Calculus of Variations

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1 Calc. Var. DOI 1.17/s y Calculus of Variations On the equilibrium set of magnetostatic energy by differential inclusion aisheng Yan Received: 13 September 211 / Accepted: 2 April 212 Springer-Verlag 212 Abstract This paper concerns the set of equilibriums of the nonlocal magnetostatic energy for saturated magnetizations. We study the stability of the equilibrium set under the weak-star convergence using methods of differential inclusion and quasiconvex analysis. The equilibrium set is shown to be unstable under the weak-star convergence and an estimate on its weak-star closure is obtained. This estimate is also shown to be accurate when the physical domain is an ellipsoid. Mathematics Subject Classification 35Q6 49J45 78A25 1 Introduction In the Landau Lifshitz theory of micromagnetics (see, e.g., [4,18,19]), the magnetostatic energy of a ferromagnetic material occupying a domain in R 3 is defined to be I(M) = 1 2 dx, 2 R 3 H M where M = M(x) represents the magnetization vector at point x and H M is the magnetostatic field induced by M on whole R 3 through the simplified Maxwell equation curl H M =, div(h M + Mχ ) = inr 3. (1.1) From this equation, it is easily seen that I(M) = 1 H M Mdx. 2 Communicated by J. all.. Yan () Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA yan@math.msu.edu

2 . Yan In this theory, under the certain temperature, the magnetization is assumed to be saturated; that is, the length M is constant over. We thus assume the rescaled saturation condition M(x) =1 (x ). (1.2) Such a magnetization vector M is called an equilibrium for energy I provided that, for all vector-fields M(x, t), smooth in t ( ɛ, ɛ) and satisfying M(x, t) =1andM(x, ) = M(x), it follows that d dt I(M(, t)) t= =. This condition can be shown to be equivalent to the condition M H M = on. Therefore, the set of equilibriums, or the equilibrium set, is given by E() = { M L ( ; R 3) : M =1, M H M = a.e. }. (1.3) Equilibrium magnetizations are closely related to the asymptotic study of the Landau Lifshitz Gilbert equations in ferromagnetism [5,15]. Note that a dynamic Landau Lifshitz equation governing the magnetization evolution under solely the magnetostatic energy I is given by t M = γ M H M + αγ M M (M H M), (1.4) where γ<isthe electron gyromagnetic ratio and α>isalandau Lifshitz phenomenological damping parameter. This equation can be written as an equivalent Landau Lifshitz Gilbert equation: t M = γ ( 1 + α 2) M H M + α M M t M. (1.5) Therefore, the equilibrium set E() defined above also represents all time-independent solutions of (1.4)or(1.5). Existence and uniqueness of global weak solution to initial value problem of the Landau Lifshitz Eq. (1.4) have been recently established in [11]; see also [14]. For the solution M = M(x, t) of (1.4), the energy identity I(M(, t)) I(M(, s)) = t s αγ M M H M 2 dxdτ (1.6) holds for all s < t <. Furthermore, if the initial datum M satisfies M (x) =1on, then M(x, t) =1onfor all t >. Hence (1.6) implies that there exists a sequence t j such that M j = M(, t j ) M and M j H M j inl 2 (). To study such a weak-star ω-limit M, the stability of the equilibrium set E() under weak-star convergence becomes relevant and critical. There are some fundamental difficulties in studying the set E() due to the nonconvexity of condition M =1, nonlocality of H M and the nonlinearity of condition M H M =. The following two sets of special magnetization configurations are included in the equilibrium set E(): (1) The set E 1 () of all M with M =1 and div(mχ ) = (hence H M = ), which, for the special energy I, is the set of energy minimizers and has been studied in the work of [8,12,13,24,27].

3 Equilibrium set of magnetostatic energy by differential inclusion (2) The set E 2 () of all M with M =1andcurl(Mχ ) = (hence H M = Mχ ), which can be studied through the Eikonal equation ϕ =1in with ϕ =. oth E 1 () and E 2 () are nonempty and, thanks to the linear differential equations, weakstar limits of equilibriums in either of them must satisfy the same linear differential equation; in particular, such a weak limit M must have the zero-average: Mdx=. However, this zero-average property may not hold for the weak-star limits of general equilibriums in E(). In general, determining the weak-star closure of E() is a challenging problem. The main purpose of this paper is to present a good estimate for such a closure. We study the set E() under a general framework of the partial differential inclusions and the vectorial calculus of variations, based on the notion of Morrey s quasiconvexity [2]; see also [2,7]. To discuss the main ideas, we observe that, by the Maxwell Eq. (1.1) above, there exist functions u : R 3 R and v : R 3 R 3 such that H M = u, Mχ = u + curl v (1.7) in the sense of distributions on R 3. The function u in the decomposition (1.7) is unique up to adding a constant and u H 1 (). The term curl v in (1.7) remains unchanged when v is replaced by v + ϕ for any function ϕ Hloc 1 (R3 ); hence function v can be adjusted to satisfy that v H 1 (; R 3 ).HereH 1 denotes the standard Sobolev ( ) space. Let U = (u,v) H 1 (; R 4 u ). The gradient matrix U =,where u R v 3, v M 3 3, is considered as a matrix in M 4 3. Introduce a linear map δ : M 3 3 R 3 such that and define K = δ( v(x)) = curl v(x) v C (R 3 ; R 3 ), {( ) a } : a δ() =, a + δ() =1. (1.8) It follows that if M E() then M = u + curl v a.e. in for some map U = (u,v) H 1 (; R 4 ) satisfying the partial differential inclusion: U(x) K a.e. x. (1.9) In this manner, the study of equilibrium set E() can be accomplished by studying partial differential inclusion (1.9) with the set K defined by (1.8). Most of the study in the paper will be devoted to the quasiconvex analysis related to the differential inclusion (1.9), following the work of [9,21 23,26]; see also a recent paper [1]. The main conclusions resulting from these studies are the following theorems. Theorem 1.1 Let M j MinL ( ; R 3) as j. If lim j then the weak-star limit M satisfies ( 1 M j 2 + M j H M j ) dx =, (1.1) M M H M 1 a.e.. (1.11) Conversely, for any M satisfying (1.11), there exists a sequence {M j } in L (; R 3 ) satisfying (1.1) such that M j Masj.

