An extended magnetostatic Born-Infeld model with a concave lower ordertermJun Chen and Xing-Bin Pan Citation: J. Math. Phys. 54, 111501 (2013); doi: 10.1063/1.4826995 View online: http://dx.doi.org/10.1063/1.4826995 View Table of Contents: http://jmp.aip.org/resource/1/JMAPAQ/v54/i11 Published by the AIP Publishing LLC. Additional information on J. Math. Phys.Journal Homepage: http://jmp.aip.org/ Journal Information: http://jmp.aip.org/about/about_the_journal Top downloads: http://jmp.aip.org/features/most_downloaded Information for Authors: http://jmp.aip.org/authors

JOURNAL OF MATHEMATICAL PHYSICS 54, 111501 (2013)

An extended magnetostatic Born-Infeld modelwith a concave lower order term

Jun Chen1,a) and Xing-Bin Pan2,b)

1College of Computer and Information Engineering, Fujian Agricultureand Forestry University, Fuzhou 350002, Fujian, People’s Republic of China2Department of Mathematics, East China Normal University, and NYU-ECNU Institute ofMathematical Sciences at NYU Shanghai, Shanghai 200062, People’s Republic of China

(Received 14 July 2013; accepted 11 October 2013; published online 11 November 2013)

This paper concerns an extended Born-Infeld model with a concave lower order termfor the magnetostatic case. Three types of boundary value problems are considered:the boundary condition prescribing the tangential component of A, the natural bound-ary condition, and the boundary condition prescribing the tangential component ofcurl A. In each case we obtain existence and regularity of solutions for small boundarydata. C© 2013 AIP Publishing LLC. [http://dx.doi.org/10.1063/1.4826995]

I. INTRODUCTION

A. Motivations

This paper concerns an extended Born-Infeld model with a concave lower order term. Themotivations of our study come from the interests both in physics and in mathematics.

The physical motivation comes from the Born-Infeld theory.2, 3, 6, 7 In the magnetostatic case theaction density in this theory is

L = S(| curl A|2),

where

S(t) = b2(√

1 + 1

b2t − 1

), t ≥ 0, (1.1)

with the scaling parameter b > 0. For the equation in the entire space R3, Yang20 has shown thatthe solutions with finite energy are trivial, namely the curl of solutions must be zero vector field.Therefore, extended Born-Infeld functionals which allow non-trivial solutions are necessary. InRef. 12 we followed the ideas of Born-Infeld,6 Yang,20, 21 Lin and Yang,16 Benci and Fortunato,4 andconsidered an extended model by introducing a convex lower order term into the magnetostatic Born-Infeld functional. It has been noticed in Ref. 12 that the extended functionals with either a convexor concave lower order term will have quite different behavior, and require different approaches. Inthis paper we study the functionals with concave lower order terms such as

S−[A] =∫

�

(S(| curl A|2) − F(x, A)

)dx (1.2)

and

H−[A] =∫

�

(S(| curl A|2) − F(x, A)

)dx + 2

∫∂�

(D0

T × AT) · ν dS. (1.3)

a)E-mail: jchen_1001@163.comb)E-mail: xbpan@math.ecnu.edu.cn

0022-2488/2013/54(11)/111501/29/$30.00 C©2013 AIP Publishing LLC54, 111501-1

111501-2 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Such type functionals are not convex, and in general they are unbounded from below, so the directmethod used in Ref. 12 for convex functionals does not work in this case. We shall look for solutionsof nonlinear eigenvalue problems associated with these functionals.

The mathematical motivation of this project mainly comes from our interest in the variationalproblems of functionals involving the operator curl. The difficulties in the study of the functionalsof the form of (1.2) and (1.3) include lack of control on divergence and degree one growth of theleading order term for large | curl A|. The effect of lower order terms and boundary conditions playsan important role in the existence and regularity theory, which needs to be studied systematically.Here we would like to mention some recent works on quasilinear systems involving the operatorcurl, see for instance Ref. 8, 15, and 17.

In this paper, we consider three types of boundary conditions.

(i) The Dirichlet problem: Find a critical point of the functional S− with F(x, A) replaced byβF(x, A) for some positive constant β, under the Dirichlet boundary condition

AT = A0T on ∂�. (1.4)

More precisely, we look for a positive constant β and a vector field A which solves the equation

curl(S′(| curl A|2) curl A

) − β

2∇z F(x, A) = 0 in � (1.5)

and satisfies the boundary condition (1.4).(ii) The Neumann problem: Find a critical point A of the functional H− with F(x, A) and D0

Treplaced by βF(x, A) and βD0

T for some positive constant β, and without prescribing boundarydata. More precisely we look for a positive constant β and a vector field A which is a solutionof (1.5) subjected to the following natural boundary condition

S′(| curl A|2)(curl A)T = βD0T on ∂�. (1.6)

(iii) The third type problem: Find solutions of the extended Born-Infeld models (see (1.14) and(1.15)) with prescribing tangential component of curl A, i.e.:

(curl A)T = B0T on ∂�. (1.7)

B. Assumptions and main results

The majority of our method is valid for a general convex function F(x, A), here we shallrepresent our main results in the special case where F(x, A) = a(x)|A|2. Let 0 < α < 1. We shallmake the following assumptions:

(A1) � is a simply-connected bounded domain in R3 without holes, and with a C4 boundary.

(A2) a(x) ∈ C1,1(�), a0 = minx∈� a(x) > 0.

(A3) A0T ∈ TC2,α(∂�,R3), and ν · curl A0

T = 0 on ∂�.

(A4) D0T ∈ TC1,α(∂�,R3), and ν · curl D0

T ∈ C1,α(∂�).

(A5) B0T ∈ TC2,α(∂�,R3).

To get the higher regularity of solutions we need a stronger condition on a(x):

(A2′) a(x) ∈ C2,α(�), a0 = minx∈� a(x) > 0.

For the nonlinear eigenvalue problem with the Dirichlet boundary condition{curl

(S′(| curl A|2) curl A

) − βa(x)A = 0 in �,

AT = A0T on ∂�,

(1.8)

111501-3 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

we have the following:Theorem 1.1. Assume that � and a(x) satisfy (A1) and (A2), respectively, with 0 < α < 1. Then

there exists R1 > 0 such that for any A0T satisfying (A3) with

‖A0T ‖C2,α(∂�) ≤ R1, (1.9)

there exists a positive constant β such that (1.8) has a solution A, and

A = v + ∇φ ∈ C2,αt (�, div0, A0

T ) + grad C2,α0 (�). (1.10)

In Theorem 1.1, if we replace the condition (A2) by (A2′), then the solution A ∈ C2,α(�,R3);see Corollary 3.10.

For the nonlinear eigenvalue problem with the natural boundary condition{curl

(S′(| curl A|2) curl A

) − βa(x)A = 0 in �,

S′(| curl A|2)(curl A)T = βD0T on ∂�,

(1.11)

we have the following:

Theorem 1.2. Assume that � and a(x) satisfy (A1) and (A2), respectively, with 0 < α < 1. Thenthere exists R2 > 0 such that for any D0

T satisfying (A4) with

‖D0T ‖C1,α(∂�) ≤ R2, (1.12)

there exists a positive constant β such that (1.11) has a solution A, and

A = v + ∇φ ∈ C2,αn0 (�, div0) + grad C2,α(�). (1.13)

For the third type problem we consider both the system{curl

(S′(| curl A|2) curl A

) + a(x)A = 0 in �,

(curl A)T = B0T on ∂�,

(1.14)

and the system {curl

(S′(| curl A|2) curl A

) − a(x)A = 0 in �,

(curl A)T = B0T on ∂�.

(1.15)

The functional associated with (1.14) has a convex lower order term, and that associated with(1.15) has a concave lower order term. Existence of critical points will be proved using the implicitfunction theorem, and it will be shown that if the boundary datum B0

T is small then (1.14) has alwaysa solution, while (1.15) has a solution provided λ = 2 is not an eigenvalue of the following problem:⎧⎪⎪⎨

⎪⎪⎩curl

(1

a(x) curl u)

− λu = 0 in �,

divu = 0 in �,

uT = 0 on ∂�.

(1.16)

Theorem 1.3. In addition to the conditions (A1) and (A2′), assume furthermore ∂� is of class C4, α

with 0 < α < 1. Then we have the following conclusions.

(i) There exists R3 > 0 such that for any B0T satisfying (A5) with

‖B0T ‖C2,α (∂�) ≤ R3,

then (1.14) has a solution A, and

A = v + ∇φ ∈ C3,αn0 (�, div0) + grad C2,α(�). (1.17)

(ii) The same conclusion is true for problem (1.15) if λ = 2 is not an eigenvalue of (1.16).

111501-4 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

C. Outlines of our approach

We start with the general functionals of the form (1.2) and (1.3), as most part of our approachis valid for them. The spaces mentioned in this subsection will be given in Subsection II A.

1. The Dirichlet problem

As in Ref. 12 we modify the function S(t) for t > K, with 0 < K < b2, to get a strictly increasingfunction SK(t) which has a linear growth in t at infinity, and we consider the corresponding modifiedfunctional

S−K [A] =

∫�

(SK (| curl A|2) − F(x, A)

)dx, (1.18)

and look for critical points of S−K in H

2,pt (�, curl, A0

T ). They turn out to be the critical points of theoriginal functional S− if

‖ curl A‖L∞(�) ≤√

K . (1.19)

The functional S−K is neither convex nor coercive, and it is lack of compactness on the

“natural admissible set” H2,pt (�, curl, A0

T ). To resume the compactness one may consider the spaceH

2,pt (�, curl, div0, A0

T ). Then a critical point A, if exists, satisfies the following integral equation∫�

(2S′

K (| curl A|2) curl A · curl w − ∇z F(x, A) · w)

dx = 0, ∀ w ∈ C1t0(�, div0),

from which we cannot derive that A is a weak solution of the Euler-Lagrange equation of S−K since

the set C1t0(�, div0) is not dense in H

2,pt (�, curl, A0

T ).To overcome this difficulty we shall adapt the method developed by Benci and Fortunato,4 and

then we can find a solution of (1.5) for some positive constant β, with A having the followingdecomposition:

A = v + ∇φ, (1.20)

where v is a divergence-free vector field.

