# A Two-Dimensional Landau-Lifshitz Model in Studying Thin Film Micromagnetics

Authors: Jingna Li
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

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.


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## Body

## 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 [1]. They can be written as(1.1)∂m∂t=−α2m×(m×ℋ(m))+βm×ℋ(m),
where × is the vector
cross product in Rn(n≥2),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.ℋ(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. Here(1.2)E=E(m)=∫Ω|∇m|2dx+∫Ωϕ(m)dx+∫Rn|∇Φ|2dxis the free energy
functional, and it is composed of three parts:(i)
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;(ii)
Ean(m)=∫Ωϕ(m)dx is the
anisotropy energy. ϕ∈C∞(R3),ϕ≥0 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;(iii)
Efi(m)=∫Rn|∇Φ|2dx is the energy
of the stray field ∇Φ induced by m. By the magnetostatics theory(1.3)−ΔΦ=divmin𝒟'(Rn).Equation (1.1) has been widely studied. In the case β=0,α≠0, (1.1)
corresponds to the heat flow for harmonic maps studied in [2, 3]; if β≠0,α≠0 (which implies
strong damping in physics), the interested readers can refer to [2, 4–7] for mathematical theory; while in the conservative
case, that is, β≠0,α=0, (1.1) corresponds to Schrödinger flow which
represents conservation of angular momentum [8]. 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 [11]). The
physical consequences of the model are discussed further in [12]. The free energy functional
is E(m)=∫R2(|ξ⋅mχΩ^|2/|ξ|)dξ. We have δE/δm=−∇(−Δ)−1/2divm (see Section 2
for detailed computation) and the Landau-Lifshitz equation (β=0) becomes,(1.4)∂m∂t−∇(−Δ)−1/2divm+∇(−Δ)−1/2divm⋅mm=0,in which m=(m1,m2):𝕋2×[0,+∞)→R2 is in-plane
component of the magnetization, 𝕋2=R2/(2πZ)2 is a flat
torus. u⋅v 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,(1.5)∂mε∂t=εΔmε+∇(−Δ)−1/2divmε+ε|∇mε|2mε−∇(−Δ)−1/2divmε⋅mεmε.Equation (1.5) is the Landau-Lifshitz
equation corresponding to the free energy  E(m)=ε∫Ω|∇m|2dx+∫R2(|ξ⋅mχΩ^|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.
Section2 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 ε→0.


## 2. Approximation Equations

In this
section, we always suppose that𝕋2=R2/(2πZ)2 is the flat
torus. We prove existence of a weak solution of the following
equations:(2.1)∂mε∂t=εΔmε+∇(−Δ)−1/2divmε+ε|∇mε|2mε−∇(−Δ)−1/2divmε⋅mεmε,in𝕋2×(0,+∞),(2.2)mε(x,0)=m0(x),on𝕋2,(2.3)mε:𝕋2×(0,∞)→R2,|mε|=1a.e.in𝕋2.
Denote Lmε=−εΔmε−∇(−Δ)−1/2divmε.  Note that the
corresponding energy is E(m)=ε∫Ω|∇m|2dx+∫R2(|ξ⋅mχΩ^|2/|ξ|)dξ. The variation of the self-induced energy islimη→0∫R2|ξ⋅(mχΩ+ηv)^|2−|ξ⋅mχΩ^|2|ξ|ηdξ=∫R22iξ⋅mχΩ^|ξ|1/2iξ⋅v^¯|ξ|1/2dξ=2∫R2((−Δ)−1/4divmχΩ)((−Δ)−1/4divv)dx=2∫R2(−Δ)−1/2divmχΩdivvdx=2∫R2−∇(−Δ)−1/2divmχΩ⋅vdx. Equation (2.1) can be written as(2.4)∂mε∂t=−Lmε+(Lmε⋅mε)mε.It is very easy to prove that
(2.1) is equivalent to(2.5)mε×∂mε∂t+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 thatmε is a solution
of (2.1)–(2.3). By the vector cross product formula(2.6)a×(b×c)=(a⋅c)b−(a⋅b)c,we have(2.7)∂mε∂t=−Lmε+(Lmε⋅mε)mε=(Lmε⋅mε)mε−(mε⋅mε)Lmε=mε×(mε×Lmε).By the cross product of mε and (2.7), we
have(2.8)mε×∂mε∂t=mε×(mε×(mε×Lmε))=−mε×Lmε.This proves that mε satisfies
(2.5).