4 . Yan Theorem 1.2 Let be an ellipsoid. Then any constant vector M R 3 satisfying the inequality (1.11) is weak-star limit of a sequence of equilibriums in E(). Clearly, condition (1.1) is satisfied if each M j E(). It is also satisfied if M j =1 and M j H M j strongly in L 2 (); this is the case when M j (x) = M(x, t j ),where M(x, t) is the solution to (1.4) or(1.5) andt j is certain time-sequence. Theorem 1.2 indicates that, for ellipsoid domains, estimate (1.11) is very accurate for the weak-star closure of E(). In general, we are unable to show that the set of M L (; R 3 ) defined by condition (1.11) is exactly the weak-star closure of equilibrium set E(). The rest of the paper is organized as follows. In Sect. 2, we review some notation and useful definitions and results in the quasiconvex analysis and differential inclusions; most materials can be found in [7]. In Sect. 3, we explicitly compute the lamination-convex hull of the set K defined by (1.8) and discover a useful quasiconvex function. In Sect. 4, we prove some existence results for differential inclusion (1.9) using the existence theorems in [7,23,26]. The proof of Theorem 1.1 isgiveninsect.5after we prove a technical lemma on the representation of divergence-free fields. Finally, in Sect. 6, we study the special case of ellipsoid domains and present the proof of Theorem Notation and preliminaries Let M m n be the space of all real m n matrices, with standard Euclidean structures. Denote by rank ξ the rank of a matrix ξ M m n. Given ξ 1 = ξ 2 in M m n,weuseξ 1 ξ 2 to denote the open line segment joining ξ 1 and ξ 2 ;thatis,ξ ξ 1 ξ 2 if and only if ξ = λξ 1 + (1 λ)ξ 2 for some <λ<1. As above, a matrix ξ M 4 3 ( = R 3 M 3 3 a ) is also identified as ξ = with a R 3 and M 3 3. Let δ : M 3 3 R 3 be the linear map defined above through δ( v(x)) = curl v(x) v C ( R 3 ; R 3), where curl v = v is the usual curl operator on vector-fields on R 3. Given p, q R 3,letp q R 3 be the vector product of p and q in R 3 ; let p q denote the matrix (p i q j ) in M 3 3. Note that a matrix R M 3 3 has rank-one if and only if R = p q for some p = andq =. It is easy to see that the linear map δ : M 3 3 R 3 enjoys the following properties: δ(p q) = q p; δ() = if and only if T =, where T stands for the matrix transpose; hence the null space of δ is the set of all symmetric matrices in M 3 3. We now review some important definitions and results in the studies of partial differential inclusions and in quasiconvex analysis, that are mostly related to our study in this paper. For complete studies and more extensive references in this direction, we refer to the newest edition of monograph [7]. Given any set K M m n,let α(k ) = { ξ M m n } : ξ 1,ξ 2 K with rank(ξ 1 ξ 2 ) = 1 such that ξ ξ 1 ξ 2. Note that a matrix ξ α(k ) if and only if there exists a rank-one matrix η and two numbers t < < t + such that ξ + t ± η K. Let L 1 (K ) = K α(k ) (2.1)

5 Equilibrium set of magnetostatic energy by differential inclusion and define, inductively, L n+1 (K ) = L 1 (L n (K )) for n = 1, 2, 3,... The set K lc = n=1 L n(k ) is called the lamination-convex hull of K. Let f : M m n R be a given function. We say that f is rank-one convex if, for arbitrary ξ, η M m n with rank η = 1, the function g(t) = f (ξ + tη) is a convex function of t R. We say that f is quasiconvex (in the sense of Morrey [2]) if, for any ξ M m n,any bounded domain R n and any ϕ C (; Rm ), it follows that f (ξ) 1 f (ξ + ϕ(x)) dx. The largest quasiconvex function below f is called the quasiconvex envelope of f and is denoted by f qc. Under some mild condition, it follows that [7] f qc 1 (ξ) = inf f (ξ + ϕ(x)) dx. (2.2) ϕ W 1, (;R m ) The quasiconvex functions are closely related to the weak lower semicontinuity of integral functionals [1] and also play a pivotal role in nonlinear elasticity [2]. It is well-known that every quasiconvex function is rank-one convex; the converse is false in general if m 3. However, we have the following result; see [7, Theorem 5.25] for a proof. Theorem 2.1 Let f : M m n R be a homogeneous quadratic polynomial. Then, f is quasiconvex if and only if f is rank-one convex. We need the following important definition to study the existence of solutions to differential inclusions; see [7, Definition 1.2], [23, Definition 1.1], or [26, Definition 1.6]. Definition 2.1 Let E, K M m n. We say that E is reducible to K (or that E has the relaxation property with respect to K ) provided that, for every bounded domain, everyɛ> and every ξ E, there exists a piecewise affine function w ξ x + W 1, (; R m ) such that w(x) E K a.e. in ; dist( w(x); K ) dx <ɛ, (2.3) where dist(η; K ) denotes the distance function to set K. The following important existence result has been proved in [7, Theorem 1.4], [23, Theorem 1.2], or [26, Theorem 1.7]; the proofs in [7,26] have been motivated by the work of [17]. Theorem 2.2 Let R n be a bounded domain. Let E, K M m n be such that E is bounded and K is compact. Assume that E is reducible to K (or E has the relaxation property with respect to K ). Then, for each ξ E and each ɛ>, there exists a function w in ξ x + W 1, (; R m ) such that w ξ x L () <ɛ, w(x) K a.e. in. Finally, the following result proved in [25, Lemma 3.4] is useful to prove the reduction principle for open sets; for related results, see [7, Corollary 1.23].

