij ∈ C0,α(Ω) and h ∈ C0,α(Ω). Thus ψK ∈ C2,α(Ω) and

‖ψK‖C2,α(Ω) ≤ C(‖ψK‖C0(Ω) + ‖h‖C0,α(Ω)) ≤ C2


where C2 depends on Ω, α, ‖fK‖C1([0,∞)), ‖QK‖C1,α(Ω) and ‖ψK‖C1,α(Ω). There-

fore we have QK + ∇ψK ∈ C1,α(Ω,R3). So the right hand side of the firstequation of (3.11) belongs to C1,α(Ω,R3). Hence from the regularity of thediv-curl system (3.11), we see that v∗

K ∈ C2,α(Ω,R3) and

‖v∗K‖C2,α(Ω) ≤ C(‖curl A∗

K‖C1,α(Ω) + ‖A0T‖C2,α(Ω)).

Therefore, from (3.12) we find φ∗K ∈ C3,α(Ω) and we can write

A∗K = v∗

K + ∇φ∗K ∈ C2,α(Ω,R3) + gradC3,α(Ω).

This completes the proof of Proposition 3.6.

Proposition 3.16. Let 0 < α < 1, 0 < K < b2 and A∗K = v∗

K + ∇φ∗K ∈

C2,α(Ω,R3) + gradC3,α(Ω) ⊂ C2,α(Ω,R3) be the minimizer of S+K in X 2. Then

for any 0 < σ < α, we have

lim‖�0

T ‖C2,α(∂Ω)+‖�‖C2,α(Ω)→0‖A∗

K‖C2,σ(Ω) = 0.

Proof. Assume that the conclusion is false. Then there exist 0 < σ < α, A0n,T ∈

C2,α(∂Ω,R3) and bn ∈ C2,α(Ω,R3) such that A0n,T → 0 in C2,α(∂Ω,R3), bn →

0 in C2,α(Ω,R3) as n→ ∞ and

lim infn→∞

‖A∗K,n‖C2,σ(Ω) > 0 (3.14)

where A∗K,n are corresponding solutions. By Proposition 3.6, A∗

K,n ∈ C2,α(Ω,R3)

and {A∗K,n} is bounded in C2,α(Ω,R3). Therefore there exist a subsequence

{A∗K,nj

} and A0 ∈ C2,σ(Ω,R3) such that A∗K,nj

→ A0 in C2,σ(Ω,R3), A0 �≡ 0

in Ω from (3.14) and A0T = 0 on ∂Ω. From (2.6), for any H ∈ C1

t0(Ω,R3) :=

{H ∈ C1(Ω,R3); HT = 0 on ∂Ω}, we have∫Ω

S ′K(|curl A∗

K,nj|2)curl A∗

K,nj· curl Hdx

+

∫Ω

(MA∗K,nj

) · Hdx +

∫Ω

bnj· Hdx = 0.

Letting j → ∞, it follows that∫Ω

S ′K(|curl A0|2)curl A0 · curl Hdx +

∫Ω

(MA0) · Hdx = 0.

If we put H = A0 and note that S ′K > 0 and the matrix M is positive definite,

we obtain A0 = 0. This leads to a contradiction.


Proof of Theorem 1.1

Fix K such that 0 < K < b2, and construct the function SK as in section2. By Lemma 3.1, for any A0

T ∈ C2,α(∂Ω,R3) and b ∈ C1,1(Ω,R3), the system(2.4) has a unique weak solution A∗

K = v∗K + ∇φ∗

K ∈ X 2. By Proposition 3.6,we see that A∗

K ∈ C2,α(Ω,R3). Moreover, by Proposition 3.16, there existsR0 > 0 such that if ‖A0

T‖C2,α(∂Ω)+‖b‖C1,1(Ω) ≤ R0, then ‖curl A∗K‖C0(Ω) ≤

√K.

Thus S ′K(|curl A∗

K |2) = S ′(|curlA∗K |2). Hence A∗

K is a classical solution of(1.8). The uniqueness of classical solution satisfying ‖curl A∗

K‖C0(Ω) ≤ √K

follows from the uniqueness of weak solution of the system (2.4).

Remark 3.17. When F (x,A) is of the form F (x,A) = a(x)|A|2 wherea(x) > 0 is a scalar function, [6] showed that if the boundary data is small,then A∗

K is a classical solution of (1.8). However, in our case where F (x,A)is of the form (1.7), in order to get a classical solution of (1.8), not only theboundary data but also the term b in (1.7) must be small.