Suppose thatmε is a solution
of (2.5). Then by the cross product of mε and (2.5), we
obtain(2.9)mε×(mε×∂mε∂t)+mε×(mε×Lmε)=0.Since |mε|=1, we have mε⋅(∂mε/∂t)=0. Hence (2.9) implies(2.10)∂mε∂t=−Lmε+(Lmε⋅mε)mε.We define a local weak solution of (2.1) as
follows.Definition 2.2.
A vector-valued functionmε(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
(1)
mε∈L∞(0,T;H1(𝕋2)) and ∂mε/∂t∈L2(𝕋2×(0,T));(2)
|mε(x,t)|=1a.e.in𝕋2×(0,T);(3)
(2.1) holds in the sense of distribution;(4)
mε(x,0)=m0(x) in the trace
sense.We state our main result in this section
as follows.Theorem 2.3.
For everym0(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 [13], we first
construct weak solutions to a penalized problem, where the constraint |mε|=1 is
relaxed: (2.11)∂mk∂t+Lmk−k2(1−|mk|2)mk=0,in𝕋2×(0,+∞),(2.12)mk(x,0)=m0(x),on𝕋2,(2.13)|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:(2.14)∂m∂t=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 thex-variable,
(2.14) are turned into(2.15)m^t+ε|ξ|2m^+(ξ⊗ξ|ξ|)m^−k2m^=0,inR2×(0,+∞),m^(ξ,0)=m^0(ξ),onR2.For each fixed ξ, the problem has a unique solution(2.16)m^(ξ,t)=e−ℬ(ξ)t⋅e−𝒜(ξ)tm^0(ξ),where(2.17)𝒜(ξ)=1|ξ|(ξ12ξ1ξ2ξ1ξ2ξ22),ℬ(ξ)=(−k2+ε|ξ|200−k2+ε|ξ|2).So the problem has the solution(2.18)m(x,t)=14π2(e−ℬ(ξ)t)∨∗(e−𝒜(ξ)t)∨∗m0(x).Now the only problem left is to
find the inverse Fourier transforms of e−𝒜(ξ)t and e−ℬ(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,(2.19)𝒪(ξ)=1|ξ|(ξ2−ξ1ξ1ξ2).Now we begin to calculate the
inverse Fourier transform of e−𝒜(ξ)t(2.20)(e−𝒜(ξ)t)∨=12π∫R2eix⋅ξe−𝒜(ξ)tdξ=12π∫R2eix⋅ξ𝒪τ(ξ)𝒪(ξ)e−𝒜(ξ)t𝒪τ(ξ)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(𝒪(ξ)𝒜(ξ)𝒪τ(ξ))n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(000|ξ|)n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+∑n=1∞(−1)ntnn!(000|ξ|n))𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+(000e−|ξ|t−1))𝒪(ξ)dξ=12π∫R2eix⋅ξ(I+1|ξ|2(ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ=δ(x)I+12π∫R2eix⋅ξ1|ξ|2((ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ.Denote(2.21)Rij(x,t)=12π∫R2eix⋅ξ(e−|ξ|t−1)ξiξj|ξ|2dξ.By the property of the Fourier
transform, we have(2.22)Rij(x,t)=−∂xi∂xj{12π∫R2eix⋅ξ(e−|ξ|t−1)1|ξ|2dξ}.Denote (1/2π)∫R2eix⋅ξ(e−|ξ|t−1)(1/|ξ|2)dξ by I(x,t). Obviously, we have(2.23)I(x,0)=0,∂I(x,t)∂t=−12π∫R2eix⋅ξe−|ξ|t1|ξ|dξ,∂I(x,t)∂t|t=0=−12π∫R2eix⋅ξ1|ξ|dξ,∂2I(x,t)∂t2=12π∫R2eix⋅ξe−|ξ|tdξ.By [14, page 15-16], we know
that(2.24)I′′(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|.Hence(2.25)∂I(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,(2.26)I(x,t)=∫0t∂I(x,τ)∂τdτ=ln|x|−ln|t+|x|2+t2|.We continue to compute other
terms,(2.27)∂I(x,t)∂xi=xi|x|2−xit2+|x|2+tt2+|x|2,i=1,2,Rij(x,t)=−∂2I(x,t)∂xi∂xj=2xixj|x|4−2xixj+t(t2+|x|2)−1/2xixj(t2+|x|2+tt2+|x|2)2,in which i,j=1,2,i≠j(2.28)Rii(x,t)=−∂2I(x,t)∂xi2=2xi2|x|4−2xi2+t(t2+|x|2)−1/2xi2(t2+|x|2+tt2+|x|2)2−1|x|2+1t2+|x|2+tt2+|x|2,i=1,2.Hence we obtain(2.29)(e−𝒜(ξ)t)∨=(δ(x)+R11R12R21δ(x)+R22).By standard procedure, we can
get(2.30)(e−ℬ(ξ)t)∨=(W(x,t)00W(x,t)),where W(x,t)=(2εt)−1e−(|x|2/4εt)+k2t. Therefore,(2.31)(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
thatm0(x)∈(L2(R2))2, then there exists a solution m(x,t)∈(C([0,T];L2(R2)))2 of (2.14) and(2.32)limt→0∥m(x,t)−m0(x)∥L2(R2)=0.Proof.