Step 1. Define functionals

F[A] =∫

�

F(x, A) dx, F[v, φ] = F[v + ∇φ]. (1.21)

For any fixed v ∈ H2,pt (�, curl, div0, A0

T ), the functional F[v, ·] has a unique minimizer φv onW 1,p

0 (�), and hence φv is a weak solution of{div

(∇z F(x, v + ∇φv)) = 0 in �,

φv = 0 on ∂�.(1.22)

Step 2. If the functional

I[v] ≡ S−K [v + ∇φv] =

∫�

SK (| curl v|2) dx − F[v, φv],

has a minimizer v in H2,pt (�, curl, div0, A0

T ), then using (1.22) we can show that A = v + ∇φv is aweak solution of the Euler-Lagrange equation for S−

K [A]. Unfortunately, the functional I may not bebounded from below. Following the idea of Benci and Fortunato,4 we introduce a bounded functionG and consider the truncated functional

S−K ,G[v] =

∫�

SK (| curl v|2) dx − G(F[v, φv]), (1.23)

111501-5 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

where G satisfies

(G) G : R → R is a C1 function, G′(t) > 0 for all t, and

M1 = supt∈R

G(t) < +∞, M2 = supt∈R

G ′(t) < +∞.

Such functions G always exist. For example, we can choose G(t) = arctan t . Note that multiplyingG by a small positive constant we can require M2 to be sufficiently small, and this fact will be usedin (3.19) and (3.28).

Now the functional S−K ,G is bounded from below, and we can show that S−

K ,G has a minimizer

vK ∈ H2,pt (�, curl, div0, A0

T ). If � has no holes, using (1.22) we can show that there exists a constantβ such that AK = vK + ∇φvK is a weak solution of the system{

curl(S′

K (| curl A|2) curl A) − β

2 ∇z F(x, A) = 0 in �,

AT = A0T on ∂�.

(1.24)

For the definition of weak solutions of (1.24) we refer to Ref. 12, Definition 2.3.The last step in our approach is the verification of (1.19). This will be done in the special case

where F(x, A) = a(x)|A|2.

2. The Neumann problem

Step 1. Define the functional

F c[v, φ] = F[v, φ] − 2∫

∂�

(D0T × (∇φ)T ) · ν dS. (1.25)

For any fixed v ∈ H2,pn (�, curl, div0, g), the functional F c[v, ·] has a unique minimizer φc

v onW 1,p(�). Hence φc

v is a weak solution of{div

(∇z F(x, v + ∇φ)) = 0 in �,

ν · ∇z F(x, v + ∇φ) = 2ν · curl D0T on ∂�.

(1.26)

Step 2. The functional

Ic[v] ≡ H−K [v + ∇φc

v] =∫

�

SK (| curl v|2) dx − F c[v, φcv] + 2

∫∂�

(D0T × vT ) · ν dS,

may also be unbounded from below, so we consider the truncated functional

H−K ,G[v] =

∫�

SK (| curl v|2) dx − G(F c[v, φc

v] − 2∫

∂�

(D0T × vT ) · ν dS

), (1.27)

where G satisfies the condition (G). Because of noncoercivity of H−K ,G in a natural admissible space

H2,p(�, curl, div0) ∩ H⊥1 (�), we consider the minimization problem of H−

K ,G in the subspace ofH2,p(�, curl, div0) ∩ H⊥

1 (�) with an additional boundary condition

ν · v = g on ∂�. (1.28)

We shall prove that H−K ,G has a minimizer vK . Moreover, if � is simply connected, using (1.26) we

can show that there exists a constant β such that AK = vK + ∇φcvK

is a weak solution of the system{curl

(S′

K (| curl A|2) curl A) − β

2 ∇z F(x, A) = 0 in �,

S′K (| curl A|2)(curl A)T = βD0

T on ∂�.(1.29)

For the definition of weak solutions of (1.29) we refer to Ref. 12, Definition 2.4.The verification of (1.19) will be also done in the special case where F(x, A) = a(x)|A|2.This paper is organized as follows. Section II contains preliminary results that will be used,

and a list of conditions on F. The Dirichlet problem is studied in Sec. III, where we first prove

111501-6 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

the existence of critical points of the truncated functional S−K ,G in Subsection III A, and give the

L2 and C2,α estimates in Subsection III B, then we obtain the solutions of (1.8) and prove Theorem1.1 in Subsection III C. The Neumann problem is treated in Sec. IV. We prove the existence andregularity of the critical points of the truncated functional H−

K ,G in Subsections IV A and IV B, andobtain the solutions of (1.11) and prove Theorem 1.2 in Subsection IV C. The connection betweenDirichlet problem and Neumann problem is shown in Subsection IV D. The third type problemand the corresponding gauge invariant problem is dealt with in Sec. V using the implicit functiontheorem.

II. PRELIMINARIES

A. Notation

We shall adapt the notation used in Ref. 12. As in Ref. 12, � is a bounded domain in R3, andν denotes the unit outer normal vector field of the domain boundary ∂�. AT denotes the tangentialcomponent of a vector field A on ∂�:

AT = A − (A · ν)ν.

For 1 < p < + ∞, p′ denotes the conjugate index of p, i.e., 1/p + 1/p′ = 1. We use Ck,α(�),W k,p(�), Hk(�), etc., to denote the usual Holder spaces and Sobolev spaces, and use Ck,α(�,R3),W k,p(�,R3), H k(�,R3), etc., to denote the corresponding Holder spaces and Sobolev spaces forvector fields. Denote

H1(�) = {v ∈ L2(�,R3) : curl v = 0, div v = 0 in �, ν · v = 0 on ∂�

},

H2(�) = {v ∈ L2(�,R3) : curl v = 0, div v = 0 in �, vT = 0 on ∂�

},

and denote by H⊥j (�), j = 1, 2, the orthogonal complement of H j (�) in L2(�,R3) with respect to

the L2 inner product. The following spaces have been introduced in Ref. 12 with 1 ≤ p, q < + ∞:

Hq,p(�, curl) = {u ∈ L p(�,R3) : curl u ∈ Lq (�,R3)

},

Hq,p(�, div) = {u ∈ L p(�,R3) : divu ∈ Lq (�)

},

Hq,p(�, curl, div) = Hq,p(�, curl) ∩ Hq,p(�, div).

For simplicity we denote

Y = H2,pt (�, curl, div0, A0

T ) ∩ H⊥2 (�),

Y0 = H2,pt0 (�, curl, div0) ∩ H⊥

2 (�),

Z = H2,pn (�, curl, div0, g) ∩ H⊥

1 (�),

Z0 = H2,pn0 (�, curl, div0) ∩ H⊥

1 (�),

which will be used in Subsections III B and IV B, respectively.If X(�, ∗) denotes a set of vector fields on �, with the asterisk stands for a description on the

set, then we use the notation X(�, ∗, div0), Xt (�, ∗, A0T ), Xn(�, ∗, g), Xt0(�, ∗), and Xn0(�, ∗) as

in Ref. 12. If Y(∂�, ∗) is a set of vector fields on ∂�, then

TY (∂�, ∗) ≡ {w ∈ Y (∂�, ∗) : ν · w = 0 on ∂�}.If X(�) denotes a space of functions defined on �, then we write

X (�) = {φ ∈ X (�) :∫

�

φ dx = 0}.

The constants c, c′, C, and C′ denote generic constants which may vary from line to line.

111501-7 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

B. On the spaces of vector fields

In this subsection we collect some useful facts about the spaces of vector fields. Part of theresults are well-known, see, for instance Refs. 1, 9–11, 13–15, and 19.

Lemma 2.1. Assume that � is a bounded domain inR3 with a C2 boundary, A0T ∈ TH 1/2(∂�,R3)

and g ∈ H1/2(∂�). Then for any 1 ≤ p ≤ 6 we have

H2,2t (�, curl, div0, A0

T ) = H2,pt (�, curl, div0, A0

T ),(2.1)

H2,2n (�, curl, div0, g) = H2,p

n (�, curl, div0, g).

Proof. The equalities (2.1) hold in the sense that the norms are equivalent. We give only theproof of the first equality. Assume v ∈ H

2,pt (�, curl, div0, A0

T ). When 2 ≤ p ≤ 6 then obviouslyv ∈ H

2,2t (�, curl, div0, A0

T ). When 1 ≤ p < 2, we use the interpolation inequality

‖u‖L2(�) ≤ ε‖∇u‖L2(�) + C(ε)‖u‖L p(�). (2.2)

and div-curl-gradient-inequalities of vector fields (see, for instance, Ref. 13, p. 212; and Ref. 9) tofind

‖v‖L2(�) ≤ C(�)(‖ curl v‖L2(�) + ‖A0

T ‖H 1/2(∂�)) + C ′(�)‖v‖L p(�).

Hence v ∈ H2,2t (�, curl, div0, A0

T ).Next assume v ∈ H

2,2t (�, curl, div0, A0

T ). Applying div-curl-gradient-inequalities of vectorfields we find v ∈ H 1(�,R3), thus v ∈ L p(�,R3) by the Sobolev embedding, and hence v ∈H

2,pt (�, curl, div0, A0

T ).We also see that the norms in the two spaces are equivalent. �From Lemma 2.1 we know that for 1 ≤ p ≤ 6, Y0 and Z0 are both well-defined as subspaces of

L2(�,R3).Recall the Poincare inequality for functions

‖φ‖L2(�) ≤ C p(�)‖∇φ‖L2(�), ∀φ ∈ H 10 (�). (2.3)

For vector fields we have a similar inequality which may also be called a variation of Poincareinequality.

Lemma 2.2. Let 1 ≤ p ≤ 6. Assume that � is a bounded domain in R3 with a C2 boundary. Thenthere exist positive constants Cvp(�) and C(�) such that if either u ∈ Y0, or u ∈ Z0, we have

‖u‖L2(�) ≤ Cvp(�)‖ curl u‖L2(�), (2.4)

‖∇u‖L2(�) ≤ C(�)‖ curl u‖L2(�). (2.5)

Proof. We only prove (2.4) for u ∈ Y0. Suppose it is not true, applying Lemma 2.1, thenthere exists a sequence {un} ⊂ Y0 = H

2,2t0 (�, curl, div0) ∩ H⊥

2 (�), such that ‖un‖L2(�) = 1, but‖ curl un‖L2(�) → 0. Hence {un} is bounded in H 1(�,R3) by the div-curl-gradient inequalities ofvector fields, and there exists a subsequence {un j }, such that as j → ∞, un j → u0 weakly inH 1(�,R3) and strongly in L2(�,R3). Thus we have

‖u0‖L2(�) = 1, (2.6)

and u0 ∈ H2(�). On the other hand, H⊥2 (�) is weakly closed in H

2,pt0 (�, curl, div0), thus u0 ∈

H⊥2 (�). Hence u0 = 0 a.e. in �, which contradicts with (2.6). Hence (2.4) is proved.

Applying the div-curl-gradient inequalities of vector fields to u ∈ H2,pt0 (�, curl, div0) we have

‖∇u‖L2(�) ≤ C(�)(‖ curl u‖L2(�) + ‖u‖L2(�)

).