4 The Neumann problem.

In this section, we consider the Neumann problem (2.5). Remember that thecorresponding extended functional is

H+K [A] =

∫Ω

(SK(|curlA|2) + 〈M(x)A,A〉 + b(x) · A + c(x))dx

+ 2

∫∂Ω

(DT × AT ) · νdS (4.1)

and the Euler-Lagrange equation is{curl (S ′

K(|curlA|2)curl A) +M(x)A + b(x) = 0 in Ω,S ′K(|curlA|2)(curl A)T = DT on ∂Ω.

(4.2)

Define

a(H+K ,DT ) = inf

�∈H2(Ω,curl )H+K [A]. (4.3)

We can prove the following proposition as similar as [6].

Proposition 4.1. Let Ω be a bounded domain in R3 with a C2 boundary

and DT ∈ H1/2(∂Ω). Then H+K has a unique minimizer AK ∈ H2(Ω, curl ).

Therefore the minimizer AK is a weak solution of (4.2) in the sense of (2.7).We shall examine the regularity of the minimizer.


Proposition 4.2. Assume that Ω,M(x), b(x), c(x) and DT satisfy (A1),(A2) and (A4) with 0 < α < 1, respectively, and let 0 < K < b2 and AK ∈H2(Ω, curl ) be a minimizer given in Proposition 4.1. . Then

AK = vK + ∇φK ∈ C2,αt0 (Ω, div 0,R3) + gradC2,α(Ω)

where

C2,αt0 (Ω, div 0,R3) = {A ∈ C2,α(Ω); div A = 0 in Ω,ν · A = 0 on ∂Ω}

and

‖vK‖C2,α(Ω) + ‖∇φK‖C1,α(Ω) ≤ C

where the constant C depends on Ω, α,m0, ‖M‖C1,α(Ω), K, δ, ‖DT‖C2,α(Ω) and‖b‖C1,α(Ω).

Corollary 4.3. In addition to the conditions of Proposition 4.2, assumethat SK ∈ C4([0,∞)), fK ∈ C3([0,∞)) and ∂Ω is of class C4,α, and M ∈C2,α(Ω,R9), b ∈ C2,α(Ω,R3) and ν · curl DT ∈ C2,α(Ω). Then

AK = vK + ∇φK ∈ C3,α(Ω, div 0,R3) + gradC3,α(Ω)

and

‖vK‖C3,α(Ω) + ‖∇φK‖C2,α(Ω) ≤ C

where C depends on Ω, α,m0, ‖M‖C2,α(Ω), ‖b‖C2,α(Ω), K, δ, ‖DT‖C2,α(Ω) and ‖ν ·curl DT‖C2,α(Ω).

The proof of Proposition 4.2 consists of several lemmas.The following two lemma are due to [6].

Lemma 4.4. Let Ω be a bounded domain in R3 with a C2 boundary.

(i) Let w ∈ L2(Ω,R3). Then w ∈ H2(Ω, div ) if and only if there exists aconstant C > 0 such that ∣∣∣∣∫

Ω

∇η · wdx∣∣∣∣ ≤ C‖η‖L2(Ω)

for all η ∈ C∞0 (Ω).

(ii) Let f ∈ L2(Ω), g ∈ H−1/2(∂Ω) and w ∈ H2(Ω, div ). Then w satisfiesthe equation div w = f in Ω, ν · w = g on ∂Ω if and only if∫

Ω

∇η · wdx = −∫

Ω

ηfdx+ 〈η, g〉H1/2(∂Ω),H−1/2(∂Ω)

for all η ∈ C1(Ω) where 〈·, ·〉H1/2(∂Ω),H−1/2(∂Ω) denotes the duality of H1/2(∂Ω)

and H−1/2(∂Ω). In this case the above equality also holds for any η ∈ H1(Ω).


Lemma 4.5. Let Ω be a bounded domain in R3 without holes and with a

C2 boundary, and let DT ∈ H1/2(Ω). Let De ∈ H1(Ω,R3) be a divergence-free extension of DT . If AK ∈ H2(Ω, curl ) is a minimizer of H+

K, thenM(x)AK(x) + b(x) ∈ H2(Ω, div ), ν · (MAK + b),ν · curl De ∈ H−1/2(∂Ω),and {

div (MAK + b) = 0 in Ω,ν · (MAK + b) + ν · curl De = 0 on ∂Ω.