From (2.21) and (2.30), we know W^R^ij∈L∞(R2)=ℳ22(R2), so Rij∗W∈L22(R2) and Rij∗W∗m0∈L2(R2).ℳ22 is a Hörmander space
(see [14], page 49-50).
Moreover,(2.33)∫R2|Rij∗W∗m0i|2dx=∫R2|W^|2|R^ij|2|m0i^|2dξ=∫{ξ∈R2||ξ|≤A}|W^|2|R^ij|2|m0i^|2dξ+∫{ξ∈R2||ξ|>A}|W^|2|R^ij|2|m0i^|2dξ=I+II.Notice that |W^||R^ij|=|e−|ξ|2t||(e−|ξ|t−1)ξiξj/|ξ|2|≤C.
For anyε>0, choosing A large enough
such that ∫{ξ∈R2||ξ|>A}|m0i^|2dξ<(ε/2C), we have II<ε/2.
For aboveε, there exists a t0>0 such that |R^ij|<(ε/∥W^∥L∞∥m0i^∥L2) as t<t0 and |ξ|≤A. Hence I<(ε/2), that is(2.34)∫R2|Rij∗W∗m0i|2dx→0,ast→0.By standard procedure (see
[14]), we can prove
that(2.35)limt→0∥W(⋅,t)4π2∗m0i−m0i∥L2=0.Therefore the proof is
completely finished.Remark 2.5.
Consider(2.36)∂m∂t=εΔm+∇(−Δ)−1/2divm+k2m,in𝕋2×(0,+∞),m(x,0)=m0(x),on𝕋2,where m0(x+2πn→)=m0(x),∀n→∈Z2 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(2.37)(m1m2)=((W˜+W*R11˜)*m01(x)+W*R12˜*m02(x)W*R21˜*m01(x)+(W˜+W*R22˜)*m02(x)),in which(2.38)W˜(x,t)=∑n→∈Z2W(x+2πn→,t),W∗R˜ij=∑n→∈Z2(W∗Rij)(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 [15]).
LetN(u):X→X 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 problem(2.39)∂u∂t=Lu+N(u),t∈[0,∞),u(x,0)=u0(x),has a
unique mild solution u on [0,T).
Moreover, if T<∞, then
limt→T∥u(t)∥=∞.Applying Theorem 2.6
to (2.11)–(2.13), we get the following theorem.Theorem 2.7.
For
everym0∈H1(𝕋2), there exists a unique mild solution mk of (2.11)–(2.13).Proof.
HereLmk=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
inequality(2.40)∥N(u)−N(v)∥H1≤C(∥u∥H12+∥v∥H12)∥u−v∥H1.Letting B(u,v,w)=k2uvw, we have(2.41)N(u)−N(v)=B(u−v,u,u)+B(v,u−v,u)+B(v,v,u−v).So it is sufficient to
prove(2.42)∥B(u,v,w)∥H1≤C(∥u∥H12+∥v∥H12)∥w∥H1.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 (2.1)–(2.3)

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 Theorem2.3.