6 . Yan Lemma 2.3 Let E be an open set in M m n and let η E and η = tη 1 + (1 t)η 2 with t (, 1) and rank(η 1 η 2 ) = 1. Then, for each ɛ>, there exists a piece-wise affine map w ηx + W 1, (; R m ) and finitely many points η 3,...,η s in E such that { w(x) {η1,η 2,η 3,...,η s } a.e. x ; {x : w(x) {η 1,η 2 }} <ɛ. 3 Quasiconvex analysis of set K Let K be the set defined by (1.8) above; that is, {( ) } a K = M 4 3 : a δ() =, a + δ() =1. (3.1) The following result is obtained through explicit computation of L 1 (K ). Theorem 3.1 Let K be defined by (3.1). Then the set L 1 (K ) defined by (2.1) above is given by {( ) } a S = : a + δ() a δ() 1. (3.2) Moreover, for every ξ S \ K, there exist two matrices ξ ± K with rank(ξ + ξ ) = 1 such that where C > is a uniform constant independent of ξ. ξ ξ ξ +, ξ ± C ( ξ +1), (3.3) Proof We proceed in two steps. Step 1. We first show that L 1 (K ) S. Obviously, K S. Take ξ α(k )\ K. Then there exists a rank-one matrix η and two numbers t ±, with t < < t +, such that ξ + t ± η K. Observe that rank-one matrix η M 4 3 can be written as η = and m, p R 3, with s 2 + m 2 = andp =. Let ξ = to ( sp m p ) for some s ( a ). Then ξ +tη K is equivalent (a + tsp) (δ() + tp m) =, (3.4) a + tsp + δ() + tp m =1, (3.5) where t = t ± with t < < t +. Since replacing m by rp + m, foranyr R, does not change both equations above, we assume that p m =. Eq. (3.4) then becomes a δ() + t[sp δ() + a (p m)] t 2 s p 2 m =. (3.6) To prove that L 1 (K ) S it remains to show that a + δ() a δ() 1. We divide the proof in two cases. Case 1 a δ() =. In this case, in Eq. (3.5), letting sp + p m = b, it follows that a + δ() + t ± b =1. Let λ = t + t + t. Then <λ<1and a + δ() = λ[a + δ() + t b]+(1 λ)[a + δ() + t + b]. Hence, by convexity, a + δ() 1 and so ξ S.

7 Equilibrium set of magnetostatic energy by differential inclusion Case 2 a δ() =. In this case, since Eq. (3.6) has two distinct nonzero solutions t = t ± with t < < t +, it follows that s > andm = ; hence p m =. Dot-product of (3.6) with p m yields (a δ()) (p m) + ts(p δ()) (p m) =, which has two distinct solutions t = t ±, and hence it follows that (a δ()) (p m) = and (p δ()) (p m) =. So both a δ() and p δ() are in the plane spanned by m and p; hence p δ() = xm for some x R. Now dot-product of (3.6) with a must yield a m = since the equation deduced has two distinct solutions t = t ± ; hence a (p m) = (a p)m. Therefore, every vector in (3.6) is a multiple of vector m. Let a δ() = ym with y R. Then Eq. (3.6) becomes which has two roots t ± satisfying t t + = written as a quadratic equation s p 2 t 2 +[(a p) sx]t y =, (3.7) y s p 2 < ; hence y >. Now Eq. (3.5) can be p 2 ( s 2 + m 2) t 2 + 2[(a + δ()) (sp + p m)]t + a + δ() 2 1 =, which has the same roots as the quadratic Eq. (3.7) above; hence 1 a + δ() 2 s 2 + m 2 = y s, (3.8) 2(a + δ()) (sp + p m) s 2 + m 2 Note that y 2 m 2 = a δ() 2 ; hence, (3.8) is equivalent to = a p sx. (3.9) s 1 a + δ() 2 a δ() 2 = ys +, (3.1) ys which yields 1 a + δ() 2 2 a δ() and hence proves ξ S. Therefore, L 1 (K ) S. Step 2. We show that, for every ξ S \ K, there exist two matrices ξ ± K with rank(ξ + ξ ) = 1 such that ξ ξ ξ +, ξ ± C ( ξ +1), (3.11) where C > is a uniform constant independent of ξ. This will prove the second part of the theorem and will also imply that S \ K α(k ); hence S L 1 (K ), completing the proof of the theorem. ( a ) Let ξ = S \ K. We also consider several cases. Case 1 δ() =. First, assume a = ; then ( ) ξ = = 1 ( ) e + 1 ( ) e, 2 2 ( ) where e R 3 ±e is any unit vector. Hence ξ ξ ξ +,whereξ ± = K, have rank-one difference and, clearly, satisfy ξ ± 2( ξ +1). So (3.11) is proved in this case.