(4.4)

Proof. Since from (A2), MAK + b ∈ L2(Ω,R3) and from (4.2), we can seethat div (MAK + b) = 0 in D′(Ω) and (MAK + b) ∈ H2(Ω, div 0). Henceby Lemma 2.1, ν · (MAK + b) ∈ H−1/2(∂Ω). Since curl De ∈ L2(Ω,R3) anddiv (curl De) = 0 in Ω, we also have ν · curl De ∈ H−1/2(∂Ω). From (4.2), wehave

ν · (MAK + b) = −ν · curl (S ′K(|curl AK |2)curl AK)

= −ν · curl (S ′K(|curl AK |2)(curl AK)T )

= −ν · curl De.

We consider the following div-curl system.⎧⎨⎩curl P = −M(x)AK(x) − b(x) in Ω,div P = 0 in Ω,P T = DT on ∂Ω.

(4.5)

Lemma 4.6. The system (4.5) has a unique solution PK ∈ H1(Ω, div 0).

Proof. If we put P = P − De, then we see that (4.5) has a solution PK ∈H1(Ω, div 0) if and only if the system⎧⎨⎩

curl P = −M(x)AK(x) − b(x) − curl De in Ω,div P = 0 in Ω,P T = 0 on ∂Ω

(4.6)

has a solution PK ∈ H1t0(Ω, div 0). In fact, Since Ω is simply connected, it

follows from [8, Proposition 4 and Remark 5] that

{u ∈ L2(Ω,R3); div u = 0 in Ω,ν · u = 0 on ∂Ω}= curlH1

t0(Ω,R3) = curlH1

t0(Ω, div 0).

If we note that from Lemma 4.5,

div (MAK + b + curlDe) = div (MAK + b) = 0 in Ω,

and ν · (MAK + b + curlDe) = 0 on ∂Ω, then we see that (4.6) has a solution

PK ∈ H1t0(Ω, div 0). Since Ω has no holes, the uniqueness of solution of (4.5)

follows.


Let AK ∈ H2(Ω, curl ) be a minimizer of H+K . Then we can decompose

AK so that AK = vK + ∇φK where vK ∈ H2n0(Ω, curl , div 0) := {v ∈

H2(Ω, curl , div ); div v = 0 in Ω,ν · v = 0 on ∂Ω} and φK ∈ H1(Ω). In fact,since Ω is simply connected and has no holes, the following div-curl system⎧⎨⎩

curl v = curlAK in Ω,div v = 0 in Ω,ν · v = 0 on ∂Ω

(4.7)

has a unique solution vK ∈ H1(Ω,R3) (cf. [15, Lemma 5.7] or [1, Theorem3.3]). Since curl (AK − vK) = 0 in Ω and Ω is simply connected, there existsa function φK ∈ H1(Ω) such that AK = vK + ∇φK .

Lemma 4.7. If ν · curl DT ∈ H1/2(∂Ω), then φK ∈ H2(Ω) and

‖∇φK‖H1(Ω) ≤ C(‖vK‖L2(Ω) + ‖ν · curl DT‖H1/2(∂Ω) + ‖b‖H1(Ω))


Proof. Taking (4.2) into consideration, we see that φK satisfies the followingNeumann equation{

div (M(x)∇φ) = −div (M(x)vK) − div b in Ω,ν · (M(x)∇φ) = −ν · curl DT − ν · (MvK) − ν · b on ∂Ω.

(4.8)

We note that the compatibility condition holds. By [10, p. 160] or Murataand Kurata [14, Theorem 2.38], we see that φK ∈ H2(Ω) and

‖φK‖H2(Ω) ≤ C(‖φK‖L2(Ω) + ‖div (MvK)‖L2(Ω) + ‖div b‖L2(Ω)

+‖ν · curl DT‖H1/2(∂Ω) + ‖ν · (MvK)‖H1/2(∂Ω)

+‖ν · b‖H1/2(∂Ω))

≤ C1(‖φK‖L2(Ω) + ‖vK‖H1(Ω) + ‖b‖H1(Ω)

+‖ν · curl DT‖H1/2(∂Ω))

where C1 depends on Ω,m0 and ‖M‖C1(Ω). We may assume that∫ΩφKdx = 0.

