Landau-Lifshitz equations

Frontiers of Research with the Chinese Academy of Sciences – Vol. 1

LANDAU–LIFSHITZ EQUATIONS Boling Guo Beijing Institute of Applied Physics and Computational Mathematics, China

Shijin Ding South China Normal University, China

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Library of Congress Cataloging-in-Publication Data Guo, Boling. Landau-Lifshitz equations / by Boling Guo & Shijin Ding. p. cm. -- (Frontiers of Research with the Chinese Academy of Sciences ; v. 1) Includes bibliographical references. ISBN-13: 978-981-277-875-8 (hardcover : alk. paper) ISBN-10: 981-277-875-6 (hardcover : alk. paper) 1. Differential equations, Partial--Numerical solutions. 2. Maxwell equations--Numerical solutions. 3. Geometry. 4. Mathematical physics. I. Ding, Shijin, 1959– II. Title. QC776.G86 2008 518'.64--dc22 2007044801

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Preface In studying the dispersive theory of magnetization of ferromagnets in 1935, Landau– Lifshitz [102] proposed the equations of ferromagnetic spin chain which are important magnetization equations, called Landau–Lifshitz equations now. Later on, such equations were also found in the condensed matter physics. In the 1960s, Soviet physicists A. Z. Akhiezer, V. G. Beryahltar, S. V. Peletninskii studied spin wave, the equations of ferromagnetic chain and the traveling wave solutions in detail in their book “Spin Waves”[3]. In 1974, K. Nakamura, T. Sasada [122] first observed that there is a soliton solution to the one-dimensional Landau–Lifshitz equations without Gilbert damping. Then, many mathematicians and physicists studied the soliton theory of Landau–Lifshitz equations using the approaches including inverse scattering method, infinite many conservation laws, geometry expression method and gauge equivalence of nonlinear Schr¨odinger equations and so on. Early in 1957, Suhl [131] had studied the infinite dimensional dynamic system of the Landau–Lifshitz equations with Gilbert damping term. A series of further studies on the theory of dynamics and numerical results have appeared since then. In recent years, the ferromagnetic materials have been widely applied in the video and recording apparatus. This is one of the applications for Landau–Lifshitz equations. From 1982, mathematicians began their studies on the well-posedness for Landau– Lifshitz equations. In China, a group headed by Yulin Zhou and Boling Guo proved the existence of the global weak solutions to the initial value problems and initial boundary value problems for Landau–Lifshitz equations from one dimension to multidimensions [150–157]. Alouges and Soyeur [4] proved similar results by penalty method in 1992. We also refer the readers to the result by P L. Sulem, C. Sulem and C. Bardos [132]. Since then, many other results on the global existence were obtained [20–22]. However, the regularity and the uniqueness were unsolved in the 1980s due to the complexity of Landau–Lifshitz equations. However, in 1991, Zhou, Guo and Tan [158] obtained the existence and uniqueness of global smooth solution to one-dimensional Landau–Lifshitz equations with or without Gilbert damping by using a mobile frame on S 2 and some fine a priori estimates. In 1993, Guo and Hong began the studies on two-dimensional Landau–Lifshitz equations. They established in [77] the relations between two-dimensional Landau– Lifshitz equations and harmonic maps and applied the approaches studying harmonic maps to get the global existence and uniqueness of partially regular weak solution. v

vi

Landau-Lifshitz Equations

This conclusion has been cited by many others up to now and gives rise to many successive works (see, for example, [47–49, 52, 89, 113, 115, 143]). Later on, in 1998, Chen, Ding and Guo [29] further proved that all the weak solutions with finite energy must be the Chen–Struwe solutions [34]. The uniqueness was also given. This says that the weak solution with finite energy is globally smooth with exception of finitely many singular points at most. From 1998 to 2001, Guo and Ding discussed many other Landau–Lifshitz equations such as inhomogeneous equations, unsaturated equations and compressible equations [50, 51, 75, 108]. From the beginning of the new century, more and more mathematicians are interested in the researches of Landau–Lifshitz equations. We refer the readers to the works by Guo, Su, Carbou and Harpes, et al. [20–23, 80–84, 89] on Landau– Lifshitz equations and Landau–Lifshitz–Maxwell equations. A natural question is the regularity of weak solutions to the higher dimensional Landau–Lifshitz equations. In this aspect, in 2004, Liu [109] proved that the “stationary” weak solutions of higher dimensional Landau–Lifshitz equations are partially regular. The Hausdorff dimensions and the Hausdorff measures of the singular set were estimated. These extend the results on harmonic map heat flow by Feldman [58] to Landau–Lifshitz equations. At the same time, Moser [115] obtained the similar results for lower dimensional Landau–Lifshitz equations by different methods. We know that the “stationary” conditions are hard to verify. So, in 2005, Melcher [113] proved the partial regularity for the weak solutions to the initial value problems of Landau–Lifshitz equations. However, as stated by Melcher, his method does not fit the other dimensional problems and, the partial regularity of weak solutions to the boundary value problems are still unsolved. This attracted the attention of Changyou Wang at the University of Kentucky. Wang [143], using the method of [142], proved the partial regularity for the weak solutions of the initial value problems and initial boundary value problems on threeand four-dimensional manifolds. However, all the results on the partial regularity only answered the questions on the singular set such as how many points there are in the set or how large the set is, provided that the singularity does exist. But, the existence of finite time singularity of weak solutions is not answered. For the harmonic map heat flow, the similar questions were answered by Chen and Ding [30] in 1990 (n ≥ 3) and by Chang, Ding and Ye [26] (n = 2) in 1992 (see also [39] and many others). Does the weak solution of Landau–Lifshitz equations really blow-up at finite time? Pistella and Valente [123] in 2002, Bartels, Ko and Prohl [13] in 2005, gave positive answers respectively by numerical analysis. In 2007, Ding and Wang [52] rigorously proved that in three and four dimensions, some Dirichlet problems and Neumann problems for Landau–Lifshitz equations indeed admit finite time blow-up solutions. Comparing with the similar proofs for harmonic map heat flows, we do not have the monotonicity inequality and the Bochner identity.

Preface

vii

Unfortunately, our method does not apply to two-dimensional problems and higher dimensional problems. We do not know the types of the singularities either. In recent years, there are many papers discussing the other problems such as domain wall, energy concentrations and vortices. We refer to [44–46, 98–100, 116, 117] and references therein. The aim of this book is to introduce the readers the key works of the group headed by Yulin Zhou and Boling Guo in China from 1980s. There is a bibliography comment at the end of every chapter to introduce other mathematicians’ works in this field and briefly state the development in recent years. However, it is not possible to include all the works and achievements throughout the world in such a short comment behind every chapter. The authors are deeply grateful to Professor Yulin Zhou for his support to this work. We also want to thank Professors Yongqian Han, Jianqing Chen for their help in the preparation of this book. This work was partially supported by the projects of Natural Science Foundation of China (Grant No. 10471050), the National 973 Program of China (Grant No. 2006CB805902), Natural Science Foundation of Guangdong Province (Grant No. 7005795) and University Special Research Fund for Ph.D. Program of China (Grant No. 20060574002). The Authors March 2007

Contents Preface

v

1 Spin Waves and Equations of Ferromagnetic Spin Chain 1.1 Physics Background for the Equations of Ferromagnetic Spin 1.2 A Simple Derivation of Landau–Lifshitz Equation . . . . . . 1.3 Equations for the Antiferromagnets . . . . . . . . . . . . . . 1.4 Spin Waves in Ferromagnets . . . . . . . . . . . . . . . . . . 1.5 Spin Waves in Antiferromagnets . . . . . . . . . . . . . . . . 1.6 Bibliography Comments . . . . . . . . . . . . . . . . . . . . 2 Integrability of Heisenberg Chain 2.1 Spin Waves and Solitary Waves . . . . . . . . . . . . . . . . 2.2 Geometric Representation for the Landau–Lifshitz Equations 2.3 Inhomogeneous Heisenberg Chain . . . . . . . . . . . . . . . 2.4 Spherical (Cylindrical) Symmetric Heisenberg Equations of Ferromagnetic Spin Chain . . . . . . . . . . . . . . . . . . . 2.5 Bibliography Comments . . . . . . . . . . . . . . . . . . . .

. . . . . .

1 1 3 10 12 22 34

. . . . . . . . . . . . . . .

35 35 40 43

. . . . . . . . . .

46 52

Chain . . . . . . . . . . . . . . . . . . . .

3 One-Dimensional Landau–Lifshitz Equations 3.1 Initial Boundary Value Problem of One-dimensional Ferromagnetic Spin Chain Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Nonlinear Initial-boundary Value Problem for the System of Ferromagnetic Spin Chain . . . . . . . . . . . . . . . . . . . . . . . 3.3 Smooth Solution for the Ferromagnetic Spin Chain Systems . . . . 3.4 Smooth Solution for the 1D Inhomogeneous Heisenberg Chain Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Measure-Valued Solution to the Strongly Degenerate Compressible Heisenberg Chain Equations . . . . . . . . . . . . . . . . . . . . . . 3.6 Bibliography Comments . . . . . . . . . . . . . . . . . . . . . . . .

53 .

53

. .

65 87

.

99

. 114 . 121

4 Landau–Lifshitz Equations and Harmonic Maps 123 4.1 Weak Solution to Multidimensional Ferromagnetic Spin Chain Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 ix

Landau-Lifshitz Equations

x

4.2 4.3 4.4 4.5 4.6 4.7 4.8

Landau–Lifshitz Equations on Riemannian Manifold and Harmonic Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalized L–L Systems and Harmonic Maps . . . . . . . . . . . . . Regularity of Weak Solutions to the Two-Dimensional Landau–Lifshitz Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ginzburg–Landau Approximation to Landau–Lifshitz Equations . . . Smooth Solution and Decay Estimates to the L–L System with Small Initial Data in Higher Dimensions . . . . . . . . . . . . . . . . Radial Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography Comments . . . . . . . . . . . . . . . . . . . . . . . . .

5 Landau–Lifshitz–Maxwell Equations 5.1 Global Weak Solution in Three Dimensions . . . . . . . . . . . . . 5.2 Global Smooth Solution in One or Two Dimensions with Small Initial Data . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Global Smooth Solution to One-Dimensional L–L–M with Large Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Global Weak Solution to L–L–M System on Riemannian Manifold 5.5 Partial Regularity for Stationary Solutions to L–L–M Equations . 5.6 Weak Solutions to Landau–Lifshitz–Maxwell Equations with Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Bibliography Comments . . . . . . . . . . . . . . . . . . . . . . .

172 181 202 217 239

241 . . 241 . . 255 . . 268 . . 281 . . 291 . . 309 . . 331

6 Long Time Behavior of Solutions to the System of Ferromagnetic Spin Chain 6.1 Existence and Stability of Steady State Solutions . . . . . . . . . . 6.2 Asymptotic Behavior of L–L Equations . . . . . . . . . . . . . . . . 6.3 Approximate Inertial Manifold for One-Dimensional L–L Equations 6.4 Attractor of Landau–Lifshitz Equations on Riemannian Manifold . 6.5 The Attractors for Landau–Lifshitz–Maxwell Equations . . . . . . . 6.6 Bibliography Comments . . . . . . . . . . . . . . . . . . . . . . . . Bibliography

137 158

. . . . . .

333 333 336 344 354 368 391 393

Chapter 1 Spin Waves and Equations of Ferromagnetic Spin Chain 1.1 1.1.1

Physics Background for the Equations of Ferromagnetic Spin Chain Motion Equations for Magnetization

Studying the dispersive theory of magnetization of ferromagnets, Landau–Lifshitz proposed the following motion equation of magnetization
×H
e − λ2 S
× (S
×H
e ),
t = λ1 S S

(1.1.1)


= (S1 , S2 , S3 ) is the vector of magnetization, H
e is the effective magnetic where S field applied to magnetic moment,
).
e = ∂ emag (S H
∂S

(1.1.2)


) denotes the density of the total magnetic energy, H
e is also related to Here emag (S Maxwell equations, λ1 , λ2 are constants, λ2 > 0. If a bounded multiply connected domain Ω ⊂ R3 is occupied by a ferromagnet under the constant temperature below Curie temperature, and if mechanics effects are not taking into account, one can consider the following magnetic energy functional: 1. Anisotropic energy Anisotropic energy reads as Ean =




ω


)dx, Φ(S

(1.1.3)

in which Φ : R3 → R+ is a convex function and depends on the crystal structure of the materials. Near the Curie temperature, taking first order approximation, one has
) := Φ(S

 lm

1

blm Sl Sm ,

(1.1.4)

2

Landau–Lifshitz Equations


) = k|S
|2 (sin θ)2 for exwhere {blm } is a symmetric, positively definite tensor. Φ(S
and the direction of ample for the uniaxial crystal, where θ is the angle between S “easy” magnetization, k is a positive constant. 2. Exchange energy The behavior of ferromagnets is that the quantum force makes the magnetic field of molecules arrange in order. the most important quantity is the exchange energy Eex

1 := alm 2 l,m


Ω


∂S
∂S dx, ∂xl ∂xm

(1.1.5)

in which {alm } is a symmetric, positive definite tensor.
3. The energy to the magnetic field H
is: The energy of magnetic field H
:= 1 EH (S) 8π


R3

H 2 dx,

(1.1.6)


and S
being given by Maxwell’s equations. Note that here the integral is where H
does not vanish outside the domain Ω. extended to the whole space R3 since H The total magnetic energy has the form
:= Ean (S
) + Eex (S
) + E H (S
) Emag (S)

(1.1.7)

and at equilibrium state Emag attains an absolute minimum. The magnetostatic Maxwell equation is:
+ 4π S)
= 0, ∇ · (H
= 0, ∇×H

in R3 , in R3 ,

(1.1.8) (1.1.9)

where ∇· = div, ∇× = curl, “·” denotes the inner product and “×” denotes the
satisfies the non-convex condition: vector product. S
|S(x)| = S0 ,

1.1.2

in Ω.

(1.1.10)

Landau–Lifshitz Equations

One model of dynamical system is the Landau–Lifshitz equation:
∂S
×H
e − λ2 S
× (S
×H
e ), = λ1 S ∂t

 ∂2S
,
e := − ∂Φ(S ) + alm +H H
∂x ∂x ∂S l m l,m

in Ω × (0, T ),

(1.1.11)

in Ω × (0, T ),

(1.1.12)

in which λ1 , λ2 are constants in physics, λ2 > 0. The first term on the right-hand side
around H
e , has of (1.1.11) which is not dissipative but resulted from the motion of S

Spin Waves and Equations of Ferromagnetic Spin Chain

3


according a constant angle; the second term expresses the ordered arrangement of S e
to H and it is due to the viscosity and then dissipative. The initial condition is as follows
(x, 0) = S
0 (x), S

x ∈ Ω.

(1.1.13)

It follows from (1.1.11) that
∂
2
∂ S = 0. |S| = 2S ∂t ∂t

(1.1.14)


0 (x) satisfies (1.1.10), one has |S
(x, t)| = S0 , x ∈ Ω, t ≥ 0; if S
×H
e = 0, Then, if S
× (S
×H
e ) is an orthogonal base which are on the tangential plane of
×H
e and S S
| = S0 . Hence, (1.1.11) is a dissipative nonlinear evolutionary equation the sphere |S
on the surface of sphere. of S
and S
solve the Maxwell equation The fields H

= ε ∂ E + 4π J,
∇×H c ∂t c
+ 4π S
),
= − 1 ∂ (µ0 H ∇×E c ∂t

in R3 × [0, T ],

(1.1.15)

in R3 × [0, T ].

(1.1.16)

It follows from Ohm law that:
+ f
), J
= σ(E (1.1.17)
represents the electric field, J
the density of current, σ the conductivity, ε where E magnetization rate of the electric medium, c the speed of light and f
is the given
+ 4π S.

:= µ0 H non-induction electric force. Usually there holds: B The initial data are

0 (x), E(x, 0) = E
0 = 0.

0 (x), ∇ · B B(x, 0) = B

1.2 1.2.1

(1.1.18) (1.1.19)

A Simple Derivation of Landau–Lifshitz Equation Magnetically Ordered Crystals

Many crystals have an ordered magnetic structure. This means that in the absence of an external magnetic field, the mean magnetic moment of at least one of the atoms in each unit of cell of the crystal is non-zero. In the simplest type of magnetically ordered crystals, i.e. ferromagnets such as Fe, Ni, Co and Dy, the mean magnetic moments of all the atoms have the same orientation provided that the temperature of the ferromagnet does not exceed a critical value, i.e. the Curie temperature. For this reason ferromagnets have spontaneous magnetic moments, i.e. non-zero macroscopic magnetic moments, even in the absence of an external magnetic field.

4

Landau–Lifshitz Equations

In antiferromagnets, these include carbonates, anhydrous sulphates, oxides and fluorides of transition metals Mn, Ni, Co and Fe, the mean atomic magnetic moments compensate each other within each unit cell (in zero external magnetic field). In other words, an antiferromagnet consists of a set of sublattices (called magnetic sublattices), each of which has a non-zero mean moment. This type of magnetic order occurs if the temperature of the antiferromagnets is less than a critical temperature, known as the Neel temperature. Finally, there is one further type of magnetically ordered crystal — that of the ferrites — which consists of a number of magnetic sublattices whose magnetic moments are uncompensated (in contrast to antiferromagnets); thus ferrites exhibit spontaneous magnetic moments. Examples of this type are compounds of transition metals such as the salts MnO.Fe2 O3 , 3Y2 O3 .5Fe2 O3 .

1.2.2

The Wave Function and Spin Operator for the System of Two Electrons

Let us consider a simple molecule model. Assume the molecule of Argon has two electrons and two protons between which there is no interaction with each other since the mass of the protons is much larger. The interaction of this system is as in Figure 1.2.1 in which a, b denote protons, 1, 2 represent electrons provided that there is Coulomb force between them.

Figure 1.2.1. Interaction of oxygen molecular system.

1. Wave function of electrons Consider two-body problem:  



h ¯2 2 e2 ∇1 − − ϕ(ra1 ) = E0 ϕ(ra1 ), 2m ra1

(1.2.1)

h ¯2 2 e2 ∇2 − − ϕ(rb2 ) = E0 ϕ(rb2 ), 2m rb2

(1.2.2)



where ϕ(ra1 ) is the wave function of electron “1”, ϕ(rb2 ) is the wave function of electron “2”, h ¯ is the Planck constant, m is the mass of the electron and e is the

Spin Waves and Equations of Ferromagnetic Spin Chain

5

ˆ=H ˆ0 + H ˆ 1 where: charge. The Hamiltonian for this system is H ¯2 e2 e2 ˆ0 = − h (∇21 + ∇22 ) − H − , 2m ra1 rb1 2 2 2 2 ˆ1 = e + e − e − e , H R r ra2 rb2

(1.2.3) (1.2.4)

ˆ = EΨ, we take ˆ 1 is a small perturbation. To find the eigenvalue E such that HΨ H single electron approximation: ϕ1 = ϕ(ra1 )ϕ(rb2 ),

ϕ2 = ϕ(ra2 )ϕ(rb1 ),

(1.2.5)

which is the wave function of the system. Let ψ = C 1 ϕ1 + C2 ϕ2 ,

(1.2.6)

and assume that ϕ1 , ϕ2 are normalized. It follows from the identical principle that electrons “1” and “2” are identical or are invariant after two exchanges. Thus we have


ˆ ϕ∗1 Hψdx




=

Eϕ∗1 ψdx




=E

ϕ∗1 (C1 ϕ1 + C2 ϕ2 )dx

= C1 E0 + C2 E0 ν 2 , where




2

ν =

(1.2.7)

ϕ∗ (rb2 )ϕ∗ (ra1 )ϕ(ra2 )ϕ(rb1 )dr1 dr2 .

The left-hand side of (1.2.7) is


ˆ ϕ∗1 Hψdx

with




= C1


ˆ 1 dx ϕ∗1 Hϕ




+ C2

ˆ 2 dx = C1 α11 + C2 α12 , ϕ∗1 Hϕ

(1.2.8)

ˆ 0 ϕ(ra1 )ϕ(rb2 )dr1 dr2 ϕ∗ (rb2 )ϕ∗ (ra1 )H

α11 =




e2 ϕ(ra1 )ϕ(rb2 )dr1 dr2 R  
2 2 2 e e e − + ϕ∗ (rb2 )ϕ∗ (ra1 ) − ϕ(ra1 )ϕ(rb2 )dr1 dr2 r ra2 rb1 +

ϕ∗ (rb2 )ϕ∗ (ra1 )

= 2E0 +

e2 + A(r), R

here


A(r) =



(1.2.9) 

e2 e2 e2 − ϕ (rb2 )ϕ (ra1 ) − ϕ(ra1 )ϕ(rb2 )dr1 dr2 , r ra2 rb1 ∗

∗

(1.2.10)

6

Landau–Lifshitz Equations

and


α12 =

ˆ 0 ϕ(ra )ϕ(rb )dr1 dr2 ϕ∗ (rb2 )ϕ∗ (ra1 )H 2 1

e2
∗ + ϕ (rb2 )ϕ∗ (ra2 )ϕ(ra2 )ϕ(rb1 )dr1 dr2 R  
e2 e2 e2 ∗ ∗ − + ϕ (rb2 )ϕ (ra1 ) − ϕ(rb1 )ϕ(rb2 )dr1 dr2 r ra2 rb1





e2 2 = 2E0 + ν + B(r), R

(1.2.11)

here


B(r) =





e2 e2 e2 ϕ∗ (rb2 )ϕ∗ (ra1 ) − ϕ(rb1 )ϕ(rb2 )dr1 dr2 . − r ra2 rb1

(1.2.12)

Hence we have from (1.2.7) and (1.2.8) that C1 α11 + C2 α12 = C1 E0 + C2 E0 ν 2 .

(1.2.13)

ˆ = E0 ϕ∗2 ψ that Similarly it follows from ϕ∗2 Hψ C1 α21 + C2 α22 = C1 E0 ν 2 + C2 E0 , where

   α11  α

22

= (2E0 + = 2E0 +

e2 )ν 2 R

e2 R

+ B(r),

(1.2.14)

(1.2.15)

+ A(r).

We have from (1.2.13) and (1.2.14) that C1 (E0 − α11 ) + C2 (E0 ν 2 − α12 ) = 0,

 

  e2 e2 2 C1 E0 − 2E0 + ν + B(r) + C2 E0 − 2E0 + + A(r) = 0. R R If C1 , C2 ≡ 0, one has



E0 − α11



E0 ν 2 − α21

E0 ν − α12

E0 − α22 2

=0

(1.2.16)

and the corresponding eigenfunctions are (1)

C1 = −C2 = Ca ,

ψa = Ca [ϕ(ra1 )ϕ(rb2 ) − ϕ(ra2 )ϕ(rb1 )]. (2)

(1.2.17)

C 1 = C2 = Cs ,

ψs = Cs [ϕ(ra1 )ϕ(rb2 ) + ϕ(ra2 )ϕ(rb1 )].

(1.2.18)

Spin Waves and Equations of Ferromagnetic Spin Chain

Since



7



ψa ψa∗ = ψs ψs∗ = 1, we have 1 Ca =  , 2(1 − γ 2 )

Ca = 

1 2(1 + γ 2 )

,

(1.2.19)




where

ϕ(ra1 )ϕ(rb1 )dr1 .

γ=

(1.2.20)

2. Spin wave function The wave function ψ for the system of two electrons can be written as a product of the space and the spin wave functions ψ(r1 σ1 , r2 σ2 ) = ϕ(r1 , r2 )χ(σ1 , σ2 ),

(1.2.21)

where σ1 , σ2 are the projections of the electron spins along a given axis. In accordance with the Pauli principle, the wave function ψ must be antisymmetric with respect to the simultaneous interchange of the coordinates and of the spin variables of electrons. This means that an antisymmetric space function will be associated with symmetric spin function, and conversely, a symmetric space function will be associated with an antisymmetric spin function. The function χ will be symmetric if the resultant spin S of the two electrons is equal to unity (S = 1) and antisymmetric if S = 0. Therefore the space wave function will be antisymmetric for S = 1 and symmetric for S = 0. We shall denote these wave functions by ϕa (for S = 1) and ϕs (for S = 0): ψa = Ca [ϕ(ra1 )ϕ(rb2 ) − ϕ(ra2 )ϕ(rb1 )],

S = 1.

(1.2.22)

ψs = Cs [ϕ(ra1 )ϕ(rb2 ) + ϕ(ra2 )ϕ(rb1 )],

S = 0.

(1.2.23)

The energies of the molecules in states corresponding to S = 1 and S = 0 are related to the functions ϕa and ϕs by


E↑↑ (r) = E↑↓ (r) =




ˆ a (r1 , r2 )dr1 dr2 , ϕa (r1 , r2 )Hϕ

(1.2.24)

ˆ 1 ϕs (r1 , r2 )dr1 dr2 ϕs (r1 , r2 )H

(1.2.25)

in which we have omitted the symbol indicating complex conjugate in the integrals, since ϕa and ϕs are real functions (the atoms are assumed to be in the ground states). Substituting the expressions of ψa and ψs into (1.2.24) and (1.2.25), we find that e2 A(r) − B(r) , E↑↑ (r) = 2E0 + + R 1 − γ2 e2 A(r) + B(r) E↑↓ (r) = 2E0 + + , R 1 + γ2

S = 1,

(1.2.26)

S = 0.

(1.2.27)

Choosing wave functions such that γ = 0, we have e2 + (A(r) − B(r)), R e2 E↑↓ (r) = 2E0 + + (A(r) + B(r)), R E↑↑ (r) = 2E0 +

S = 1,

(1.2.28)

S = 0.

(1.2.29)

8

Landau–Lifshitz Equations

Figure 1.2.2. Total energy of a system of two electrons.

As shown in Figure 1.2.2, the total energy of the system can be represented by 2
1 · S
2 ˆ = 2E0 + e + A − B − 2B S E=H R 2 ˆ0 + H ˆ ex , =H

(1.2.30)


1 · S
2 is called exchange Hamiltonian which was first obtained by ˆ ex = −2B S where H
2 are electron spin operators. We want to prove that (1.2.30)
1 and S Dirac in 1929. S is identical to (1.2.28) and (1.2.29). In fact,
2 + S
2 = 1S
2)
2 − 1 (S
1 · S S 1 2 2 2 3 1 = S(S + 1) − 2 4  =

  −3,

S=0

  1,

S = 1,

4

4

(1.2.31)


1 · S
2 therefore we get E↑↑ (S = 1) and E↑↓ (S = 0). We see that the term −2B S denote the multibody effect. For any operator Fˆ we can obtain the following motion equation: dFˆ ∂ Fˆ 1 ˆ ˆ = + [H, H], (1.2.32) dt ∂t h ˆ is the Hamiltonian operator, [H, ˆ Fˆ ] = H ˆ Fˆ − Fˆ H. ˆ For the spin operator S
1, where H we have
1
1 ∂S 1 ˆ
dS ˆ S
1] = + [H, S 1 ] = [H, dt ∂t

ih  h2 1 1 2 2
1 · S
2, S
1 (∇ + ∇2 ) − A − B − 2B S − = ih 2m 1 2 1
1 · S
1 · S
2, S
1 ] = −2B [S
2, S
1 ], = [−2B S ih ih
2, S
1 ] = [S
1 · S
2, S
1x ]
i + [S
1 · S
2, S
1y ]
j + [S
1 · S
2, S
1z ]
k,
1 · S [S

9

Spin Waves and Equations of Ferromagnetic Spin Chain

where
1 · S
2, S
1x ] = [S
1x · S
2x + S
1y · S
2y + S
1z · S
2z , S
1x ] [S
2y + S
1z · S
2z , S
1x ]
1y · S = [S
1x ]S
2y + [S
1z , S
1x ]S
2z
1y , S = [S
1y S
1z S
2z − S
2y ) = ih(S and the other terms can be derived in the similar manner. Hence we finally obtain

where

1.2.3


1 dS
1 × (−2B S
2) = S
1 × H
eff , =S dt

(1.2.33)


1 · S
2 = S
1 (−2B S
2) = S
1 · H
eff . ˆ e = −2B S H

(1.2.34)

Multi-electron Wave Function and Spin Operator

1. Equation for the isotropic ferromagnetic chain Now we consider the spin operator for multi-electron system, i.e. one-dimensional homogeneous Heisenberg model (see Figure 1.2.3): Assume that the ferromagnets are isotropic.
i · ˆ e = −2B S H

k 


i (S
i−1 + S
i+1 ).
j = −2B S S

(1.2.35)

j=1

Figure 1.2.3. One-dimensional homogeneous Heisenberg model.



i + ∂S i a +
i+1 = S S ∂x

i−1 = S
i − ∂S i a + S ∂x therefore


i ∂2S a2 , ∂x2
i ∂2S a2 , ∂x2 



2

i dS
i × H ˆe = S
i × (−2B) 2S
i + a2 ∂ S i =S dt ∂x2 2

i × ∂ Si .
i · S = −2Ba2 S (1.2.36) ∂x2
i → S
(x, t) and setting the total energy of the Passing to the continuous case: S system to be

2 1 ∂S dx, (1.2.37) H= 2 ∂x

10

Landau–Lifshitz Equations

we get the magnetization spin motion equation as follows:
∂S
× Heff , =S ∂t
eff = − ∂H . H ∂S

(1.2.38) (1.2.39)

For the nonhomogeneous isotropic Heisenberg chain, the nonhomogeneous isotropic Heisenberg exchange Hamiltonian is H = −J

N −1 


i+1 ,
i · S fi S

(1.2.40)

i=1


i is The motion equation of S
i ∂S
i × S
i+1 ) + Jfi−1 (S
i × S
i−1 ). = Jfi (S ∂t

(1.2.41)


(x, t), fi → f (x, t) and S
i , fi varies
i → S Considering the continuous situation: S
(x + a, t), slowly in the same lattices (with length a), taking Taylor expansions for S
f (x − a, t), we have from (1.2.41) that S (x, t) meets the motion equation
×S
xx ) + fx (x)(S
×S
x ),
t (x, t) = f (x)(S S

(1.2.42)

where the variable of times has the scaling factor Ja2 . 2. Anisotropic ferromagnetic chain equations Now let us consider the continuous anisotropic chain equation for the nonhomogeneous ferromagnets: 1 H= 2 It follows from this that


eff H

1.3.1


∂S ∂x

2



−

J1 S12

−

J2 S22

−

J3 S32

.


∂S
×H
eff , =S ∂t 

in which

1.3




(1.2.43)

(1.2.44)  



0   S1   J1 0    ∂H      =− =  0 J 2 0  ·  S2  .    
∂S     S3 0 0 J3

(1.2.45)

Equations for the Antiferromagnets Antiferromagnetic Moments and Magnetic Energy

Some ferromagnets exhibit the so-called antiferromagnetic property if placed under the temperature much less than the Curie temperature. These ferromagnets are called

Spin Waves and Equations of Ferromagnetic Spin Chain

11

Figure 1.3.1. Orientation of electron spin.

antiferromagnets. Their electron spins are in different directions and compensate with each other as shown in Figure 1.3.1. We can divide antiferromagnets into two systems: moment densities M1 (r, t) and M2 (r, t), and |M1 (r, t)| = |M2 (r, t)|. The energy of magnetic dipole interaction Wm and the energy WH of the ferromagnet in the external magnetic field are given by the following formulae −1 Wm = 2
WH = −


V

(M1 + M2 )H (m) d
r,

(1.3.1)

(e)

V

(M1 + M2 )H0 d
r,

(1.3.2)

where H (m) represents the magnetic field due to the magnetic moment of the atoms in the antiferromagnets and the integrals are evaluated over the volume of the ferromagnet.

1.3.2

Equations for the Antiferromagnets

The density of the exchange energy We of an antiferromagnet is therefore of the form 

We =

f (M1 , M2 , M21, M22 )  + αik

∂M1 ∂M2 , ∂xi ∂xk

1 ∂M1 ∂M1 ∂M2 ∂M2 + αik + 2 ∂xi ∂xk ∂xi ∂xk



(1.3.3)

 are where f is a symmetric function of the magnetic moments M1 , M2 and αik , αik tensors. The first term in this expression represents the exchange energy density of uniformly magnetized sublattices, and the second and the third terms represent the exchange energy density connected with the non-uniformity of the magnetic moments. At the same time, the second term describes the exchange interaction in each of the sublattices, and the third term the exchange interaction between the sublattices. At sufficiently low temperatures the squares of the magnetic moment densities are practically constant and the function f can be regarded as depending only on the single variable M1 · M2 . In the simplest model of an antiferromagnet it is assumed that (1.3.4) f (M1 , M2 , M21 , M22) = δM· M2 , δ > 0.

12

Landau–Lifshitz Equations

The total exchange energy of an antiferromagnet is therefore of the form


We =

V

we dr




=

V



1 ∂M1 ∂M1 ∂M2 ∂M2 δM · M2 + αik + 2 ∂xi ∂xk ∂xi ∂xk





 ∂M1 ∂M2 + αik dr, ∂xi ∂xk (1.3.5)

and the total exchange energy of an isotropic antiferromagnet is H=


V





1 ∂M1 ∂M1 ∂M2 ∂M2 d3 x αik + 2 ∂xi ∂xk ∂xi ∂xk

  + αik

∂M1 ∂M2 ∂xi ∂xk 

1 (e) + δM1 · M2 − (M1 + M2 )H (m) − (M1 + M2 )H0 , 2

(1.3.6)

In some simplified situations, we have H=


V

d3 x[k1 |∇M1 |2 + k12 |∇M2 |2 + k12 ∇M1 · ∇M2 ].

(1.3.7)

In this case the motion equation is of the form     

∂M1 = M1 × [2k1 ∇2 M1 + k12 ∇2 M2 ], ∂t   ∂M2   = M2 × [2k2 ∇2 M2 + k12 ∇2 M1 ]. ∂t

1.4 1.4.1

(1.3.8)

Spin Waves in Ferromagnets Equilibrium State of Ferromagnets

1. Equilibrium state conditions of ferromagnets Consider a uniaxial ferromagnet. Assume that
(r, t), M(r, t) = M0 (r, t) + m
(i) (r, t) = H
0(i) +
h(r, t), H

(1.4.1) (1.4.2)


0(i) , M0 is the equilibrium magnewhere m,

h are small derivations from M0 and H
0(i) denotes the magnetic field inside the ferromagnet. According to the tization, H equilibrium conditions one has

or


0(i) + β
n(M0 ·
n) − 2M0 f  (M2 ) = 0, H 0

(1.4.3)

ˆ · M0 − 2M0 f  (M2 ) = 0,
(e) n(M0 ·
n) − 4π N H 0 + β
0

(1.4.4)

13

Spin Waves and Equations of Ferromagnetic Spin Chain

ˆ =N ˆ (r) is the where β is a constant,
n is a unit vector along the anisotropy axis, N demagnetization tensor with elements dr  1 ∂2
. Nik (r) = 4π ∂xi ∂xk V |r − r  | When the ferromagnet is an ellipsoid, M = constant, and then 


V

(1.4.5) 


∞ x2 dr  ds y2 z2 1 − , = πabc − − |r − r  | Rs a2 + s b2 + s c2 + s 0

(1.4.6)



where Rs = (a2 + s)(−b2 + s)(c2 + s), a, b, c are the semi-axes of the ellipsoid, x, y, z are the projection of the radial vector
r in any point of the ellipsoid onto the main
are axses. The elements on the diagonal of the tensor N    N1        


∞ ds 1 = abc , 2 (a2 + s)Rs 0
∞ ds 1 N2 = abc , 2  2 (b + s)Rs 0    
∞   ds 1   .  N3 = abc 2 2 (c + s)Rs 0

(1.4.7)

It is clear that N1 + N2 + N3 = 1. If the ferromagnet occupies a sphere, then N1 = N2 = N3 = 1/3. If it occupies a cylinder (a = ∞, b = c), N1 = 0, N2 = N3 = 1/2. What we are concerned are the equilibrium states when (1.4.7) admits solution: N3 > N2 > N1 , if β > 0;

N3 < N2 < N1 , if β < 0,

see Figures 1.4.1 and 1.4.2. 2. The equation of motion for the magnetization The expression of effective magnetic field related to (1.4.1) and (1.4.2) is ∂2m
1 (i) − {M0 · H0 + β(M0 ·
n )2 }m
∂xi ∂xk M20 + β
n (m
·
n ) − 4M0 f  (M20 )(M0 · m).


H =
h + αik

(1.4.8)

Therefore the equation of motion for the magnetization in the case of small departures from the equilibrium value, which we shall refer to as the linearized equation of motion, will be of the form



1 ∂2m
∂m
(i) = g M0 ,
h + αik + β
n (m
·
n ) − {M0 · H0 + β(M0 ·
n )2 }m
. 2 ∂t ∂xi ∂xk M0 If the ferromagnet exhibits magnetic anisotropy of the “easy axis” type (β > 0) and (i) H0 is parallel to
n , the linearized effective magnetic field is given by 

(i) 

H ∂2m
H =
h − β + 0 m
+ αik + (β − 4M0 f  (M20 ))(m
·
n )n. M0 ∂xi ∂xk

(1.4.9)

14

Landau–Lifshitz Equations

Figure 1.4.1. Equilibrium states of a ferromagnet-1 N3 > N2 > N1 , if β > 0.

Spin Waves and Equations of Ferromagnetic Spin Chain

Figure 1.4.2. Equilibrium states of a ferromagnet-2 N3 < N2 < N1 , if β < 0.

15

16

Landau–Lifshitz Equations

If the ferromagnet exhibits magnetic anisotropy of the “easy plane” type (β < 0) and (i) H0 is perpendicular to
n , the linearized effective magnetic field is given by 

(i) 

∂2m
H
+ αik + β
n (m
·
n ) − 4M0f  (M20 )(M0 · m).
H =
h − 0 m M0 ∂xi ∂xk

(1.4.10)

The linearized equation of motion for the magnetization (1.4.8) must be augmented by the boundary conditions for the magnetization. The general boundary condition for the magnetization was formulated as

∂F

νk

= 0, ∂(∂M/∂xk ) S

(1.4.11)

where F is the energy density of the ferromagnet: 

∂Mi F M, ∂xk



=

1 ∂Mi ∂Mk ∂Mi + γik (M) αik,lm (M) 2 ∂xk ∂xm ∂xk 1 + βik Mi · Mk + f (M2 ), 2

(1.4.12)

where ν is the unit vector along the outward normal to the surface of the ferromagnet. Since we are interested in small departures of the magnetic moment from the equilibrium value and small magnetization gradients, we have 

∂m
i ∂F = αkj + γik,l m
l, ∂(∂Mi /∂xk ) ∂xj

γik,l =

∂γik ∂Ml





,

M=M0

and the boundary condition assumes the form 





∂mi

αkj + γik,l ml νk

= 0 . ∂xj S

(1.4.13)

The linearized equation of motion for the magnetization is then of the form

where

∂m
= g[M0 , H], ∂t

(1.4.14)

H =
h − β(z)m
+ α∆m.


(1.4.15)

Assuming that the function β(z) increases rapidly in the thin layer of thickness δ in which m
and
h are practically constant, we obtain after integrating this equation with respect to z between 0 and δ,




δ ∂m


∂m


= g M0 ,
h · δ − m
β(z)dz + α δ ∂t ∂z z=δ 0 

 

∂2m ∂2m

+α + δ , 2 ∂x ∂y 2

(1.4.16)

Spin Waves and Equations of Ferromagnetic Spin Chain

17

where we have taken into account the fact that

∂m


= 0. ∂z z=0

(1.4.17)

Assuming further that as δ → 0, the integral
δ 0

β(z)dz < ∞,

(1.4.18)

we obtain from the last equation the following effective boundary condition: d

∂m
− m|
z=0 = 0, ∂z

where d = δ 0

(1.4.19)

α . β(z)dz

(1.4.20)

In order to be able to neglect in (1.4.16) terms proportional to δ, it is clearly necessary that the frequency ω of the variation in the magnetization, the wavelength λ and the quantity β(0) must satisfy the conditions: 

ω gM0β(0), Then α

λ

α , β(0)

β(0) 1.

(1.4.21)

∂2m
− β(0)m
= 0, ∂z 2

it can readily be concluded that 

δ

α β(0)

or δ d. √ If the wavelength λ satisfies the inequalities δd λ d, the effective boundary condition assumes the form

∂m


= 0, ∂z z=0

√

δd λ d.

(1.4.22)

If on the other hand λ d, then m|
z=0 = 0,

λ d.

(1.4.23)

18

Landau–Lifshitz Equations

1.4.2

Spin Waves in Ferromagnets

1. Motion of spin waves in ferromagnets Now we consider the propagations of spin waves in ferromagnets. Using the Fourier representations


m(

k, ω) exp[i(
k ·
r − ωt)]d
kdω ,

m(r,
t) =
h(r, t) =




(1.4.24)


h(
k, ω) exp[i(
k ·
r − ωt)]d
kdω ,

we have from (1.4.8) that





(i)

M 0 · H0 (M0 ·
n )2 −iω m(

k, ω) = g M0 , h(
k, ω) − αij kikj + + β m
(
k, ω) M20 M20 

+ β
n (
n · m(

k, ω)) .

(1.4.25)

This equation gives the relationship between the Fourier components m(

k, ω) and h(
k, ω) which we shall write in the form mi (
k, ω) = χij (
k, ω)hj (
k, ω), 

where

(1.4.26)



 χxx χxy 0    



χij (
k, ω) =  χyx χyy 0  ,     0 0 0 χxx =

(1.4.27)

gM0Ω1 gM0Ω2 iωgM0 , χyy = , χxy = −χyx = , 2 2 Ω1 Ω2 − ω Ω1 Ω2 − ω Ω1 Ω2 − ω 2 

Ω1 = gM0



Ω2 = gM0

(i)



M 0 · H0 αij ki kj + + β cos2 ψ , 2 M0 (i)



m0 · H0 αij ki kj + + β cos 2ψ , M20

(1.4.28)

and ψ is the angle between the anisotropy axis
n and the vector M0 ; the z-axis lies along M0 and the x-axis lies in the plane containing the vector
n and M0 . The quantities χij (
k, ω) form a tensor called the high-frequency magnetic susceptibility tensor of the ferromagnet. If the ferromagnet exhibits anisotropy of the “easy axis” type (β > 0) and m0 ,
n , (i) H0 are parallel with each other, then Ω1 = Ω2 = Ω,

(1.4.29)

Spin Waves and Equations of Ferromagnetic Spin Chain

where



Ω = gM0

(i)

19



H αij ki kj + 0 + β . M0

(1.4.30)

The formulae corresponding to this case are valid also for ferromagnets with cubic symmetry. All that is required is to replace β by 2β  M20 , if the easy magnetization axis lies along the edge of the cube, and by 43 |β  |M20, if the easy magnetization axis lies along the space diagonal of the cube, where β  is the anisotropy constant. If the ferromagnet exhibits magnetic anisotropy of the “easy plane” type (β < 0) (i) and M0 ⊥
n , M0 is parallel to H0 ,      Ω   1       Ω2

= gM0 = gM0



(i) 



(i)

H αij ki kj + 0 , M0



H αij ki kj + 0 + |β| . M0

(1.4.31)

2. Dispersion of spin waves We must now establish the dependence of the frequency ω of the spin wave on its wave vector
k. This requires the use of both the equation of motion for the magnetic moment and the Maxwell equations. However, spin waves are low-frequency magnetic waves, so that the electric field can be neglected and the magnetic field may be assumed to be irrotational. In other words, spin waves can be treated in terms of the magnetostatic approximation, i.e. we can assume that m
(r, t) and
h(r, t) satisfy the equations    rot
h(r, t) = 0, (1.4.32)   div
h(r, t) = 4π div m(r,
t). Transforming to the Fourier components in these equations, we obtain    [
k,
h(
k, ω)]  
k

= 0,

·
h(
k, ω) = 4π
k · m(

k, ω).

(1.4.33)

From the first equation it follows that the magnetic field
h(
k, w) is parallel to the wave vector
k :
h(
k, ω) = −i
k · ϕ(
k, ω), where ϕ(
k, ω) is the Fourier component of the magnetic potential. Since m(

k, ω) = χ(
k, ω)
h(
k, ω), we can write the second equation in (1.4.33) in the form (k 2 + 4πki kj χij (
k, ω))ϕ(
k, ω) = 0 and hence

k 2 + 4πkikj χij (
k, ω) = 0.

(1.4.34)

20

Landau–Lifshitz Equations

This equation relates ω and the wave vector
k of the spin wave (it is called the dispersion relation). Using (1.4.27) we can reduce the dispersion relation (1.4.34) to the form 4πgM0Ω2 ky2 4πgM0Ω1 kx2 + =0 1+ Ω1 Ω2 − ω 2 k 2 k 2 Ω1 Ω2 − ω 2k 2 k 2 and hence ωs (
k) =



Ω1 Ω2 + 4πgM0(Ω1 cos2 ϕk + Ω2 sin2 ϕk ) sin2 θk ,

(1.4.35)

where θk and ϕk are the polar and azimuthal angles of the wave vector
k. We recall that the z-axis lies along the vector M0 and the x-axis lies in the plane containing M0 and
n . For wave vectors with αk 2 1, the equation for the frequency of the spin wave becomes much simpler: ωS (
k) = gM0αij ki kj .

(1.4.36)

In the isotropic case this formula assumes the form θC (ak)2 , ωS (
k) = h ¯

(1.4.37)

¯ gM0 α/a2 (θC is of the order of the Curie temperature). It follows that where θC = h for large wave vectors (αk 2 1), the spin wave frequency is proportional to the square of the wave vector.

1.4.3

Damping of Spin Waves

1. Expression of the damping of spin waves Now we consider the damping of spin waves. Damping is due to the interaction of spin waves with each other and also with lattice vibrations and conduction electrons. The phenomenological description can be obtained from the equation of motion for the magnetization, containing the relaxation term R: R=

1 1 H − [
n , [
n , H]], τ2 τ1

(1.4.38)

M0 where
n = |M and τ1 , τ2 are constants which have the dimensions of time, ( τ12 > 0, 0| 1 + τ12 > 0). Then the equation for M will be the form τ1

∂M 1 1 = g[M, H] + H − [
n , [
n , H]], ∂t τ2 τ1

(1.4.39)

the effective magnetic field is 

(i) 

H ∂2m
H =
h − β + 0 m
+ αik + (β − 4M0 f  (M20 ))(m
·
n )
n , (1.4.40) M0 ∂xi ∂xk

Spin Waves and Equations of Ferromagnetic Spin Chain

21 (i)

here we mainly focus on the uniaxial ferromagnet and assume that the field H0 is parallel to the easy magnetization axis. The quantity f  (M20 ) in the expression for H can readily be related to the static susceptibility of the ferromagnet, χ0zz = (i) ∂M0 /∂H0 . We have seen that in the state of equilibrium (i)

H0 + βM0 − 2M0 f  (M20 ) = 0 and hence 2 

(2M0 ) f

(M20 )

(1.4.41)

(i)

1 H = 0 − 0 . χzz M0

(1.4.42)

From these formulae we obtain the following expression for the high-frequency susceptibility tensor χ:   0   χxx χxy     χyx χyy (1.4.43) χ(
k, ω) =  0 ,  

in which χxx = χyy =

0

0

χzz

 

gM0 Ω − (iω/τ ) + (Ω/gM0τ 2 ) , Ω2 − (ω + iΩ/gM0τ )2

χzz =

χ0zz , 1 + χ0zz (αij ki kj − iωτ2 )

χxy = −χyx = 1 1 1 = + , τ τ1 τ2

iωgM0 , Ω2 − (ω + iΩ/gM0 τ )2 

Ω = gM0

(i)

(1.4.44) 

H αij ki kj + 0 + β . M0

We note that 1/gM0τ 1. We also note that the component χzz on the tensor χ(
k, ω) is not zero, whereas it does vanish when τ2 = ∞. Substituting (1.4.44) into (1.4.34) we obtain the equation for the frequency of the spin wave as a function of its wave vector. When R = 0, this equation has complex roots whose real part determines the frequencies of the spin waves and whose imaginary part determines the damping rate. 2. Damping rate Since Ω/gM0 τ |ω|, the damping rate γS (
k) is given by γS (
k) = If αk 2 1, then

1 (Ω + 2M0 sin2 θk ). gM0τ 

β2 µ0 M0 T gM0 γS (
k) = 24π θC θC

(1.4.45)

2

;

(1.4.46)

22

Landau–Lifshitz Equations

if, on the other hand αk 2 1, then 



θC T , γS (
k) ≈ (ak)3 h ¯ θC

h ¯ ωS (
k) T.

(1.4.47)

Substitute
M = M0 + m into (1.4.39), here m
is a small addition to the equilibrium magnetization M0 , which we shall assume that M0 = M0 (t) is independent of r. The effective magnetic field is given by, to within terms linear in m,




(e)

4π H H=− +β+ 0 3 M

0



(e)

4π H0 m
+ β− + 3 M



1 − 0 (m
·
n )
n , χzz 0

and the solution of (1.4.39) is of the form    mz

= mz0 exp(−γz t),

(1.4.48)

 m

x + imy = (mx0 + imy0 ) exp(−γ⊥ t) exp(iω0 t),

where





1 1 4π + 0 , γz = τ2 3 χzz



γ⊥ =

1 1 + τ1 τ2



(e) 

H β+ 0 , M0

(1.4.49)

and mz0 and mx0 + imy0 are the initial values of the longitudinal and transverse
from the (relative to M0 ) components of the deviation of the magnetization m equilibrium value M0 . The quantities 1 1 , (1.4.50) τz = , τ⊥ = γz γ⊥ are the relaxation times for the longitudinal and transverse components of the mag(i) netic moment. Since M0 is only slightly dependent on H0 , it follows that χ0zz 1 (e) and consequently γz ≈ 1/τ2 χ0zz . If β + H0 /M0 1, we have 1 , γz = τ2 χ0zz

1.5 1.5.1



γ⊥ =

1 1 + τ1 τ2



(e) 

H β+ 0 . M0

(1.4.51)

Spin Waves in Antiferromagnets Equilibrium States of Antiferromagnets

1. Motion equations of magnetic moments If we do not take dissipation of energy into account, the equation of motion for the magnetizations M1 (r, t) and M2 (r, t) of the two magnetic sublattices will

23

Spin Waves and Equations of Ferromagnetic Spin Chain

be of the form

     

∂M1 = g[M1, H1 ], ∂t  ∂M2    = g[M2, H2 ],  ∂t

(1.5.1)

where g is the gyromagnetic ratio and H1 and H2 are the effective magnetic fields acting on the moments M1 (r, t) and M2 (r, t) H1 =

δW , δM1

H2 =

δW , δM2

(1.5.2)

W is the energy of the antiferromagnets. Using the expression of W , we obtain       H1

∂F ∂F ∂ , + ∂M1 ∂xk ∂(∂M1 /∂xk )  ∂F ∂F ∂  (i)   , +  H2 = H − ∂M2 ∂xk ∂(∂M2 /∂xk ) = H (i) −

(1.5.3)

where H (i) is the magnetic field inside the antiferromagnets. We note that the equations of motion for the magnetizations are consistent with the conservation of energy and lead to the following expression for the energy flux density in a ferromagnet 

c
(m) ∂M1 ∂M1 ∂M2 ∂M2 [E, H ]i − αij + πi = 4π ∂xj ∂t ∂xj ∂t 

 − αij





∂M1 ∂M2 ∂M2 ∂M1 + . ∂xj ∂t ∂xj ∂t

(1.5.4)

2. Equilibrium states of antiferromagnets Consider the equilibrium values of the magnetizations M10 , M20 of the two sublattices about which the oscillations of the magnetizations M1 and M2 take place. To do this we must equate the effective magnetic fields to zero. The equilibrium state of the antiferromagnets corresponding to the direction of magnetic moments of sublattices for which the energy density of the antiferromagnets 1 w = δM1 · M2 − β[(M1 ·
n )2 + (M2 ·
n )2 ] 2 (e)  − β (M1 ·
n ) · (M2 ·
n ) − H0 (M1 + M2 ) reaches a minimum. We have neglected in this expression the energy wd which is responsible for the appearance of weak ferromagnetism and the energy ˆ (M1 + M2 ) 2π(M1 − M2 )N which depends on the shape of the body. Let us assume that the external magnetic field is zero. If at the same time, β − β  > 0, it is readily to verify that the minimum

24

Landau–Lifshitz Equations

of w is reached when the magnetic moments of the sublattices lie along the anisotropy axis and M10 +M20 = 0. Such antiferromagnets are said to have magnetic anisotropy of the “easy axis” type. When β − β  < 0, the minimum of w is reached when the magnetic moments of the sublattices are perpendicular to the anisotropy axis and M10 + M20 = 0. In this case it is said that the antiferromagnets has a magnetic anisotropy of “easy plane” type. Examples of the “easy axis” anisotropy are CuCl2 .2H2 O, Cr2 O3 and FeCO3 ; antiferromagnets with magnetic anisotropy of the “easy plane” type are hematite in its low temperature phase, the carbonates and fluorides of transition metals. The equilibrium directions of M10 and M20 are readily determined even in the (e) presence of an external magnetic field H0 . The results of the corresponding calculations are given in Figures 1.5.1 and 1.5.2.

1.5.2

Spin Waves in Antiferromagnets

1. Motion of spin waves in antiferromagnets Consider the spin waves in antiferromagnets. Substitute     M1 (r, t)   

= M10 + m
1 (r, t),


2 (r, t), M2 (r, t) = M20 + m       H (i) (r, t) = H0(i) +
h(r, t).

(1.5.5)


2 and
h are small deviations from the equilibrium values M10 , into (1.5.1) where m
1, m (i) M20 and H0 , and then linearize these equations. This results in a set of two linear
2 . We can now express the Fourier component of differential equations for m
1 and m
2 (
k, ω) and
h(
k, ω) the deviations of the magnetic moments by m
1 (
k, ω), m m(

k, ω) = m
1 (
k, ω) + m
2 (
k, ω) = χ(
k, ω)
h(
k, ω),

(1.5.6)

where χ(
k, ω) is a tensor which depends on
k and ω and on the quantities char(e) acterizing the equilibrium state of the antiferromagnets. If the  magnetic field H0 (e) is parallel to the anisotropy axis and H0 < H1 , H1 = M0 2δ(β − β  ), then the
2 (
k, ω) are of the form linearized equations for m
1 (
k, ω) and m      −iω m
1 (
k, ω)                  −iω m
2 (
k, ω)             





(e)



H = g M10 ,
h(
k, ω) − δ + 0 + β − β  + αij ki kj · m
1 (
k, ω) M0

− (δ +

 k i k j )m
2 (
k, ω) αij

,



(e)



H = g M20 ,
h(
k, ω) − δ + 0 + β − β  + αij ki kj · m
2 (
k, ω) M0

− (δ +

 k i k j )m
1 (
k, ω) αij

, (1.5.7)

Spin Waves and Equations of Ferromagnetic Spin Chain

Figure 1.5.1. Numerical results for the equilibrium states of an antiferromagnet (1).

25

26

Landau–Lifshitz Equations

Figure 1.5.2. Computational results for the equilibrium states of an antiferromagnet (2).

Spin Waves and Equations of Ferromagnetic Spin Chain 

χ( ˆ
k, ω) =



 χxx    χyx  

χxy 0  χyy

0

where

     χxx        



= χyy

0

  , 0  

(1.5.8)

0

(e)

(e)



Ω+ (Ω+ − gH0 ) Ω− (Ω− + gH0 )  1 = χ0  + , 2 Ω2+ − ω 2 Ω2− − ω 2 

(e)

(e)



(Ω+ − gH0 ) (Ω− + gH0 )  = iωχ0  + , Ω2+ − ω 2 Ω2− − ω 2

 χxy = −χyx             Ω = gM 2δ(α ±

27

0

ij

(1.5.9)

(e)

− αij
)ki kj + (H1 /M0 )2 ± gH0 ,

in which χ0 = δ −1 . Proceeding in an analogous fashion, we can determine the high-frequency magnetic (i) susceptibility tensor of an antiferromagnet when the magnetic field H0 is at right angles to the anisotropy axis: 

χ( ˆ
k, ω) =



 χxx    0  

0

where

  Ω22   χ = χ  xx 0   Ω22 − ω 2      (e) 2     χ = χ (gH0 ) zz

      Ω1          Ω2

0

Ω21 − ω 2

0

0   

, χyy χyz    χzy χzz

(1.5.10)

Ω21 , Ω21 − ω 2

,

χyy = χ0

,

χyz = −χzy = iωχ0

(e)



gH0 , Ω21 − ω 2

(1.5.11)

(e)

= gM0 2δ(αij − αij
)ki kj + (H1 /M0 )2 + (H0 /M0 )2 , 

= gM0 2δ(αij − αij
)ki kj + (H1 /M0 )2 . (e)

It is assumed that H0 H2 and H2 = 2δM0 (the z-axis lies along the anisotropy (e) axis and the x-axis lies along H0 ). Consider now an antiferromagnet with magnetic anisotropy of the “easy plane” (e) (e) type. If the field H0 is perpendicular to the anisotropy axis and H0 H2 , the high-frequency magnetic susceptibility tensor is given by 



 χxx 

 χ( ˆ
k, ω) =  0  

0

0

0   

, χyy χyz    χzy χzz

(1.5.12)

28

where

Landau–Lifshitz Equations

  Ω22    χ = χ ,  xx 0 2   Ω2 − ω 2       Ω12  χ = χ , 0

zz

      Ω1          

Ω12 − ω 2

(e)

χyy = χ0

(gH0 )2 , Ω12 − ω 2

χyz = −χzy



(e)

gH = −iωχ0  2 0 2 , Ω1 − ω

(1.5.13)

(e)

= gM0 2δ(αij − αij
)ki kj + (H0 /M0 )2 , 

Ω2 = gM0 2δ(αij − αij
)ki kj + (H1 /M0 )2

(the z-axis lies along the anisotropy axis and the x-axis lies along the magnetic field (e) H0 ). (e) If the magnetic field H0 is parallel to the anisotropy axis, then 



 χxx χxy 

0  





 χ( ˆ
k, ω) =  χyx χyy 

0

0



, 0   χzz

(1.5.14)

where  (e)  (gH0 )2    χxx = χ0  2 ,    Ω1 − ω 2       Ω2 2  χ = χ , zz

      Ω1          

0

Ω2 2 − ω 2 

χyy

Ω1 2 = χ0  2 , Ω1 − ω 2

χyz = −χzy = iωχ0

(e)

gH0 ,  2 Ω1 − ω 2

(1.5.15)

(e)

= gM0 2δ(αij − αij
)ki kj + (H1 /M0 )2 + (H0 /M0 )2 , 

Ω2 = gM0 2δ(αij − αij
)ki kj (1 − (H1 /M0 )2 ).

Here the z-axis lies along the anisotropy axis and the x-axis lies in the plane of the magnetic moments of the sublattices. 2. The spin-wave spectrum in antiferromagnets If we know the high-frequency magnetic susceptibility tensor of an antiferromagnet, we can readily find its spin-wave spectrum. To do this we must use the general dispersion relation (1.4.42) which determines the spin-wave spectrum in the magnetostatic approximation of both ferromagnets and antiferromagnets, k 2 + 4πkikj χij (
k, ω) = 0.

(1.5.16)

However, in the case of an antiferromagnet, this equation needs not be solved since the components of the tensor χij (
k, ω) for an antiferromagnet are proportional to a small parameter χ0 and therefore to within terms of the order of gM0χ0 , the spinwave frequencies must coincide with the poles of the tensor χ( ˆ
k, ω). Hence it follows,

Spin Waves and Equations of Ferromagnetic Spin Chain

29

for example, that in the case of antiferromagnets with magnetic anisotropy of the “easy axis” type, the spin-wave frequencies are given by    ωS1

 ≡ Ω+ = gM0 2δ(αij − αij )ki kj + (H1 /M0 )2 + gH0 ,

 ω

 ≡ Ω− = gM0 2δ(αij − αij )ki kj + (H1 /M0 )2 − gH0 ,

S2



(e)



(e)

(e)

(1.5.17)

(e)

if the field H0 is parallel to the anisotropy axis and H0 < H1 , and by    ωS1

 ≡ Ω1 = gM0 2δ(αij − αij )ki kj + (H1 /M0 )2 + (H0 /M0 )2 ,

 ω

 ≡ Ω2 = gM0 2δ(αij − αij )ki kj + (H1 /M0 )2 ,

S2



(e)



(1.5.18)

(e)

if the field H0 is perpendicular to the anisotropy axis. In the case of antiferromagnets with magnetic anisotropy of the “easy plane” type, the spin-wave frequencies are given by    ωS1

 ≡ Ω1 = gM0 2δ(αij − αij )kikj + (H1 /M0 )2 + (H0 /M0)2 ,

 ω

 ≡ Ω2 = gM0 2δ(αij − αij )kikj (1 − (H1 /M0 )2 ),

S2



(e)



(1.5.19)

(e)

if the magnetic field H0 is parallel to the anisotropy axis, and by 

ωS1 ≡

Ω1

(e)

 = gM0 2δ(αij − αij )ki kj + (H0 /M0 )2 ,



 )ki kj + (H1 /M0 )2 ωS2 ≡ Ω2 = gM0 2δ(αij − αij (e)

if the magnetic field H0 is perpendicular to the anisotropy axis.