8 . Yan Now assume a = ; then < a < 1. Let λ = 1+ a 2 (, 1). Then ( ) ( ) ( ) a a/ a a/ a ξ = = λ + (1 λ). ( ) ±a/ a Hence ξ ξ ξ +,whereξ ± = K, have rank-one difference and, clearly, satisfy ξ ± 2( ξ +1). So (3.11) is proved in this case. Case 2 δ() =, a δ() =. In this case, let a = tδ(). Since ξ S \ K, we have 1 + t δ() < 1. Let a ± = ( 1 ± δ() 1 )δ(). Then a ± δ() +1 C( +1) and ( ) a± the matrices ξ ± = K and have rank-one difference. Let λ = 1+(1+t) δ() 2 (, 1). Then ξ = λξ + + (1 λ)ξ ; hence ξ ξ ξ +. Clearly, ξ ± C ( ξ +1); hence (3.11) is proved in this case. Case 3 a δ() =. In this case, conditions a+δ() 2 +2 a δ() 1and a+δ() = 1 cannot hold simultaneously; hence, a + δ() a δ() 1, a + δ() < 1. (3.12) We first show that there exists a rank-one matrix η such that ξ ± = ξ + t ± η K for two numbers t ± with t < < t + ; hence, ξ ξ ξ +. We then estimate ξ ± to complete the proof of (3.11) in this case. ( ) sp Assume that such a rank-one matrix η is given by η = for some s and m p m, p R 3, with s 2 + m 2 = andp =. Then, as seen in the case 2 of Step 1, one has that s > andm = ; hence, by rescaling p and m, wemayassumes = 1. In this case, again from the case 2 of Step 1, m = a δ()/y and p = x y a + zδ(), where y > is determined by Eq. (3.1) (with s = 1); hence p δ() = x y a δ() = xm. Let y be the larger root of Eq. (3.1) (with s = 1), which always exists, thanks to the condition a + δ() a δ() 1. This y satisfies Hence, by (3.12), m = 1 a + δ() 2 y 1. 2 a δ() y 2 a δ() 1. (3.13) 1 a + δ() 2 Once y and m are selected, we solve Eq. (3.9), again with s = 1, for nonzero (x, z) R 2. We can uniquely solve z/x from a simple nontrivial linear equation. Using x, y, z found, we obtain p = x y a + zδ() (up to constant multiples) and then solve t ± from Eq. (3.7), which always has solutions t ± with t < < t + ; in fact, t ± = x a p ± (x a p) p 2 y 2 p 2. (3.14) ( ) p This gives the rank-one matrix η = and the numbers t m p ± such that ξ ± = ξ +t ± η K ; hence ξ ξ ξ +.

9 Equilibrium set of magnetostatic energy by differential inclusion Finally, we estimate ξ ±. From (3.9), since m 1, it follows that x 9 ξ p. Using < y 1 and the formula (3.14), it follows that t ± C( ξ +1)/ p for some constant C >. Since m 1, it follows that η 2 p. Therefore, t ± η C( ξ +1); hence ξ ± ξ + t ± η C( ξ +1) for a constant C >. This proves (3.11) in this case and hence completes the proof of the theorem. The previous theorem motivates the following result. Theorem 3.2 Let f : M 4 3 R be the function defined by f (ξ) = a + δ() a δ(), ξ = Then f is quasiconvex (in the sense of Morrey [2]). Proof Note that f (ξ) = where, for each h R 3 with h =1, max f h (ξ), h R 3, h =1 f h (ξ) = a + δ() 2 + 2(a δ()) h, ξ = ( a ), (3.15) ( a ). If we show that each f h is quasiconvex, then so is f. To show that each f h is quasiconvex, note that each f h is a homogeneous quadratic polynomial; hence f h will be quasiconvex if it is rank-one convex (see Theorem 2.1). Therefore, we need only to prove that each f h is rank-one convex on M 4 3 ; namely, given any ξ and η with rank η = 1, the function ( a ) ( ) sp g(t) = f h (ξ + tη) is a convex function on t R. Let ξ = and η =,where m p s R and p, m R 3. It is easily seen that function g(t) = f h (ξ + tη) = βt is a quadratic function of t with the coefficient of t 2 given by β = sp + p m 2 + 2s(p (p m)) h = s 2 p 2 + p m 2 + 2s(p (p m)) h. Since h =1and p (p m) p p m, it follows that β s 2 p 2 + p m 2 2 s p p m =( s p p m ) 2. Hence g(t) = βt is convex. This completes the proof. Remark 3.1 We point out that function f is not convex. For example, with ( ) ( ) e2 e ξ = 1 e, η = 1, e 3 e 2 e 2 e 3 + e 1 e 3 one can easily check that function g(t) = f (ξ + tη) = (1 + t) t 2, which is clearly not convex on R.