Then we can remove the term ‖φK‖L2(Ω) in the right hand side of the aboveinequality by the same reason as Remark 3.4.

We examine the regularity of PK .

Lemma 4.8. Assume that Ω is simply connected bounded domain in R3

without holes, and with a C3 boundary. Let PK ∈ H1(Ω, div 0) be a uniquesolution of (4.5). Then PK ∈ H2(Ω, div 0) and

‖PK‖H2(Ω) ≤ C(‖curl AK‖L2(Ω) + ‖DT‖H3/2(∂Ω) + ‖ν · curl DT‖H1/2(∂Ω))



Proof. Since vK ∈ H1n0(Ω, div 0) and Ω is simply connected, it follows from [8]

that‖vK‖H1(Ω) ≤ C(Ω)‖curlvK‖L2(Ω) = C(Ω)‖curlAK‖L2(Ω). (4.9)

By Lemma 4.7 and (4.9), MAK + b = M(vK + ∇φK) + b ∈ H1(Ω,R3), and

‖MAK + b‖H1(Ω) ≤ C(Ω, ‖M‖C1(Ω))(‖vK‖H1(Ω) + ‖∇φK‖H1(Ω)

+‖b‖H1(Ω))

≤ C(Ω, ‖M‖C1(Ω))(‖AK‖H1(Ω) + ‖b‖H1(Ω)

+‖DT‖H3/2(∂Ω)).

By the regularity of the div-curl system (4.5), we see that PK ∈ H2(Ω,R3)and the concluding estimate holds.

By the similar arguments as section 3, there exists a function ψK ∈ H1(Ω)such that

S ′K(|curl vK |2)curl vK = PK + ∇ψK .

Since

DT = S ′K(|curl AK |2)(curl AK)T = (PK + ∇ψK)T = DT + (∇ψ)T ,

we have (∇ψK)T = 0. Since ∂Ω is connected as Ω has no holes, we may assumethat ψK = 0 on ∂Ω. Thus from (2.11) ψK is a weak solution of the followingequation. {

div (fK(|PK + ∇ψ|2)(PK + ∇ψ)) = 0 in Ω,ψ = 0 on ∂Ω.

(4.10)

Similarly as (3.10), we have⎧⎨⎩curl vK = curlAK = fK(|PK + ∇ψK)|2)(PK + ∇ψK) in Ω,div vK = 0 in Ω,ν · vK = 0 on ∂Ω.

(4.11)

We examine W 1,q regularity of ψK .

Lemma 4.9. Under the condition of Lemma 4.8, we have ψK ∈ W 1,q(Ω)for any 1 < q <∞ and

‖ψK‖W 1,q(Ω) ≤ C(aK + ‖curl AK‖L2(Ω) + ‖DT‖H3/2(∂Ω)

+ ‖ν · curl DT‖H1/2(∂Ω) + ‖b‖H1(Ω))

where C depends on Ω, q,K, δ, d0, ‖M‖C1(Ω) and m0.


Since the proof is similar as that of Lemma 3.8, we omit it.

We note that by the Sobolev embedding theorem W 1,q(Ω) ↪→ C0,τ (Ω) withτ = 1−3/q > 0, so we have ψK ∈ C0,τ (Ω) for any τ ∈ (0, 1) and ‖ψK‖C0,τ (Ω) ≤C(Ω, τ)‖ψK‖W 1,q(Ω).

We examine W 1,q regularity of vK .

Lemma 4.10. Under the condition of Lemma 4.8, we have vK ∈W 1,q(Ω,R3)for any 1 < q <∞ and

‖vK‖W 1,q(Ω) ≤ C(Ω, q)‖curlAK‖L2(Ω) ≤ C(Ω, q)d0‖PK + ∇ψK‖Lq(Ω).

In particular, vK ∈ C0,τ (Ω,R3) for any τ ∈ (0, 1) and

‖vK‖C0,τ (Ω) ≤ C(Ω, τ)‖vK‖W 1,q(Ω).

Proof. By Lemma 4.8 and 4.9, we see that PK + ∇ψK ∈ Lq(Ω) for any 1 <q <∞. Thus

curl AK = fK(|PK + ∇ψK |2)(PK + ∇ψK) ∈ Lq(Ω,R3).