1.5.3

Electromagnetic Waves in Magnetically Ordered Crystals

1. Dispersion relation for electromagnetic waves First of all we discuss the dispersion relation for electromagnetic waves. Assume that the magnetic moments have magnetostatic oscillations. This leads to m(

k, ω) = χ( ˆ
k, ω) ·
h(
k, ω),

(1.5.20)

where m(

k, ω) and
h(
k, ω) are the amplitudes of the oscillating components of the magnetization and the the magnetic field, and χ( ˆ
k, ω) is the high-frequency magnetic susceptibility tensor of a ferromagnet or an antiferromagnets. Maxwell’s equations for plane waves with allowance for (1.5.20) are of the form [k, e] =

ω b, c

ω [k, h] = − d, c

(1.5.21)

30

Landau–Lifshitz Equations

where b = µ ˆ
h and d = ˆe are the amplitudes of the oscillating components of the magnetic and electric induction, µ ˆ(
k, ω) = 1 + 4π χ( ˆ
k, ω) is the magnetic permeability tensor and ˆ is the permittivity tensor. We shall assume for simplicity that ij = δij and that  is independent of
k and ω. Equating the determinant of (1.5.21) to zero, we obtain the dispersion relation connecting the frequency and the wave vectors of the electromagnetic waves in magnetically ordered crystals D(
k, ω) = A(
k, ω)n4 + B(
k, ω)n2 + C(
k, ω),

(1.5.22)

√ where n = ck/ω  is the refractive index, 4π A(
k, ω) = 1 + 2 ki kj χij (
k, ω), k   k k i j
− δij ∆ij (
k, ω), B(k, ω) = k2 C(
k, ω) = detµij (
k, ω),

(1.5.23)

and ∆ij (
k, ω) are the minors of the determinant |µij (
k, ω)|. This dispersion relation will in general define not one but several frequencies for a given wave vector
k. For a ferromagnet there are three such frequencies, whereas for an antiferromagnet there are four. Different frequencies corresponding to the same
k define different branches of the oscillations. In fact, dividing (1.5.22) by n4 , and allowing n to tend to infinity, we obtain A(
k, ω) = 0, (1.5.24) which is the same as the dispersion relation 1+

4π ki kj χij (
k, ω) k2

(1.5.25)

for the spin waves. Since the spin wave frequency ωS (
k) is of the order of gM0, it may be said that spin waves correspond to wave vector
k satisfying the inequality k

gM0 c

λ

c . gM0

or in terms of wave length

(1.5.26)

0 , in addition to the spin wave there are also two proper electromagnetic When k gM c waves characterized by the dispersion relation

ck ω=√ . 

(1.5.27)

Our problem now is to determine the properties of the branches of electromagnetic 0 . oscillations for k gM c

31

Spin Waves and Equations of Ferromagnetic Spin Chain

0 Since √1α gM , it follows that in this region of wave vectors αk 1 and, consec quently, the spatial dispersion of the high-frequency magnetic susceptibility tensor is unimportant. This means that the coefficients A, B and C in the dispersion relation (1.5.22) will depend only on the frequency in the direction of the wave vector when 0 , but not on its magnitude. k gM c Under these conditions the dispersion relation can conveniently be looked upon as the equation for n or, what amounts to the same thing, as an equation for the modulus of the wave vector
k for given frequency and directions of propagation. The solution of this equation is of the form



n21,2

=

ck1,2 √ ω 

2

=

−B(κ, ω) ±



B 2 (κ, ω) − 4A(κ, ω)C(κ, ω) 2A(κ, ω)

,

(1.5.28)



where κ = kk . Real refractive indices ni correspond to waves propagating with phase velocity vi (κ, ω) = √

c , ni (κ, ω)

(1.5.29)

which is a function of κ and ω. 2. Interaction between proper electromagnetic waves and spin waves 0 requires Further study of the branches of electromagnetic oscillations for k ≤ gM c detailed knowledge of the high-frequency magnetic susceptibility tensor. Here we shall confine our attention to a uniaxial ferromagnet and will assume that the external magnetic field lies along the anisotropy axis. The tensor µ ˆ(
k, ω) is then of the form       

µ ˆ=

where

µ

−iµ       0

    µ(ω)         

iµ 

µ 0



   0   

0,     1

4πgM0Ω0 , Ω0 − ω 2 4πgM0ω µ (ω) = 1 + , Ω0 − ω 2            Ω0

(1.5.30)

=1+



= gM0

(1.5.31)

(i) 

H β+ 0 , M0

and the z-axis lies along the anisotropy axis. Using the unit vectors
k [
n ,
k] [
k, [
n ,
k]] J
1 = , J
2 = , J
3 = , k |[
n ,
k]| |[
k, [
n ,
k]]|

(1.5.32)

32

Landau–Lifshitz Equations

where
n is a unit vector along the anisotropy axis, we can write the vectors
b,
h and
e in the form   
b = b1 J
1 + b2 J
2 ,    


= h1 J
1 + h2 J
2 + h3 J
3 ,

h      

(1.5.33)


e = e1 J
1 + e2 J
2 .

Eliminating the vector
e from (1.5.21) we obtain 2
k(
k ·
h) − k 2
h = − ω 
b, c2

and hence

b1 = n2 h1 ,

b2 = n2 h2 .

(1.5.34)

(1.5.35)

Since
b = µ ˆ
h, it follows that     n2  



µ cos2 θk + (µ2 − µ2 ) sin2 θk µ cos θk − b1 − i b2 = 0, µ sin2 θk + cos2 θk µ sin2 θk + cos2 θk    cos θ µ µ k  2  b1 + n − b2 = 0, i µ sin2 θk + cos2 θk µ sin2 θk + cos2 θk

(1.5.36)

where θk is the angle between the wave vector
k and the anisotropy axis. If we eliminate the amplitudes b1 and b2 from these equations, we obtain the following values for the refractive index n21,2 =

1 (µ sin2 θk + cos2 θk )−1 {µ(1 + cos2 θk ) + (µ2 − µ2 ) sin2 θk 2  ±

(µ2 − µ2 − µ) sin4 θk + 4µ2 cos2 θk }.

(1.5.37)

These formulae are a special case of (1.4.79) for a uniaxial crystal. The value of b1 and b2 for waves with refractive index nj are related by (j)

b1

(j)

b2 where

= iρj ,

(1.5.38)

n2j cos2 θk − µ(1 − n2j sin2 θk ) ρj = . µ cos θk

(1.5.39)

ρ1 ρ2 = −1

(1.5.40)

We note that and consequently

(1)

b1

(1)

b2

(2)

=−

b2

(2)

b1

= iρ1 .

(1.5.41)

Spin Waves and Equations of Ferromagnetic Spin Chain

33

It follows from (1.5.36) that (j)

(j)

h1

= iρj ,

(j)

h2

h3

(j)

h2

=

µ sin θk + (µ − 1) sin θk cos θk ) , µ sin2 θk + cos2 θk

where as before the subscript j represents waves with refractive index nj . Let us now return to the expressions given by (1.5.37) for the refractive indices and find the values of ω for which the wave vector is zero. Since the right-hand sides of (1.5.37) have finite limits, which are equal to µ(0) and µ(0) , cos2 θk + µ(0) sin2 θk

(1.5.42)

it follows that k will be zero together with ω, and when ω gM0 , 

k1 = ω



µ(0) c

, k2 = ω

µ(0) c



1

cos2 θk + µ(0) sin2 θk

.

(1.5.43)

Moreover, the wave vector will vanish for a certain value of ω of the order of gM0 and, in particular, for (i)

ω = ω0 = g(H0 + 4πM0 + βM0).

(1.5.44)

3. Properties of the branches of electromagnetic oscillations in ferromagnets The above results can be used to obtain a schematic representation of the properties of the branches of electromagnetic oscillations in uniaxial ferromagnets. In Figure 1.5.3 spin waves correspond to the broken curve ω = ωS (
k), whilst the √ proper electromagnetic waves correspond to the broken curve ω = ck/ . This curve tends asymptotically to a part of branch I for k gM0/c, and the straight line is the common asymptote of branches II and III (also for k gM0 /c). It is readily shown √ that the deviation of these curves from the straight line ω = ck/  for k gM0/c is given by   ck 2πgM0 √  cos θk . (1.5.45) ωII,III = √ 1 ±  ck Consider now the polarization properties of the above branches of electromagnetic oscillations. It is readily shown that as k → 0, the induction vector
b lies along the vector J
1 for oscillations in branch I, and along the vector J
2 for oscillations in branch II. Branch III is characterized by elliptical polarization for k → 0, and 1 bIII 1 = . III b2 cos θk For large k gM0 /, branches II and II have circular polarizations. For large k the magnetic field for branches II and III is transverse and for branch I longitudinal. For small k, on the other hand, there are both transverse and longitudinal magnetic field components, and both are of the same order of magnitude.

34

Landau–Lifshitz Equations

Figure 1.5.3. Electromagnetic oscillating bifurcation property of a uni-axial ferromagnet.

1.6

Bibliography Comments

In this chapter, we give the physics backgrounds and the derivations for the Landau– Lifshitz equations. There are two different approaches to derive Landau–Lifshitz equations. One is to consider the macroscopic motion of the ferromagnets and evaluate the total energy, then derive Landau–Lifshitz equations from the Hamiltonian (see the original paper by Landau and Lifshitz [102], [140]). The other is to begin with the microscopic aspects. Applying the quantum-mechanical spin theory, one can make out the Hamiltonian for the Heisenberg chain, and then get the Landau–Lifshitz equations and at the same time the equations of ferromagnets. For the ferromagnetic equations for multi-media and the propagations and interactions of spin waves in ferromagnets, we refer to the book “Spin waves” by Akhiezer et al. [3] and references therein.

Chapter 2 Integrability of Heisenberg Chain This chapter is concerned with the integrability of the Heisenberg chain without Gilbert damping including spin waves, solitary waves, the geometry representation for the Landau–Lifshitz equations, nonhomogeneous Heisenberg chain and the spherical (cylindrical) symmetric Heisenberg chain equations.

2.1 2.1.1

Spin Waves and Solitary Waves Spin Waves

In this section, we shall consider the spin waves and solitary waves, the approximate solutions and the equivalence to Schr¨odinger equations of the following Landau– Lifshitz equation without Gilbert damping 2~ ~ ∂S ~ × ∂ S. =S ∂t ∂x2

(2.1.1)

Suppose that equation (2.1.1) admits a solution of the form ~ t) = A ~ cos α + {B ~ cos(kx − ωt) + C ~ sin(kx − ωt)} sin α, S(x,

(2.1.2)

~ B, ~ C) ~ forms a unit right-hand orthogonal system, α, k are any real numbers, where (A, 2 ω = k sin α. The solutions of such form are called spin waves. Their energy density ε and density flow j are constants of the form 1 ε = k 2 sin2 α, 2

j = k 3 sin2 α cos α = vε,

(2.1.3)

in which v = 2k cos α = dω dk is the group velocity of the wave. The solution (2.1.2) ~ t) = S(u) ~ of spin wave is in fact the form S(x, with u = x − ct. In fact, substituting this form into (2.1.1), we have ~0 = S ~ ×S ~ 00 , −cS 35

(2.1.4)

Landau–Lifshitz Equations

36

~0 × S ~ 0 = 0 one has where “ 0 ” denotes the differential with respect to u. Noting that S from (2.1.4) that d ~ ~0 ~ = 0, (S × S + cS) (2.1.5) du and integrating this equality one further has ~ ×S ~ 0 + cS ~=S ~0 , S

(2.1.6)

~0 is a constant vector. Taking inner product of (2.1.6) with S, ~ we have by where S ~ = 1 that: noting |S| ~ ·S ~0 = c. S (2.1.7) ~ be the unit vector along the direction S ~0 and construct the other two vectors Let A ~ C ~ such that they are orthogonal to A ~ and orthogonal with each other in the rightB, ~ hand system. S(u) can be written as ~ ~ cos α + {B ~ cos ϕ(u) + C ~ sin ϕ(u)} sin α, S(u) =A

(2.1.8)

~ and S ~0 , ϕ(u) is the angle between the where α is the constant angle between S ~ ~ ~ ~ Substituting (2.1.8) into projection of vector S onto (B, C)-plane and vector B. (2.1.4), we get c ϕ0 (u) = . (2.1.9) cos α This implies that ϕ(u) = kx − ωt, where k = c/cos α, ω = k 2 cos α. This is the spin wave (2.1.2).

2.1.2

Solitary Waves

In order to study equation (2.1.1) in detail, we introduce the polar coordinates as follows:    u = sin θ(x, t) cos ϕ(x, t), v = sin θ(x, t) sin ϕ(x, t), (2.1.10)   w = cos θ(x, t) ~ t) = (u(x, t), v(x, t), w(x, t)). It follows from (2.1.1) that where S(x,  dθ   = −2θx ϕx cos θ − ϕxx sin θ,

dt   ϕt sin θ = θxx − (ϕx )2 sin θ cos θ.

(2.1.11) (2.1.12)

The energy density is of the form

1 ε(x, t) = (θx2 + χ2 ), 2

χ = ϕx sin θ.

(2.1.13)

First we assume that θ and ϕ are both the functions of u = x − ct which corresponds to the case of spin wave. Then we assume that θ(x, t) be of the function of u but ϕ take the form ϕ(x, t) = ϕ(u) + Ωt. (2.1.14)

Integrability of heisenberg chain

37

Define x(u) = ϕ0 (u) sin θ(u).

(2.1.15)

It follows from (2.1.11) and (2.1.12) that (

x0 = θ 0 (c − χ cos θ),

(2.1.16)

θ 00 = −x(c − χ cos θ) + Ω sin θ,

(2.1.17)

where “ 0 ” denote the differential with respect to u. It follows from this equation that (θ 0 )2 + x2 + 2Ω cos θ = 2α,

(2.1.18)

where α is a quantity independent of u. Now the energy density (2.1.13) reads as follows: ε(u) = −Ω cos θ(u) + α. (2.1.19) If we define

(

Z(u) = cos θ(u), g(u) = ϕ0 (u) sin2 θ(u),

(2.1.20)

then it follows from (2.1.16) that y 0 = −cz 0 and Z 00 − cy + Ω(1 − Z 2 ) + From (2.1.18), we have

Z(Z 0 + y 2 ) = 0. 1 − Z2

(2.1.21)

(2.1.22)

Z 0 + y 2 = 2(α − ΩZ)(1 − Z 2 ).

(2.1.23)

Z 00 − cy − 3ΩZ 2 + 2αZ + Ω = 0.

(2.1.24)

And, (2.1.22) becomes

It is not difficult to get the general solutions to (2.1.21) and (2.1.22). The solitary solution is then of the form

where

  Z(u) = (tanh 1 cu)2 , 2

(2.1.25)

1 α = Ω = c2 . 2

(2.1.26)

dϕ c = . du 1 + tanh2 21 cu

(2.1.27)

 y(u) = c sech2 1 cu, 2

It follows from (2.1.20) that

Landau–Lifshitz Equations

38

Then we have by integrating this equation that ϕ = tan−1

"

#

1 1 tanh cu + cu. 2 2

(2.1.28)

It also follows from (2.1.14) and (2.1.26) that ϕ = tan−1

"

#

1 1 tanh c(x − ct) + cx, 2 2

(2.1.29)

and the energy density and the flow density are c2 , ε(u) = 2 cosh2 21 cu

2.1.3

j(u) = cε(u).

(2.1.30)

Approximate Solutions

~ t) = S(η), ~ Let S(x, η = xtα with α = − 12 , i.e. 1

η = xt− 2 .

(2.1.31)

~ 0 = −2S ~ ×S ~ 00 . ηS

(2.1.32)

Substituting this into (2.1.1) we have

For large η, the approximate solution to (2.1.32) is of the form ~ S(η) =

!

1 2q 1 2q sin η 2 , − cos η 2 , 1 + O(η −2 ), η 4 η 4

(2.1.33)

~ 0 (η)| = q. In this case, the energy density and the flow density are of the form with |S 1 ~0 |S (η)|2 , 2t 1 ~ ~ 0 ~ 00 j(x, t) = 3/2 S · (S × S ), t ε(x, t) =

or, from (2.1.32) j(x, t) =

η ~0 |S (η)|2 . 3/2 2t

(2.1.34)

(2.1.35)

~0 · S ~ 00 = 0, we get Since it follows from (2.1.32) that S ~ 0 (η)|2 = q 2 = constant. |S Hence ε(x, t) =

q2 , 2t

j(x, t) =

q2x x = ε(x, t). 2 2t t

(2.1.36)

Integrability of heisenberg chain

2.1.4

39

Equivalence to Nonlinear Schr¨ odinger Equations

1. Vector representation for the L–L equations without damping ~ t) by ~e1 (x, t) to rewrite equation (2.1.1) as Denote the unit vector S(x, ~e1t = ~e1 × ~e 00 .

(2.1.37)

In order to have an equation independent of the coordinate system, we let ~e1 be a tangential vector of a curve with curvature k and torsion τ : 1

k(x, t) = (~e10 · ~e10 ) 2 , −2

τ (x, t) = k ~e1 ·

(~e10

(2.1.38) × ~e100 ).

(2.1.39)

Now establish an orthogonal three-side-body ~e1 , ~e2 , ~e3 where ~e2 lies along the direction of ~e10 , ~e3 = ~e1 × ~e2 . For such vectors we have from the Serret–Frenet formula that   e10 = k~e2 , ~ ~e20 = −k~e1 + τ~e3 , (2.1.40)   0 ~e3 = −τ~e3 . It follows from these equations and (2.1.37) that  0 e1 = −kτ~e2 + k 0~e3 ,  ~     k 00   0

~e2 = kτ~e1 +

with compatible conditions

       e30 = −k 0~e1 + ~

d ˙ ∂ ˙ ~ej = ~ej , dt ∂x

k

−τ

τ2 −

2

!

~e3 ,

00

(2.1.41)

!

k ~e3 , k

j = 1, 2, 3.

(2.1.42)

These conditions lead to the following equations of k and τ :    kt = −2kx τ − kτx ,  ! k xx 2  τt =  −τ + kk 0 . 

k

(2.1.43)

x

Equation (2.1.43) is equivalent to the continuum equation ∂ε ∂j + = 0, ∂t ∂x

(2.1.44)

~ ~ · (S ~x × S ~xx ) and where ε = 21 | ∂∂xS |2 , j(x, t) = S

1 ε(x, t) = x2 , 2

j(x, t) = k 2 τ.

(2.1.45)

Landau–Lifshitz Equations

40

2. Equivalence to nonlinear Schr¨ odinger equations Using the complex change of variables ( Z

Ψ(x, t) = k(x, t) exp i

)

τ (x, t)dx ,

(2.1.46)

we may change system (2.1.43) into the nonlinear Schr¨odinger equations 1 iΨt + Ψxx + |Ψ|2 Ψ = 0 2

(2.1.47)

which admits solitary solution of the form: Ψ(x, t) = 4η exp{−4(ξ 2 − η 2 )t − 2iξx}sech[2η(x − x0 ) + 8ηξt],

(2.1.48)

where ζ = ξ + iη is the characteristic parameter, x0 is a constant. It follows from this that the energy density is 1 ε(x, t) = k 2 (x, t) = 8η 2 sech2 [2η(x − x0 ) + 8ηξt], 2

(2.1.49)

and the density of momentum is j(x, t) = k 2 (x, t)τ (x, t) = 32ξη 2 sech2 [2η(x − x0 ) + 8ηξt]. Take ξ = η, τ = −2ξ =

c 2

= const. to give "

#

c2 c ε(x, t) = sech2 (x − x0 − ct) . 2 2

2.2 2.2.1

(2.1.50)

(2.1.51)

Geometric Representation for the Landau–Lifshitz Equations Establishing the Natural Coordinate System

Consider the following Landau–Lifshitz equation: ~ ∂S ~ ×S ~xx + λ[S ~xx − (S ~ ·S ~xx )S], ~ =S ∂t ~ 2 (x, t) = 1, S ~ = (S1 , S2 , S3 ), S

(2.2.1) (2.2.2)

where λ ≥ 0 is the Gilbert damping constant. In order to get the geometric representation of this system, we let [~e1 , ~e2 , ~e3 ] be the natural coordinate consisting of the usual unit tangential vector, unit normal vector and unit conormal vector and ~e1t = ~e1 × ~e1xx + λ[~e1xx − (~e1 · ~e1xx )~e1 ].

(2.2.3)

Integrability of heisenberg chain

41

It follows from Serret–Frenet formula that ~eix = d~ × ~ei , d~ = k~e3 + τ~e1 ,

(2.2.4)

where k, τ are the curvature and torsion of the space curve with tangential vector ~e 1 , and ~eit = ω ~ × ~ei ,

ω ~ = ω1~e1 + ω2~e2 + ω3~e3 ,

(2.2.5)

in which #

"

λ kxx − τ 2 + (2kx τ + kτx ), ω1 = k k ω2 = −(kx + λkτ ),

(2.2.6) (2.2.7)

ω3 = −kτ + λkx .

(2.2.8)

It follows from (2.2.4) and (2.2.5) and the compatible condition (~eit )x = (~eix )t that kt = −2kx τ − kτx + λ(kxx − kτ 2 ), # " ! # " 1 2 kxx 2 2 + kkx + λ −τ (k τ )x + k τ , τt = k k2 x x

(2.2.9) (2.2.10)

or by the densities of energy and flow: "

#

ε2 j2 εt = −jx + λ εxx − x − , 2ε 2ε "

#

"

ε2 j2 2εx εxx − x + ε2 − jt = εxx − ε 2ε ε "

#

(2.2.11)

+ 2εj x

#

jεxx jε2x j3 jx ε x +λ − 2 − 2 + jxx − . ε 2ε 2ε ε

(2.2.12)

Hence the total energy of the ferromegnet is E(t) =

2.2.2

Z

+∞ −∞

1 ε(x, t)dx = 2

Z

+∞ −∞

Z ∂S ~ 2 1 +∞ 2 dx = k (x, t)dx. ∂x 2 −∞

Geometric Representation of Landau–Lifshitz Equation

It follows from (2.2.9)–(2.2.12) that Z +∞ dE(t) d Z +∞ 1 2 = k dx = −λ [(kx )2 + k 2 τ 2 ]dx ≤ 0. dt dt −∞ 2 −∞

(2.2.13)

Landau–Lifshitz Equations

42

Making complex change for the unknowns " Z

Ψ(x, t) = k(x, t) exp i

+∞

τ (x, t)dx

−∞

" Z

= [2E(x, t)]1/2 exp i

+∞ −∞

# #

j(x, t) dx , 2ε(x, t)

(2.2.14)

we can rewrite (2.2.9) and (2.2.10) into the Schr¨odinger with damping 1 iΨt + (1 − iλ)Ψxx + |Ψ|2 Ψ 2 Z +∞ iλ + Ψ (ΨΨ∗x − Ψ∗ Ψx )dx = 0. 2 −∞

(2.2.15)

It follows from (2.2.15) and its complex conjugate that i

d(ΨΨ∗ ) + (1 − iλ)Ψ∗ Ψxx − (1 + iλ)Ψ∗xx Ψ = 0. dt

(2.2.16)

Integrate (2.2.14) and (2.2.16) to give iτ

dE = (1 + iλ) dt

Z

+∞ −∞

Ψ∗xx Ψdx − (1 − iλ)

Z

+∞

Ψ∗ Ψxx dx.

(2.2.17)

−∞

This equation is of the same form as (2.2.13). We know that the single solitary solution to (2.2.15) without damping (λ = 0) is of the form "

c Ψ(x, t) = c sech x 2 then

c k(x, t) = c sech x, 2

#"

#

c exp i x , 2

(2.2.18)

c τ (x, t) = , 2

(2.2.19)

where c is a constant. In the case of λ > 0, c should be regarded as a function of t, i.e. c = c(t). Then it follows from (2.2.7) and (2.2.18) that dc 2 = − λc3 , dt 3 and therefore c = c(t) =

√

3 (λt + δ)−1/2 , 2

where δ is the integral constant. Setting El =

e1l + ie2l , 1 − e3l

e21l + e22l + e23l = 1,

(2.2.20)

(2.2.21)

Integrability of heisenberg chain

43

where e1l is the lth component of ~e1 , one get the following two Riccati equations 1 Zlx = −ikZl + iτ (Zi2 − 1), (2.2.22) 2 " ! # kxx λ Zlt = −i − τ 2 + (2kx τ + kτx ) Zi k k 1 − [(kτ − λkx ) + i(kx + λτ k)]Zi2 2 1 − [(kτ − λkx ) − i(kx + λτ k)]. (2.2.23) 2 If we know the curvature k and the torsion τ , then we can solve the spin equation by virtue of (2.2.22) and (2.2.23). In fact, we have  √ !  1 3 3x    ε(x, t) = k 2 = (λt + δ)−1 sech2 (λt + δ)−1/2 ,   2 8 4 (2.2.24) √ ! √   3 3x 3 1  2 −3/2 −1/2 2   (λt + δ) sech (λt + δ) .  j(x, t) = k τ = 16 4 2 And, we also have

#( " √ ! # " √ !  3x 3x  −1/2 −1/2   tan h S x = sech (λt + δ) (λt + δ)    4 4    # " √ ! #) " √ !    3x 3x  −1/2 −1/2   − cos sin (λt + δ) (λt + δ)    4 4    " √ ! #( " √ ! #   3x 3x

S y = − sech

(λt + δ)−1/2

tan h (λt + δ)−1/2 4 √ ! #) 3x −1/2 (λt + δ) 4

 4     # " √ !    3x  −1/2  − sin cos (λt + δ)    4    " √ ! ( #    3x   2 −1/2 z  . (λt + δ)  S = tan h

(2.2.25)

4

2.3

2.3.1

Inhomogeneous Heisenberg Chain Inhomogeneous Ferromagnetic Equations

Consider the nonhomogeneous ferromagnetic chain equation ~t (x, t) = f (x)S ~ ×S ~xx + fx (x)(S ~ ×S ~x ). S

(2.3.1)

Let ~e1 (x, t), ~e2 (x, t), ~e3 (x, t) be the tangential vector, normal vector and co-normal vector of a moving space curve and form a natural coordinate system. From the Serret–Frenet formula   e1x = k~e2 , ~ ~e2x = −k~e1 + τ~e3 , (2.3.2)   ~e3x = −τ~e3 ,

Landau–Lifshitz Equations

44

where the curvature k = (~e1x · ~e1x )1/2 and the torsion τ = k −2~e1 · (~e1x × ~e1xx ). Set " Z #  +∞    N = (~e2 + i~e3 ) exp i τ (x, t)dx ,   −∞ " Z #  +∞  k   τ (x, t)dx ,  g = exp i

2

then Nx = −2q~e1 . Let

(2.3.3)

−∞

(

~ t = αN ~ + βN ~ ∗ + γ~e1 , N ~ + µN ~ ∗ + ν~e1 , ~e1t = λN

(2.3.4)

we have µ = − 21 = λ∗ , β = 0, α = iR, R is real number. Hence  N ~ t = iRN ~ + γ~e1 , ~ ~ ∗ + γN ~ ), e1t = − 21 (γ ∗ N

(2.3.5)

and it follows from Nxt = Ntx that

  q + 1 γ − iRq = 0, t 2 x  R = i(γq ∗ − γ ∗ q).

(2.3.6)

x

~ 2 = 1. We let the vector S ~ be ~e1 . Then equation (2.3.1) becomes Note that |S| ~ ∗ − (q ∗ f )x N ~ ]. ~e1t = −kτ f~e2 + (kf )x~e3 = i[(qf )x N

(2.3.7)

Comparing this equation with (2.3.5), we may take, if f be a real function, γ = −2i(gf )x .

(2.3.8)

Substituting (2.3.8) into (2.3.6), replacing q by q exp(i(ax + bt)) and eliminating the integral constant, one may get the Schr¨odinger equation with coefficients depending on x: Z +∞ iqt + f qxx + 2f |q|2 q + 2q fx |q|2 dx + qfxx + 2fx qx = 0. (2.3.9) −∞

From the Hamiltonian change

H = −J

N −1 X i=1

~i · S ~i+1 , fi S

(2.3.10)

~ 2 and the continuum equation we get the energy density E(x, t) = 21 f S x Et + Px = 0, ~ · (S ~x × S ~xx ) or by k, τ where P (x, t) = f 2 S 1 E = f k2, 2

P = f 2 k 2 τ.

For equation (2.3.9), we have 1 E = f |q|2 , 2

P = 4f 2 |q|2 (arg q)x .

(2.3.11)

Integrability of heisenberg chain

2.3.2

45

Inhomogeneous Heisenberg Chain

Now consider another ferromagnetic chain equation:  S ~t (x, t) = (γ2 + µ2 x)S ~ ×S ~xx + µ2 (S ~×S ~x ) − (γ1 + µ1 x)S ~x , S ~ = (S1 , S2 , S3 ), ~ 2 = 1. |S|

(2.3.12)

Using the Serret–Frenet formula:

~eix = d ∧ ~ei ,

d = −k~e3 + τ~e1 ,

(2.3.13)

~ in (2.3.12), we have and taking ~e1 = S ~e1t = (γ2 + µ2 x)~e1 × ~e1xx + µ2 (~e1 × ~e1x ) − (γ1 + µ1 x)~e1x . It follows from (2.3.13) that ~e1t = −[(γ2 + µ2 x)kτ + (γ2 + µ1 x)k]~e2 + [(γ2 + µ2 x)k]x~e3 . The equations for ~e2t , ~e3t are similar as above. If we take solid body motion, we have ~eit = ω ~ ∧ ~ei ,

ω ~ = ω1~e1 + ω2~e2 + ω3~e3 ,

(2.3.14)

where  −1 2  e1x − (γ1 + µ1 x)τ,   ω1 = k [(γ2 + µ2 x)k]xx − (γ2 + µ2 x)τ , −(γ1 + µ1 x)~

ω = −[(γ + µ x)k] ,

2 2 2 x    ω = −[(γ + µ x)kτ + (γ + µ x)k]. 3 2 2 1 1

(2.3.15)

Since

Rx = ~e1 ,

Rt = (γ1 + µ2 x)~e1 ∧ ~e1x − (γ1 + µ1 x)~e1 + µ1 R,

we have from (2.3.13), (2.3.14) and the compatible condition (~eit )x = (~eix )t , i = 1, 2, 3 that  kt = −2µ2 kτ − (γ2 + µ2 x)[kτx + 2kx τ ] − (γ1 + µ1 x)kx − µ1 k,      τ = µ [k −1 k − τ 2 + 1 k 2 ] t 2 xx 2 1 2 −1 2 −1  + (γ + µ x)(k k  2 2 xx − τ + 2 k )x + 2µ2 (k kx )x    1 2

(2.3.16)

− µ1 τ − (γ1 + µ1 x)τx + 2 µ2 k .

Setting q = 21 k exp(i

R +∞ −∞

τ dx), we have

iqt + iµ1 q + i(γ1 + µ1 x)qx + (γ2 + µ2 x)(qxx + 2q|q|2 ) + 2µ2 qx + q

Z

+∞ −∞

0

2

|q(x , t)| dx

0

!

= 0.

(2.3.17)

Landau–Lifshitz Equations

46

2.4 2.4.1

Spherical (Cylindrical) Symmetric Heisenberg Equations of Ferromagnetic Spin Chain Radial Symmetric Equations

Consider n-dimensional Landau–Lifshitz equation  ~t (~r, t) = S ~ × ∆S, ~ ~r = (r1 , r2 , . . . , rn ),  S 

 ~ 2 = 1,  S

~ = (S x , S y , S z ), S

∆=

∂2 ∂2 + · · · + . ∂r12 ∂rn2

The radial symmetric equation is of the form    ~t = S ~ ×S ~rr + n − 1 S ~ ×S ~r , S r    r = (r 2 + · · · + r 2 )1/2 , 1 n

(2.4.1)

(2.4.2)

0 < r < ∞.

In equation (2.4.2), the curvature of the space curve is ~r · S ~r ]1/2 , k(r, t) = [S

(2.4.3)

and the torsion is

~ · (S ~r × S ~rr )]. τ (x, t) = k −2 [S

(2.4.4)

~ = k~e3 + τ~e1 , Ω = ω1~e1 + ω2~e2 + ω3~e3 , D ! ! krr n−1 kr 1 2 ω1 = −τ + − , k r r r ω2 = −kr − [(n − 1)/r]k,

(2.4.6)

Using Serret–Frenet relation, equation (2.4.2), in the natural coordinate system, exhibits in the form ~ ∧ ~ei , ~eit = ~ei ∧ Ω, i = 1, 2, 3, ~eir = D (2.4.5)

where

ω3 = −kτ.

(2.4.7) (2.4.8) (2.4.9)

It follows from the compatible condition (~eir )t = (~eit )r , i = 1, 2, 3 and complex change of unknowns that " Z # r k(r, t) 0 0 q(r, t) = exp i τ (r , t)dr , (2.4.10) 2 0 and then the following radial symmetric nonlinear Schr¨odinger equation is derived n−1 n−1 qr − q iqt + qrr + 2q|q|2 + r r2 Z r |q|2 0 + 4(n − 1)q dr = 0, if n = 1. (2.4.11) r0 0 This explains the equivalence of one-dimensional Heisenberg chain equation to nonlinear Schr¨odinger equation. If n 6= 1, we need to study the integrability of system (2.4.2) and equation (2.4.11).

Integrability of heisenberg chain

2.4.2

47

Painleve Property of the Radial Symmetric Nonlinear Schr¨ odinger Equation

First of all, we give the Painleve analysis for equation (2.4.11). For this reason we rewrite (2.4.2) as n−1 n−1 qr + 2kq − q = 0, r r2 n−1 2 Rr − (|q|2 )r − 2 |q| = 0. r

iqt + qrr +

(2.4.12) (2.4.13)

Let q = a + ib, where a, b are real numbers. It follows from (2.4.12), (2.4.13) that n−1 n−1 br + 2kb − b = 0, r r2 n−1 n−1 a = 0, −bt + arr + ar + 2ka − r r2 n−1 2 Rr − (a2 + b2 )r − 2 (a + b2 ) = 0. r at + brr +

(2.4.14) (2.4.15) (2.4.16)

Suppose that the first order solution of (2.4.14)–(2.4.16) are a ∼ a 0 ϕα ,

b ∼ b 0 ϕβ ,

R ∼ R 0 ϕγ ,

(2.4.17)

where a0 , b0 , R0 are analytical functions of (r, t), α, β, γ are determined by (2.4.14)– (2.4.16): α = β = −1, γ = −2, (2.4.18) and a20 + b20 = −ϕ2r ,

k0 = −ϕ2r .

(2.4.19)

It is clear that one of the functions is arbitrary, a0 or b0 . In order to derive the resonance conditions, we take Laurant expansion a=

X j

aj ϕj−1 ,

b=

X

bj ϕj−1 ,

R=

j

X

ϕj−1 .

(2.4.20)

j

Reserving the terms with highest order in equation (2.4.14)–(2.4.16), we get the matrix equation for the coefficients of the terms with lower order as 





(j 2 − 3j)ϕ2r 0 2a0 aj    2 2    0 (j − 3j)ϕr 2b0     bj  = 0. −2a0 (j − 2) −2b0 (j − 2) j − 2 Rj

(2.4.21)

It follows from (2.4.19) that the resonance conditions are j = −1, 0, 2, 3, 4.

(2.4.22)

Landau–Lifshitz Equations

48

Obviously, resonance j = −1 implies the arbitrariness of the singular flow ϕ(x, t) = 0, resonance j = 0 indicates the arbitrariness of the functions a0 or b0 . In order to verify the existence of sufficiently many functions with respect to the resonance j = 2, 3, 4, we substitute the expansions (2.4.20) into (2.4.14)–(2.4.16) and first combine the coefficients of (ϕ−2 , ϕ−2 , ϕ−2 ) to get the equation



2R0

  0 

2a0

0 2R0 2b0





n−1  −b0 ϕt +   a0 ϕr + 2a0 ϕr + a0 ϕrr    r a1 2a0       n − 1     2b0   b1  =  a0 ϕt + b0 ϕr + 2b0 ϕr + b0 ϕrr  .   r   R1 1   n−1 −2 ϕr r

(2.4.23)

It follows from (2.4.23) that 1 n−1 a1 = b0 ϕt − 2a0r ϕrr + a0 ϕrr + ϕrr , 2 2ϕr r

!

(2.4.24)

!

(2.4.25)

1 n−1 b1 = b0 ϕrr + b0 ϕr − a0 ϕt − 2b0r ϕr , 2 2ϕr r n−1 R1 = ϕr + ϕrr . r

(2.4.26)

Similarly we have from the coefficients of (ϕ−1 , ϕ−1 , ϕ−1 ) that 2ϕ2r [R2 + a0 a2 + b0 b2 ] = b0 a0t − a0 b0t − ϕr ϕrr /r

+ b0 b0rr + a0 a0rr + ϕ2rr , 1 2ϕ2r (a0 b2 + b0 a2 ) = −ϕr ϕrr + (a0 b0r − b0 a0r ) r + (a0 b0rr − b0 a0rr ) + 2R1 (a0 b1 − b0 a1 ),

(2.4.27)

(2.4.28)

4(n − 1) (a0 a1 + b0 b1 ) = 0. (2.4.29) r It follows from (2.4.27) and (2.4.28) that one of the functions a2 , b2 , R2 is arbitrary which corresponds to the resonance j = 2. On the other hand, It also follows from (2.4.29), (2.4.19) and (2.4.24)–(2.4.26) that R1r − 2(a0 a1 + b0 b1 )x −

(n − 1) = (n − 1)2 .

(2.4.30)

We should introduce exponential movable singular manifold except for n = 1, 2. Similarly, it follows from the coefficients of (ϕ0 , ϕ0 , ϕ0 ) that 1 (b1r + b2 ϕr ) + b1rr + 2b2r ϕr + b2 ϕrr r + 2(R1 b2 + R2 b1 + R3 b0 ) − b1 /r 3 = 0,

a1t + a2 ϕt +

(2.4.31)

Integrability of heisenberg chain

1 (a1r + a2 ϕr ) + a1rr + 2a2r ϕr + a2 ϕrr r + 2(R1 a2 + R2 a1 + R3 a0 ) − a1 /r 3 = 0,

49

−b1t − b2 ϕt +

R3 ϕr − 2(a0 a3 + b0 b3 )ϕr + R2r − 2(a0 a2 + +a21 /2 + b0 b2 + b21 /2) − 2(a1 a2 + b1 b2 )ϕr = 0.

(2.4.32) (2.4.33)

It can be inferred from (2.4.31) and (2.4.32) that R3 is identical (n = 1, 2), and then it follows from (2.4.33) that one of the functions a3 , b3 is arbitrary. Finally, it can be verified from the coefficients of (ϕ1 , ϕ1 , ϕ1 ) that a4 or b4 is arbitrary. Therefore, the general solutions (a(r, t), b(r, t), R(r, t)) of (2.4.14)–(2.4.16) admits the required number of arbitrary functions. For n = 1, 2, the elements lead to the movable critical manifolds. Hence, the Painleve property is satisfied for n = 1 or n = 2. The conclusion is obvious for n = 1. If n = 2, (2.4.11) is of the form q qr iqt + qrr + 2|q| q + − 2 + 4 r r 2

2.4.3

Z

0

r

|q|2 0 dr = 0. r0

(2.4.34)

Normal Change for Radial Symmetric Equation and Matrix Form for the Radial Symmetric Heisenberg Equations

Now consider the normal change. Let aj = bj = 0, j ≥ 2, Rj = 0, j ≥ 3 in (2.4.20). Then ( a = a0 ϕ−1 + a1 , b = b0 ϕ−1 + b1 , (2.4.35) R = R0 ϕ−2 + R1 ϕ−1 + R2 . Substituting (2.4.35) into (2.4.34), we get the differential system of a1 , b1 , R2 . Define new variables ~ 0 = a0 + ib0 , A ~ 0 = a0 − ib0 , B

~ = a + ib, q=A ~ = a − ib. q∗ = B

(2.4.36)

It follows from simple computations that "

~ 0r ~ r A 2Aϕ + ~0 ~0 rA rA

#

r

~ 0r ~ r B 2Bϕ = + ~0 ~0 rB rB "

#

= 0.

(2.4.37)

r

Let iλ be the integral constant, we have ~ 0r + iλA ~ 0 = −2qϕr , A

~ 0r − iλB ~ 0 = 2Bϕ ~ r. B

(2.4.38)

Define square characteristic functions as follows: ~ 0 = iΨ2 , A 1

~ 0 = iΨ2 , B 2

ϕr = −iΨ1 Ψ2 .

(2.4.39)

We obtain the dispersion problem Ψr = V1 Ψ,

Ψ = (Ψ1 , Ψ2 )T ,

(2.4.40)

Landau–Lifshitz Equations

50

where

"

#

−iλr/2 q V1 = . ∗ −q iλr/2

(2.4.41)

"

A B Ψt = V1 Ψ = , −B ∗ −A

(2.4.42)

"Z # t |q|2 iλ2 r 2 |q|2 0 A=− + 2i dr + , 2 2 0 r0

(2.4.43)

Similarly one can also get the evolutionary part of the dispersion problem from (2.4.24) and (2.4.25) as follows: #

where

!

q B = λrq + i qr + , r

(2.4.44)

and the compatible condition is U1t − V1r + [U1 , V1 ] = 0. These lead to the non-identical spectrum representation for equation (2.4.34): λt = 2λ2 .

(2.4.45)

Introduce normal change: ˆ = g −1 Ψ, Ψ

g(r, t) ∈ GL(2, C).

(2.4.46)

Substitute (2.4.40) into (2.4.42) to give new eigenvalue equation ˆ r = Uˆr Ψ, ˆ Ψ where

(

ˆ t = Vˆ1 Ψ, ˆ Ψ

Uˆ1 = g −1 U1 g − g −1 gr , Vˆ1 = g −1 V1 g − g −1 gt , gr = U1 (λ = 0)g, gt = V1 (λ = 0)g.

(2.4.47)

(2.4.48)

Define spin matrix of the form ~ = −g −1 σ3 g = S ~ · ~σ , S

~σ = (σ1 , σ2 , σ3 ) is the Pauli matrix. We have

iλr ~ iλ2 r 2 ~ λr ~ ~ Uˆ1 = S, Vˆ1 = S + [S, Sr ]. 2 2 4 The compatible condition Uˆ1t − Vˆ1t + [Uˆ1 , Vˆ1 ] = 0,

and λ1 = 2λ2 lead to

~ t = 1 [S, ~ S ~ r ] + 1 [S, ~ S ~ r ]. S τi τi r This is the matrix form of radial symmetric Heisenberg equation (2.4.2).

(2.4.49)

Integrability of heisenberg chain

2.4.4

51

B¨ acklund Change for Radial Symmetric Nonlinear Schr¨ odinger Equation and Solitary Solutions

Now we consider the B¨acklund change for equation (2.4.34) and its solitary solution. Let Γ = Ψ1 /Ψ2 . (2.4.50) From (2.4.40) and (2.4.42) we can get the following Riccati equation of Γ (

Γr = −iλrΓ + q/2 + (q ∗ /2)Γ2 , Γt = B + 2AΓ + B ∗ Γ2 .

(2.4.51)

We can also construct Γ0 which meets the same equations (2.4.51) and has potential q 0 (r), and q 0 (r, t) = q(r, t) + f (Γ, λ). (2.4.52) On the other hand, eliminating Γ from (2.4.51), one gets the B¨acklund change for (2.4.34). Take, for example, Γ0 and q 0 as follows: (

Γ0 = 1/Γ∗ , q 0 + q = −4βrΓ/(1 + |Γ|2 ).

(2.4.53)

Then Γ0 satisfies (2.4.51) with potential q 0 (r, t) where λ(t) = α(t) + iβ(t), β > 0. Solve Γ from (2.4.53) as Γ = −[2βγ + 4β 2 r 2 − |q 0 + q|1/2 ]/(q 0∗ + q ∗ ).

(2.4.54)

Substitute (2.4.54) into (2.4.51) to give the BΓ of (2.4.34): qr + qr0 = (q + q 0 )/r 1 + (q 0 + q)(4β 2 r 2 − |q 0 + q|2 )1/2 − iαr(q 0 + q), 2 ! iq qt + iqt0 = 4α(q + q 0 ) + iqr + − αrq 0 (4β 2 r 2 − |q 0 + q|2 )1/2 r iq 0 |q|2 |q + q|2 + i(q 0 + q) r 0 dr 0 − ir 2 q(α2 − β 2 ) 4 r 0 i − ir 2 q(α2 + β 2 ) − (q 02 q r + q|2 q). 4 Z

+

(2.4.55)

Now we may construct the solitary solution of (2.4.34). Assuming q(0) = 0, we get from (2.4.40), (2.4.42) and (2.4.53) that "

#

"

#

βr 2 iαr 2 q(1) = −2βγ sech + δ1 exp − + δ2 , 2 2

(2.4.56)

where δ1 , δ2 are phase constants. It follows from αt = 2(α2 − β 2 ),

βt = 4αβ,

(2.4.57)

52

Landau–Lifshitz Equations

that α(0) + 2t , (α(0) + 2t)2 + β(0)2 β(0) β(t) = − . (α(0) + 2t)2 + β(0)2

α(t) = −

(2.4.58)

we can then construct q(2) from (2.4.53) and Ψ(1), and so on. In this way we obtain a series of solitary solutions.

2.5

Bibliography Comments

This chapter is mainly concerned with the theory of integrability of Heisenberg chain without Gilbert damping. Since Nakamura and Sasada [122] found the solitary solution to L–L equations in 1974, there have been a series studies on the solitary solution to L–L equation, scattering method, geometric representation, infinite conservation law and the normal equivalence to nonlinear Schr¨odinger equations, for the details we refer to references [140], [71]. At the same time, we refer to references [12, 59, 110] and [124] for the discussions on the one-dimensional inhomogeneous and compressible flow Heisenberg chain, the Painleve properties for the solitary waves of spherical symmetric Heisenberg ferromagnetic systems.

Chapter 3 One-Dimensional Landau–Lifshitz Equations In this chapter, we discuss the boundary value problems of one-dimensional Landau– Lifshitz equations. We state and prove some results on the existence and uniqueness of weak solutions and smooth solution.

3.1

Initial Boundary Value Problem of One-dimensional Ferromagnetic Spin Chain Equations

This section deals with the ferromagnetic spin chain system of the form ~t = Z ~ ×Z ~ xx + f~(x, t, Z) ~ Z

(3.1.1)

with initial-boundary condition of the form   Z(0, ~ t) = Z(l, ~ t) = 0,  Z(x, ~ 0) = Z ~ 0 (x), 0 ≤ x ≤ l,

(3.1.2)

~ t) = (u(x, t), v(x, t), w(x, t)) is a three-dimensional unknown, f~(x, t, Z) ~ is where Z(x, ~ × denotes the cross product a known three-dimensional vector of variables x, t, Z, of two three-dimensional vectors. In the classical studies, isotropic Heisenberg chain equations and Landau–Lifshitz equations are the special cases of (3.1.1).

3.1.1

Initial Boundary Value Problem of Ferromagnetic Spin Chain Equations

System (3.1.1) is a strongly coupling and strongly degenerate equation. In fact, (3.1.1) can be rewritten as ~ t = A(Z) ~ Z ~ xx + f~(x, t, Z), ~ Z (3.1.3) 53

Landau–Lifshitz Equations

54

where





u   ~ Z =  v , w





0 −w v  ~ 0 −u  A(Z) =  w . −v u 0

(3.1.4)

It is not difficult to verify that ~ is zero definite, i.e. (1) The matrix A(Z) ~ = 0, ξ · A(Z)ξ

~ ∈ R3 . ∀ ξ, Z

~ is singular, i.e. (2) A(Z) ~ = 0, det A(Z)

~ ∈ R3 . ∀Z

This implies that (3.1.1) is strongly degenerate.

3.1.2

Theory on Quasilinear Parabolic Equations

1. Initial boundary value problem of linear parabolic systems We consider the following linear parabolic systems ~ut − A(x, t)~uxx + B(x, t)~ux + C(x, t)~u = f~(x, t)

(3.1.5)

with initial boundary condition (

~u|x=0 = ~u|x=l = 0, ~u|t=0 = u0 (x),

(3.1.6)

where ~u(x, t) and ~u0 (x) are two N -dimensional vector-valued functions. Lemma 3.1.1 If (3.1.5) and ~u0 (x) satisfy the following conditions (i) A(x, t) is a positively definite n × n matrix over QT = [0, l] × [0, T ]; (ii) A(x, t), B(x, t) and C(x, t) are measurable bounded N × N matrix ; (iii) f~(x, t) is quadratically integrable N -dimensional vector-valued function over QT ; (iv) ~u0 (x) ∈ H01 (x), then problem (3.1.5)–(3.1.6) admits a unique solution ~u(x, t) ∈ L∞ (0, T ; H01 (0, l)) and the following estimate holds

\

(2,1)

W2

(QT ),

(3.1.7)

sup k~u(·, t)kH01 (0,l) + k~ut kL2 (QT ) + k~uxx kL2 (QT )

0≤t≤T

≤ K(k~u0 kH01 (0,l) + kf~kL2 (QT ) ), where K depends on A(x, t), B(x, t), C(x, t), but is independent of ~u(x, t).

(3.1.8)

One-dimensional Landau–Lifshitz equations

55

Proof. Taking inner products of (3.1.5) with vectors ~u and ~uxx respectively, we obtain (~u, ~ut ) − (~u, A(x, t)~uxx ) + (~u, B(x, t)~ux ) + (~u, C(x, t)~u) = (~u, f~(x, t)); (~uxx , ~ut ) − (~uxx , A(x, t)~uxx ) + (~uxx , B(x, t)~ux ) + (~uxx , C(x, t)~u) = (~uxx , f~(x, t)). Subtracting one from the other in the above two equations, we get Z

l 0

(~uxx (x, t), ~ut (x, t))dx = (~ux (x, t), ~ut (x, t))|x=l x=0 −

Z

l 0

(~ux (x, t), ~uxt (x, t))dx.

Using boundary condition (3.1.6), we see that the first term on the right-hand side of the above equality is zero. Therefore we have Z

l 0

(~uxx (x, t), ~ut (x, t))dx = −

1d k~ux (·, t)k2L2 (0,l) . 2 dt

In the rectangular domain Qτ = {0 ≤ x ≤ l, 0 ≤ t ≤ τ }, (0 ≤ τ ≤ T ), making subtraction of these two equalities leads to k~u(·, t)k2H01 (0,l) =2

Z Z

+2

−

Qτ

k~u0 k2H01 (0,l)

+2

Z Z

Qτ

(~uxx , A~uxx )dxdt

(~uxx , B~ux + C~u − f~ )dxdt

Z Z

Qτ

(~u, A~uxx − B~ux − C~u + f~ )dxdt.

It is easy to derive (3.1.8) from the above equality by the conditions of this lemma. And, using the parameter extension method, one gets the existence of the solution to the above problem. The uniqueness can be derived from (3.1.8). Then, the proof is complete. 2 2. Initial boundary value problem of spin systems Now we turn to equation (3.1.1) by viscosity method, that is, we consider the ~ xx following system with small diffusion term εZ ~ t = εZ ~ xx + Z ~ ×Z ~ xx + f~(x, t, Z) ~ Z

(3.1.9)

which is called spin system. It is clear that this is a parabolic system whose initial boundary condition is (3.1.6). Take three-dimensional vector-valued function space B = L∞ (QT ) as the base space of fixed point argument. We define a functional mapping with parameter 0 ≤ λ ≤ 1 from the base space onto itself by Tλ : B → B : For every ~u ∈ B, the image ~ = Tλ (~u) is the solution of the linear parabolic system Z ~ t = εZ ~ xx + ~u × Z ~ xx + f~(x, t, ~u) Z

(3.1.10)

Landau–Lifshitz Equations

56

subject to the condition (3.1.6), where 0 ≤ λ ≤ 1. It follows from Lemma 3.1.1 ~ = Tλ ~u is the unique solution of (3.1.10) under condition (3.1.6). This unique that Z T (2,1) solution is in the space G where G = L∞ (0, T ; H01 (0, l)) W2 (QT ). It is easy to see that for every λ the operator Tλ is completely continuous and for every bounded set M in B the operator Tλ is uniformly continuous in λ : 0 ≤ λ ≤ 1. In order to prove the existence of weak solution to spin equation (3.1.9) under (3.1.6) by fixed point theorem, we should make out the a priori uniform estimate in λ ∈ (0, 1) for all possible fixed point of the mapping Tλ . To this aim, we assume ~ is continuously differentiable in x, Z, ~ and, 3 × 3 Jacobi matrix (1) Vector f~(x, t, Z) ~ ~ fZ~ (x, t, Z) is semi-bounded, i.e., there is a constant b > 0 such that for any ξ ∈ R 3 there holds ~ · ξ ≤ b|ξ|2 , (3.1.11) ξ · f~Z~ (x, t, Z) ~ ∈ QT × R3 . Moreover, f~0 (x, t) ≡ f~(x, t, 0) ∈ L2 (QT ). for any (x, t, Z) ~ ∈ QT × R3 , there holds (2) For (x, t, Z) ~ ≤ c(x, t)|Z| ~ 3 + d(x, t), |f~x (x, t, Z)|

(3.1.12)

where c(x, t) ∈ L∞ (QT ), d(x, t) ∈ L2 (QT ). ~ 0 (x) ∈ H 1 (0, l). (3) Z 0 Construct the following vector system ~ t = εZ ~ xx + λZ ~ ×Z ~ xx + λf~(x, t, Z) ~ Z

(3.1.13)

~ t) to give and take inner product of this system with Z(x, ~ ·Z ~ t = εZ ~ xx · Z ~ + λZ ~ · (Z ~ ×Z ~ xx ) + λZ ~ · f~(x, t, Z). ~ Z

(3.1.14)

Simple calculations give that Z

~ · f~(x, t, Z) ~ =Z ~· Z

1 0

!

~ ~ +Z ~ · f~0 (x, t) f~Z~ (x, t, τ Z)dτ Z

~ 2 + 1 |f~0 (x, t)|2 , ≤ (b + δ)|Z| 4δ where δ > 0. Integrate (3.1.14) over the rectangular domain Qτ (0 ≤ τ ≤ T ) to get ~ τ )k2 2 ~ 2 kZ(·, L (0,l) − kZ0 kL2 (0,l) ≤ 2λ(b + δ)

Z

τ

0

~ t)k2L2 (0,l) dt + kZ(·,

λ ~ 2 kf0 kL2 (QT ) 2δ

and hence there holds !

λ ~ 2 ~ t)k2 2 ~ 2 kZ(·, kf0 kL2 (0,l) e2λ(b+δ)t , L (0,l) ≤ kZ0 kL2 (0,l) + 2δ

(3.1.15)

One-dimensional Landau–Lifshitz equations

57

where 0 ≤ λ ≤ 1, 0 ≤ t ≤ T , δ > 0. Therefore we have ~ t)k2 2 sup kZ(·, L (0,l) ≤ K1 ,

(3.1.16)

0≤t≤T

where K1 is a constant independent of ε, λ, l. ~ xx , we have Now taking the inner product of (3.1.13) with Z ~ xx · Z ~ t = εZ ~ xx · Z ~ xx + λZ ~ xx · (Z ~ ×Z ~ xx ) + λZ ~ xx · f~(x, t, Z). ~ Z

(3.1.17)

For the left-hand side of (3.1.17), we have Z

~ xx · Z ~ t )dx = − 1 d kZ ~ x (·, t)k2 2 . (Z L (0,l) 2 dt 0 l

(3.1.18)

For the last term on the right of (3.1.17), we have Z

l 0

~ xx · f~(x, t, Z))dx ~ ~ x (l, t) · f~(l, t, Z(l, ~ t)) (Z =Z ~ x (0, t) · f~(0, t, Z(0, ~ t)) − −Z

Z

l 0

~ x · Dx f~ )dx. (Z

It follows from assumption (1) that Z Z

Qτ

~ x · Dx f~ )dxdt = (Z

Z Z

+

Qτ

~ x · f~x (x, t, Z) ~ Z ~ x )dxdt (Z

Z Z

Qτ

~ x · f~x (x, t, Z))dxdt ~ (Z

!Z

l 1 ~ x (·, t)k2 2 dt ≤ b+ kZ L (0,l) 2 0 Z Z 1 ~ 2 dxdt. |f~x (x, t, Z)| + 2 Qτ

It follows from assumption (2) and interpolation formula that 1 2

Z Z

and

Qτ

Z

|f~x |2 dxdt ≤ kc(x, t)k2L∞ (QT ) l

0

Z Z

Qτ

~ 6 dxdt + kdk2 2 |Z| L (QT )

2 ~ t)|6 dx ≤ C1 kZ(·, ~ t)k4 2 ~ |Z(x, L (0,l) · kZ(·, t)kH01 (0,l) .

Therefore (3.1.19) can be rewritten in the following inequality: Z Z

Qτ

~ xx · f~(x, t, Z))dx ~ (Z

≤

Z

τ 0

~ x (l, t) · f~(l, t, Z(l, ~ t)) − Z ~ x (0, t) · f~(0, t, Z(0, ~ t))]dt [Z

+ C2

Z

τ

0

~ x (·, t)k2 2 dt + C3 , kZ L (0,l)

(3.1.19)

Landau–Lifshitz Equations

58

where 0 ≤ τ ≤ T , and !

1 C2 = b + + C1 kck2L∞ (QT ) K14 , 2

C3 = kdk2L2 (QT ) .

When system (3.1.9) is homogeneous, i.e. f~(x, t, 0) ≡ 0, we can obtain Z Z

Qτ

~ xx · f~(x, t, Z))dx ~ (Z ≤ C2

Z

τ 0

~ x (·, t)k2 2 dt + C3 . kZ L (0,l)

(3.1.20)

Integrating (3.1.17) over Qτ , 0 ≤ τ ≤ T , we finally get ~ x (·, t)kL2 (0,l) ≤ K2 , sup kZ

(3.1.21)

0≤t≤T

where K2 is independent of λ and ε, and also of l if kckL∞ (Qτ ) and kdkL2 (Qτ ) are independent of the width l, then K2 is also independent of l. Therefore we have Theorem 3.1.1 Let assumptions (1)–(3) hold, (3.1.9) is homogeneous, i.e. f~(x, t, 0) ≡ 0. Then the spin system (3.1.9) with (3.1.6) admits unique global weak solution \ ~ t) ∈ L∞ (0, T ; H 1(0, l)) W2(2,1) (QT ). Z(x, 0 If b < 0, we take δ > 0 such that b + δ < 0. If T = ∞, denote (1)–(3) by (1∞ )– ~ t)kL2 (0,l) of the global weak (3∞ ), respectively; hence, when b < 0, the norm kZ(·, ~ t) of the spin system (3.1.9) (with (3.1.6)) tend to zero when t → ∞, solution Z(x, ~ t)kL2 (0,l) = 0. i.e. limt→∞ kZ(·,

3.1.3

Approximate Solution to the Initial Boundary Problem of the System of Ferromagnetic Spin Chain

Now we begin to consider the initial-boundary problem (3.1.1)–(3.1.2). In the proof of the above theorem, we see that the a priori estimates for the approximate solutions are independent of λ. Denote the approximate solution obtained in the above ~ε = Z ~ ε (x, t). subsection by Z 1. Approximate solutions to initial-boundary value problem ~ ε (x, t) obtained in Theorem 3.1.1 meets the estimates Lemma 3.1.2 The solution Z ~ ε (·, t)kL2 (0,l) ≤ K3 (kZ ~ 0 kL2 (0,l) + kf~0 kL2 (Q ) )e(b+δ)t , kZ T ~ sup0≤t≤T kZε (·, t)kH 1 (0,l) ≤ K4 , 0

(3.1.22) (3.1.23)

~ 0 kH 1 (0,l) , where 0 ≤ t ≤ T, δ > 0, K3 , K4 are independent of ε and of l if kZ 0 kf~0 kL2 (QT ) , kckL∞ (QT ) and kdkL2 (QT ) are independent of l.