10 . Yan Corollary 3.3 It follows that K lc = L 1 (K ) = S. Proof We only need to prove that L 1 (S) = S. For this, it suffices to show α(s) S. Let ξ α(s). Then there exists a rank-one matrix η and two numbers t ± with t < < t + such that ξ ± = ξ + t ± η S; hence f (ξ ± ) 1. Let λ = t + t + t.then<λ<1and ξ = λξ + (1 λ)ξ +. Since function f is also rank-one convex, it follows that f (ξ) = f (λξ + (1 λ)ξ + ) λf (ξ ) + (1 λ) f (ξ + ) 1; henceξ S. This proves α(s) S. Corollary 3.4 Let F : M 4 3 R be the function defined by F(ξ) = 1 a + δ() 2 +2 a δ(), ξ = If we denote t + = max{t, } for t R, then ( ) a. (3.16) ( f (ξ) 1) + F qc (ξ) F(ξ). (3.17) Proof Clearly, F(ξ) ( f (ξ) 1) + and ( f (ξ) 1) + is quasiconvex. Therefore, (3.17) follows from the definition of F qc. 4 Existence results for differential inclusion (1.9) The main result of this section is the following theorem. Theorem 4.1 Let f : M 4 3 R be the quasiconvex function defined by (3.15). Then, given any ξ M 4 3 with f (ξ )<1, there exists a Lipschitz map U : R 4 such that U(x) K a.e. in, U(x) = ξ x on. (4.1) Proof Select a constant Q such that ξ < Q.Definesets E ={ξ M 4 3 : ξ < Q, f (ξ) < 1}, K L ={ξ K : ξ L}, where L = C(Q + 1), with C being the constant in Theorem 3.1. Note that E is bounded and open, K L is compact, and ξ E. Therefore, by Theorem 2.2,toprovethetheorem,we only need to verify that E is reducible to K L or E has the relaxation property with respect to K L as defined in Definition 2.1 above; namely, for every ξ E, ɛ>, there exists a piecewise affine function w ξ x + W 1, (; R 4 ) such that w(x) E K L a.e. x ; dist( w(x); K L ) dx <ɛ. (4.2) Let ξ E. Then ξ S \ K ; hence, by Theorem 3.1, thereexistη 1,η 2 K such that rank(η 1 η 2 ) = 1andξ = tη 1 + (1 t)η 2 for some t (, 1). y (3.3), it follows that η 1, η 2 C( ξ +1) C(Q + 1) = L and hence η 1,η 2 K L. y Lemma 2.3, for each ɛ>, there exists a piece-wise affine map w ξ x + W 1, (; R 4 ) and finitely many points η 3,...,η s in E such that w(x) {η 1,η 2,η 3,...,η s } E K L a.e. ; {x : w(x) / {η 1,η 2 }} < ɛ L + Q.

11 Equilibrium set of magnetostatic energy by differential inclusion Since X Y X + Y < Q + L for all X E and Y K L, it follows that dist(η; K L )< Q + L for all η E, and hence dist( w(x); K L ) dx = dist( w(x); K L ) dx { w(x)/ {η 1,η 2 }} ɛ <(Q + L) L + Q = ɛ. This establishes the reduction principle and hence completes the proof. Corollary 4.2 Given any ξ M 4 3 with f (ξ )<1, there exists a sequence of Lipschitz maps U j ξ x + W 1, (; R 4 ) such that U j (x) K a.e.in ; U j ξ x in W 1, ( ; R 4). (4.3) Proof The general case under the reduction principle follows directly from Theorem 2.2 with ɛ = 1/j. The result as stated here can also be easily derived from Theorem 4.1 using a standard argument similar to the Vitali covering lemma; see the proof of Theorem 2.2 in the references cited above. Corollary 4.3 Let f, F be defined by (3.15), (3.16) above. Then F qc (ξ) = if and only if f (ξ) 1. Proof y (3.17), it is sufficient to show that F qc (ξ ) = if f (ξ ) 1. First, if f (ξ )<1, by Theorem 4.1, there exists a function U ξ x + W 1, (; R 4 ) such that U(x) K and thus F( U(x)) = a.e.. Hence, by the formula (2.2), F qc (ξ ) ; this proves F qc (ξ ) = if f (ξ )<1. Now let f (ξ ) = 1. Then, for all < t < 1, f (tξ ) = t 2 f (ξ )< 1 and hence F qc (tξ ) =. Letting t 1 and using the continuity of F qc,wehave F qc (ξ ) =. The following result, which is not needed in this paper, indicates that for certain special ξ the differential inclusion (4.1) may have some special solutions by other existence theorems. ( ) Proposition 4.4 If ξ = M 4 3 with δ( ) < 1, then there exists a function U ξ x + W 1, (; R 4 ) such that U(x) K = {( ) } M 4 3 : δ() =1 K. Proof We only need to find solutions of the form U = (,v).let w(x) = v(x) x. It suffices to solve for w W 1, (; R 3 ) with curl w + δ( ) =1. Let = { η R 3 : η + δ( ) =1 }. Since δ( ) < 1, it follows that is in the interior of the convex hull of. Hence, by a result in [3, Theorem 1.1] (see also [8, Theorem 6.2]), there exists a function w W 1, (; R 3 ) such that curl w(x) a.e. in. This completes the proof.

12 . Yan 5 Representation of divergence-free fields: Proof of Theorem 1.1 Although the following result seems standard, we could not find a direct proof elsewhere, especially on the specific estimate (b) given below. So we include it here and present an elementary proof for the convenience of the readers. Lemma 5.1 Let G L 2 (R 3 ; R 3 ) be such that div G = in the sense of distributions. Then there exists a function w Lloc 2 (R3 ; R 3 ) with w e 3 = w 3 = such that 1 (a) G = curl w; (b) sup R 1 R 3 w 2 dx C G 2 dx, (5.1) where Q R ={x R 3 : x i < R, i = 1, 2, 3} and C is an absolute constant independent of G and R. Proof Let G = (γ 1,γ 2,γ 3 ) and let G ɛ = G ρ ɛ be the standard smooth mollifiers of G. Then G ɛ C (R 3 ; R 3 ),divg ɛ = and G ɛ L 2 (R 3 ) G L 2 (R 3 ). For each c R,define w ɛ (x, c) = (p ɛ (x, c), q ɛ (x, c), ) on x R 3, where p ɛ (x, c) = x 3 c q ɛ (x, c) = Q R γ 2 ɛ (x 1, x 2, s) ds x 3 c x 2 R 3 γ 3 ɛ (x 1, t, c) dt, (5.2) γ 1 ɛ (x 1, x 2, s) ds. (5.3) Using div G ɛ =, it follows that curl w ɛ (x, c) = G ɛ (x) on R 3 for each c R. We first estimate q ɛ. For each c 1, by Hölder s inequality, q ɛ (x, c) 2 ( x 3 +1) γɛ 1 (x 1, x 2, s) 2 ds. R Integrating this inequality over the cube Q R, we obtain Q R R 3 q ɛ (x, c) 2 1 dx 2R(R + 1) γɛ dx c 1. (5.4) Next we write p ɛ (x, c) = g ɛ (x, c) f ɛ (x, c) with x = (x 1, x 2 ), where g ɛ (x, c) = x 3 c γ 2 ɛ (x 1, x 2, s) ds, f ɛ (x, c) = x 2 γ 3 ɛ (x 1, t, c) dt. (5.5) For g ɛ (x, c), we have the same estimate as q ɛ (x, c): Q R R 3 g ɛ (x, c) 2 2 dx 2R(R + 1) γɛ dx c 1. (5.6) For f ɛ (x, c), similarly, we estimate that f ɛ (x, c) 2 dx 2R 2 3 γɛ (x, c) 2 dx = 2R 2 H ɛ (c), (5.7) x i <R R 2