Multiplying (2.11) with ∂mk/∂t, and integrating over 𝕋2×(0,T), we have(2.43)ε2∫𝕋2|∇mk|2dx+k24∫𝕋2(|mk|2−1)2dx+∫0T∫𝕋2|∂mk∂t|2dxdt≤∫𝕋2|(−Δ)−1/4divm0|22dx+ε2∫𝕋2|∇m0|2dx.We now take the limit as k goes to
infinite: from (2.43), we deduce that(2.44)mkisbounded inL∞(0,T;H1(𝕋2)),∂mk∂tisbounded inL2(0,T;L2(𝕋2)),|mk|2−1→0,inL∞(0,T;L2(𝕋2)).
Therefore, up to a subsequence,
we have(2.45)mk⇀mεinL∞(0,T;H1(𝕋2))weak∗,(2.46)∂mk∂t⇀∂mε∂tinL2(0,T;L2(𝕋2))weakly,(2.47)mk→mεinL2(0,T;L2(𝕋2))strongly,(2.48)|mk|2−1→0inL∞(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 holds(2.49)∫0T∫𝕋2(mk×∂mk∂t)⋅Φdxdt+∫0T∫𝕋2(mk×Lmk)⋅Φdxdt=0.From (2.45), (2.46), and (2.47),
as k goes to
infinite, we have(2.50)∫0T∫𝕋2mk×∂mk∂t⋅Φdxdt→∫0T∫𝕋2mε×∂mε∂t⋅Φdxdt,∫0T∫𝕋2mk×∇(−Δ)−1/2divmk⋅Φdxdt→∫0T∫𝕋2mε×∇(−Δ)−1/2divmε⋅Φdxdt,∫0T∫𝕋2mk×εΔmk⋅Φdxdt=−∫0T∫𝕋2mk×ε∇mk⋅∇Φdxdt→−∫0T∫𝕋2mε×ε∇mε⋅∇Φdxdt.Namely, (2.49) is convergent
to(2.51)∫0T∫𝕋2mε×∂mε∂t⋅Φdxdt+∫0T∫𝕋2mε×(−∇(−Δ)−1/2divmε)⋅Φdxdt−∫0T∫𝕋2mε×ε∇mε⋅∇Φdxdt=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.


## 2.1. The Penalized Equation

In the spirit
of [13], we first
construct weak solutions to a penalized problem, where the constraint |mε|=1 is
relaxed: (2.11)∂mk∂t+Lmk−k2(1−|mk|2)mk=0,in𝕋2×(0,+∞),(2.12)mk(x,0)=m0(x),on𝕋2,(2.13)|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:(2.14)∂m∂t=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 thex-variable,
(2.14) are turned into(2.15)m^t+ε|ξ|2m^+(ξ⊗ξ|ξ|)m^−k2m^=0,inR2×(0,+∞),m^(ξ,0)=m^0(ξ),onR2.For each fixed ξ, the problem has a unique solution(2.16)m^(ξ,t)=e−ℬ(ξ)t⋅e−𝒜(ξ)tm^0(ξ),where(2.17)𝒜(ξ)=1|ξ|(ξ12ξ1ξ2ξ1ξ2ξ22),ℬ(ξ)=(−k2+ε|ξ|200−k2+ε|ξ|2).So the problem has the solution(2.18)m(x,t)=14π2(e−ℬ(ξ)t)∨∗(e−𝒜(ξ)t)∨∗m0(x).Now the only problem left is to
find the inverse Fourier transforms of e−𝒜(ξ)t and e−ℬ(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,(2.19)𝒪(ξ)=1|ξ|(ξ2−ξ1ξ1ξ2).Now we begin to calculate the
inverse Fourier transform of e−𝒜(ξ)t(2.20)(e−𝒜(ξ)t)∨=12π∫R2eix⋅ξe−𝒜(ξ)tdξ=12π∫R2eix⋅ξ𝒪τ(ξ)𝒪(ξ)e−𝒜(ξ)t𝒪τ(ξ)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(𝒪(ξ)𝒜(ξ)𝒪τ(ξ))n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(000|ξ|)n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+∑n=1∞(−1)ntnn!(000|ξ|n))𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+(000e−|ξ|t−1))𝒪(ξ)dξ=12π∫R2eix⋅ξ(I+1|ξ|2(ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ=δ(x)I+12π∫R2eix⋅ξ1|ξ|2((ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ.Denote(2.21)Rij(x,t)=12π∫R2eix⋅ξ(e−|ξ|t−1)ξiξj|ξ|2dξ.By the property of the Fourier