Since vK ∈ H1(Ω,R3) is a unique solution of (4.7) and Ω is simply connected,it follows from the regularity of div-curl system (4.7) (cf. [1]) that vK ∈W 1,q(Ω,R3) and

‖vK‖W 1,q(Ω) ≤ C(Ω, q)‖curl AK‖Lq(Ω) ≤ C(Ω, q)d0‖PK + ∇ψK‖Lq(Ω).

Proof of Proposition 4.2.

Step 1. C1,τ estimate of φK .

We know that φK ∈ H2(Ω) is a solution of (4.8). Since vK ∈ W 1,q(Ω,R3)for any 1 < q < ∞ from Lemma 4.10, we can see that div (MvK) + div b ∈Lq(Ω), and by the hypotheses and Lemma 4.10, we see that ν · curl DT − ν ·(MvK)−ν ·b ∈W 1−1/q,q(∂Ω). Therefore we get φK ∈W 2,q(Ω) ↪→ C1,1−3/q(Ω)for 1 − 3/q > 0. Since q is arbitrary, φK ∈ C1,τ (Ω) for any τ ∈ (0, 1), and

‖φK‖C1,τ (Ω) ≤ C(Ω, τ, q)(‖φK‖Lq(Ω) + ‖vK‖W 1,q(Ω) + ‖b‖W 1,q(Ω)

+ ‖ν · curl DT‖W 1−1/q,q(∂Ω)). (4.12)

We can remove ‖φK‖Lq(Ω) in the right hand side by using Lemma 4.7 and thePoincare inequality.

Step 2. C1,τ estimate of PK .


By Lemma 4.10 and Step 1, for any 0 < τ < 1, MAK = M(vK + ∇φK) ∈C0,τ (Ω,R3). Since PK is a solution of the div-curl system (4.5), it follows from[15] that PK ∈ C1,τ (Ω,R3) and

‖PK‖C1,τ (Ω) ≤ C(Ω, τ)(‖MAK + b‖C0,τ (Ω) + ‖DT‖C1,τ (∂Ω))

≤ C(‖vK‖C0,τ (Ω) + ‖∇φK‖C0,τ (Ω) + ‖b‖C0,τ (Ω)

+‖DT‖C1,τ (∂Ω))

where C depends on Ω, τ and ‖M‖C0,τ (Ω).

Step 3. C1,θ estimate of ψK for some θ ∈ (0, 1).We consider the equation (4.10). By Step 2, we see that

A(x,z) = fK(|PK(x) + z|2)(PK(x) + z) ∈ C1,τ (Ω × R3,R3).

By (2.14), we have

3∑i,j=1

∂Ai∂zj

(x,z)ξIξj ≥ λ|ξ|2

for all ξ = (ξ1, ξ2, ξ3) ∈ R3 where λ = λ(K, δ), and by the properties of fK , we

have

(1 + |z|2)∣∣∣∣∂Ai∂zj

(x,z)

∣∣∣∣ + (1 + |z|)(∣∣∣∣∂Ai∂xj

(x,z)

∣∣∣∣ + |Ai(x,z)|)

≤ Λ(1 + |z|2)

for some Λ > 0. According to [10, Chapter 4, Theorem 6.5], there existsθ ∈ (0, 1) and C > 0 depending on Ω, ‖PK‖C1(Ω), ‖ψK‖C0(Ω), λ and Λ such

that ψK ∈ C1,θ(Ω) and ‖ψK‖C1,θ(Ω) ≤ C.

Step 4. C1,θ estimate of AK .Since by Step 2 and 3, PK + ∇ψK ∈ C0,θ(Ω,R3), it follows from the

regularity of the div-curl system (4.11) that we see that vK ∈ C1,θ(Ω,R3) and

‖vK‖C1,θ(Ω) ≤ C(Ω, θ)‖curl AK‖C0,θ(Ω)

≤ C(‖PK‖C0,θ(Ω) + ‖∇ψK‖C0,θ(Ω))

where C depends on Ω, θ, fK . Here we used the first equation of (4.11). Ifwe return to the equation (4.8) and apply the Schauder theory, then we seeφK ∈ C2,θ(Ω) and

‖φK‖C2,θ(Ω) ≤ C(‖vK‖C1,θ(Ω) + ‖b‖C1,θ(Ω) + ‖ν · curl DT‖C1,θ(∂Ω) + ‖φK‖C0(Ω)).