One-dimensional Landau–Lifshitz equations

59

~ ε (x, t) obtained in Theorem 3.1.1, there holds Lemma 3.1.3 For the solution Z ~ εt kH −1 (0,l) ≤ K5 , sup kZ (3.1.24) 0≤t≤T

~ 0 kH 1 (0,l) , kf~0 kL2 (Q ) , kckL∞ (Q ) and where K5 is independent of ε and of l if kZ T T 0 kdkL2 (QT ) are independent of l. Proof. For any ψ(x) ∈ H01 (0, l) we have Z

l

0

~ εt dx = −ε ψ(x)Z

It follows from (3.1.23) that

Z

−

Z

+

Z

l

0 l

0

0

l

~ εx (x, t)dx ψx (x)Z

~ ε (x, t) × Z ~ εx (x, t)]dx ψx (x)[Z ~ ε (x, t))dx. ψ(x)f~(x, t, Z

Z l ~ εt dx ≤ C4 kψkH 1 (0,l) , ψ(x)Z 0 0

where C4 is independent of ε and l. Lemma 3.1.3 follows from the definition of Sobolev space of negative order. 2. Estimates for approximate solutions Lemma 3.1.4 For the above approximate solutions there holds the estimate ~ ε (x, t)k 1 , 1 kZ ≤ (1 + l)K6 , (3.1.25) C 2 4 (Q ) T

~ 0 kH 1 (0,l) , kf~0 kL2 (Q ) , kckL∞ (Q ) and kdkL2 (Q ) are with K6 independent of ε and if kZ T T T 0 independent of l, then K6 is also independent of l. R R ~ ε (ξ, t)dξ. Then w ~ εt (ξ, t)dξ, w Proof. Let w ~ ε (x, t) = 0x Z ~ εt (x, t) = 0x Z ~ εxt (x, t) = ~ εt (x, t), w ~ ε (x, t) and w ~ εx (x, t). For any ψ ∈ H 1 (0, l), the Z ~ εx (x, t) = Z ~ εxx (x, t) = Z 0 following holds Z Z l l ~ εt (x, t)dx ψ 0 (x)w ~ εt (x, t)dx = ψ(x)Z 0 0

≤ C4 kψkH01 (0,l) ≤ C4 (1 + l)kψ 0 kL2 (0,l) ,

this implies sup kw ~ εt (·, t)kL2 (0,l) ≤ (1 + l)K4 ,

0≤t≤T

and then for any given l > 0, {w ~ ε (x, t)} is uniformly bounded in ε in the space T (1) L∞ (0, T ; H 2 (0, l)) W∞ (0, T ; L2 (0, l)) so that ~ ε (x, t2 ) − Z ~ ε (x, t1 )| = |w |Z ~ εx (x, t2 ) − w ~ εx (x, t1 )| 1/2

3/4

≤ C5 kw ~ ε (·, t2 ) − w ~ ε (·, t1 )kL2 (0,l) kw ~ ε (·, t2 ) − w ~ ε (·, t1 )kH 2 (0,l) 1/4

3/4

≤ C6 |t2 − t1 |1/2 sup kw ~ εt (·, t)kL2 (0,l) sup kw ~ ε (·, t)kH 2 (0,l) . 0≤t≤T

This completes the proof of Lemma 3.1.4.

0≤t≤T

Landau–Lifshitz Equations

60

3.1.4

Weak Solution to the Ferromagnetic Spin Chain Equation

1. Weak solution to the initial-boundary value problem of ferromagnetic spin chain system In this subsection we discuss the limit of the approximate solutions when ε → 0. From this procedure, we can get the weak solution to the initial-boundary value problem (3.1.1) and (3.1.2). ~ t) ∈ L2 (0, T ; Definition 3.1.1 A three-dimensional vector-valued function Z(x, T 1 H (0, l)) C(QT ) is called a weak solution of (3.1.1) and (3.1.2) if for any test function ϕ(x, t) ∈ Φ := {ϕ : ϕ ∈ C 1 (QT ), ϕ(x, T ) ≡ 0} there holds Z Z

QT

~ − ϕ x (Z ~ ×Z ~ x ) + ϕf~(x, t, Z)]dxdt ~ [ϕt Z +

Z

l 0

~ 0 (x)dx = 0, ϕ(x, 0)Z

(3.1.26)

~ t) = 0 in which ϕ(0, t) = 0 (or ϕ(l, t) = 0) if the condition on x = 0 (or x = l) is Z(0, ~ (or Z(l, t) = 0). ~ ε (x, t) and any test function ϕ ∈ Φ, it is clear that For the approximate solution Z the following integral identity holds Z Z

QT

~ ε − εϕx Z ~ εx − ϕx (Z ~ε × Z ~ εx ) [ϕt Z

~ ε )]dxdt + + ϕf~(x, t, Z

Z

l 0

~ 0 (x)dx = 0. ϕ(x, 0)Z

(3.1.27)

It follows from Lemmas 3.1.2–3.1.4 that the set of approximate solutions {Z¯ε (x, t)} to problem (3.1.1) and (3.1.2) is uniformly bounded in ε in the space T 1 1 L∞ (0, T ; H 1 (0, l)) C ( 2 , 4 ) (QT ). It follows from the a priori estimates that there ~ ε } of {Z ~ ε } such that it uniformly converges on QT to a is a subsequence {Z i ~ t). Hence {f~(x, t, Z ~ ε )} uniformly converges on QT to vector-valued function Z(x, i ~ ~ ~ ~ x (x, t). This means that f (x, t, Z(x, t)). Moreover, {Zεi x (x, t)} converges weakly to Z T 1 1 ~ t) ∈ L∞ (0, T ; H 1 (0, l)) C ( 2 , 4 ) (QT ). Z(x, In order to verify the convergence of the third term of (3.1.27), we estimate as follows: Z Z

QT

~ε × Z ~ ε x )dxdt − ϕx (Z i i

=

Z Z

QT

Z Z

QT

~ ×Z ~ x )dxdt ϕx (Z

~ ε − Z) ~ ×Z ~ ε x ]dxdt + ϕx [(Z i i

Z Z

QT

~ × (Z ~ε x − Z ~ x )]dxdt. ϕ x [Z i

~ ε x (x, t)} converges weakly to Z ~ x (x, t), the second integral on the right tends Since {Z i to zero, and for the first integral, when εi → 0, we have Z Z ~ ~ ~ ϕ x [(Zεi − Z) × Zεi x ]dxdt QT

~ ε x kL2 (Q ) kZ ~ ε − Zk ~ L∞ (Q ) → 0. ≤ kϕx kL2 (QT ) kZ i i T T

One-dimensional Landau–Lifshitz equations

61

Hence, as εi → 0, the integral identity (3.1.27) tends to (3.1.26). This means that ~ t) is a weak solution of (3.1.1) and (3.1.2). the limit Z(x, Theorem 3.1.2 Suppose that the homogeneous ferromagnetic spin chain system ~ 0 (x) meet the conditions (1)–(3). (3.1.1) (i.e. f~(x, t, 0) ≡ 0) and the initial data Z Then problem (3.1.1)–(3.1.2) admits at least one global weak solution: ~ t) ∈ L∞ (0, T ; H 1 (0, l)) Z(x, From estimates (3.1.22), we have

\

1 1

C ( 2 , 4 ) (QT ).

~ t) Theorem 3.1.3 If b < 0 and T = ∞, then the L2 -norm (in x) of the solution Z(x, obtained in Theorem 3.1.2 tends to 0 when t → ∞, i.e. ~ t)kL2 (0,l) = 0. lim kZ(·,

t→∞

Now we consider the first problem of spin equation (3.1.9) over the domain Q∗T = {x ∈ R+ , 0 ≤ t ≤ T } as follows ~ t) = 0, Z(0,

~ 0) = Z ~ 0 (x), Z(x,

(3.1.28)

~ 0 (x) is the initial data defined on the infinite interval R+ = [0, ∞). Denote where Z the assumptions (1), (2), (3) for l = ∞ by (1∗ ), (2∗ ), (3∗ ). 2. Weak solutions to spin equations ~ and Z ~ 0 (x) meet conditions (1∗ ), (2∗ ), (3∗ ). Theorem 3.1.4 Assume that f~(x, t, Z) Then the homogeneous spin equation (3.1.9) with the first boundary condition (3.1.28) admits unique global weak solution ~ t) ∈ L∞ (0, T ; H 1 (R+ )) Z(x,

\

(2,1)

W2

(Q∗T ).

Proof. Select a subsequence ls such that when s → ∞, ls → ∞ and for every s, ~ 0(s) (x) defined on [0, ls ] satisfyconstruct a three-dimensional vector-valued function Z ~ 0(s) (x) ≡ ing the boundary condition at x = 0 and x = ls such that for x ∈ [0, ls−1 ], Z ~ 0 (x). At the same time, the H 1 (0, ls )-norm of Z ~ 0(s) (x) is bounded uniformly in s. Z Therefore we can first consider the first boundary problem of (3.1.9) on the rectangular (s) (s) domain QT = {0 ≤ x ≤ ls , 0 ≤ t ≤ T } whose unique global weak solution on QT will ~ (s) (x, t). It is clear that Z ~ (s) (x, t) ∈ L∞ (0, T ; H 1 (0, ls )) T W2(2,1) (Q(s) be denoted by Z T ). Therefore we have ~ (s) (·, t)kH 1 (0,l ) + kZ ~ t(s) k 2 (s) + kZ ~ (s) k 2 (s) ≤ K7 , sup kZ xx L (Q ) s L (Q ) 0

0≤t≤T

T

(3.1.29)

T

where K7 is independent of ls but depends on ε. ~ (si ) (x, t)} from {Z ~ (s) (x, t)} such that Therefore we can choose a subsequence {Z ~ (si ) (x, t) and Z ~ (si ) (x, t) uniformly converge to Z(x, ~ t) and Z ~ x (x, t) on any rectangular Z x ˜ T = {0 ≤ x ≤ l, 0 ≤ t ≤ T } (l > 0), and, at the same time, Z ~ (si ) (x, t) and domain Q xx (si ) ~ ~ ~ Zt (x, t) weakly converge to Zxx (x, t) and Zt (x, t). It is easy to see that this vector ~ t) is the global solution of (3.1.9) and (3.1.28). This finishes the proof of the Z(x, theorem.

Landau–Lifshitz Equations

62

3. Estimates of approximate solutions and the solvability Theorem 3.1.5 If b < 0 and T = ∞, then for the solution obtained in Theorem 3.1.4, there holds ~ t)kL2 (0,+∞) = 0. lim kZ(·, (3.1.30) t→∞

Now we intend to give the uniform estimates for the approximate solutions of (3.1.9) and (3.1.28) on the unbounded domain Q∗T = {x ∈ R+ , 0 ≤ t ≤ T }. Lemma 3.1.5 Suppose that (1∗ ), (2∗ ), (3∗ ) hold. Then for the solution of (3.1.9) and (3.1.28), there holds ~ ε (·, t)kH 1 (R+ ) ≤ K8 , sup kZ

(3.1.31)

~ ε (x, t)kL∞ (R+ ) ≤ K9 , kZ

(3.1.32)

0≤t≤T

and hence where K8 , K9 are independent of ε. Corollary 3.1.1 Under the conditions of Lemma 3.1.5 and that f~(x, t, 0) ∈ Ln (Q∗T ), ~ 0 (x) ∈ Ln (R+ ), there holds Z ~ ε (·, t)kLn (R+ ) ≤ K10 (n), sup kZ

(3.1.33)

0≤t≤T

where K10 is independent of ε, but depends on n (≥ 2). Lemma 3.1.6 Suppose that (1∗ ), (2∗ ), (3∗ ) and one of the following conditions hold. ~ ∈ Q∗T × R3 : (4∗1 ). For any (x, t, Z) ~ ≤ c(x, t)F (Z) ~ + d(x, t), |f~(x, t, Z)|

(3.1.34)

~ is a continuous function in Z ~ ∈ R3 , c(x, t), d(x, t) ∈ L∞ (0, T ; L2 (R+ )); where F (Z) ~ ∈ Q∗T × R3 : (4∗2 ). For any (x, t, Z) ~ ≤ c(x, t)|Z| ~ l + d(x, t), |f~(x, t, Z)|

(3.1.35)

where l ≥ 0, c(x, t) ∈ L∞ (Q∗T ), d(x, t) ∈ L∞ (0, T ; Ls (R+ )), 1 < s ≤ 2. ~ ε (x, t) of (3.1.9) and (3.1.28), there holds Then for the solution Z ~ εt (·, t)kH −1 (R+ ) ≤ K11 , sup kZ

0≤t≤T

where K11 is independent of ε.

(3.1.36)

One-dimensional Landau–Lifshitz equations

63

Proof. Testing the equation (3.1.9) by ψ ∈ H01 (R+ ), we have Z

∞ 0

~ εt (x, t)dx = −ε ψ(x)Z − +

Z

Z

Z

∞ 0 ∞

0

0

∞

~ εx (x, t)dx ψx (x)Z

~ ε (x, t) × Z ~ εx (x, t)]dx ψx (x)[Z ~ ε (x, t))dx. ψ(x)f~(x, t, Z

For the first two terms on the right-hand side of the above equality, we have

and

Z ∞ ~ ψ (x) Z (x, t)dx ≤ K7 kψx kL2 (R+ ) x εx 0

Z ∞ ~ ε (x, t) × Z ~ εx (x, t)]dx ≤ K7 K8 kψx kL2 (R+ ) . ψx (x)[Z 0

Under condition (4∗1 ), the third term on the right-hand side can be estimated as follows Z ∞ ~ ~ ~ ε (x, t))kL2 (R+ ) . ψ(x) f (x, t, Z (x, t))dx ≤ kψkL2 (R+ ) kf~(x, t, Z ε 0

It follows from (3.1.34) that

~ ε )| ≤ M c(x, t) + d(x, t), |f~(x, t, Z

(3.1.37)

~ ~ ε (x, t)) is uniformly (in ε) where M = sup|Z|≤K |F (Z)|. This means that f~(x, t, Z ~ 8 ∞ 2 + bounded in the space L (0, T ; L (R )). ~ ε (x, t)} is Under the condition (4∗2 ), it follows from Corollary 3.1.1 that {Z uniformly bounded (in ε) in L∞ (0, T ; Lsl (R+ )). Then Z ∞ ~ ~ ~ ε (x, t))kLs (R+ ) , ψ(x)f (x, t, Zε (x, t))dx ≤ C5 kψkLr (R+ ) kf~(x, t, Z 0

where 1r + 1s , 1 < s ≤ 2, 2 ≤ r < ∞, C5 is independent of ε and t ∈ [0, T ]. Using (3.1.35), we have that ~ ε (x, t))kLs (R+ ) ≤ C6 kckL∞ (Q∗ ) [K10 (ls)]l kf~(x, t, Z T + kdkL∞ (0,T ;Ls (R+ )) ,

where C6 is independent of ε and t ∈ [0, T ]. Therefore in both cases, we have Z

+∞ 0



~ εt (x, t)dx ≤ C7 kψkH 1 (R+ ) ϕ(x)Z

where C7 is independent of ε and t ∈ [0, T ].

Landau–Lifshitz Equations

64

Lemma 3.1.6 follows. Now consider the solvability of problem (3.1.1) and (3.1.28) over the infinite rectangular domain Q∗T = {x ∈ R+ , 0 ≤ t ≤ T }. ~ t) ∈ L∞ (0, T ; H 1 (R+ )) T C(Q∗ ) is called a weak Definition 3.1.2 A vector Z(x, T solution of (3.1.1) and (3.1.28) if for any test function ϕ(x, t) ∈ Φ∗ = {ϕ : ϕ ∈ ~ t) meets C(Q∗T ), ϕ(x, T ) = 0, supp ϕ < ∞}, Z(x, Z Z

Q∗T

~ − ϕ x (Z ~ ×Z ~ x ) + ϕf~(x, t, Z(x, ~ t))]dxdt [ϕt Z

+

Z

∞ 0

~ 0 (x)dx = 0, ϕ(x, t)Z

(3.1.38)

where ϕ(0, t) = 0. 4. Global weak solutions ~ ε (x, t) of spin equation Now we prove that when ε → 0, the approximate solutions Z (3.1.9) and (3.1.28) tend to the global weak solution of (3.1.1) and (3.1.28). First we ~ ε (x, t) of spin equation (3.1.9) and (3.1.8), note that for the approximate solutions Z there hold Z Z

Q∗T

~ ε − εϕx Z ~ εx − ϕx (Z ~ε × Z ~ εx ) + ϕf~(x, t, Z ~ ε (x, t))]dxdt [ϕt Z

+

Z

∞ 0

~ 0 (x)dx = 0 ϕ(x, t)Z

(3.1.39)

for all ϕ ∈ Φ∗ with ϕ(0, t) = 0. ~ ε (x, t) is a Lemmas 3.1.1 and 3.1.6 imply that the set of approximate solutions Z T (1) uniformly (in ε) bounded set in the space L∞ (0, T ; H 1 (R+ )) W∞ (0, T ; H −1 (R+ )). Hence, in any rectangular domain QT with width L: QT = {0 ≤ x ≤ L, 0 ≤ t ≤ T }, ~ ε (x, t)} is also unformly bounded in L∞ (0, T ; H 1 (0, L)) T W (1) (0, T ; the set {Z ∞ H −1 (0, L)). It follows from the similar proof as that of Lemma 3.1.5 that for any rectangular domain QT , there holds ~ εk ( 1 , 1 ) kZ C 2 4 (Q

T)

≤ (1 + L)K12 ,

(3.1.40)

where K12 is independent of ε and L. ~ ε (x, t)} from {Z ~ ε (x, t)} such that for some Choose a suitable subsequence {Z i ~ ~ ~ vector-valued function Z(x, t) there holds Zεi → Z uniformly in any QT for any ~ ε ) → f~(x, t, Z) ~ uniformly in any QT too. Moreover, we may width L and f~(x, t, Z i ~ ε x weakly converges to Z ~ x and verify that Z(x, ~ t) ∈ L∞ (0, T ; H 1(R+ )) T have that Z i (1/2,1/4) Cloc (Q∗T ). By the similar method we can see that the integrale relation (3.1.38) ~ t) is a global weak solution is the limit of the relation (3.1.39). These imply that Z(x, of (3.1.1) and (3.1.28). Therefore we have the following: ~ and Z ~ 0 (x) meet conditions (1∗ ), (2∗ ), (3∗ ) Theorem 3.1.6 Suppose that f~(x, t, Z) and one of (4∗1 ) and (4∗2 ) and assume that (3.1.1) is homogeneous. Then the first

One-dimensional Landau–Lifshitz equations

65

initial value problem (3.1.1) and (3.1.28) admits at least one global weak solution L∞ (0, T ; H 1(R+ ))

\

(1,1)

Cloc2 4 (Q∗T ).

Theorem 3.1.7 Suppose that conditions of Theorem 3.1.6 hold and b < 0. Then ~ t) to the first initial boundary value problem (3.1.1) and for the weak solution Z(x, (3.1.28), there holds ~ t)kL2 (R+ ) = 0. lim kZ(·, t→∞

Corollary 3.1.2 For the system of ferromagnetic spin chain ~t = Z ~ ×Z ~ xx + f~(x, t, Z), ~ Z

(3.1.41)

we can prove by similar method that there exists at least one global weak solution with the same properties as above under one of the following conditions: (1) the second boundary value problem: ~ x (0, t) = Z ~ x (l, t) = 0, Z ~ t=0 = Z ~ 0 (x), 0 ≤ x ≤ l; Z|

(3.1.42)

~ t) = Z ~ x (l, t) = 0, or Z ~ x (0, t) = Z(l, ~ t) = 0, Z(0, ~ t=0 = Z ~ 0 (x), 0 ≤ x ≤ l; Z|

(3.1.43)

(2) the mixed condition:

(3) the periodic boundary condition ~ t) = Z(x ~ + 2D, t) = 0, Z(x,

~ t=0 = Z ~ 0 (x), Z|

D > 0;

(3.1.44)

(4) the initial condition ~ t=0 = Z ~ 0 (x), Z|

x ∈ R.

(3.1.45)

Remark Under the above assumption (2), the condition (3.1.12) is not essential. ~ on Z ~ can be eliminated. For example we In fact, the growth condition of f~(x, t, Z) may make use of the cutoff method to obtain the proof.

3.2

Nonlinear Initial-boundary Value Problem for the System of Ferromagnetic Spin Chain

We continue to consider the equation in the above section, but with nonlinear boundary conditions. This time, we will use the difference method to prove the existence of global weak solutions.

Landau–Lifshitz Equations

66

3.2.1

Nonlinear Initial-boundary Value Problem for the System of Ferromagnetic Spin Chain

1. Equations, boundary conditions and initial conditions Consider the following system of ferromagnetic spin chain: ~t = Z ~ ×Z ~ xx + f~(x, t, Z) ~ Z

(3.2.1)

and the relative spin equation: ~ t = εZ ~ xx + Z ~ ×Z ~ xx + f~(x, t, Z) ~ Z

(3.2.2)

with the nonlinear boundary condition  Z ~ x (0, t) = grad ψ0 (t, Z(0, ~ t)) ,  −Z ~ (l, t) = grad ψ (t, Z(l, ~ t)) x

and initial condition

(3.2.3)

1

~ 0) = ϕ(x) , Z(x,

(3.2.4)

~ and ψ1 (t, Z) ~ are vector-valued functions of variables t ∈ [0, T ] and where ψ0 (t, Z) 3 ~ ∈ R , “grad” denotes the gradient operator with respect to Z. ~ On the other hand Z we also consider, in this section, the global solution for problem (3.2.1) and (3.2.2) over the semi-infinite domain Q∗T = {x ∈ R+ , 0 ≤ t ≤ T } with the nonlinear boundary condition:  Z ~ x (0, t) = grad ψ0 (t, Z(0, ~ t)) , (3.2.5) Z ~ x (x, 0) = ϕ(x) .

To this aim, we will use the difference (in spatial variable) method to prove the existence of global weak solution. We suppose: ~ and ψ1 (t, Z) ~ are two scalar functions which are continuously first(1) ψ0 (t, Z) ~ order differentiable in t ∈ [0, T ] and continuously second-order differentiable in t, Z 2 2 ~ = ∇ ψ0 (t, Z) ~ and H1 (t, Z) ~ = ∇ ψ1 (t, Z) ~ are mixed. 3 × 3 Hessian matrix H0 (t, Z) non-negatively definite and grad ψ0 (t, 0) = grad ψ1 (t, 0) = 0. ~ is a three-dimensional vector-valued function which is continuously (2) f~(x, t, Z) ~ ∈ R3 . Moreover, condition (3.1.11) differentiable with respect to (x, t) ∈ QT and Z holds, that is ~ · ξ ≤ b|ξ|2 , ∀ ξ ∈ R3 , ξ · f~Z~ (x, t, Z) (3.2.6) (3) ϕ(x) ∈ H 2 (0, l) is the three-dimensional vector-valued initial data which, on the end points of the interval [0, l], is compatible with the nonlinear condition (3.2.3). Now divide the interval [0, l] into J small intervals by points xj = jh (j = 0, 1, . . . , J) where Jh = l, J is a positive integer, h is the length of the step. Vector ~ h (t) = {Z ~ j (t) : j = 0, 1, . . . , J} is function defined on the points xj (j = 0, 1, . . . , J). Z

One-dimensional Landau–Lifshitz equations

67

2. Construct the nonlinear ordinary differential systems We establish the following difference-differential equation: ~j ~j ~ dZ ∆+ ∆− Z ~ j × ∆+ ∆− Z j =ε + Z dt h2 h2 + f~(xj , t, zj ), j = 1, 2, . . . , J − 1,

(3.2.7)

and the nonlinear boundary condition relative to (3.2.3): ~ 0 (t) ∆+ Z ~ 0 (t)) , = grad ψ0 (t, Z (3.2.8) h ~ J (t) ∆− Z ~ J (t)) − = grad ψ1 (t, Z h ! J−1 ~j ~j XZ 1 ∆+ Z ∆+ Z ~ ~ ~ h − · fz (xj , t, τ Zj+1 + (1 − τ )Zj ) dτ h h j=0 0 ~ 0 )grad ψ0 (t, Z ~ 0 ) − f~(l, t, Z ~ J ) · grad ψ1 (t, Z ~ J ). − f~(0, t, Z

(3.2.9)

~ j h and summing from j = 1 to J and integrating over [0, t], Multiplying (3.2.7) by Z we have ~ h k2 + 2ε kδ Z 2

Z

t

~ h (t)k2 dt ≤ (2b + 1) kδ 2 Z 2

0

Z

t

0

~ h k2 dt + 2G, kδ Z 2

where 1 G= 2

Z

+

Z

+

t J−1 X Z 1

0 j=0

Z

t

0 t 0



0

2 ~ j+1 )dτ hdt + 1 kδψh k2 f~x (xj+2 , t, Z 2 2

~ 0 ) · grad ψ0 (t, Z ~ 0 ) + f~(l, t, Z ~ J ) · grad ψ1 (t, Z ~ J )]dt [f~(0, t, Z ~ 0 ) + ψ1 (t, Z ~ J )]dt [ψ0 (t, Z

~ 0 (t)) + ψ1 (t, Z ~ J (t))] + [ψ0 (0, ϕ0 (t)) + ψ1 (l, ϕJ (t))] − [ψ0 (t, Z which is clearly bounded. Since J−1 X

~ h (t)k2 = kδ 2 Z 2

J−1 X

Therefore, we have ~ h k2 + 2ε kδ Z 2

Z

~ 2 ∆+ Z j h, h

~ h (t)k2 = kδ Z 2

0

t

j=1

j=1

~ 2 ∆+ ∆− Z j h. h2

~ h (t)k2 dt ≤ (2b + 1) kδ 2 Z 2

where C2 is independent of h, ε ≥ 0.

Z

t 0

~ h k2 dt + C2 , kδ Z 2

Landau–Lifshitz Equations

68

3.2.2

Discrete Solution of Nonlinear Ordinary Differential Systems

1. Estimates for discrete solutions It follows from the above discussions that we have ~ j of the nonlinear Lemma 3.2.1 Let assumption (1)–(3) hold. For the solution Z ordinary differential system (3.2.7) with (3.2.8) and (3.2.9), there holds ~ h k∞ + sup kδ Z ~ h k2 + sup kδ Z

0≤t≤T

√

Z

ε

0≤t≤T

+

√

ε

Z

T 0

~ h0 k22 dt kδ Z

!1/2

T 0

kδ

2~

Zh (t)k22 dt

!1/2

≤ K1 ,

(3.2.10)

where K1 is a constant independent of h, ε > 0. ~ 0 (t) = {Z ~ 0 (t) : j = 0, 1, . . . , J}. Differentiating (3.2.7) Now we estimate y(t) = Z h j with respect to t, we have ∆+ ∆− y j ∆+ ∆− y j ~ j × ∆+ ∆− y j + yj × +Z 2 2 h h h2 ~ j ) + f~z (xj , t, Z ~ j )yj , + f~t (xj , t, Z j = 1, 2, . . . , J − 1.

yj0 = ε

(3.2.11)

The relative boundary conditions are  ∆+ y 0  ~ ~    h = grad ψ0t (t, Z0 ) + H0 (t, Z0 )y0 ,  ∆ y   ~ J ) + H1 (t, Z ~ J )yJ ,  − − J = grad ψ1t (t, Z

(3.2.12)

h

and

 ∆+ ∆− ϕ j ∆+ ∆− ϕ j    + ϕj × + f~(xj , 0, ϕj ), yj (0) = ε  2   h h2

(3.2.13)

 (E + hH0 (0, ϕ0 ))y0 (0) = y1 (0) − h grad ψ0t (0, ψ~0 ),      (E + hH (0, ϕ )y (0) = y ~ 1 J−1 (0) − h grad ψ1t (0, ψJ ), J J

where E is a 3 × 3 unit matrix. Taking inner product of system (3.2.11) with yj h and summing from j = 1 to J − 1, we have J−1 X j=1

yj yj0 h

J−1 X

J−1 X ∆+ ∆− y j ~ j × ∆+ ∆− y j h =ε yj h + yj Z 2 h h2 j=1 j=1

+

J−1 X j=1

!

~ j )h + yj f~t (xj , t, Z

J−1 X j=1

~ j )yj h. yj f~z (xj , t, Z

(3.2.14)

One-dimensional Landau–Lifshitz equations

69

It is obvious that J−1 X

yj yj0 h

j=1

1d = 2 dt

J−1 X j=1

2

!

|yj | h ,

J−1 X X 1 J−1 1 J−1 X ~ 2 ~ ~ j )|2 h, y f (x , t, Z )h ≤ |y | h + |f~t (xj , t, Z j t j j j 2 2 j=1 j=1 j=1

J−1 X j=1

~ j )yj h ≤ b yj f~z (xj , t, Z

J−1 X j=1

|yj |2 h.

For the first term on the right-hand side of (3.2.14), we have J−1 X

yj

j=1

∆+ ∆− y j ~ 0 ) + H0 (t, Z ~ 0 )y0 ) h = −kδyh (t)k22 − y0 (grad ψ0t (t, Z 2 h ~ J ) + H1 (t, Z ~ J )yJ ) , − yJ (grad ψ1t (t, Z

where ~ 0 )| ≤ C3 kyh k∞ ≤ 1 kδyh k2 + C4 kyh (t)k2 + C5 , |y0 grad ψ0t (t, Z 2 2 8 ~ J )| ≤ 1 kδyh k22 + C4 kyh (t)k22 + C5 . |yJ grad ψ1t (t, Z 8 It follows from (3.2.12) that |y0 | ≤ C6 |y1 | + C7 ,

|yJ | ≤ C6 |yJ−1 | + C7 .

Hence kyh (t)k22

≤ C8

J−1 X j=1

|yj |2 h + C9 .

(3.2.15)

For the second term on the right-hand side of (3.2.14), direct computations give J−1 X j=1

yj

J−1 ~ j ∆+ y j X ∆+ Z ~ j × ∆+ ∆− y j h = − Z y × h j h2 h h j=1

!

!

~ 0 × ∆+ y 0 − y0 · Z h

!

!

~ J × ∆− y J , + yJ · Z h

where y0 ·

~ 0 × ∆+ y 0 Z h

y J ·

~ J × ∆− y J Z h

! ~ 0 × (grad ψ0t (t, Z ~ 0 ) + H0 (t, Z ~ 0 )y0 ))| = |y0 · (Z

!

≤ C10 kyh (t)k2∞ + C11 ε ≤ kδyh (t)k22 + C12 kyh (t)k22 + C13 , 8 ε ≤ kδyh (t)k22 + C12 kyh (t)k22 + C13 . 8

(3.2.16)

Landau–Lifshitz Equations

70

For the first term on the right-hand side of (3.2.16), we have ~ j ∆+ y j ∆+ Z yj × hdt h h j=1

J−1 X

Z t 0

Z

≤

t

0

~ h (t)k2 kδyh (t)k2 dt kyh (t)k∞ kδ Z

≤ C14 K1 ε ≤ 4



!

Z

t 0

Z

t 0

1/2

3/2

(kyh (t)k2 kδyh (t)k2 + kyh (t)k2 kδyh (t)k2 )dt

kδyh (t)k22 dt

+ C15

Z

t 0

kyh (t)k22 dt,

where C15 is a constant independent of h but depends on ε > 0. Therefore we have from (3.2.14) that kyh (t)k22

kyh (0)k22

−

Z t εZ t 2 kδyh (t)k2 dt ≤ C16 kyh (t)k22 dt + C17 , + 2 0 0

where C16 , C17 are positive constants independent of h but depending on ε. This ~ 0 (t)k2 and R t kδ Z ~ 0 (t)k2 dt are uniformly bounded in h. implies that sup0≤t≤T kZ h h 0

~ h (t) of (3.2.7)–(3.2.9), there holds Lemma 3.2.2 Let (1)–(3) hold. For the solution Z sup 0≤t≤T

~ 0 (t)k2 kZ h

Z

+

t 0

~ 0 (t)k2 dt kδ Z h 2

!1/2

≤ K2 (ε).

(3.2.17)

~ ∆ ∆ Z

Since the coefficient matrix of + h2− j in (3.2.7) is positively definite, sup0≤t≤T ~ h (t)k2 is uniformly bounded. Similarly, for the differential-difference equation, kδ 2 Z we have 2 ~ 0 (t) ~ ~ j (t) ∆+ Z ∆2 ∆− Z j ~ j+1 (t) × ∆+ ∆− Zj (t) +Z =ε + 3 h h h3 Z ~ j (t) ∆+ ∆− Z ~ j (t) 1 ∆+ Z ~ j+1 )dz + × + f~x (xj+1 , t, Z h h2 0 Z 1 ~ ~ j+1 + (1 − τ )Z ~ j )dτ ) ∆+ Zj (t) . + f~z (xj , t, τ Z h 0

This implies that

RT 0

~ h (t)k2 dt is uniformly bounded in h. Hence, we have kδ 2 Z 2

~ h (t) of (3.2.7)–(3.2.9), there holds Lemma 3.2.3 Let (1)–(3) hold. For the solution Z 2~

sup kδ Zh (t)k2 +

0≤t≤T

Z

T 0

kδ

3~

where K3 (ε) is a constant independent of h.

Zh (t)k22 dt

!1/2

≤ K3 (ε),

(3.2.18)

One-dimensional Landau–Lifshitz equations

71

2. Further estimates for the discrete solutions Using the discrete interpolation formula, we get from the above estimates the following lemma. ~ h (t) of (3.2.7)–(3.2.9), there holds Lemma 3.2.4 Let (1)–(3) hold. For the solution Z for t ∈ [0, T − ∆t] ~ j (t)| ≤ K4 h, sup |∆+ Z

0≤t≤T

j = 0, 1, . . . , J − 1,

~ j (t)| ≤ K5 h1/2 , sup |∆2+ Z

j = 0, 1, . . . , J − 1,

(3.2.20)

~ j (t + ∆t) − Z ~ j (t)| ≤ K6 ∆t1/2 , max |Z

(3.2.21)

~ j (t + ∆t) − ∆+ Z ~ j (t)| ≤ K7 h∆t1/4 , |∆+ Z

(3.2.22)

0≤t≤T

j=0,1,...,J

max

j=0,1,...,J−1

(3.2.19)

where Kj , (j = 4, 5, 6, 7) are constants independent of h but depending on ε.

3.2.3

Global Weak Solution for the Spin System

1. Construct three-dimensional vector-valued functions In order to establish the existence of global weak solution of (3.2.2) with nonlinear boundary condition (3.2.3) and initial condition (3.2.4), we approximate it by the discrete solution of (3.2.7)–(3.2.9). For this reason, we construct a vector-valued ~ ∗ (x, t) defined on QT as follows: on Qj = {xj ≤ x ≤ xj+1 , 0 ≤ t ≤ T }, function Z h T j = 0, 1, . . . , J − 1: ~ j+1 (t) + xj+1 − x Z ~ j (t). ~ ∗ (x, t) = x − xj Z Z h h h ~ ∗ (x, t)} is uniformly bounded It follows from (3.2.10), (3.2.19) and (3.2.21) that {Z h and equi-continuous over QjT = {xj ≤ x ≤ xj+1 , 0 ≤ t ≤ T } (j = 0, 1, . . . , J − 1), ~ ∗ (x, t)} uniformly converge over h > 0. We may choose a sequence {hj } such that {Z hj ~ t) as hj → 0. Similarly, construct on QjT QT to a vector-valued function Z(x, ~ ~ ∗ ~ (x, t) = x − xj ∆+ Zj+1 (t) + xj+1 − x ∆+ Zj (t) , j = 0, 1, . . . , J − 2, Z h h h h h and ~ ∗ ~ (x, t) = ∆− ZJ−1 , Z h h

on QJ−1 T .

It follows from (3.2.10), (3.2.20) and (3.2.22) that we may choose a subsequence of ∗ ~ (x, t)} uniformly converges over QT to a {hj }, still denoted by {hj }, such that {Z hj ~ vector-valued function Zx (x, t) as hj → 0.

Landau–Lifshitz Equations

72

~ h (t) = {Z ~ j (t) : j = 0, 1, . . . , J} let For the discrete vector-valued function Z ~ ~ (1) (x, t) = ∆+ Zj (t) , on Q(j) , j = 0, 1, . . . , J − 1 Z T h h ~0 (1) ~ 0 (t), ~˜h (x, t) = ∆+ Zj (t) , on Q(j) , j = 0, 1, . . . , J − 1 Z Z T j h ~ j (t) ∆+ ∆− Z (j) , on QT , j = 1, 2, . . . , J − 1, h2 ~ ∆+ ∆− Z(t) (0) , on QT , h2 ~ j (t) ∆2+ ∆− Z (j) , on QT , j = 1, 2, . . . , J − 2, 3 h 2 ~ 1 (t) ∆+ ∆− Z (0) , on QT , 3 h 2 ~ J−2 (t) ∆+ ∆− Z (J−2) , on QT . h3

~ h (x, t) = Z ~ j (t), Z ~˜h (x, t) = Z ~ (2) (x, t) = Z h ~ (2) (x, t) = Z h ~ (3) (x, t) = Z h ~ (3) (x, t) = Z h ~ (3) (x, t) = Z h

These functions are piecewise constant in direction of the variable x on QT . It is clear that ~ ∗ (x, t) − Z ~ h (x, t)| ≤ K4 h, |Z h ∗ ~˜h (x, t) − Z ~ (1) (x, t)| ≤ K5 h1/2 . |Z h On the other hand, it follows from Lemmas 3.2.2 and 3.2.3 that ~ h kL∞ (Q ) + kZ ~ (1) kL∞ (Q ) + sup kZ ~ (2) (·, t)kL2 (0,l) kZ h h T T 0≤t≤T

(1) ~˜h (·, t)kL2 (0,l) + kZ ~˜h kL2 (Q ) + kZ ~ (3) kL2 (Q ) + sup kZ h T T 0≤t≤T

≤ K8 , where K8 is a constant independent of h but depending on ε. ~ h } and {Z ~ (1) } uniformly converge We may choose {hj } such that when hj → 0, {Z j hj (2) (1) ~ ~ ~ ~ to Z(x, t) and Zx (x, t) in QT respectively; and {Zhj } and {Zhj } weakly converge in ˜~ ~ (2) (x, t) and Z(x, ~ (3) } and Lp (0, T ; L2 (0, l)) (2 ≤ p < ∞) to Z t) respectively; and, {Z hj (1) (1) ~˜ } weakly converge in L2 (QT ) to Z ~ (3) (x, t) and Z ~˜ (x, t) respectively. The norms {Z hj ˜~ ~ (2) (x, t) and Z(x, of Z t) in Lp (0, T ; L2 (0, l)) are uniformly bounded in p. This implies ˜~ ~ (2) and Z(x, that Z t) ∈ L∞ (0, T ; L2 (0, l)). 2. The characteristics of vector-valued functions ˜~ ~ (2) = Z ~ xx , Z ~ (3) = Z ~ xxx , Z ~t By the standard method we can prove that Z = Z (1) ˜~ ~ xt . That is, the limit functions Z(x, ~ t) is weakly differentiable and and Z = Z ∞ 2 2 ~ ~ ~ ~ ~ Zx , Zxx , Zt ∈ L (0, T ; L (0, l)), Zxt , Zxxx ∈ L (QT ).

One-dimensional Landau–Lifshitz equations

73

(0)

Let Φ(x, t) be any test function, and Φh (x, t) = Φj (t) = Φ(xj , t) in QT , ~ j (t)) in Q(j) , j = 0, 1, . . . , J − 1. Similarly we can define Fh (x, t) = f~(xj , t, Z T j = 0, 1, . . . , J − 1. It follows from (3.2.7) that Z

0

~ ~ 0 (t) − ε ∆+ ∆− Zj (t) Φj (t) Z j h2 j=0

T J−1 X

"

~ ~ j (t) × ∆+ ∆− Zj (t) − f~(xj , t, Z ~ j (t)) hdt = 0. −Z h2 #

This is equivalent to the following integral identity: Z

T 0

Z

l 0

~˜h (x, t) − εZ ~ (2) (x, t) Φh (x, t)[Z h (2)

~ h (x, t) × Z ~ (x, t) − Fh (x, t)]dxdt = 0. −Z h

~ h (x, t), Fh (x, t) uniformly converge on QT to Sending hi → 0, we have that Φh (x, t), Z ˜~ 2 ~ t), f~(x, t, Z(x, ~ t)); Z ~ (2) Φ(x, t), Z(x, h (x, t) and Zh (x, t) weakly converge in L (QT ) to ~ t (x, t) and Z ~ xx , respectively. Therefore by sending hi → 0, we get Z Z

T 0

Z

l 0

~ t (x, t) − εZ ~ xx (x, t) Φ(x, t)[Z

~ ~ xx (x, t) − f~(x, t, Z(x, ~ t))]dxdt = 0. −Z(x, t) × Z

3. Nonlinear weak solution to the spin equations ~ t) solves the spin equation (3.2.2) in It follows from above discussion that Z(x, the sense of distribution. ~ h (x, t)} uniformly converges on QT to Z(x, ~ t), and Z ~ (1) (x, t) uniformly Since {Z i hi ~ converges on QT to Zx (x, t) when hi → 0, the nonlinear boundary condition (3.2.3) ~ t) in the classical sense. and initial condition (3.2.4) are satisfied by Z(x, Theorem 3.2.1 Under conditions (1)–(3), problem (3.2.2)–(3.2.4) admits at least ~ t) which has classical first order differential Z ~ x (x, t) one global weak solution Z(x, ~ xx , Z ~ t ∈ L∞ (0, T ; L2 (0, l)) and Z ~ xt (x, t), Z ~ xxx (x, t) ∈ L2 (QT ). and weak derivatives Z Moreover , the equation is met in the sense of distribution and the boundary condition (3.2.3) and initial condition (3.2.4) in the classical sense. Theorem 3.2.2 Under conditions (1)–(3), the solution to problem (3.2.2)–(3.2.4) is unique. ~ t) and Z(x, ~ t) to problem (3.2.2)–(3.2.4), Proof. If there are two solutions Z(x, ~ t) − Z(x, ~ t) solves the following equation then w(x, t) = Z(x, Z Z

QT

~ xx − Z ~ × wxx Φ[wt − εwxx − w × Z

~ t)) − f~(x, t, Z(x, ~ t))]dxdt = 0, − [f~(x, t, Z(x,

Landau–Lifshitz Equations

74

˜~ wx (0, t) = H0 (t, Z(0, t))w(0, t), ˜~ −wx (l, t) = H1 (t, Z(l, t))w(l, t), w(x, 0) = 0,

where Φ(x, t) ∈ L2 (QT ) is any test function and ˜~ H0 (t, Z(0, t)) = ˜~ H1 (t, Z(l, t)) =

Z

Z

1 0 1

0

˜~ ~ t))dτ, H0 (t, τ Z(0, t) + (1 − τ )Z(0, ˜~ ~ t))dτ. H1 (t, τ Z(l, t) + (1 − τ )Z(l,

Take test function Φ = w(x, t) to give kw(·, t)k2L2 (0,l) + εkwx k2L2 (QT ) ≤ε

t

Z

+ −

0

Z

[w(l, t)wx (l, t) − w(0, t)wx(0, t)]dt l

0

~ t) × wx (l, t) − w(0, t)Z(0, ~ t) × wx (0, t)]dt [w(l, t)Z(l,

Z Z

QT

~ x × wx dxdt + bkwk2 2 wZ L (QT ) .

Since w(l, t)wx (l, t) − w(0, t)wx (0, t) ˜~ ˜~ = −w(l, t) · H1 (t, Z(l, t))w(l, t) − w(0, t) · H0 (t, Z(0, t))w(0, t) ≤ 0; ~ t) × wx (l, t) − w(0, t)Z(0, ~ t) × wx (0, t)| |w(l, t)Z(l, ˜~ ~ t) × H1 (t, Z(l, = |w(l, t) · Z(l, t))w(l, t) ˜~ ~ t) × H0 (t, Z(0, + w(0, t) · Z(0, t))w(0, t)| ≤ C18 kw(·, t)k2L∞ (0,l) ε ≤ kwx (·, t)k2L2 (0,l) + C19 (ε)kw(·, t)k2L2 (0,l) 2

and

Hence

Z Z ε ~ w · Zx × wx dxdt ≤ kwx k2L2 (QT ) + C20 (ε)kwk2L2 (QT ) . 2 QT

kw(·, t)k2L2 (0,l)

≤ C21 (ε)

Z

t 0

kw(·, t)k2L2 (0,l) dt.

This implies kw(·, t)kL2 (0,l) = 0, ∀ t ∈ [0, T ]. The proof of the theorem is complete.

One-dimensional Landau–Lifshitz equations

3.2.4

75

Global Weak Solution to the Equations of Ferromagnetic Spin Chain

1. Global weak solutions to spin equations Now we prove the existence of global weak solution to equation of ferromagnetic spin chain (3.2.1) with nonlinear boundary condition (3.2.3) and initial condition (3.2.4). We have obtained the existence of global weak solution of (3.2.2)–(3.2.4) for ε > 0 and have proved some uniform a priori estimates for such solutions. Lemma 3.2.5 Let condition (1), (2) and (3) hold. Then the following estimate for ~ ε (x, t) (ε > 0) to problem (3.2.2)–(3.2.4) holds the solution Z ~ ε kL∞ (Q ) + sup kZ ~ εx kL2 (0,l) ≤ K9 , kZ T

(3.2.23)

0≤t≤T

where K9 is independent of ε > 0. Let g(x, t) ∈ H 1 (0, l) be any test function. Simple computations give Z

l 0

~ εt (x, t) = −ε g(x)Z

Z

−

Z

+

Z

l 0 l

0

0

l

~ εx (x, t)dx gx (x)Z

~ ε (x, t) × Z ~ εx ))dx gx (x)(Z ~ ε (x, t))dx, g(x)f~(x, t, Z

where g(0) = g(l) = 0. It follows from (3.2.23) that there is a constant C22 independent of t ∈ [0, T ] and ε > 0 such that Z l ~ εx (x, t)dx ≤ C22 kgkH 1 (0,l) . g(x) Z 0

~ εt (x, t)} is bounded in L∞ (0, T ; H −1 (0, l)) uniform in ε. Then we This means that {Z have the following lemma. Lemma 3.2.6 Let (1)–(3) hold. Then for the global weak solution of (3.2.2)–(3.2.4) there holds ~ εt (·, t)kH −1 (0,l) ≤ K10 , sup kZ (3.2.24) 0≤t≤T

where K10 is independent of ε. Lemma 3.2.7 Let (1)–(3) hold. Then for the global weak solution of (3.2.2)–(3.2.4), there holds   |Z ~ ε (x1 , t) − Z ~ ε (x2 , t)| ≤ K11 |x1 − x2 |1/2 , (3.2.25)  |Z ~ ε (x, t1 ) − Z ~ ε (x, t2 )| ≤ K12 |t1 − t2 |1/4 , where K11 , K12 are independent of ε, x1 , x2 ∈ [0, l] and t1 , t2 ∈ [0, T ].

Landau–Lifshitz Equations

76

~ ε (x, t)dx, then Sεt (x, t) = x Z ~ εt (x, t)dx, Sεx (x, t) = Proof. Let Sε (x, t) = 0x Z 0 1 ~ ~ Zε (x, t), Sεxx (x, t) = Zεx (x, t). For any g(x) ∈ H0 (0, l) there holds R

R

Z Z l l 0 ~ εt (x, t)dx g (x)S g(x) Z εt (x, t) = 0 0

≤ K10 kgkH01 (0,l) ≤ C23 kgkL2 (0,l) ,

where g(0) = g(l) = 0, C23 is independent of t ∈ [0, T ] and ε > 0. This implies sup kSεt (·, t)kL2 (0,l) ≤ C24 .

0≤t≤T

Then it is clear that sup kSε (·, t)kH 2 (0,l) ≤ C25

0≤t≤T

with C25 independent of ε > 0. It follows from the interpolation inequality that |Sεx (x, t1 ) − Sεx (x, t2 )|

1/4

3/4

≤ C26 kSε (x, t1 ) − Sε (x, t2 )kL2 kSε (x, t1 ) − Sε (x, t2 )kH 2 ,

≤ C26 k

Z

t2

t1

1/4

3/4

Sεt (x, t)dtkL2 kSε (x, t1 ) − Sε (x, t2 )kH 2 1/4

3/4

≤ C27 |t1 − t2 |1/4 · sup kSεt (x, t)kL2 kSε (·, t)kH 2 , ∀ x ∈ [0, l]. 0≤t≤T

Similarly, we have |Sεx (x1 , t) − Sεx (x2 , t)| ≤ C28 |x1 − x2 |1/2 kSεxx (·, t)kL2 (0,l) , ∀ x ∈ [0, l]. The constants C26 , C27 , C28 are independent of ε. Therefore we get the first estimate of (3.2.25). The second one can be proved in the similar manner. 2. Global weak solution ~ t) ∈ L2 (0, T ; H01 (0, l)) C(QT ) is called a weak soluDefinition 3.2.1 Vector Z(x, tion of problem (3.2.1) and (3.2.3)–(3.2.4), if for any test function g(x, t) ∈ H 1 (QT ), g(x, T ) = 0, there holds T

Z Z

QT

~ t) − gx (x, t)(Z(x, ~ t) × Z ~ x (x, t)) [gt (x, t)Z(x,

~ t))]dxdt + + g(x, t)f~(x, t, Z(x, −

Z

T 0

Z

l

g(x, 0)ϕ(x)dx

0

~ t) × grad ψ1 (t, Z(l, ~ t))) [g(l, t)(Z(l,

~ t) × grad ψ0 (t, Z(0, ~ t)))]dt = 0. + g(0, t)(Z(0,

(3.2.26)

One-dimensional Landau–Lifshitz equations

77

Theorem 3.2.3 Let (1)–(3) hold. Problem (3.2.1) and (3.2.3)–(3.2.4) admits at least ~ t) ∈ L∞ (0, T ; H 1 (0, l)) T C ( 12 , 41 ) (QT ). one weak solution Z(x, 0

~ ε (x, t) be the unique weak solution of problem (3.2.2)–(3.2.4) as Proof. Let Z ~ ε (x, t)} is uniabove. Since the estimates (3.2.23) and (3.2.25) is uniform in ε, {Z ~ ε (x, t)} formly bounded and equi-continuous. We may choose a subsequence {Z i ~ ~ from {Zε (x, t)} such that {Zεi (x, t)} uniformly converges on QT to some vector ~ t). Moreover, Z(x, ~ t) ∈ C ( 21 , 41 ) (QT ) since {Z ~ ε (x, t)} is uniformly bounded in Z(x, ( 21 , 41 ) ~ ε x (x, t)} weakly converges C (QT ). On the other hand, we may assume that {Z i p 2 ~ ~ ε x (x, t)} is bounded in in L (0, T ; L (0, l)) to Zx (x, t) for any 2 ≤ p ≤ ∞. Since {Z i ~ x (x, t) in Lp (0, T ; L2 (0, l)) is bounded Lp (0, T ; L2 (0, l)) uniformly in p, the norm of Z ~ x (x, t) ∈ L∞ (0, T ; L2 (0, l)). uniformly in p. Therefore, Z ~ ε (x, t) there holds Furthermore, for Z Z Z

QT

~ ε (x, t) − εgx (x, t)Z ~ εx (x, t) [gt (x, t)Z

~ ε (x, t) × Z ~ εx (x, t)) − gx (x, t)(Z ~ ε (x, t))]dxdt + + g(x, t)f~(x, t, Z −ε

Z

T

0

Z

l

g(x, 0)ϕ(x)dx

0

~ ε (l, t)) [g(l, t)grad ψ1 (t, Z

~ ε (0, t))]dt + g(0, t)grad ψ0 (t, Z −

Z

T 0

~ ε (l, t) × grad ψ1 (t, Z ~ ε (l, t))) [g(l, t)(Z

~ ε (0, t) × grad ψ0 (t, Z ~ ε (0, t)))]dt = 0, + g(0, t)(Z where g(x, t) ∈ H 1 (QT ), g(x, T ) = 0. ~ ε (x, t)} uniformly converges on QT to Z(x, ~ t), f~(x, t, Z ~ ε (x, t)) uniformly Since {Z i i ~ t)). Similarly, {Z ~ ε (0, t)}, {Z ~ ε (l, t)}, {grad ψ0 (t, Z ~ ε (0, t))} converges to f~(x, t, Z(x, i i i ~ ε (l, t))} uniformly converge on QT to Z(0, ~ t), Z(l, ~ t), and {grad ψ1 (t, Z i ~ ~ grad ψ0 (t, Z(0, t)) and grad ψ1 (t, Z(l, t)) respectively. The corresponding integrals are also bounded. The diffusion term with small parameter ε tends to zero. Therefore by sending εi → 0, we see that the above integral equality tends to ~ t) is a global weak solution of the integral equality (3.2.26). This indicates that Z(x, problem (3.2.1) and (3.2.3)–(3.2.4).

3.2.5

Mixed Boundary Value Problem

Now consider the mixed boundary value problem: (3.2.1) with ~ x (0, t) = grad ψ0 (t, Z(0, ~ t)), Z ~ 0) = ϕ(x), x ∈ [0, l]. Z(x,

~ t) = 0, Z(l,

Let conditions (2) and (3) hold and replace (1) by

(3.2.27) (3.2.28)

Landau–Lifshitz Equations

78

~ is scalar function with continuous derivative with respect to t ∈ (10 ). ψ0 (t, Z) ~ ∈ R3 , and the second [0, T ], and the mixed derivative with respect to t ∈ [0, T ] and Z ~ ∈ R3 are continuous. The Hessian matrix H0 (t, Z) ~ of derivative with respect to Z ~ is assumed to be non-negative definite and grad ψ0 (t, 0) = 0, ϕ(l) = 0. ψ0 (t, Z) ~ ε (x, t) to problem (3.2.2), (3.2.27) We first consider the existence of the solution Z and (3.2.28). Theorem 3.2.4 Let (10 ), (2) and (3) hold. Then the mixed boundary value problem (3.2.2), (3.2.27) and (3.2.28) admits unique weak solution ~ ε (x, t) ∈ Zl = L∞ (0, T ; H 2 (0, l)) Z

\

(1) W∞ (0, T ; L2 (0, l))

\

L2 (0, T ; H 3 (0, l)).

Proof. The proof is similar to that of Theorems 3.2.1 and 3.2.2. We construct the nonlinear boundary condition for the ordinary differential equation (3.2.7) ~0 ∆+ Z ~ 0 ), ~ J = 0, = grad ψ0 (t, Z Z (3.2.29) h and initial condition ~ j (0) = ϕj , j = 0, 1, . . . , J, Z (3.2.30) ~ j (t), where ϕj = ϕj = ϕj (xj ), whose solution (approximate solution) is denoted by Z (j = 1, 2, . . . , J − 1), ϕJ = ϕJ = ϕ(l) = 0, ϕ0 is subject to: ϕ1 = ϕ0 + h grad ψ0 (0, ϕ0 ). Before applying fixed point theorem, we must prove that problem (3.2.7), (3.2.29) ~ h = {Z ~ j (t) : j = 0, 1, . . . , J} with Z ~ j (t) ∈ and (3.2.30) admits at least one solution Z (2) (1) ~ ~ C ([0, T ]) (j = 1, 2, . . . , J) and Z0 (t) ∈ C ([0, T ]). In order to get solution Zε (t) of ~ h (t) of (3.2.2), (3.2.27) and (3.2.28), we first give the a priori estimates for solution Z (3.2.7), (3.2.29) and (3.2.30) and then send h to zero. ~ j |p−2 Z ~ j h, 2 ≤ p < ∞, and summing from j = 1 to j = J, Multiplying (3.2.7) by |Z and integrating over t ∈ [0, t], 0 ≤ t ≤ T , we have J−1 X j=1

~ j |p h − |Z

J−1 X j=1

|ϕj |h

~ 2 X pε Z t J−1 p−2 p−2 ∆+ Zj ~ ~ =− (|Zj+1 | + | Zj | ) hdt h 2 0 j=0

pε 2

Z

− pε

Z

−

+p

Z

t J−1 X

0 j=0 t

0

~ j+1 |p−2 − |Z ~ j |p−2 )(|Z ~ j+1 |2 − |Z ~ j |2 ) (|Z

~ 0 |p−2 Z ~ 0 grad ψ0 (t, Z ~ 0 )dt |Z

t J−1 X

0 j=1



~ j |p−2 Z ~ j · f~(xj , t, Z ~ j )hdt, |Z

dt h

One-dimensional Landau–Lifshitz equations

79

~ J (t) = 0. As before we may conclude that zh (t) = {Z ~ j (t) : j = 0, 1, . . . , J} is where Z uniform bounded in j = 0, 1, . . . , J and t ∈ [0, T ]. ~ − Zj Then we test (3.2.7) by ∆+ ∆ h, and sum from j = 1 to j = J − 1, and integrate h2 over t ∈ [0, t], 0 ≤ t ≤ T to give 2 Z t J−1 ~ ~ j 2 X ∆+ ∆− Z ∆+ Zj (t) ~ 0 (t)) h+2 hdt + 2ψ0 (t, Z h h2 0

J−1 X j=0

j=1

=

2 ∆+ ϕ j h + 2ψ0 (0, ϕ0 ) h

J−1 X j=0

+2

Z

t

0

+2

~ 0 (t))dt ψ0t (t, Z

Z

t J−1 X

Z

t J−1 X

0 j=0

+2

0 j=0

+2

Z

0

t

~j Z 1 ∆+ Z ~ j+1 )hdτ dt f~x (xj+2 , t, Z h 0 ~j ∆+ Z h

Z

1 0

~ ~ j+1 + (1 − τ )Z ~ j ∆+ Zj hdtdτ f~z (xj , t, τ Z h !

~ 0 ) · grad ψ0 (t, Z ~ 0 )dt − 2 f~(0, t, Z

Z

t

0

~J ∆− Z dt, f~(l, t, 0) h

~ 0 (t) = 0. All terms except for the last term on the right-hand side can be where Z J R ~ h k2 2 dt + B, where A, B are independent of h > 0, ε ≥ 0. For controlled by A 0t kδ Z L the last term, we have Z Z t Z t ~ J t ∆ Z − ~ h (t)k∞ dt + ε kδ 2 Z ~ h (t)k2 2 dt + 1 C29 . dt ≤ C28 2 f~(l, t, 0) kδ Z L 0 h ε 0 0

The above estimates are independent of h but depending on ε. If f~(l, t, 0) = 0, then these estimates are independent of both h and ε. ~ 0 (t) = {Z ~ 0 (t) : j = 0, 1, . . . , J}. We have Next we estimate yh (t) = Z h j ∆+ y 0 ~ 0 ) + H0 (t, Z ~ 0 )y0 , yJ = 0; = grad ψ0t (t, Z h ∆+ ∆− ϕ j ∆+ ∆− ϕ j yj (0) = ε + ϕj × + f~(xj , t, ϕj ), j = 1, 2, . . . , J − 1; 2 h h2 (E + hH0 (0, ϕ0 ))y0 (0) = y1 (0) − h grad ψ0 (0, ϕ0 ), yJ (0) = 0, ~ j0 (t)k22 where E is a 3×3 unit matrix. From this we can get the estimates of sup0≤t≤T kZ and

~0 ~ j ∆+ Z ∆+ ∆− Z h , 2 h h

and

~h ∆2+ ∆− Z . 3 h

we have

~ h (t)k2 + sup kδ Z ~ h (t)k2 + sup kδ 2 Z ~ h (t)k2 sup kZ

0≤t≤T

0≤t≤T

+ sup 0≤t≤T

~ 0 (t)k2 kZ h

0≤t≤T

+

≤ K13 (ε), where K13 (ε) is independent of h.

Z

t 0

~ 0 (t)k2 dt kδ 2 Z h 2

!1/2

+

Z

t 0

kδ

3~

Zh (t)k22 dt

!1/2

Landau–Lifshitz Equations

80

Now, as before, we may send h → 0 (for some subsequence if necessary) to get a ~ ε (x, t) which has weak derivatives Z ~ εx , Z ~ εx x and Z ~ εt ∈ L∞ (0, T ; L2 (0, l)) limit vector Z ~ εx t, Z ~ εx xx ∈ L2 (QT ). Moreover, Z ~ ε (x, t) solves (3.2.2) in the sense of distribuand Z tion and satisfies (3.2.27) and (3.2.28) in the classical sense. The uniqueness of solu~ ε (x, t) to the problem (3.2.2) and (3.2.27)–(3.2.28) can be given by standard tion Z argument. The theorem is proved. 2. Approximate solution of the mixed initial-boundary problem of the spin equation In order to prove the existence of equation (3.2.1) with nonlinear boundary condition (3.2.27) and initial condition (3.2.28), we must give the uniform (in ε) a priori ~ ε (x, t) obtained above. estimates for the approximate solution Z Lemma 3.2.8 Let (10 ), (2), (3) hold and f~(l, t, 0) = 0. Then for the approximate ~ ε (x, t) obtained above, we have solution Z ~ ε kL∞ + sup kZ ~ εx (·, t)kL2 + sup kZ ~ εt (·, t)kH −1 (0,l) ≤ K14 , kZ 0≤t≤T

(3.2.31)

0≤t≤T

where K14 is a constant independent of ε > 0. Lemma 3.2.9 Let (10 ), (2), (3) hold and f~(l, t, 0) = 0. Then for the approximate ~ ε (x, t) of problem (3.2.2) and (3.2.27)–(3.2.28), we have solution Z ~ ε (x1 , t) − Z ~ ε (x2 , t)| ≤ K15 |x1 − x2 |1/2 , |Z ~ ε (x, t1 ) − Z ~ ε (x, t2 )| ≤ K16 |t1 − t2 |1/4 , |Z

(3.2.32) (3.2.33)

where K15 , K16 are constants independent of ε > 0.