transform, we have(2.22)Rij(x,t)=−∂xi∂xj{12π∫R2eix⋅ξ(e−|ξ|t−1)1|ξ|2dξ}.Denote (1/2π)∫R2eix⋅ξ(e−|ξ|t−1)(1/|ξ|2)dξ by I(x,t). Obviously, we have(2.23)I(x,0)=0,∂I(x,t)∂t=−12π∫R2eix⋅ξe−|ξ|t1|ξ|dξ,∂I(x,t)∂t|t=0=−12π∫R2eix⋅ξ1|ξ|dξ,∂2I(x,t)∂t2=12π∫R2eix⋅ξe−|ξ|tdξ.By [14, page 15-16], we know
that(2.24)I′′(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|.Hence(2.25)∂I(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,(2.26)I(x,t)=∫0t∂I(x,τ)∂τdτ=ln|x|−ln|t+|x|2+t2|.We continue to compute other
terms,(2.27)∂I(x,t)∂xi=xi|x|2−xit2+|x|2+tt2+|x|2,i=1,2,Rij(x,t)=−∂2I(x,t)∂xi∂xj=2xixj|x|4−2xixj+t(t2+|x|2)−1/2xixj(t2+|x|2+tt2+|x|2)2,in which i,j=1,2,i≠j(2.28)Rii(x,t)=−∂2I(x,t)∂xi2=2xi2|x|4−2xi2+t(t2+|x|2)−1/2xi2(t2+|x|2+tt2+|x|2)2−1|x|2+1t2+|x|2+tt2+|x|2,i=1,2.Hence we obtain(2.29)(e−𝒜(ξ)t)∨=(δ(x)+R11R12R21δ(x)+R22).By standard procedure, we can
get(2.30)(e−ℬ(ξ)t)∨=(W(x,t)00W(x,t)),where W(x,t)=(2εt)−1e−(|x|2/4εt)+k2t. Therefore,(2.31)(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
thatm0(x)∈(L2(R2))2, then there exists a solution m(x,t)∈(C([0,T];L2(R2)))2 of (2.14) and(2.32)limt→0∥m(x,t)−m0(x)∥L2(R2)=0.Proof.
From (2.21) and (2.30), we know W^R^ij∈L∞(R2)=ℳ22(R2), so Rij∗W∈L22(R2) and Rij∗W∗m0∈L2(R2).ℳ22 is a Hörmander space
(see [14], page 49-50).
Moreover,(2.33)∫R2|Rij∗W∗m0i|2dx=∫R2|W^|2|R^ij|2|m0i^|2dξ=∫{ξ∈R2||ξ|≤A}|W^|2|R^ij|2|m0i^|2dξ+∫{ξ∈R2||ξ|>A}|W^|2|R^ij|2|m0i^|2dξ=I+II.Notice that |W^||R^ij|=|e−|ξ|2t||(e−|ξ|t−1)ξiξj/|ξ|2|≤C.
For anyε>0, choosing A large enough
such that ∫{ξ∈R2||ξ|>A}|m0i^|2dξ<(ε/2C), we have II<ε/2.
For aboveε, there exists a t0>0 such that |R^ij|<(ε/∥W^∥L∞∥m0i^∥L2) as t<t0 and |ξ|≤A. Hence I<(ε/2), that is(2.34)∫R2|Rij∗W∗m0i|2dx→0,ast→0.By standard procedure (see
[14]), we can prove
that(2.35)limt→0∥W(⋅,t)4π2∗m0i−m0i∥L2=0.Therefore the proof is
completely finished.Remark 2.5.
Consider(2.36)∂m∂t=εΔm+∇(−Δ)−1/2divm+k2m,in𝕋2×(0,+∞),m(x,0)=m0(x),on𝕋2,where m0(x+2πn→)=m0(x),∀n→∈Z2 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(2.37)(m1m2)=((W˜+W*R11˜)*m01(x)+W*R12˜*m02(x)W*R21˜*m01(x)+(W˜+W*R22˜)*m02(x)),in which(2.38)W˜(x,t)=∑n→∈Z2W(x+2πn→,t),W∗R˜ij=∑n→∈Z2(W∗Rij)(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 [15]).