13 Equilibrium set of magnetostatic energy by differential inclusion where H ɛ (c) = γɛ 3(x, c) 2 dx ; hence H ɛ (c) dc = γɛ 3 2 <. Note that L 2 (R 3 ) R 2 R 1 {c R : H ɛ (c) >α} 1 α R H ɛ (c) dc = 1 α γ 3 ɛ 2 L 2 (R 3 ). Let E ɛ ={c [ 1, 1] : H ɛ (c) α }, where α = 2 γ 3 ɛ 2 L 2 (R 3 ). Then [ 1, 1]\E ɛ 2 and hence E ɛ 2 1 for all ɛ. Therefore, there exists a sequence ɛ k and a point c [ 1, 1] such that c k=1 E ɛ k ; that is, H ɛk (c ) 2 γɛ 3 2 for all k = 1, 2,... L 2 (R 3 ) Combining this with (5.7), we have f ɛk (x, c ) 2 dx dx 3 4R 3 H ɛk (c ) 8R 3 3 γɛ k (x) 2 dx, Q R R 3 which, together with (5.6), yields, for all R 1, p ɛk (x, c ) 2 dx cr 3 ( γ 2 ɛk (x) 2 + γɛ 3 k (x) 2) dx. (5.8) Q R R 3 Consider the sequence μ k (x) = w ɛk (x, c ). Then G ɛk = curl μ k and, for all R 1, by (5.4)and(5.8), μ k (x) 2 dx CR 3 G 2 L 2 (R 3 ). (5.9) Q R Using a diagonalization argument, we find a subsequence μ k j and a function w Lloc 2 (R3 ; R 3 ) such that μ k j wweakly as k j on all cubes Q R ={ x i < R}, R >. Clearly w e 3 = w 3 = andg = curl w in the sense of distributions on R 3. Moreover, by (5.9), 1 sup R 1 R 3 w(x) 2 dx C G(x) 2 dx. Q R This completes the proof. We now split Theorem 1.1 into two separate theorems. Theorem 5.2 Let M j lim R 3 MinL (; R 3 ) as j. Assume j Then M M H M 1 a.e. in. ( 1 M j 2 +2 M j H M j ) dx =. (5.1) Proof y the Helmholtz decomposition, M j χ = H M j + G j,whereg j L 2 (R 3 ; R 3 ) satisfies div G j = in the sense of distributions on R 3. The curl-free part H M j can be written as H M j = u j for a unique function u j Lloc 2 (R3 ) with u j dx = ; hence u j H 1 () with u j H 1 () C H M j L 2 (R 3 ) C M j χ L 2 (R 3 ) C M j L 1/2. (5.11)

14 . Yan For the divergence-free part G j, by Lemma 5.1, it follows that G j = curl w j for some w j L 2 loc (R3 ; R 3 ) satisfying the estimate (b) in (5.1). We now modify w j so that w j H 1 (; R 3 ).LetQ R, R 1, be a cube containing. We solve the Dirichlet problem: { ψ = div w j on Q R ψ Q R =. Since div w j H 1 (Q R ), this problem has a unique solution ψ j H 1 (Q R) satisfying ψ j L 2 (Q R ) w j L 2 (Q R ). Let ψ j = ψ j χ Q R. Then ψ j H 1 (R 3 ). Let v j = w j + ψ j. Then G j = curl v j and v j L 2 (Q R ) C w j L 2 (Q R ) CR 3/2 G j L 2 (R 3 ) CR 2 M j L 1/2. Moreover, v j solves the elliptic system: div v j = andcurlv j = G j on Q R (in the sense of distributions). Standard elliptic (interior) estimates imply that v j H 1 (; R 3 ), with the estimate v j H 1 () CR 2 M j L 1/2. (5.12) Let U j = (u j,v j ) H 1 (; R 4 ). Then M j = u j + curl v j, H M j = u j a.e. in and, hence, 1 M j 2 +2 M j H M j =F( U j ) a.e. in. Therefore, condition (5.1) implies that F( U j )dx as j. y (5.11, 5.12), let U j U = (u,v)weakly in H 1 (; R 4 ). Hence, by the lower semicontinuity theorem, F qc ( U(x)) dx lim inf F qc ( U j (x)) dx =. j Hence F qc ( U(x)) = a.e. in. y Corollary 4.3, f ( U(x)) 1 a.e. in. Finally, note that M j M in L (; R 3 ); hence M j χ Mχ weakly in L 2 (R 3 ; R 3 ) and H M j H M, G j G M weakly in L 2 ( R 3 ; R 3), where M = H M + G M with div G M =. Since H M j = u j u and G j = curl v j curl v weakly in L 2 (; R 3 ), it follows that u = H M and curl v = G M a.e. in. Hence f ( U(x)) = M M H M 1a.e. in. This completes the proof. The following result is stronger than the conclusion in Theorem 1.1. Theorem 5.3 Let M L (; R 3 ) satisfy M M H M 1a.e. in. Then there exists a sequence {M j } in L (; R 3 ) satisfying M j (x) =1 a.e. in such that M j H M j dx and M j MinL (; R 3 ) as j. Proof As above, let Mχ = u + curl v for some functions u,v on R 3 with U = (u,v) H 1 (; R 4 ). Moreover, H M = uon R 3. In this case, we have F qc ( U(x)) = a.e. on. Hence, by the relaxation theorem of [1] (seealso[7, Theorem 9.8]), there exists a sequence {U j } in U + H 1(; R4 ) such that U j U in L 2 (; R 4 ) and lim F( U j (x)) dx = F qc ( U(x)) dx =. j

