AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation60359110.1155/2009/603591603591Research ArticleA Two-Dimensional Landau-Lifshitz Model in Studying Thin Film MicromagneticsLiJingnaAlikakosNicholasDepartment of MathematicsJinan UniversityGuangzhou 510632Chinajnu.edu.cn200924022009200926112008060220092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The paper is concerned with a two-dimensional Landau-Lifshitz equation which was first raised by A. DeSimone and F. Otto, and so fourth, when studying thin film micromagnetics. We get the existence of a local weak solution by approximating it with a higher-order equation. Penalty approximation and semigroup theory are employed to deal with the higher-order equation.

1. Introduction

Landau-Lifshitz equations are fundamental equations in the theory of ferromagnetism. They describe how the magnetization field inside ferromagnetic material evolves in time. The study of these equations is a very challenging mathematical problem, and is rewarded by the great amount of applications of magnetic devices, such as recording media, computer memory chips, and computer disks. The equations were first derived by Landau and Lifshitz on a phenomenological ground in . They can be written asmt=α2m×(m×(m))+βm×(m), where × is the vector cross product in Rn(n2),m=(m1,m2,,mn):Ω×[0,+)Rn is the magnetization and α2 is a Gilbert damping constant. The system (1.1) is implied by the conservation of energy and magnitude of m.(m)=δE/δm is the unconstrained first variation of the energy functional E(m). The magnitude of the magnetization is finite, that is, |m|2=i=1nmi2=1. HereE=E(m)=Ω|m|2dx+Ωϕ(m)dx+Rn|Φ|2dxis the free energy functional, and it is composed of three parts:

Eex(m)=Ω|m|2dx is the exchange energy. It tends to align m in the same direction and prevents m from being discontinuous in space;

Ean(m)=Ωϕ(m)dx is the anisotropy energy. ϕC(R3),ϕ0 depends on the crystal structure of the material. It arises from the fact that the material has some preferred magnetization direction, for example, if (1,0,0) is the preferred magnetization direction, ϕ(m)=m22+m32 for |m|=1;

Efi(m)=Rn|Φ|2dx is the energy of the stray field Φ induced by m. By the magnetostatics theoryΔΦ=divmin𝒟'(Rn).

Equation (1.1) has been widely studied. In the case β=0,α0, (1.1) corresponds to the heat flow for harmonic maps studied in [2, 3]; if β0,α0 (which implies strong damping in physics), the interested readers can refer to [2, 47] for mathematical theory; while in the conservative case, that is, β0,α=0, (1.1) corresponds to Schrödinger flow which represents conservation of angular momentum . The numerical treatment to the problem can be found in [9, 10].

Recently, the study of the theory of ferromagnetism, especially the theory on thin film, is one of the focuses for both physicists and mathematicians. In the asymptotic regime which is readily accessible experimentally, DeSimone and Otto, and so forth, deduced a thin film micromagnetics model in which self-induced energy is the leading term of the free energy functional (see ). The physical consequences of the model are discussed further in . The free energy functional is E(m)=R2(|ξmχΩ^|2/|ξ|)dξ. We have δE/δm=(Δ)1/2divm (see Section 2 for detailed computation) and the Landau-Lifshitz equation (β=0) becomes,mt(Δ)1/2divm+(Δ)1/2divmmm=0,in which m=(m1,m2):𝕋2×[0,+)R2 is in-plane component of the magnetization, 𝕋2=R2/(2πZ)2 is a flat torus. uv is the inner product. To the best knowledge of ours, this is the first time a new model has been raised. Equation (1.4) is not easy to deal with because of lower order of differential operator with respect to x-variable and its strong nonlinear term. Inspired by physical prototype of the problem, we approximate it by a second-order equation,mεt=εΔmε+(Δ)1/2divmε+ε|mε|2mε(Δ)1/2divmεmεmε.Equation (1.5) is the Landau-Lifshitz equation corresponding to the free energy E(m)=εΩ|m|2dx+R2(|ξmχΩ^|2/|ξ|)dξ, sum of exchange and self-induced energy. One difficulty in dealing with (1.5) lies in the nonconvex constraint |mε|=1, which is overcame by considering a penalty approximation mimicking treatment of harmonic maps. To get existence of a unique mild solution of the penalized equation, we first give the formal solution of the corresponding linear equation, which requires special tricks and techniques. In the convergence process, a compensated compactness principle is applied.