LetN(u):X→X 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 problem(2.39)∂u∂t=Lu+N(u),t∈[0,∞),u(x,0)=u0(x),has a
unique mild solution u on [0,T).
Moreover, if T<∞, then
limt→T∥u(t)∥=∞.Applying Theorem 2.6
to (2.11)–(2.13), we get the following theorem.Theorem 2.7.
For
everym0∈H1(𝕋2), there exists a unique mild solution mk of (2.11)–(2.13).Proof.
HereLmk=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
inequality(2.40)∥N(u)−N(v)∥H1≤C(∥u∥H12+∥v∥H12)∥u−v∥H1.Letting B(u,v,w)=k2uvw, we have(2.41)N(u)−N(v)=B(u−v,u,u)+B(v,u−v,u)+B(v,v,u−v).So it is sufficient to
prove(2.42)∥B(u,v,w)∥H1≤C(∥u∥H12+∥v∥H12)∥w∥H1.This last result is an easy
consequence of Sobolev embedding theorem. Therefore, Theorem 2.6 gives us the
desired result.


## 2.1.1. The Corresponding Linear Equation

First, we
consider the corresponding linear equation of (2.11)–(2.13) in the whole
space:(2.14)∂m∂t=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 thex-variable,
(2.14) are turned into(2.15)m^t+ε|ξ|2m^+(ξ⊗ξ|ξ|)m^−k2m^=0,inR2×(0,+∞),m^(ξ,0)=m^0(ξ),onR2.For each fixed ξ, the problem has a unique solution(2.16)m^(ξ,t)=e−ℬ(ξ)t⋅e−𝒜(ξ)tm^0(ξ),where(2.17)𝒜(ξ)=1|ξ|(ξ12ξ1ξ2ξ1ξ2ξ22),ℬ(ξ)=(−k2+ε|ξ|200−k2+ε|ξ|2).So the problem has the solution(2.18)m(x,t)=14π2(e−ℬ(ξ)t)∨∗(e−𝒜(ξ)t)∨∗m0(x).Now the only problem left is to
find the inverse Fourier transforms of e−𝒜(ξ)t and e−ℬ(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,(2.19)𝒪(ξ)=1|ξ|(ξ2−ξ1ξ1ξ2).Now we begin to calculate the
inverse Fourier transform of e−𝒜(ξ)t(2.20)(e−𝒜(ξ)t)∨=12π∫R2eix⋅ξe−𝒜(ξ)tdξ=12π∫R2eix⋅ξ𝒪τ(ξ)𝒪(ξ)e−𝒜(ξ)t𝒪τ(ξ)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(𝒪(ξ)𝒜(ξ)𝒪τ(ξ))n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(∑n=0∞(−1)ntnn!(000|ξ|)n)𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+∑n=1∞(−1)ntnn!(000|ξ|n))𝒪(ξ)dξ=12π∫R2eix⋅ξ𝒪τ(ξ)(I+(000e−|ξ|t−1))𝒪(ξ)dξ=12π∫R2eix⋅ξ(I+1|ξ|2(ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ=δ(x)I+12π∫R2eix⋅ξ1|ξ|2((ξ12ξ1ξ2ξ1ξ2ξ22)(e−|ξ|t−1))dξ.Denote(2.21)Rij(x,t)=12π∫R2eix⋅ξ(e−|ξ|t−1)ξiξj|ξ|2dξ.By the property of the Fourier
transform, we have(2.22)Rij(x,t)=−∂xi∂xj{12π∫R2eix⋅ξ(e−|ξ|t−1)1|ξ|2dξ}.Denote (1/2π)∫R2eix⋅ξ(e−|ξ|t−1)(1/|ξ|2)dξ by I(x,t). Obviously, we have(2.23)I(x,0)=0,∂I(x,t)∂t=−12π∫R2eix⋅ξe−|ξ|t1|ξ|dξ,∂I(x,t)∂t|t=0=−12π∫R2eix⋅ξ1|ξ|dξ,∂2I(x,t)∂t2=12π∫R2eix⋅ξe−|ξ|tdξ.By [14, page 15-16], we know