15 Equilibrium set of magnetostatic energy by differential inclusion Let U j = (u j,v j ) on and define N j = u j + curl v j on. Then N j χ = ũ j + curl ṽ j on R 3, where (ũ j, ṽ j ) = (u j,v j )χ + (u,v)χ R 3 \. This shows that H N j = u j on ; so, F( U j ) = 1 N j 2 +2 N j H N j a.e. in. Hence, lim 1 N j 2 dx = ; lim N j H N j dx =. j j From this and U j U in L 2 (; R 4 ), it can be shown that N j M in L 2 (; R 3 ). Let M j (x) = N j (x)/ N j (x). Then M j L (; R 3 ) and M j N j 2 = 1 N j 2 1 N j 2. So M j N j and hence H M j H N j strongly in L 2 (; R 3 ). Therefore M j in L (; R 3 ). One also easily sees that lim M j H M j dx = lim N j H N j dx =. j This completes the proof. j M 6 Linear boundary traces on ellipsoids: Proof of Theorem 1.2 Note that a solution U H 1 (; R 4 ) to differential inclusion (1.9) can define a decomposition for a vector-field of the form Mχ on whole R 3 only when its boundary trace U = (φ, ψ) satisfies the following linear differential equation on the exterior domain c = R 3 \ : { u = curl v a.e. c (6.1) u(x) = φ(x), v(x) = ψ(x) on. Again admissible solutions are sought for in the spaces with u L 2 ( c ) and curl v L 2 ( c ). The differential equation in (6.1) implies that u is harmonic on c and is hence uniquely determined by its boundary trace φ H 1/2 ( ). Once u is known, the solution v is solved from equation curl v = uon c. Although there exist many solutions of v on c,the boundary traces of v from exterior domain c must satisfy certain conditions solely determined by φ; therefore, the boundary trace ψ given from the interior domain cannot be arbitrarily prescribed. In this section, we assume is an ellipsoid and look for solutions of (6.1) with linear data (φ, ψ). We then use the results obtained to prove Theorem 1.2. Assume is an ellipsoid in R 3 defined by { } 3 = x R 3 xi 2 /a i < 1, (6.2) i=1 where a i > are some constants. We follow some well-known methods in potential theory for ellipsoidal domains; see, e.g., [16,24].

16 . Yan For i = 1, 2, 3, define elliptic integral functions P i (r) by P i (r) = r (a i + t) 1 [(a 1 + t)(a 2 + t)(a 3 + t)] 1/2 dt, r. (6.3) Let β i = P i (). For simplicity, let Q(r) =[(a 1 + r)(a 2 + r)(a 3 + r)] 1/2. It is easy to verify that, for all r, { P1 (r) + P 2 (r) + P 3 (r) = 2Q(r), P i (r) = Q(r) a i +r, i = 1, 2, 3. (6.4) Let F(x, r) = 3 i=1 x 2 i a i + r 1. Given x c, there exists a unique r = r(x) such that F(x, r(x)) =. This function r(x) is known to be an ellipsoidal coordinate of x and is smooth on c and vanishes exactly on. Differentiating equation F(x, r(x)) = with respect to x i yields that 2 x i r xi =, with r = r(x), (6.5) G(x, r) a i + r for i = 1, 2, 3, where G(x, r) = 3 i=1 We define the following functions: x 2 i (a i +r) 2. w i (x) = P i (r(x))x i i = 1, 2, 3. (6.6) Then one can verify that w i satisfies w i L 2 ( c ) and solves w i = in c, w i (x) = β i x i on. (6.7) We now find a function v i : c R 3 satisfying curl v i = w i on c. One may try for functions v with v e i =. For example, for w 3, one solves for functions of the form v = ( f, g, ) to satisfy v = w 3. This amounts to solving g x3 = P 3 (r)r x 1 x 3, f x3 = P 3 (r)r x 2 x 3, (6.8) f x2 g x1 = P 3 (r) + P 3 (r)r x 3 x 3. (6.9) a y (6.5), r xi x 3 = r 3 +r x3 a i +r x i for i = 1, 2; so the first equation in (6.8) becomes g x3 = P 3 (r)a 3 + r a 1 + r r x 3 x 1 = (A(r)x 1 ) x3, where A (r) = P 3 (r) a 3+r a 1 +r. If A( ) =, it follows easily that A(r) = P 1(r); hence, we can choose g(x) = P 1 (r)x 1 = w 1 (x). Similarly, for the second equation in (6.8), we can choose f (x) = w 2 (x). With f = w 2, g = w 1, condition (6.9) above is equivalent to the identity 3 i=1 [P i (r(x)) + P i (r(x))r x i x i ]=, which can be easily established using (6.4) and(6.5). Therefore, in this way, we have obtained w 2 w 3 = w 1 on c. (6.1)