The rest of this paper is organized as follows. Section 2 is devoted to studying (1.5). More precisely, we first study the penalized equation. In order to do this, we consider the corresponding linear equation and get its formal solution and well-posedness, then we get the existence of a unique mild solution of the penalized equation using semigroup theory. Second, we get the existence of a weak solution of (1.5) by passing to the limit in the penalized equation. The key point in the convergence process relies on a compensated compactness principle. In Section 3, we get existence of weak solution of (1.4) in Theorem 3.1 by passing to the limit in (1.5) as ε0.

2. Approximation Equations

In this section, we always suppose that 𝕋2=R2/(2πZ)2 is the flat torus. We prove existence of a weak solution of the following equations:mεt=εΔmε+(Δ)1/2divmε+ε|mε|2mε(Δ)1/2divmεmεmε,in𝕋2×(0,+),mε(x,0)=m0(x),on𝕋2,mε:𝕋2×(0,)R2,|mε|=1 a.e.in𝕋2. Denote Lmε=εΔmε(Δ)1/2divmε. Note that the corresponding energy is E(m)=εΩ|m|2dx+R2(|ξmχΩ^|2/|ξ|)dξ. The variation of the self-induced energy islimη0R2|ξ(mχΩ+ηv)^|2|ξmχΩ^|2|ξ|ηdξ=R22iξmχΩ^|ξ|1/2iξv^¯|ξ|1/2dξ=2R2((Δ)1/4divmχΩ)((Δ)1/4divv)dx=2R2(Δ)1/2divmχΩdivvdx=2R2(Δ)1/2divmχΩvdx. Equation (2.1) can be written asmεt=Lmε+(Lmεmε)mε.It is very easy to prove that (2.1) is equivalent tomε×mεt+mε×Lmε=0.The equivalence follows from the following.

Lemma 2.1.

In the classical sense, mε is a solution of (2.1)–(2.3) if and only if mε is a solution of (2.5).

Proof.

Suppose that mε is a solution of (2.1)–(2.3). By the vector cross product formulaa×(b×c)=(ac)b(ab)c,we havemεt=Lmε+(Lmεmε)mε=(Lmεmε)mε(mεmε)Lmε=mε×(mε×Lmε).By the cross product of mε and (2.7), we havemε×mεt=mε×(mε×(mε×Lmε))=mε×Lmε.This proves that mε satisfies (2.5).

Suppose that mε is a solution of (2.5). Then by the cross product of mε and (2.5), we obtainmε×(mε×mεt)+mε×(mε×Lmε)=0.Since |mε|=1, we have mε(mε/t)=0. Hence (2.9) impliesmεt=Lmε+(Lmεmε)mε.

We define a local weak solution of (2.1) as follows.Definition 2.2.

A vector-valued function mε(x,t) is said to be a local weak solution of (2.1), if mε is defined a.e. in 𝕋2×(0,T) such that

mεL(0,T;H1(𝕋2)) and mε/tL2(𝕋2×(0,T));

|mε(x,t)|=1a.e.in𝕋2×(0,T);

(2.1) holds in the sense of distribution;

mε(x,0)=m0(x) in the trace sense.

We state our main result in this section as follows.Theorem 2.3.

For every m0(x)H1(𝕋2) and |m0(x)|=1, a.e. in 𝕋2, there exists a weak solution of (2.1)–(2.3).

To prove Theorem 2.3, we have to consider a penalized equation.

2.1. The Penalized Equation

In the spirit of , we first construct weak solutions to a penalized problem, where the constraint |mε|=1 is relaxed: mkt+Lmkk2(1|mk|2)mk=0,in𝕋2×(0,+),mk(x,0)=m0(x),on𝕋2,|m0(x)|=1,on𝕋2. Here mk:𝕋2×(0,)R2. In order to prove the existence of a mild solution of semilinear system (2.11)–(2.13), we consider the corresponding linear equation.

2.1.1. The Corresponding Linear Equation

First, we consider the corresponding linear equation of (2.11)–(2.13) in the whole space:mt=εΔm+(Δ)1/2divm+k2m,inR2×(0,+),m(x,0)=m0(x),onR2,where m0(x)=(m01(x),m02(x)). While dealing with linear equation (2.14), we just write m instead of mk unless there may be some confusion.