that(2.24)I′′(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|.Hence(2.25)∂I(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,(2.26)I(x,t)=∫0t∂I(x,τ)∂τdτ=ln|x|−ln|t+|x|2+t2|.We continue to compute other
terms,(2.27)∂I(x,t)∂xi=xi|x|2−xit2+|x|2+tt2+|x|2,i=1,2,Rij(x,t)=−∂2I(x,t)∂xi∂xj=2xixj|x|4−2xixj+t(t2+|x|2)−1/2xixj(t2+|x|2+tt2+|x|2)2,in which i,j=1,2,i≠j(2.28)Rii(x,t)=−∂2I(x,t)∂xi2=2xi2|x|4−2xi2+t(t2+|x|2)−1/2xi2(t2+|x|2+tt2+|x|2)2−1|x|2+1t2+|x|2+tt2+|x|2,i=1,2.Hence we obtain(2.29)(e−𝒜(ξ)t)∨=(δ(x)+R11R12R21δ(x)+R22).By standard procedure, we can
get(2.30)(e−ℬ(ξ)t)∨=(W(x,t)00W(x,t)),where W(x,t)=(2εt)−1e−(|x|2/4εt)+k2t. Therefore,(2.31)(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
thatm0(x)∈(L2(R2))2, then there exists a solution m(x,t)∈(C([0,T];L2(R2)))2 of (2.14) and(2.32)limt→0∥m(x,t)−m0(x)∥L2(R2)=0.Proof.
From (2.21) and (2.30), we know W^R^ij∈L∞(R2)=ℳ22(R2), so Rij∗W∈L22(R2) and Rij∗W∗m0∈L2(R2).ℳ22 is a Hörmander space
(see [14], page 49-50).
Moreover,(2.33)∫R2|Rij∗W∗m0i|2dx=∫R2|W^|2|R^ij|2|m0i^|2dξ=∫{ξ∈R2||ξ|≤A}|W^|2|R^ij|2|m0i^|2dξ+∫{ξ∈R2||ξ|>A}|W^|2|R^ij|2|m0i^|2dξ=I+II.Notice that |W^||R^ij|=|e−|ξ|2t||(e−|ξ|t−1)ξiξj/|ξ|2|≤C.
For anyε>0, choosing A large enough
such that ∫{ξ∈R2||ξ|>A}|m0i^|2dξ<(ε/2C), we have II<ε/2.
For aboveε, there exists a t0>0 such that |R^ij|<(ε/∥W^∥L∞∥m0i^∥L2) as t<t0 and |ξ|≤A. Hence I<(ε/2), that is(2.34)∫R2|Rij∗W∗m0i|2dx→0,ast→0.By standard procedure (see
[14]), we can prove
that(2.35)limt→0∥W(⋅,t)4π2∗m0i−m0i∥L2=0.Therefore the proof is
completely finished.Remark 2.5.
Consider(2.36)∂m∂t=εΔm+∇(−Δ)−1/2divm+k2m,in𝕋2×(0,+∞),m(x,0)=m0(x),on𝕋2,where m0(x+2πn→)=m0(x),∀n→∈Z2 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(2.37)(m1m2)=((W˜+W*R11˜)*m01(x)+W*R12˜*m02(x)W*R21˜*m01(x)+(W˜+W*R22˜)*m02(x)),in which(2.38)W˜(x,t)=∑n→∈Z2W(x+2πn→,t),W∗R˜ij=∑n→∈Z2(W∗Rij)(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 [15]).
LetN(u):X→X 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 problem(2.39)∂u∂t=Lu+N(u),t∈[0,∞),u(x,0)=u0(x),has a
unique mild solution u on [0,T).
Moreover, if T<∞, then
limt→T∥u(t)∥=∞.Applying Theorem 2.6
to (2.11)–(2.13), we get the following theorem.Theorem 2.7.
For
everym0∈H1(𝕋2), there exists a unique mild solution mk of (2.11)–(2.13).Proof.