Using this and (2.4) we get (2.5). �

111501-8 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Lemma 2.3. Let 1 ≤ p ≤ 6. Assume that � is a bounded domain in R3 with a C2 boundary. Thenthere exist positive constants c = c(p, �) and C = C(p, �) such that for any u ∈ H

2,pt0 (�, curl, div0)

we have

c‖u‖H

2,pt0 (�,curl,div0) ≤ ‖∇u‖L2(�) ≤ C‖u‖

H2,pt0 (�,curl,div0), (2.7)

and for any u ∈ H2,pn0 (�, curl, div0) we have

c‖u‖H

2,pn0 (�,curl,div0) ≤ ‖∇u‖L2(�) ≤ C‖u‖

H2,pn0 (�,curl,div0). (2.8)

Hence ‖∇u‖L2(�) is an equivalent norm in H2,pt0 (�, curl, div0) and in H

2,pn0 (�, curl, div0).

Proof. We only prove (2.7). For any u ∈ H2,pt0 (�, curl, div0), we apply the div-curl-gradient

inequalities of vector fields. Noting that divu = 0 and u × ν = 0 on ∂�, we get

‖∇u‖L2(�) ≤ C1(�)(‖ curl u‖L2(�) + ‖u‖L2(�)

). (2.9)

If p ≥ 2 then the right inequality of (2.7) follows from (2.9). If 1 ≤ p < 2, then the right inequalityof (2.7) follows from (2.9) and the interpolation inequality (2.2).

On the other hand we have a Poincare type inequality: For 1 ≤ p ≤ 6 and any u ∈H

2,pt0 (�, curl, div0), we have

‖u‖L p(�) ≤ C(p,�)‖∇u‖L2(�).

This inequality can be proved by contradiction, see for instance the proof of Lemma 2.2. So the leftinequality is true. �C. Conditions on F

We list the conditions on F(x, z) that will be used later. Let 1 < p < + ∞.

(F1) F : � × R3 → R is strictly convex in z and has continuous partial derivatives ∇2xz F

and ∇2z F .

(F2) There exist positive constants c1, c2, C1, C2, such that for any x ∈ � and z ∈ R3,

c1|z|p − c2 ≤ F(x, z) ≤ C1(|z|p + 1),

|∇z F(x, z)| ≤ C2(|z|p−1 + 1).

D. The modified functionals

Let S(t) be the function defined in (1.1) and let (t) = t(S′(t))2. Then (t) ∈ C∞([0, + ∞)) is apositive, strictly increasing and strictly concave function on (0, + ∞). Let

f (ρ) ≡ 1

S′ (−1(ρ)) = 2

(1 − 4

b2ρ)−1/2

.

Lemma 2.4 (Ref. 12, Lemma 3.2). Given a constant K with 0 < K < b2, we can find a small δ

> 0 and construct a positive function SK(t) ∈ C3([0, + ∞)) such that the following conclusions aretrue:

(i) SK(t) = S(t) for t ∈ [0, K] and SK(t) = aKt + b1 for t ≥ K + δ, where aK and b1 are positiveconstants. Moreover, we have

S′K (t) > 0, S′

K (t) + 2t S′′K (t) > 0, ∀ t ≥ 0. (2.10)

(ii) Let

K (t) = t(S′

K (t))2

, fK (ρ) = 1

S′K

(−1

K (ρ)) . (2.11)

111501-9 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Then K is C2 on [0, + ∞) and it is a strictly increasing function. fK is C2 on [0, + ∞),positive-valued, and fK(ρ) = 1/aK for all ρ ≥ a2

K (K + δ). Moreover,

2′K (t) − K (t)

t≥ l(K ) > 0, ∀ t ∈ [K , K + δ],

fK (ρ) − 2 f ′K (ρ)ρ ≥ λ(K , δ) > 0, ∀ ρ ∈ [0,+∞),

(2.12)

where l(K) is a constant depending on K, and λ(K, δ) is a constant depending on K and δ.

Lemma 2.5. Let f and fK be the functions stated above. Then we have the following conclusions:

(i) u = S′(|B|2)B if and only if B = f (|u|2)u.(ii) u = S′

K (|B|2)B if and only if B = fK (|u|2)u.

Proof. (i) If u = S′(|B|2)B then we have |u|2 = (|B|2), which is equivalent to

|B|2 = −1(|u|2) = 4|u|21 − 4

b2 |u|2 .

Hence we obtain

B = uS′(|B|2)

= uS′(−1(|u|2))

= f (|u|2)u = 2(1 − 4

b2|u|2)−1/2

u.

If B = f (|u|2)u, we only need to calculate backward.(ii) Replacing S, and f by SK, K and fK in the proof of (i), the conclusion is followed. �

III. THE DIRICHLET PROBLEM

A. Existence of critical points of S−K,G

Let F[A] and F[v, φ] be the functionals introduced in (1.21), and S−K ,G be the truncated

functional introduced in (1.23). We show that S−K ,G has critical points.

1. Minimization of F

Proposition 3.1. Let 1 < p < + ∞. Assume that � is a bounded domain in R3 with a C2

boundary, and F satisfies (F1) and (F2). Then for any v ∈ H2,pt (�, curl, div0, A0

T ), the functionalF[v, ·] has a unique minimizer φv in W 1,p

0 (�) with

F[v, φv] = a(F , v) ≡ infφ∈W 1,p

0 (�)F[v, φ].

Proof. Let {φn} ⊂ W 1,p0 (�) be a minimizing sequence. By (F2), we have∫

�

(c1|v + ∇φn|p − c2) dx ≤∫

�

F(x, v + ∇φn) dx ≤ a(F , v) + o(1).

Thus {∇φn} is bounded in L p(�,R3), and hence {φn} is bounded in W 1,p0 (�). After passing to a

subsequence we may assume that, φn → φv weakly in W 1,p0 (�). By the definition of a(F , v) we

have F[v, φv] ≥ a(F , v).The weakly lower semi-continuity of the functional F[v, φ] is derived from the convexity of F

by (F1). Therefore, we have

F[v, φv] ≤ lim infn→∞ F[v, φn] = a(F , v).

So φv is a minimizer. The uniqueness follows from the strict convexity of F. �Since φv is the minimizer of the functional F[v, φ] in W 1,p

0 (�),∫�

∇z F(x, v + ∇φv) · ∇ψ dx = 0, ∀ψ ∈ W 1,p0 (�). (3.1)

111501-10 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Lemma 3.2. Let 1 < p < + ∞. Assume that F satisfies (F1) and (F2). Then the mappingA �→ ∇z F(x, A(x)) is a bounded mapping from L p(�,R3) into L p′

(�,R3), and the functionalF[A] is continuous in L p(�,R3).

Lemma 3.3. Let 1 < p < 6. Assume that � is a bounded domain in R3 with a C2 boundary, andA0

T ∈ TH 1/2(∂�,R3). Assume that F satisfies (F1) and (F2). Then the functional v �→ F[v, φv] isweakly continuous on H

2,pt (�, curl, div0, A0

T ).

Proof. Since A0T ∈ TH 1/2(∂�,R3), using the div-curl-gradient inequalities of vector fields,

Lemma 2.1, and Sobolev imbedding theorem, we have

H2,pt (�, curl, div0, A0

T ) ⊂ H 1(�,R3) ↪→↪→ L p(�,R3).

Let vn → v weakly in H2,pt (�, curl, div0, A0

T ). We shall show that F[vn, φvn ] → F[v, φv]. Supposenot, there exist ε > 0 and a subsequence, still denoted by {vn}, such that∣∣F[vn, φvn ] − F[v, φv]

∣∣ ≥ ε. (3.2)

After passing to another subsequence, we may assume that vn → v strongly in L p(�,R3). Henceby Lemma 3.2, F[vn] → F[v]. By (F2) and Proposition 3.1, for each vn , the functional F[vn, φ]has a unique minimizer φvn , and

−c2 meas(�) ≤ F[vn, φvn ] ≤ F[vn, 0] = F[vn].

Hence {F[vn, φvn ]} is bounded. So we can find a subsequence such that

F[vn j , φvn j] → a as j → ∞, (3.3)

where a is a constant.Next we show that F[v, φv] = a. By (3.3) and (F2) we know that {vn j + ∇φvn j

} is bounded

in L p(�,R3). Since {vn j } is bounded in L p(�,R3), so {∇φvn j} is also bounded in L p(�,R3), and

{φvn j} is bounded in W 1,p

0 (�). Passing to another subsequence if necessary, we may assume that

φvn j→ φ weakly in W 1,p

0 (�) as j → ∞. By (3.3), the convexity of F and Proposition 3.1 we knowthat

a = limj→∞

F[vn j , φvn j] ≥ F[v, φ] ≥ F[v, φv]. (3.4)

On the other hand, by (F1), and using the Taylor expansion, we have

F[v, φv] =∫

�

F(x, vn j + ∇φvn j) dx

+∫

�

∇z F(x, vn j + ∇φvn j) · [

v + ∇φv − (vn j + ∇φvn j

)]dx

+ 1

2

∫�

⟨∇2z F(x, Dn j )

[v + ∇φv − (

vn j + ∇φvn j

)],

v + ∇φv − (vn j + ∇φvn j

)⟩dx,

(3.5)

where Dn j ∈ L p(�,R3) is a measurable vector field (by the Caratheodody theorem), and Dn j (x) isa suitable convex combination of v(x) + ∇φv(x) and vn j (x) + ∇φvn j

(x) for a.e. x ∈ �. From theconvexity condition of F we know that∫

�

⟨∇2z F(x, Dn j )

[v + ∇φv − (

vn j + ∇φvn j

)], v + ∇φv − (

vn j + ∇φvn j

)⟩dx ≥ 0.

Applying (3.1) to v = vn j with ψ = φv − φvn j, we get∫

�

∇z F(x, vn j + ∇φvn j) · ∇(φv − φvn j

) dx = 0.

111501-11 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Since {vn j + ∇φvn j} is bounded in L p(�,R3) and vn j − v → 0 in L p(�,R3), using Lemma 3.2 we

have ∫�

∇z F(x, vn j + ∇φvn j) · (

v − vn j

)dx → 0, as j → ∞.

So we can let j → ∞ in (3.5) and obtain F[v, φv] ≥ a. From this and (3.4) we see thatF[v, φv] = a. This and (3.3) contradict (3.2). �2. Minimization of S−

K,G

Now we shall verify the existence of the minimizer of the truncated functional S−K ,G

in H2,pt (�, curl, div0, A0

T ), where G is a truncation function satisfying the condition (G) inSubsection I C.

For A0T ∈ TH 1/2(∂�,R3), let Ae ∈ H 1

t (�, div0, A0T ) be the divergence-free extension of A0

Tsuch that

‖Ae‖H 1(�) ≤ C(�)‖A0T ‖H 1/2(∂�), (3.6)

see Ref. 18, Theorem 1.3. By the Sobolev embedding theorem we have Ae ∈ L p(�,R3) for any1 < p < 6. Thus Ae ∈ H

2,pt (�, curl, div0, A0

T ). Note that any v ∈ H2,pt (�, curl, div0, A0

T ) can bewritten as v = u + Ae, where u ∈ H

2,pt (�, curl, div0, A0

T ). Set

J [u] = S−K ,G[u + Ae].