Thus we have AK = vK + ∇φK ∈ C1,θ(Ω,R3) and

‖AK‖C1,θ(Ω) ≤ C(‖PK‖C0,θ(Ω) + ‖∇ψK‖C0,θ(Ω)

+ ‖b‖C1,θ(Ω) + ‖ν · curl DT‖C1,θ(∂Ω))


where C depends on Ω, θ, fK , m0 and ‖M‖C1,1(Ω).

Step 5. C2,θ estimate of ψK .By Step 4 and the regularity of the div-curl system (4.5), we see that

PK ∈ C2,θ(Ω,R3) and

‖PK‖C2,θ(Ω) ≤ C(‖AK‖C1,θ(Ω) + ‖b‖C1,θ(Ω) + ‖DT‖C2,{θ,α}(∂Ω))

where C depends on Ω, θ and ‖M‖C1,1(Ω). We rewrite the equation (4.10) intothe form { ∑3

i,j=1 aij(x)∂2ψK

∂xi∂xj= h(x) in Ω,

ψK = 0 on ∂Ω(4.13)

where

aij(x) = fK(|PK + ∇ψK |2)δij+2f ′

K(|PK + ∇ψK |2)(PK,i + ∂iψK)(PK,j + ∂jψK),

h(x) = −2f ′K(|PK + ∇ψK |2)〈∇PK(PK + ∇ψK),PK + ∇ψK〉

−fK(|PK + ∇ψK |2)div PK .

Since aij , h ∈ C0,θ(Ω) and (4.13) is uniformly elliptic with lower bound λ =λ(K, δ) > 0 from (2.14), we see that ψK ∈ C2,θ(Ω) and

‖ψK‖C2,θ(Ω) ≤ C‖h‖C0,θ(Ω) ≤ C1

where C1 depends on Ω, θ, ‖PK‖C1,θ(Ω), ‖∇ψK‖C0,θ(Ω), fK and f ′K .

Step 6. C2,θ estimate of vK .By Step 2 and 5, PK+∇ψK ∈ C1,θ(Ω,R3). By the regularity of div-curl sys-

tem (4.11), we see that vK ∈ C2,θ(Ω,R3) and ‖vK‖C2,θ(Ω) ≤ C(Ω, θ)‖curl AK‖C1,θ(Ω).

In particular, since vK ∈ C1,α(Ω,R3), it follows form (4.8) that φK ∈ C2,α(Ω)and

‖φK‖C2,α(Ω) ≤ C(‖vK‖C1,α(Ω) + ‖b‖C1,α(Ω) + ‖ν · curl DT‖C1,α(Ω))

where C depends on Ω, α and ‖M‖C1,α(Ω).End of the proof of Proposition 4.2.We have AK = vK + ∇φK ∈ C2,θ(Ω,R3) + gradC2,α(Ω) ⊂ C1,α(Ω,R3).When {θ, α} = α, the proof is done.When {θ, α} = θ, since AK ∈ C1,α(Ω,R3) and DT ∈ C2,α(∂Ω,R3), taking

(4.5) into consideration we see that PK ∈ C2,α(Ω,R3) and

‖PK‖C2,α(Ω) ≤ C(‖vK‖C1,α(Ω) + ‖∇φK‖C1,α(Ω) + ‖DT‖C2,α(∂Ω))

where C depends on Ω, α and ‖M‖C1,α(Ω). Therefore from (4.13), ψK ∈ C2,α(Ω)and ‖ψK‖C2,α(Ω) ≤ C‖h‖C0,α(Ω) ≤ C1 where C1 depends Ω, ‖PK‖C1,α(Ω), ‖∇ψK‖C0,α(Ω), λ


and Λ. Thus PK + ∇ψK ∈ C1,α(Ω,R3). From (4.7), we see that vK ∈C2,α(Ω,R3) and

‖vK‖C2,α(Ω) ≤ C(Ω, α)‖curl AK‖C1,α(Ω). (4.14)

Hence we can write AK = vK + ∇φK ∈ C2,α(Ω,R3) + gradC2,α(Ω). Thiscompletes the proof of Proposition 4.2.