HereLmk=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
inequality(2.40)∥N(u)−N(v)∥H1≤C(∥u∥H12+∥v∥H12)∥u−v∥H1.Letting B(u,v,w)=k2uvw, we have(2.41)N(u)−N(v)=B(u−v,u,u)+B(v,u−v,u)+B(v,v,u−v).So it is sufficient to
prove(2.42)∥B(u,v,w)∥H1≤C(∥u∥H12+∥v∥H12)∥w∥H1.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 (2.1)–(2.3)

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 Theorem2.3.
Multiplying (2.11) with ∂mk/∂t, and integrating over 𝕋2×(0,T), we have(2.43)ε2∫𝕋2|∇mk|2dx+k24∫𝕋2(|mk|2−1)2dx+∫0T∫𝕋2|∂mk∂t|2dxdt≤∫𝕋2|(−Δ)−1/4divm0|22dx+ε2∫𝕋2|∇m0|2dx.We now take the limit as k goes to
infinite: from (2.43), we deduce that(2.44)mkisbounded inL∞(0,T;H1(𝕋2)),∂mk∂tisbounded inL2(0,T;L2(𝕋2)),|mk|2−1→0,inL∞(0,T;L2(𝕋2)).
Therefore, up to a subsequence,
we have(2.45)mk⇀mεinL∞(0,T;H1(𝕋2))weak∗,(2.46)∂mk∂t⇀∂mε∂tinL2(0,T;L2(𝕋2))weakly,(2.47)mk→mεinL2(0,T;L2(𝕋2))strongly,(2.48)|mk|2−1→0inL∞(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 holds(2.49)∫0T∫𝕋2(mk×∂mk∂t)⋅Φdxdt+∫0T∫𝕋2(mk×Lmk)⋅Φdxdt=0.From (2.45), (2.46), and (2.47),
as k goes to
infinite, we have(2.50)∫0T∫𝕋2mk×∂mk∂t⋅Φdxdt→∫0T∫𝕋2mε×∂mε∂t⋅Φdxdt,∫0T∫𝕋2mk×∇(−Δ)−1/2divmk⋅Φdxdt→∫0T∫𝕋2mε×∇(−Δ)−1/2divmε⋅Φdxdt,∫0T∫𝕋2mk×εΔmk⋅Φdxdt=−∫0T∫𝕋2mk×ε∇mk⋅∇Φdxdt→−∫0T∫𝕋2mε×ε∇mε⋅∇Φdxdt.Namely, (2.49) is convergent
to(2.51)∫0T∫𝕋2mε×∂mε∂t⋅Φdxdt+∫0T∫𝕋2mε×(−∇(−Δ)−1/2divmε)⋅Φdxdt−∫0T∫𝕋2mε×ε∇mε⋅∇Φdxdt=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 (1.4)

From above
section, we know that for each fixedε>0, (2.1)–(2.3) admit weak solutions mε∈L∞(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 thatm0(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 have(3.1)ε2∫Ω|∇mk|2dx+∫0T∫Ω|∂mk∂t|2dxdt≤ε2∫Ω|∇m0|2dx.Passing to the limit as k→∞ and taking
(2.45), (2.46) into consideration, we have(3.2)ε2∫Ω|∇mε|2dx+∫0T∫Ω|∂mε∂t|2dxdt≤ε2∫Ω|∇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,(3.3)mε⇀minL∞(0,T;H1(𝕋2))weak∗,∂mε∂t⇀∂m∂tinL2(0,T;L2(𝕋2))weakly.
By
[16, Chapter 1, Theorem 5.1, pages 56–60], we
know that(3.4)mε→mstronglyinL2(𝕋2×(0,T)),a.e.in𝕋2×(0,T).Passing to the limit as ε goes to zero in
(2.51), we have,(3.5)∫0T∫𝕋2mε×∂mε∂t⋅Φdxdt→∫0T∫𝕋2m×∂m∂t⋅Φdxdt,∫0T∫𝕋2mε×∇(−Δ)−1/2divmε⋅Φdxdt→∫0T∫𝕋2m×∇(−Δ)−1/2divm⋅Φdxdt,∫0T∫𝕋2mε×ε∇mε⋅∇Φdxdt→0.That is to say, m is the weak
solution of(3.6)m×∂m∂t−m×∇(−Δ)−1/2divm=0.By an argument analogous to
Lemma 2.1, (3.6) is equivalent to (1.4).



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*Source: 603591-2009-03-24.xml*