Obviously

infv∈Y

S−K , G[v] = inf

u∈Y0

J [u].

Now we show that J attains its minimum on Y0.

Proposition 3.4. Let 1 < p < 6 and 0 < K < + ∞. Assume that � is a bounded domain inR3 with a C2 boundary, and A0

T ∈ TH 1/2(∂�,R3). Assume that F satisfies (F1) and (F2), and Gsatisfies (G). Then J [u] attains its minimum on Y0.

Proof. Step 1. We show that the functional J is coercive in Y0 with equivalent norm ‖∇u‖L2(�)

(see Lemma 2.3). For any v ∈ H2,pt (�, curl, div0, A0

T ) we denote

�v = {x ∈ � : | curl v(x)|2 ≥ K + δ

},

where δ > 0 is the number given in Ref. 12, Lemma 3.2, see also Lemma 2.4 above. Since SK(t) >

0 for any t ∈ [0, + ∞) and SK(t) = aKt + b1 for t ≥ K + δ with aK, b1 > 0, and G(t) ≤ M1 for allt, we have

S−K ,G[v] ≥

∫�v

(aK | curl v|2 + b1

)dx − M1

= aK

∫�

| curl v|2 dx − aK

∫�\�v

| curl v|2 dx + b1 meas(�v) − M1

≥ aK

∫�

| curl v|2 dx − aK (K + δ) meas(�) − M1.

Using this and the Young’s inequality we have

J [u] ≥aK

∫�

| curl(u + Ae)|2 dx − aK (K + δ) meas(�) − M1

≥aK

2

∫�

| curl u|2 dx − aK

∫�

| curl Ae|2 dx − aK (K + δ) meas(�) − M1.

111501-12 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

From this and applying (2.5) to u then yields

J [u] ≥ aK

2C(�)2‖∇u‖2

L2(�) − aK

∫�

| curl Ae|2 dx − aK (K + δ) meas(�) − M1,

which shows the coercivity of J in Y0.Step 2. Since F[v, φv] is weakly continuous in H

2,pt (�, curl, div0, A0

T ) by Lemma 3.3, thefunctional

u �→ F[u + Ae, φu+Ae ],

is weakly continuous in H2,pt0 (�, curl, div0). This together with the convexity of the functional

v �→∫

�

SK (| curl v|2) dx,

imply that J [u] is weakly lower semi-continuous in Y0. Combining this with step 1 we see that Jhas a minimizer uK ∈ Y0. �

Let

vK = uK + Ae, AK = vK + ∇φvK . (3.7)

Then vK is a minimizer of S−K ,G in the set

H2,pt (�, curl, div0, A0

T ) ∩ (H⊥2 (�) + Ae).

We shall derive the Euler-Lagrange equations for vK and AK . Note that the functional G(F[v, φv])may not be differentiable in v, thus we cannot derive these equations by differentiation. Here weshall borrow the method from Refs. 4 and 5.

Lemma 3.5. Under the conditions of Proposition 3.4, let vK be given in (3.7). Then for anyw ∈ Y0, the following equality holds:∫

�

2S′K (| curl vK |2) curl vK · curl w dx − βK

∫�

∇z F(x, vK + ∇φvK ) · w dx = 0, (3.8)

where

βK = G ′ (F[vK , φvK ]), 0 < βK ≤ M2, (3.9)

with the positive constant M2 given in the condition (G).

Proof. For any w ∈ Y0, we use the Taylor expansions for SK, G, and F to get

0 ≤J [uK + εw] − J [uK ] = S−K ,G[vK + εw] − S−

K ,G[vK ]

=∫

�

SK (| curl vK + ε curl w|2) dx −∫

�

SK (| curl vK |2) dx

− G(F[vK + εw, φvK +εw]) + G(F[vK , φvK ])

= ε

∫�

2S′K (| curl vK |2) curl vK · curl w dx + ε2

∫�

S′K (| curl vK |2)| curl w|2 dx

+ 1

2ε2

∫�

S′′K (tε(x))

(2 curl vK · curl w + ε | curl w|2)2

dx

− G ′(ξε)∫

�

⟨∇z F(x, vK + ∇φvK ), εw + ∇φvK +εw − ∇φvK

⟩dx

− 1

2G ′(ξε)

∫�

⟨∇2z F(x, Dε(x))

(εw + ∇φvK +εw − ∇φvK

),

εw + ∇φvK +εw − ∇φvK

⟩dx .

(3.10)

111501-13 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

In the above,

tε(x) = αε(x) |curl vK (x)|2 + (1 − αε(x)) |curl vK (x) + ε curl w(x)|2 ,

ξε = γεF[vK , φvK ] + (1 − γε)F[vK + εw, φvK +εw],

where αε(x) is a measurable function and 0 ≤ αε ≤ 1 for a.e. x ∈ �, γ ε is a constant and 0 ≤ γ ε ≤ 1.Dε(x) ∈ L p(�,R3) is also a measurable vector field which is a suitable convex combination of

vK (x) + ∇φvK (x) and vK (x) + εw(x) + ∇φvK +εw(x)

for a.e. x ∈ �.Applying (3.1) with v = vK and ψ = φvK +εw − φvK , we obtain∫

�

⟨∇z F(x, vK + ∇φvK ),∇φvK +εw − ∇φvK

⟩dx = 0. (3.11)

Using the convexity of F and the condition (G) we see that the last term in the right side of (3.10) isnon-positive. We drop this term, substitute (3.11) into (3.10), and then cancel ε from both sides toget

0 ≤∫

�

2S′K (| curl vK |2) curl vK · curl w dx + ε

∫�

S′K (| curl vK |2)| curl w|2 dx

+ ε

2

∫�

S′′K (tε)

(2 curl vK · curl w + ε | curl w|2)2

dx

− G ′(ξε)∫

�

⟨∇z F(x, vK + ∇φvK ), w⟩

dx .

(3.12)

After passing to a subsequence we may assume that γ ε → γ as ε → 0, where 0 ≤ γ ≤ 1. By Lemma3.3 we know that

F[vK + εw, φvK +εw] → F[vK , φvK ] as ε → 0.

Hence ξε → F[vK , φvK ] as ε → 0. Thus by (G) we have G ′(ξε) → G ′(F[vK , φvK ]) as ε → 0.Letting ε → 0 in (3.12) we obtain

0 ≤∫

�

2S′K (| curl vK |2) curl vK · curl w dx

− G ′(F[vK , φvK ])∫

�

∇z F(x, vK + ∇φvK ) · w dx .

Replacing w by −w we deduce that for any w ∈ Y0,

0 =∫

�

2S′K (| curl vK |2) curl vK · curl w dx

− G ′ (F[vK , φvK ]) ∫

�

∇z F(x, vK + ∇φvK ) · w dx .

Denote βK = G ′ (F[vK , φvK ]). From (G) we see that 0 < βK ≤ M2. Now (3.8) and (3.9) follow. �

Let us mention that the constant βK in (3.8) depends not only on K , A0T , but also on the solution

AK .Now we show that AK satisfies the following:∫�

2S′K (| curl A|2) curl A · curl H dx − βK

∫�

∇z F(x, A) · H dx = 0, ∀ H ∈ H⊥2 (�). (3.13)

Proposition 3.6. Under the conditions of Proposition 3.4, let AK ∈ H2,pt (�, curl, A0

T ) be given in(3.7). Then AK satisfies (3.13) for all H ∈ H⊥

2 (�), where βK is given in (3.9).

111501-14 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Proof. For any H ∈ C1t0(�,R3) ∩ H⊥

2 (�), we decompose it into H = w + ∇ψ , whereψ ∈ W 2,p(�) ∩ W 1,p

0 (�) is the solution of

�ψ = divH in �, ψ = 0 on ∂�,

and w ∈ Y0. Then (3.8) holds for this test field w. Applying (3.1) to v = vK with this ψ as a testfunction, we get ∫

�

∇z F(x, vK + ∇φvK ) · ∇ψ dx = 0.

Adding this equality to (3.8) we obtain∫�

2S′K (| curl vK |2) curl vK · curl H dx − βK

∫�

∇z F(x, vK + ∇φvK ) · H dx = 0.

Note that curl(∇φvK ) = 0 holds in the distribution sense (see Ref. 13), and by the representationformula of H−1(�,R3), we know that this is also true in L2(�,R3). Hence in the first term in theabove equality we can replace curl vK by curl AK , and get, for any H ∈ C1

t0(�,R3) ∩ H⊥2 (�),∫

�

2S′K (| curl AK |2) curl AK · curl H dx − βK

∫�

∇z F(x, AK ) · H dx = 0. (3.14)

Since C1t0(�,R3) ∩ H⊥

2 (�) is dense in H⊥2 (�), (3.14) holds for all H ∈ H⊥

2 (�). Hence AK satisfies(3.13). �

Using the structure of the space H2(�) (see Ref. 13, p. 222), we may view (3.13) as a weakformula of the system{

curl(S′

K (| curl A|2) curl A) − βK

2 ∇z F(x, A) = ∇ in �,

AT = A0T on ∂�,

where ∇ ∈ H2(�) is a vector field, and βK is a constant, both being determined by the solutionAK . If ∇ = 0, then the above equation is reduced to (1.24).

Corollary 3.7. Under the conditions of Proposition 3.4, if furthermore assume that � has noholes, then AK ∈ H

2,pt (�, curl, A0

T ) given in (3.7) is a weak solution of system (1.24) with β = βK.

Proof. If � has no holes, then H2(�) = {0} and H⊥2 (�) = L2(�,R3). Then from the proof of

Proposition 3.6 we see that (3.14) holds for all H ∈ C1t0(�,R3). Thus AK ∈ H

2,pt (�, curl, A0

T ) is aweak solution of system (1.24). �

B. Estimates of the critical points of S−K,G

In this subsection we look for L2 and C2,α estimates of the weak solutions of (1.24) which havebeen obtained in last subsection. We shall only consider the special case where

F(x, z) = a(x)|z|2, (3.15)

where a(x) satisfies the condition (A2). Obviously this F satisfies (F1) and (F2) with p = 2. Let 0< K < + ∞. According to the discussion in Subsection III A we know that, if � has no holes, thenS−

K ,G has a critical point AK ∈ H2,2t (�, curl, A0

T ), and it is a solution of the Euler-Lagrange equation

{curl

(S′

K

(| curl A|2) curl A) − βK a(x)A = 0 in �,

AT = A0T on ∂�.