3.2.6

The Mixed Boundary Problem of Equations of Ferromagnetic Spin Chain

1. Weak solution for the mixed boundary problem of equations of ferromagnetic spin chain ~ t) ∈ L2 (0, T ; H 1 (0, l)) T C(QT ) with Z(l, ~ t) = 0 is Definition 3.2.2 Vector Z(x, called a weak solution of equation (3.2.1) subject to mixed nonlinear boundary condition (3.2.27) and initial condition (3.2.28), if for any test function g(x, t) ∈ H 1 (QT ), g(l, t) = g(x, T ) = 0 there holds Z Z

QT

~ t) − gx (x, t)(Z(x, ~ t) × Z ~ x (x, t)) [gt (x, t)Z(x,

~ t))]dxdt + + g(x, t)f~(x, t, Z(x, −

Z

T 0

Z

l

g(x, 0)ϕ(x)dx

0

~ t) × grad ψ0 (t, Z ~ 0 (0, t)))dt = 0. g(0, t)(Z(0,

(3.2.34)

One-dimensional Landau–Lifshitz equations

81

Theorem 3.2.5 Let (10 ), (2), (3) hold and f~(l, t, 0) = 0. Then equation (3.2.1) with the mixed nonlinear boundary condition (3.2.27) and initial condition (3.2.28) admits at least one weak solution: ~ t) ∈ L∞ (0, T ; H 1 (0, l)) Z(x,

\

1 1

C ( 2 , 4 ) (QT ).

Proof. The proof is similar to that of Theorem 3.2.3. In the following we consider another problem. Let Q∗T = {x ∈ R+ , 0 ≤ t ≤ T }. Consider the nonlinear boundary condition ~ x (0, t) = grad ψ0 (t, Z(0, ~ t)), Z ~ 0) = ϕ(x). Z(x,

(3.2.35) (3.2.36)

~ is a vector-valued function and continuously differentiable with (2∗ ) f~(x, t, Z) ~ ∈ R3 and the Jacobi matrix f~~ (x, t, Z) ~ is semi-bounded, respect to (x, t) ∈ Q∗T and Z Z that is, there is a constant b > 0 such that ~ · ξ ≤ b|ξ|2 , ξ ∈ R3 , (x, t, ξ) ∈ Q∗ × R3 ; ξ · f~Z~ (x, t, Z) T (3∗ ) ϕ(x) ∈ H 1 (R+ ), ϕ0 (x) = grad ψ0 (0, ϕ). ~ f~x (x, t, Z), ~ f~t (x, t, Z) ~ meet one of the following conditions: (4∗ ) f~(x, t, Z), ~ |f~x (x, t, Z)|, ~ |f~t (x, t, Z)| ~ |f~(x, t, Z)|, ~ + b(x, t); ≤ a(x, t)F (Z)

(3.2.37)

or ~ |f~x (x, t, Z)|, ~ |f~t (x, t, Z)| ~ |f~(x, t, Z)|, ~ k + d(x, t), ≤ c(x, t)|Z|

(3.2.38)

where a(x, t), b(x, t), d(x, t) ∈ L∞ (0, T ; L2 (R+ )), c(x, t) ∈ L∞ (Q∗T ), k ≥ 0 is a con~ is a continuous scalar function. stant, F (Z) 2. Weak solution to mixed boundary value problem of system of ferromagnetic spin chain Theorem 3.2.6 Let (10 ), (2∗ ), (3∗ ), (4∗ ) hold. Then the mixed boundary value problem of system (3.2.2) defined on Q∗T with (3.2.35) and (3.2.36) admits unique weak solution: ~ ε (x, t) ∈ Z∞ = L∞ (0, T ; H 2 (R+ )) Z

\

(1) W∞ (0, T ; L2 (R+ ))

\

L2 (0, T ; H 3 (R+ )).

Proof. For any l > 0, construct a continuously differentiable vector-valued func3 ~ for (x, t) ∈ Q(l) ~ tion f~ (l) (x, t, Z), T = {0 ≤ x ≤ l, 0 ≤ t ≤ T } and Z ∈ R such that ~ ∈ R3 , f~ (l) (x, t, Z) ~ ≡ f~(x, t, Z), ~ f~(x, t, 0) ≡ 0. (i) For x ∈ [0, l − 1], t ∈ [0, T ], Z

Landau–Lifshitz Equations

82

(l) ~ is semi-bounded, that is, there is a constant b > 0 (ii) Jacobi matrix f~Z~ (x, t, Z) such that (l) ~ · ξ ≤ b|ξ|2 , l > 0, ξ · f~Z~ (x, t, Z) (3.2.39)

where b is independent of l > 0. ~ f~ (l) (x, t, Z), ~ f~t(l) (x, t, Z) ~ meet one of the following conditions: (iii) f~ (l) (x, t, Z), x ~ |f~ (l) (x, t, Z)|, ~ |f~t(l) (x, t, Z)| ~ |f~ (l) (x, t, Z)|, x ~ + b(x, t); ≤ a(x, t)F (Z)

(3.2.40)

or ~ |f~ (l) (x, t, Z)|, ~ |f~t (x, t, Z)| ~ |f~ (l) (x, t, Z)|, x ~ k + d(x, t), ≤ c(x, t)|Z| (l)

(3.2.41)

where the norms of a(x, t), b(x, t), d(x, t) in L∞ (0, T ; L2 (0, l)), the norm of c(x, t) in (l) L∞ (QT ), and k, F are all independent of l > 0. We also construct for any l > 0 the vector ϕ(l) (x) ∈ H 1 (0, l) such that ϕ(l) (x) = ϕ(x) for x ∈ [0, l − 1], ϕ(l) (l) = 0 and kϕ(l) kH 1 (0,l) is uniform bounded in l > 0. For any l > 0, consider mixed nonlinear boundary condition (3.2.27) and the initial condition: ~ 0) = ϕ(l) (x). Z(x, (3.2.42) (l)

Over the rectangular domain QT , we consider the spin equation: ~ t = εZ ~ xx + Z ~ ×Z ~ xx + f~ (l) (x, t, Z). ~ Z

(3.2.43)

It follows from Theorem 3.2.4 that problem (3.2.43), (3.2.27), (3.2.42) admits unique weak solution: ~ (l) (x, t) ∈ Zl = L∞ (0, T ; H 2(0, l)) Z \

\

(1) W∞ (0, T ; L2 (0, l))

L2 (0, T ; H 3(0, l)).

~ p−2 Z, ~ (2 ≤ p ≤ ∞) and integrating it over Q(l) Multiplying (3.2.43) by |Z| t , we have (l) ~ t)kp p kZ(·, L (0,l) − kϕ kLp (0,l)

= −pε

Z Z

(l)

Qt

~ p−2 (Z ~x · Z ~ x )dxdt |Z|

− p(p − 2)ε − pε +p

Z

t 0

Z Z

Z Z

(l)

Qt

~ p−2 |Z|

~ Z ~x ·Z ~ |Z|

!2

dxdt

~ t)|p−2 Z(0, ~ t) · grad ψ0 (t, Z(0, ~ t))dt |Z(0, (l)

Qt

~ p−2 Z ~ · f~(x, t, Z)dxdt. ~ |Z|

One-dimensional Landau–Lifshitz equations

83

This yields ~ t)kLp (0,l) ≤ K17 e(b+δ)T [kϕ(l) kLp (0,l) + kf~0 k p (l) ], sup kZ(·, L (Q ) (l)

T

0≤t≤T

(3.2.44)

(l) where f~0 ≡ f~ (l) (x, t, 0), δ > 0, and constant K7 depends on δ > 0. Therefore, 1/2+1/p

1/2−1/p

kϕ(l) kLp (0,l) ≤ C30 kϕ(l) kL2 (0,l) kϕ(l) kH 1 (0,l) ,

kf~0 kLp (Q(l) ) ≤ C31 [ sup kf~x(l) (·, t, 0)kL2 (0,l) (l)

T

0≤t≤T

+ sup kf~x(l) (·, t, 0)kLp (0,l) ].

(3.2.45)

0≤t≤T

It follows from (3.2.40) and (3.2.41) that the right-hand side of (3.2.45) is uniformly bounded in l > 0. Then ~ (l) (·, t)kLp (0,l) ≤ C32 , sup kZ

0≤t≤T

where C32 is uniform in both 2 ≤ p < ∞ and l > 0, hence ~ (l) (·, t)kL∞ (0,l) ≤ C32 . sup kZ

0≤t≤T

~ xx and then integrating it over Q(l) , we have Multiplying equation (3.2.43) by Z T 2 ~ x (·, t)k2L2 (0,l) − kϕ(l) ~ 2 kZ x kL2 (0,l) + 2εkZxx kL2 (0,l)

~ t)) − ψ0 (0, ϕ(l) (0)) + ψ0 (t, Z(0,

=2

Z

l

0

+2

~ t))dt ψ0t (t, Z(0, Z

t 0

~ t)) · f~ (l) (0, t, Z(0, ~ t))dt grad ψ0 (t, Z(0,

+2

Z Z

Qt

~ x · f~ (l) (x, t, Z)dxdt ~ Z x

+2

Z Z

(l) QT

~ x · f~ (l) (x, t, Z)dxdt, ~ Z x

(l)

~ t) = Z ~ t (l, t) ≡ 0, f~ (l) (x, t, 0) ≡ 0. It follows from the above that where Z(l, ~ (l) (·, t)k2 2 ~ (l) 2 kZ x L (0,l) + 2εkZxx kL2 (Q(l) ) t

~ x(l) (·, t)k2 (l) ≤ (2b + 1)kZ L2 (Q ) t

+2

Z Z

(l)

Qt

~ (l) )|2 dxdt |f~x(l) (x, t, Z

where the constant C33 is independent of l > 0.

+ C33 ,

(3.2.46)

Landau–Lifshitz Equations

84

~ meets condition (3.2.37), f~ (l) (x, t, Z) ~ satisfies condition (3.2.40). We If f~x (x, t, Z) x have Z Z

(l)

Qt

~ (l) (x, t))|2 dxdt |f~x(l) (x, t, Z

~ 2 kak2 (l) + 2kbk2 (l) . ≤ 2 sup |F (Z)| L2 (Q ) L2 (Q ) ~ |Z|≤C 32

t

t

~ meets condition (3.2.38), f~x(l) (x, t, Z) ~ satisfies condition (3.2.41). We If f~x (x, t, Z) have Z Z

(l)

Qt

~ (l) (x, t))|2 dxdt |f~x(l) (x, t, Z

~ (l) (x, t)k2k2k + 2kdk2L2 (Q(l) ) , ≤ 2kckL∞ (Q(l) ) sup kZ L (0,l) t

t

0≤t≤T

whose right-hand side is independent of l > 0. Combining these inequalities, we have √ ~ (l) k 2 (l) ≤ C34 , ~ (l) (·, t)kL2 (0,l) + εkZ sup kZ xx L (Q ) x 0≤t≤T

T

where C34 is independent of l > 0. ~ t(l) (x, t) in L∞ (0, T ; L2 (0, l)), we In order to get the estimate for y (l) (x, t) = Z differentiate (3.2.43) with respect to t to obtain ~ xx + Z ~ × yxx + f~t(l) (x, t, Z) ~ + f~ (l) (x, t, Z)y. ~ yt = εyxx + y × Z ~ Z

(3.2.47)

RR ~ (l) (x, t))|2 dxdt is bounded uniform in l. Assumption (4∗ ) implies that Q(l) |f~t (x, t, Z t By the similar method as above we have √ ~ (l) k 2 (l) ≤ C35 , ~ t(l) (·, t)kL2 (0,l) + εkZ (3.2.48) sup kZ xx L (Q ) 0≤t≤T

T

where C35 is independent of l > 0. Differentiating (3.2.43) with respect to x, we obtain ~ xt = εZ ~ xxx + Z ~ ×Z ~ xxx + Z ~x × Z ~ xx + f~x(l) (x, t, Z) ~ + f~ (l) (x, t, Z) ~ Z ~ x. Z ~ Z (l)

(3.2.49)

~ (l) (x, t) in L2 (QT ) can be derived from the Then the uniform boundedness in l of Z xxx (l) (l) ~ ~ xxx uniform boundedness in l of Zxt (x, t) in L2 (QT ) since the coefficient matrix of Z is positively definite and non-singular. ~ (lj ) } and {Z ~ x(lj ) } uniformly Hence, we may choose {lj } such that when lj → ∞, {Z ~ t) and Z ~ x (x, t) on any rectangular domain {0 ≤ x ≤ l, 0 ≤ t ≤ converge to Z(x, (lj ) ~ xx ~ t(lj ) } weakly converge in L∞ (0, T ; L2 (0, l)) to T } respectively; and, {Z } and {Z (lj ) (lj ) ~ xx (x, t) and Z ~ t (x, t) respectively; and, {Z ~ xxx ~ xt Z } and {Z } weakly converge in L2 (Q∗T ) ~ xxx (x, t) and Z ~ xt (x, t) (ε > 0) respectively. This means that the limit Z(x, ~ t) is to Z a weak solution of (3.2.2) and (3.2.35) and (3.2.36), and it satisfies the conditions (3.2.35) and (3.2.36) in the classical sense. By the same method as above, we can prove the uniqueness. Let l → ∞, we have

One-dimensional Landau–Lifshitz equations

85

Lemma 3.2.10 Under the condition (1∗ ), (2∗ ), (3∗ ) and (4∗ ), for the weak solution ~ ε (x, t) of the problem (3.2.2) and (3.2.35) and (3.2.36) (ε > 0), there holds Z ~ ε (·, t)kL2 (R+ ) sup kZ

0≤t≤T

~ εx (·, t)kL2 (R+ ) + sup kZ ~ εt (·, t)kH −1 (R+ ) + sup kZ 0≤t≤T

0≤t≤T

≤ K18 .

(3.2.50)

The constant K18 is independent of ε > 0. Proof. Let g(x) ∈ H01 (R+ ) be any test function. We have Z

∞ 0

Z

~ εt (x, t)dx = −ε g(x)Z − +

Z

Z

∞

0 ∞

0 ∞

0

~ εx (x, t)dx gx (x)Z

~ ε (x, t) × Z ~ εx (x, t))dx gx (x)(Z ~ ε (x, t))dx. g(x)f~(x, t, Z

And we have Z ∞ ~ ε (x, t))dx g(x)f~(x, t, Z 0

≤ kgkL2 (R+ )

Z

∞

0

~ ε (x, t))|2 dx |f~(x, t, Z

!1/2

.

In the case of (3.2.37) or (3.2.38), we have Z

∞ 0

~ εt (x, t))|2 dx |f~(x, t, Z ≤2

~ 2 sup ka(·, t)k2L2 (R+ ) + sup kb(·, t)k2L2 (R+ ) sup |F (Z)|

~ |Z|≤K 18

0≤t≤T

0≤t≤T

!

and Z

∞ 0

~ ε (x, t))|2 dx |f~(x, t, Z ≤2

!

~ ε (·, t)k2k2k + kckL∞ (Q∗ ) + sup kd(·, t)k2 2 + . sup kZ L (R ) L (R ) T

0≤t≤T

0≤t≤T

This implies that (3.2.50) is true. Lemma 3.2.11 Under the condition (10 ), (2∗ ), (3∗ ) and (4∗ ), for the weak solution ~ ε (x, t) of the problem (3.2.2) and (3.2.38) and (3.2.39) (ε > 0), there holds Z ~ ε (x, t1 ) − Z ~ ε (x, t2 )| ≤ K19 (x)|t1 − t2 |1/4 , |Z ~ ε (x1 , t) − Z ~ ε (x2 , t)| ≤ K20 |x1 − x2 |1/2 , |Z where K19 (x), K20 are independent of ε, K19 (x) depends on x.

(3.2.51) (3.2.52)

Landau–Lifshitz Equations

86

It follows from the discussion of Lemma 3.2.10 that the uniform boundedness in R ~ ε (x, t)dx and its derivative Sεt (x, t) = R x Z ~ εt (x, t)dx ε > 0 of the vector Sε (x, t) = 0x Z 0 is only true on the bounded domain, therefore the estimate (3.2.51) holds only for finite x. Now we are in the position to consider the existence of the weak solution to problem (3.2.1) and (3.2.35) and (3.2.36). 3. Existence of weak solution to the nonlinear boundary problem of the equations of ferromagnetic spin chain ~ t) ∈ L2 (0, T ; H 1 (R+ )) T C(Q∗ ) is called a weak Definition 3.2.3 A vector Z(x, T solution to problem (3.2.1) and (3.2.35) and (3.2.36), if for any test function g(x, t) ∈ H 1 (Q∗T ), g(x, T ) = 0, supp g(x, t) < ∞ there holds Z Z

Q∗T

~ t) − gx (x, t)(Z(x, ~ t) × Z ~ x (x, t)) [gt (x, t)Z(x,

~ t))]dxdt + + g(x, t)f~(x, t, Z(x, −

Z

T

Z

∞

g(x, 0)ϕ(x)dx 0

~ t) × grad ψ0 (t, Z ~ 0 (0, t)))dt = 0. g(0, t)(Z(0,

0

Theorem 3.2.7 Let (10 ), (2∗ ), (3∗ ) and (4∗ ) hold. Then there is at least one weak ~ t) of the problem (3.2.1) and (3.2.35) and (3.2.36) on Q∗ : solution Z(x, T ~ t) ∈ L∞ (0, T ; H 1(R+ )) Z(x,

\

(1,1)

Cloc2 4 (Q∗T ).

4. Properties of weak solution In the following we discuss the “blow-up” problem and the asymptotic properties for the solution. Theorem 3.2.8 Let (1)–(3) hold for any T > 0 and f~(x, t, 0) ∈ Lp (Q∞ ) for any 2 ≤ p < ∞. If in (3.2.6), b < 0, then the solution to spin equation (3.2.2) with (3.2.3) and (3.2.4) satisfies ~ ε (·, t)kLp (0,l) = 0. lim kZ t→∞

Proof. For any fixed 2 ≤ p < ∞, take δ > 0 such that b + δ < 0 to give from the ~ ε (x, t) that L -estimates of Z p

~ ε (·, t)kLp (0,l) kZ ≤e

(b+δ)t

(

kϕkLp (0,l) + (δ(p − 1))

1/p

!

)

p−1 ~ kf0 kLp (Q∞ ) . 2p

This implies the conclusion. Similarly we have Theorem 3.2.9 Under the same conditions as in Theorem 3.2.8, the solution to equation (3.2.1) with (3.2.3) and (3.2.4) satisfies ~ t)kLp (0,l) = 0, lim kZ(·,

t→∞

2 ≤ p < ∞.

One-dimensional Landau–Lifshitz equations

87

Theorem 3.2.10 Let (10 ), (2) and (3) hold and f~(x, t, 0) ∈ Lp (Q∞ ) for any 2 ≤ p < ∞, b < 0. Then the solution to equation (3.2.1) with (3.2.35) and (3.2.36) satisfies ~ t)kLp (0,l) = 0, lim kZ(·,

2 ≤ p < ∞.

t→∞

Theorem 3.2.11 Let (10 ), (2∗ ), (3∗ ) and (4∗ ) hold and f~(x, t, 0) ∈ Lp (Q∗∞ ) for any 2 ≤ p < ∞, b < 0. Then the solution to equation (3.2.1) and (3.2.2) with (3.2.35) and (3.2.36) satisfies ~ t)kLp (0,l) = 0, lim kZ(·,

2 ≤ p < ∞.

t→∞

Now consider the “blow-up” problem. If there holds ~ · f~(x, t, Z) ~ ≥ C0 |Z| ~ 2+δ , Z

~ ∈ R3 (x, t) ∈ QT , Z

(3.2.53)

~ p−2 Z ~ and integrating over where C0 > 0, δ > 0, we have by multiplying (3.2.1) by |Z| [0, l] with respect to x that Z l 1d ~ ~ t)|p−2 Z(x, ~ t) · f~(x, t, Z)dx. ~ kZ(·, t)kpLp (0,l) = |Z(x, p dt 0

This combined with (3.2.53) yields δ d ~ ~ t)k1+δ kZ(·, t)kpLp (0,l) ≥ C0 l− p kZ(·, Lp (0,l) , dt

and hence δ

1

−p −δ ~ t)kLp (0,l) ≥ (kϕk−δp kZ(·, ) , 2 ≤ p < ∞. L (0,l) − C0 tδl

(3.2.54)

It follows from (3.2.54) that ~ satisfies (3.2.53) and kϕkLp (0,l) 6= 0, 2 ≤ p < ∞, then Theorem 3.2.12 If f~(x, t, Z) the solution to (3.2.1) blows up at finite time, i.e. ~ t)kLp (0,l) = +∞. lim kZ(·,

t→∞

3.3

Smooth Solution for the Ferromagnetic Spin Chain Systems

In the above sections, we have obtained the existence and uniqueness of the global weak solution for one-dimensional problems. In this section, we shall prove the existence and uniqueness of smooth solution for the one-dimensional periodic initialboundary value problem and initial value problem. Our method is to use the mobile framework on S 2 in the procedure of viscosity vanishing to get the uniform estimate.

Landau–Lifshitz Equations

88

3.3.1

Smooth Solution to the Nonlinear Systems with Periodic Initial Boundary Conditions

1. The problem In order to establish the existence and uniqueness of smooth solution to the equation of ferromagnetic spin chain ~t = Z ~ ×Z ~ xx Z

(3.3.1)

with periodic initial-boundary conditions ~ 0) = Z ~ 0 (x), Z(x,

~ + D, t) = Z(x ~ − D, t), Z(x

~ 0 (x)| ≡ 1, |Z

x ∈ R1 ,

(3.3.2)

or initial condition ~ 0) = Z ~ 0 (x), Z(x,

~ 0 (x)| ≡ 1, |Z

x ∈ R1 ,

(3.3.3)

we consider the following nonlinear equations with small parameter ε > 0 ~ t = −εZ ~ × (Z ~ ×Z ~ xx ) + Z ~ ×Z ~ xx , Z

(3.3.4)

with (3.3.2) to approximate the desired solution. It can be proved that:  ~ ~ ~ ~ ~ 2~    Zt = εZxx + Z × Zxx + ε|Zx | Z,

~ 0) = Z ~ (x), Z(x,

0    Z(x ~ + D, t) = Z(x ~ − D, t)

(3.3.5)

is equivalent to (3.3.4) with (3.3.2) in the classical sense. Here we assume that at ~ 0 (x)| = C0 (positive constant, the initial time the ferromagnet is saturated, that is, |Z C0 = 1, for example). We use the difference method to prove the existence of local smooth solution of periodic initial value problem (3.3.5). This also means the existence of local smooth solution of (3.3.4) and (3.3.2) (ε > 0). Then we use the uniform estimates in ε and in D for solutions of (3.3.4) and (3.3.2) to get the existence of solution to (3.3.1) and (3.3.2) by sending ε → 0, and next to obtain the existence of solution to (3.3.1) and (3.3.3) by sending D → ∞. The uniqueness of smooth solution can be easily given. 2. Smooth solution to the difference-differential system Using difference in the spatial direction, we prove that (3.3.5) admits at least one local smooth solution. For simplicity, we let ε = 1 and establish the following difference-differential equation:  ~ j ∆+ Z ~ j 2 ~j ~j  ∆ ∆ Z d Z ∆ ∆ Z + − + −  ~j × ~ ,  = +Z + Z   h j  dt h2 h2

~ j |t=0 = Z ~ 0j = Z ~ 0 (jh),  Z      ~ j+J = Z ~j , Z

j = 0, ±1, . . . , ±J,

(3.3.6)

One-dimensional Landau–Lifshitz equations

89

~ j = Z(x ~ j , t), xj = jh, j = 0, ±1, . . . , ±J, h = 2D/J, J > 0, ∆+ , ∆− denote where Z the forward and backward difference respectively. It is not difficult to know that the initial value problem (3.3.6) admits a local smooth solution. For such solution, we shall give some estimates uniformly in h. In ~j . this section we always denote the smooth solution of (3.3.6) by Z ~ 0 (x) ∈ H 1 (Ω) (Ω = (−D, D)), then there are constants T0 > 0, Lemma 3.3.1 If Z C > 0 independent of h such that ~ h (t)k2 ≤ C, sup kZ

~ h (t)k2 ≤ C, sup kδ Z

0≤t≤T0

(3.3.7)

0≤t≤T0

where uh = {uj = u(xj ) | j = 0, 1, 2, . . . , J}, xj = jh, h = 2D/J, and k · kp = k · kLp (Ω) ,

kδ k uh kp =

p ! p1 k u ∆ + i h . hk

J−k X i=0

~ j h and summing from j = 1 to J, we have Proof. Multiplying (3.3.6) by Z 2



J J J ~ j ~ X X ∆+ Z 1d X ~ j |2 h = ~ j ∆+ ∆− Z j h + |Z ~ j |2 h |Z Z 2 2 dt j=1 h h j=1 j=1

2



2



J ~ ~ X ∆ Z ∆ Z + j h+ + j |Z ~ j |2 h. =− h h j=0 j=1 J−1 X

Since

1 ~ h k∞ ≤ CkZ ~ h k22 kδ Z ~ h k2 + 1 kZ ~ h k2 kZ 2D



we have

1 2

,

d ~ 2 ~ h k2 ≤ C(kZ ~ h k4 + kδ Z ~ h k4 ). kZh k2 + kδ Z 2 2 2 dt

Moreover, multiplying (3.3.6) by

~j ∆+ ∆− Z h



2

2



and summing from j = 1 to J, we get

J J ~j ~ j ~j X X ∆+ ∆− Z ∆+ ∆− Z ∆+ ∆− Z ~ Zjt = h + h h2 h j=1 j=1 j=1 J X

Therefore, one gets





2

2 ~ j ∆+ Z Z ~ j. h

2

J ~ ~ X 1d ∆ Z ∆ ∆ Z + j h+ + − j h 2 dt j=0 h h2 j=1

J−1 X

(3.3.8)



4

J J ~ ~ X 1X ∆ ∆ Z ∆ Z ~ j |2 + − j h + C max |Z + j h. ≤ 1≤j≤J 4 j=1 h2 h j=1

Landau–Lifshitz Equations

90

Applying interpolation inequalities 3

~ h k∞ ≤ CkZ ~ h k24 (kδ 2 Z ~ h k2 + k Z ~ h k2 ) 14 , kZ 3

~ h k4 ≤ Ckδ Z ~ h k24 (kδ 2 Z ~ h k2 + kδ Z ~ h k2 ) 14 kδ Z and H¨older inequality, we have d ~ 2 ~ h k2 ≤ C(1 + kZ ~ h k8 + kδ Z ~ h k8 ). kδ Zh k2 + kδ 2 Z 2 2 2 dt

(3.3.9)

Combining (3.3.8) with (3.3.9), we have d ~ 2 ~ h k2 ) + kδ 2 Z ~ h |2 ≤ C + C(kZ ~ h k2 + kδ Z ~ h k 2 )4 . (kZh k2 + kδ Z 2 2 2 2 dt This inequality combined with Gronwall inequality implies that there exist constants T0 , C > 0 independent of h such that ~ h (t)k2 + kδ Z ~ h (t)k2 ≤ C, kZ and

Z

The lemma is proved.

T0 0

∀ t ∈ [0, T0 ],

~ h (t)|2 ≤ C. kδ 2 Z 2

Corollary 3.3.1 Under the conditions in Lemma 3.3.1, we have, for some constant C independent of h, ~ j | ≤ C. sup |Z (3.3.10) 0≤t≤T0 ;0≤j≤J

~ 0 (x) ∈ H 2 (Ω), there are constants T0 > 0, C > 0 independent of Lemma 3.3.2 If Z h such that Z T0 ~ h (t)k2 ≤ C, ~ h (t)k2 ≤ C. sup kδ 2 Z kδ 3 Z (3.3.11) 0

0≤t≤T0

Proof. It follows from (3.3.6) that d ~j = ∆+ Z dt

~j ∆2+ ∆− Z h2

Multiplying this equality by J ~j X ∆2+ ∆− Z

j=1

h3

J ~j X ∆2+ ∆− Z

j=1

h3



~j ∆2+ ∆− Z , h3



summing it from j = 1 to J and noting that 2



~ j X ∆2+ Z 1 d J−1 ~ h, · ∆+ Zjt = − 2 dt j=0 h2 

2



~ j ∆+ Z ~ j ∆+ ∆− Z ~ Z ~  + ∆ + Z j × + j . 2 h h







J ~j ~ ~ X ∆2+ ∆− Z ~ j × ∆+ ∆− Z j  = ~ j × ∆+ ∆− Z j  ∆+ Z · ∆+  Z · h2 h3 h2 j=1

One-dimensional Landau–Lifshitz equations



J X

j=1

2

2





91

2

J 2 J ~ j ~ j X ~ j ∆+ Z 1X ∆+ ∆− Z ∆+ ∆− Z h, ≤ max h + 1≤j≤J 4 j=1 h3 h2 h j=1

 2  2 J 2 ~ ~ X ∆ ∆ Z ∆ Z 1 − j + j   + ~ h · ∆+  Zj  ≤ 3 4 j=1 h h  2 2 6  X J 2 ~ J ~ ~ X ∆+ Z j ∆+ Z j ∆+ Z j   h , + C  max h+ 1≤j≤J h j=1 h2 h j=1

~j ∆2+ ∆− Z h3

as well as the following inequalities

2~ 1/2 ~ h k∞ ≤ Ckδ Z ~ h k1/2 ~ kδ Z , 2 (kδ Zh k2 + kδ Zh k2 ) 2/3 2 1/3 ~ h k6 ≤ Ckδ Z ~ h k2 (kδ Z ~ h k2 + kδ Z ~ h k2 ) , kδ Z

we finally get d 2~ 2 ~ h k22 ≤ C + Ckδ 2 Z ~ h k32 . kδ Zh k2 + kδ 3 Z dt Lemma 3.3.2 follows from the Gronwall’s inequality. Corollary 3.3.2 Under the conditions in Lemma 3.3.2, we have, for some C independent of h,  ~  ∆ Z (t)  + j  ≤ C,  sup  h 0≤t≤T0 ;1≤j≤J (3.3.12) Z  T0   2  ~ ht (t)k dt ≤ C.  kδ Z 0

2

3. Local smooth solution to the nonlinear equation with periodic initial data By the similar method as in the proof of Lemmas 3.3.1 and 3.3.2 and using induction argument, we have

~ 0 (x) ∈ H k (Ω), then there are constants T0 > 0, C > 0 independent Lemma 3.3.3 If Z of h such that ~ h k2 ≤ C, sup kδ k Z

0≤t≤T0

~ ht k2 ≤ C, sup kδ k−2 Z

k ≥ 2,

~ htt k2 ≤ C, sup kδ k−4 Z

k ≥ 4.

0≤t≤T0 0≤t≤T0

(3.3.13)

~ j (t) and from the standard method, From the above uniform estimates in h of Z we can prove that the solutions of (3.3.6) approximates the solution of (3.3.5), and then we get the existence of local smooth solution to (3.3.5). That is, we have

Landau–Lifshitz Equations

92

~ 0 (x) ∈ H k (Ω) and Z ~ 0 (x − D) = Z ~ 0 (x + D). Then Theorem 3.3.1 Let ε > 0, Z ~ problem (3.3.5) admits at least one local smooth solution Z(x, t) in ~ t) ∈ G(T0 ) Z(x, 

  k+1  [ 2 ] \ \ \     s =  W∞ (0, T0 ; H k−2s (Ω))  H s (0, T0 ; H k+1−2s(Ω)) , [ k2 ]

s=0

(3.3.14)

s=0

where T0 > 0 is independent of k, s and k are non-negative integers with k − 2s ≥ 0. ~ 0 (x)| = 1, x ∈ R1 . Then Theorem 3.3.2 Let conditions in Theorem 3.3.1 hold and |Z problem (3.3.5) is equivalent to the following problem:  ~ ~ ~ ~ ~ ~   Zt = −εZ × (Z × Zxx ) + Z × Zxx , 

~ 0) = Z ~ (x), Z(x,

(3.3.15)

0   ~ ~ − D, t). Z(x + D, t) = Z(x

~ t) be a classical solution of (3.3.15) (ε > 0) and prove Proof. First we let Z(x, that it is also a solution of (3.3.5). ~ t)| ≡ 1. Then we have In fact it is easy to get that |Z(x, ~ × (Z ~ ×Z ~ xx ) = ε|Z| ~ 2Z ~ xx − ε(Z ~ ·Z ~ xx )Z ~ −εZ ~ xx + ε|Z ~ x |2 Z, ~ = εZ ~ solves (3.3.5) in the classical sense. this implies that Z ~ is a classical solution of (3.3.5), we want to prove On the other hand, if Z ~ t)| ≡ 1 in QT . In fact, denoting u(x, t) = |Z(x, ~ t)|2 , we have from (3.3.5) |Z(x, ~ x |2 (u − 1), ut = εuxx + 2ε|Z u(x, 0) = 1,

u(x − D, t) = u(x + D, t). ~ t)|2 − 1. Then It is clear that u = 1 also solves this problem. Let w = u − u = |Z(x, ~ x |2 w, wt = εwxx + 2ε|Z

(3.3.16)

w(x, 0) = 0,

(3.3.17)

w(x − D, t) = w(x + D, t).

(3.3.18)

Multiplying (3.3.16) by w and integrating over [−D, D], we have 1 2

Z

D −D

|w|2 dx + ε

Z

D

−D

~ x |2 |wx |2 dx ≤ 2ε max |Z x,t

Z

D −D

|w|2 dx.

This combined with Gronwall inequality yields the conclusion. 4. Global smooth solution to nonlinear system with periodic initial data It follows from Theorems 3.3.1 and 3.3.2 that the problem (3.3.5) also admits a local smooth solution. In order to establish the global existence of smooth solution for problem (3.3.15). We must derive the a priori estimates for the solutions of (3.3.5).

One-dimensional Landau–Lifshitz equations

93

~ t) be a smooth solution of (3.3.5) |Z ~ 0 (x)| = 1. Then we have Lemma 3.3.4 Let Z(x,  ~ t)| = 1,   |Z(x,

∀ (x, t) ∈ R1 × [0, T ], ~   sup kZx (·, t)k2 ≤ C,

(3.3.19)

0≤t≤T

where C is independent of T and D.

Proof. Since (3.3.15) is equivalent to (3.3.5), the lemma can be proved by multi~ t) and Z ~ xx and integrating by parts. plying (3.3.5) by Z(x, ~ t) be the solution of (3.3.5). Then under the conditions of Lemma 3.3.5 Let Z(x, Lemma 3.3.4 there hold  ~ xx (·, t)kL2 (Ω) ≤ C,  sup kZ     0≤t≤T2

~ xxx (x, t)kL2 (Q ) ≤ C, kZ  T     ~ 2 kZxt (·, t)kL

(QT )

(3.3.20)

≤ C.

Proof. It follows from (3.3.5) (let ε = 1 for simplicity): ~ xxt · Z ~ xx = Z ~ xxxx · Z ~ xx + (Z ~ ×Z ~ xx )xx + (|Z ~ x |2 Z) ~ xx · Z ~ xx . Z Integrating this equality over Ω, we get 1d 2 dt

Z

D −D

≤

Z

~ xx |2 dx + |Z D

−D

+2 

Z

D −D

~ xxx |2 dx |Z

~ x ||Z ~ xx ||Z ~ xxx |dx + |Z Z

D −D

Z

D −D

~ x |3 |Z ~ xxx|dx |Z

~ Z ~ x ||Z ~ xx||Z ~ xxx |dx |Z|| 

~ x | kZ ~ xx kL2 (Ω) kZ ~ xxx kL2 (Ω) ≤ 3 sup |Z x

~ x k3 6 kZ ~ xxx kL2 (Ω) . + kZ L (Ω)

(3.3.21)

From the interpolation inequalities 1/4 ~ x kL∞ (Ω) ≤ CkZ ~ x k3/4 ~ kZ L2 (Ω) kZxxx kL2 (Ω) , 1/6 ~ x kL6 (Ω) ≤ CkZ ~ x k5/6 ~ kZ L2 (Ω) kZxxx kL2 (Ω) , 1/2

1/2

~ xx kL2 (Ω) ≤ CkZ ~ x k 2 kZ ~ xxx k 2 , kZ L (Ω) L (Ω)

we have from (3.3.21) and Lemma 3.3.4 and H¨older inequality that d dt

Z

~ xx |2 dx + 1 |Z 2 −D D

Z

D −D

~ xxx |2 dx ≤ C. |Z

This inequality and Gronwall inequality yields the desired conclusion.

Landau–Lifshitz Equations

94

Corollary 3.3.3 Under the conditions of Lemma 3.3.5, there holds ~ x | ≤ C, sup |Z

(3.3.22)

QT

where C is independent of D and T . By the similar method in proving Lemma 3.3.5, we have from induction argument ~ t) be a classical solution of (3.3.5). Under the conditions of Lemma 3.3.6 Let Z(x, ~ 0 (x)| = 1, there holds Theorem 3.3.1 and |Z  ~ xk−2s ts (·, t)kL2 (Ω) ≤ C,   sup kZ 0≤t≤T

  kZ ~

xk+1−2s ts kL2 (QT )

k − 2s > 0,

(3.3.23)

≤ C, k + 1 − 2s ≥ 0,

where k, s are non-negative integers and C is independent of D. It follows from Lemma 3.3.6 and extension method that problem (3.3.5) and (3.3.15) admits a global smooth solution. That is ~ 0 (x) ∈ H k (Ω) (k ≥ 4), |Z ~ 0 (x)| = 1, Z ~ 0 (x − D) = Z ~ 0 (x + D). Theorem 3.3.3 Let Z ~ t) ∈ G(T ), where T is Problem (3.3.15) (ε > 0) admits a global smooth solution Z(x, any positive number.

3.3.2

Smooth Solution to the Equations of Ferromagnetic Spin Chain

1. Estimates of smooth solution In order to get the existence of global smooth solution to the periodic initial value problem (3.3.1) and (3.3.2), we must have the a priori estimates uniform in ε for the smooth solutions of problem (3.3.15) with ε > 0. ~ ε (x, t) be a classical solution of (3.3.5). Under the conditions of Lemma 3.3.7 Let Z Theorem 3.3.3, there holds  ~ ε (x, t)| = 1,   |Z

∀ (x, t) ∈ R × [0, T ],

~   sup kZεx (·, t)kL2 (Ω) ≤ K1 ,

(3.3.24)

0≤t≤T

where K1 is independent of D, ε.

Lemma 3.3.8 Under the conditions of Theorem 3.3.3 (k ≥ 3), for the classical solutions of (3.3.5), there holds  ~    sup kZεx x(·, t)kL2 (Ω) ≤ K2 , 0≤t≤T

~    sup kZεt (·, t)kL2 (Ω) ≤ K2 , 0≤t≤T

where K2 is independent of D, ε.

(3.3.25)

One-dimensional Landau–Lifshitz equations

95

~ ε for simplicity. Since |Z(x, ~ t)| ≡ 1, the vectors Proof. Drop the subscript ε for Z 3 ~ ~ ~ ~ ~ ~ + βZ ~x + γZ ~ ×Z ~x. Z, Zx , Z × Zx form a unit orthogonal basis of R . Let Zxx = αZ ~ x |2 , β = Z ~x · Z ~ xx /|Z ~ x |2 , γ = ( Z ~ ×Z ~ x) · Z ~ xx /|Z ~ x |2 . Since Then α = −|Z Z D 1d ZD ~ 2 ~ xx · (εZ ~ xx + Z ~ ×Z ~ xx + ε|Z ~ xx |2 Z) ~ xx dx |Zxx | dx = Z 2 dt −D −D

= −ε

Z

−ε

D −D

Z

~ xxx|2 dx − |Z

D −D

Z

D −D

~x × Z ~ xx ) · Z ~ xxx dx (Z

~ xxx · (|Z ~ x |2 Z ~ x + 2Z( ~ Z ~x · Z ~ xx ))dx, Z

where Z Z

D −D

D −D

~ xxx · |Z ~ x |2 Z ~ x dx = − Z

Z

~ xxx · (Z( ~ Z ~x · Z ~ xx ))dx = −3 Z

Z

D −D

D −D

Z

~ x |2 |Z ~ xx |2 dx − 2 |Z

D −D

Z

D −D

~x · Z ~ xx |2 dx, |Z

~x · Z ~ xx |2 dx, |Z

~x × Z ~ xx ) · Z ~ xxx dx (Z ~ ~ ~ ~ ×Z ~ x ))] · Z ~ xxx dx ~ x × (−|Z ~ x |2 |Z ~ + (Z × Zx ) · Zxx (Z [Z ~ x |2 −D |Z

=

Z

=

Z

D

D

−D

+ =

5 2

Z

Z

~ x |2 ( Z ~ ×Z ~ x) · Z ~ xxx dx |Z D −D D

−D

~ ×Z ~x) · Z ~ xx (Z ~ x |2 Z ~ ·Z ~ xxx dx |Z ~ x |2 |Z

~ x |2 ( Z ~ ×Z ~x) · Z ~ xxx dx. |Z

In the above relations we have used ~ ·Z ~ xxx = − 3 (|Z ~ x |2 ) x . Z 2 In summary we have 1d 2 dt

Z

D −D

=ε

~ xx |2 dx + ε |Z Z

D

−D

−

5 2

Z

D −D

~ xxx |2 dx |Z

~ x |2 |Z ~ xx |2 dx + 8ε |Z

Z

D −D

Z

D

−D

~x · Z ~ xx |2 dx |Z

~ x | · (Z ~ ×Z ~x) · Z ~ xxx dx. |Z

(3.3.26)

Landau–Lifshitz Equations

96

On the other hand, 1d 4 dt

Z

D

~ x |4 dx |Z

−D

D

Z

=

−D Z D

=

−D

~ x |2 Z ~ x · (εZ ~ xx + Z ~ ×Z ~ xx + ε|Z ~ x |2 Z) ~ x dx |Z ~ x |2 Z ~ x · (εZ ~ xxx + Z ~x × Z ~ xx + Z ~ ×Z ~ xxx |Z

~ x |2 Z ~ x + 2εZ( ~ Z ~x · Z ~ xx ))dx + ε|Z

= −ε

Z

+ε

D

−D Z D

~ x |2 |Z ~ xx |2 dx − 2ε |Z

−D

~ x |6 dx − |Z

Z

D −D

Z

D

−D

~x · Z ~ xx |2 dx |Z

~ x |2 ( Z ~ ×Z ~ x) · Z ~ xxx dx, |Z

and hence 1d 4 dt

Z

D −D

=ε

Z

~ x |4 dx + ε |Z D

Z

−D

D

~ x |2 |Z ~ xx |2 + 2ε |Z

−D

~ x |6 dx − |Z

Z

D −D

Z

D

~x · Z ~ xx |2 dx |Z

−D

~ x |2 ( Z ~ ×Z ~x) · Z ~ xxx dx. |Z

It follows from (3.3.26) and (3.3.27) that d 4 dt

Z

D −D

~ x |2 dx + 8ε |Z

d =5 dt

Z

D −D

−D

~ xxx |2 dx + 20ε |Z

~ x |4 dx + 28ε |Z

Z

+ 104ε

D

Z

D

−D

D

Z

−D

D

Z

−D

~ x |6 dx |Z

~ x |2 |Z ~ xx |2 dx |Z

~x · Z ~ xx |2 dx. |Z

It follows from Lemma 3.3.7 that 28

Z

D

−D

~ x |2 |Z ~ xx |2 dx + 104 |Z

≤ 132 ≤δ ≤δ

Z

D

−D

D

Z

≤δ

Z

−D D

−D Z D

−D

D

Z

−D

~x · Z ~ xx |2 dx |Z

~ x |2 |Z ~ xx |2 dx |Z

~ x |6 dx + C(δ) |Z

Z

D −D

~ xx |3 dx |Z

7/4 ~ x |6 dx + C(δ)kZ ~ x k5/4 ~ |Z L2 (Ω) kZxxx kL( Ω)

~ x |6 dx + C(δ)kZ ~ xxx k2 ( + C 0 (δ). |Z L Ω)

Taking δ = 41 , we have d 4 dt

Z

D −D

~ xx |4 dx + 4ε |Z

Z

D

−D

~ xxx|2 dx + 16ε |Z

d ZD ~ 4 ≤5 |Zx | dx + C, dt −D

Z

D

−D

~ x |6 dx |Z

(3.3.27)

One-dimensional Landau–Lifshitz equations

97

where C is independent of ε. Integrating this inequality over t ∈ [0, T ] and using the following imbedding inequality ~ x kL4 (Ω) ≤ CkZ ~ x k3/4 ~ 1/4 kZ L2 (Ω) kZxx kL2 (Ω) , ~ xx kL2 (Ω) immediately. Using equation (3.3.5), we also we can get the estimates on kZ ~ t kL2 (Ω) . obtain the estimates for kZ

~ ε (x, t) be a smooth solution to problem (3.3.5). We have Corollary 3.3.4 Let Z ~ εx (x, t)| ≤ K3 , sup |Z

(3.3.28)

x,t

where K3 is independent of ε, D. 2. Uniform boundedness Lemma 3.3.9 Under the conditions of Theorem 3.3.2 (k ≥ 4), for the smooth solu~ ε (x, t), there holds tion of (3.3.5) Z  ~    sup kZεx xx(·, t)kL2 (Ω) ≤ K4 , 0≤t≤T

~    sup kZεx t(·, t)kL2 (Ω) ≤ K4 ,

(3.3.29)

0≤t≤T

where K4 is independent of ε, D.

Proof. Differentiating equation (3.3.5) three times with respect to x, and multi~ xxx , then integrating in x, we have plying it by Z 1d 2 dt

Z

D −D

~ xxx |2 dx + ε |Z

= −2

Z

D −D

Z

D −D

~ xxxx |2 dx |Z

~x × Z ~ xxx ) · Z ~ xxxx dx − ε (Z

Z

D

−D

~ x |2 Z) ~ xx · Z ~ xxxx dx. (|Z

Since Z D 2 ~ x | Z) ~ xx · Z ~ xxxxdx (|Z −D Z Z D Z D 1 D ~ 2 2 4 ~ ~ |Zxxxx| dx + C ≤C+ |Zxxx| dx + C |Zxx | dx 4 −D −D −D

and the following imbedding inequality

1/8 ~ xx k4 ≤ CkZ ~ xx k7/8 ~ kZ 2 kZxxxx k2 ,

we have from H¨older inequality that 1d 2 dt

Z

D

~ xxx |2 dx + ε |Z 2 −D

≤C +C

Z

D −D

Z

D −D

~ xxxx|2 dx |Z

~ xxx |2 dx − 2 |Z

Z

D

−D

~x × Z ~ xxx ) · Z ~ xxxx dx. (Z

(3.3.30)

Landau–Lifshitz Equations

98

~ ·Z ~ x = 0, Z ~ ·Z ~ xxx = −3|Z ~ xx |2 − 4Z ~x · Z ~ xxx , assuming Z ~ xxx = α0 Z ~ + β 0Z ~x + Since Z ~ ~ ~ ( Z× Z )· Z x xxx ~ ×Z ~ x , where α0 = Z ~ ·Z ~ xxx = −3Z ~ ·Z ~ xx , β 0 = Z ~x · Z ~ xxx /|Z ~ x |2 , γ 0 = γ0Z , we ~ x |2 |Z have Z

D −D

~x × Z ~ xxx ) · Z ~ xxxx (Z =

Z

"

D −D

= −3

Z

−4 −3 ≤C

~ ~ ~ ~x · Z ~ xx ) · (Z ~ x × Z) ~ + (Z × Zx ) · Zxxx |Z ~ x |2 · Z ~ ·Z ~ xxxx dx − 3(Z ~ x |2 |Z #

D

~x · Z ~ xx ) · (Z ~ ×Z ~ x )]x · Z ~ xxx dx [(Z

−D Z D

−D Z D

Z

−D D

−D

~x × Z ~x) · Z ~ xxx )Z ~x · Z ~ xxx dx ((Z ~ ×Z ~x) · Z ~ xxx )|Z ~ xx |2 dx ((Z ~ xxx |2 dx + |Z

Z

D −D

!

~ xx |4 dx . |Z

Here we have used the following Sobolev inequality 1/4 ~ ~ xx kL4 (Ω) ≤ CkZ ~ xx k3/4 kZ L2 (Ω) kZxxx kL2 (Ω)

and the H¨older inequality. From (3.3.30), we finally get Z D εZ D ~ 1d ZD ~ ~ xxx |2 dx. |Zxxx |2 dx + |Zxxxx |2 dx ≤ C + C |Z 2 dt −D 2 −D −D

The lemma then follows from Gronwall inequality. Corollary 3.3.5 Under the conditions of Lemma 3.3.9, for the smooth solution of ~ ε (x, t), there holds (3.3.5) Z   ~   sup |Zεx x(x, t)| ≤ K5 , x,t

 ~   sup kZεt (x, t)| ≤ K5 ,

(3.3.31)

x,t

where K5 is independent of ε, D. By induction, we also have

Lemma 3.3.10 Under the conditions of Theorem 3.3.3, for the smooth solution of ~ ε (x, t), there holds (3.3.5) Z ~ εxk−2s ts (·, t)| ≤ K6 , sup kZ

(3.3.32)

0≤t≤T

where K6 is independent of ε, D, k, s are any integers such that k − 2s > 0, s ≥ 0, k ≥ 3.

One-dimensional Landau–Lifshitz equations

99

3. Smooth solution to the equations of ferromagnetic spin chain From the above estimates uniform in ε and choosing subsequence by the standard method, sending ε → 0 in (3.3.5), we obtain the global smooth solution to problem (3.3.1) and (3.3.2). That is we have ~ 0 (x) ∈ H k (Ω), k ≥ 4 and |Z ~ 0 (x)| ≡ 1, Z ~ 0 (x − D) = Z ~ 0 (x + D), Theorem 3.3.4 Let Z ~ t) ∈ D > 0. Then problem (3.3.1) and (3.3.2) admits a global smooth solution Z(x,

[k/2] T s=0

s W∞ (0, T ; H k−2s(Ω)).

Similarly, sending ε → 0 and D → ∞, we know that the initial value problem (3.3.1) and (3.3.3) admits a global smooth solution. By the standard method, one can prove the uniqueness of such smooth solution.

3.4 3.4.1

Smooth Solution for the 1D Inhomogeneous Heisenberg Chain Equations Inhomogeneous Heisenberg Chain Equations

In this section, we are concerned with the existence and uniqueness of smooth solution to the one-dimensional periodic initial value problem of the inhomogeneous nonautomorphic Landau–Lifshitz equation ~ t = f (x, t)Z ~ ×Z ~ xx + ∂f (x,t) Z ~ ×Z ~ x, Z ∂x ~ 0) = Z ~ 0 (x), Z(x ~ + D, t) = Z(x ~ − D, t), |Z ~ 0 (x)| ≡ 1, Z(x,

(3.4.1) x ∈ R1 ,

(3.4.2)

~ 0 (x) are smooth functions and f (x, t) ≥ f0 > 0 where D > 0 is a constant, f (x, t) and Z ~ t) = (Z1 (x, t), Z2 (x, t), Z3 (x, t)). We assume that f (x, t) for some constant f0 , Z(x, ~ 0 (x) are periodic functions with period 2D. and Z Equation (3.4.1) is related to the generalized model of inhomogeneous ferromagnetisms and the simplified compressible ferromagnetisms which reads as ~ t = (G(Z ~ x )Z ~ ×Z ~ x )x , Z

(3.4.3)

in which G(ξ) = A + B|ξ|2 . Equation (3.4.3) with B = 0, A = f (x) corresponds to the inhomogeneous Heisenberg chain equation; B = 0, A = f (x, t) is just equation (3.4.1). In order to prove the existence of smooth solution to problem (3.4.1)–(3.4.2), we use the viscosity vanishing method as in the above section. First, by difference method, we establish the existence and uniqueness of smooth solution to the following problem ~ t = εZ ~ xx + ε|Z ~ x |2 Z ~ + f (x, t)Z ~ ×Z ~ xx + ∂f (x,t) Z ~ ×Z ~ x , ε > 0, Z ∂x ~ 0) = Z ~ 0 (x), Z(x ~ + D, t) = Z(x ~ − D, t), |Z ~ 0 (x)| ≡ 1, x ∈ R1 . Z(x,

(3.4.4) (3.4.5)

Landau–Lifshitz Equations

100

Then we prove some a priori estimates for the solution of (3.4.4)–(3.4.5) uniform in ε by constructing some new energy laws, then we send ε to zero to obtain the existence of smooth solution to problem (3.4.1)–(3.4.2). In this procedure, as in the ~ Z ~x, Z ~ ×Z ~ x } in our proof. above section, we use the orthogonal base {Z, In the classical sense, equation (3.4.4) is equivalent to ~ t = −εZ ~ × (Z ~ ×Z ~ xx ) + f (x, t)Z ~ ×Z ~ xx + ∂f (x, t) Z ~ ×Z ~ x, Z ∂x

(3.4.6)

~ × (B ~ × C) ~ = (A ~ · C) ~ B ~ − (A ~ · B) ~ C ~ and |Z(x, ~ t)| ≡ 1. since A

3.4.2

ε > 0: Local Smooth Solution

Lemma 3.4.1 Let q, r be real numbers and j, m be integers such that 1 ≤ q, r ≤ ∞, 0 ≤ j < m. If u ∈ W m,r (Ω) ∩ Lq (Ω), then kD j ukp ≤ Ckuk1−α kD m ukαr , q j m

where k · kp = k · kLp (Ω) , p ≥ 1,

≤ α ≤ 1 and !

1−α 1 1 −j = +α −m , p q r

Ω ⊂ R1 .

Lemma 3.4.2 Let p be real number and j, m be integers such that 2 ≤ p ≤ ∞, 0 ≤ j < m. Then j

kδ uh kp ≤

Ckuh k1−α 2

kuh k2 kδ uh k2 + (2D)m m

!α

,

where uh = {uj = u(xj ) | j = 0, 1, 2, . . . , J}, xj = jh, h = 2D/J, α = kδ k uh kp =

1 (j m

p ! p1 k ∆ u + i h . hk

+

1 2

− 1p ),

J−k X i=0

Lemma 3.4.3 Let uh = {uj | j = 0, ±1, ±2, . . . , ±J, . . .}, vh = {vj | j = 0, ±1, ±2, . . . , ±J, . . .} such that uj+J = uj , vj+J = vj . We have (i)

J−1 X

u j ∆+ v j = −

J X

u j ∆+ ∆− v j = −

j=0

(ii)

j=1

(iii)

J X

v j ∆− u j ,

j=1 J−1 X

∆+ u j ∆+ v j ,

j=0

∆+ (uj vj ) = uj+1 ∆+ vj + vj ∆+ uj ,

where ∆+ , ∆− denote the forward and backward difference respectively.

One-dimensional Landau–Lifshitz equations

101

We use the differential-difference method to prove the local existence of smooth solution of (3.4.4)–(3.4.5). We establish the following difference-differential equation: 2

~j ~ j ∆+ Z ~ j ~ ~ dZ ∆+ ∆− Z ~ j + fj Z ~ j × ∆+ ∆− Z j + ∆+ f j Z ~ j × ∆+ Z j , = + Z h dt h2 h2 h h ~ j |t=0 = Z ~ 0j = Z ~ 0 (jh), Z

(3.4.7) (3.4.8)

~ j+J = Z ~j , Z

(3.4.9)

, J > 0. where j = 0, ±1, . . . , ±J, . . ., h = 2D J It is clear that the initial value problem for ordinary differential equation (3.4.7)– (3.4.9) admits a local smooth solution. For such solution, we shall give some estimates uniformly in h and then get a local smooth solution to problem (3.4.4)–(3.4.5). ~j . In this section we always denote a smooth solution of (3.4.7)–(3.4.9) by Z ~ 0 (x) ∈ H 1 (Ω), Lemma 3.4.4 If Z C > 0 independent of h such that

∂f ∂x

∈ L∞ (Q) then there are constants T0 > 0,

~ h (t)k2 ≤ C, sup kZ

(3.4.10)

~ h (t)k2 ≤ C. sup kδ Z

(3.4.11)

0≤t≤T0 0≤t≤T0

~ j h and summing from j = 1 to J, we have Proof. Multiplying (3.4.7) by Z 2



J J J ~j ~ j X X ∆+ ∆− Z ∆+ Z 1d X 2 ~ ~ |Z ~ j |2 h |Z j | h = Zj · h+ 2 2 dt j=1 h h j=1 j=1

2





2

J ~ j ~ j X ∆+ Z ∆+ Z ~ j |2 h. |Z =− h+ 2 h h j=0 j=1

J−1 X

It follows from Lemma 3.4.2 that

1 2

~ h k∞ ≤ CkZ ~ h k2 kZ

~ h k2 + 1 kZ ~ h k2 kδ Z 2D

!1 2

.

Therefore we have d ~ 2 ~ h k2 ≤ C(kZ ~ h k4 + kδ Z ~ h k4 ). kZh k2 + kδ Z 2 2 2 dt Moreover, multiplying (3.4.7) by ~j ∆+ ∆− Z ~ jt ·Z h j=1 J X

~j ∆+ ∆− Z h

and summing from j = 1 to J, we get

  2 2 J ~j ~ j  ∆+ Z ~j X ∆+ ∆− Z ∆+ ∆− Z h+ Z ~  = ·  j 2 h h h j=1 j=1   J ~j ~ X ∆+ Z ∆ ∆ Z ∆+ f j  ~  · + − j h. + Zj × J X

j=1

h

(3.4.12)

h

h2

Landau–Lifshitz Equations

102

Therefore, one gets 2



2



J ~j ~ X ∆+ Z X 1 d J−1 ∆ ∆ Z h+ + − j h 2 dt j=0 h h2 j=1



2



4



2

J J J 2 X 2 X ~ ~ ~ ∆ ∆ Z ∆ Z ∆ Z 1X ~ j ~ j + − j h + C max Z + j h + C max Z + j h. ≤ 1≤j≤J 1≤j≤J 4 j=1 h2 h h j=1 j=1

Applying Lemma 3.4.2, we have

3

~ h k∞ ≤ CkZ ~ h k24 (kδ 2 Z ~ h k2 + k Z ~ h k) 14 , kZ ~ h k4 ≤ Ckδ Z ~ h k 34 (kδ 2 Z ~ h k2 + kδ Z ~ h k2 ) 14 . kδ Z Then it follows from the Young’s inequality that   d  ~ 2 ~ h k2 ≤ C 1 + k Z ~ h k18 + kδ Z ~ h k18 . kδ Zh k2 + kδ 2 Z 2 2 2 dt

(3.4.13)

Hence, putting (3.4.12) and (3.4.13) together, we have    d  ~ 2 ~ h k2 ≤ C 1 + k Z ~ h k2 + kδ Z ~ h k2 9 . kδ Zh k2 + kδ Z 2 2 2 dt

This inequality, combined with Gronwall’s inequality, implies that there are constants T0 , C > 0, independent of h such that ~ h (t)k2 + kδ Z ~ h (t)k2 ≤ C, kZ Z

T0

0

∀ t ∈ [0, T0 ] ,

~ h (t)k2 dt ≤ C. kδ 2 Z 2

Lemma 3.4.4 is proved. Corollary 3.4.1 Under the conditions in Lemma 3.4.4 and f (x, t) ∈ L∞ (Q), we have, for some constant C independent of h, sup 0≤t≤T0 ;1≤j≤J

~ j | ≤ C. |Z

~ 0 (x) ∈ H 2 (Ω), f (x, t), ∂f ∈ L∞ (Q), Lemma 3.4.5 If Z ∂x constants T0 > 0, C > 0 independent of h such that ~ h (t)k2 ≤ C, sup kδ 2 Z

(3.4.14) ∂2f ∂x2

∈ L∞ (Q), then there are (3.4.15)

0≤t≤T0

Z

T0 0

~ h (t)k2 ≤ C. kδ 3 Z

(3.4.16)

One-dimensional Landau–Lifshitz equations

103

Proof. It follows from (3.4.7) that d ~j = ∆+ Z dt

 2    ~j ~ ∆ ∆ Z + −  ∆+ Zj ~  ~  + ∆ +  Zj  + fj+1 ∆+ Zj × h2 h    

~j ∆2+ ∆− Z h2

~j × + ∆ + f j Z 

~j ~ ∆+ ∆− Z ∆ f ~ j × ∆+ Z j   + + j+1 ∆+ Z h2 h h 

~j ∆+ Z ∆2 f j ~ . + + Z j × h h Multiplying this equality by

~j ∆2+ ∆− Z , h3

J ~j X ∆2+ ∆− Z

h3

j=1

h3



2





~ ∆2 Z + j h2

~ jt = − 1 d · ∆+ Z 2 dt

J−1 X





j=0

~ ~ j × ∆+ ∆− Z j  · ∆+ f j  Z h2

J ~j X ∆2+ ∆− Z

j=1

summing it from j = 1 to J and noting that h,

2 2 J 2 J ~ ~ X X ∆+ ∆− Z j ∆+ ∆− Zj 1 ~ 2 max Zj ≤ h + C 1≤j≤J h 10 j=1 h3 h2 j=1

and

J ~j X ∆2+ ∆− Z

h3

j=1

=



J ~j X ∆2+ ∆− Z

h3

j=1



~ ~ j × ∆+ ∆− Z j  · fj+1 ∆+ Z h2 

2





~ ~ j × ∆+ ∆− Z j  · fj+1 ∆+ Z h2

2

2



J 2 J ~ j ~ j X ~ j ∆+ Z ∆+ ∆− Z ∆+ ∆− Z 1 X h, ≤ max h + 1≤j≤J 10 j=1 h3 h2 h j=1

as well as

1

~ h k∞ ≤ Ckδ Z ~ h k22 (kδ 2 Z ~ h k2 + kδ Z ~ h k2 ) 12 , kδ Z 2

~ h k6 ≤ Ckδ Z ~ h k23 (kδ 2 Z ~ h k2 + kδ Z ~ h k2 ) 13 kδ Z we finally get d dt



2





~ ∆2 Z + j h2

J−1 X j=0

h+



2





~ ∆2 ∆ Z + − j h3

J−1 X j=0



h≤C +C 

2 2 ~ ∆ Z + j h2

J−1 X j=0

Then, Lemma 3.4.5 follows from the Gronwall’s inequality.