Proof of Corollary 4.3.First we consider the equation (4.8). By the hypotheses and (4.14), −div (MvK)−

div b ∈ C1,α(Ω) and −ν ·DT − ν · b ∈ C2,α(∂Ω). Since ∂Ω is of class C4,α, wehave φK ∈ C3,α(Ω) and

‖φK‖C3,α(Ω) ≤ C(‖vK‖C2,α(Ω) + ‖b‖C2,α(Ω) + ‖ν · curl DT‖C2,α(∂Ω)).

Next consider the equation (4.13). Since PK ∈ C2,α(Ω,R3) and ψK ∈C2,α(Ω), we have h ∈ C1,α(Ω). Therefore by the Schauder theory we haveψK ∈ C3,α(Ω) and ‖ψK‖C3,α(Ω) ≤ C‖h‖C1,α(Ω) ≤ C1 where C1 depends onΩ, α, ‖PK‖C2,α(Ω), ‖ψK‖C2,α(Ω), fK , f

′K and f ′′

K .

Finally we consider the equation (4.11). Since fK(|PK + ∇ψK |2)(PK +∇ψK) ∈ C2,α(Ω,R3), it follows from the regularity estimate of the div-curlsystem (4.12) that we get vK ∈ C3,α(Ω,R3) and

‖vK‖C3,α(Ω) ≤ C(‖PK‖C2,α(Ω) + ‖∇ψK‖C2,α(Ω))

where C depends on Ω, α and ‖fK‖C2,α . Hence we can write

AK = vK + ∇φK ∈ C3,αt0 (Ω, div 0,R3) + gradC3,α(Ω).

This completes the proof of Corollary 4.3.

Proposition 4.11. Assume that Ω,M(x), b(x) and DT satisfy (A1), (A2)and (A4) with 0 < α < 1, respectively. Let AK = vK+∇φK ∈ C2,α

t0 (Ω, div 0,R3)+gradC2,α(Ω) be a solution of (4.2). Then for any 0 < σ < α, we have

lim‖�T ‖C2,α(∂Ω)+‖�‖C1,α(Ω)→0

‖AK‖C1,σ(Ω) = 0.

Proof. Assume that the conclusion is false. Then there exist 0 < σ < α,Dn,T ∈ C2,α(∂Ω,R3) and bn ∈ C1,α(Ω,R3) such that Dn,T → 0 in C2,α(∂Ω,R3),bn → 0 in C1,α(Ω,R3) and

lim infn→∞

‖AK,n‖C1,σ(Ω) > 0 (4.15)

where AK,n = vK,n + ∇φK,n are corresponding solutions of (4.2) with b = bnand DT = Dn,T , and vK,n ∈ C2,α

t0 (Ω, div 0,R3) and φK,n ∈ C2,α(Ω) Then from


the above estimates, we see that {vK,n} and {φK,n} are bounded in C2,α(Ω,R3)and C2,α(Ω), respectively. Since 0 < σ < α, passing to a subsequences, we mayassume that AK,n → A1 in C1,σ(Ω,R3). From (4.15), we have ‖A1‖C1,σ(Ω) > 0.If we take the inner product of the first equation of (4.2) with any H ∈C1(Ω,R3), and then using the integration of parts, we have∫

Ω

S ′K(|curl AK,n|2)curl AK,n · curl Hdx +

∫Ω

MAK,n · Hdx

+

∫Ω

bn · Hdx+

∫∂Ω

(Dn,T × HT ) · νdS = 0.

Letting n→ ∞, we get∫Ω

S ′K(|curlA1|2)curl A1 · curl Hdx+

∫Ω

MA1 · Hdx = 0.

Choosing H = A1, it follows from the positivity of S ′K and positively definite-

ness of the matrix M that A1 = 0. This contradicts (4.15).

Proof of Theorem 1.3.By Proposition 4.11, there exists a constant R2 > 0 such that if ‖DT‖C2,α(∂Ω)+

‖b‖C1,α(Ω) ≤ R2, then ‖curl AK‖C0(Ω) ≤√K. Therefore AK = vK + ∇φK ∈

C2,α(Ω, div 0,R3) + gradC2,α(Ω) is a solution of (1.9). The uniqueness followsfrom the uniqueness of the weak solution of (4.2).

Remark 4.12. As similar as Remark 3.17, in our case where F (x,A) isof the form (1.7), in order to get a classical solution of (1.9), not only theboundary data but also the term b in (1.7) must be small.

References

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Received: March 29, 2014

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