(3.16)

111501-15 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Recall that AK = vK + ∇φvK with vK ∈ H2,2t (�, curl, div0, A0

T ) and φvK ∈ H 10 (�), where vK is a

weak solution of{curl

(S′

K (| curl v|2) curl v) − βK a(x)

(v + ∇φvK

) = 0 in �,

vT = A0T on ∂�,

(3.17)

and φK is a weak solution of {div (a(x) (vK + ∇φ)) = 0 in �,

φ = 0 on ∂�.(3.18)

Now we derive the L2 estimate of vK and ∇φvK . Define a positive constant

N1(�, a) = 16‖a‖C(�)Cvp(�)2[2 + (

1 + C p(�))a−1

0 ‖a‖C(�)

],

where Cp(�) and Cvp are the constants in the Poincare inequality (2.3) and in the variation ofPoincare inequality (2.4), respectively. As mentioned in Subsection I C, we can always choose thefunction G to satisfy (G) and also

M2 ≡ supt∈R

G ′(t) ≤ min S′K

N1(�, a). (3.19)

As an immediate consequence, the constant βK in (3.16) satisfies

0 < βK ≤ min S′K

N1(�, a). (3.20)

Lemma 3.8. Let 0 < K < + ∞. Assume that � is a bounded domain in R3 with a C2 boundary,and A0

T ∈ TH 1/2(∂�,R3). Assume that a(x) ∈ C(�) is a positive function, and G is chosen to satisfy(G) and (3.19). Let vK and φvK be the weak solution of (3.17) and (3.18), respectively. Then

‖vK ‖H 1(�) + ‖φvK ‖H 1(�) ≤ C‖A0T ‖H 1/2(∂�), (3.21)

where C depends on �, a0, ‖a‖C(�), βK , and K.

Proof. In the proof, for simplicity of notation we shall drop the subscripts and write vK , φvK ,and βK as v, φv, and β, and denote Cp(�) and Cvp(�) by Cp and Cvp, respectively.

Step 1. Since φv ∈ H 10 (�) is a weak solution of (3.18), we have∥∥∇φv

∥∥L2(�) ≤ a−1

0 ‖a‖C(�)

∥∥v∥∥

L2(�).

Using the Poincare inequality (2.3), we obtain

‖φv‖H 1(�) ≤ C3‖v‖L2(�), (3.22)

where C3 = (1 + C p)a−10 ‖a‖C(�).

Step 2. Let us denote by Ae the divergence-free extension of A0T satisfying (3.6). Write v =

u + Ae. Then u ∈ H2,pt0 (�, curl, div0). Taking u as a test field for the weak form of (3.17), and using

the Cauchy inequality we get

‖ curl v‖2L2(�) ≤ max S′

K

min S′K

‖ curl Ae‖2L2(�) + 2β‖a‖C(�)

min S′K

[(2 + C3)‖v‖2

L2(�) + ‖Ae‖2L2(�)

]. (3.23)

Using (2.4) we get

‖u‖L2(�) ≤ Cvp‖ curl u‖L2(�) ≤ Cvp(‖ curl v‖L2(�) + ‖ curl Ae‖L2(�)

),

hence

‖v‖L2(�) ≤ Cvp(‖ curl v‖L2(�) + ‖ curl Ae‖L2(�)

) + ‖Ae‖L2(�).

111501-16 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

From this and (3.23) we have

‖v‖2L2(�) ≤8(2 + C3)C2

vp

β‖a‖C(�)

min S′K

‖v‖2L2(�) + 4C2

vp

(1 + max S′

K

min S′K

)‖ curl Ae‖2

L2(�)

+ 2[4(2 + C3)C2

vp

β‖a‖C(�)

min S′K

+ 1]‖Ae‖2

L2(�).

(3.24)

From (3.20) we see that

β ‖a‖C(�) ≤ min S′K

16(2 + C3)C2vp

.

So we get from (3.24)

‖v‖2L2(�) ≤8C2

vp

(1 + max S′

K

min S′K

)‖ curl Ae‖2

L2(�)

+ 4[4(2 + C3)C2

vp

β‖a‖C(�)

min S′K

+ 1]‖Ae‖2

L2(�).

(3.21) follows from this, (3.6), (3.22), and Ref. 13, Corollary 1, p. 212 immediately. �The C2,α estimate of vK can be derived as in the proof of Proposition 4.5 in Ref. 12, with a(x)

there replaced by − βKa(x).

Proposition 3.9. Let 0 < K < b2. Assume that �, a(x), A0T satisfy (A1), (A2), and (A3), respec-

tively with 0 < α < 1, and G is chosen to satisfy (G) and (3.19). Let vK and φvK be the weak solutionof (3.17) and (3.18), respectively. Then vK ∈ C2,α

t (�, div0, A0T ), φvK ∈ C2,α

0 (�), and

‖vK ‖C2,α (�) + ‖∇φvK ‖C1,α (�) ≤ C,

where C depends on �,α, a0, ‖a‖C1,1(�), K , δ, M2, and ‖A0T ‖C2,α (∂�).

Let us mention that we can furthermore require

SK ∈ C4([0,+∞)), fK ∈ C3([0,+∞)).

In this case we have

Corollary 3.10. In addition to the conditions of Theorem 3.9, assume furthermore that a(x) ∈C2,α(�). Then

AK = vK + ∇φK ∈ C3,αt (�, div0, A0

T ) + grad C3,α0 (�) ⊂ C2,α(�,R3),

and

‖vK ‖C3,α(�) + ‖∇φK ‖C2,α (�) ≤ C,

where C depends on �,α, a0, ‖a‖C2,α (�), K , δ, M2, and ‖A0T ‖C2,α (∂�).

C. Solutions of the Dirichlet problem (1.8)

In Subsections III A and III B we have proved the existence and regularity of the solutionsAK of the modified system (1.24). In this subsection, for F(x, A) given in (3.15) we shall showthe smallness of ‖curl AK ‖L∞(�) for small boundary datum A0

T , hence AK is exactly the classicalsolution of the Dirichlet problem (1.8). For this purpose we need the first eigenvalue of the followingproblem

curl2 u = λ u in �, uT = 0 on ∂�. (3.25)

We say that u is an eigenvector of (3.25) associated with an eigenvalue λ if u ∈ H2,2t0 (�, curl), u �≡ 0

and it is a weak solution of (3.25).

111501-17 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Lemma 3.11. Assume that � is a bounded domain in R3 with a C2 boundary and without holes.Then the first eigenvalue λ1 of (3.25) is positive and is given by

λ1 = infu∈H

2,2t0 (�,curl,div0), u �≡0

‖ curl u‖2L2(�)

‖u‖2L2(�)

. (3.26)

Proof. Obviously the number λ1 defined by (3.26) is the lowest eigenvalue. Now we show λ1

> 0. Otherwise, there exists u ∈ H2,2t0 (�, curl, div0), u �≡ 0, such that curl u = 0. Then u ∈ H2(�).

But H2(�) = {0} since � has no holes. Thus u = 0, a contradiction. �For the given positive function a(x) with a0 = minx∈� a(x) > 0, we define

N2(�, a) = 2‖a‖C(�)

(1 + a−2

0 ‖a‖2C(�)

). (3.27)

As mentioned before we can always choose the function G to be small. So we assume G satisfies(G), (3.19), and

M2 <λ1 min S′

K

N2(�, a). (3.28)

Proposition 3.12. Let 0 < K < b2. Assume that � and a(x) satisfy (A1) and (A2), respectively,with 0 < α < 1, and G is chosen to satisfy (G), (3.19), and (3.28). Let

AK = vK + ∇φvK ∈ C2,αt (�, div0, A0

T ) + grad C2,α0 (�)

be the solution of system (3.16) obtained in Proposition 3.6. Then for any ε > 0 and any σ ∈ [0, α),there exists a constant η = η(ε, σ ) > 0 such that if A0

T satisfies (A3) with

‖A0T ‖C2,α (∂�) ≤ η,

then we have ‖A‖C1,σ (�) ≤ ε.

Proof. Suppose the conclusion were false. Then there exists 0 < σ < α, ε0 > 0 and a sequence{A0

n,T } with each of them satisfying (A3) and ‖A0n,T ‖C2,α (∂�) → 0, and for each n the solution

AK ,n = vK ,n + ∇φvK ,n of (3.16) with the constant βK = βK, n, such that

‖AK ,n‖C1,σ (�) ≥ ε0.

For simplicity, in the proof we denote AK ,n , vK ,n , φvK ,n , and βK, n by An , vn , φvn , and βn, respectively.From (G) we know that 0 < βn ≤ M2 for all n. According to Proposition 3.9, ‖vn‖C2,α (�),

‖∇φvn ‖C1,α (�) and ‖An‖C1,α (�) are uniformly bounded in n. We choose subsequences of {vn} and{φvn

}, denoted as

{vn j

}and

{φvn j

}, such that as j → ∞,

‖An j ‖C1,σ (�) → lim supn→∞

‖An‖C1,σ (�).

After passing to a subsequence again, we may also assume that

vn j → v in C2,σ (�,R3), vT = 0 on ∂�,

φvn j→ φ in C2,σ (�), φ = 0 on ∂�,

βn j → β ≤ M2.

(3.29)

Hence An j → A ≡ v + ∇φ in C1,σ (�,R3) and AT = 0 on ∂�, and

‖A‖C1,σ (�) ≥ ε0. (3.30)

Recall the integral form of the equation for An j : For any H ∈ C1t0(�,R3),∫

�

S′K (| curl An j |2) curl An j · curl H dx − βn j

∫�

a(x)An j · H dx = 0.

111501-18 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Letting j → ∞ and then setting H = A as the test field in the resulted equality, we find

(min S′K )‖ curl v‖2

L2(�) = (min S′K )‖ curl A‖2

L2(�) ≤ β‖a‖C(�)‖A‖2L2(�). (3.31)

Similarly, φvn jsatisfies, for any ψ ∈ H 1

0 (�),

∫�

a(x)(vn j + ∇φvn j

) · ∇ψ dx = 0.

Letting j → ∞ and then setting ψ = φ as the test function in the resulted equality we deduce that

‖∇φ‖L2(�) ≤ a−10 ‖a‖C(�)‖v‖L2(�). (3.32)

Hence we have

‖A‖2L2(�) ≤ 2

(‖v‖2

L2(�) + ‖∇φ‖2L2(�)

)≤ 2

(1 + a−2

0 ‖a‖2C(�)

)‖v‖2

L2(�). (3.33)

From (3.31) and (3.33) we get

‖curl v‖2L2(�) ≤ 2β

min S′K

‖a‖C(�)

(1 + a−2

0 ‖a‖2C(�)

)‖v‖2

L2(�) . (3.34)

Using (3.29), (3.27), and (3.28) we find that

2β

min S′K

‖a‖C(�)

(1 + a−2

0 ‖a‖2C(�)

)< λ1.