3

h . 

Landau–Lifshitz Equations

104

Corollary 3.4.2 Under the conditions in Lemma 3.4.5, we have, for some C independent of h, ~ j (t) ∆+ Z ≤ C, sup h 0≤t≤T0 ;1≤j≤J Z T0 0

(3.4.17)

~ ht (t)k22 dt ≤ C. kδ Z

(3.4.18)

~ 0 (x) ∈ H 3 (Ω), f (x, t), ∂ j fj ∈ L∞ (Q), j = 1, 2, 3, then there are Lemma 3.4.6 If Z ∂x constants T0 > 0, C > 0 independent of h such that ~ h (t)k2 ≤ C, sup kδ 3 Z

(3.4.19)

0≤t≤T0

Z

T0 0

~ h (t)k2 ≤ C. kδ 4 Z

(3.4.20)

Proof. From (3.4.7) we have J ~j X ∆3+ ∆− Z

j=1

h4

·

~ jt ∆2+ Z h h2

2 2 J 3 J ~ j ~j ~ ~ j ∆3+ ∆− Z X X ∆+ Z ∆ ∆ Z ∆ Z − + j+2 + = h + · h h4 h h2 h4 j=1 j=1   J 2 ~ ~ ~ ~ ∆3 ∆ Z X ∆ Z ∆ Z ∆ Z  + j+1 · + j+2  + j · + − j h +

h2

j=1

h

h4

h





~ ~ ~ ~ ~ ~ ∆3 ∆ Z ∆2 Z ∆2 Z ∆ Z ∆ Z ∆ Z  + j+2 · + j+1 + + j+1 · + j+1  + j · + − j h + h h2 h2 h h h4 j=1 J X





3 ~ ~ ~ ~ ~ ∆3 Z ∆2 Z ∆2 Z ∆ Z ~ j · ∆+ ∆− Z j h  + j+2 · + j + + j+1 · + j  Z + h h3 h2 h2 h4 j=1 J X







~ ~ ~ ∆3 ∆ Z ~ ∆2 Z ∆ Z ∆ Z  + j+1 · + j+1   + j · + − j  + h h2 h h3 j=1 J X

   2 ~ 2 J 3 ~ 3 ~j ~ X ∆+ Z j  ∆ ∆ Z −  ∆+ Zj+1 ∆+ Zj ~j · + h + · + 2  Z  3 4 h h h h j=1   J ~ j+1 ∆2+ ∆− Z ~j ~ X ∆ Z ∆3 ∆ Z + · + − jh +2 fj+2  ×

h

j=1

h3

h4





~ j ∆+ ∆− Z ~ ~j ∆2 Z ∆3 ∆ Z · + − jh + fj+2  +2 × h h2 h4 j=1 J X

+2

J X ∆+ fj+1

j=1

h





~ ~ ~ ∆ Z ∆3 ∆ Z ∆ ∆ Z  + j × + − j· + − jh h h3 h4

One-dimensional Landau–Lifshitz equations

+2

J X ∆+ fj+1

h

j=1

+

J X ∆2+ fj

h2

j=1

+

J X ∆+ fj+2



h

h

J X ∆2+ fj+1

h2

j=1

+

J X ∆2+ fj+1

h2

j=1

+

J X ∆3+ fj

h3

j=1



~j ∆2+ ∆− Z  h3

~j ∆3+ ∆− Z · h h4



3 ~ ~ ~ j × ∆+ ∆− Z j  · ∆+ ∆− Z j h Z h2 h4



j=1

+

~ j+1 × Z

J X ∆+ fj+2

j=1

+





105





3 3 ~ ~ ~ j+2 × ∆+ Zj  · ∆+ ∆− Zj h Z h3 h4



~ ~ ~ ∆3 ∆ Z ∆2 Z ∆ Z  + j+1 × + j  · + − j h h h2 h4 



3 ~ ~ ~ j × ∆+ Z j  · ∆+ ∆− Z j h Z h h4





2 ~ 3 ~ ~ j+1 × ∆+ Zj+1  · ∆+ ∆− Zj h Z h2 h4



3 ~ ~ ~ j × ∆+ Zj  · ∆+ ∆− Zj h. Z h h4

From Lemma 3.4.2, we have 1



~ h k4 ≤ CkZ ~ h k24 kδ 3 Z ~ h k2 + k Z ~ h k2 kδ 2 Z

3

4

.

By Lemmas 3.4.3–3.4.5, one gets J ~j X ∆3+ ∆− Z

h4

j=1



2



~ jt ~ j X ∆3+ Z ∆2+ Z 1 d J−1 h, · h=− h2 2 dt j=0 h3 

~ ~ ∆3 ∆ Z ~ ~ ∆ Z ∆ Z ∆2 Z  + j+1 · + j+2  + j · + − j h h2 h h h4 j=1 J X

2



2



J 3 J 3 ~ ~ j X ∆+ ∆− Z ∆ Z 1 X + j h + C. ≤ h+C 4 3 32 j=1 h j=1 h

The others are given in the same way. Therefore we have

2



2



2

J 3 J−1 ~ j ~ j ~ X ∆3+ Z X ∆3+ Z 1 d J−1 1X ∆ ∆ Z h+ + − j h ≤ C h + C. 3 2 dt j=0 h3 2 j=1 h4 h j=0

Lemma 3.4.6 follows from the Gronwall’s inequality.

Landau–Lifshitz Equations

106

By induction method we have ~ 0 (x) ∈ H k (Ω, S 2 ), f (x, t) ∈ L∞ (Q), ∂ j fj ∈ L∞ (Q), j = 1, 2, . . . , k, Lemma 3.4.7 If Z ∂x then there are constants T0 > 0, C > 0 independent of h such that ~ h k2 ≤ C, sup kδ k Z

(3.4.21)

0≤t≤T0

~ ht k2 ≤ C, sup kδ k−2 Z

k ≥ 2,

(3.4.22)

~ htt k2 ≤ C, sup kδ k−4 Z

k ≥ 4.

(3.4.23)

0≤t≤T0 0≤t≤T0

From Lemma 3.4.7 and the a priori estimates for solutions to the differentialdifference equation (3.4.7)–(3.4.9), we conclude that there exists a constant T0 > 0 such that problem (3.4.4)–(3.4.5) admits a smooth solution in Ω × [0, T0 ]. This result is stated as follows. ~ 0 (x) ∈ H k (Ω, S 2 ), f (x, t) ∈ L∞ (Q), ∂ j fj ∈ L∞ (Q), Theorem 3.4.1 Let  > 0, Z ∂x j = 1, 2, . . . , k, f (x, t) ≥ f0 > 0. Then (3.4.4)–(3.4.5) admits a local smooth solution ~ t)in [0, T0 ] with T0 depending on k : Z(x,   k+1  [ 2 ] \ \ \    s ~ t) ∈  W∞ (0, T0 ; H k−2s(Ω))  H s (0, T0 ; H k+1−2s(Ω)) . Z(x,  

3.4.3

[ k2 ]

s=0

s=0

ε > 0: Global Solution and Uniform Estimates

In the above subsection, we have obtained a local smooth solution for (3.4.4)–(3.4.5) when ε > 0 is fixed. In this subsection, we intend to prove the global existence of smooth solution to problem (3.4.4)–(3.4.5) for fixed ε > 0 by deriving the global (in time) estimates. To meet the need of sending ε to zero, these estimates have to be equal both global in time and uniform in ε. ~ 0 (x) ∈ H 1 (Ω, S 2 ) and Z(x, ~ t) is a global smooth solution of probLemma 3.4.8 If Z lem (3.4.4)–(3.4.5), then we have ~ t)| = 1, |Z(x,

∀ (x, t) ∈ R1 × [0, +∞).

(3.4.24)

Lemma 3.4.9 Under the same conditions of Lemma 3.4.8 and ∂f , ∂f ∈ L∞ (Q), we ∂t ∂x have, for any given T > 0, there exists C > 0 independent of ε and D, ~ x (·, t)k2 ≤ C. sup kZ

(3.4.25)

0≤t≤T

Proof. From (3.4.6), we have 



~ xt = −ε Z ~× Z ~ ×Z ~ xx Z



x

+

∂2f ~ ~ ∂f ~ ~ ~x × Z ~ xx + f Z ~ ×Z ~ xxx . Z × Zx + 2 Z × Zxx + f Z 2 ∂x ∂x

One-dimensional Landau–Lifshitz equations

107

~ × (B ~ × C)) ~ ·B ~ = (A ~ × B) ~ · (C ~ × B), ~ we get By (A Z

Ω

   ∂f  ~  ~ ~  ~ ~× Z ~ xx × Z ~ ·Z ~ xx Z × Z × Zxx · Zx − ε f Z Ω Ω ∂x Z Z   ∂f  ~ ~  ~ ~x ~ ×Z ~ xx · Z +2 f Z × Zxx · Zx + f 2 Z x Ω ∂x Ω Z Z ∂f  ~  ~ ~  ~ ~ ~ xx 2 . =ε ×Z Z × Z × Zxx · Zx − ε f Z Ω ∂x Ω Z

Z

~x · Z ~ xt = ε fZ

Therefore, one gets 1d 2 dt

Z

Ω



2

~x + ε f Z

Z

Ω

2



1 ∂f ~ 2 ∂f  ~  ~ ~  ~ Z x + ε Z × Z × Zxx · Zx 2 ZΩ ∂t Ω ∂x 2 ε Z ~ ~ 2 ~ ≤ C f Z dx + f Z × Zxx . x 2 Ω Ω Z

~ ×Z ~ xx = f Z

It is easy to see that

Z

Z 2 1 d Z ~ 2 ~ f Zx dx ≤ C f Z x dx. 2 dt Ω Ω

Then lemma 3.4.9 follows.

2

~ 0 (x) ∈ H 2 (Ω, S 2 ), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, 3, then Lemma 3.4.10 Let Z ∂t ∂x for any given T > 0, there are constants C > 0 independent of ε, D, such that ~ xx (·, t)k2 ≤ C, sup kZ

(3.4.26)

~ t (·, t)k2 ≤ C. sup kZ

(3.4.27)

0≤t≤T

0≤t≤T

Proof. In the proof, we shall use the following identities which follow from the ~ t)| = 1: fact |Z(x, ~ ·Z ~ t = 0, Z

~ ·Z ~ x = 0, Z

~ ·Z ~ xx = −|Z ~ x |2 , Z

~ ·Z ~ xxx = −3Z ~x · Z ~ xx . Z

(3.4.28)

It follows from (3.4.4) that ~ xxt = εZ ~ xxxx + ε Z ~ x 2 Z ~ xx + 2ε Z ~x · Z ~ xx Z ~x Z + 2ε

+3

h







~x · Z ~ xx Z ~ Z

i



x

+



∂3f ~ ~ ∂2f ~ ~ ∂f ~ ~ Z × Z + 3 Z × Zxx + 3 Z x x × Zxx 3 2 ∂x ∂x ∂x

∂f ~ ~ ~x × Z ~ xxx + f Z ~ ×Z ~ xxxx . Z × Zxxx + 2f Z ∂x

(3.4.29)

Landau–Lifshitz Equations

108

From (3.4.29) we get 1d 2 dt

Z

Ω

~ 2 f 2 Z xx dx

∂f = f ∂t Ω Z



Z ~ 2 ~ xx · Z ~ xxt dx Zxx dx + f 2Z Ω

∂f ~ ~ = −2ε f Z xx · Zxxx dx − ε Ω ∂x Z

+ 8ε

Z

Z

Ω

2



~ xxx dx + ε f 2 Z

Z

Ω

~ x 2 Z ~ xx 2 dx f 2 Z



Z Z 2 ∂f ∂f  ~ ~   ~ ~  2 ~ ~ f Zx · Zxx dx − 4ε f Zx · Zxx Z · Zxx dx + f

Ω ∂x Z 3     ∂ f ~ xxx dx. ~ xx dx − f 3 Z ~x × Z ~ xx · Z ~ ×Z ~x · Z + f2 3 Z ∂x Ω Ω Ω

Z



∂t

Ω

~ 2 Zxx dx

(3.4.30)

~ x | 6= 0, then the vectors Z, ~ Z ~x, Z ~ ×Z ~ x form an orthogonal basis Note that, if |Z 3 of R . Let ~ xx = αZ ~ + βZ ~x + γZ ~ ×Z ~ x, Z and it is easy to see that ~ x |2 , α = −|Z

β=

~x · Z ~ xx Z , ~ x |2 |Z

γ=

~ ×Z ~x) · Z ~ xx (Z . ~ x |2 |Z

Therefore we have −

Z

Ω





~x × Z ~ xx · Z ~ xxx f3 Z  





 ~ ~ ~ ~ x × −|Z ~ x |2 Z ~ + (Z × Zx ) · Zxx (Z ~ ×Z ~ x ) · Z ~ xxx = − f3 Z   ~ x |2 Ω |Z Z

=−

Z

Ω

5 =− 2

Z

~ 2 ~ ~x · Z ~ xxx − f 3 Z Z×Z x

Ω





Z

f3

Ω

h

Z   9 ∂f 3 ~ 2 ~ ~ ~ f Zx Z × Zx · Zxxx − f2

2

Ω



~ ×Z ~x · Z ~ xx Z ∂x

i

3 ~ 2 − |Z x| 2



2   ~ ~ ~x · Z ~ xx , Z x Z × Z

x

(3.4.31)

~ × (B ~ × C) ~ = (A ~ · C) ~ B ~ − (A ~ · B) ~ C. ~ where we have used the fact that A Combining (3.4.30) with (3.4.31), we get d 2 dt

Z

Z 2 ~ 2 ~ f Zxx + 4ε f 2 Z xxx 2

Ω

Ω

2 ∂f ~ 2 ~ 2 ~ 2 2 ~ ~ ~ = −8ε f Z · Z + 4ε f Z Z + 32ε f Z · Z xx xxx x xx x xx Ω ∂x Ω Ω Z Z ∂f  ~ ~   ~ ~  ∂f ~ 2 Zx · Zxx Z · Zxx + 4 f Zxx − 16ε f ∂t Ω Ω ∂x Z Z 2   ∂3f  ~ ~  ~ ~ ~ ~ ~ + 4 f2 3 Z × Zx · Zxx − 10 f 3 Z Z × Z x x · Zxxx ∂x Ω Ω Z ∂f ~ 2  ~ ~  ~ − 18 f 2 Zx Z × Zx · Zxx . (3.4.32) ∂x Ω Z

Z

Z

One-dimensional Landau–Lifshitz equations

109

On the other hand, from (3.4.4) we get that Z 5 d Z 2 ~ 4 ∂f f Zx dx = 5 f 2 dt Ω ∂t Ω Z ∂f =5 f ∂t Ω

+ 10ε

Z

Ω

− 10

Z

Ω

Z 4 ~ ~ x |2 Z ~x · Z ~ xt dx Zx dx + 10 f 2 |Z Ω

Z 4 2 ~ ~ x Z ~x · Z ~ xxx dx Zx dx + 10ε f 2 Z Ω

Z  6  ~ 2 ∂f ~ 2 ~ ~ ~ Z Z × Z f 2 Z dx − 20 f x x · Zxx dx x Ω

∂x

2   ~ ~ ~ ~ f 3 Z Z × Z x x · Zxxx dx.

(3.4.33)

Putting (3.4.32) and (3.4.33) together, we have d 2 dt

Z

Z Z 2 ~ 2 2 ~ f Zxx dx + 4ε f Zxxx dx + 10ε f 2

Z 6 4 5d ~ ~ Zx dx − f 2 Z x dx

2 dt Ω Ω Ω Z Z ∂f ~ 4 ∂f ~ 2 ∂f ~ ~ xxx dx = −5 f Zx dx + 4 Zxx dx − 8ε f f Zxx · Z ∂t ∂t Ω Ω Ω ∂x Z Z Z 2 ∂f ~ 2 ~ ~ 2 ~ 2 ~ 2 2 ~ ~ + 4ε f Zx Zxx dx + 32ε f Zx · Zxx dx + 16ε f Zx Zx · Zxx dx Ω Ω Ω ∂x Z Z 2   ~ ~ 3 ∂f ~ 2 ~ ~ ~ ~ − 10ε f 2 Z Z · Z dx + 2 f Z Z × Z x x xxx x x · Zxx dx ∂x Ω Ω Z ∂3f  ~ ~  ~ + 4 f2 3 Z × Zx · Zxx dx. (3.4.34) ∂x Ω Ω

Z

By Lemmas 3.4.1 and 3.4.8, we have Z

Z 2 ~ ~ 2 ~ ~ 2 4ε f Zx Zxx dx + 32ε f 2 Z x · Zxx dx Ω Ω Z 5 7 2 ~ 6 ~ x k24 kZ ~ xxx k24 ≤ Cδ1 ε f Zx dx + CεkZ ZΩ Z 6 2 ~ 2 ~ ≤ Cδ1 ε f 2 Z dx + Cεδ f Z x 2 xxx dx + C. 2

Ω

Letting δ1 = 2

d dt

Z

Ω

2 , δ2 C



=

2

1 , C

we get

~ xx + 3ε f 2 Z

∂f ≤4 f ∂t Ω Z

− 10ε

Z

Ω

Ω

Z

Ω



2

~ xxx + 8ε f 2 Z

Z ∂f ~ 2 Zxx − 5 f Ω

∂t

Z

Ω



6

Ω

∂x

~ x − f 2 Z

5d 2 dt

Z

Ω

Z 4 ∂f ~ ~ ~ Zx − 8ε f Z xx · Zxxx

Z 2 ∂f ~ 2 ~ ~ ~ ~ ~ f 2 Z Z · Z + 16ε f Zx Zx · Zxx x x xxx



4

~ x f 2 Z

∂x

Ω

Z ∂f ~ 2  ~ ~  ~ ∂3f  ~ ~  ~ + 2 f3 Zx Z × Zx · Zxx + 4 × Zx · Zxx + C f2 3 Z ∂x ∂x Ω Ω Z Z Z 2 ~ 2 2 ~ 6 2 ~ ≤ C f 2 Z + 6ε f Z + ε f Z (3.4.35) xx x xxx + C, Z

Ω

Ω

Ω

2

1

~ x k6 ≤ k Z ~ x k23 kZ ~ xx k23 . where we have used the fact that kZ

Landau–Lifshitz Equations

110

Therefore one gets d dt

Z

Z Z d ~ 2 2 ~ 4 ~ xx 2 dx + C. f Zxx dx ≤ C f Zx dx + C f 2 Z 2

Ω

dt

Ω

(3.4.36)

Ω

3

1

~ x k4 ≤ CkZ ~ x k24 kZ ~ xx k24 , we have Integrating (3.4.36), by Lemma 3.4.9 and kZ Z

Z tZ ~ 2 ~ 2 f Zxx dx ≤ C f 2 Z xx dxdt + C. 2

Ω

0

Ω

The conclusion of Lemma 3.4.10 follows from this inequality and Gronwall’s inequality. Similarly, we have the following lemma: ~ 0 (x) ∈ H 3 (Ω), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, 3, 4. For any Lemma 3.4.11 Let Z ∂t ∂x given T > 0, there are C > 0 independent of ε, D such that ~ xxx (·, t)k2 ≤ C, sup kZ

(3.4.37)

~ xt (·, t)k2 ≤ C. sup kZ

(3.4.38)

0≤t≤T

0≤t≤T

Proof. It follows from (3.4.29) that 1d 2 dt

Z

Ω

~ xxx|2 dx f 3 |Z

3 ∂f ~ 2 ~ xxx · Z ~ xxxt dx f 3Z f |Z xxx | dx + 2 Ω ∂t Ω Z Z Z 3 ∂f ~ ∂f ~ 2 3 ~ 2 ~ = f |Zxxx | dx − ε f |Zxxxx| dx − 3ε f 2 Z xxx · Zxxxx dx 2 Ω ∂t ∂x Ω Ω Z Z    3 ~ 2 ~ 2 ~x · Z ~ xx Z ~ xx · Z ~ xxx dx + ε f |Zx | |Zxxx | dx + 4ε f 3 Z =

Z

Z

Ω

+ 2ε

Z

Ω

− 2ε

Ω





~x · Z ~ xxx dx + 2ε ~ xx |2 Z f 3 |Z

Z

Ω

~ ~ 2 f 3 Z x · Zxxx dx



Z

~ xx |2 Z ~+ Z ~x · Z ~ xxx Z ~+ Z ~x · Z ~ xx Z ~x · Z ~ xxxx dx f 3 |Z

Z

f2

Ω

h









i

∂f h ~ 2 ~  ~ ~  ~  ~ ~  ~ i ~ |Zxx | Z + Zx · Zxxx Z + Zx · Zxx Zx · Zxxx dx ∂x Ω Z Z ∂3f  ~ ~  ~ ∂4f  ~ ~  ~ × Zx · Zxxx dx + 4 f 3 3 Z × Zxx · Zxxx dx + f3 4 Z ∂x ∂x Ω Ω Z Z    ∂2f  ~ 4 ~ ~ ~ ~ ~ + 6 f3 2 Z × Z · Z dx − 2 f Z × Z x xx xxx x xxx · Zxxxx dx. ∂x Ω Ω − 6ε

~ x | 6= 0, let Z ~ xxx = α0 Z ~ + β 0Z ~x + γ0Z ~ ×Z ~x. Then if |Z It is easy to see that ~x · Z ~ xx , α0 = −3Z

β0 =

~x · Z ~ xxx Z , ~ x |2 |Z

γ0 =

~ ×Z ~ x) · Z ~ xxx (Z . ~ x |2 |Z

(3.4.39)

One-dimensional Landau–Lifshitz equations

111

Thus −2

Z



Ω

= −2



~ xxxx dx ~x × Z ~ xxx · Z f4 Z Z

Ω

 





~ x ×  −3Z ~x · Z ~ xx f 4 Z





   ~ ~ ~ ~ + (Z × Zx ) · Zxxx Z ~ xxxx dx ~ ×Z ~x  · Z Z  2 ~ |Z x |

  ∂f  ~ ~   ~ ~  ~ ~ 2 ~ ~ ~ Z × Z Zx · Zxx Z × Zx · Zxxx dx + 12 f 4 Z xx x · Zxxx dx ∂x Ω Ω Z Z       ~ xxx dx ~ ×Z ~x · Z ~ xxx dx + 6 f 4 Z ~x · Z ~ xxx Z ~ ×Z ~ xx · Z ~x · Z ~ xx Z + 6 f4 Z

= 24

Z

+8

f3

ZΩ

Ω

f4

Ω

≤C

Z

Ω

Z

h



~ xxx ~ ×Z ~x · Z Z

~ 2 f 3 Z xxx dx + C,



i



~x · Z ~ xxx dx Z

(3.4.40)

~ ·Z ~ xxxx = −3|Z ~ xx |2 − 4Z ~x · Z ~ xxx . where we have used the fact that Z Putting (3.4.39) and (3.4.40) together, we have 1d 2 dt

Z

Z Z Z 2 2 2 ε ~ ~ 3 ~ 2 3 ~ f Zxxx + ε f |Zxxxx | ≤ C f Zxxx + f 3 Z xxxx + C. 3

Ω

Ω

Ω

2

Ω

Employing Gronwall’s inequality, we obtain (3.4.37). Using

2 ~ xt = εZ ~ xxx + ε|Z ~ x |2 Z ~ x + 2ε(Z ~x · Z ~ xx )Z ~ + ∂ fZ ~ ×Z ~x Z ∂x2 ∂f ~ ~ ~x × Z ~ xx + f Z ~ ×Z ~ xxx , +2 Z × Zxx + f Z ∂x

we get (3.4.38). By induction, we have ~ 0 (x) ∈ H k (Ω), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, . . . , k + 1. Lemma 3.4.12 Let Z ∂t ∂x For any given T > 0, there is C > 0 independent of ε, D such that ~ t)k2 ≤ C, sup k∂ts ∂xk−2s Z(·,

0≤t≤T

0 ≤ s ≤ [k/2].

(3.4.41)

Combining the local existence obtained above and the global in time estimates in Lemmas 3.4.8–3.4.12, we can get the global existence of smooth solution to problem (3.4.4)–(3.4.5) for fixed ε > 0 in the following sense: ~ 0 (x) ∈ H k (Ω), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, . . . , k + 1, Theorem 3.4.2 Let Z ∂t ∂x f (x, t) ≥ f0 > 0. Then, for any given T > 0, problem (3.4.4)–(3.4.5) admits at least ~ t) in [0, T ] : one smooth solution Z(x, ~ t) ∈ Z(x,

[ k2 ]

\

s W∞ (0, T ; H k−2s(Ω)).

s=0

All the a priori estimates in Lemmas 3.4.8–3.4.12 are uniform in ε.

(3.4.42)

Landau–Lifshitz Equations

112

3.4.4

ε = 0: Global Solution and Uniqueness

In the above section, we have obtained a global smooth solution for (3.4.4)–(3.4.5) for fixed ε > 0, and the estimates are all uniform in ε. These uniform estimates allow us to pass to the limit ε → 0 in equation (3.4.4) and then get the global smooth solution of problem (3.4.1)–(3.4.2). Therefore, we have the following theorem: ~ 0 (x) ∈ H k (Ω), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, . . . , k + 1, Theorem 3.4.3 Let Z ∂t ∂x ~ t) : f (x, t) ≥ f0 > 0. Then (1.1), (1.2) admits a global smooth solution Z(x, ~ t) ∈ Z(x,

[ k2 ]

\

s=0

s W∞ (0, ∞; H k−2s(Ω)).

On the other hand, all the estimates in Sec. 3 are independent of D. Thus, by sending D to ∞, we get the global existence of smooth solution to the Cauchy problem of (3.4.1). ~ 0 (x) ∈ H k (Ω), f (x, t), ∂f , ∂ j fj ∈ L∞ (Q), j = 1, 2, . . . , k + 1, Theorem 3.4.4 Let Z ∂t ∂x f (x, t) ≥ f0 > 0. Then the Cauchy problem of (3.4.1) admits a global smooth solution ~ t) : Z(x, ~ t) ∈ Z(x,

[ k2 ]

\

s=0

s W∞ (0, ∞; H k−2s(Ω)).

Remark 3.4.1 For any given ε > 0, it is not difficult to see that the problem (3.4.4)–(3.4.5) (or the Cauchy problem) admits at least one global smooth solution. Now we turn to prove that the global smooth solution of (3.4.1) and (3.4.2) obtained in Theorem 3.4.3 is unique. That is, we prove Theorem 3.4.5 The global smooth solution of (3.4.1)–(3.4.2) obtained in Theorem 3.4.3 is unique. Proof. First of all, we prove that the smooth solution to periodic problem (3.4.4)– ~ 1 (x, t) and Z ~ 2 (x, t) be smooth solutions of (3.4.4) and (3.4.5). (3.4.5) is unique. Let Z ~ ~ ~ Let W (x, t) = Z1 − Z2 . Then, we have ~ t = εW ~ xx + εZ ~2 W

h

~ 1x + Z ~ 2x · W ~ + ε Z ~ 1x 2 W ~ Z 

i





~2 × W ~ xx + f W ~ ×Z ~ 1xx + ∂f Z ~2 × W ~ x + ∂f W ~ ×Z ~ 1x , + fZ ∂x ∂x

(3.4.43)

~ , using with homogeneous initial boundary conditions. Multiplying (3.4.43) by W ~1, Z ~ 2 are smooth, we have H¨older inequality and noting that Z 1d 2 dt

Z

D −D

ε ≤ 4

Z 2 ~ W dx + ε

D

−D

Z

D

−D

~ 2 Wx dx

Z ~ 2 Wxx dx + C

D

−D

Z ~ 2 Wx dx + C

D −D

2 ~ W dx.

(3.4.44)

One-dimensional Landau–Lifshitz equations

113

~ xx , we have On the other hand, multiplying (3.4.43) by W 1d 2 dt

Z

D −D

ε ≤ 4

Z ~ 2 Wx dx + ε Z

D

D −D

~ 2 Wxx dx

Z ~ 2 Wxx dx + C

−D

D

Z ~ 2 Wx dx + C

−D

D −D

2 ~ W dx.

(3.4.45)

Combining (3.4.44) and (3.4.45), using Gronwall’s inequality and noting that ~ W (x, 0) = 0, Wx (x, 0) = 0, we can get the uniqueness when ε > 0. In the second step, we let ε = 0 in (3.4.43) to get ~ t = f (x, t)Z ~2 × W ~ xx + f (x, t)W ~ ×Z ~ 1xx + ∂f Z ~2 × W ~ x + ∂f W ~ ×Z ~ 1x . W ∂x ∂x

(3.4.46)

Therefore we have, 1d 2 dt

Z

D −D

~ |2 dx = |W

Z

D −D D





~2 × W ~ xx · W ~ dx + f (x, t) Z

Z

D −D

 ∂f  ~ ~x ·W ~ dx Z2 × W ∂x

Z D    ∂f  ~ ~ ·W ~ x dx + ~ 2x × W ~ ·W ~ x dx = Z2 × W f Z −D ∂x −D Z D  ∂f  ~ ~ ·W ~ x dx − Z2 × W −D ∂x Z

≤C

Z

D

Z 2 ~ W dx + C

−D

D

−D

On the other hand, from (3.4.46) we have



2

~ x dx. f W

(3.4.47)

~ tx = 2 ∂f Z ~2 × W ~ xx + f Z ~ 2x × W ~ xx + f Z ~2 × W ~ xxx + 2 ∂f W ~ ×Z ~ 1xx W ∂x ∂x ∂2f ~ ~ ~ ~ ~ ~ x + ∂f Z ~ 2x × W ~x + f Wx × Z1xx + f W × Z1xxx + 2 Z2 × W ∂x ∂x ∂2f ~ ~ 1x + ∂f W ~x×Z ~ 1x . + 2W ×Z ∂x ∂x ~1, Z ~ 2 are smooth, we have Therefore using H¨older inequality and noting that Z 1d 2 dt

Z

D D

Z 1 ~ 2 f Wx dx =

D

2 −D 1Z D = 2 −D +2

Z

∂f ∂t ∂f ∂t D

f

−D

+ ≤C

Z

Z

D

f −D D

−D

2

D

Z ~ 2 Wx dx + ~ 2 Wx dx +

−D Z D −D

~x·W ~ xt dx fW h



 ∂f  ~ ~ 1xx · W ~ x dx W ×Z ∂x 



~ ×Z ~ 1xxx · W ~ x dx + W

Z ~ 2 W dx + C

D

−D



~2 × W ~ xx · W ~x f2 Z

Z

~ 2 f W x dx.

D

f −D

i

dx x

 ∂2f  ~ ~ 1x · W ~ x dx W × Z ∂x2

(3.4.48)

Landau–Lifshitz Equations

114

Putting (3.4.47) and (3.4.48) together, we get d dt

Z

D −D

Z 2 ~ W dx +

D −D

~ f W x dx

!

≤C

Z

D −D

Z 2 ~ W dx +

D −D

! ~ 2 f Wx dx .

~ (x, 0) = 0, Wx (x, 0) = 0, we can Using Gronwall’s inequality and noting that W get the conclusion of Theorem 3.4.5.

3.5

3.5.1

Measure-Valued Solution to the Strongly Degenerate Compressible Heisenberg Chain Equations Compressible Heisenberg Chain Model and Compressible Heisenberg Chain System

In this section, we intend to establish the existence of measure-valued solution to the following periodic initial value problem: ~ t = (G(Z ~ x )Z ~ ×Z ~ x )x , x ∈ R 1 , t ∈ R + , Z ~ 0) = Z ~ 0 (x), Z(x ~ + D, t) = Z(x ~ − D, t), |Z ~ 0 (x)| ≡ 1, x ∈ R1 , Z(x,

(3.5.1) (3.5.2)

where G(ξ) = A + B|ξ|2 and A, B, D > 0 are constants by considering the viscosity problem ~ t = ε(G(Z ~ x )Z ~ x )x + (G(Z ~ x )Z ~ ×Z ~ x )x , x ∈ R 1 , t ∈ R + , Z ~ 0) = Z ~ 0 (x), Z(x ~ + D, t) = Z(x ~ − D, t), |Z ~ 0 (x)| ≡ 1, x ∈ R1 . Z(x,

(3.5.3) (3.5.4)

We first prove that problem (3.5.3)–(3.5.4) admits at least one global weak solution, and then give the a priori estimates uniform in ε for such solutions to obtain the existence of the measure-valued solution to (3.5.1)–(3.5.2) by letting ε → 0. To get the existence of local solution of (3.5.3)–(3.5.4), we apply the differencedifferential method. In the sequel, we denote Ω = (−D, D). For simplicity, we let ε = 1 in this subsection. We establish the following difference-differential equation: ∆− ~j dZ = dt

  ~ j  ∆+ Z ~j  ~ j  ∆+ Z ~j  ∆+ Z ∆+ Z ~ G ∆− Z j × G h h h h + , h h ~ j |t=0 = Z ~ 0j = Z ~ 0 (jh), Z ~ j+J = Z ~j , Z

 

(3.5.5) (3.5.6) (3.5.7)

where j = 0, ±1, . . . , ±J, . . . , h = 2D/J, J > 0. It is clear that the initial value problem of ordinary differential equation (3.5.5)– (3.5.7) admits a local smooth solution. We shall give some estimates uniformly in h

One-dimensional Landau–Lifshitz equations

115

for such solution, and then get the local solution to problem (3.5.3)–(3.5.4). In this ~j . section we always denote the solution of (3.5.5)–(3.5.7) by Z ~ 0 (x) ∈ W 1,4 (Ω), there are constants T0 > 0, C > 0 independent Lemma 3.5.1 If Z of h such that sup 0≤t≤T0





~ h (t)k2 + kδ Z ~ h (t)k2 + kδ Z ~ h (t)k4 ≤ C, kZ Z tZ 0

Ω

(3.5.8)

~ h (t)k2 ≤ C. kδ 2 Z

(3.5.9)

~ j h and summing from j = 1 to J, we have Proof. Multiplying (3.5.5) by Z J J−1 ~ j  ∆+ Z ~ j 2 X  ∆+ Z 1d X ~ j |2 h = − h |Z G 2 dt j=1 h h j=0



+

J−1 X j=0

=−

J−1 X j=0

Therefore, we have

~ ~ ~ j × G ∆+ Z j ∆+ Z j Z h h 





·

~j ∆+ Z h h

2



~ j  ∆+ Z ~ j ∆+ Z 2 h. A + B| | h h

1d ~ 2 ~ h k2 + Bkδ Z ~ h k4 = 0. kZh k2 + Akδ Z 2 4 2 dt It follows from (3.5.5) that ∆+ ∆− ~j d∆+ Z = dt

  ~j  ~j  ~ j  ∆+ Z ~ j  ∆+ Z ∆+ Z ∆+ Z ~ G ∆+ ∆− Z j × G h h h h + . h h

 

Multiplying this equation by G give J−1 X j=0

G



+



~j ~ j  ∆+ Z ~ j d∆+ Z ∆+ Z h

J−1 X

h

∆+ ∆−

j=0



dt

~ j  ∆+ Z ~j ∆+ Z h h

=

J ∆− X

j=1

and summing from j = 0 to j = J − 1 to  

J−1 X ∆+ ∆− G j=0

~ j ∆+ Z ~j ∆+ Z  ~ j  ∆+ Z ~j ∆+ Z h h G h h h 



~j  ~ j  ∆+ Z ∆+ Z ~  Zj × G ~ j  ∆+ Z ~j ∆+ Z h h G h h h 

  ~  2 ~  ∆ G ∆+ Z j ∆+ Z j J − X h h =− h h j=1    ~ ~ 

−

(3.5.10)

~ ~ ~ j × G ∆+ Z j ∆+ Z j ∆− G ∆+ Z j ∆+ Z j Z h h h h h. h h  





(3.5.11)

Landau–Lifshitz Equations

116

We claim J X

~ ~ ~ j × G ∆+ Z j ∆+ Z j Z h h !

∆−

!

~ j ∆+ Z ~j ∆+ Z G h h !

∆−

h

j=1

!

h

h = 0.

(3.5.12)

In fact, since J X

~ ~ ~ j × G ∆+ Z j ∆+ Z j Z h h !

∆−

!

∆−

~ j ∆+ Z ~j ∆+ Z G h h !

h

j=1

h

~j ~ j−1 ∆+ Z ~ j−1 ∆− ∆− Z ∆+ Z   = ×G h h h j=1 =

J X



J X



j=1

= 0,





!

~ j−1 ∆+ Z h

!

h

~ j ∆+ Z ~j ∆+ Z G h h !

h

~j ∆+ Z ∆ ! G − ~ j−1 ~ j−1 ∆+ Z h ∆+ Z  ×G h

!

h

!

~j ∆+ Z h

h

h !

h

~ j = ∆+ Z ~ j−1 , the claim is where we have used ~a × ~a = 0, (~a × ~b) · ~b = 0 and ∆− Z proved. Therefore we have

2



4



2

J ~ j ~ j ~ j X X ∆+ Z X ∆+ Z ∆+ ∆− Z 1 d J−1 d J−1 2 h h+ B h+A A 2 dt j=0 h 2 dt j=0 h h j=1

2 2 2 J J ~ ~ ~ ~ ~ X X ∆+ Z j ∆+ ∆− Z j ∆+ Zj ∆+ ∆− Zj ∆+ Zj−1 + AB G( ) h h+B h h2 h h2 h j=1 j=1 2 2 J ~ ~ ~ ~ ~ X ∆ Z ∆ Z ∆ ∆ Z ∆ Z ∆ ∆ Z + j−1 + j + − j + + j−1 + − j h ≤ 0. + B2 (3.5.13) h h h2 h h2 j=1

This inequality combined with (3.5.10) leads to

 d  ~ 2 ~ h k2 + kδ Z ~ h k4 +kZ ~ h k2 +kδ Z ~ h k2 +kδ Z ~ h k4 +kδ 2 Z ~ h k2 ≤ 0. kZh k2 + kδ Z 2 4 2 2 4 2 dt

(3.5.14)

Inequality (3.5.14) combined with Gronwall inequality implies that there exist constants T0 , C > 0 independent of h such that ~ h (t)k2 + kδ Z ~ h (t)k2 + kδ Z ~ h (t)k4 ≤ C, kZ 4 Z

T0 0

~ h (t)k2 ≤ C. kδ 2 Z 2

∀ t ∈ [0, T0 ],

One-dimensional Landau–Lifshitz equations

117

The lemma is proved. Corollary 3.5.1 Under the conditions in Lemma 3.5.1, we have, for some constant C independent of h, ~ j | ≤ C. sup |Z (3.5.15) 0≤t≤T0 ;1≤j≤J

Sending h → 0, we obtain the local existence of the solution to (3.5.3)–(3.5.4). For the global solution to the viscosity problem, we prove ~ 0 (x) ∈ W 1,4 (Ω). Then (3.5.3)–(3.5.4) admits a local Theorem 3.5.1 Let ε = 1, Z ~ t) in [0, T0 ] in the space solution Z(x, ~ t) ∈ L∞ (0, T0 ; W 1,4 (Ω)) Z(x, and the following estimates hold sup 0≤t≤T0 ;x∈Ω

\

L2 (0, T0 ; H 2 (Ω))

~ ≤ C, |Z|

~ ~ ~ 4 kZ(t)k 2 + kZx (t)k2 + kZx k4 ≤ C, Z

T0

0

(3.5.16)

(3.5.17) ∀ t ∈ [0, T0 ],

~ xx (t)k22 ≤ C. kZ

(3.5.18) (3.5.19)

In order to prove the global existence of weak solution to (3.5.3)-(3.5.4), we need the following a priori estimates for the solution of (3.5.3)-(3.5.4). ~ 0 (x) ∈ W 1,4 (Ω), T > 0 and Z(x, ~ t) ∈ L∞ (0, T ; W 1,4 (Ω)) ∩ Lemma 3.5.2 Let ε = 1, Z L2 (0, T ; H 2 (Ω)) is a solution of (3.5.3)–(3.5.4). Then the following estimates hold sup 0≤t≤T ;1≤j≤J

~ ≤ C, |Z|

~ ~ ~ 4 kZ(t)k 2 + kZx (t)k2 + kZx k4 ≤ C, Z

T

0

(3.5.20) ∀ t ∈ [0, T ],

~ xx (t)k2 ≤ C. kZ 2

(3.5.21) (3.5.22)

~ t) and integrating it over Ω, we get Proof. Multiplying (3.5.3) by Z(x, 1d ~ ~ x k22 + BkZ ~ x k44 = 0. kZ(·, t)k22 + AkZ 2 dt

(3.5.23)

~ x )Z ~ x , one has Differentiating (3.5.3) with respect to x and then testing it by G(Z ~ x )Z ~xZ ~ xt = (G(Z ~ x )Z ~ x )xx G(Z ~ x )Z ~ x + (G(Z ~ x )Z ~ ×Z ~ x )xx G(Z ~ x )Z ~x. G(Z Integrating this equation by parts, we have Z Ad ~ 2 B d ~ 4 Z 2 ~ ~ ~ x )Z ~ ×Z ~ x )x (G(Z ~ x )Z ~ x )x k Zx k 2 + kZx k4 + |(G(Zx )Zx )x | = − (G(Z 2 dt 4 dt Ω Ω

Landau–Lifshitz Equations

118

which implies 1 d ~ 2 1 d ~ 4 Z ~ x )Z ~ x )x |2 = 0. A kZx k2 + B kZx k4 + |(G(Z 2 dt 4 dt Ω

(3.5.24)

Since ~ x )Z ~ x )x |2 = |G(Z ~ x )Z ~ xx + 2B(Z ~x · Z ~ xx )Z ~ x |2 |(G(Z ~ x )|Z ~ xx |2 + 4B 2 |(Z ~x · Z ~ xx )Z ~ x |2 + 4BG(Z ~ x )|Z ~x · Z ~ xx |2 = G 2 (Z ~ xx |2 , ≥ A 2 |Z

~ 0 kW 1,4 (Ω) , it follows from (3.5.24) that where C depends only on kZ Z 1 d ~ 2 1 d ~ 4 ~ xx |2 ≤ 0. A k Z x k 2 + B k Z x k 4 + A 2 |Z 2 dt 4 dt Ω

Putting this inequality and (3.5.23) together, we get from the Gronwall’s inequality that (3.5.21) and (3.5.22) hold. (3.5.20) can be derived from these inequalities. The lemma is proved. Now, we can use the extension method to give ~ 0 ∈ W 1,4 (Ω). Then problem (3.5.7)–(3.5.8) Theorem 3.5.2 Let ε > 0 be fixed and Z ~ ε (x, t) in the space admits a global solution Z ~ ε (x, t) ∈ L∞ (0, ∞; W 1,4(Ω)) Z and the following estimates hold sup 0≤t<∞;x∈Ω

\

L2 (0, ∞; H 2 (Ω))

~ ε (x, t)| ≤ C1 , |Z

~ ε (t)k2 + kZ ~ εx (t)k2 + kZ ~ εx (t)k4 ≤ C1 , kZ 4 ~ εt kL4/3 (0,∞;W −1,4/3 (Ω)) ≤ C1 , kZ Z

∞

0

(3.5.25)

(3.5.26) ∀ t ∈ [0, ∞),

~ εxx (t)k2 ≤ Cε , kZ 2

(3.5.27) (3.5.28) (3.5.29)

~ 0 kW 1,4 (Ω) . where C1 depends only on kZ

3.5.2

Measure-Valued Solution to the Strongly Degenerate Equations

The a priori estimates we have obtained are not enough for us to obtain the weak solution for (3.5.1)–(3.5.2), we apply the notion of measure-valued solution. Denote M = C1 where C1 is given in Theorem 3.5.2 which depends only on ~ ~ε, Z ~ εx ), τ (ξ) = B|(ξ4 , ξ5 , ξ6 )|2 (ξ1 , ξ2 , ξ3 ) × (ξ4 , ξ5 , ξ6 ) : (R3 ∩ kZ0 kW 1,4 (Ω) . Let Z ε = (Z 3 {|(ξ1 , ξ2 , ξ3 )| ≤ M }) × R → R3 . Then Z ε is uniformly bounded in L4 (Q)6 where Q = Ω × (0, T ) ∈ R2 and |τ (ξ)| ≤ CM (1 + |ξ|)3 ,

∀ ξ ∈ (R3 ∩ {|(ξ1 , ξ2 , ξ3 )| ≤ M }) × R3 .

One-dimensional Landau–Lifshitz equations

119

Lemma 3.5.3 Let Q ⊂ R2 be a bounded open set. Let Z εn be uniformly bounded in L4 (Q)6 . Then there exists a subsequence, still denoted by Z εn , and a measure-valued function ν such that for all τ : (R3 ∩ {|(ξ1 , ξ2 , ξ3 )| ≤ M }) × R3 → R3 satisfying for some q > 0 the growth condition |τ (ξ)| ≤ C(1 + |ξ|)3 , ∀ ξ ∈ (R3 ∩ {|(ξ1 , ξ2 , ξ3 )| ≤ M }) × R3 , we have τ (Z εn ) * τ weakly in Lr (Q), τ (y) = hνy , τ i a.e. y ∈ Q, provided that

4 1
Proof. It suffices to verify the condition (2.7) of Theorem 4.2.1 of [34, Chap. 4]. Taking the Young function Ψ(u) = ur , we have Z

εn

Q

Ψ(τ |Z |) =

Z

εn r

Q

|Z | ≤ C

r

Z

Q

(1 + |Z εn |)3r

and the last term is uniformly bounded (with respect to n) if 3r ≤ 4. The lower bound r > 1 follows from the properties of Orlicz functions, namely from lims→∞ Ψ(s)/s = ∞. Definition 3.5.1 (3.5.2) if

~ ν) is called a measure-valued solution of (3.5.1)– A pair (Z,

~ ∈ L∞ (0, ∞; W 1,4(Ω)), Z

ν∈

6 L∞ ω (Q; Prob(R )),

(3.5.30) (3.5.31)

∞ and if for any ϕ ∈ D(−∞, T ; Cper (Ω)) there holds

Z

Q

~0ϕ = Z

Z

Q

~ t−A Zϕ

Z

Q

~ ×Z ~x − ϕx Z

Z

Q

ϕx

Z

R6

τ (λ)dνt,x (λ)dxdt

(3.5.32)

where τ is defined as above, Q = Ω × (0, T ). ~ 0 ∈ W 1,4 (Ω). Then problem (3.5.1)–(3.5.2) admits a measureTheorem 3.5.3 Let Z valued solution. Proof. It follows from Lemma 3.5.3 that there exists a subsequence Z εn and a measure-valued function ν such that 4 τ (Z εn ) * τ weakly in Lr (Q), 1 < r ≤ , 3 τ (x, t) = hνt,x , τ i a.e. (x, t) ∈ Q.

(3.5.33) (3.5.34)

Landau–Lifshitz Equations

120

~ εn , that To complete the proof, we only need to prove, for some subsequence Z Z Z

Z

Q

~ εn t ϕ → − Z

Q

~ εn × Z ~ εn x )ϕx → (Z Z

Zxi =

R6

Q

Z

~tϕ + Z Z

Q

~ 0 ϕ(x, 0), Z

(3.5.35)

~ ×Z ~ x )ϕx , (Z

(3.5.36)

Ω

λi+3 dνt,x (λ), a.e. (x, t) ∈ Q, i = 1, 2, 3.

(3.5.37)

~ εn that In view of (3.5.27), we have for some subsequence Z ~ εn * Z ~ weakly in Lr (0, T ; L2 (Ω)), ∀ r > 1, Z ~ εn x * Z ~ x weakly in Lr (0, T ; L4 (Ω)), ∀ r > 1. Z

(3.5.38) (3.5.39)

∞ (Ω)) to give To prove (3.5.35), we take ϕ ∈ D(−∞, T ; Cper

Z

T 0

Z

Ω

~ εn t ϕ = − Z

−→ −

Z

T 0

Z

Ω

Z

T 0

Z

Ω

~ ε n ϕt + Z

~ t+ Zϕ

Z

Ω

Z

Ω

~ 0 (x)ϕ(x, 0) Z

~ 0 (x)ϕ(x, 0), Z

this proves (3.5.35). Now we prove (3.5.36) and (3.5.37). ~ εn is uniformly bounded in the space Since Z 1,4 {v : v ∈ Lr (0, T ; Wper (Ω)), vt ∈ L4/3 (0, T ; (W 1,4 (Ω))∗ )}

(3.5.40)

∗

1,4 for any r > 1 and since Wper (Ω) ,→,→ Lr (Ω) ,→ (W 1,4 (Ω)) , applying Aubin–Lions ∗ 1,4 Lemma [8, Lemma 1.2.48] with X0 = Wper (Ω), X = Lr (Ω), X1 = (W 1,4 (Ω)) , α = r (r > 1), β = 4/3, we know that the space defined in (3.5.40) is compactly imbedded into Lr (0, T ; Lr (Ω)), that is

~ εn → Z ~ strongly in Lr (0, T ; Lr (Ω)). Z Since Z  Q

~ εn × Z ~ εn x )ϕx − (Z ~ ×Z ~ x )ϕx (Z =

Z

Q

~ εn − Z) ~ ×Z ~ εn x )ϕx + ((Z

= I1 + I2 ,



Z

Q

~ × (Z ~ εn x − Z ~ x ))ϕx (Z

it follows from (3.5.26) and (3.5.39) that I2 → 0 as n → ∞,

(3.5.41)

One-dimensional Landau–Lifshitz equations

121

and it follows from (3.5.27) and (3.5.41) that |I1 | ≤

Z

T 0

Z

Z

≤C

Ω T

0

~ εn − Z| ~ |Z Z

Ω

4

!1/4

~ εn − Z| ~ 4 |Z

Z

T 0

Z

Ω

|ϕx |

4

!1/4

Z

T 0

Z

Ω

~ εn x | |Z

2

!1/2

!1/4

→ 0 as n → ∞. The proof of (3.5.36) is complete. Since Lemma 3.5.3 is true for all τ , if we let τ = Id, then for r = 4, q = 1, ~ εn , Z ~ εn x ) that ∀ ψ ∈ L4/3 (Q), we have for Z εn = (Z Z

εn

Q

Z ψdxdt →

Z

ψ Q

Z

R6

λdνt,x (λ)dxdt.

~ εn → Z ~ strongly in Lr (Q) and Z ~ εn x * Z ~ x in L4 (Q), we know However, Z Zxi

=

Z

R6

λi+3 νt,x (λ), a.e. (x, t) ∈ Q, i = 1, 2, 3.

This verifies (3.5.37). Corollary 3.5.2 Since the former estimates are independent of D, we get by letting ~ t) : D → ∞ that the Cauchy problem of (3.5.1)–(3.5.2) admits a solution Z(x, ~ t) ∈ L∞ (R+ ; W 1,4 (R1 )) Z(x,

3.6

\

L2 (R+ ; H 2 (R1 )).

Bibliography Comments

In this chapter we have introduced the mathematical theory of the global solutions to the equations of ferromagnetic spin chain with or without Gilbert damping term in one dimension. Applying Leray–Schauder principle, we proved the global existence of weak solution to one-dimensional ferromagnetic spin chain equation. By the differential-difference method we get the existence of global weak solution to the one-dimensional nonlinear initial boundary value problem for the ferromagnetic spin chain. To get the existence and uniqueness of smooth solution, Zhou, Guo and Tan [158] created a particular mobile frame on S 2 , this method was first used by them and enabled one to get the existence and uniqueness of the smooth solution by the special estimates for the higher derivatives, which is a challenging problem for a long time. Applying the similar method, Ding, Guo and Su [51] obtained the smooth solution for the system of inhomogeneous Heisenberg chain. However, the frame on S 2 was not used for the inhomogeneous problem in [51] until the work by Lin and Ding [108] in 2006. Introducing the notion of measure-valued solution, Ding, Guo and Su [50] discussed the compressible Heisenberg chain equations.

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Chapter 4 Landau–Lifshitz Equations and Harmonic Maps In Chapter 3, we have proved some global existence and uniqueness results for the one-dimensional Landau–Lifshitz equations. It is natural to ask how about the higherdimensional problems. We will find, in general, that there exist global weak solutions for the initialboundary value problems. However, in higher dimensions, there exists global smooth solution to the Landau Lifshitz equations with the Gilbert damping term only for small initial data. If the initial data is large, one can only prove the short time existence of smooth solution. Generally speaking, higher-dimensional Landau–Lifshitz equations with large initial data may develop singularities. In this aspect, the Landau–Lifshitz equations behave like the heat flow of harmonic maps.

4.1 4.1.1

Weak Solution to Multidimensional Ferromagnetic Spin Chain Equations Existence of Weak Solution to Multidimensional Ferromagnetic Spin Chain Equations

1. Initial boundary value problem for multidimensional ferromagnetic spin chain equations Let Ω ∈ Rm be a bounded domain with boundary ∂Ω ∈ C 2 . We study the ferromagnetic spin chain system with m variables of the form ~t = Z ~ × ∆Z ~ + f~(x, t, Z), ~ Z

(4.1.1)

~ t) = (u(x, t), v(x, t), w(x, t)) is a three-dimensional unknown and f~(x, t) is where Z(x, ~ ∈ R3 , ∆ = Pm ∂ 22 . a known three dimensional vector of variables x ∈ Rm , t ∈ R+ , Z i=1 ∂x i In the cylinder QT = {x ∈ Ω, 0 ≤ t ≤ T }, we give the homogeneous boundary 123

Landau–Lifshitz Equations

124

condition of the form ~ t) = 0, Z(x,

x ∈ ∂Ω,

0 ≤ t ≤ T,

(4.1.2)

and initial condition ~ 0) = ϕ(x), Z(x,

x ∈ Ω,

(4.1.3)

where ϕ(x) is a given vector. We also consider the spin system: ~ t = ε∆Z ~ +Z ~ × ∆Z ~ + f~(x, t, Z). ~ Z

(4.1.4)

We establish the existence of a global weak solution to initial boundary value problem (4.1.1)–(4.1.3) and problem (4.1.4), (4.1.2) and (4.1.3) by the Galerkin method. 2. Approximate solution to the initial boundary value problem of spin systems In order to prove the existence of global weak solution to the spin system, we use the Galerkin method. To this aim we construct the Galerkin approximate solution. Let {wn }∞ n=1 be the normal eigenfunctions of the equation ∆u + λu = 0, u|∂Ω = 0, corresponding to eigenvalue {λn }∞ n=1 . It is well known that {wn } forms a normal 1 orthogonal basis of H0 (Ω). We denote the approximate solution of (4.1.4), (4.1.2), and (4.1.3) by ~ N (x, t) = Z

N X

αnN (t)wn (x).

(4.1.5)

n=1

According to the Galerkin method, coefficients αnN solve the first order ordinary differential system as follows: Z

Ω

~ N t (x, t)ws (x)dx = Z

Z

Ω

~ N (x, t) × ∆Z ~ N (x, t))ws (x)dx (Z

+ε

Z

Ω

~ N (x, t)ws (x)dx + ∆Z

Z

Ω

~ N )ws (x)dx, f~(x, t, Z s = 1, 2, . . . , N

(4.1.6)

with the initial condition Z

Ω

~ N (x, 0)ws (x)dx = Z

It is clear that

Z

Ω

ϕ(x)ws (x)dx, s = 1, 2, . . . , N.

R 0 ~  Ω ZN t (x, t)ws (x)dx = αsN (t), R

~ Ω ZN (x, 0)ws (x)dx = αsN (0),

(4.1.7)

(4.1.8)

Landau–Lifshitz Equations and Harmonic Maps

and ϕs =

Z

ϕ(x)ws (x)dx,

Ω

125

s = 1, 2, . . . , N

are the coefficients of expansion ϕ(x) = N s=1 ϕs ws (x). 3. Estimates for the approximate solutions In order to prove the existence of a global weak solution, we assume the following: ~ is semi-bounded, i.e., there is a constant (1) The 3 × 3 Jacobi matrix f~Z~ (x, t, Z) 3 b > 0 such that for any ξ ∈ R P

ξ · f~Z~ · ξ ≤ b|ξ|2

(4.1.9)

holds, where “ · ” denotes the three-dimensional inner product. (2) (4.1.1) and (4.1.4) are homogeneous, that is, f~(x, t, 0) ≡ 0. Moreover, it is assumed that for some constants A and B:  ~ ≤ A|Z| ~ l + B, 2 ≤ l ≤ 2 +  |f~(x, t, Z)|

4 , m−2

 |∇ f~(x, t, Z)| ~ ≤ A|Z| ~ 1+ m2 + B. x

m ≥ 2,

(4.1.10)

(3) Vector ϕ(x) ∈ H01 (Ω).

Lemma 4.1.1 Let condition (1) hold , ϕ(x) ∈ L2 (Ω) and f~(x, t, 0) ∈ L2 (QT ). Then ~ N (x, t) meets the estimate the approximate solution Z ~ N (·, t)kL2 (Ω) ≤ K1 , sup kZ

(4.1.11)

0≤t≤T

where K1 is independent of ε and N . Proof. Multiplying (4.1.6) by αnN (t) and summing from s = 1 to s = N , we have Z

Ω

~N t · Z ~ N dx = Z

Z

Ω

~ N × ∆Z ~N ) · Z ~ N dx (Z

+ε

Z

Ω

~N · Z ~ N dx + ∆Z

Z

Ω

~N ) · Z ~ N dx, f~(x, t, Z

(4.1.12)

or simply write as Z

Ω

~N t · Z ~ N dx = Z

Z

Ω

~ N × ∆Z ~N ) · Z ~ N dx (Z

+ε

Z

Ω

~N · Z ~ N dx + ∆Z

Z

Ω

~N ) · Z ~ N dx. f~(Z

Z

~ N ∗ ∇Z ~N , ∇Z

We have ~N t · Z ~ N dx = − 1 d kZ ~ N (·, t)k2 2 , Z L (Ω) 2 dt Ω Z ~ N × ∆Z ~N ) · Z ~ N dx = 0, (Z

Z

Ω

Z

Ω

~N · Z ~N = ∆Z

Z

∂Ω

~N · Z ~ N ) ∗ νm − (∇Z

Ω

(4.1.13)

Landau–Lifshitz Equations

126

~ N is an m × 3 tensor and ∗ is the where νm is the unit outer normal vector to ∂Ω, ∇Z inner product in an m-dimensional space. It follows from the homogeneous condition that Z ~N · Z ~ N = −k∇Z ~ N k2 2 . ∆Z L (Ω) Ω

For the last term on the right-hand side of (4.1.13), we have Z

Ω

~N ) · Z ~N = f~(Z

Z

1 0

+





~ N (x, t)) ∂ f~(x, t, τ Z ~ N (x, t) · Z ~ N (x, t)dx ·Z dτ  ~ Ω ∂Z

Z

Z

Ω

~ N (x, t)dx, f~(x, t, 0) · Z

and hence !

~N ) · Z ~ N ≤ b + 1 kZ ~ N (·, t)k2 2 f~(Z L (Ω) 2 Ω 1 + kf~(·, t, 0)k2L2 (Ω) . 2 Therefore it follows from (4.1.13) that Z

d ~ ~ N (·, t)k2 2 kZN (·, t)k2L2 (Ω) + 2εk∇Z L (Ω) dt ~ N (·, t)k2 2 + kf~(·, t, 0)k2 2 . ≤ (2b + 1)kZ L (Ω) L (Ω)

(4.1.14)

This proves the lemma. ~ N (x, t), Lemma 4.1.2 Let conditions (1)–(3) hold. For the approximate solution Z ~ N kL2 (Ω) ≤ K2 sup k∇Z

(4.1.15)

0≤t≤T

holds, where K2 is independent of ε and N . Proof. Multiplying (4.1.6) by −λs αsN (t) and summing from s = 1 to s = N , we have Z

Ω

~ N t · ∆Z ~ N dx = Z

Z

Ω

~ N × ∆Z ~ N ) · ∆Z ~ N dx (Z

+ε

Z

Ω

~ N · ∆Z ~ N dx + ∆Z

Z

Ω

~ N ) · ∆Z ~ N dx. f~(Z

(4.1.16)

The first term in the above equality is Z

~ N t · ∆Z ~ N = − 1 d k∇Z ~ N (·, t)k2 2 . Z L (Ω) 2 dt Ω

(4.1.17)

The last term on the right-hand side of (4.1.16) is Z

Ω

~ N ) · ∆Z ~N = f~(Z

Z

∂Ω

=−

Z

~ N ) ∗ ∇Z ~ N ) ∗ νm − (f~(Z Ω

~N − ∇x f~ ∗ ∇Z

Z

Ω

Z

Ω

~ N ) ∗ ∆Z ~N D f~(Z

∂ f~ ~ N ∗ ∇Z ~N . · ∇Z ~N ∂Z !