By Fourier transform in the x-variable, (2.14) are turned intom^t+ε|ξ|2m^+(ξξ|ξ|)m^k2m^=0,inR2×(0,+),m^(ξ,0)=m^0(ξ),onR2.For each fixed ξ, the problem has a unique solutionm^(ξ,t)=e(ξ)te𝒜(ξ)tm^0(ξ),where𝒜(ξ)=1|ξ|(ξ12ξ1ξ2ξ1ξ2ξ22),(ξ)=(k2+ε|ξ|200k2+ε|ξ|2).So the problem has the solutionm(x,t)=14π2(e(ξ)t)(e𝒜(ξ)t)m0(x).Now the only problem left is to find the inverse Fourier transforms of e𝒜(ξ)t and e(ξ)t. First, we need to find an orthogonal matrix 𝒪(ξ) such that 𝒪(ξ)𝒜(ξ)𝒪τ(ξ) is the Jordan normal form of 𝒜(ξ). In fact,𝒪(ξ)=1|ξ|(ξ2ξ1ξ1ξ2).Now we begin to calculate the inverse Fourier transform of e𝒜(ξ)t(e𝒜(ξ)t)=12πR2eixξe𝒜(ξ)tdξ=12πR2eixξ𝒪τ(ξ)𝒪(ξ)e𝒜(ξ)t𝒪τ(ξ)𝒪(ξ)dξ=12πR2eixξ𝒪τ(ξ)(n=0(1)ntnn!(𝒪(ξ)𝒜(ξ)𝒪τ(ξ))n)𝒪(ξ)dξ=12πR2eixξ𝒪τ(ξ)(n=0(1)ntnn!(000|ξ|)n)𝒪(ξ)dξ=12πR2eixξ𝒪τ(ξ)(I+n=1(1)ntnn!(000|ξ|n))𝒪(ξ)dξ=12πR2eixξ𝒪τ(ξ)(I+(000e|ξ|t1))𝒪(ξ)dξ=12πR2eixξ(I+1|ξ|2(ξ12ξ1ξ2ξ1ξ2ξ22)(e|ξ|t1))dξ=δ(x)I+12πR2eixξ1|ξ|2((ξ12ξ1ξ2ξ1ξ2ξ22)(e|ξ|t1))dξ.DenoteRij(x,t)=12πR2eixξ(e|ξ|t1)ξiξj|ξ|2dξ.By the property of the Fourier transform, we haveRij(x,t)=xixj{12πR2eixξ(e|ξ|t1)1|ξ|2dξ}.Denote (1/2π)R2eixξ(e|ξ|t1)(1/|ξ|2)dξ by I(x,t). Obviously, we haveI(x,0)=0,I(x,t)t=12πR2eixξe|ξ|t1|ξ|dξ,I(x,t)t|t=0=12πR2eixξ1|ξ|dξ,2I(x,t)t2=12πR2eixξe|ξ|tdξ.By [14, page 15-16], we know thatI(t)=t(t2+|x|2)3/2.In harmonic analysis, (2.24) is known as Poisson kernel.

Also by [14, page 107], we have I(0)=1/|x|.

HenceI(x,t)t=0tτ(τ2+|x|2)3/2dτ+I(0)=(t2+|x|2)1/2+|x|1|x|1=(t2+|x|2)1/2.Therefore,I(x,t)=0tI(x,τ)τdτ=ln|x|ln|t+|x|2+t2|.We continue to compute other terms,I(x,t)xi=xi|x|2xit2+|x|2+tt2+|x|2,i=1,2,Rij(x,t)=2I(x,t)xixj=2xixj|x|42xixj+t(t2+|x|2)1/2xixj(t2+|x|2+tt2+|x|2)2,in which i,j=1,2,ijRii(x,t)=2I(x,t)xi2=2xi2|x|42xi2+t(t2+|x|2)1/2xi2(t2+|x|2+tt2+|x|2)21|x|2+1t2+|x|2+tt2+|x|2,i=1,2.Hence we obtain(e𝒜(ξ)t)=(δ(x)+R11R12R21δ(x)+R22).By standard procedure, we can get(e(ξ)t)=(W(x,t)00W(x,t)),where W(x,t)=(2εt)1e(|x|2/4εt)+k2t. Therefore,(m1m2)=14π2(W+W*R11W*R12W*R21W+W*R22)*(m01(x)m02(x))=14π2((W+W*R11)*m01(x)+W*R12*m02(x)W*R21*m01(x)+(W+W*R22)*m02(x)).Theorem 2.4.