From this and (3.34), and by the definition of λ1, we know that v = 0. Then from (3.33) we haveA = 0, which contradicts (3.30). �

Proof of Theorem 1.1. Let us fix a constant K such that 0 < K < b2 and construct the function SK

as in Subsection II D. Then we find a function G satisfying (G), (3.19), and (3.28). From Proposition3.6, for any A0

T satisfying (A3), the system (3.16) with this SK has a weak solution AK = vK + ∇φvK .From Proposition 3.9 we know that AK ∈ C1,α(�,R3). From Proposition 3.12 we know that thereexists η > 0 such that if ‖A0

T ‖C2,α (∂�) ≤ η, then ‖ curl AK ‖C(�) ≤ √K . Hence

S′K (| curl AK (x)|2) = S′(| curl AK (x)|2), ∀ x ∈ �.

Therefore, AK is a solution of (1.8). Since K is fixed now, we can denote AK by A and βK given in(3.9) by β. Then A is a solution of (1.8) with the constant β. �

In Subsection III A we obtained critical points for the truncated functionals with fairly generalnonlinear functions F. In order to show that they are solutions of the extended magnetostatic Born-Infeld systems we need to prove the C1,α regularity and uniform estimates of the solution, which areestablished in Theorem 1.1 for the special form of F. For the system with a general function F, inorder to overcome the difficulties in proving the C1, α regularity, one may try to modify the functionF(x, z) and apply the method in this paper to deal with the modified functional. We wish to examinethis approach in the future.

IV. THE NEUMANN PROBLEM

In Subsection IV A we assume that

� is a bounded domain in R3 with a C4 boundary,

D0T ∈ TW 1−1/r ′,r ′

(∂�,R3), g ∈ H 1/2(∂�),

F satisfies (F1) and (F2), 1 < p < 6, r = min{2,p}.(4.1)

111501-19 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

A. Critical points of H−K,G

1. Minimization of Fc

Let F c be the functional defined in (1.25).

Lemma 4.1. Assume that �, D0T , g, F, p, r satisfy (4.1). Then for any v ∈ H

2,pn (�, curl, div0, g)

the functional F c[v, ·] has a unique minimizer φcv ∈ W 1,p(�).

Proof. Using the condition on p, g in (4.1) and Lemma 2.1 we see that

H2,pn (�, curl, div0, g) = H2,2

n (�, curl, div0, g) = H 1n (�, div0, g),

where

H 1n (�, div0, g) = {v ∈ H 1(�,R3) : div v = 0 in �, ν · v = g on ∂�}.

We also know that H 1n (�, div0, g) is compactly imbedded into L p(�,R3). If φ ∈ W 1,p(�),

then ∇φ ∈ H2,p(�, curl), thus (∇φ)T ∈ W −1/r,r (∂�,R3). Using the condition on D0T in (4.1)

we see that the boundary integral part of F c is well defined for φ ∈ W 1,p(�). Thus for anyv ∈ H

2,pn (�, curl, div0, g), the functional F c[v, ·] is well-defined on W 1,p(�). Denote

a(F c, v) = infφ∈W 1,p(�)

F c[v, φ].

Let {φn} ⊂ W 1,p(�) be a minimizing sequence. Using the Cauchy inequality we have∣∣∣∣∫

∂�

(D0T × (∇φn)T ) · ν dS

∣∣∣∣ ≤ ‖D0T ‖W 1−1/r ′ ,r ′ (∂�)‖(∇φn)T ‖W −1/r,r (∂�)

≤ εC(p,�)

2‖∇φn‖p

L p(�) + C(ε)

2‖D0

T ‖p′

W 1−1/r ′ ,r ′ (∂�),

where ε is a small positive constant to be determined later. This together with (F2) yields

a(F c, v) + o(1) ≥∫

�

F(x, v + ∇φn) dx − 2∫

∂�

(D0T × (∇φn)T ) · ν dS

≥∫

�

(2−pc1|∇φn|p − c1|v|p − c2) dx − εC(p,�)‖∇φn‖pL p(�) − C(ε)‖D0

T ‖p′

W 1−1/r ′ ,r ′ (∂�).

By choosing ε = 2− p − 1c1/C(p, �) we see that {∇φn} is bounded in L p(�,R3), hence {φn}is bounded in W 1,p(�). Since W 1,p(�) is compactly imbedded in Lp(�) and W 1,p(�) is weaklyclosed, there exists a subsequence {φn j } and φc

v ∈ W 1,p(�) such that φn j → φcv weakly in W 1,p(�)

and strongly in Lp(�) as j → + ∞. By the strict convexity of F(x, z) in z we see that φcv is the

unique minimizer of F c[v, ·]. �Now we show that the Euler-Lagrange equation of φc

v is (1.26). Since φcv is a minimizer of

F c[v, ·] in W 1,p(�), a direct computation shows that for any ψ ∈ C1(�),∫�

∇z F(x, v + ∇φcv) · ∇ψ dx − 2

∫∂�

(D0T × (∇ψ)T ) · ν dS = 0. (4.2)

For any ψ ∈ C1(�) there exists a constant c such that ψ − c ∈ C1(�), and hence (4.2) holds for allψ ∈ C1(�). Let De ∈ H 1

t (�, div0, D0T ) be the divergence-free extension of D0

T . Using integrationby parts we get∫

∂�

(D0T × (∇ψ)T ) · ν dS =

∫�

curl De · ∇ψ dx =∫

∂�

(ν · curl D0T ) ψ dS, (4.3)

111501-20 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

where in the last equality we have used the identity ν · curl De = ν · curl DeT on ∂�. From (4.2) and

(4.3) we obtain, for all ψ ∈ C1(�), and hence for all ψ ∈ W 1,p(�),∫�

∇z F(x, v + ∇φcv) · ∇ψ dx − 2

∫∂�

(ν · curl D0T ) ψ dS = 0.

Therefore φcv is a weak solution of (1.26).

Remark 4.2. If F(x, z) = b(x, |z|2), where b(x, s) is C1 in s and

∂sb(x, s) �= 0 f or x ∈ ∂�, s ∈ R,

and if ν · curl D0T = 0 on ∂�, then for any v ∈ H

2,pn (�, curl, div0, g) Eq. (1.26) can be written as{

div(∂sb(x, |v + ∇φ|2)(v + ∇φ)

) = 0 in �,

∂φ

∂ν= −g on ∂�.

Arguing as in the proof of Lemma 3.3 we can prove the following conclusion.

Lemma 4.3. Assume that �, D0T , g, F, p, r satisfy (4.1). Then the functional F c[v, φ] is contin-

uous on H2,pn (�, curl, div0, g) × W 1,p(�), and the functional

v �→ F c[v, φcv]

is weakly continuous in H2,pn (�, curl, div0, g).

2. Minimization of H−K,G

Now we consider the functional H−K ,G introduced in (1.27), where G is a truncation function

satisfying (G). Due to the non-coercivity of the leading order term of the functional H−K ,G in

H2,p(�, curl, div0) ∩ H⊥1 (�), we look for a minimizer vK of H−

K ,G on Z. From (4.1), Lemma 2.1and the div-curl-gradient inequalities we see that

Z ⊂ H2,2n (�, curl, div0, g) ⊂ H 1(�,R3).

Hence the functional H−K ,G is well defined on Z. We shall also show that there exists a constant βc

Ksuch that the minimizer vK satisfies the following equality:∫

�

(S′

K (| curl vK |2) curl vK · curl w − βcK

2∇z F(x, vK + ∇φvK ) · w

)dx

+ βcK

∫∂�

(D0T × wT ) · ν dS = 0, ∀ w ∈ Z0.

(4.4)

Proposition 4.4. Let 0 < K < + ∞. Assume that �, D0T , g, F, p, r satisfy (4.1), and G satisfies

(G). Then the functional H−K ,G has a minimizer vK on Z, and vK satisfies (4.4) with

0 < βcK = G ′

(F c[vK , φc

vK] − 2

∫∂�

(D0T × vK ,T ) · ν dS

)≤ M2, (4.5)

where M2 > 0 is the constant given in (G).

Proof. Step 1. For g ∈ H1/2(∂�), let ge ∈ H 1n (�, div0, g) be a divergence-free extension of g in

the sense that

div ge = 0 in �, ν · ge = g on ∂�.

Moreover,

‖ge‖H 1(�) ≤ C(�)‖g‖H 1/2(∂�).

111501-21 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

In fact, we can choose ge = ∇ζ , where ζ ∈ H2(�) is a weak solution of

�ζ = 0 in �,∂ζ

∂ν= g on ∂�.

By the Sobolev embedding theorem we have ge ∈ L p(�,R3) for any 1 < p < 6. Thus ge ∈H

2,pn (�, curl, div0, g).

For any v ∈ H2,pn (�, curl, div0, g) we can write v = u + ge with u ∈ Z0. If we define

J c[u] = H−K ,G[u + ge + ∇φc

u+ge ],

then

infu∈Z0

J c[u] = infv∈Z

H−K , G[v].

Using the condition on g in (4.1) and Lemma 2.1, for u ∈ Z0 we have uT ∈ TH 1/2(∂�,R3). Bythe condition on D0

T in (4.1) we see that the surface integral∫∂�

(D0T × uT ) · ν dS

is well-defined, and hence it defines a weakly continuous functional on Z0. Combining this fact withLemma 4.3 we see that the functional

u �→ G(F c[u + ge, φu+ge ] − 2

∫∂�

(D0

T × (uT + geT )

) · ν dS)

is weakly continuous on Z0. So the functional J c is weakly lower semi-continuous on Z0. Similar tothe proof of Proposition 3.4, we can show that J c is coercive in Z0 with the equivalent norm shownin Lemma 2.3. Hence J c has a minimizer uK ∈ Z0. Let

vK = uK + ge. (4.6)

Then vK is a minimizer of H−K ,G on Z.

Step 2. Similar to the proof of Lemma 3.5, we can prove that vK satisfies (4.4) with βcK given

in (4.5). Here we omit the details. �Let AK = vK + ∇φc

vK. Then AK ∈ H2,p(�, curl). Now we show that AK is a weak solution of

the system {curl

(S′

K (| curl A|2) curl A) − βc

K2 ∇z F(x, A) = U in �,

S′K (| curl A|2)(curl A)T = βc

K D0T on ∂�,

(4.7)

for some U ∈ H1(�), with βcK a constant given in (4.5).

Definition 4.5. A ∈ H2,p(�, curl) is called a weak solution of (4.7) if the following equalityholds for all H ∈ H⊥

1 (�):∫�

(S′

K (| curl A|2) curl A · curl H − βcK

2∇z F(x, A) · H

)dx

+ βcK

∫∂�

(ν × D0T ) · HT dS = 0.

(4.8)

Let us mention that, if A satisfies (4.8) and if furthermore A ∈ C2(�,R3), then there exists a vectorfield U with U ∈ H1(�) such that (4.7) holds.