(4.1.18)

Landau–Lifshitz Equations and Harmonic Maps

Since

∂ f~ ~ ∂Z

127

is semi-bounded, one has Z

Ω

∂ f~ ~ N ≤ bk∇Z ~ N (·, t)k2 2 . ~ N ∗ ∇Z · ∇Z L (Ω) ~ ∂ ZN !

The first term on the right of (4.1.18) can be estimated by Z 1 1 2 2 ~ ~ ~ ~ ~ ∇ f ∗ ∇ Z N ≤ k∇ZN (·, t)kL2 (Ω) + k∇f (·, t, ZN (·, t))kL2 (Ω) . Ω 2 2

It follows from (4.1.10) that

~ N (·, t))k2L2 (Ω) ≤ C1 k∇f~(·, t, Z

Z

4

Ω

~ N (x, t)|2+ m dx + C2 , |Z

where C1 , C2 are constants. From the Sobolev inequality, we have Z

Ω

2 ~ N (x, t)|2+ m4 dx ≤ C3 kZ ~ N (·, t)k4/m ~ |Z L2 (Ω) k∇ZN (·, t))kL2 (Ω) 4

~ N (·, t))k2+2 m , + C 4 kZ L (Ω) where C3 , C4 are constants. Finally, we have from (4.1.16) that d ~ N (·, t)k2L2 (Ω) + 2εk∆Z ~ N (·, t)k2L2 (Ω) k∇Z dt ~ N (·, t)k2 2 + C6 , ≤ (2b + C5 )k∇Z L (Ω) ~ N (·, t)kL2 (Ω) but indepenwhere C5 , C6 are constants depending only on sup0≤t≤T kZ dent of N. Applying the Gronwall inequality to the last inequality, we can finish the proof of this lemma. Lemma 4.1.3 Let (1)–(3) hold. For above approximate solutions, ~ N t (·, t)k −(2+[ m ]) ≤ K3 sup kZ 2 (Ω) H

0≤t≤T

holds with K3 independent of ε and N. 2+[ m ] 2

Proof. Let ν(x) ∈ H0

(Ω) be a test function. We have ν(x) = νN (x) + ν N (x), νN (x) = ν N (x) =

N X

βn wn (x),

n=1 ∞ X

n=N +1

βn wn (x),

(4.1.19)

Landau–Lifshitz Equations

128

where βn =

R

Ω

ν(x)wn (x)dx, n = 1, 2, . . . . If s ≥ N + 1, then Z

Ω

~ N t ws = 0. Z

If 1 ≤ s ≤ N , one has Z

Ω

~ N t ws = Z

Z

∂Ω

~ N × ∇Z ~ N ) ∗ νm ]ws − [(Z

+ε +

Z

Z

∂Ω

Ω

~ N ∗ νm )ws − ε (Z

Z

Ω

Z

Ω

~ N × ∇Z ~ N ) ∗ ∇ws (Z

~ N ∗ ∇ws ∇Z

~ N (x, t))ws (x)dx. f~(x, t, Z

(4.1.20)

We estimate every term on the right-hand side of (4.1.20). The first and the third term are equal to zero. For the second term, we have Z ~ ~ ~ N (·, t)kL2 (Ω) k∇Z ~ N (·, t)kL2 (Ω) . (ZN × ∇ZN ) ∗ ∇ws ≤ C7 k∇ws kL∞ (Ω) kZ Ω

It follows from the Sobolev inequality that

k∇ws kL∞ (Ω) ≤ C8 kws kH 2+[ m2 ] (Ω) . Therefore, we have Z ~ N × ∇Z ~ N ) ∗ ∇ws ≤ C9 kws k 2+[ m ] , (Z 2 (Ω) H Ω

~ N (·, t)kH 1 (Ω) but is independent of N . where C9 depends only on sup0≤t≤T kZ For the fourth term on the right-hand side of (4.1.20), we have Z ~ N ∗ ∇ws ≤ C10 k∇Z ~ N (·, t)kL2 (Ω) k∇ws kL2 (Ω) . ∇ Z Ω

For the last term on the right-hand side of (4.1.20), we have Z ~ ~ f (x, t, Z N (x, t))ws (x)dx Ω Z

≤ kws kL∞ (Ω)

Ω

~ N (x, t))|dx |f~(x, t, Z

≤ kws kL∞ (Ω) A

Z

Ω

~ N (x, t)|l dx + B |Z

!

≤ C11 kws kL∞ (Ω) ≤ C12 kws kH 2+[ m2 ] (Ω) , where we have used Z

α=

m 2

−

m , l

Ω

αl ~ N (x, t)|l dx ≤ C13 kZ ~ N (·, t)k(1−α)l ~ |Z L2 (Ω) kZN (·, t)kH 1 (Ω) ,

2≤l≤

2m . m−2

Landau–Lifshitz Equations and Harmonic Maps

129

In summary we have Z ~ N t ws ≤ C14 kws k 2+[ m ] , Z 2 (Ω) H Ω

~ N (·, t)kH 1 (Ω) but is independent of ε, where constant C14 depends on sup0≤t≤T kZ 0 ≤ t ≤ T , and N . Similarly, we also have

and hence

Z ~ ZN t νN ≤ C14 kνN kH 2+[ m2 ] (Ω) , Ω

Z 2+[ m ](Ω) ~ . ZN t ν ≤ C14 kνkH 2+[ m2 ] (Ω) , ∀ ν ∈ H0 2 Ω

The proof is complete. 4. Existence of global weak solution Applying the fixed point theorem and the a priori estimates on the approximate solutions, we can prove the existence of global solution αsN (t) of system (4.1.6) and (4.1.7) (0 ≤ t ≤ T , s = 1, 2, . . . , N). Lemma 4.1.4 Let (1)–(3) hold. Then the ordinary differential system (4.1.6) and (4.1.7) admits at least one global continuously differentiable solution α sN (t) (0 ≤ t ≤ T , s = 1, 2, . . . , N). Lemma 4.1.5 Let (1)–(3) hold. Then for the approximate solution, 1

~ N (·, t + ∆t) − Z ~ N (·, t)kL2 (Ω) ≤ K4 ∆t 3+[ m2 ] kZ

(4.1.21)

holds, where K4 is independent of ε, N, and 0 ≤ t, t + ∆t ≤ T . Proof. It follows from the Sobolev inequality with negative index that ~ N (·, t)kL2 (Ω) ≤ C15 kZ ~ N (·, t)k kZ

1 3+[ m ] 2 m H −(2+[ 2 ]) (Ω)

~ N (·, t)k kZ

2+[ m 2 ] 3+[ m ] 2 H 1 (Ω)

.

~ N (x, t + ∆t) − Z ~ N (x, t), we get Applying this inequality to Z ~ N (·, t + ∆t) − Z ~ N (·, t)kL2 (Ω) kZ

~ N (·, t + ∆t) − Z ~ N (·, t)k ≤ C15 kZ

1 3+[ m ] 2 m H −(2+[ 2 ]) (Ω)

~ N (·, t + ∆t) − Z ~ N (·, t)k · kZ

where 0 ≤ t, t + ∆t ≤ T . Thus, ~ N (·, t + ∆t) − Z ~ N (·, t)kL2 (Ω) kZ ≤ C15 ∆t

1 3+[ m ] 2

~ N (·, t)k sup kZ

0≤t≤T

holds. This proves the lemma.

1 3+[ m 2 ]

H −(2+[

m ]) 2 (Ω)

~ N (·, t)k · sup kZ 0≤t≤T

2+[ m ] 2 3+[ m 2 ] 1 H (Ω)

2+[ m 2 ] 3+[ m ] 2 H 1 (Ω)

,

Landau–Lifshitz Equations

130

4.1.2

Weak Solution to Multidimensional System of Ferromagnetic Spin Chain

1. Convergence of approximate solutions In order to get a weak solution to system (4.1.4) or (4.1.1) with homogeneous boundary condition (4.1.2) and initial condition (4.1.3), we allow N → ∞. ~ t) ∈ L∞ (0, T ; H 1(Ω)) is called Definition 4.1.1 A vector-valued function Z(x, a weak solution to system (4.1.4) (ε > 0) or (4.1.1) (ε = 0) with homogeneous boundary condition (4.1.2) and initial condition (4.1.3), if for any test function ν(x, t) ∈ C 1 (QT ): ν(x, t) = 0 for x ∈ Ω and t = T and x ∈ ∂Ω and 0 ≤ t ≤ T, then Z

T 0

Z

Ω

~ t) − ∇ν(x, t) ∗ (Z(x, ~ t) × ∇Z(x, ~ t)) [νt (x, t)Z(x,

~ t) + ν(x, t)f~(x, t, Z(x, ~ t))]dxdt − ε∇ν(x, t) ∗ ∇Z(x, +

Z

Ω

ν(x, 0)ϕ(x)dx = 0, ε ≥ 0

(4.1.22)

~ t) satisfies the homogeneous boundary condition on manifold ∂Ω × holds and Z(x, [0, T ]. The Galerkin approximate function of ν(x, t) is νN (x, t) =

N X

βnN (t)wn (x),

n=1

which uniformly converges in C 1 (QT ) to ν(x, t) as N → ∞, where βsN (t) =

Z

Ω

ν(x, t)ws (x)dx, s = 1, 2, . . . , N.

It follows from (4.1.6) that the following integral equality holds: Z

T 0

Z

Ω

~ N (x, t) − ∇νN (x, t) ∗ (Z ~ N (x, t) × ∇Z ~ N (x, t)) [νN t (x, t)Z

~ N (x, t) + νN (x, t)f~(x, t, Z ~ N (x, t))]dxdt − ε∇νN (x, t) ∗ ∇Z +

Z

Ω

νN (x, 0)ϕN (x)dx = 0, ε ≥ 0.

(4.1.23)

2. Estimates of uniform boundedness and convergence ~ N (x, t) is uniformly bounded in the space: It follows from former estimates that Z G = L∞ (0, T ; H01 (Ω))

\

m

(1) W∞ (0, T ; H −(2+[ 2 ]) (Ω)).

~ N (x, t) from Z ~ N (x, t) such that Z ~ N (x, t) conThen we may choose a subsequence Z i i p 2 p 1 ~ verges in L (0, T ; L (Ω)) to a vector Z(x, t) ∈ L (0, T ; H (Ω)) (1 < p < ∞),

Landau–Lifshitz Equations and Harmonic Maps

131

~ N (x, t) converges in Lp (0, T ; L2 (Ω)) to ∇Z(x, ~ t). On the other hand, since and ∇Z i −(2+[ m ]) p ~ 2 (Ω)) (1 < p < ∞), we may asZNi t (x, t) is uniformly bounded in L (0, T ; H ~ N t (x, t) weakly converges to a vector w(x, t) ∈ Lp (0, T ; H −(2+[ m2 ]) (Ω)) sume that Z i ~ t (x, t). In fact, we have (1 < p < ∞). Now we claim w(x, t) = Z Z

T

Z

0

Ω

~ N t ν(x, t)dxdt = − Z i

Z

T 0

Z

Ω

~ N νt (x, t)dxdt, Z i

where ν(x, t) is any test function with compact support set in QT . Let Ni → ∞, we get Z TZ Z TZ ~ t)νt (x, t)dxdt. w(x, t)ν(x, t)dxdt = − Z(x, 0

Ω

0

Ω

The claim follows. This means that

~ t) ∈ Gp = Lp (0, T ; H 1 (Ω)) Z(x,

\

m

Wp(1) (0, T ; H −(2+[ 2 ]) (Ω)), 1 < p < ∞,

~ t) in Gp is uniformly bounded in p. Then Z(x, ~ t) ∈ G∞ . We and the norm of Z(x, have ~ ε (x, t) of the Lemma 4.1.6 Suppose that (1)–(3) hold. Then for the limit function Z ~ N (x, t), approximate solution Z ~ ε (·, t)kH 1 (Ω) ≤ K5 , sup kZ

(4.1.24)

~ εt (·, t)k −(2+[ m ]) ≤ K6 , sup kZ 2 (Ω) H

(4.1.25)

0≤t≤T

0≤t≤T

1

~ ε (·, t + ∆t) − Z ~ ε (·, t)kL2 (Ω) ≤ K7 ∆t 3+[ m2 ] kZ

(4.1.26)

hold, where K5 , K6 , K7 are independent of ε and 0 ≤ t, t + ∆t ≤ T . ~ N (x, t)}, still denoted by {Z ~ N (x, t)}, such Choose a suitable subsequence from {Z i i ~ ~ ~ that for some vector-valued function Z(x, t) ZNi (x, t) → Z(x, t) holds strongly in ~ N (·, t)kL2 (Ω) → kZ(·, ~ t)kL2 (Ω) uniformly in 0 ≤ t ≤ L2 (Ω) for every 0 ≤ t ≤ T , and kZ i 1 ~ N (x, t) converges to Z(x, ~ t) in C (0, 3+[ m2 ]+δ ) (0, T ; L2 (Ω)), δ > 0. T . More precisely, Z i ~ t) satisfies the homogeneous boundary condition almost evIt is clear that Z(x, erywhere on ∂Ω × [0, T ]. Now we consider the limit of the integral (4.1.23) as Ni → ∞. Since as Ni → ∞, {νNi t (x, t)} and {∇νNi (x, t)} uniformly converge in QT to νt (x, t) and ∇ν(x, t), {νNi (x, 0)} uniformly converge in Ω to ν(x, 0); {ϕNi (x)} converge in L2 (Ω) to ϕ(x), for the second term of (4.1.23), we have Z Z T ~ ~ ~ ~ [∇νNi ∗ (ZNi × ∇ZNi ) − ∇ν ∗ (Z × ∇Z)] 0 Ω Z Z T ~ N × ∇Z ~ N ) ≤ ∇(νNi − ν) ∗ (Z i i 0 Ω

Landau–Lifshitz Equations

132

Z Z T ~ N − Z) ~ × ∇Z ~ N ) + ∇ν ∗ ((Z i i 0 Ω Z Z T ~ × ∇(Z ~ N − Z)) ~ + ∇ν ∗ (Z i 0 Ω

~ N kL2 (Q ) k∇Z ~ N kL2 (Q ) ≤ k∇(νNi − ν)kL∞ (QT ) kZ i i T T ~ N − Zk ~ L2 (Q ) k∇Z ~ N kL2 (Q ) + k∇νkL∞ (QT ) kZ i i T T Z Z T ~ ~ ~ + ∇ν ∗ (Z × ∇(ZNi − Z)) . 0 Ω

Every term of the above inequality tends to zero as Ni → ∞. ~ N (x, t)} strongly converges to Z(x, ~ t), {Z ~ N (x, t)} converges to Z(x, ~ t) Since {Z i i ~ N (x, t))} converges to f~(x, t, Z(x, ~ t)) a.e. on QT , too. a.e. on QT , and {f~(x, t, Z i Therefore it follows from (4.1.10) that ~ N (·, t))kq q kf~(x, t, Z i L (Ω) =

Z

Ω

~ N (x, t))|q dx |f~(x, t, Z i

≤ C17

Z

4

Ω

~ N (x, t)|2+ m−2 dx + C18 , |Z i

4 4 ~ N (x, t))} weakly )/l > 1 since l < 2 + m−2 . Hence {f~(x, t, Z where q = (2 + m−2 i ∞ q ~ ~ converges to f (x, t, Z(x, t)) in L (0, T ; L (Ω)) as Ni → ∞. Now we may send Ni → ∞ so that (4.1.23) tends to (4.1.22). This means that ~ t) is a weak solution of (4.1.4) (ε > 0) or (4.1.1) (ε = 0) and the limit vector Z(x, satisfies the homogeneous boundary condition (4.1.2) and initial condition (4.1.3). 3. The global weak solution with homogeneous boundary condition

Theorem 4.1.1 Let (1)–(3) hold. Then Eqs. (4.1.4) (ε > 0) and (4.1.1) (ε = 0) with conditions (4.1.2) and (4.1.3) admit at least one global weak solution ~ t) ∈ L∞ (0, T ; H01(Ω)) Z(x,

\

C

(0, 3+[1m ] ) 2

(0, T ; L2 (Ω)).

~ t) be any limit function of any converTheorem 4.1.2 Let (1)–(3) hold. Let Z(x, ~ gent sequence of weak solutions {Zε (x, t)} to the spin equation (4.1.4) (ε > 0) with (0,

1 m

)

conditions (4.1.2) and (4.1.3) in C 3+[ 2 ] (0, T ; L2 (Ω)) ∩ L∞ (0, T ; H01 (Ω)) as ε → 0. ~ t) is a weak solution of problem (4.1.1)–(4.1.3). Then Z(x, ~ ε (x, t) ∈ G such that Proof. For any ε > 0, there exists at least one Z Z

T 0

Z

Ω

~ ε (x, t) − ∇ν(x, t) ∗ (Z ~ ε (x, t) × ∇Z ~ ε (x, t)) [νt (x, t)Z

~ ε (x, t) + ν(x, t)f~(x, t, Z ~ ε (x, t))]dxdt − ε∇ν(x, t) ∗ ∇Z +

Z

ν(x, 0)ϕ(x)dx = 0, Ω

ε ≥ 0,

(4.1.27)

Landau–Lifshitz Equations and Harmonic Maps

133

where ν(x, t) is any test function. Since ε

Z

T 0

Z

Ω

~ ε (x, t)dxdt → 0, as ε → 0, ∇ν(x, t) ∗ Z

~ t) ∈ G meets limit Z(x, Z

T 0

Z

Ω

~ t) − ∇ν(x, t) ∗ (Z(x, ~ t) × ∇Z(x, ~ t)) + ν(x, t)f~(x, t, Z ~ ε (x, t))]dxdt [νt (x, t)Z(x,

+

Z

ν(x, 0)ϕ(x)dx = 0. Ω

That is, ~ t) ∈ L∞ (0, T ; H 1 (Ω)) Z(x, 0

\

C

(0, 3+[1m ] ) 2

(0, T ; L2 (Ω))

is a weak solution of problem (4.1.1)–(4.1.3). 4. Global weak solution in the cylinder with infinite length In former discussion, T may be arbitrary. We denote the cylinder with infinite length by Q∞ = {x ∈ Ω, 0 ≤ t ≤ ∞}. We see that the ordinary system (4.1.6) and (4.1.7) of coefficients αnN (t) admits at least one continuously differentiable solution αnN (t) which exists on [0, ∞] so that ~ N (x, t) exists on Q∞ too. the approximate solution Z ~ N ,i (x, t)} Let Tk be a number sequence such that Tk → ∞ as k → ∞. Let {Z k ~ N (x, t) such that (k = 1, 2, . . . ; i = 1, 2, . . .) be the kth subsequence of Z ~ N ,i (x, t)} ⊂ {Z ~ N ,i (x, t)}; (1) {Z k+1 k ~ t) defined on Q∞ such that the subse(2) there is a vector-valued function Z(x, ~ ~ t) in quence {ZNk ,i (x, t)} weakly converges to Z(x, G(Tk ) = L∞ (0, Tk ; H01 (Ω)) \

C

(0, 3+[1m ] ) 2

\

m

(1) W∞ (0, Tk ; H −(2+[ 2 ]) (Ω))

(0, Tk ; L2 (Ω)),

k = 1, 2, . . . .

~ N ,k (x, t) that, on Hence we may take the diagonal method to choose a subsequence Z k ~ t). Hence Z(x, ~ t) is in the space any cylinder QT , weakly converges in G to Z(x, 1 G∞ = L ∞ loc (0, ∞; H0 (Ω))

\

1 ⊂ L∞ loc (0, ∞; H0 (Ω))

and meets the integral relation (4.1.22).

m

1 Hloc (0, ∞; H −(2+[ 2 ]) (Ω))

\

C

(0, 3+[1m ] ) 2

(0, ∞; L2 (Ω)),

Theorem 4.1.3 Suppose ~ is semi-bounded , i.e. for (x, t) ∈ Q∞ and (i) the 3 × 3 Jacobi matrix f~Z~ (x, t, Z) ~ ∈ R3 , (4.1.9) holds. Z

Landau–Lifshitz Equations

134

~ ∈ R3 , (ii) for any 0 < T < ∞ and (x, t) ∈ QT and Z  ~ ≤ A(T )|Z| ~ l + B(T ),  |f~(x, t, Z)|  

~ ≤ A(T )|Z| ~ 1+ m2 + B(T ), |∇ f~(x, t, Z)|

x   ~ f (x, t, 0) ≡ 0

(4.1.28)

holds, where A(T ), B(T ) are positive constant depending on T, and 2≤l ≤2+

4 , m ≥ 2. m−2

(iii) Vector ϕ(x) ∈ H01 (Ω). Then both the spin equation (4.1.4) (ε > 0) and the ferromagnetic spin chain equation (4.1.1) (ε = 0) with conditions (4.1.2) and (4.1.3) have a global weak solution 1 ~ t) ∈ L∞ Z(x, loc (0, ∞; H0 (Ω))

If b < 0, it follows from (4.1.14) that

\

(0, 3+[1m ] )

Cloc

2

(0, ∞; L2 (Ω)).

d ~ ~ N (·, t)k2 2 , kZN (·, t)k2L2 (Ω) ≤ −2|b|kZ L (Ω) dt and if (4.1.4) and (4.1.1) are both homogeneous, i.e. f~(x, t, 0) ≡ 0, then ~ N (·, t)kL2 (Ω) ≤ kϕkL2 (Ω) e−|b|t , 0 ≤ t < ∞. kZ

~ t), that is, we We conclude that the similar inequality also holds for limit Z(x, have ~ t) to the spin Theorem 4.1.4 Let (i)–(iii) hold. Then for the weak solution Z(x, equation (4.1.4) (ε > 0) or the ferromagnetic spin chain equation (4.1.1) (ε = 0) in Q∞ with conditions (4.1.2) and (4.1.3), ~ t)kL2 (Ω) = 0 lim kZ(·,

t→∞

holds. 5. Uniqueness of smooth solution ~ is twice continuously differentiable with reTheorem 4.1.5 Assume that f~(x, t, Z) ~ ~ is semi-bounded. Then the classical spect to x, and Z, and the Jacobi matrix f~Z~ (x, t, Z) solution to problem (4.1.1)–(4.1.3) is unique. ~ t) be two solutions. Set w(x, t) = ~u (x, t) − Z(x, ~ t). We Proof. Let ~u (x, t), Z(x, have Z d 2 kw(·, t)kL2 (Ω) = 2 (∇w · (w × ~u )) ∗ νm dt ∂Ω Z Z ∂ f˜ · w, − 2 ∇w ∗ (w × ∇~u ) + 2 w · ~ Ω Ω ∂Z

Landau–Lifshitz Equations and Harmonic Maps

d k∇w(·, t)k2L2 (Ω) = 2 dt

Z

∂Ω

+2 −2

~ ∗ νm − 2 (∇w · (w × ∆Z))

Z

∂ f˜ · w ∗ νm − 2 ∇w · ~ ∂Z !

∂Ω

Z

Ω

Z

∇w ∗

Ω

Z

∂ 2 f˜ · w · ∇~u − 2 ~2 ∂Z !

135

~ ∇w ∗ (w × ∇∆Z)

Ω

∇w ∗ ∇

Z

∇w ∗

Ω

∂ f˜ · ∇w ~ ∂Z

~ ∂ f~(x, t, Z) · ∇w, ~ ∂Z

where ~ ∂ f˜ Z 1 ∂ f~(x, t, τ~u + (1 − τ )Z) = dτ , ~ ~ 0 ∂Z ∂Z Z 1 2~ ~ ∂ 2 f˜ ∂ f (x, t, τ~u + (1 − τ )Z) = dτ , ~2 ~2 0 ∂Z ∂Z Z 1 ~ ∂ f˜ ∂ f~(x, t, τ~u + (1 − τ )Z) = dτ. ∇ ∇ ~ ~ 0 ∂Z ∂Z Hence w(x, t) ∈ C (3,1) (QT ) satisfies the homogeneous equation and the homogeneous boundary and initial conditions. Thus d k∇w(·, t)k2H 1 (Ω) ≤ C18 kw(·, t)k2H 1 (Ω) . dt This proves the theorem. ~ t) of (4.1.1). Finally we consider the “blow-up” problem for the weak solution Z(x, Theorem 4.1.6 If ~ · f~(x, t., Z) ~ ≥ C0 |Z| ~ 2+δ , Z

~ ∈ R3 , (x, t) ∈ QT , Z

(4.1.29)

~ t) ∈ holds, where C0 > 0, δ > 0, and kϕ(x)kL2 (Ω) > 0, then the weak solution Z(x, (2,1) W2 (QT ) to (4.1.1) blows up at finite time, i.e. for a finite t0 > 0 ~ t)kLp (Ω) → ∞, 2 ≤ p < ∞. lim kZ(·,

t→t0 −0

~ p−2 Z ~ and integrating over [0, l] with Proof. we have by multiplying (4.1.1) by |Z| respect to x that Z 1d ~ ~ t)|p−2 Z(x, ~ t) · f~(x, t, Z))dx ~ kZ(·, t)kpLp (Ω) = (|Z(x, p dt Ω

≥ C0

Z

Ω

~ t)|p+δ dx ≥ C0 (mes Ω)− pδ kZ(·, ~ t)kp+δ |Z(x, Lp (Ω) ,

or δ d ~ ~ t)k1+δ kZ(·, t)kLp (Ω) ≥ C0 (mes Ω)− p kZ(·, Lp (Ω) , dt

Landau–Lifshitz Equations

136

and then δ

1

~ t)kLp (Ω) ≥ (kϕk−δp − C0 tδ(mes Ω)− p )− δ , 2 ≤ p < ∞. kZ(·, L (Ω) This proves the theorem. In the following we consider a more general blow-up result. For example, we consider a general system ut = ∇ ∗ A(x, t, u) · ∇u + f (x, t, u),

(4.1.30)

where u(x, t) = (u1 (x, t), . . . , un (x, t)) is an n-dimensional vector-valued unknown defined on cylinder QT = {x ∈ Ω ⊂ Rm , 0 ≤ t ≤ T } and · and ∗ denote the inner product in Rn and Rm , respectively. A(x, t, u) is a nonsingular and zero-definite matrix, and f (x, t, u) is an n-dimensional vector-valued function satisfying u · f (x, t, u) ≥ C0 |u|2+δ

(4.1.31)

for (x, t) ∈ QT and u ∈ Rn , C0 > 0, δ > 0. Multiplying (4.1.30) by u and integrating over Ω, one has 1d ku(·, t)k2L2 (Ω) = 2 dt

Z

∂Ω

−

(u · A · ∇u) ∗ νm

Z

Ω

∇u ∗ A · ∇u +

Z

Ω

u · f.

Consider the first initial boundary condition u(x, t) = 0,

x ∈ ∂Ω,

u(x, 0) = ϕ(x),

0 ≤ t ≤ T,

(4.1.32)

x ∈ Ω;

(4.1.33)

or the second boundary condition A(x, t, u(x, t)) · ∇u(x, t) ∗ νm = 0, u(x, 0) = ϕ(x),

x ∈ ∂Ω,

x ∈ Ω.

0 ≤ t ≤ T,

(4.1.34) (4.1.35)

We have Theorem 4.1.7 Under above condition on A, f, for solution u(x, t) to problem (4.1.30), (4.1.32) and (4.1.33), or problem (4.1.30), (4.1.34) and (4.1.35), we have ku(x, t)kL2 (Ω) → ∞, t → t1 − 0, where t1 > 0 is finite and kϕkL2 (Ω) > 0.

Landau–Lifshitz Equations and Harmonic Maps

4.2 4.2.1

137

Landau–Lifshitz Equations on Riemannian Manifold and Harmonic Maps Landau–Lifshitz Equations and Harmonic Maps

1. Landau–Lifshitz equations on Riemannian manifold Let M be an m-dimensional Riemannian manifold of metric γ without boundary, and S 2 be the surface of a unit sphere in R3 centered at x = 0. In local coordinate (x1 , x2 , . . . , xm ) on M , the Laplace–Beltrami operator and norm |Du(x)| are expressed by ! ! ∂2 1 ∂ ∂ ∂ αβ √ αβ k γ ∆M := √ γ α =γ − Γαβ k γ ∂xβ ∂x ∂xα ∂xβ ∂x and |Du(x)|2 =

XX

γ αβ

i

α,β

∂ui ∂ui , ∂xα ∂xβ

where (γ αβ ) is the inverse of (γαβ ). The Landau–Lifshitz equations with the Gilbert damping term on M is given by 1 ∂ ~u t = −α1 ~u × ~u × √ γ ∂xβ 1 ∂ + α2~u × √ γ ∂xβ

γ

γ

αβ √

αβ √

∂~u γ α ∂x

!!

!

∂~u γ α , ∂x

(4.2.1)

where α1 > 0 is the Gilbert constant and α2 is a constant too, with initial value ~u 0 (x) such that |~u 0 (x)|2 = 1, ∀ x ∈ M. (4.2.2) 2. Harmonic maps Lemma 4.2.1 (Gronwall Inequality) Let u, v, w be continuous functions for t ≥ 0 and suppose that w ≥ 0. Then the inequality u(t) ≤ v(t) +

Z

t

w(s)u(s)ds 0

implies the inequality u(t) ≤ v(t) +

Z

t

w(s)v(s)exp 0

Z

t



w(r)dr ds. 0

Now we can consider an equivalent form of (4.2.1) under condition (4.2.2) 1 ∂ ~u t = α1 √ γ ∂xβ

γ

αβ √

∂~u γ α ∂x

!

1 ∂ + α1 |∇~u | ~u + α2~u × √ γ ∂xβ 2

γ

αβ √

!

∂~u γ α , ∂x

(4.2.3)

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138

with initial value ~u 0 (x) such that |~u 0 (x)|2 = 1,

∀ x ∈ M.

(4.2.4)

The equivalence follows from Theorem 4.2.1 In the classical sense, ~u is a solution of (4.2.1) and (4.2.2) if and only if ~u is a solution of (4.2.3) and (4.2.4). Proof. Suppose that ~u is a solution of (4.2.1) and (4.2.2). By the vector product formula ~a × (~b × ~c ) = (~a · ~c )~b − (~a · ~b)~c, we have ~u × (~u × ∆M ~u ) = (~u · ∆M ~u ) − |~u |2 ∆M ~u . Multiplying (4.2.1) by ~u , we have ~u · ~u t = 0,

∀ (x, t) ∈ M × [0, ∞).

Then from (4.2.2) we get |~u (x, t)|2 = 1,

∀ (x, t) ∈ M × [0, ∞).

This implies ~u · D~u = 0,

~u · ∆M u = −|D~u |2 .

Thus ~u × (~u × ∆M ~u ) = −∆M ~u − |D~u |2~u . This proves that ~u solves (4.2.3) and (4.2.4). Suppose that ~u is a solution of (4.2.3) and (4.2.4). Set z(x, t) = |~u (x, t)|2 on MT := M × [0, T ], where T is a finite time T < ∞. We have zt = 2~u · ~u t ,

Dz = 2~u · D~u .

By simple calculations we have ∆M z = 2~u · ∆M ~u + 2|D~u |2 . Thus we have zt = α1 ∆M z + 2α1 |D~u |2 (|~u |2 − 1),

z(x, 0) = 1, x ∈ M.

Setting w = z − 1, we have wt = α1 ∆M w + 2α1 |D~u |2 w, w(x, 0) = 0.

(4.2.5) (4.2.6)

Landau–Lifshitz Equations and Harmonic Maps

139

Multiplying (4.2.5) by w and integrating it on M we get 1d 2 dt

Z

M

|w|2 dx + α1

Z

M

|Dw|2 dx ≤ 2α1 max |D~u |2 x,t

Z

M

|w|2 dx.

It follows from Gronwall’s and H¨older’s inequalities and (4.2.6) that w(x, t) = 0 for any (x, t) ∈ MT , i.e. |~u (x, t)|2 = 1. This proves that ~u solves (4.2.1) and (4.2.2). Corollary 4.2.1 Let α2 = 0 in (4.2.1). In the classical sense the Landau–Lifshitz equation is equivalent to the heat flow of harmonic maps ~u t = ∆M ~u + |∆~u |2~u . Now we establish a relation between harmonic maps and the solutions to the elliptic type Landau–Lifshitz equations. Theorem 4.2.2 In the classical sense, ~u : M → S 2 is a harmonic map if and only if ~u solves (4.2.1) and ~u t (x, t) = 0 for t ≥ 0. Proof. First, suppose ~u : M → S 2 is a harmonic map, i.e. −∆M ~u = |D~u |2~u . By the property of vector cross products, we get ~u t = −α1~u × (~u × ∆M ~u ) + α2~u × ∆M ~u = 0. Second, suppose that ~u (x, t) solves (4.2.1) and ~u t = 0, i.e. α1 ∆M ~u + α1 |∆~u |2~u + α2 ~u × ∆M ~u = 0.

(4.2.7)

By the cross product of ~u and (4.2.7), we have α1 ~u × ∆M ~u + α2~u × (~u × ∆M ~u ) = 0. By an argument similar to that of Theorem 4.2.1, we have α1 ~u × ∆M ~u − α2 ∆M ~u − α2 |D~u |2~u = 0. It implies that ~u × ∆M ~u =

α2 (∆M ~u + |D~u |2~u ). α1

(4.2.8)

Combining (4.2.8) with (4.2.7), we get α1 ∆M ~u + α1 |D~u |2~u +

α2 (∆M ~u + |D~u |2~u ) = 0, α1

which implies that ~u is a harmonic map. This proves Theorem 4.2.2.

(4.2.9)

Landau–Lifshitz Equations

140

Corollary 4.2.2. Suppose that ~u 0 : M → S 2 is a given initial data. Then in the classical sense, solution ~u of the Landau–Lifshitz equation (4.2.1) (α 2 6= 0) is usually not equal to a solution of the equation of the harmonic map heat flow. It equals to one solution of the equation of the harmonic map heat flow if and only if ~u (x, t) ≡ ~u 0 (x) for all (x, t) ∈ M × R+ and ~u 0 (x) is a harmonic map from M to S 2 . Suppose that ~u (x, t) ≡ ~u 0 (x) for (x, t) ∈ M ×R+ and ~u 0 (x) is a harmonic map, i.e. −∆M ~u = |D~u |2~u . Then it is easy to see that ~u 0 (x) is a solution to both (4.2.1) and the equation of harmonic map heat flow. Suppose that ~u 0 (x), a solution to the equation of harmonic map heat flow, is also a solution of (4.2.1) with α2 6= 0. Then we have (α1 − 1)∆M ~u + (α1 − 1)|D~u |2~u + α2 × ∆M ~u = 0. Taking the cross product of ~u with the above equation, we can prove as that of Theorem 4.2.2 that ∆M ~u + |D~u |2~u = 0. Then it follows from the equation of heat flow that ~u t = 0. Therefore, ~u (x, t) ≡ ~u 0 (x) is a harmonic map.

4.2.2

Local Smooth Solution of L–L Equation

1. Strong parabolicity Now we consider a family of second order parabolic differential operators of the form ~u t =

m X

Dα (aαβ (x, t, ~u )Dβ ~u ) + f (x, t, ~u , D~u ),

α,β=1

0 ≤ t ≤ T < ∞,

(4.2.10)

where Dα = ∂x∂α , acting on N -vector valued real functions ~u : Ω → RN . We assume that the coefficient functions are smooth, that is, aαβ = aβα ,

aαβ ∈ C ∞ (Ω × [0, T ] × RN , Z(RN )),

where Z(RN ) is the space of all endomorphisms of RN . Like in [74], we say that Eq. (4.2.10) is a strongly parabolic system, i.e. N X

m X

i,j=1 α,β=1

α β aij αβ (x, t, η)ξ ξ ζi ζj > 0

Landau–Lifshitz Equations and Harmonic Maps

141

for all (x, t, η) ∈ Ω × [0, T ] × RN , for all ξ = (ξ 1 , ξ 2 , . . . , ξ m ) ∈ Rm \ {0}, and for all ζ = (ζ1 , ζ2 , . . . , ζN ) ∈ RN \ {0}, where aij αβ are the elements of matrices ajk . We denote matrix g = (gij ) by α1 −α2 u3 α2 u2 g11 g12 g13    α1 −α2 u1  .  g21 g22 g23  =  α2 u3 −α2 u2 α2 u1 α1 g31 g32 g33 







By Theorem 4.2.1, Eq. (4.2.1) is equivalent to (4.2.3) if the initial data ~u 0 (x) satisfies condition (4.2.2). For the Landau–Lifshitz equation (4.2.3), the principle part is given by m X

Dα (aαβ (x, t, ~u )Dβ ~u ) =

α,β=1

m X

α,β=1

Dα (γ αβ Dβ ~u + ~u × γ αβ Dβ ~u ).

The corresponding coefficients aij αβ are given by αβ aij αβ = γ gij .

Therefore the Landau–Lifshitz equation is strongly parabolic by the following relation: N m X X

i,j=1 α,β=1

α β 2 aij αβ (x, t, η)ξ ξ ζi ζj = α1 |ζ|

m X

γ αβ ξ α ξ β > 0

α,β=1

for all (x, t, η) ∈ Ω × [0, T ] × RN , for all ξ = (ξ 1 , ξ 2 , ξ 3 ) ∈ R3 \ {0}, and for all ζ = (ζ1 , ζ2 , ζ3 ) ∈ R3 \ {0}. 2. Local smooth solution for L–L equation Now consider the Landau–Lifshitz equation from R2 → S 2 . First, let us consider the Cauchy problem of the Landau–Lifshitz equation from the flat torus. In order to formulate the problem, let Q be the flat torus R 2 /Z 2 . We suppose that ~u 0 is a map from Q to S 2 . Consider the Cauchy problem Q into R3 ~u t = −α1~u × (~u × ∆~u ) + α2~u × ∆~u ,

(4.2.11)

∀ x ∈ R2 .

(4.2.12)

with initial condition ~u |t=0 = ~u (x, 0) = ~u 0 (x),

It is easy to see from Theorem 4.2.1 that if ~u is a smooth solution of (4.2.11) and (4.2.12), then |~u (x, t)|2 = 1. Lemma 4.2.2 Suppose that ~u is a smooth solution of the Cauchy problem (4.2.11) and (4.2.12). Then k∇~u (·, t)kL2 (Q) ≤ k∇~u 0 kL2 (Q) ,

∀ t ≥ 0.

Landau–Lifshitz Equations

142

Proof. Multiplying (4.2.11) by ∆~u , we have ∆~u · ~u t = −α1 ∆~u · (~u × (~u × ∆~u ))

= −α1 (~u × ∆~u ) · (∆~u × ~u ) = α1 |~u × ∆~u |2 .

Then

1d k∇~u kL2 (Q) + α1 k~u × ∆~u kL2 (Q) = 0. 2 dt

This implies that

1d k∇~u kL2 (Q) ≤ 0. 2 dt

The lemma follows. Lemma 4.2.3 (Gagliardo–Nirenberg Inequality) Let Ω be R n or a bounded Lipschitz domain in Rn with boundary ∂Ω, and let u be any function in W m,r (Ω) ∩ Lq (Ω), 1 ≤ r, q ≤ ∞. For any integer j: 0 ≤ j < m, and for any number α in the interval j ≤ α ≤ 1, set m   1 j 1 1 m = +α − + (1 − α) . p n r n q n If m − j − r is not a nonnegative integer, then k∇j uk0,p ≤ C(kukm,r )α (kuk0,q )1−α .

(4.2.13)

If m − j − nr is a nonnegative integer , then (4.2.13) holds for α = mj , where k · kh,k = k · kW h,k (Ω) . Constant C depends only on r, q, m, j, α and the shape of Ω. Lemma 4.2.4 Suppose that ∇~u 0 ∈ H 1,2 (Q). Let T > 0 be a constant. Let ~u be a smooth solution of (4.2.11) and (4.2.12) on QT = Q × [0, T ]. Then there is a constant C > 0 such that k∇2~u (·, t)kL2 (Q) ≤ C,

∀ T ≥ t ≥ 0;

k∇3~u kL2 (QT ) ≤ C,

provided that k∇~u 0 kL2 (Q) ≤ λ for some small λ > 0. Proof. Denote D α = ∂xα11 ∂xα22 , |α| = α1 + α2 , where α1 , α2 are nonnegative integers. For simplicity, let D k denote any kind of D α , where |α| = k. Differentiating (4.2.11) with D 2 , multiplying it by D 2~u , and integrating with respect to x on Q, we have (∂t D 2~u , D 2~u ) = α1 (∆D 2~u , D 2~u ) + α1 (D 2 (|∇~u |2~u ), D 2~u ) + α2 (D 2 (~u × ∆~u ), D 2~u ).

(4.2.14)

Since Q has no boundary we get  1d   (∂t D 2~u , D 2~u ) = kD 2~u kL2 (Q) ,    2 dt

 (∆D 2~u , D 2~u ) = −k∇D 2 ~u kL2 (Q) ,     2 2 2 2

|(D (|∇~u | ~u ), D ~u )| = |(D(|∇~u | ~u ), D 3~u )|.

(4.2.15)

Landau–Lifshitz Equations and Harmonic Maps

143

Due to D(|∇~u |2~u ) = |∇~u |2 D~u + 2~u ∇~u · D∇~u , we have kD(|∇~u |2~u )kL2 (Q) ≤ k∇~u k2L∞ (Q) kD~u kL2 (Q)

+ 2k~u kL∞ (Q) k∇~u kL∞ (Q) kD∇~u kL2 (Q) .

From the Gagliardo–Nirenberg inequality, we have 1/4

3/4

(4.2.16)

3/4

1/4

(4.2.17)

1/3

2/3

(4.2.18)

k∇~u kL4 (Q) ≤ C1 k∇~u kH 2,2 (Q) k∇~u kL2 (Q) , k∇2 ~u kL4 (Q) ≤ C1 k∇~u kH 2,2 (Q) k∇~u kL2 (Q) , k∇~u kL6 (Q) ≤ C1 k∇~u kH 2,2 (Q) k∇~u kL2 (Q) . Then from (4.2.16)–(4.2.18) |(D 2 (|∇~u |2~u ), D 2~u )|

≤ k∇3~u kL2 (Q) (2k~u kL∞(Q) k∇~u kL4 (Q) kD∇2~u kL4 (Q) + k∇~u k3L6 (Q) ) ≤ C2 (C1 k∇~u kL2 (Q) + C13 k∇~u k2L2 (Q) )k∇~u kH 2,2 (Q)

≤ C2 (2C1 k∇~u 0 kL2 (Q) + C13 k∇~u 0 k2L2 (Q) )k∇~u kH 2,2 (Q) ,

(4.2.19)

where C2 > 0 is a constant. On the other hand, from (4.2.16)–(4.2.18) |(D 2 (~u × ∆~u ), D 2~u )|

= |(D 2 ∇ · (~u × ∇~u ), D 2~u )| = |(D 2 (~u × ∇~u ), ∇D 2~u )|

= |(D 2~u × ∇~u + 2D~u × D∇~u + ~u × ∇D 2~u , ∇D 2~u )| ≤ C3 k∇~u kL4 (Q) k∇2~u kL4 (Q) k∇3~u kL4 (Q)

≤ C12 C3 k∇~u kL2 (Q) k∇~u kH 2,2 (Q)

≤ C12 C3 k∇~u 0 kL2 (Q) k∇~u kH 2,2 (Q) ,

(4.2.20)

where C3 > 0 is an absolute constant and ~u × ∇~u := (~u × ∇1~u , ~u × ∇2~u ). Therefore from (4.2.14)–(4.2.20) we have d k∇2~u (·, t)k2L2 (Q) + α1 k∇3~u (·, t)k2L2 (Q) dt ≤ α1 k∇3~u (·, t)~u k2L2 (Q) (C13 C2 k∇~u 0 k2L2 (Q) α2 + 2C1 C2 k∇~u 0 kL2 (Q) + C12 C3 k∇~u 0 kL2 (Q) ) α1 + Ck∇2~u (·, t)k2L2 (Q) + C,

(4.2.21)

Landau–Lifshitz Equations

144

where C > 0 depends on k∇~u 0 kL2 (Q) . When λ is chosen small enough such that 

ε := C13 C2 λ2 + 2C1 C2 λ +

α2 2 C C3 λ < 1, α1 1 

we have d 2 |∇ ~u (·, t)|2L2 (Q) + α1 (1 − ε)k∇3~u (·, t)k2L2 (Q) ≤ Ck∇2~u (·, t)k2L2 (Q) + C. dt By the Gronwall inequality we have k∇2~u (·, t)k2L2 (Q) ≤ C,

∀ T ≥ t > 0,

and k∇3~u k2L2 (QT ) ≤ C. The lemma follows. Lemma 4.2.5 Let Ω be R2 or bounded Lipschitz domain in R2 with boundary Ω. Let u be any function in H 2,2 (Ω). Then we have 3/4

1/4

kukL∞ (Ω) ≤ CkukH 2,2 (Ω) kukL2 (Ω) , where C > 0 is a constant. Proof. Without loss of generality, we assume that Ω be R 2 , u ∈ C0∞ (R2 ). Then for any x = (x1 , x2 ) ∈ R2 we have Z ∂u(x1 , x2 ) |u(x1 , x2 )| ≤ dx1 . ∂x1

For any fixed x1 , we define

∂u(x1 , x2 ) ∂x1 which is a one-dimensional function. By using Theorem 2.2 of Chapter 2 of [39], we have 1/2 1/2 |v(x2 )| ≤ Ckvx2 kL2(R) kvkL2 (R) . v(x2 ) =

Therefore using the above inequalities and the Gagliardo–Nirenberg inequality we have

Z

∂ 2 u(·, x ) 1/2 ∂u(·, x ) 1/2 2 2

|u(x1 , x2 )| ≤ C

dx1

∂x1 ∂x2 2 ∂x1 2

≤ Ck∇

2

L 1/2 1/2 ukL2 (Ω) k∇ukL2 (Ω) 3/4

1/4

≤ Ck∇2 ukL2 (Ω) kukL2 (Ω) . This proves the lemma.

L

Landau–Lifshitz Equations and Harmonic Maps

145

Lemma 4.2.6 Suppose that ∇~u 0 ∈ H k,2 (Q; S 2 ), k ≥ 2 and the conditions of Lemma 4.2.4 hold. Then there exists a constant C such that sup k∇~u (·, t)k2H m,2 (Q) ≤ C

for

0≤t≤T

2 ≤ m ≤ k.

Proof. We prove this lemma by induction. First, we know from Lemma 4.2.4 that k∇~u (·, t)kH 1,2 (Q) ≤ C for all t ≤ T . Second, from (4.2.3) we have d 3 D ~u = α1 ∆D 3~u + α1 D 3 (|∇~u |2~u ) + α2 D 3 (~u × ∆~u ). dt Since Q has no boundary, we have (D 3~u t , D3~u ) = 3

3

1d kD 3~u (·, t)k2L2 (Q) , 2 dt

(∆D ~u , D ~u ) = −k∇D

3

(4.2.22)

~u (·, t)k2L2 (Q) .

Since D 2 (|∇~u |2~u ) = |∇~u |2 D 2~u + 4∇~u D∇~u · D~u

+ 2|D∇~u |2~u + 2∇~u D 2 ∇~u · ~u ,

we have from Lemmas 4.2.3 and 4.2.5 |(D 3 (|∇~u |2~u ), D 3~u )| = |(D 2 (|∇~u |2~u ), D 4~u )|

≤ C(k∇~u k2L∞ (Q) kD∇~u kL2 (Q) + k∇~u k2L2 (Q) + k∇~u kL∞ (Q) kD 2 ∇~u k2L2 (Q) )kD 4~u kL2 (Q) ε ≤ k∇~u k2H 3,2 (Q) + Ck∇3~u k2L2 (Q) + C 2

and |α2 (D 3 (~u × ∆~u ), D 3~u )|

= |α2 (∇ · D 3 (~u × ∇~u ), D 3~u )| = |α2 (D 3 (~u × ∇~u ), ∇D 3~u )|

= |α2 (D 3~u × ∇~u + 5D 2~u × D∇~u + 4D~u × D 2 ∇~u + ~u × ∇D 3~u , ∇D 3~u )|

≤ C(k∇~u k2L∞ (Q) kD 3~u kL2 (Q) + kD 2~u kL4 (Q) kD∇~u kL4 (Q) + k∇~u kL∞ (Q) )kD 3 ∇~u kL2 (Q) α1 ε ≤ k∇~u k2H 3,2 (Q) + Ck∇3~u k2L2 (Q) + C. 2

Choosing ε small enough, we have d α1 k∇3~u k2L2 (Q) + k∇4~u k2L2 (Q) ≤ Ck∇3~u k2L2 (Q) + C. dt 2

Landau–Lifshitz Equations

146

By the Gronwall inequality, we have k∇~u (·, t)k2H 2,2 (Q) ≤ C,

∀ 0 ≤ t ≤ T,

where C depends on k∇~u 0 kH 2,2 (Q) . In fact, k∇~u kL∞ (Q) ≤ C. Second, suppose that k∇~u kH m,2 (Q) ≤ C, m ≥ 2. From (4.2.3) we have

D m+1 ~u t = α1 ∆D m+1~u + α1 D m+1 (|∇~u |2~u ) + α2 D m+1 (~u × ∆~u ).

Since Q has no boundary, we have 1d kD m+1~u (·, t)k2L2 (Q) , 2 dt m+1 m+1 (∆D ~u , D ~u ) = −k∇D m+1~u (·, t)k2L2 (Q) . (D m+1 ~u t , Dm+1~u ) =

By using the well-known Kato’s inequality (see Lemma 3.1 of [97]), we have kD s (f g) − f D s gkLp (Q)

≤ C(kDf kL∞ (Q) kD s−1 gkLp (Q) + kD s f kLp (Q) kgkL∞ (Q) )

for any two functions f, g. Setting f = ~u , g = |∇~u |2 , s = m, we have kD m (|∇~u |2~u )kL2 (Q)

≤ k~u D m (|∇~u |2 )kL2 (Q) + C(k∇~u kL∞ (Q) kD m−1 |∇~u |2 kL2 (Q) + kD m~u kL2 (Q) k|∇~u |2 kL∞ (Q) )

≤ kD m |∇~u |2 kL2 (Q) + C(kD m−1 |∇~u |2 kL2 (Q) + kD m~u kL2 (Q) ) ≤ C1 k∇m+1 ~u kL2 (Q) ) + C2 .

Then |(D m+1 (|∇~u |2~u ), D m+1~u )| = |(D m (|∇~u |2~u ), D m+2~u )| 1 ≤ k∇D m+1~u k2L2 (Q) + k∇m+1~u k2L2 (Q) . 4 On the other hand, we have |(D m+1 (~u × ∆~u ), D m+1~u )| = |(D m+1 (~u × ∇~u ), ∇D m+1~u )|. Since D m+1 (~u × ∇~u ) = D m+1~u × ∇~u + ~u × D m+1 ∇~u + α2 |(D m+1 (~u × ∇~u ), ∇D m+1~u )| ≤ α2 |(D m+1~u × ∇~u , ∇D m+1~u )| + α2 ≤ ≤

k∇~u k2L∞ (Q) kD m+1~u kL2 (Q)

m

m X

h=1

m X

h=1

ch D h~u × D m+1−h ∇~u ,

ch |(D h~u × D m+1−h ∇~u , ∇D m+1~u )|

+ CkD ~u kL4 (Q) kD m ∇~u kL4 (Q) k∇D m+1~u kL2 (Q)

α1 k∇D m+1~u k2L2 (Q) + Ck∇m+1~u k2L2 (Q) . 4

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147

Therefore, we have d α1 k∇m+1~u (·, t)k2L2 (Q) + k∇m+2~u (·, t)k2L2 (Q) ≤ Ck∇m+1 ~u (·, t)k2L2 (Q) . dt 2 By the Gronwall inequality, we have k∇~u (·, t)k2H m+1,2 (Q) ≤ C,

m ≥ 2,

This prove the lemma.

4.2.3

∀ 0 ≤ t ≤ T.

Global Smooth Solution to L–L Equation

Due to the strong parabolicity of the Landau-lifshitz equation with Gilbert damping (4.2.3), the local existence of smooth solution has been proved by Amann in [5]. Combining this local existence with Lemma 4.2.2–Lemma 4.2.6, we conclude that the local smooth solution can be extended to a large t to obtain the existence of global solution. Theorem 4.2.3 Let Q be the flat torus R2 /Z 2 . Suppose that ∇~u 0 (x) is a given initial value in H k (Q; S 2 ), where k is large enough. Then there exists a constant λ such that the periodic value problem (4.2.11) and (4.2.12) with the initial value ~u 0 admits a smooth global solution ~u (x, t) provided that k∇~u 0 k2 ≤ λ.

4.2.4

On the Landau–Lifshitz Equation on Riemannian Surface

1. Local estimates Now we consider the Landau–Lifshitz equations on Riemannian manifold. Let M be a closed Riemannian surface, BRM (x) = {y ∈ M : |y − x|M < R} for a domain Ω and −∞ < s < t < ∞, and let Ωts = Ω × [s, t]. The energy on BRM (x) is given by Z 1 ER (~u ; x) = M e(~u )dM, where e(~u ) = |∇~u |2 . 2 BR (x) We denote V

(MτT ; S 2 )

(

= ~u : M × [τ, T ] → S 2 | ~u is measurable and ess sup τ ≤t≤T

Z

2

M

|∇~u (·, t)| dM +

Z

T τ

2

2

)

2

(|∇ ~u | + |~u t | )dMdt < ∞ .

Like in [133], we have Lemma 4.2.7 There exist constants C and R0 such that for any u ∈ V (M T ; S 2 ), and any R ∈ (0, R0 ) there holds the estimate Z

MT

|∇u|4 dM dt ≤ C ess sup (x,t)∈M T

×

Z

Z

M (x) BR

2

MT

2

|∇u(x, t)|2 dM

|∇ ~u | dMdt + R

−2

Z

2

MT



|∇~u | dMdt .

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Lemma 4.2.8 There exists a constant C = C(S 2 ) such that for any solution ~u ∈ V (M T ; S 2 ) of (4.2.11) there holds the estimate Z

MT

|~u t |2 dMdt ≤ CE(~u 0 ).

Moreover, E(~u (·, t)) is absolutely continuous on [0, T ] and non-increasing. Proof. By Lemma 4.2.7, we may multiply (4.2.3) by ~u t and integrate. Then for any s, t ∈ [0, T ] Z

Mst

Z

|~u t |2 dM dt + α1

t s

Z d E(~u (·, t))dt − α2 ~u t · (~u × ∆M ~u )dMdt = 0. (4.2.23) dt Mst

In the distribution sense we have ~u t = α1 (∆M ~u + |D~u |2~u ) + α2 (~u × ∆M ~u ). By the cross product of the above equation with ~u , we get ~u × ~u t = α1 (~u × ∆M ~u ) + α2 (~u × (~u × ∆M ~u ))

= α1 (~u × ∆M ~u ) − α2 ∆M ~u − α2 |Du|2~u .

Since −∆M ~u − |Du|2~u = − we have

1 α2 ~u t + (~u × ∆M ~u ), α1 α1 !

α2 α2 ~u × ~u t + ~u t = α1 + 2 (~u × ∆M ~u ). α1 α1

(4.2.24)

Multiplying the above equation by ~u t , we have ~u t · (~u × ∆M ~u ) = α2 (α12 + α22 )−1 |~u t |2 . Then we have α2

Z

MT

α2 ~u t (~u × ∆M ~u )dMdt = 2 2 2 α1 + α 2

Z

MT

|~u t |2 dMdt.

(4.2.25)

The conclusion of the lemma follows by (4.2.24) and (4.2.25). Corollary 4.2.3 Combining Lemma 4.2.7 and Lemma 4.2.8 we obtain the estimate Z

4

MT

|∇~u | dM dt ≤ C

sup ER (~u (·, t), x)

(x,t)∈M T

Z

T |∇ ~u | dMdt + 2 E(~u 0 ) R 2

MT

2

for any solution ~u ∈ V (M T ; S 2 ) of (4.2.3) and any R ∈ (0, R0 ].



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This corollary makes it important to control energy locally. 2. Estimates for the solution to the equivalent equation Like in [133], we have Lemma 4.2.9 There exists a constant C1 = C1 (M ; S 2 ) such that for any solution ~u ∈ V (M T ; S 2 ) of (4.2.3), any R ∈ (0, R0 ], and any (x, t) ∈ M T , there holds the estimate t ER (~u (·, t), x) ≤ E2R (~u (·, 0), x) + C1 2 E(~u 0 ). R ∞ M Proof. Let φ ∈ C0 (B2R (x0 )) satisfy 0 ≤ φ ≤ 1, φ ≡ 1 on BRM (x0 ), |∇φ| ≤ CR . By multiplying Eq. (4.2.3) by φ2~u t , and by the same arguments used in Lemma 4.2.8, we have Z Z α22 2 ~u t · (~u × ∆M ~u )φ dMdt = 2 α2 |~u t |2 φ2 dMdt. 2 T T α1 + α 2 M M Hence by the same steps of Lemma 3.6 in [133], we obtain ER (~u (·, t), x) ≤

Z

M

e(~u (·, t))φ2 dM ≤ E2R (~u 0 , x) + C

t E(~u 0 ) R2

as claimed. For a solution ~u ∈ V (M T ; S 2 ) of (4.2.3) and R ∈ (0, R0 ], let ε(R) := ε(R; ~u , T ) =

sup (x,t)∈M T

ER (~u (·, t); x).

We give a priori estimates for the V -norm and the H¨older norm of ~u in terms of initial energy E(~u 0 ), T , and the number R = sup{R > 0 : ε(R; ~u , T ) ≤ ε1 }, where ε1 > 0 is a parameter depending on M which is to be determined. Then we have Lemma 4.2.10 There is a constant ε1 > 0 such that for any solution ~u ∈ V (M T ; S 2 ) of (4.2.3) and any number R ∈ (0, R0 ], there holds the estimate Z

MT

|∇2~u |2 dMdt ≤ CE(~u 0 )(1 + T R−2 ),

provided ε(R) ≤ ε1 . Proof. We only note that Z

MT

∆M ~u · (α1~u × ∆M ~u )dMdt = 0.

The other steps follow as those of Lemma 3.7 in [133]. Corollary 4.2.4 In the above proof we have the estimate Z

2

MT

|∆M ~u | dM dt ≥

Z

2

MT

2

|∇ ~u | dMdt − C

Z

MT

|∇~u |2 dMdt.

By the same proof as that of Lemma 3.8 in [133], we obtain

(4.2.26)

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Lemma 4.2.11 For any number ε, τ, E0 > 0, R1 ∈ (0, R0 ], there exists a number δ > 0 such that for any solution ~u ∈ V (M T ; S 2 ) of (4.2.3) and any I ⊂ [τ, T ] with measure |I| < δ there holds the estimate Z Z I

M

|∇~u |4 dMdt < ε,

provided ε(R1 ) ≤ ε1 , E(~u 0 ) ≤ E0 . Lemma 4.2.12 For numbers E0 > 0, R0 > 0, let ~u m ∈ V (M T ; S 2 ) be solutions to Eq. (4.2.3) with initial value ~u m (0) provided that ε(R; ~u m , T ) ≤ ε1 for R ∈ (0, R0 ], and ~u m (0) → ~u 0 as m → ∞. Then ~u m converges to a solution of (4.2.3) in V (M T ; S 2 ) as m → ∞. Proof. The proof of Lemma 4.2.10 and [133, Lemma 3.8] then show that Z

Z

!

4

M

|∆M ~u m | dM dt

is uniformly absolutely continuous on [0, T ]. We may suppose that ~u m → ~u a.e. and ∂t ~u m * ∂t ~u , ∆2 ~u m * ∆2~u weakly in L2 (M T ), and ∇~u m → ∇~u strongly in L2 (M T ). Let vm := ~u m − ~u ; |∇Um | := |∇~u m | + |∇~u |. Then (∂t vm − α1 ∆M vm + α2~u m × ∆M vm + α2 vm × ∆M ~u m ) ≤ Cm (|vm ||∇Um |2 + |∇vm ||∇Um |).