Suppose that m0(x)(L2(R2))2, then there exists a solution m(x,t)(C([0,T];L2(R2)))2 of (2.14) andlimt0m(x,t)m0(x)L2(R2)=0.

Proof.

From (2.21) and (2.30), we know W^R^ijL(R2)=22(R2), so RijWL22(R2) and RijWm0L2(R2).22 is a Hörmander space (see , page 49-50). Moreover,R2|RijWm0i|2dx=R2|W^|2|R^ij|2|m0i^|2dξ={ξR2||ξ|A}|W^|2|R^ij|2|m0i^|2dξ+{ξR2||ξ|>A}|W^|2|R^ij|2|m0i^|2dξ=I+II.Notice that |W^||R^ij|=|e|ξ|2t||(e|ξ|t1)ξiξj/|ξ|2|C.

For any ε>0, choosing A large enough such that {ξR2||ξ|>A}|m0i^|2dξ<(ε/2C), we have II<ε/2.

For above ε, there exists a t0>0 such that |R^ij|<(ε/W^Lm0i^L2) as t<t0 and |ξ|A. Hence I<(ε/2), that isR2|RijWm0i|2dx0,ast0.By standard procedure (see ), we can prove thatlimt0W(,t)4π2m0im0iL2=0.Therefore the proof is completely finished.

Remark 2.5.

Considermt=εΔm+(Δ)1/2divm+k2m,in𝕋2×(0,+),m(x,0)=m0(x),on𝕋2,where m0(x+2πn)=m0(x),nZ2 and 𝕋2 is a flat torus R2/(2πZ)2. By extending the equations periodically with respect to variable x to the whole space, and using Fourier transform, we obtain(m1m2)=((W˜+W*R11˜)*m01(x)+W*R12˜*m02(x)W*R21˜*m01(x)+(W˜+W*R22˜)*m02(x)),in whichW˜(x,t)=nZ2W(x+2πn,t),WR˜ij=nZ2(WRij)(x+2πn,t),i,j=1,2.

2.1.2. Existence of a Unique Mild Solution of the Penalized Equation

First, let us recall a classical theorem in the theory of semigroup.Theorem 2.6 (see [<xref ref-type="bibr" rid="B18">15</xref>]).

Let N(u):XX be locally Lipschitz continuous in u. If L is the infinitesimal generator of a C0 semigroup S(t) on X, then for every u0(x)X there is a T such that the initial value problemut=Lu+N(u),t[0,),u(x,0)=u0(x),has a unique mild solution u on [0,T). Moreover, if T<, then limtTu(t)=.

Applying Theorem 2.6 to (2.11)–(2.13), we get the following theorem.Theorem 2.7.

For every m0H1(𝕋2), there exists a unique mild solution mk of (2.11)–(2.13).

Proof.

Here Lmk=εΔmk+(Δ)1/2divmk+k2mk,N(mk)=k2|mk|2mk. By Theorem 2.4 and Remark 2.5, we know that L is the infinitesimal generator of a C0 semigroup on H1(𝕋2). Next, we want to check the inequalityN(u)N(v)H1C(uH12+vH12)uvH1.Letting B(u,v,w)=k2uvw, we haveN(u)N(v)=B(uv,u,u)+B(v,uv,u)+B(v,v,uv).So it is sufficient to proveB(u,v,w)H1C(uH12+vH12)wH1.This last result is an easy consequence of Sobolev embedding theorem. Therefore, Theorem 2.6 gives us the desired result.

2.2. Existence of Weak Solution of Approximate Equation (<xref ref-type="disp-formula" rid="eq2.1">2.1</xref>)–(<xref ref-type="disp-formula" rid="eq2.3">2.3</xref>)

In this section, we establish our main results about the approximate equations (2.1)–(2.3) by passing to the limit in the penalized equation (2.11) as k.

Proof of Theorem <xref ref-type="statement" rid="thm2.1">2.3</xref>.