Proposition 4.6. Let 0 < K < +∞. Assume that �, D0T , g, F, p, r satisfy (4.1), and G satisfies

(G). Let vK be the minimizer of H−K ,G obtained in Proposition 4.4. Then AK = vK + ∇φc

vKis a weak

solution of (4.7). If furthermore � is simply connected, then AK is a weak solution of (1.29) withβ = βc

K .

111501-22 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Proof. Step 1. From Proposition 4.4 we know that vK satisfies (4.4) for any w ∈ Z0. Now weshow that (4.8) holds for any H ∈ C1(�,R3) ∩ H⊥

1 (�). To prove, we decompose H as H = w + ∇ψ ,where ψ ∈ W 2,p(�) satisfies

�ψ = divH in �,∂ψ

∂ν= ν · H on ∂�,

and w ∈ Z0. Note that (4.4) holds for this w. And the equality curl(∇φcvK

) = 0 holds in the sense ofL2(�,R3). Because φc

vKis the minimizer of F c[v, ·] in W 1,p(�), it satisfies (4.2) with v = vK and

with this ψ as a test function. Adding all these together we get (4.8) for all H ∈ C1(�,R3) ∩ H⊥1 (�).

The conclusion is followed since C1(�,R3) ∩ H⊥1 (�) is dense in H⊥

1 (�).Step 2. If � is simply connected, then H1(�) = {0} and H⊥

1 (�) = L2(�,R3). Then (4.8) holdsfor all H ∈ C1(�,R3). Thus AK ∈ H2,p(�, curl) is a weak solution of system (1.29). �

Remark 4.7. In addition to the conditions of Proposition 4.6, if F and D0 satisfy the assump-tions in Remark 4.2, then the weak solution AK = vK + ∇φc

vKof (4.7) satisfies an extra boundary

condition

ν · A = 0 on ∂�.

B. Estimates of the critical points of H−K,G

In this subsection we only consider the special case where F(x, z) is given in (3.15) witha(x) satisfying (A2). Let 0 < K < + ∞. From Proposition 4.6, if � is simply connected, thenH−

K ,G has a critical point AK on H2,2(�, curl), which is a weak solution of the Euler-Lagrangeequation {

curl(S′

K (| curl A|2) curl A) − βc

K a(x)A = 0 in �,

S′K (| curl A|2)(curl A)T = βc

K D0T on ∂�,

(4.9)

and

AK = vK + ∇φcvK

∈ H2,2n (�, curl, div0, g) + grad H 1(�).

Notice that g could be arbitrarily given, in the following for simplicity we set g = 0, thus vK ∈H

2,2n0 (�, curl, div0).

We shall study the higher regularity of AK and derive some estimates.

Proposition 4.8. Let 0 < K < b2. Assume that �, a(x), D0T satisfy (A1), (A2), and (A4), re-

spectively, with 0 < α < 1, and G satisfies (G). Let AK ∈ H2,2(�, curl) be a weak solution of (4.9)obtained in Proposition 4.6. Then

AK = vK + ∇φcvK

∈ C2,αn0 (�, div0) + grad C2,α(�),

and

‖vK ‖C2,α (�) + ‖∇φcvK

‖C1,α (�) ≤ C,

where C depends on �,α, a0, ‖a‖C1,α (�), K , δ, M2, ‖D0T ‖C1,α (∂�), and ‖ν · curl D0

T ‖C1,α (∂�).

Proof. The proof of Proposition 4.8 is identical to the proof of Proposition 5.2 in Ref. 12 withφK in Ref. 12 replaced by φc

vK, Eq. (5.7) of PK in Ref. 12 replaced by⎧⎪⎪⎨

⎪⎪⎩curl P = βc

K a(x)AK in �,

div P = 0 in �,

PT = βcK D0

T on ∂�,

111501-23 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

and Eq. (5.13) for φK in Ref. 12 replaced by the following equation for φcvK

:{div(a(x)∇φ) = −vK · ∇a(x) in �,

a(x) ∂φ

∂ν= ν · curl D0

T on ∂�,

and accordingly the estimates are altered, and using the fact 0 < βcK ≤ M2. �

Similar with Corollary 3.10 for Dirichlet problem, we can also derive higher regularity of thesolution AK .

Corollary 4.9. In addition to the conditions of Proposition 4.8, assume furthermore that∂� ∈ C4,α , and

a(x) ∈ C2,α(�), D0T ∈ C2,α(∂�,R3), ν · curl D0

T ∈ C2,α(∂�).

Then

AK = vK +∇φcvK

∈ C3,αn0 (�, div0) + grad C3,α(�) ⊂ C2,α(�,R3),

‖vK ‖C3,α (�) + ‖∇φcvK

‖C2,α(�) ≤ C,

where C depends on �,α, a0, ‖a‖C2,α (�), K , δ, M2, ‖D0T ‖C2,α (∂�) and ‖ν · curl D0

T ‖C2,α (∂�).

C. Solutions of the Neumann problem (1.11)

In this subsection, we shall show the smallness of ‖curl AK ‖L∞(�) for small boundary datumD0

T , hence AK is exactly the classical solution of Neumann problem (1.11). For this purpose weneed the first eigenvalue of the following problem:

curl2 u = μ u in �, ν · u = 0 on ∂�. (4.10)

We say that u is an eigenvector of (4.10) associated with an eigenvalue μ if u ∈ H2,2n0 (�, curl), u �≡ 0

and it is a weak solution of (4.10).

Lemma 4.10. Assume that � is a simply connected bounded domain in R3 with a C2 boundary.Then the first eigenvalue μ1 of (4.10) is positive and is given by

μ1 = infu∈H

2,2n0 (�,curl,div0), u �≡0

‖ curl u‖2L2(�)

‖u‖2L2(�)

.

Proof. We show μ1 > 0. Otherwise, there exists u ∈ H2,2n0 (�, curl, div0), u �≡ 0, such that

curl u = 0. Then u ∈ H1(�). But H1(�) = {0} since � is simply connected. Thus u = 0, acontradiction. �

As mentioned before we can always choose the function G to be small. So we assume G satisfies(G) and

0 < M2 <μ1 min S′

K

N2(�, a), (4.11)

where N(�, a) is given in (3.27).

Proposition 4.11. Let 0 < K < b2. Assume that � and a(x) satisfy (A1) and (A2), respectively,with 0 < α < 1, and G is chosen to satisfy (G) and (4.11). Let

AK = vK + ∇φcvK

∈ C2,αn0 (�, div0) + grad C2,α(�)

be the solution of system (4.9) obtained in Proposition 4.6. Then for any ε > 0 and σ ∈ [0, α), thereexists a constant δ = δ(ε, σ , M2) > 0 such that if D0

T satisfies (A4) with

‖D0T ‖C1,α (∂�) ≤ δ,

then ‖AK ‖C1,σ (�) ≤ ε.

111501-24 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Proof. The proof is similar to the proof of Proposition 3.12, with λ1 replaced by μ1. Notice thatβc

K appears in the boundary condition for Neumann problem (4.9), so the constant δ also dependson M2 given in (G). �

Proof of Theorem 1.2. Let us fix a constant K such that 0 < K < b2. Then we find a functionG satisfying (G) and (4.11). From Proposition 4.6, for any D0

T satisfying (A4), the system (4.9)with this SK has a weak solution AK = vK + ∇φc

vK. From Proposition 4.8 we know that AK ∈

C1,α(�,R3). From Proposition 4.11 we know that there exists δ > 0 such that if ‖D0T ‖C1,α (∂�) ≤ δ,

then ‖ curl AK ‖C(�) ≤ √K . Hence

S′K (| curl AK (x)|2) = S′(| curl AK (x)|2), ∀ x ∈ �.

Therefore AK is a solution of (1.11). Since K is fixed now, we denote AK by A and denote βcK given

in (4.5) by β. Then A is a solution of (1.11) with the constant β. �

D. Connection between the Dirichlet and Neumann problems

In this subsection we show that the quasilinear Neumann problem{curl

(S′(| curl A|2) curl A

) + ε a(x)A = 0 in �,

S′(| curl A|2)(curl A)T = D0T on ∂�,

(4.12)

where ε = ± 1, is equivalent to the semilinear Dirichlet problem{curl

(1

a(x) curl u) + ε f (|u|2)u = 0 in �,

uT = D0T on ∂�,

(4.13)

where f(ρ) is the function given in Lemma 2.5, and the explicit meaning of “equivalence” will beclear below.

If A ∈ C2,αn0 (�, div0) + grad C2,α(�) is a solution of (4.12), we let B = curl A and taking curl

on the both sides of (4.12) to obtain a new system{curl

[1

a(x) curl(S′(|B|2)B

)] + ε B = 0 in �,

S′(|B|2)BT = D0T on ∂�,

(4.14)

which can be written as (4.13), using the relation between B and u given in Lemma 2.5 (i).On the other hand, if u ∈ C2,α

t0 (�, div0) + grad C2,α(�) is a solution of (4.13) satisfying‖u‖C0(�) < b/2, then we can find a solution A of (4.12) using this u. In fact we first get B = f (|u|2)ufrom this u using Lemma 2.5 (i), and hence B ∈ C1,α(�,R3) and satisfies (4.14). Now we look for avector field A such that curl A = B and A solves (4.12). For this purpose we first look for a solutionH of the following div-curl system:{

curl H = B, div H = 0 in �,

ν · H = 0 on ∂�,(4.15)

where the solvability of H is guaranteed since (4.13) implies that

B = − curl( 1

a(x)curl u

) ∈ curl H 1(�,R3).

From (4.14) we know that

1

a(x)curl(S′(|B|2)B) + ε H ∈ Ker(curl) = H1(�) ⊕ grad H 1(�),

111501-25 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

and hence there exist U ∈ H1(�) and φ ∈ H1(�) such that

−[ 1

a(x)curl(S′(|B|2)B) + ε H

]= U + ∇φ.

Then A = ε H + U + ∇φ is a solution of (4.12).Let us mention that the regularity of the solution H of (4.15) is determined by B and �, and the

regularity of U is determined by �. From (4.12) we see that div(a(x)A) = 0, so φ satisfies{div(a(x)∇φ) = −(ε H + U) · ∇a(x) in �,

a(x) ∂φ

∂ν= −ν · curl D0

T on ∂�,

and hence the regularity of φ is determined by �, a(x), B, U, ν · curl D0T .

By a similar discussion and using the relation between B and u given in Lemma 2.5 (ii), we seethat the modified quasilinear equation{

curl(S′

K (| curl A|2) curl A) + ε a(x)A = 0 in �,

S′K (| curl A|2)(curl A)T = D0

T on ∂�,(4.16)

is equivalent to the following semilinear equation:⎧⎨⎩

curl(

1a(x) curl u

) + ε fK (|u|2)u = 0 in �,

uT = D0T on ∂�.

(4.17)

We believe that the methods in Subsections 4.1 and 4.2 of Ref. 12 can be used to obtain the existenceand regularity of the solutions of (4.17).