Multiplying it by ∆M vm and integrating it give 1 2

Z

d (∇vm , ∇vm )M dMdt + α1 |∆M vm |2 dMdt M dt MT Z Z α1 2 ≤ |∆M vm | dMdt + |vm |2 |∆M ~u |2 dMdt 2 MZ T MT Z

+C

MT

(|vm |2 |∇Um |4 + |∇vm |2 |∇Um |2 )dMdt.

Then by (4.2.26) and Vitali’s theorem, we have sup 0≤t≤T

Z

2

M

|∇vm (·, t)| dM +

Z

MT

|∇2 vm |2 dMdt → 0

since ∇vm (·, 0) → 0 in L2 (M ). Similarly, Z

MT

|∂t vm |2 dM dt ≤ C

Z

MT

(|∆M vm |2 + |vm |2 |∇Um |4

+ |∇vm |2 |∇Um |2 + |vm |2 |∆M ~u |2 )dMdt → 0, (m → ∞),

i.e. ~u m → ~u strongly in V (M T ; S 2 ).

Landau–Lifshitz Equations and Harmonic Maps

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3. Regularity of solutions to L–L equation Lemma 4.2.13 Let ~u ∈ V (M T ; S 2 ) τ >0 C 2 (MτT ; S 2 ) be a regular solution to (4.2.3). Then for any τ > 0 the H¨ older norms of ~u and its derivatives may be estimated T uniformly on Mτ by quantities involving only E(~u 0 ), τ, T and R, provided ε(R) ≤ ε1 . T

Proof. Multiplying (4.2.3) by ∆M ~u and integrating it give Z

M

|∆M ~u (·, t)|2 dM ≤

1 2

Z

M

+C

|∆M ~u (·, t)|2 dM

Z

M

|∂t~u (·, t)|2 dM + C

Z

M

|∇~u (·, t)|4 dM,

which implies Z

2

M

|∆M ~u (·, t)| dM ≤ C

Z

M

(|∇~u (·, t)|4 + |∂t~u (·, t)|2 )dM, a.e. t ∈ [0, T ].

From ~u × ∆M ~u = ∇ · (~u × ∇~u ),

we have Z

MT

(~u × ∆M ~u )t · ~u t dMdt =

Z

MT

=− 1 ≤ 4

Z Z

(∇ · [∂t ~u × ∇~u + ~u × ∇∂t ~u ]) · ~u t dMdt

MT

(∂t ~u × ∇~u ) · ∇~u t dMdt 2

MT

|∇∂t~u | dMdt + C

Z

|∂t~u |2 |∇~u |2 dMdt.

MT

In order to estimate the right-hand side of the above inequality, we differentiate (4.2.3) with respect to t, and multiply with ∂t ~u , and then integrate over MsT , where τ ≤ s ≤ t ≤ T , to give Z 1Z 2 ∂t |~u t | dM dt + α1 |∇∂t ~u |2 dMdt 2 MsT MsT

≤C

≤

Z

MsT

(|∂t ~u |2 |∇~u |2 + |∂t ~u ||∇~u ||∇∂t ~u |)dMdt

Z + α2 (~u × ∆M ~u )t · ~u t dMdt MsT Z Z

α 2

MsT

|∇∂t~u |2 dMdt + C

MsT

|∂t ~u |2 |∇~u |2 dMdt.

Repeating the proof of Lemma 3.10 of [133], we have  Z   sup   τ ≤t≤T

Z    

2

M

M

|∂t~u (·, t)|2 dM ≤ C(1 + τ −1 )E(~u 0 ); 2

|∇ ~u (·, t)| dM ≤ C(1 + τ

−1

−2

+ R )E(~u 0 ),

(4.2.27)

Landau–Lifshitz Equations

152

where ε(R) ≤ ε1 . On the other hand, it follows from Lemma 4.2.5 and (4.2.27) that sup k~u (·, t1 ) − ~u (·, t2 )kL∞ (M )

x∈M

3/4

1/4

≤ Ck~u (·, t1 ) − ~u (·, t2 )kH 2,2 (M ) k~u (·, t1 ) − ~u (·, t2 )kL2 (M ) ≤ 2C ≤C



Z t2 3/4

sup k~u (·, t)kH 2,2 (M )

t1 t

3/4 sup k~u (·, t)kH 2,2 (M ) |t1 t

1/4

∂t ~u dt

2

− t2 |

L (M )

1/4

Z

2

|~u t | dMdt

MτT

≤ C|t1 − t2 |1/4 .

!1/4

(4.2.28) 1 1 , 2 4

Combining (4.2.27) with (4.2.28), we know that the C -H¨older norm of ~u can be estimated uniformly on MτT by quantities in E(~u 0 ) and ε(R) ≤ ε1 . Since (4.2.3) is parabolic, using [101, Chap. VII, Theorem 10.4] and by the standard bootstrap method, we may get the higher regularity about the solution. 4. Uniqueness of solution to L–L equation Now we consider the uniqueness of solution to (4.2.3) in V (M T ; S 2 ). Theorem 4.2.4 Suppose ~u 1 , ~u 2 ∈ V (M T ; S 2 ) are two solutions to (4.2.3) with ~u 1 (·, 0) = ~u 2 (·, 0) = ~u 0 (·). Then ~u 1 = ~u 2 in M T . Proof. Let v = ~u 1 − ~u 2 and |∇U | := |∇~u 1 | + |∇~u 2 |. From (4.2.1) we obtain

|vt − α1 ∆M v + α2~u 1 × ∆M ~u 1 − α2~u 2 × ∆M ~u 2 | ≤ C(|v||∇U |2 + |∇v||∇U |).

If we multiply this by v and integrate over M T , we have Z Z 1 |v(·, t)|2 dM + α1 (∇v, ∇v)M dMdt 2 M MT Z ≤C

MT

(|v|2 |∇U |2 + |v||∇v||∇U |)dMdt

Z

+ C

Since

MT



(~u 1 × ∆M ~u 1 − ~u 2 × ∆M ~u 2 )vdMdt .

(~u 1 × ∆M ~u 1 − ~u 2 × ∆M ~u 2 )v = (v × ∆M ~u 1 + ~u 2 × ∆M v)v

= (~u 2 × ∆M v) · v = −(~u 2 × v) · ∆M v,

Z

MT



(~u 1 × ∆M ~u 1 − ~u 2 × ∆M ~u 2 )vdMdt

Z = (~u 2 × v)∆M vdMdt MT Z = (∇~u 2 × v + ~u 2 × ∇v) · ∇vdMdt MT Z = (∇~u 2 × v) · ∇vdMdt MT Z Z

≤C

MT

|v|2 |∇U |2 dMdt +

1 4

MT

|∇v|2 dMdt.

Landau–Lifshitz Equations and Harmonic Maps

153

Therefore 1 2

Z

M

|v(·, t)|2 dM + α1

≤

α1 2

Z

MT

Z

MT

(∇v, ∇v)M dMdt

(∇v, ∇v)M dMdt + C

Z

MT

|v|2 |∇U |2 dMdt.

By the same steps as that of the proof of Lemma 3.12 in [133], we complete the proof of the theorem. In order to prove a local existence of solution to (4.2.3), we improve Hamilton’s idea and derive the following Lemma 4.2.14 For every function g ∈ Lp (M × [α, ω]) there exists a unique f ∈ W 2,p (M × [α, ω]) such that ∂t f = α1 ∆M f + a∇f + bf + α2 z × ∆M f + g on M × [α, ω], where a, b, z are given smooth functions. Proof. Let Hf = ∂t f − α1 ∆M f − α2 z × ∆M f and Kf = a∇f + bf . Then map f → Hf defines an isomorphism from W 2,p (M × [α, ω]) onto Lp (M × [α, ω]) since H is a strongly parabolic linear operator. Moreover, since K has weight 1, map K : W 2,p (M × [α, ω]) → Lp is compact as it factors through the compact inclusion of W 2,p into Lp . By the theory of Fredholm mappings map W 2,p → Lp given by f → (Hf − Kf ) has finite dimensional kernel and cokernel. Moreover, its index is zero. Then to show that it is an isomorphism it suffices to show that its kernel is zero. Let f ∈ W 2,p (M × [α, ω]) we know that f |[M ×α] = 0 since ∂t f − α1 ∆M f − α2 z × ∆M f − a · ∇f − bf = 0. Taking the scalar product of the above equation with f , we obtain Z

∂t |f |2 dM + α1 |∇f |2 dM − α2

M

=

Z

M

Z

M

f · (z × ∆M f )dM

af · ∇f dM + bf 2 dM.

Since Z

M

Z

M

f ∇ · (z × ∇f )dM = − f · (z × ∆M f )dM = −

Z

Z

M

∇f · (z × ∇f )dM = 0,

M

f · ∇z × ∇f dM,

we have Z

α1 ∂t |f | dM + α1 |∇f | dM ≤ 2 M 2

2

Z

2

M

|∇f | dM + C

Z

bf 2 dM. M

Using the Gronwall inequality, we have f = 0. This proves the lemma.

Landau–Lifshitz Equations

154

From Proposition 2.1 of [25], we have Lemma 4.2.15 Let M be a two-dimensional compact Riemannian manifold. Then for all u ∈ Wp1,2 (M T ), 4 < p < ∞, there exists constant C > 0 such that sup ku(·, t)kC 1+α (M ) ≤ CkukWp1,2 (M T ) ,

t∈[0,T ]

where α = 1 − p4 . We now prove the existence of solutions of (4.2.3) for short periods of time. Theorem 4.2.5 Let ~u 0 : M → S 2 be a smooth map. There exists a constant ε > 0 and a map ~u : M × [0, ε] → S 2 with ~u ∈ Lp2 (M ε ) solving the equation ∂t ~u = α1 ∆M ~u + α1 |∇~u |2~u + α2 ~u × ∆M ~u , ~u = ~u 0 ,

on M × [0, ε], on M × {0} .

Moreover , ~u is unique and smooth. Proof. First, we define a nonlinear map L : W 2,p (M × [0, ω]) → Lp (M × [0, ω]) as follows: L := α1 ∆M ~u + α1 |∇~u |2~u + α2 ~u × ∆M ~u . For the smooth map ~u , the derivative of L at ~u is given by DL(~u )k = α1 ∆M k + α2 ~u × ∆M k + a(~u )k + b(~u )k, where a and b are smooth matrices of functions. Choosing ~u b : M × [0, ω] → R3 to be a smooth map with ~u b = ~u 0 on M × {0}, we denote ~u as sum ~u b + ~u ∗ . Let H(~u ) = ∂t ~u − L(~u ). The derivative of H(~u ) is given by DH(~u )k = ∂t k − α1 ∆M k − α2~u × ∆M k − a∇k − bk. Fix ~u b and consider ~u ∗ as a variable function. Then ~u ∗ → H(~u b + ~u ∗ ) defines a continuously differentiable map of W 2,p (M × [0, ω]) → Lp (M × [0, ω]). Its derivative at ~u ∗ = 0, DH(~u b ) : W 2,p (M × [0, ω]) → Lp (M × [0, ω]) is given by DH(~u b )k = ∂t k − α1 ∆M k − α2~u × ∆M k − a∇k − bk, which by Lemma 4.2.14 is an isomorphism. Therefore by the inverse function theorem, the set of all H(~u b + ~u ∗ ) for ~u ∗ is a neighborhood of H(~u b ) in Lp (M × [0, ω]). If we choose ε > 0 small enough, the function equals to 0 for 0 ≤ t ≤ ε, and equals to H(~u b ) for ε < t ≤ ω will be in this neighborhood. So ~u |M ε is in W 2,p (M × [0, ε]), and it solves the Landau–Lifshitz equation (4.2.3). Second, using [133, Chap. VII, Theorem 10.4] and the standard bootstrap method, we conclude that ~u is smooth.

Landau–Lifshitz Equations and Harmonic Maps

155

5. Uniqueness Using Lemma 4.2.15, we have sup k~u (·, t)kC 1+α (M ) ≤ C.

0≤t≤s

From the above estimate and Lemma 4.2.5 we know that the local solution is unique. We thus have Theorem 4.2.6 For any initial value ~u 0 ∈ H 1,2 (M ; S 2 ) there exists a unique solution ~u of (4.2.3) on M × [0, ∞) which is regular on M × [0, ∞) with exception of at most finitely many points (xl , T l ), 1 ≤ l ≤ L, characterized by the condition that lim l sup ER (~u (·, T ), xl ) > ε1 for all R ∈ (0, R0 ].

T →T T
The proof of this theorem, which is based on Lemma 4.2.8, Lemma 4.2.9, Lemma 4.2.12, Theorem 4.2.4, and Theorem 4.2.5, is the same as the proof of Theorem 4.2 of [133].

4.2.5

The Landau–Lifshitz Equation in Higher Dimensions

1. Global weak solution to L–L equation in higher dimensions Let M be a compact m-dimensional Riemannian manifold without boundary and m ≥ 3. Using a similar way as before, we can prove that Eq. (4.2.11) is equivalent to the following equation: α12

α2 α1 ∂ ~u − 2 ~u × ∂t ~u = ∆~u + |∇~u |2~u . 2 t α1 + α22 + α2

(4.2.29)

So we define a global weak solution for (4.2.29) as follows: Definition 4.2.1 A vector function ~u is said to be a global weak solution to (4.2.29), if ~u is defined a.e. on M × R+ such that (1) ~u ∈ L∞ ((0, ∞); H 1,2 (M )) and ∂t~u ∈ L2 ((0, ∞); L2 (M )); (2) |~u (x, t)|2 = 1 a.e. on M × R+ ; (3) (4.2.29) holds in the sense of distribution; (4) ~u (x, 0) = ~u 0 (x) in the trace sense. We consider the following penalized equation: α12

α1 α1 ∂ ~u k − 2 ~u k × ∂t ~u k − ∆~u k + k 2 (|~u k |2 − 1)~u k = 0, 2 t + α2 α1 + α22

(4.2.30)

where t > 0, x ∈ M . We construct approximate solutions of (4.2.30) by using the Galerkin method. Let wj (x) (j = 1, 2, . . .) be the normalized eigenfunction of equation ∆~u + λ~u = 0 on M corresponding to eigenvalue λj (j = 1, 2, . . .). {wj (x)} forms a normalized orthogonal basis in H 1,2 (M, R3 ).

Landau–Lifshitz Equations

156

Denote the approximate solutions ~u kN (x, t) of Eq. (4.2.30) by the following form: ~u kN (x, t) =

N X

αkj (t)wj (x),

j=1

where αkj (t) (j = 1, 2, . . .) are vector-valued functions of t ∈ R+ . By the standard procedure of the Galerkin approximation, there exists a global weak solution of (4.2.30) that satisfies Z

M

"

α12

α1 α1 ∂ ~u k · ws − 2 (~u k × ∂t ~u kN ) · ws 2 t N + α2 α1 + α22 N #

+ ∇~u kN · ∇ws + k 2 (|~u kN |2 − 1)~u kN · ws dM = 0,

(4.2.31)

for all s = 1, 2, . . . , N, with the initial value ~u N (x, 0) satisfying Z

M

~u N (x, 0)ws (x)dM =

Z

~u 0 (x)ws (x)dM,

M

s = 1, 2, . . . N,

(4.2.32)

where ~u 0 (x) ∈ H 1,2 (M ; S 2 ) is the initial value. Moreover ~u kN satisfies Z

t 0

k∂t ~u kN (τ )k2L2 (M ) dτ + k∇~u kN (·, t)k2L2 (M ) +

k2 4

Z

M

(|~u kN (·, t)|2 − 1)2 dM ≤ C

(4.2.33)

for all t ∈ R+ . Since k is fixed, using (4.2.31) and (4.2.33) and letting N → ∞, there exists a function f k (x) ∈ (H 1,2 (M T ))∗ such that lim

Z

N →∞ M T

~u kN

×

∂t ~u kN

· φdx =

Z

MT

f k (x) · φdx, ∀ φ ∈ H 1,2 (M T ; R3 ).

On the other hand, from (4.2.33) we have lim

Z

N →∞ M T

~u kN

×

∂t ~u kN

· φdx =

Z

MT

~u k × ∂t ~u k · φdx, ∀ φ ∈ H 1,2 (M T ; R3 ).

By a density argument, we get f k (x) = ~u k × ∂t ~u k . Therefore using (4.2.31) and (4.2.33) there exists a global weak solution ~u k of (4.2.30) such that Z

t 0

k∂τ ~u

k

(τ )k2L2 (M ) dτ

for all t ∈ R+ .

+ k∇~u

k

(·, t)k2L2 (M )

k2 + 4

Z

M

(|~u k (·, t)|2 − 1)2 dM ≤ C (4.2.34)

Landau–Lifshitz Equations and Harmonic Maps

157

From (4.2.31) we have Lemma 4.2.16 Let ~u k be a global weak solution of (4.2.30). Then we have |~u k (x, t)| ≤ 1. k

Proof. Let φ = ~u k − |~~uu k | min{1, |~u k |} be a test function for Eq. (4.2.30) like in [73]. Then by a simple calculation, we get 1 α1 ∂t 2 α12 + α22

Z

k 2

|~u k |≥1

|~u |

!

1 1 − k dM + |~u |

Z

k 2

|~u k |≥1

|∇~u |

!

1 1 − k dM ≤ 0. |~u |

(4.2.35)

Since |~u k (x, 0)| = |~u 0 | = 1, we get that |~u k | ≤ 1 for all t ≥ 0 and a.e. on M . Using Lemma 4.2.16 and (4.2.34), we prove the following: Theorem 4.2.7 Let α1 > 0 in (4.2.29). For all ~u 0 ∈ H 1,2 (M ; S 2 ), there exists a global weak solution of Eq. (4.2.29). Proof. From (4.2.34) and Lemma 4.2.16 we can choose a subsequence of ~u k , still denoted by ~u k , such that as k → ∞ ~u k ∂t ~u k ~u k × ∂t ~u k ~u k

* * * →

~u , ∂t ~u , ~u × ∂t ~u , ~u ,

weakly-* in L∞ (0, ∞; H 1,2 (M )); weakly in L2 (0, ∞; L2 (M )); weakly in L2 (0, ∞; L2 (M )); strongly in L2 (0, ∞; L2 (M ));

and |~u | = 1 a.e. on R+ × M.

By taking the edge product of (4.2.30) with ~u k , we get 0=

!

α1 α2 ∂ ~u k − 2 ~u k × ∂t ~u k − ∆~u k × ~u k . 2 2 t 2 α1 + α 2 α1 + α 2

Then the conclusion of theorem follows by allowing k → ∞. 2. Examples of solution to L–L equation in higher dimensions From the theory of harmonic maps we give the examples of solution to (4.2.29). x Example 4.2.1 Let M = R3 . Let ~u = |x| : R3 → S 2 . By the well-known result, we have −∆~u = |∇~u |2~u , ∀ x ∈ R3 \ {0}.

Then from the well-known result of De Giorgi, ~u is a solution of (4.2.3) from R 3 → S 2 , but it is not smooth at the origin in R3 . Example 4.2.2 Let M = R3 . For any constant K, we set ~u (x, t) ≡ ~u (x) = (cos Kx3 , sin Kx3 , 0), where x = (x1 , x2 , x3 ). Then ~u is a smooth solution of (4.2.3) from R3 × R+ into S 2 .

Landau–Lifshitz Equations

158

4.3 4.3.1

Generalized L–L Systems and Harmonic Maps Generalized Landau–Lifshitz Systems

1. Generalized Landau–Lifshitz systems and harmonic maps Let (M, γ) be an m-dimensional Riemannian manifold of metric γ. Then generalized Landau–Lifshitz system from M to S n−1 is of the form ~u t = α1 (∆M ~u + |∇~u |2~u )

+ α2 ∗ [~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ],

(4.3.1)

where “∧” denotes the outer product in Rn , “∗” is the Hodge-∗ operator in Euclid space Rn , ai (~u ) : S n−1 → Rn (i = 1, 2, . . . , n − 2) are smooth vector-valued functions which are linear independent. If n = 3 and M is a bounded domain, (4.3.1) is the usual Landau–Lifshitz system discussed in above section. Note that (∆M ~u + |∇~u |2~u ) · (∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ]) = 0,

that is, vector ∆M ~u +|∇~u |2~u is orthogonal to vector ∗[~u ∧a2 (~u )∧· · ·∧an−2 (~u )∧∆M ~u ]. We may see that the harmonic map from M into S n−1 , i.e. the solution of ∆M ~u + |∇~u |2~u = 0

is also a solution from M to S n−1 of the following elliptic Landau–Lifshitz system: α1 ∆M ~u + |∇~u |2~u + α2 ∗ [~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ] = 0. This establishes the relation between harmonic maps and the generalized Landau– Lifshitz system and makes that some approaches for harmonic maps can be applied to the generalized Landau–Lifshitz system. First, we want to prove the existence of a global smooth solution to (4.3.1) from a closed Riemannian manifold into S n−1 with initial data of small energy. Then we discuss the global existence of a weak solution to (4.3.1) if the initial data ~u 0 ∈ H 1 (M ; S n−1 ) and ai (~u ) : S n−1 → Rn (i = 1, 2, . . . , n − 2) are linear independent. If m = 2, we have known that such weak solutions are in fact regular with exception of at most finitely many points. In local coordinate x = (x1 , x2 , . . . , xm ) on M , (4.3.1) can be expressed as follows: 1 ∂ ~u t = α1 √ γ ∂xβ

γ

αβ √

∂~u γ α ∂x

!

+ α1 |∇~u |2~u

"

1 ∂ + α2 ∗ ~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ √ γ ∂xβ where ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ] is 

b11 b12   b21 b22  b  31 b32  . ..  .. .  bn1 bn2

b13 b23 b33 .. . bn3



γ



· · · b1n ∆M u 1    ∆ u2  · · · b2n   M 3   · · · b3n    ∆M u  .    . . ..  .  . ..   .  · · · bnn ∆M u n

αβ √

∂~u γ α ∂x

!#

,

(4.3.2)

Landau–Lifshitz Equations and Harmonic Maps

159

For simplicity, let α1 = α2 = 1. From the definitions of outer product and Hodge-∗ operator, we have ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ] 

e1 e2 e3 e4 .. .

u1 u2 u3 u4 .. .

en

un−1 un

      =      en−1

a21 (~u ) a22 (~u ) a23 (~u ) a24 (~u ) .. .

··· ··· ··· ··· .. .

an−2,1 (~u ) an−2,2 (~u ) an−2,3 (~u ) an−2,4 (~u ) .. .

∆ M u1 ∆ M u2 ∆ M u3 ∆ M u4 .. .

a2n (~u )

···

an−2,n (~u )

∆M un−2



      ,    n−1   a2,n−1 (~u ) · · · an−2,n−1 (~u ) ∆M u

where {e1 , e2 , . . . , en } is a normal orthogonal basis of Rn , ~u = (u1 , u2 , . . . , un ). Direct computation shows that (bij ) is antisymmetric. Let S = (sij ) denote the following matrix:   1 b12 b13 · · · b1n    −b12 1 b23 · · · b2n     1 · · · b3n  . S = (sij ) =  −b13 −b23  . .. ..  .. ..  .. . .  . .   −b1n −b2n −b3n · · · 1 It is easy to see that matrix (sij ) is an n × n nondegenerate matrix. In fact, n X

sij ηi ηj =

i,j=1

n X i=1

ηi2 ,

∀ η ∈ Rn \ {0}.

Therefore in the local coordinates, the principle part of the generalized Landau– Lifshitz system is m X

Di (aij (·, ~u )Dj ~u ) =

i,j=1

m X

i,j=1

Di (γ ij Dj ~u + ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ γ ij Dj ~u ]).

The corresponding coefficients aαβ ij are given by ij aαβ ij = γ sαβ .

It is clear that (aαβ ij ) satisfies the Legender–Hadamard condition: n m X X

α,β=1 i,j=1

i j 2 aαβ ij (x, p)ξ ξ ηα ηβ = |η|

m X

γ ij ξ i ξ j > 0

i,j=1

for all (x, p) ∈ M × Rn , η ∈ Rn \ {0}, and ξ ∈ Rm \ {0}. 2. Initial boundary value problem for the generalized Landau–Lifshitz equations Consider the following initial boundary value problem for the Landau–Lifshitz equation ~u t = ∆M ~u + |∇~u |2~u + ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ], ~u = ψ, x ∈ ∂M, t ≥ 0; ~u (·, 0) = ~u 0 , x ∈ M,

(

(4.3.3)

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160

where M is a Riemannian manifold with boundary, and ψ : ∂M → S n is a smooth map. ~u 0 : M → S n belongs to Cψ2 (M ; S n ) ≡ {~u : ~u ∈ C 2 (M ; S n ), ~u |∂M = ψ}. In order to prove the local existence of solutions for (IBP) we shall use the results of Amann [5]. Let Ω be a bounded smooth domain in Rm and consider the second order parabolic systems of the form ~u t =

m X

i,j=1

Di (aij (x, t, u)Dj u) + f (x, t, Du), 0 < t < T, x ∈ Ω,

(4.3.4)

where Dj = ∂x∂ j is acting on the N -dimensional vector u : Ω → RN . Suppose that the coefficients of system (4.3.4) are smooth, that is, aij = aji , aij ∈ C ∞ (Ω × [t, T ] × RN ; l(RN )), where l(RN ) is the self-isomorphism space. Furthermore, suppose that n m X X

r,s=1 j,k=1

N n m i j ars jk ξ ξ ηr ηs > 0, ∀ (x, t, p) ∈ Ω × [t, T ] × R , η ∈ R \ {0}, ξ ∈ R \ {0},

where ars jk are the elements of matrix ajk . Let f ∈ C ∞ (Ω × [t, T ] × Rm × RmN ; RN ). 3. Initial boundary value problem for the quasilinear parabolic systems Consider the following initial boundary value problem:  m X   ~ ut = Dj (ajk (x, t, u)Dk u) + f (x, t, u, Du), s < t ≤ T, x ∈ Ω, j,k=1    u = 0, on ∂Ω × (s, T ]; u(·, s) = u ,

(4.3.5)

0

where 0 ≤ s < T . The classical solution u of this problem is defined on interval J ⊂ [s, T ] such that u ∈ C(Ω × J; RN ) ∩ C 1 (Ω × J; RN ) ∩ C 2,0 (Ω × J; RN ), and it satisfies (4.3.5) pointwise. Let m < p < ∞, and denote Wpr (Ω, RN ) by Wpr , the usual Sobolev space, where r ∈ [0, ∞]. H. Amann in [5] proved the following local existence result: Theorem 4.3.1 Let 0 ≤ s < T, mp < τ < ∞, u0 ∈ Wpτ . Then the initial-boundary value problem (4.3.5) admits a classical solution u defined on an open interval J ⊂ [s, T ]. If sup ku((t)kWpτ < ∞, t∈J

then J = [s, T ] and u is a global solution.

Landau–Lifshitz Equations and Harmonic Maps

161

4. Local smooth solution Applying Amann Theorem 4.3.1 to problem (4.3.3), we have Theorem 4.3.2 Let ψ(x, t) ∈ C ∞ (∂M ; S n ), ~u 0 ∈ Cψ2 (M ; S n ) and ai (~u ) (i = 2, 3, . . . , n − 2) be smooth functions. Then the initial boundary value problem (4.3.3) admits a unique classical local solution ~u (x, t) defined on [0, ω) ⊂ [0, T ]. If M is a Riemannian manifold without boundary, we can consider the initial value problem (IP) for the generalized Landau–Lifshitz systems from M to S n−1 as follows: ~u t = ∆M ~u + |∇~u |2~u + ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ], t > 0, x ∈ M, ~u (x, 0) = ~u 0 , x ∈ M,

(4.3.6)

where ~u 0 ∈ C 2 (M, S n−1 ) and |~u 0 | = 1. Theorem 4.3.3 Let M be a closed smooth Riemannian manifold, ~u 0 ∈ C 2 (M, S n−1 ), and ai (~u ) (i = 2, 3, . . . , n − 2) be smooth functions. Then the initial boundary value problem (4.3.6) admits a unique classical solution ~u (x, t) defined on t ∈ [0, ω) ⊂ [0, T ]. Proof. From Theorem 4.3.2, it suffices to prove that the smooth solution of (4.3.6) satisfies |~u (x, t)| = 1. In fact, setting z(x, t) = |~u (x, t)|2 on MT := M × [0, T ], where T is a finite time T < ∞, we have zt = 2~u · ~u t ,

∇z = 2~u · ∇~u .

By simple calculations we have ∆M z = 2~u · ∆M ~u + 2|D~u |2 . Thus we have zt = ∆M z + 2|D~u |2 (|~u |2 − 1),

z(x, 0) = 1, x ∈ M.

Setting w = z − 1, we have wt = ∆M w + 2|D~u |2 w,

(4.3.7)

w(x, 0) = 0.

(4.3.8)

Multiplying (4.3.7) by w and integrating it on M , we get 1d 2 dt

Z

M

|w|2 dx +

Z

M

|Dw|2 dx ≤ 2 max |D~u |2 x,t

Z

M

|w|2 dx.

It follows from Gronwall’s and (4.3.8) that w(x, t) = 0 for any (x, t) ∈ MT , i.e. |~u (x, t)|2 = 1. This proves the theorem.

Landau–Lifshitz Equations

162

Now consider the following Cauchy problem for the Landau–Lifshitz equation from the flat torus T 2 = R2 /Z ⊕ Z to S n−1 . Let ~u 0 : T 2 → S n−1 be a mapping. Consider the following problem: ~u t = ∆M ~u + |∇~u |2~u + ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ], ~u (x, 0) = ~u 0 (x), x ∈ T 2 ,

(4.3.9) (4.3.10)

where ai (~u ) (i = 2, 3, . . . , n − 2) are smooth functions and |ai (~u )| = 1. Lemma 4.3.1 Let ~u ∈ C ∞ (T 2 × R+ ; S n−1 ) be a smooth solution to the Cauchy problem (4.3.9) and (4.3.10). Then we have E(~u (·, t)) ≤ E0 (~u 0 ) = E0 , where E(~u ) =

R

∀ t ≥ 0,

(4.3.11)

e(~u )dx, e(~u ) = 21 |∇~u |2 .

T2

Proof. Multiplying (4.3.9) by ∂t ~u and integrating it over T 2 × [0, T ], we have Z

T

Z

0

2

|~u t | dxdt −

T2

Z

=

T 0

Z

T2

Z

T 0

Z

T2

~u t · (∆~u + |∇~u |2~u )dxdt

~u t · {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}dxdt.

(4.3.12)

Since ∆~u + |∇~u |2~u and ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ] are orthogonal with each other, we have 2



|~u t |2 = ∆~u + |∇~u |2~u |∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]|2 .

Then it follows from the definition of outer product that ~u t · {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}

= |∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]|2

2

= ∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ (∆~u + |∇~u |2~u )] ≤ |∆~u + |∇~u |2~u |2 ,

where we have used |ai (~u )| = 1 (i = 2, . . . , n − 2). Then we have 1 |~u t · ∗{[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}| ≤ |~u t |2 , 2

(4.3.13)

and therefore Z

T 0

Z

T2

~u t · (∆~u + |∇~u |2~u )dxdt = −

Z

T 0

d E(~u (t))dt. dt

It then follows from (4.3.12)–(4.3.14) that Z

The lemma follows.

T 0

Z

T2

|~u t |2 dxdt + 2

Z

T 0

d E(~u (t))dt ≤ 0. dt

(4.3.14)

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163

Lemma 4.3.2 Suppose that ~u 0 ∈ H 2 (T 2 ). Let ~u be a smooth solution of (4.3.9) and (4.3.10) on T 2 × [0, T ]. Then there is a constant C > 0 depending only on E0 (if E0 is small enough) such that k∇2~u (·, t)kL2 (T 2 ) ≤ C, ∀ 0 < t ≤ T ;

k∇3~u (x, t)kL2 (T 2 ×[0,T ]) ≤ C.

Proof. Acting on (4.3.9) by ∆, multiplying it by ∆~u , and integrating it with respect to x on T 2 , we have Z

T2

Z

∆~u t · ∆~u dx =

2

∆ ~u · ∆~u dx +

T2

+

Z

T2

Z

T2

∆(|∇~u |2~u ) · ∆~u dx

∆{∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}∆~u dx.

It is clear that Z

Z

1d k∆~u k2L2 (T 2 ) ; 2 dt

T2

∆~u t · ∆~u dx =

T2

∆2~u · ∆~u dx = −k∇∆~u k2L2 (T 2 ) .

Hence Z

∆(|∇~u | ~u ) · ∆~u )dx = 2

T2

Z

∇(|∇~u | ~u ) · ∇∆~u )dx 2

T2

≤ k∇∆~u kL2 (T 2 ) (2k~u kL∞(T 2 ) k∇~u kL4 (T 2 ) · k∆~u kL4 (T 2 ) + k∇~u k3L6 (T 2 ) ).

(4.3.15)

From the following Gagliardo–Nirenberg inequalities 3/4

1/4

(4.3.16)

2/3

(4.3.17)

3/4

(4.3.18)

k∇~u kL4 (T 2 ) ≤ C1 k∇~u kL2 (T 2 ) k∇∆~u kL2 (T 2 ) , 1/3

k∇~u kL6 (T 2 ) ≤ C1 k∇∆~u kL2 (T 2 ) k∇~u kL2 (T 2 ) , 1/4

k∇2~u kL4 (T 2 ) ≤ C1 k∇~u kL2 (T 2 ) k∇∆~u kL2 (T 2 ) , we have Z q 2 ∇(|∇~u | ~u ) · ∇∆~u dx ≤ (23/2 C12 E0 + 2C13 E0 )k∇∆~u k2L2 (T 2 ) . T2

On the other hand, using (4.3.15) and (4.3.17), we have

Z ∆~ u · ∆{∗[~ u ∧ a u ) ∧ · · · ∧ a u ) ∧ ∆~ u ]}dx 2 (~ n−2 (~ T2 Z ≤ ∇∆~u · {∗[∇~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}dx T2 Z n−2 X 0 + ∇∆~ u · {∗[~ u ∧ a u ) ∧ · · · ∧ a u )∇~ u ∧ · · · ∧ a u ) ∧ ∆~ u ]}dx 2 (~ i (~ n−2 (~ T2 i=2

Landau–Lifshitz Equations

164

Z + ∇∆~u · {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∇∆~u ]}dx T2

≤ k∇∆~u kL2 (T 2 ) (C12 k∇~u kL2 (T 2 ) k∇∆~u kL2 (T 2 )

+ (n − 2)C12 max ka0i (~u )kL∞ (T 2 ×(0,T ]) k∇~u kL2 (T 2 ) k∇∆~u kL2 (T 2 ) ). i

We finally get d k∆~u (·, t)k2L2 (T 2 ) + k∇∆~u (·, t)k2L2 (T 2 ) ≤ εk∇∆~u k2L2 (T 2 ) , dt where



3

1

1

ε := C12 2 2 + 2 2 + 2 2 (n − 2) max ka0i (~u )kL∞ i

We have that when E0 is small enough such that ε < 1

q

E0 + 2C13 E0 .

d k∆~u (·, t)k2L2 (T 2 ) + (1 − ε)k∇∆~u k2L2 (T 2 ) ≤ 0. dt By the Gronwall inequality we have k∇2~u (·, t)k2L2 (T 2 ) ≤ C,

∀ T ≥ t > 0,

and k∇∆~u kL2 (T 2 ×(0,T )) ≤ C.

The lemma follows.

Lemma 4.3.3 Let ~u 0 ∈ H 3 (T 2 ; S n−1 ) and the conditions of Lemma 4.3.2 hold. Then there exists a constant C such that sup k∇(·, t)kL2 (T 2 ) ≤ C.

(4.3.19)

0≤t≤T

Proof. Acting on (4.3.9) by ∇∆, multiplying it by ∇∆~u , and integrating it over T , we have 2

Z 1d k∇∆~u k2L2 (T 2 ) + k∆2~u k2L2 (T 2 ) = ∇∆(|∇~u |2~u ) · ∇∆~u dx 2 dt T2 Z

+

T2

∇∆~u · ∇∆{∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}dx.

(4.3.20)

It follows from Lemma 4.3.4 and the Sobolev inequality that the first term on the right-hand side of (4.3.20) can be estimated as follows: Z

Z 2 2 ∇∆(|∇~u | ~u ) · ∇∆~u dx = ∆(|∇~u | ~u ) · ∆ ~u dx T2 Z = {5|∇~u |2 ∆~u + 2|∇~u |2~u + 2(∆~u · ∇∆~u )}∆2~u dx 2 2

T2

T

≤ C(k∇~u k2L∞ (T 2 ) k∆~u kL2 (T 2 ) + k∆~u kL2 (T 2 )

+ k∇~u kL∞ (T 2 ) k∇∆~u kL2 (T 2 ) )k∆2~u kL2 (T 2 ) 1 ≤ k∆2~u k2L2 (T 2 ) + Ck∇∆~u k2L2 (T 2 ) + C1 . 2

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165

On the other hand, we estimate the second term on the right-hand side of (4.3.20) as follows Z

T2

∇∆~u · ∇∆{∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}dx Z = ∆2~u · ∆{∗[∇~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}dx 2 T

≤ C(k∇~u k2L∞ (T 2 ) k∆~u k2L2 (T 2 ) + k∇~u kL∞ (T 2 ) k∇∆~u kL2 (T 2 ) + k∆~u k2L4 (T 2 ) )k∆2 ~u kL2 (T 2 )

1 ≤ k∆2~u k2L2 (T 2 ) + Ck∇∆~u k2L2 (T 2 ) + C1 . 2 Since E0 is small enough, we finally get d k∇∆~u k2L2 (T 2 ) ≤ Ck∇∆~u k2L2 (T 2 ) + C1 . dt By the Gronwall inequality, we have sup k∇∆~u (·, t)k2L2 (T 2 ) ≤ C2 ,

0≤t≤T

where C2 depends on k∇∆~u 0 kL2 (T 2 ) and E0 . The lemma follows. 5. Global smooth solution Combining the local existence Theorem 4.3.2 and the a priori estimate Lemma 4.3.6, we know that the local solution can be extended to the global solution. Theorem 4.3.4 Suppose that ~u 0 ∈ H k (T 2 ; S n−1 ), k ≥ 4, and E0 is small enough. Then the initial value problem (4.3.9) and (4.3.10) admits at least one global classical solution.

4.3.2

The Global Weak Solution to the Generalized L–L Equations

Now we consider the global weak solution to the generalized Landau–Lifshitz equations from a two-dimensional Riemannian manifold to S n−1 . We first give some notions. Let M be a compact Riemannian manifold of metric γ = (γij )1≤i,j≤2 , ~u : M → n−1 S be a smooth map, and ∂uα ∂uα 1 e(~u ) = γ ij (x) 2 ∂xi ∂xj be the density of energy on M where (γ ij ) = (γij )−1 . The energy on M is given by E(~u ) =

Z

M

e(~u )dM.

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The stationary point of E(~u ) is called the harmonic map from M to S n−1 which satisfies the Euler–Lagrange equation ∆M ~u + |∇~u |2~u = 0. Let Lp , Hpm , and C ∞ be the Lebesgue, Sobolev, and H¨older spaces, respectively. Hpm (M , S n−1 ) is the space of function ~u : M → S n−1 , ~u |Ω ∈ Hpm (Ω, Rn ), Ω ⊂ M , and |~u | = 1, a.e. x ∈ M . dM (·, ·) is the geodesic distance, BRM (x) = {y : y ∈ M, dM (x, y) < R}, ER (~u ; x) =

Z

M (x) BR

e(~u )dM.

1. Generalized Landau–Lifshitz equations on Riemannian manifold Now consider the following problem: ~u t = ∆M ~u + |∇~u |1~u + {∗[∇~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}, ~u (x, 0) = ~u 0 (x),

(4.3.21)

x ∈ M,

(4.3.22)

where ai (~u ) (i = 2, . . . , n − 2) are n-dimensional linear independent vectors, ~u 0 ∈ C 2 (M ), and M is a compact closed Riemannian manifold. 1 Lemma 4.3.4 ([133]) For any ~u ∈ Hloc (R2 ) and any function ϕ ∈ C0∞ (BR ), 0 ≤ 2 ϕ ≤ 1 and |∇ϕ| ≤ R ,

Z

4

R2

2

|~u | ϕ dx ≤ C

Z

2

BR

!

|~u | dx ·

Z

2

BR

2

|∆~u | ϕ dx + R

−2

Z

2

BR

|~u | dx

!

holds, where C is independent of ~u and R. Lemma 4.3.5 Let ~u ∈ C 2 (M × [0, T ]; S n−1 ) be a solution to problem (4.3.21) and (4.3.22). Then Z tZ 1 sup |~u t |2 dxdt + sup E(~u (t)) ≤ E(~u 0 ). 2 0≤t≤T 0 M 0≤t≤T

The proof of this lemma is just the same as that of Lemma 4.3.1; hence we omit it. Lemma 4.3.6 There exists an absolute constant C3 > 0 such that for any R < 21 iM M E(~u (T ); BRM ) ≤ E(~u 0 ; B2R ) + C3

for any solution of (4.3.21) and (4.3.22).

T E(~u 0 ). R2

Landau–Lifshitz Equations and Harmonic Maps

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M Proof. Multiplying (4.3.21) by ϕ2~u t , where ϕ ∈ C0∞ (B2R ), 0 ≤ ϕ ≤ 1, ϕ(x) ≡ 1 2 M in BR and |∇ϕ| ≤ R , and integrating by parts by noting

1 |~u t |2 ϕ2 ≥ ~u t ϕ2 · {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆M ~u ]}, 2 we have ( ) 1Z TZ d 2 2 2 2 |~u t | ϕ + (|∇~u | ϕ ) dMdt M 2 0 B2R dt Z

≤ −2 ≤

1 2

Z

T 0

T 0

Z

Z

M B2R

M B2R

∇~u · ~u t ∇ϕ · ϕdMdt

|~u t |2 ϕ2 dMdt + 2

Z

T 0

Z

M B2R

|∇~u |2 ϕ2 dMdt.

Hence we have E(~u (T ); BRM )

−

M E(~u 0 ; B2R )

1 ≤ 2

Z

M B2R

|∇~u |2 ϕ2 dM |T0 ≤ C2

T E(~u 0 ). R2

This completes the proof. M Lemma 4.3.7 There exists ε1 = ε1 (M ; S n−1 ) > 0 such that if ~u ∈ C 2 (B2R × n−1 M (0, T ); S ) be a solution of (4.3.21) and (4.3.22) on B2R × [0, T ) and if for some 0 < R < 21 iM sup E(~u (t); BRM (x)) < ε1 , M ;0≤t
then Z

T 0

Z

M BR

!

T |∇2~u |2 dMdt ≤ CE(~u 0 ) 1 + 2 , R

where C depends only on M and S n−1 . The proof is similar to that of Lemma 3.7 of [133] if we note that ∆M ~u · {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}. 2. Existence of global weak solution Let V 0 (M × [τ, T ); S n−1 ) (

= ~u : M × [τ, T ) → S n−1 , ~u is measurable ess sup 0≤t
Z

2

M

|∇~u (· · · , t)| dM +

Z

T

τ

Z

2

M

2

2

)

(|∇ ~u | + |~u t | )dMdt < ∞ .

Modifying the proof of Lemma 3.8 in [133], we can get

Landau–Lifshitz Equations

168

Lemma 4.3.8 For any numbers ε, τ, E01 > 0, 0 < R < 21 iM , there exists a positive M number δ such that if ~u ∈ V 0 (B2R × [0, T ]; S n−1 ) is a solution of (4.3.21) and (4.3.22), then for any interval I ⊂ [τ, T ) with |I| < δ there holds the estimate Z Z I

M BR

|∆~u |4 dMdt < ε,

where E0 < E01 and M ε(2R; x) = ε(~u ; 2R, x, T ) = sup E(~u (·, t); B2R (x)) < ε1 . 0≤t
M M Lemma 4.3.9 Let ~u ∈ C 2 (B2R (x) × [τ, T ]; S n−1 ) V 0 (B2R (x) × [τ, T ]; S n−1 ) be a regular solution of (4.3.21) and (4.3.22), where 0 < R ≤ 21 iM . Then for any τ > 0 when ε(4R, x) < ε1 , the norms of ~u and its derivatives on BRM ×[τ, T ) can be uniformly estimated by a quantity depending only on E(~u 0 ), τ, T, and R.

T

M Proof. Multiplying (4.3.21) by ϕ2 ∆M ~u , where ϕ ∈ C0∞ (B2R ), 0 ≤ ϕ ≤ 1, ϕ(x) ≡ 1 2 M in BR , and |∇ϕ| ≤ R , and integrating by parts, we have

Z

M B2R

|∆M ~u |2 ϕ2 dM ≤ C

Z

M B2R

(|∇~u (·, t)|4 ϕ2 + |~u t |2 ϕ2 )dM.

(4.3.23)

In order to estimate the bound for the right-hand side of above inequality, differentiating (4.3.21) with respect to t and then multiplying it by ϕ2~u t and integrating it M over B2R × [s, t], we have 1 2

Z tZ s

M B2R

≤C

∂t |~u t |2 ϕ2 dM dt +

(Z Z t

+

s

Z

t s

Z

M B2R

M B2R

Z tZ s

M B2R

|∇~u t |2 ϕ2 dMdt

(|∇~u |2 |~u t |2 ϕ2 + |~u t ||∇~u ||∇~u t |ϕ2 )dMdt

|~u t ||∇ϕ||∇~u t |ϕdMdt

Z Z ) t 2 . (4.3.24) + ϕ ~ u u ∧ a u ) ∧ · · · ∧ a u ) ∧ ∆ u ]}dMdt t · ∂t {∗[~ 2 (~ n−2 (~ M~ s BM 2R

The last term in the above inequality can be estimated by

Z Z t 2 ϕ ~ u u ∧ a u ) ∧ · · · ∧ a u ) ∧ ∆ u ]}dMdt t · ∂t {∗[~ 2 (~ n−2 (~ M~ s BM 2R ! Z Z t

≤C

s

+C ≤

1 4

M B2R

Z

t s

Z tZ s

Z

(2|∇ϕ||~u t ||∇~u t |ϕ + |~u t ||∇~u |∇~u t |ϕ2 )dMdt

M B2R

M B2R

(|~u t |2 + |∇~u |2 )dMdt

|∇~u t |2 ϕ2 + C

Z tZ s

M B2R

|∇~u |2 |~u t |2 ϕ2 dMdt + C.

(4.3.25)

Landau–Lifshitz Equations and Harmonic Maps

169

Hence 1 2

Z

t s

Z

M B2R

≤C

1 2

∂t |~u t |2 ϕ2 dMdt + Z tZ s

M B2R

Z tZ s

|∇~u t |2 ϕ2 dMdt

M B2R

|~u t |2 |∇~u |2 ϕ2 dMdt + C.

Since Z

t s

Z

2

M B2R

2

Z tZ

2

|~u t | |∇~u | ϕ dM dt ≤ Z

≤

t s

Z

·

Z t

ess sup

s

s≤θ≤t

Z tZ

≤C

s

+

Z

|∇~u | dM dt

M B2R

Z

t s

Z

4

M B2R

!1 2

2

|~u t | ϕ dMdt

· 2

M B2R

2

|~u t (·, θ)| ϕ dM !1 2

4

M B2R

!1

M B2R

|∇~u | dMdt

Z

2

4

M B2R

s

4

|∇~u | dM dt

Z

t s

Z

!

2

M B2R

ess sup s≤θ≤t

Z

2

|∇(ϕ|~u t |)| dM dt 2

M B2R

2

!1

|~u t (·, θ)| ϕ dM

!

ϕ2 |∇~u t |2 dM dt) + C.

It follows from Lemma 4.3.8 that if t − s < δ is small enough, the right-hand side of (4.3.25) can be controlled by the left-hand side. Therefore, Z

2

M B2R

2

|~u t (·, t)| ϕ dM ≤

inf

Z

ϕ2 |~u t (·, s)|2 dM + C,

t−δ≤s≤t B M 2R

where C depends only on E(~u 0 ), τ , T , and R, and then ess sup τ ≤t
Z

|~u t (·, t)|2 ϕ2 dM ≤ C(1 + τ −1 )

M B2R

Z

T

τ

Z

M B2R

|~u t |2 dMdt + C,

that is ess sup τ ≤t
Z

M B2R

|~u t (·, t)|2 dM ≤ C(1 + τ −1 )E0 + C(E0 , R, T ),

(4.3.26)

where C(E0 , R, T ) depends only on E0 , R, T . Applying Lemma 4.3.8 to v = ∇~u (·, t), we have Z

4

M BR

|∇~u | dM ≤

Z

·

4

M B2R

Z

2

|∇~u | ϕ dM ≤ C ess sup τ ≤t
2

M B2R

2

2

Z

ϕ |∇ ~u (·, t)| dM + R

≤ C ess sup ER (~u (·, t), x) τ ≤t
Z

M B2R

−2

2

M B2R

Z

|∇~u (·, t)|2 dM · 2

M B2R

2

|∇~u (·, t)| dM 2

! −2

!

ϕ |∇ ~u (·, t)| dM + R E(~u 0 ) . (4.3.27)

Landau–Lifshitz Equations

170

If ε(4R; x) ≤ ε1 is small enough, then from (4.3.23), (4.3.26), and (4.3.27) we have the uniform estimate in t ∈ [τ, T ) as follows: Z

M BR





|∇2~u (·, t)|2 dM ≤ CE(~u 0 ) 1 + τ −1 + R−2 ,

(4.3.28)

where we have used in (4.3.28) the following inequality: Z

1 |∆~u | ϕ dM ≥ M 2 B2R 2

2

Z

2

M B2R

2

2

|∇ ~u | ϕ dM − C

Z

M B2R

|∇~u |2 |∇ϕ|2 dM.

On the other hand, it follows from the Gagliardo–Nirenberg inequality and (4.3.28) that sup k~u (x0 , t2 ) − ~u (x0 , t1 )kL∞

M x0 ∈BR

3

1

≤ Ck~u (·, t2 ) − ~u (·, t1 )kH4 2 (B M ) · k~u (·, t2 ) − ~u (·, t1 )kL4 2 (B M ) R

3

≤ C sup k~u (·, t)kH4 2 (B M ) R

τ ≤t
1 4

≤ C|t2 − t1 | ;

Z

R

T τ

Z

!1 4

M BR

|∂t ~u |dMdt

1

|t2 − t1 | 4

(t1 , t2 ) ⊂ [τ, T ).

(4.3.29) 1

Then we deduce from (4.3.28) and (4.3.29) that the C 1−δ, 4 -H¨older norm on BRM × [τ, T ) of ~u can be uniformly estimated by E(~u 0 ), R, τ, T and ε1 . This proves the lemma. M Lemma 4.3.10 Let ~u ∈ V 0 (B2R × [τ, T ); S n−1 ) be a weak solution of (4.3.21) and 1 1 (4.3.22), where τ > 0, 0 < R ≤ 4 iM . Then ~u ∈ C 1−δ, 4 (BRM × [τ, T ); S n−1 ) and its H¨ older norm on BRM × [τ, T ) are uniformly bounded. M Lemma 4.3.11 Let Q = B2R ×(0, T ), where 0 < R ≤ 14 iM . If ~u ∈ C ∞ (Q; S n−1 ) τ >0 M V 0 (B2R ×[τ, T ); S n−1 ) is a weak solution of (4.3.21) and (4.3.22), and if ε(4R; x) < ε1 , M then ~u can be smoothly extended to BRM × (0, T ), where x is the center of B2R .

T

M M Proof. From Lemma 4.3.9 and the imbedding of H 2 (B2R ) into Wp1 (B2R ) for any p < ∞, we have M |~u t − ∆M ~u − {∗[~u ∧ a2 (~u ) ∧ · · · ∧ an−2 (~u ) ∧ ∆~u ]}| ∈ Lp (B2R × [τ, T )). 1

On the other hand, we know from Lemma 4.3.9 that ~u ∈ C 1−δ, 4 (BRM ×[τ, T ); S n−1 ) and its H¨older norm on BRM × [τ, T ) is uniformly bounded. Hence it follows from the Lp theory of a strongly parabolic system [23, 24] that ~u ∈ Wp2,1 (BRM × [τ, T ) for any p < ∞. The higher regularity can be proved by Eq. (4.3.21) and the bootstrap method.

Landau–Lifshitz Equations and Harmonic Maps

171

Lemma 4.3.12 Let V (M × [0, T ]; S

n−1

(

) = ~u : M × [0, T ) → S n−1 , ~u is measurable ess sup τ ≤t≤T

Z

2

M

|∇~u (·, t)| dM +

Z

0

TZ

2

M

2

2

)

(|∇ ~u | +|~u t | )dMdt < ∞ .

Let ~u 1 , ~u 2 ∈ V (M × [0, T ]; S n−1 ) be two weak solutions of (4.3.21) with ~u 1 (·, 0) = ~u 2 (·, 0) = ~u 0 . Then ~u 1 = ~u 2 = ~u 0 on M × [0, T ]. Proof. Set v = ~u 1 − ~u 2 , |∇U | = |∇~u 1 | + |∇~u 2 |. It follows from (4.3.21) that |vt − ∆M v + {∗[~u 1 ∧ a2 (~u 1 ) ∧ · · · ∧ an−2 (~u 1 ) ∧ ∆~u 1 ]} − {∗[~u 2 ∧ a2 (~u 2 ) ∧ · · · ∧ an−2 (~u 2 ) ∧ ∆~u 2 ]}|

≤ C(|v||∇U |2 + |∇v||∇U |). It follows from (4.3.21) that

Z tZ 1Z 2 |v(·, t)| dM + (∇v, ∇v)M dMdt 2 M 0 M

≤C

Z tZ 0

M Z t

+ C

Therefore, we have 1 2

Z

0

(|v||∇v||∇U | + |v|2 |∇U |2 )dMdt Z

M

v · ({∗[~u 1 ∧ a2 (~u 1 ) ∧ · · · ∧ an−2 (~u 1 ) ∧ ∆~u 1 ]}

− {∗[~u 2 ∧ a2 (~u 2 ) ∧ · · · ∧ an−2 (~u 2 ) ∧ ∆~u 2 ]})dMdt .

1 |v(·, t)| dM + 2 M 2

Z tZ 0

2

M

|∇v| dMdt ≤ C

Z tZ 0

M

|v|2 |∇U |2 dMdt.

The remaining of the proof is similar to that of Lemma 3.12 of [133]. Lemma 4.3.13 Let ~u m ∈ V (M × [0, T ]; S n−1 ) be weak solutions of (4.3.21) with ~u m (·, 0) = ~u m (0). If ε(~u m , R, T ) = sup(x,t)∈M ×[0,T ] ER (~u m ) ≤ ε1 , R ∈ [0, R0 ], where R0 be a positive constant, and ~u m (0) → ~u 0 in H 1 (M ) (m → ∞), then ~u m → ~u (m → ∞) in V (M × [0, T ]; S n−1 ), where ~u is a solution of (4.3.21). Theorem 4.3.5 For any ~u 0 ∈ H 1 (M ; S n−1 ) there exists a unique global solution to (4.3.21) and (4.3.22) on M × [0, ∞) which is regular with exception of at least finitely many points (xl , T l ) (1 ≤ l ≤ L) characterized by lim sup ER (~u (·, T ), xl ) > ε, ∀ R ∈ (0, R0 ].

T →T l ,T
Theorem 4.3.6 Let b = inf{E(~u ) | ~u : M → S n−1 is nonconstant smooth harmonic map}. If ~u 0 ∈ H 1 (M ; S n−1 ) with E(~u 0 ) < b, then the solution to (4.3.21) and (4.3.22) is globally regular on M × [0, ∞).

172

4.4

Landau–Lifshitz Equations

Regularity of Weak Solutions to the Two-Dimensional Landau–Lifshitz Equations

In the above sections, we have obtained the global existence of weak solutions to higher-dimensional Landau–Lifshitz equations. The relations with harmonic maps were proposed, and the partially regular solution to two-dimensional problem was constructed. In this section, we will show that all weak solutions, with finite energy, to twodimensional Landau–Lifshitz equations must be partially regular and then unique.

4.4.1

Cauchy Problem to Two-Dimensional L–L Equation

1. The problem Let M be a two-dimensional compact Riemannian manifold without boundary. Consider the following Cauchy problem of the Landau–Lifshitz equation on M × R + : ∂t u = αu × ∂t u + β∆M u + β|du|2 u, u(x, 0) = u0 ,

x ∈ M,

x ∈ M, t > 0,

(4.4.1) (4.4.2)

where u0 ∈ H 1 (M, S 2 ), and dim(M ) = 2. We want to prove that any weak solution with finite energy, i.e., E(u(t)) =

Z

M

|∇u(·, t)|2 dM ≤ E0 ,

(4.4.3)

is “almost smooth.” The weak solution of this problem is defined as follows: Definition 4.4.1 A map u : M × [0, T ] → S 2 is said to be a weak solution of (4.4.1) and (4.4.2) if u ∈ WT , where WT =: {u : ∂t u ∈ L2 (0, T ; L2 (M )), du ∈ L∞ ((0, T ), L2 (M )), |u| = 1, a.e. on M × [0, T ]}, and if u satisfies (4.4.1) and (4.4.2) in the sense of distribution. The existence, uniqueness, and regularity for the strong solution have been obtained in Sec. 4.2 (Theorems 4.2.4 and 4.2.6); now we prove the following theorem: Theorem 4.4.1 Let M be a compact two-dimensional Riemannian manifold without boundary. For any T > 0, if u ∈ WT is a weak solution of (4.4.1) and (4.4.2) R with finite energy E(u(t)) = M |du|2 (·, t) ≤ E0 for some constant E0 and for a.e. t ∈ [0, T ], then u is unique and is smooth on M × [0, T ] with the exception of at most finitely many points.

Landau–Lifshitz Equations and Harmonic Maps

173

From now on, we call such solution as the “almost smooth” solution. Our proof is completed by first showing that any weak solution of (4.4.1) and (4.4.2) with finite energy is in space L2 (0, T ; W 1,4(M )), and second by proving that such a solution in fact is in space L4 (0, T ; W 1,4 (M )). Then, we can prove the uniqueness for the weak solutions of (4.4.1) and (4.4.2). At last, the regularity result is the consequence of the uniqueness result and the existence result. The Sobolev space W k,p(M, S 2 ) is defined by W k,p(M, S 2 ) = {u ∈ W k,P (M, R3 )|u ∈ S 2 a.e. x ∈ M }. Norm kukW k,p(M,S 2 ) is simply denoted by kukW k,p ; here domain M and target S 2 are usually omitted. We use W k,p(M, ∧` ) to denote the space of differential `-forms on M with coefficients in the Sobolev space W k,p(M ). The exterior derivative operator is denoted by d and the conjugate operator of the exterior differential operator d in the metric of M is denoted by d∗ . It is well known that dd = 0, d∗ d∗ = 0, and the Laplace operator for differential forms is given by ∆M = dd∗ + d∗ d. Lemma 4.4.1 (Hodge decomposition theorem) [92] Let M be an m-dimensional compact Riemannian manifold without boundary and W be an `-form on M in Lp (M, ∧` ). Then there is an (` − 1)-form A and (` + 1)-form B such that W = dA + d∗ B, and A ∈ W 1,p (M, ∧`−1 ), B ∈ W 1,p (M, ∧`+1 ) satisfy kAkW 1,p + kBkW 1,p ≤ CkW kLP ,

(4.4.4)

where C > 0 is a constant depending on M and p. Moreover A and B are unique and satisfy d∗ A = dB = 0. This result is due to Iwaniec and Martin [92], and the original proof is for the case that M = Rm , but it is not difficult to see that the lemma is also true in this case. Lemma 4.4.2 (Wente’s theorem) [144] Let M be a two-dimensional compact Riemannian manifold without boundary. If g ∈ H 1 (M, ∧2 ) and h ∈ H 1 (M, ∧0 ), then hd∗ g, dhi ∈ H −1 (M, ∧0 ) and khd∗ g, dhikH −1 ≤ Ckd∗ gkL2 kdhkL2 ,

(4.4.5)

where C > 0 is a constant depending only on M . This result is due to Wente; the original proof is for the case that M = R 2 , but the argument can also be applied to this case.