Multiplying (2.11) with mk/t, and integrating over 𝕋2×(0,T), we haveε2𝕋2|mk|2dx+k24𝕋2(|mk|21)2dx+0T𝕋2|mkt|2dxdt𝕋2|(Δ)1/4divm0|22dx+ε2𝕋2|m0|2dx.We now take the limit as k goes to infinite: from (2.43), we deduce thatmkisbounded inL(0,T;H1(𝕋2)),mktisbounded inL2(0,T;L2(𝕋2)),|mk|210,inL(0,T;L2(𝕋2)). Therefore, up to a subsequence, we havemkmεinL(0,T;H1(𝕋2))weak,mktmεtinL2(0,T;L2(𝕋2))weakly,mkmεinL2(0,T;L2(𝕋2))strongly,|mk|210inL(0,T;L2(𝕋2))stronglyanda.e.in𝕋2×(0,T), and |mε|=1a.e.in𝕋2×(0,T).

In order to pass to the limit in (2.11), let Φ be in (C(𝕋2×(0,T)))3, and let the test function ψ=mk×Φ, there holds0T𝕋2(mk×mkt)Φdxdt+0T𝕋2(mk×Lmk)Φdxdt=0.From (2.45), (2.46), and (2.47), as k goes to infinite, we have0T𝕋2mk×mktΦdxdt0T𝕋2mε×mεtΦdxdt,0T𝕋2mk×(Δ)1/2divmkΦdxdt0T𝕋2mε×(Δ)1/2divmεΦdxdt,0T𝕋2mk×εΔmkΦdxdt=0T𝕋2mk×εmkΦdxdt0T𝕋2mε×εmεΦdxdt.Namely, (2.49) is convergent to0T𝕋2mε×mεtΦdxdt+0T𝕋2mε×((Δ)1/2divmε)Φdxdt0T𝕋2mε×εmεΦdxdt=0.Hence by Lemma 2.1, we know that (2.1)–(2.3) has a weak solution.

Remark 2.8.

From (2.43) and Theorem 2.6, we know that the unique mild solution of the penalized equation (2.11) globally exists.

3. Existence of Weak Solution of (<xref ref-type="disp-formula" rid="eq1.2">1.4</xref>)

From above section, we know that for each fixed ε>0, (2.1)–(2.3) admit weak solutions mεL(0,T;H1(𝕋2)). In this section, we will prove that there exists a subsequence of mε (still denoted by mε) strongly converging to m in L2(0,T;L2(𝕋2)), which is the weak solution of (1.4). More precisely, we state our main result of this section in the following theorem.Theorem 3.1.

Suppose that m0(x)H1(𝕋2),|m0(x)|=1, a.e. in 𝕋2, and divm0=0, there exists a weak solution m(x,t)L(0,T;H1(𝕋2)) and (m/t)L2(0,T;L2(𝕋2)) of (1.4).

Proof.

Form (2.43), we haveε2Ω|mk|2dx+0TΩ|mkt|2dxdtε2Ω|m0|2dx.Passing to the limit as k and taking (2.45), (2.46) into consideration, we haveε2Ω|mε|2dx+0TΩ|mεt|2dxdtε2Ω|m0|2dx. So we conclude that mε is bounded in L(0,T;H1(Ω)), and mε/t is bounded in L2(0,T;L2(Ω)).

Therefore, up to subsequence,mεminL(0,T;H1(𝕋2))weak,mεtmtinL2(0,T;L2(𝕋2))weakly. By [16, Chapter 1, Theorem 5.1, pages 56–60], we know thatmεmstronglyinL2(𝕋2×(0,T)),a.e.in𝕋2×(0,T).Passing to the limit as ε goes to zero in (2.51), we have,0T𝕋2mε×mεtΦdxdt0T𝕋2m×mtΦdxdt,0T𝕋2mε×(Δ)1/2divmεΦdxdt0T𝕋2m×(Δ)1/2divmΦdxdt,0T𝕋2mε×εmεΦdxdt0.That is to say, m is the weak solution ofm×mtm×(Δ)1/2divm=0.By an argument analogous to Lemma 2.1, (3.6) is equivalent to (1.4).

Acknowledgment

The project is supported by NNSFC (10171113) (10471156) and NSFGD (4009793).

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