17 Equilibrium set of magnetostatic energy by differential inclusion In a completely similar fashion, we also obtain w 3 w 1 = w 3, w 2 = (6.11) w 2 w 1 on c. Therefore, if b = (b 1, b 2, b 3 ) R 3, then the unique admissible solution u in (6.1) with φ(x) = b x is given by u(x) = 3 i=1 b i β i w i (x) = 3 i=1 b i β i P i (r(x))x i (x c ). A corresponding solution v in (6.1) is thus given by v(x) = b 1 w 3 (x) + b w 3 (x) 2 + b w 2 (x) 3 w 1 (x) β 1 w 2 (x) β 2 w 1 (x) β 3 with linear boundary data given by v = ψ(x) = x, where β 2 β β 3 b 3 3 β 2 b 2 β = 1 β 3 b 3 β 3 β 1 b 1 ; hence δ() = β 1 β 2 b 2 β 2 β 1 b 1 β 2 +β 3 β 1 b 1 β 1 +β 3 β 2 b 2 β 1 +β 2 β 3 b 3 on c 2 y (6.4) with r =, it follows that β 1 + β 2 + β 3 = a1 = κ. Let λ a 2 a 3 i = β κ i 1 >. Then we can write δ() = b, where = diag(λ 1,λ 2,λ 3 ) is the diagonal matrix. We have thus proved the following result. Proposition 6.1 Let be the ellipsoid defined above. Given any b R 3, the problem { u = curl v a.e. c (6.12) u(x) = b x, v(x) = x on. has admissible solutions (u,v)if δ() = b, where = diag(λ 1,λ 2,λ 3 ) is the diagonal matrix defined above. Remark 6.1 (i) Note that 1 1+λ λ D = diag ( λ 1, 1+λ 3 = 1 and diagonal matrix ) 1 1, 1 + λ λ 3. (6.13) is the so-called demagnetizing matrix of the ellipsoid. Thismeansthat,forany constant vector M R 3, H M (x) = DM x ; (ii) see, e.g., [24] and the references therein. Let q : R 3 R be a function defined by q(m) = M M DM ( M R 3 ). (6.14) Then M 2 +2 M H M =q(m) on for all M R 3. Note that, unless the ellipsoid is a ball, in which case q(m) = M 2, the function q is not convex and the set {M R 3 : q(m) 1} is nonconvex.

18 . Yan We prove the following result before completing the proof of Theorem 1.2. Theorem 6.2 Let be the ellipsoid defined above. Then any constant M R 3 satisfying q(m) <1 is the weak-star limit of a sequence of equilibriums in E(). Proof Let M R 3 satisfy q(m) <1. Define a = DM R 3.ThenM = D 1 a = (I + )a = a + a (. Let ) M 3 3 be such that δ( ) = a.then H M (x) = DM = a a on.letξ =.Then f (ξ ) = a + δ( ) a δ( ) =q(m) <1. Hence, by Corollary 4.2, there exists a sequence of functions U + j = (u + j,v+ j ) ξ x +W 1, (; R 4 ) such that U + j (x) K a.e. in ; U + j ξ x in W 1, ( ; R 4). (6.15) Since δ( ) = a, by Proposition 6.1, there exists a function U = (u,v ) on c satisfying { u = curl v a.e. c u (x) = a x, v (6.16) (x) = x on. Define Ũ j = (ũ j, ṽ j ) = χ U + j +χ cu. Then Ũ j H 1 (R 3 ; R 4 ) and Ũ j χ (ξ x)+ χ cu in H 1 (R 3 ; R 4 ). Let M j (x) = ũ j (x) + curl ṽ j (x) on. Then M j χ = ũ j + curl ṽ j a.e. R 3. This implies that H M j = ũ j ; hence, H M j (x) = u + j (x) a.e. x. So (6.15) implies that M j E(). Clearly M j = u + j + curl v + j a + δ( ) = a + a = M in L (; R 3 ). This completes the proof. Finally, we restate and prove Theorem 1.2 in the following theorem. Theorem 6.3 Let be the ellipsoid defined above. Then any constant M R 3 satisfying M M H M 1 (6.17) is the weak-star limit of a sequence of equilibriums in E(). Proof Let M R 3 satisfy (6.17). The case q(m) = M M H M < 1 has already been proved in the previous theorem. So we only need to prove the case q(m) = 1. Let M ɛ = (1 ɛ)m with <ɛ<1. Then q(m ɛ ) = (1 ɛ) 2 < 1. y Theorem 6.2, for each <ɛ<1, there exists a sequence M ɛ j E() such that M ɛ j M ɛ in the weak-star topology of L (; R 3 ) as j. Since all these M ɛ j are in the unit ball of L (; R 3 ),by a well-known result in functional analysis, the weak-star topology restricted to the unit ball of L (; R 3 ) is metrizable, say, by a metric ρ (see, e.g., [6, Theorem 5.1, p. 134]). Hence the convergences M ɛ j M ɛ and M ɛ M can be written as ( ) lim ρ M ɛ j, M ɛ =, lim ρ(m ɛ, M) =. j ɛ + For each n = 1, 2,..., with ɛ n = 1/n, wecanfind j n such that ρ(m ɛ n jn, M ɛn )<1/n. Define M n = M ɛ n jn E(). Then, the sequence { M n } in E() satisfies that ρ( M n, M), which implies M n M in the weak-star topology of L (; R 3 ). This completes the proof.

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