V. THE THIRD TYPE PROBLEM

A. Proof of Theorem 1.3

1. Problem (1.14)

Step 1. We start with the modified equation{curl(S′

K (| curl A|2) curl A) + a(x)A = 0 in �,

(curl A)T = B0T on ∂�.

which can be rewritten as the following equation for B = curl A:⎧⎪⎪⎨⎪⎪⎩

curl[

1a(x) curl

(S′

K (|B|2)B)] + B = 0 in �,

div B = 0 in �,

BT = B0T on ∂�.

(5.1)

Since � is a simply connected bounded domain in R3 without holes, and with a C3, α boundary, andB0

T ∈ C2,α(∂�,R3), we can find a divergence-free extension Be of B0T and Be ∈ C2,α(�,R3), see

Refs. 8 and 18.Step 2. Let us write B = Be + w in (5.1) and define a map

NK : C2,α(�, div0) × C2,αt0 (�, div0) → Cα(�, div0),

NK [Be, w] = MK [Be + w],

MK [B] = curl[ 1

a(x)curl

(S′

K (|B|2)B)] + B.

111501-26 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Note that MK is a continuously differentiable map, with the Frechet derivative

M′K [B](u) = curl

[ 1

a(x)curl GK (B, u)

]+ u,

where

GK (B, u) = S′K (|B|2)u + 2S′′

K (|B|2)(B · u)B.

Hence

∂wNK [Be, w] = M′K [B]

is a bounded linear map from C2,αt0 (�, div0) to Cα(�, div0). Note that

GK (0, u) = S′K (0)u = S′(0)u = 1

2u.

For a given f ∈ Cα(�, div0), the vector field u ∈ C2,αt0 (�, div0) satisfies the equation M′

K [0]u = fif and only if u is a solution of the linear problem⎧⎪⎪⎨

⎪⎪⎩12 curl

(1

a(x) curl u)

+ u = f in �,

divu = 0 in �,

uT = 0 on ∂�.

Using the conditions (A1) and (A2′) we know that the map

L+u ≡ 1

2curl

( 1

a(x)curl u

)+ u

is a homeomorphism from C2,αt0 (�, div0) onto Cα(�, div0). Since

NK [0, 0] = 0, ∂wNK [0, 0] = M′K [0] = L+,

hence M′K [0] has a bounded inverse.

So we can apply the implicit function theorem to conclude that there exist positive numbers r1,r2, with the sets

V = {Be ∈ C2,α(�, div0) : ‖Be‖C2,α(�) < r1},W = {w ∈ C2,α

t0 (�, div0) : ‖w‖C2,α (�) < r2},such that for any Be ∈ V, there exists w ∈ W such that

NK [Be, w] = 0,

and this w is the unique solution in W. Moreover the map Be �→ w = W[Be] is a continuouslydifferentiable map from V to W, and W[0] = 0. By the continuity of the map W we can deducer1 hence deduce r2 such that r1 + r2 ≤ √

K . Then we use the C2, α estimate of the divergence-freeextension (see Ref. 8) to find a constant r0 > 0 with

U = {B0T ∈ TC2,α(∂�,R3) : ‖B0

T ‖C2,α (∂�) < r0},such that for any B0

T ∈ U the divergence-free extension Be ∈ V. Then B = Be + W[Be] is a solutionof (5.1). Since ‖B‖C2,α (�) < r1 + r2 ≤ √

K , B is a solution of (5.1) with SK replaced by S.Step 3. Let B ∈ C2,α(�, div0) be a solution of (5.1) with SK replaced by S. We look for a vector

field A such that curl A = B and A is a solution of (1.14). Since � is simply connected and has noholes, and div B = 0, the div-curl system (4.15) has a unique solution H for this B. Since ∂� is ofclass C4, α and B ∈ C2,α(�, div0), hence H ∈ C3,α(�, div0). Using (5.1) with SK replaced by S, wesee that

curl[ 1

a(x)curl

(S′(|B|2)B

) + H]

= 0.

111501-27 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Since � is simply connected, there exists φ ∈ C2,α(�) such that

−[ 1

a(x)curl(S′(|B|2)B) + H

]= ∇φ.

Then A = H + ∇φ has the regularity shown in (1.17), and it is a solution of (1.14). Conclusion (i)is proved.

2. Problem (1.15)

By the same spirit we consider the following equation corresponding to the modified functionalwith a concave lower order term{

curl(S′K (| curl A|2) curl A) − a(x)A = 0 in �,

(curl A)T = B0T on ∂�.

(5.2)

Most of analysis in the above works still with necessary modification. For instance, instead of theoperator L+, now the linearization leads to an linear operator

L−u ≡ 1

2curl

( 1

a(x)curl u

)− u.

If a(x) satisfies the condition (A2′) then L− is a homeomorphism from C2,αt0 (�, div0) onto

Cα(�, div0) if and only if 2 is not an eigenvalue of L+, namely λ = 2 is not an eigenvalue of(1.16). So we get the conclusion (ii). �

We mention that if �, B0T and a has higher regularity then the solution A of (1.14) obtained in

Theorem 1.3 has higher regularity.

Corollary 5.1. In addition to the conditions of Theorem 1.3, assume B0T ∈ TC3,α(∂�,R3). Then

the solutions of (1.14) and (1.15) obtained in Theorem 1.3 belong to C2,α(�,R3).

Proof. We give the proof for Eq. (1.14). We keep the notation in the proof of Theorem 1.3.Under the conditions of this corollary, the solution B of (5.1) belongs to C3,α(�,R3). The solutionH of (4.15) belongs to C3,α(�,R3) (which belongs to C4,α(�, div0) if ∂� is of class C 5,α). From(1.14) we furthermore see that φ satisfies{

div(a(x)∇φ) = −∇a(x) · H in �,

a(x) ∂φ

∂ν= −ν · curl(S′(|B|2)B) on ∂�.

Now we have

a(x) ∈ C2,α(�), ∇a(x) · H ∈ C1,α(�, div0), ν · curl(S′(|B|2)B) ∈ C2,α(∂�).

From the regularity of higher order derivatives of linear Neumann problems we see that φ ∈ C3,α(�).Thus A ∈ C3,α(�, div0) + grad C3,α(�) ⊂ C2,α(�,R3). �B. Gauge invariant problem: The third type boundary condition

We say a problem is gauge invariant if A is a solution then A + ∇φ is also a solution, whereφ has certain regularity, see Ref. 12, Sec. VI. In this subsection we examine the gauge invariantequation with the third type boundary condition:{

curl(S′(| curl A|2) curl A) + j = 0 in �,

(curl A)T = B0T on ∂�.

(5.3)

Here we only require B0T ∈ TC1,α(∂�,R3), and j satisfies the following condition:

(A6) j ∈ Cα(�, div0) with 0 < α < 1.

111501-28 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

We shall show that this system has infinitely many solutions, among them there is a solution satisfyingextra conditions

divA = 0 in �, ν · A = 0 on ∂�.

Proposition 5.2. Assume that � and j satisfy (A1) and (A6), respectively, with 0 < α < 1. Thenthere exists R4 > 0 such that if

B0,T ∈ TC1,α(∂�,R3), ‖B0T ‖C1,α (∂�) < R4,

the problem (5.3) has a solution A ∈ C2,αn0 (�, div0). For any φ ∈ H1(�), A + ∇φ is also a weak

solution of (5.3).

Proof. Step 1. We first consider a solution A of the modification of (5.3) with S replaced by SK.Then B = curl A is a solution of⎧⎪⎪⎨

⎪⎪⎩curl(S′

K (|B|2)B) + j = 0 in �,

div B = 0 in �,

BT = B0T on ∂�.

(5.4)

Since � is a bounded domain in R3 without holes and with a C2, α boundary and B0T ∈ C1,α(∂�,R3),

we can find a divergence-free extension Be ∈ C1,α(�,R3) of B0T , and

‖Be‖C1+α (�) ≤ C(�,α)‖B0T ‖C1+α(∂�), (5.5)

see, for instance, Ref. 18. Write B = Be + w. Then (5.4) is reduced to⎧⎪⎪⎨⎪⎪⎩

curl[S′

K (|Be + w|2)(Be + w)] + j = 0 in �,

div w = 0 in �,

wT = 0 on ∂�.

(5.6)

We define a map

TK : C1,α(�, div0) × C1,αt0 (�, div0) → Cα(�, div0),

TK [Be, w] = RK [Be + w],

RK [B] = curl(S′K (|B|2)B).

Note that RK is a continuously differentiable map, with the Frechet derivative

R′K [B](u) = curl(GK (B, u)),

where

GK (B, u) = S′K (|B|2)u + 2S′′

K (|B|2)(B · u)B,

and ∂wTK [Be, w] = R′K [B] is a bounded linear map from C1,α

t0 (�, div0) to Cα(�, div0).We show that R′

K [0] has a bounded inverse. Given f ∈ Cα(�, div0), the vector field u ∈C2,α

t0 (�, div0) satisfies the equation R′K [0]u = f if and only if u is a solution of the linear problem

curl u = 2f, divu = 0 in �, uT = 0 on ∂�.

Since � is simply connected and without holes, the above system has a unique solution u and

‖u‖C1,α (�) ≤ C(�,α)‖f‖Cα (�).

Thus R′K [0] is a homeomorphism from C1,α

t0 (�, div0) to Cα(�, div0).Step 2. Since TK [0, 0] = 0 and ∂wTK [0, 0] = R′

K [0], we can apply the implicit function theoremin the way similar to the first part of the proof of Theorem 1.3, and conclude that (5.4) has a solutionB = Be + W1[Be], where w = W1[Be] is a solution of

TK [Be, w] = 0.

111501-29 J. Chen and X.-B. Pan J. Math. Phys. 54, 111501 (2013)

Moreover, ‖B‖C1,α (�) ≤ √K if ‖B0

T ‖C1,α (∂�) is small. So B is a solution of (5.4) with SK replacedby S.

Step 3. Let B ∈ C1,α(�, div0) be a solution of (5.4) with SK replaced by S. Since � is simply-connected and has no holes, and div B = 0, (4.15) has a unique solution H. Since ∂� is of class C3, α

and B ∈ C1,α(�, div0), hence H ∈ C2,αn0 (�, div0). Then H is a solution of (5.3). For any φ ∈ H1(�),

A = H + ∇φ is also a weak solution of (5.3). �ACKNOWLEDGMENTS

X.-B.P. would like to thank Professor Fanghua Lin and Professor Yisong Yang of New YorkUniversity for introducing the Born-Infeld theory to him during a workshop in 2004. Both theauthors would like to thank the referee for the valuable comments which helped us to revise thispaper. This work was partially supported by the National Natural Science Foundation of China(Grant No. 11171111) and the Science Foundations of the Ministry of Education of China (GrantNo. 20110076110001).

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