Landau–Lifshitz Equations

174

2. Regularity and uniqueness to the Cauchy problem The next lemma is on the uniqueness and regularity for weak solutions to the Cauchy problem of the following system: ∂t ui − β∆M ui = β

3 X

j=1

d∗ g ij · dui + f i ,

ui (x, 0) = ui0 (x),

x ∈ M,

x ∈ M, t > 0, i = 1, 2, 3,

(4.4.6) (4.4.7)

where β > 0 is a absolute constant, g ij (i, j = 1, 2, 3) are differential two-forms on M for each t ∈ (0, T ), and f i (i = 1, 2, 3) are functions of x and t. Let XT = {u : M → R3 | u ∈ L2 ([0, T ], H 1,2 (M )), ∂t u ∈ L2 ([0, T ], H −1 (M ))}, and for u ∈ XT , define the norm of u in XT by kukXT = k∂t ukL2 ([0,T ],H −1 ) + kukL2 ([0,T ],H 1,2 ) , and 4

YT = {u : M → R3 | u ∈ L2 ([0, T ], W 1,4 (M )), ∂t u ∈ L2 ([0, T ], L 3 (M ))},

(4.4.8)

and for u ∈ YT , define the norm of u in YT by kukYT = k∂t ukL2 ([0,T ],L 34 ) + kukL2 ([0,T ],W 1,4 ) . We have the following lemma: Lemma 4.4.3 Let M be a two-dimensional compact Riemannian manifold without 4 boundary, and in (4.4.6) and (4.4.7) f i (i = 1, 2, 3) ∈ L2 ([0, T ], L 3 (M, ∧0 )), g ij (1 ≤ i, j ≤ 3) ∈ L∞ ([0, T ], H 1 (M, ∧2 )), and u0 ∈ H 1 (M, S 2 ). Then there exists an ε0 > 0, depending only on M, T, and β, such that if kgkL∞ ([0,T ],H 1 ) = max kg ij kL∞ ([0,T ],H 1 ) < ε0 , 1≤i,j≤3

(4.4.9)

then any solution u of (4.4.6) and (4.4.7) in space XT must belong to space YT . Proof. It is sufficient to prove Claim 1: Problem (4.4.6) and (4.4.7) has a unique solution u in XT . Claim 2: Problem (4.4.6) and (4.4.7) has a unique solution w in YT . In fact, from the second claim, the problem (4.4.6) and (4.4.7) admits a unique solution w in YT . Since YT ⊂ XT , so w ∈ XT . Then, from the first claim, if u is a solution of (4.4.6) and (4.4.7) in XT , u must coincide with w. Therefore, u ∈ YT . Now let us prove these claims by the contraction mapping principle. Endowed with metric ρ1 (u1 , u2 ) = ku1 − u2 kXT on XT and with metric ρ2 (u1 , u2 ) = ku1 − u2 kYT

Landau–Lifshitz Equations and Harmonic Maps

175

on YT , XT and YT are nonempty complete metric spaces. For any v ∈ XT , by solving the Cauchy problem of the following parabolic system ∂t ui − β∆M ui = β

3 X

j=1

d∗ g ij · dv j + f i , in M × (0, T ],

ui (x, 0) = ui0 (x) on M,

(4.4.10)

i = 1, 2, 3,

(4.4.11)

we define a map L : v → u = Lv. By the theory for linear parabolic equations [101], problem (4.4.10) and (4.4.11) admits a unique solution u ∈ XT and ku||XT = k∂t ukL2 ([0,T ],H −1 ) + kdukL2 ([0,T ],L2 ) 

≤ C

3 X

j=1



kd∗ g ij · dv j kL2 ([0,T ],H −1 ) + kf kL2 ([0,T ],H −1 ) + ku0 kL2  .

Applying Lemma 4.4.3 and the Sobolev embedding theorem to the right of this inequality, we obtain that kukXT ≤ C(kd∗ gkL∞ ([0,T ],L2 ) kdvkL2 ([0,T ],L2 ) + kf kL2 ([0,T ],L 34 ) + ku0 kH 1 ),

(4.4.12)

where C > 0 is a constant depending only on M and β. Therefore, the mapping L maps XT into itself. Moreover, we would like to show that L is a contraction ¯ = Lv¯, mapping with respect to the metric of XT . For any v¯, v¯ ∈ XT , let u ¯ = L¯ v, u ¯ , v = v¯ − v¯. Since L maps XT into itself, u ¯ ∈ XT , and u satisfies and u = u¯ − u ¯, u ∂t ui − β∆M ui = β ui (x, 0) = 0

3 X

j=1

d∗ g ij · dv j , in M × (0, T ],

(4.4.13)

on M,

(4.4.14)

i = 1, 2, 3.

Applying estimate (4.4.12) to problem (4.4.13) and (4.4.14) and noticing that kdvkL2 ([0,T ],L2 ) ≤ T kvkXT , we find that under assumption (4.4.9), kukXT ≤ Cε0 kvkXT , where C > 0 is a constant depending only on M , T , and β. Therefore, if ε0 is chosen 1 to be equal to 2C , L is a contraction mapping on XT . By the contraction mapping principle, there exists a unique solution u ∈ XT to problem (4.4.6) and (4.4.7). The proof of Claim 1 is complete.

176

Landau–Lifshitz Equations

Claim 2 can be proved by the same method. In fact, by the regularity theory for linear parabolic equations [101], problem (4.4.10) and (4.4.11) admits a unique solution u ∈ YT and ku||YT ≤ k∂t ukL2 ([0,T ],L4/3 ) + kd2 ukL2 ([0,T ],L4/3 ) ≤ C(kd∗ gkL∞ ([0,T ],L2 ) kdvkL2 ([0,T ],L4 ) + kf kL2 ([0,T ],L 34 ) + ku0 kH 1 ).

(4.4.15)

By estimate (4.4.15), it easy to see that if 0 in (4.4.9) is sufficiently small, then map L is a contraction mapping on YT . Therefore, there exists a unique solution u ∈ YT to problem (4.4.6) and (4.4.7). The last lemma in this subsection is regarding the following Cauchy problem: ∂t ui − G1 (x, t)∆u = G2 (x, t)∆u + g(x, t) u(x, 0) = u0 (x),

x ∈ M,

x ∈ M, t > 0,

(4.4.16) (4.4.17)

where Gi (x, t) (i = 1, 2) are matrices and g(x, t) is a vector. Lemma 4.4.4 Let M be a two-dimensional compact Riemannian manifold without boundary. Suppose that in (4.4.16) and (4.4.17) (1) G1 (x, t)∆u is strongly elliptic. (2) G1 ∈ C ∞ (M × (0, T )), G2 ∈ L∞ (M × (0, T )), g ∈ L4 (0, T ; L4/3 (M )) and u0 ∈ H 1 (M ). Then there exists a constant 1 > 0 depending only on M and T such that if kG2 kL∞ (M ×(0,T )) ≤ 1 ,

(4.4.18)

then problem (4.4.16) and (4.4.17) has a unique solution in Ls (0, T ; W 2,4/3 (M )) for any s ∈ [2, 4]. Proof. Again we will use the contraction mapping principle to prove this lemma. Fix an s ∈ [2, 4]. For any v ∈ Ls (0, T ; W 2,4/3 (M )) by solving the following linear strongly parabolic system ∂t u − G1 (x, t)∆u = G2 (x, t)∆v + g(x, t),

x ∈ M, t > 0,

(4.4.19)

with the initial condition (4.4.17), we define a map L : v → u = Lv. By the theory for linear parabolic equations (Theorem 9.3, Remark 9.14, and Remark 9.15 of [39]), problem (4.4.19) and (4.4.17) admits a unique solution u ∈ Ls (0, T ; W 2,4/3 (M )) and ku||Ls (0,T ;W 2,4/3 (M )) = C{kG2 kL∞ (M ×[0,T ]) kvkLs (0,T ;W 2,4/3 (M )) + kgkL4 (0,T ;L4/3 (M )) + ku0 kH 1 (M ) }

≤ C{1 kvkLs (0,T ;W 2,4/3 (M )) + kgkL4 (0,T ;L4/3 (M )) + ku0 kH 1 (M ) }, (4.4.20)

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177

where C > 0 depends only on M and T , and in the last inequality we have used (4.4.18). Therefore, mapping L maps Ls (0, T ; W 2,4/3 (M )) into itself. Next we show that L is a contraction mapping with respect to the metric of Ls (0, T ; W 2,4/3 (M )). For ¯ = Lv¯, and u = u ¯ , v = v¯ − v¯. Since any v¯, v¯ ∈ Ls (0, T ; W 2,4/3 (M )), let u ¯ = L¯ v, u ¯−u s 2,4/3 s 2,4/3 ¯ ∈ L (0, T ; W L maps L (0, T ; W (M )) into itself, u¯, u (M )), and u satisfies (4.4.19) and (4.4.17) with g = 0 and u0 = 0. It is easy to see from (4.4.20) that 1 kukLs (0,T ;W 2,4/3 (M )) ≤ kvkLs (0,T ;W 2,4/3 (M )) 2 if 1 is sufficiently small. Therefore, L is a contraction mapping. By the contraction mapping principle, there exists a unique solution u ∈ Ls (0, T ; W 2,4/3 (M )) to problem (4.4.16) and (4.4.17) for any s ∈ [2, 4].

4.4.2

Proof of Regularity of Solution to L–L Equation

1. Weak solution with finite energy In this section we shall prove Theorem 4.4.1. In the first step, we will show that if u ∈ WT is a weak solution of (4.4.1) and (4.4.2) with finite energy, then u ∈ YT ∩ WT . Let W ij = ui duj − uj dui , i, j = 1, 2, 3. (4.4.21) Since E(u(t)) ≤ E0 for a.e. t ∈ [0, T ], we have

W ij ∈ L∞ ([0, T ], L2 (M, ∧1 )) and ij

kW k

L∞ ([0,T ],L2 )

q

≤ 2 E0 .

By using the fact that |u| = 1 and H´elein’s trick [90], Eq. (4.4.1) can be written as ∂t ui − β∆M ui = α(u × ∂t u)i + β

3 X

j=1

W ij · duj .

(4.4.22)

Applying the Hodge decomposition theorem (Lemma 4.4.1) to W ij at each time slice t ∈ (0, T ), one may find that Aij ∈ L∞ ([0, T ], H 1 (M, ∧0 )) and B ij ∈ L∞ ([0, T ], H 1 (M, ∧2 )) such that W ij = dAij + d∗ B ij

for a.e. t ∈ [0, T ],

(4.4.23)

and ij

ij

ij

q

kA kL∞ ([0,T ],H 1 ) + kB kL∞ ([0,T ],H 1 ) ≤ CkW kL∞ ([0,T ],L2 ) ≤ C E0 . From (4.4.23), (4.4.21) and (4.4.1), we get

∆Aij = d∗ W ij = ui ∆M uj − uj ∆M ui α + β −1 (ui ∂t uj − uj ∂t ui ) − (ui (u × ∂t u)j − uj (u × ∂t u)i ) β 2 2 0 ∈ L ([0, T ], L (M, ∧ )).

(4.4.24)

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178

By the Calderon–Zygmund inequality, dAij ∈ L2 ([0, T ], H 1 (M, ∧0 )) and kdAij kL2 ([0,T ],H 1 ) ≤ Ck∂t ukL2 (M ×[0,T ]) ,

(4.4.25)

where C > 0 is a constant depending only on α, β and M . On the other hand, since B ij ∈ L∞ ([0, T ], H 1 (M, ∧2 )), for any ε > 0, we always can find B1ij ∈ L∞ ([0, T ], C ∞ (M, ∧2 )) and B2ij ∈ L∞ ([0, T ], H 1 (M, ∧2 )) such that B ij = B1ij + B2ij ,

(4.4.26)

kB2ij kL∞ ([0,T ],H 1 ) < ε.

(4.4.27)

and Now, by using (4.4.23) and (4.4.26), we may rewrite Eq. (4.4.22) in the following form: i

i

∂t u − β∆M u = β where f i = α(u × ∂t u)i + β

3 X

j=1

3 X

j=1

d∗ B2ij · duj + f i ,

dAij · duj + β

3 X

j=1

(4.4.28)

d∗ B1ij · duj .

(4.4.29)

By the Sobolev embedding theorem and (4.4.25),

3

X



dAij · duj

j=1

≤

≤

4

L2 ([0,T ],L 3 )

3 X

kdAij kL2 ([0,T ],L4 ) kduj kL∞ ([0,T ],L2 )

3 X

kdAij kL2 ([0,T ],H 1 ) kduj kL∞ ([0,T ],L2 ) ≤ C1 E0 ,

j=1

j=1

q

(4.4.30)

where C1 > 0 is a constant depending only on M and k∂t ukL2 (M ×[0,T ]) . On the other hand, noticing that B1ij ∈ L∞ ([0, T ], C ∞(M, ∧2 )), we have kd∗ B1ij ||L2 ([0,T ],L4 ) ≤ C(T, M )kd∗ B1ij kL∞ (M ×[0,T ]) ≤ C(T, M ). This leads to the estimate

3

X

ij

d∗ B1 · duj

j=1

≤

3 X

j=1

kd

∗

4

L2 ([0,T ],L 3 )

B1ij kL2 ([0,T ],L4 ) kduj kL∞ ([0,T ],L2 )

q

≤ C2 E0 ,

(4.4.31)

where C2 > 0 is a constant depending only on M , T . It is obvious that 4

u × ∂t u ∈ L2 ([0, T ], L2 ) ⊂ L2 ([0, T ], L 3 ).

(4.4.32)

Landau–Lifshitz Equations and Harmonic Maps

179

From (4.4.29)–(4.4.32), we get that 4

f i ∈ L2 ([0, T ], L 3 ),

i = 1, 2, 3,

(4.4.33)

and kf i kL2 ([0,T ],L 34 ) is bounded by a constant depending only on M , T , ε, E0 and k∂t ukL2 (M ×[0,T ]) . Applying Lemma 4.4.4 to problem (4.4.28) and (4.4.2), from (4.4.27) and (4.4.33), we can conclude that if ε ≤ ε0 , where ε0 is determined in Lemma 4.4.4, then any solution of (4.4.28) and (4.4.2) in XT is in YT . On the other hand, if u is a weak solution of (4.4.28) and (4.4.2), then u is also a solution of (4.4.28) and (4.4.2) in XT . Therefore, u ∈ YT . 2. Uniqueness of weak solution In this step we will show that if u ∈ YT ∩ WT is a weak solution of (4.4.1) and (4.4.2), then u ∈ L4 (0, T ; W 2,4/3 ) ∩ WT . For u ∈ YT ∩ WT , u is defined a.e. on M × [0, T ]. From (4.4.1), one can have u × ∂t u = βu × ∆u − α∂t u. Inserting this into (4.4.1), we have (1 + α2 )∂t u = β∆u + β|du|2 u + αβu × ∆u, i.e. where β˜ = β/(1 + α2 ) and

2 ˜ ∂t u − G(u)∆u = β|du| u,

(4.4.34)

β −αβu3 αβu2 1  β −αβu1  G(u) =  αβu3 . 2 1+α −αβu2 αβu1 β 



(4.4.35)

Since |u| = 1, for any  > 0 we can decompose u into the form u = m + n,

(4.4.36)

where m ∈ C ∞ (M × (0, T )), n ∈ L∞ (M × (0, T )) with kmkL∞ (M ×(0,T )) ≤ CkukL∞ (M ×(0,T )) ,

(4.4.37)

knkL∞ (M ×(0,T )) ≤ .

(4.4.38)

2 ˜ ∂t u − G(m)∆u = G(n)∆u + β|du| u.

(4.4.39)

Inserting (4.4.36) into (4.4.34), we get that

2 ˜ Letting G1 (x, t) = G(m(x, t)), G2 (x, t) = G(n(x, t)), and g(x, t) = β|du| u, (4.4.39) takes the same form as (4.4.16). Moreover,

kG2 kL∞ (M ×(0,T )) ≤ CknkL∞ (M ×(0,T )) ,

kgkL4 (0,T ;L4/3 ) ≤ CkdukL∞ ([0,T ],L2 ) kdukL2 ([0,T ],L4 ) .

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180

We can see from (4.4.35), (4.4.37) and (4.4.38), that if  is sufficiently small, then the assumptions in Lemma 4.4.5 and (4.4.18) are satisfied. Therefore, by Lemma 4.4.4, problem (4.4.39) with the initial condition (4.4.2) admits an unique solution v ∈ L4 (0, T ; W 2,4/3 ) ⊂ L2 (0, T ; W 2,4/3 ) and the solution in space L2 (0, T ; W 2,4/3 ) is also unique. On the other hand, it is obvious that if u ∈ YT ∩ WT is a weak solution of (4.4.1) and (4.4.2), then u is a solution of (4.4.39) and (4.4.2) in L2 (0, T ; W 2,4/3 ). Therefore, u = v ∈ L4 (0, T ; W 2,4/3 ). 3. Almost everywhere smoothness In this step we will complete our proof for Theorem 4.4.1 by showing the following lemma: Lemma 4.4.5 Let u ∈ L4 (0, T ; W 2,4/3 ) ∩ WT and v ∈ L4 (0, T ; W 2,4/3 ) ∩ WT be two solutions of (4.4.1) and (4.4.2). Then, u = v a.e. on M × [0, T ]. Proof. For simplicity we assume α = 1, β = 1. Now w = u − v solves the equation wt − ∆M w = u × ∆M w + w × ∆M u + u(|du|2 − |dv|2 ) + |dv|2 w,

(4.4.40)

with the condition w(x, 0) = 0. Let |dU |2 = |du|2 + |dv|2 . Testing (4.4.40) by w and integrating it by parts, we obtain 1 1d kwk2L2 (M ) + 2 dt 2

Z

M

|dw|2 ≤ C

Z

M

|w|2 |dU |2 .

(4.4.41)

Using the Sobolev inequality kwk4L4 ≤ 4kwk2L2 kdwk2L2 , we have Z

2

M

Z

2

|w| |dU | ≤ C ≤

1 4

Z

M

M

|dU |

4

!1/2

|dw|2 + C

Z

Z M

M

|w|

|dU |4

2

Z

!1/2

M

Z

M

|dw|

2

!1/2

|w|2 .

This combined with (4.4.41) implies 1 f 0 (t) + g(t) ≤ Cp(t)f (t) 2 where f (t) = M |w|2 , g(t) = M |dw|2 and p(t) = M |dU |4 . Since u, v ∈ L4 (0, T ; W 2,4/3 ), p(t) ∈ L1 (0, T ). We have from the Gronwall inequality that f (t) = 0 for a.e. t ∈ [0, T ], and hence w = 0 a.e. on M × [0, T ]. The proof of the lemma is complete. The proof of Theorem 4.4.1 is complete. R

R

R

Landau–Lifshitz Equations and Harmonic Maps

4.5

181

Ginzburg–Landau Approximation to Landau–Lifshitz Equations

In this section, we will discuss the Ginzburg–Landau approximation to Landau– Lifshitz equations to obtain the existence and partial regularity for the global weak solutions to the initial and boundary value problem. We will consider the weighted Landau–Lifshitz flow in two dimensions and follow the ideas of Paul Harpes’ [89] for a(x) = 1. 1 1 ∂t u − u × ∂t u − ∇ · (a(x)∇u) = a(x)|∇u|2 u in Ω × R+ , 2 2 S u = u0 on Ω × {0} ∂Ω × R+

(4.5.1) (4.5.2)

where domain Ω ⊂ R2 is open, bounded and smooth. The initial and boundary data u0 is assumed to be a smooth map into the standard sphere S 2 ⊂ R3 . In the classical sense, Eq. (4.5.1) is equivalent to ut = u × ∇ · (a(x)∇u) − u × (u × ∇ · (a(x)∇u)).

(4.5.3)

¯ × R+ → R3 to Landau–Lifshitz flow The Ginzburg–Landau appoximations u : Ω (4.5.1) are solutions of 1 1 1 ∂t u − u × ∂t u − ∇ · (a(x)∇u ) = 2 a(x)(1 − |u |2 )u 2 2  S u = u0 on (Ω × {0}) (∂Ω × R+ ),

in Ω × R+ ,

(4.5.4) (4.5.5)

where a(x) is a positive smooth function satisfying 0 < m ≤ a(x) ≤ M . For small  > 0, we can see that maps {u } approximate the weighted Landau–Lifshitz flow in Ω × R+ . For fixed  > 0, a smooth solution to (4.5.4) and (4.5.5) on Ω × R+ exists 3 1,2 T ∞ 1,2 and if u0 ∈ H 1,2 (Ω; S 2 ) ∩ H 2 (∂Ω; S 2 ), it is unique in Hloc L (H 1,2 ) := Hloc (Ω × T R+ ; R3 ) L∞ (H 1,2 ). Existence is obtained by Galerkin’s method. C ∞ regularity follows from a standard bootstrap argument. The total energy of the approximate flow at time t ≥ 0 is defined by G (u (t)) := where

Z

Ω

g (u (x, t))dx,

"

(4.5.6) #

1 1 g (u (x, t)) = a(x) |∇u |2 + 2 (1 − |u |2 )2 . 2 4

(4.5.7)

We will see that the total energy of the  approximation always decreases. We define the local energy by G (u (t), BRΩ (x0 )) :=

Z

BR (x0 )

T

Ω

g (u (x, t))dx

(4.5.8)

which may concentrate at space–time points (x0 , t0 ) as  & 0, either for fixed t = t0 or for variable t % t0 or t & t0 . It characterizes the local “asymptotic behavior”

Landau–Lifshitz Equations

182

of the weighted flow. Here asymptotic refers to limit  & 0. We will show that all the derivatives of the family of maps {u }>0 are locally uniformly bounded on the regular set Reg{u }>0 consisting of all points z0 = (x0 , t0 ) ∈ Ω × (0, ∞) for which there is R0 = R0 (z0 ) such that lim sup

sup

&0

t0 −R20
G (u (t), BRΩ0 (x0 )) < 0

(4.5.9)

for a constant 0 > 0 that will be determined later in Lemmas 4.5.9 and 4.5.11. Complement S({u }>0 ) := Ω × R+ \Reg{u }>0 is referred to as the energy-concentration set. We will show that the approximation solutions converge to the global weak solution of (4.5.1) with the Dirichlet condition. This convergence is smooth in Reg{u }>0 , while the energy-concentration set is closed, with the locally finite parabolic Hausdorff measure. Delicate energy inequality shows that, in fact, the singular set consists of finitely many points as observed in [29, 89]. Such Ginzburg–Landau penalty method was first used to study the harmonic map heat flow in higher dimensions by Chen and Struwe in [34]. So we call the solution obtained the Chen–Struwe solution.

4.5.1

Estimates for Strong Parabolic System

In this section, we will show that under the uniform smallness condition (4.5.9) on the local energy, all higher derivatives of u are locally and uniformly bounded. We first recall some facts about Lp estimates for a strongly parabolic system and C α estimates for the parabolic system in the divergence form. Then, we derive the L∞ and Lp bounds for the right-hand side of (4.5.4) which are necessary to get the uniform bounds of 12 a(x)(1 − |u|2 ) for Lp estimates. Finally, we prove that all the derivatives of u and 12 a(x)(1 − |u |2 ) are locally uniformly bounded if the energy density satisfies the condition lim sup&0 supPR (z0 ) g (u (x, t)) < C0 , which may be verified under the uniformly smallness condition (4.5.9). For z0 = (x0 , t0 ), denote T T Ω PR (z0 ) = BR (x0 ) × (t0 − R2 , t0 ) and PδR := (BδR (x0 ) Ω) (t0 − δ 2 R2 , t0 ). We rewrite Eq. (4.5.4) in the form 1 ∂t u − M (u )a(x)4u − M (u )∇a(x) · ∇u = M (u ) 2 a(x)(1 − |u |2 )u := f (u ),  where M (u ) satisfies m|ξ|2 ≤ ξ T a(x)M (u )ξ =

2a(x) {|ξ|2 + (u · ξ)2 } ≤ 2M |ξ|2 , ∀ ξ ∈ R3 . (4.5.10) (1 + |u|2 )

We may therefore write (4.5.4) as L (u ) := ∂t u − M (u )a(x)4u − M (u )∇a(x) · ∇u = f (u ), where the coefficient-matrix M (u ) is smooth with respect to u . Note that L defines a strongly parabolic system in the Petrovskii sense [101]. So Lp global and local estimates hold for such system. We list two a priori Lp estimates concerning the strongly parabolic system in the Petrovskii sense.

Landau–Lifshitz Equations and Harmonic Maps

183

Lemma 4.5.1 (Global Lp estimates) Let f ∈ Lp (Ω × [0, T ]; R3 ) and u0 ∈ H 2,p (Ω; R3 ). A solution of L (v) = f in (Ω × (0, T ); R3 ) with v = u0 on S (Ω × {0}) (∂Ω × (0, T )) satisfies kvkHp2,1 (Ω × [0, T ]) ≤ Cp (Ω, T, ωu )(kf kLp (Ω×[0,T ]) + ku0 kH 2,p (Ω) ).

(4.5.11)

Lemma 4.5.2 (Local Lp estimates) Let f ∈ Lp (Ω×[0, T ]; R3 ) and u0 ∈ H 2,p (Ω; R3 ). S A solution of L (v) = f in (Ω × (0, T ); R3 ) with v = u0 on (Ω × {0}) (∂Ω × (0, T )) satisfies kvkHp2,1 (P Ω (z0 )) ≤ C˜p (R, Ω, T, ωu ) kf kLp (PRΩ (z0 )) 

δR

T ku k + kvkLq (PδR Ω (z )) + δB 0 ∂Ω 0 R

2− 1 ,p Ω H p (BR

T

∂Ω)



(4.5.12)

for all 1 ≤ q ≤ p. Here, δBR T ∂Ω = 1 if BR ∂Ω 6= ∅ and 0 otherwise. The trace theorem of course implies ku0 k 2− p1 ,p ≤ ku0 kH 2,p (Ω) . Constants Cp and C˜p depend H

(∂Ω)

T

on the indicated quantities and additionally on the uniform lower and upper bounds for the eigenvalues of a(x)M (u ), that is m and 2M, which are chosen to be independent of  > 0. Note that Cp and C˜p also depend on the modulus of continuity of the coefficients of the leading term-, i.e. the modulus of continuity ωu of u . The equation can also be written in the divergence form

L (v) := ∂t v−a(x)∇·(M (u )∇v)+a(x)(∂k M (u )∂k u )∂k v−M (u )∇a(x)·∇v = f (u ). Lemma 4.5.3 If we assume lim sup sup |∇u | < ∞, &0

(4.5.13)

Ω PR

γ

Ω then v ∈ C γ, 2 (PδR ; R3 ) for some γ ∈ (0, 1) and any δ ∈ (0, 1). If the right-hand side p Ω 3 f ∈ L (PR ; R ) with p > 2, we have the following estimate for the mixed H¨ older-norm Ω of v on PδR

kvkC γ, γ2 (P Ω ;R3 ) ≤ C(f ),

(4.5.14)

δR

where bound C(f ) depends on the parabolicity constants, δ, sup PRΩ |u |, kf kLp (PRΩ ) , T and also depends on ku0 kC γ (BR T ∂Ω) if BR ∂Ω = 6 ∅. If (4.5.13) holds and kf kLp (PRΩ ) or supPRΩ |∇u | is uniformly bounded with respect to  > 0, then estimate (4.5.14) holds for u and is uniform in  > 0. Assumption supPRΩ |∇u | ≤ C, however, does not include the time derivatives. (4.5.14) enables us to obtain bounds on the modulus of continuity with respect to time variable. Thus Ω the modulus of continuity of u on PδR is bounded from above independent of  > 0. Therefore estimates (4.5.11) and (4.5.12) are uniform in  > 0.

Landau–Lifshitz Equations

184

4.5.2

L∞ and Lp Bounds for

1 a(x)(1 2

− |u |2 )

Multiplying (4.5.4) by −u , we obtain 1 1 1 1 1 ∂t ρ − a(x)4ρ − ∇a(x)∇ρ + 2 a(x)ρ = a(x)|∇u |2 + a(x) 2 ρ2 , 4 2 2  

(4.5.15)

where ρ = 1 − |u|2 . On the parabolic boundary, ρ = 1 − |u0 |2 = 0. We get ρ ≥ 0 in Ω × R+ by using the maximum principle, i.e. |u | ≤ 1. Consider the following auxiliary problem: ∂t f − a(x)4f − ∇a(x) · ∇f + |f | ≤ a where ∂PR = BR (0) × {−R2 }

S

1 a(x)f = a(x)g 2 on ∂PR ,

in PR ,

(4.5.16) (4.5.17)

∂BR (0) × [−R2 , 0].

Lemma 4.5.4 Let a > 0,  ∈ (0, 1), g ∈ C 0 (PR ), with 2 supPR |g| ≤ a. Let T f ∈ C 0 (PR ) C 2 (PR ) be a solution of (4.5.16) and (4.5.17). Then there exists R0 depending on m, M, and M1 = maxx∈Ω¯ |∇a(x)| such that for any δ ∈ (0, 1), R ∈ (0, R0 ) we have ! 2a 1 1 2 2 4 |f | ≤ sup |g| + 2 exp − (1 − δ ) R on PδR . (4.5.18) Ω 2   PR Proof. Taking w(x, t) = 2a exp[− 1 (R2 − x2 )(R2 + t)], we have 2 [∂t w − a(x)4w − ∇a(x) · ∇w] + a(x)w

= w[a(x) − (R2 − x2 ) − a(x)4|x|2 (R2 + t)2

− a(x) · 4(R2 + t) − ∇a(x) · 2x(R2 + t)]

≥ w(m − R2 − 4MR4 − 4MR2 − 2RM1 R2 ). Therefore, there exists R0 depending on m, M , M1 such that if R ∈ (0, R0 ),  ∈ (0, 1) 2 [∂t w − a(x)4w − ∇a(x) · ∇w] + a(x)w > 0 w = 2a

in PR ,

on ∂PR

holds. For f1 = f − 2 supPR |g| and f2 = f + 2 supPR |g|, we have |f1 | ≤ 2a and |f2 | ≤ 2a on ∂PR and 2 [∂t f1 − a(x)4f1 − ∇a(x) · ∇f1 ] + a(x)f1 = 2 a(x)g − a(x)2 sup |g| ≤ 0 PR

2

≤  [∂t w − a(x)4w − ∇a(x) · ∇w] + a(x)w.

Landau–Lifshitz Equations and Harmonic Maps

185

Therefore, we obtain by the comparison principle that f1 − w ≤ 0

in PR .

(4.5.19)

f2 + w ≥ 0

in PR .

(4.5.20)

Similarly, we have Combining (4.5.19) and (4.5.20), one gets −w − 2 sup |g| ≤ f ≤ w + 2 sup |g|, Ω PR

Ω PR

which yields the desired conclusion !

1 2a 1 1 |f | ≤ sup |g| + 2 w ≤ sup |g| + 2 exp − (1 − δ 2 )2 R4 . 2 Ω Ω     PR PR If BR Ω 6= ∅ and f ≡ 0 on BR boundary. T

T

∂Ω, we have the local estimate near the

Lemma 4.5.5 Consider a smooth domain Ω ⊂ R2 , a > 0,  ∈ (0, 1), g ∈ C 0 (PR ), T with 2 supPR |g| ≤ a. Let f ∈ C 0 (PR ) C 2 (PR ) be a solution of (4.5.16) and (4.5.17) T and f = 0 on ∂Ω PR . Then there exist R0 , depending on m, M and M1 such that for any δ ∈ (0, 1), R ∈ (0, R0 ), we have 2a 1 |f | ≤ sup |g| + 2 exp 2 Ω   PR

1 − (1 − δ 2 )2 R4 

!

Ω on PδR .

(4.5.21)

In the sequel, we will derive a priori Lp estimates for Eqs. (4.5.16) and (4.5.17). First we give L1 estimates. Lemma 4.5.6 Let Ω ⊂ R2 be as bounded smooth domain, a > 0,  ∈ (0, 1), g ∈ T C 0 (PR ). For any nonnegative function f ∈ C 1 (Ω × (0, T )) C 2 (Ω × (0, T )) satisfying 1 a(x)f ≤ a(x)g in Ω × (0, T ), 2 S on (Ω × {0}) (∂Ω × (0, T )),

∂t f − a(x)4f − ∇a(x) · ∇f + f =0

(1) there exists some constant c > 0 only depending on M and m such that 1 k 2 f kL1 (Ω×(0,T )) ≤ ckgkL1 (Ω×(0,T )) ; 

(4.5.22)

(2) for any R, ρ > 0 and z0 = (x0 , t0 ) ∈ Ω × (0, T ) with R2 + ρ2 < t0 , we have Z

Ω (z ) PR 0

1 |f |dz ≤ c 2

Z

Ω PR+ρ (z0 )

!

c |g| + 2 |f | dz. ρ

(4.5.23)

Landau–Lifshitz Equations

186

Proof of (1). Multiplying Eq. (4.5.16) by √

f f 2 +δ 2

, we attain

f f f − a(x)4f · √ 2 − ∇a(x) · ∇f √ 2 2 2 +δ f +δ f + δ2 1 f f + 2 a(x)f √ 2 = a(x)g · √ 2 . 2  f +δ f + δ2

∂t f · √

f2

(4.5.24)

(4.5.24) can be written as |f | |∇f |2 f2 √ ∂t |f | · √ 2 + a(x) · 1 − f 2 + δ2 f + δ2 f 2 + δ2

!

+

1 f2 √ a(x) 2 f 2 + δ2

!

f f = a(x)g · √ 2 + div a(x)∇f √ 2 . 2 f +δ f + δ2

(4.5.25)

Integrating (4.5.25) over Ω×(0, t), letting δ → 0, and using the monotone convergence theorem, we have sup 0≤t≤T

Z

Ω

|f (x, t)|dx +

Z

T 0

Z

a(x) Ω

where we have used the fact that

RT R 0

Ω

1 |f (x, t)|dxdt ≤ 2 div(a(x)∇f √

Z

T 0

f f 2 +δ 2

Z

a(x)|g(x, t)|dxdt, Ω

)dxdt = 0 by the divergence

theorem. From the above inequality, we obtain Z

Taking C =

M , m

T 0

Z

Ω

MZTZ 1 |f (x, t)|dxdt ≤ |g(x, t)|dxdt. 2 m 0 Ω

we have

1

2 f



L1 (Ω×(0,T ))

≤ ckgkL1 (Ω×(0,T )) .

Proof of (2). Multiplying Eq. (4.5.16) by √

f (x, t)φ2 (x)η(t) f 2 +δ 2

where the cutoff

function φ(x) satisfies 0 ≤ φ(x) ∈ C ∞ (Ω) with spt φ ⊂ BR+ρ (x0 ) and φ ≡ 1 on BR (x0 ), η(t) satisfies η(t) ∈ C ∞ (R+ ) with 0 ≤ η(t) ≤ 1, η(t0 − R2 − ρ2 ) = 1 and η(t) ≡ 1, |∇φ| ≤ ρc , |∇2 φ| ≤ ρc2 , and |∂t η| ≤ ρc2 , we get f f f φ2 η − a(x)4f · √ 2 φ2 η − ∇a(x) · ∇f √ 2 φ2 η 2 2 +δ f +δ f + δ2 1 f f + 2 a(x)f √ 2 φ2 η = a(x)g · √ 2 φ2 η. (4.5.26) 2  f +δ f + δ2

∂t f · √

f2

(4.5.26) can be written as |f | |∇f |2 φ2 η f2 √ ∂t (|f |φ η) √ 2 + a(x) · 1 − f 2 + δ2 f + δ2 f 2 + δ2 2

f φ2 η f φ2 η √ = a(x)g · √ 2 + div a(x)∇f f + δ2 f 2 + δ2 f − a(x)∇f √ 2 2φ∇φη + |f |φ2 ∂t η. f + δ2

!

+

f 2 φ2 η 1 √ a(x) 2 f 2 + δ2

!

(4.5.27)

Landau–Lifshitz Equations and Harmonic Maps

187

Integrating (4.5.27) over Ω × (0, t) and letting δ → 0, we get sup t0 −(R2 +ρ2 )≤t≤t0

≤

Z

Ω PR+ρ

Z

|f (x, t)|dx + R

BΩ

Z

1 a(x) |f (x, t)| Ω 2 PR !

c a(x)|g(x, t)| + 2 |f (x, t)| . ρ

Recalling the assumption m = minx∈Ω |a(x)| ≤ a(x) ≤ maxx∈Ω |a(x)| = M , we arrive at the conclusion Z

Ω (z ) PR 0

1 |f |dz ≤ c 2

Z

!

c |g| + 2 |f | dz. ρ

Ω PR+ρ (z0 )

Lemma 4.5.7 Let Ω ⊂ R2 be as above, g ∈ L1 Lp (Ω × (0, T )) for p ≥ 2 and  > 0. T For any nonnegative function f ∈ C 1 (Ω × (0, T )) C 2 (Ω × (0, T )) satisfying T

1 a(x)f ≤ a(x)g in Ω × (0, T ), 2 [ on (Ω × {0}) (∂Ω × (0, T ))

∂t f − a(x)4f − ∇a(x) · ∇f + f =0

∀ δ ∈ (0, 1), z0 = (x0 , t0 ) ∈ Ω × (0, T ] with 0 < R2 < t0 , we have

1

2 f



Ω (z ) Lp (PδR 0

2/p Ω (z ) + c2  ≤ c1 kgkLp (PδR , 0

(4.5.28)

where c1 = c1 (p, m, M) and c2 = c2 (kgkLp (PRΩ ) , kf kL2p−1 (PR (z0 )) , p, δ, R, m, M). Proof. Multiplying the equation by f |f |2s−2 φ2 η, where s ≥ 1, and taking cutoff functions φ and η as in the proof of Lemma 4.5.6, we have ∂t f f |f |2s−2 φ2 η − a(x)4f f |f |2s−2φ2 η − ∇a(x) · ∇f f |f |2s−2φ2 η 1 + 2 a(x)f f |f |2s−2 φ2 η = a(x)gf |f |2s−2φ2 η. 

(4.5.29)

(4.5.29) can be written as 1 2s − 1 1 ∂t (|f |2s φ2 η) + a(x) 2 |∇|f |s |2 φ2 η + 2 a(x)|f |2s φ2 η 2s s  1 = div(a(x)∇f f |f |2s−2 φ2 η) + |f |2s φ2 ∂t η 2s 2s−2 2 + a(x)gf |f | φ η − a(x)∇|f ||f |2s−1 2∇φφη.

(4.5.30)

We now estimate the last two terms of (4.5.30). By the Young inequality we have ||g|f |f |2s−2| ≤

1 2s − 1 2s 1 |f | + (22 )2s−1 |g|2s 2 2 2s 2s

Landau–Lifshitz Equations

188

and |∇|f ||f |

2s−1

! 1 √ s √ s 2∇φφη| = 2 ∇|f | φ η · (|f | ∇φ η) s

≤

which lead to

2s − 1 2 |∇|f |s |2 φ2 η + |f |2s |∇φ|2 η 2 2s 2s − 1

2s − 1 1 1 1 ∂t (|f |2s φ2 η) + a(x) |∇|f |s|2 φ2 η + 2 a(x)|f |2s φ2 η 2 2s 2s 2 2s 1 ≤ div(a(x)∇f f |f |2s−2 φ2 η) − a(x)(22 )2s−1 |g|2s φ2 η 2s 2 + |f |2s [a(x)|∇φ|2 η + φ2 |∂t η|]. 2s − 1

(4.5.31)

Ω , we Setting p = 2s, multiplying (4.5.31) by (2s) · ( 12 )p−1 , and integrating it over PR+ρ get, for p ≥ 2,

sup t≥−R2 −ρ2

+

Z

Z

1 2

Ω BR+ρ

1 2

Ω PR+ρ

≤ C(p, m, M)

!p−1

!p−1

(Z

|f |p φ2 ηdx p 2

2 2

|∇|f | | φ ηdz +

Ω PR+ρ

|g|pφ2 η +

Z

Z

Ω PR+ρ

1 2

Ω PR+ρ

1 2

!p−1

!p

|f |pφ2 ηdz )

|f |p [|∇φ|2 η + φ2 |∂t η|] dz. (4.5.32)

Hence we obtain Z

1 2

Ω PR

Let p = implies

k 2

!p

p

|f | dz ≤ C(p, m, M)

c |g| +  Ω ρ2 PR+ρ p

2

Z

1 2

Ω PR+ρ

+ 1 for k ∈ N . Then H¨older’s inequality for q1 = Z

Ω PR+ρ

1 |f | 2

Now (4.5.33) yields

1 p

2 f

 p

(Z

Ω) L (PR

≤ C(p, m, M)

Noticing that derive that

2p−2 2p−1



1 p

2f

 p

"

!p−1

!p

2p−1 2p−2

|f |

p

)

dz. (4.5.33)

and q2 = 2p − 1



1 p−1

|f |dz ≤ 2 f

kf kL2p−1 .

 p− 1 L

kgkpLp (P Ω ) R+ρ

2

(p− 1 )· 2p−2

#

c

1

2 2p−1 kf kL2p−1 (PR+ρ Ω + 2 2 f ) . (4.5.34) ρ  Lp− 12 (P Ω ) R+ρ

< 1, we can iterate (4.5.34) for k steps and use Lemma 4.5.6 to

Ω) L (PR



≤ C kgkpLp (P Ω

R+(k+1)ρ

, kf kL2p−1 (P Ω )

R+(k+1)ρ



) , p, ρ, R .

Landau–Lifshitz Equations and Harmonic Maps

189

¯ ¯ If we use new cutoff function φ(x) and η¯(t) satisfying 0 ≤ φ(x) ∈ C ∞ (Ω) with ¯ spt φ(x) ⊂ BR+2ρ (x0 ) and φ¯ ≡ 1 on BR+ρ (x0 ) and η¯(t) satisfying η¯(t) ∈ C ∞ (R+ ) with ¯ ≤ c, 0 ≤ η¯(t) ≤ 1, η¯(t0 − (R + ρ)2 − ρ2 ) = 1, η¯(t) ≡ 1 if t ≥ t0 − (R + ρ)2 , |∇φ| ρ ¯ ≤ c2 , and |∂t η¯| ≤ c2 , we get, by the same iteration, that |∇2 φ| ρ ρ

1 p

2f

 p

Ω L (PR+ρ )

Taking ρ :=

δR k+2



≤ C kgkpLp (P Ω

R+ρ+(k+1)ρ



(4.5.35)

) , p, ρ, R .

and then substituting (4.5.35) into (4.5.34), one deduces

1 p

2 f

 p

Ω) L (PR

where c2 = c2 (kgkLp (P Ω

(1+δ)R

),

Rnew = (1 + δ)R and δnew =

4.5.3

R+ρ+(k+1)ρ

, kf kL2p−1 (P Ω )

h

≤ C(p, m, M) kgkpLp (P Ω kf kL2p−1 (P Ω

(1+δ)R

1 , 1+δ

),

(1+δ)R

i

+  2 C2 , )

(4.5.36)

p, δ, R). The lemma follows if one sets

i.e. Rnew δnew = R.

Higher Order Interior Estimates

We intend to claim that the higher derivatives of u and 12 a(x)(1 − |u |2 ) are locally and uniformly bounded in the interior point under the uniform smallness condition (4.5.9), where u are the solution to approximating equation. Lemma 4.5.8 Let u be a solution of (4.5.4) and assume lim sup sup g (u ) ≤ C0 , &0

(4.5.37)

Ω PR

where BR (x0 ) ⊂ Ω, 0 < R2 < t0 . Then for any δ ∈ (0, 1), we have lim sup ku kC k (PδR (z0 )) ≤ Ck &0

and



1

2 lim sup 2 (1 − |u | )

&0 

C k (P

δR (z0 ))

≤ C˜k

(4.5.38)

for all integers k ≥ 0. Constants Ck and C˜k depend on C0 , k, R, δ > 0, ka(x)kC k (Ω) . ∞ Proof. We prove this lemma by induction. q If k = 0, we have proved ku kL ≤ 1. 1 Using assumption (4.5.37), we obtain supPR a(x)|∇u | ≤ C0 , and 2 a(x)(1−|u|2 )2 ≤ C0 . Therefore, g = |∇u |2 + 12 ρ2 can be controlled by a multiple of C0 (m). Lemma 4.5.4 implies k 12 (1 − |u |2 )kL∞ ≤ C˜0 . For k = 1, since the conclusions for k = 0 hold, i.e., lim sup&0 ku kLp (PδR (z0 )) ≤ Ck and lim sup&0 k 12 (1−|u |2 )kLp (PδR (z0 )) ≤ C˜k , using α the C α, 2 estimate of strongly parabolic systems in the divergence form, we know there exists γ ∈ (0, 1) such that ku kC γ, γ2 ≤ C(k 12 (1 − |u |2 )kLp ). From Lemma 4.5.4, we know that k 12 1 − |u |2 kL∞ ≤ C0 . Using the Wp2,1 estimate, we get

ku kWp2,1 (PδR )



1

2 ≤ CP (Ω, T, wu ) 2 a(x)(1 − |u | )



Lp (P

R)

+ ku kLq (PR )

!

Landau–Lifshitz Equations

190

for any 1 ≤ q ≤ p. From the Sobolev inequality, we have when p > 2 + 2 = 4, k∇u kC α (PδR ) ≤ C(m, p, α, δ)kukWp2,1 (PδR ) . We can take derivatives in (4.5.16) with respect to x to obtain, ∇(a(x)[|∇u |2 + 1 (1 − |u |2 )2 ]) ∈ Lp . Using (4.5.16) and Lemma 4.5.5, we get ∇( 12 (1 − |u |2 )2 ) ∈ 2 Lp . Taking derivatives with respect to x in (4.5.4), we can get u ∈ Wp3,1 . By the Sobolev embedding theorem, we know that u ∈ C 1,1 . Using (4.5.16) again, we get 1 (1 − |u|2 ) ∈ C 1,1 . By the standard bootstrap argument, we complete the proof. 2 The following Lemma proves that the boundedness of (4.5.37) can be verified by the smallness of the energy, i.e. the energy density is uniformly bounded in the regularity points of u . √ Lemma 4.5.9 There are constants C1 > 0, 0 > 0, R0 ∈ (0, min{1, t0 }) such that for solution u of (4.5.4) satisfying sup t0 −R0 2
there holds, ∀ δ ∈ (0, 1),

Z

BR0 (x0 )

g (u (x, t))dx < 0 ,

sup g (u ) ≤

PδR0 (z0 )

C (1 − δ)2 R02

where x0 ∈ Ω and BR0 ⊂ Ω. Proof. Without loss of generality, let (x0 , t0 ) = 0. We set PR := PR (0). For fixed  > 0, since the solution u of (4.5.4) is smooth, there exist σ ∈ [0, R0 ) such that (R0 − σ )2 sup g = max (R0 − σ)2 sup g , Pσ

0≤σ≤R0

Pσ

and there is some z = (x , t ) ∈ PR (z0 ) ∈ Pσ such that e := g (u (z )) = supPσ g . Setting ρ := 12 (R0 − σ ) such that Pρ (z ) ⊂ Pσ +ρ ⊂ PR0 , we have sup g ≤

Pρ (z )

1 4 [R0 −(σ +ρ )2 ] sup g ≤ (R0 −σ )2 e ≤ 4e . 2 2 [R0 − (σ + ρ ) ] (R − σ ) Pρ +σ (z ) 0 

√ −1 Setting r = e ρ , we can consider a rescaled map v = v(y, s) = u(x + e 2 y, t + √ e−1 ˜ := e . By computation,  s), (y, s) ∈ Pr . Thus v satisfies Eq. (4.5.4) with  g√e  (v )(0, 0) = 1 and supPr g√e  (v ) = 4. We now claim that r ≤ 2. If it holds, we can use the definition of r and set σ = δR0 to finish the proof: r2

"

√ 1 = (R0 − σ ) e 2

#2

≤4

and (R0 − δR0 )2 sup g (u ) = (R0 − σ )2 sup g (u ) ≤ 16, PδR0

Pσ

Landau–Lifshitz Equations and Harmonic Maps

191

which implies sup g (u ) ≤

PδR0

C . (1 − δ)2 R02

We prove the claim by the contradiction argument. Suppose r > 2. Since BR0 (x0 ) ⊂ Ω, all the derivatives of v are then bounded on P1 independently of  > 0. Indeed, if √ lim inf &0 e  > 0, from the equation 1 1 1 ∂t v − v × ∂t v − ∇ · (a(x)∇v ) = 2 a(x)(1 − |v |2 )v 2 2 ˜ the claim holds by using the Lp estimates and the fact that |v | ≤ 1. If √ lim inf &0 e  = 0, then the claim follows from the fact that sup Pr g√e  ≤ 4 and Lemma 4.5.8. In particular, all the derivatives of v are uniformly bounded. Thus q

∂t g˜(v ), |∇g˜(v )| ≤ C < ∞ in P1 .

1 Therefore, if we choose r0 = min{ 4C , 1}, we have

1 |g˜(v )(x, t) − g˜(v )(0, 0)| = |∂t g˜(v )(x0 , t0 )||t| + |∇g˜(v )(x0 , t0 )||x| < . 2 Using the differential mean-value theorem, we get g˜(v )(x, t) > g˜(v )(0, 0) −

1 1 > , 2 2

which implies 1=

g√e  (v )(0, 0)

2 ≤ 2 πr0 ≤ C∗ ≤ C∗

sup

Z

g√e  (v )dy

−r02 <s<0 Br0

sup t −r02 e−1 
sup r2 −( e0 

+σ2 )
Z

B√e r0 (x )

Z

B √r0

e

g (u )dx

+σ (x0 )

(x )

g (u )dx.

(4.5.39)

√ Setting 1 = min{ 12 , 2C1 ∗ }, since r = e ρ > 2 > r0 , we get √re0 + σ ≤ ρ + σ ≤ R0 and ( √re0 )2 + σ2 ≤ (ρ + σ )2 ≤ R02 . Hence, the right-hand side of (4.5.39) ≤ 1 ≤ 12 . This leads to a contradiction. Therefore, r ≤ 2.

4.5.4

Boundary Estimates

In this section, we will derive local boundary sup-estimates for the energy density, thereby the Wp2,1 -estimates for u and Lp -estimates for 12 a(x)(1 − |u |2 ) near the boundary.

Landau–Lifshitz Equations

192

Lemma 4.5.10 Let u be a solution of (4.5.4) and (4.5.5) with u0 ∈ H 1,2 (Ω; S 2 ) H 2,P (∂Ω; S 2 ). Assume sup g (u ) ≤ C0

T

(4.5.40)

Ω PR

and BR (x0 )

T

∂Ω 6= ∅, 0 < R2 < t0 . Then for any δ ∈ (0, 1), we have

ku kWp2,1 (P Ω (z0 )) δR



1

≤ C1  2 (1 − |u |2 )



Ω (z )) Lp (PR 0

+ ku kL2 (PRΩ (z0 )) + ku0 k

H

1 ,p 2− p

Ω (z ) (BR 0

T

∂Ω)



,

where constant C1 depends on C0 , δ, R, p, Ω and ka(x)kL∞ (Ω) . Furthermore, for any δ ∈ (0, 1) we have

1

2

2 (1 − |u | )



Ω (z )) Lp (PδR 0 2

≤ C(p, ka(x)kL∞ )kg kLp (PRΩ ) +  p C(kg kLp (PRΩ ) , p, δ, R, ka(x)kL∞ (Ω) )

and

1

2

2 (1 − |u | )



Ω (z )) L∞ (PδR 0

≤ 4C0 + Oδ (),

where  7→ Oδ () is a function that depends on δ ∈ (0, 1) and lim&0 −k Oδ () = 0 for all k ∈ N. All the constants also depend on the parabolic constants. Proof. Assumption supPRΩ g (u ) ≤ C0 implies that lim sup&0 supPδR Ω k∇u k ≤ α C0 (m) < ∞. Therefore from the C estimate, there exists γ ∈ (0, 1) such that ku kC γ, γ2 (P Ω ) ≤ C(f ). Furthermore, supPRΩ 12 a(x)(1 − |u |2 )2 ≤ C. Therefore g = δR

1 a(x)(|∇u | 2

+

1 (1 2

− |u |2 )2 ) ∈ Lp (PRΩ ). From the Wp2,1 estimate, we have

ku kWp2,1 (P Ω (z0 )) ≤ C1 δR



1

2

2 a(x)(1 − |u | )



+ ku kL2 (PRΩ (z0 )) + ku0 k From Lemma 4.5.7, we have

1

2

2 (1 − |u | )



Ω (z )) Lp (PR 0

H

1 ,p 2− p

Ω (z ) (BR 0

T

∂Ω)

!

.

Ω (z )) Lp (PδR 0 2

≤ C(p, ka(x)kL∞ )kg kLp (PRΩ ) +  p C(kg kLp (PRΩ ) , p, δ, R, ka(x)kL∞ (Ω) ).

Landau–Lifshitz Equations and Harmonic Maps

193

From Lemma 4.5.5, we have

1

2

2 (1 − |u | )



Ω (z )) L∞ (PδR 0

!

2a 1 ≤ sup |g(u)| + 2 exp − (1 − δ 2 )2 R4 . Ω   PR

Noting that

"

1 g(u ) = |∇u | + 2 (1 − |u|2 )2  2

#

"

#

1 1 ≤ 4 |∇u |2 + 2 (1 − |u |2 )2 + Oδ () ≤ C0 (m) + Oδ (), 2 4 where lim→0 −k · 2a exp(− 1 (1 − δ 2 )2 R4 = 0, ∀ k ∈ N . The lemma follows. 2 The following lemma proves that the boundedness of (4.5.40) may be verified by the smallness of the energy, i.e. the energy density is uniformly bounded in the regularity points of u . Lemma 4.5.11 Let u be a solution to (4.5.4) and (4.5.5) with u0 ∈ H 1,2 (Ω; S 2 ) C 2 (∂Ω; S 2 ). There are constants C0 = C0 (ku0 kC 2 (∂Ω) , E0 , Ω) and 0 = 0 (ku0 kC 2 (∂Ω) , √ E0 , Ω) > 0, such that if for some z0 = (x0 , t0 ) and R0 ∈ (0, min{1, t0 }),

T

lim sup

sup

&0

t0 −R0 2
Z

BR0 (x0 )

then for any δ ∈ (0, 1), we have

T

Ω

lim sup sup g (u ) ≤ &0

Ω (z ) PδR 0 0

g (u (x, t))dx < 0 ,

C . (1 − δ)2 R02

The proof is similar to the interior case, though the computation is much more complicate. For more detail, we refer to [89].

4.5.5

Energy Estimates

In this section, we will prove that the total energy of the smooth weighted flow of (4.5.4) and (4.5.5) is decreasing. Recall that in the first section we have defined the R total energy G (u (t)) := Ω g (u (x, t))dx and the local energy G (u (t), BRΩ (x0 )) := R T BR (x0 ) Ω g (u (x, t))dx, respectively. Lemma 4.5.12 Let u be a solution of (4.5.4) and (4.5.5). Then, we have 1 G (u (T ) + 2

Z

T 0

Z

Ω

|∂t u |2 dxdt = G (u (0) = E(u0 ) = E0 .

(4.5.41)

Landau–Lifshitz Equations

194

Proof. We multiply Eq. (4.5.4) by ∂t u and integrate it over Ω to get Z

Ω

Z Z 1 1 2 |∂t u | dx − ∇ · (a(x)∇u ) · ∂t u dx = a(x)(1 − |u |2 )u · ∂t u dx 2 2 Ω Ω 

and then Z

Ω

1 ∂ Z 1 ∂ Z 1 |∂t u |2 dx + a(x)|∇u |2 = − a(x)(1 − |u |2 )2 . 2 ∂t Ω 2 ∂t Ω 42

Integrating the above equality over [0, T] leads to Z

T 0

1 1 1 |∂t u |2 dxdt + a(x) |∇u |2 + 2 (1 − |u |2 )2 (T ) 2 4 Ω Ω 2   Z 1 1 = a(x) |∇u0 |2 + 2 (1 − |u0 |2 )2 = G (u (0)) := E0 . 2 4 Ω 

Z

Z



(4.5.41) follows. The following lemma is about the estimate for the local energy. Lemma 4.5.13 Let u be a solution of (4.5.4) and (4.5.5). Then, for 0 ≤ T1 < T2 we have Ω G (u (T2 ), BRΩ (x0 )) ≤ G (u (T1 ), B2R (x0 )) +

C R2

Z

T2 T1

Ω G (u (t), B2R (x0 ))dt.

(4.5.42)

Proof. We multiply Eq. (4.5.4) by ∂t u φ2 , where φ is a cutoff function satisfying T T φ(x) ∈ Cc∞ (Ω), 0 ≤ φ(x) ≤ 1, φ ≡ 1 on BR (x0 ) Ω), φ ≡ 1 on BR (x0 ) Ω; φ ≡ 0 on T C T Ω (B2R (x0 ) Ω) , |∇φ| ≤ RC2 , and then integrate over B2R = B2R (x0 ) Ω to derive Z

1 |∂t u |2 φ2 dx − Ω ∇ · (a(x)∇u ) · ∂t u φ2 dx Ω B2R 2 B2R Z 1 = Ω 2 a(x)(1 − |u |2 )u · ∂t u φ2 dx. B2R  Z

Integrating by parts, we have Z

Ω B2R

1 ∂ |∂t u |2 φ2 dx + 2 ∂t = −2 1 ≤ 2

Z

Z

Ω B2R

Z

Z

Ω B2R

1 a(x)(1 − |u |2 )2 φ2 dx 42

a(x)∇u ∂t u ∇φφdx 2 2

Ω B2R

Ω B2R

1 ∂ a(x)|∇u |2 φ2 dx + 2 ∂t

|∂t u | φ dx + 8

Z

Ω B2R

a(x)2 |∇u |2 |∇φ|2 dx.

Integrating over [T1 , T2 ], using the assumption that a(x) ≤ maxx∈Ω |a(x)| = M and the property of the cut-off function φ, we get G (u (T2 ), BRΩ (x0 )) The lemma follows.

≤

Ω G (u (T1 ), B2R (x0 ))

C + 2 R

Z

T2 T1

Ω G (u (t), B2R (x0 ))dt.

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Lemma 4.5.14 ∀ η > 0, ∃ T0 > 0, R0 > 0, such that ∀ x0 ∈ Ω and ∀  > 0, sup G (u (t), BRΩ0 (x0 )) ≤ η

(4.5.43)

0≤t≤T0

holds. Proof. For each fixed η > 0, using the absolute continuity of integration, we can choose R0 small enough to guarantee Ω G (u (0), B2R (x0 )) = 0

Setting T1 = 0, T2 = T0 =

R20 η 2CE0

Z

B2R0 (x0 )

T

Ω

1 η a(x)|∇u0 |2 dx ≤ . 2 2

in (4.5.42) and choosing T0 small enough we attain

Ω G (u (t), BRΩ0 (x0 )) ≤ G (u (0), B2R (x0 )) 0 Z t C C η Ω + 2 G (u (t), B2R (x0 ))dt ≤ + 2 T0 E0 ≤ η. 0 R0 0 2 R0

Taking supremum for t over [0, T0 ], one has sup0≤t≤T0 G (u (t), BRΩ0 (x0 )) ≤ η. (4.5.43) is proved.

4.5.6

Hausdorff Measure Estimate for Singularity

First of all, it follows from energy estimates (4.5.41) and (4.5.42) that for any 0 ≤ s < t Ω G (u (t), BRΩ (x0 )) ≤ G (u (s), B2R (x0 )) +

C(t − s)E0 . R2

(4.5.44)

0 , where 0 is the constant from (4.5.9) and (4.5.11). We may assume Define δ0 := 2CE 0 0 < δ0 < 1, otherwise we can choose a larger C.

Lemma 4.5.15 Let u be a solution of (4.5.4) and (4.5.5) with u0 ∈ H 1,2 (Ω; S 2 ) C 2 (∂Ω; S 2 ). Then the following assertions are equivalent: (1) z0 = (x0 , t0 ) ∈ Reg({u }>0 ), (2) ∃ δ, R > 0 such that lim sup&0 supt0 −δ 0 such that limR&0 lim sup&0 supt0 −δ 0 such that lim sup&0 R12 tt00−R2 BRΩ (x0 ) g (u )dxdt < 14 δ0 0 , (5) ∃ δ, R > 0 such that lim sup&0 supt0 −δ
T

Sketch of the proof. We can easily prove them by using the energy estimates Lemma 4.5.9 and 4.5.11, which characterize the supnorm of energy density under the smallness energy condition. For the details, we refer to [89]. Corollary 4.5.1 Let u be a solution of (4.5.4) and (4.5.5) with u0 ∈ H 1,2 (Ω; S 2 ) C 2 (∂Ω; S 2 ). Let {i }i be a sequence with i & 0 as i → ∞. Then the following hold : (1) Reg({u } ) and Reg({ui }i ) are open in Ω × R+ . (2) There exists some T0 > 0 such that Ω × (0, T0 ) ⊂ Reg({u } ).

T

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196

Proof of Corollary 4.5.1. (1) follows from Lemma 4.5.15(5) which is the characterization of a regularity point. (2) follows from Lemma 4.5.12 and Lemma 4.5.14. We can set η = 0 , where 0 is determined in Lemma 4.5.9 and Lemma 4.5.11 to obtain a corresponding T0 . Then Lemma 4.1(2) implies that T0 satisfies the conclusion. This completes the proof. Set QR (z) := BR (x) × (t − R2 , t + R2 ) for z = (x, t). Let H2 denotes the twodimensional parabolic Hausdorff measure. We give in the following theorem the Hausdorff measure estimate for a singularity set. Theorem 4.5.1 Let u be a solution of (4.5.4) and (4.5.5) with u0 ∈ H 1,2 (Ω; S 2 ) C 2 (∂Ω; S 2 ). Let {i }i be a sequence with i & 0 as i → ∞. Then the following hold : (1) S({u } ) has locally finite two-dimensional parabolic Hausdorff measure. More precisely there is a constant K1 = K1 (E0 , 0 ) > 0, such that for any compact interval T I ⊂ R+ , H2 (S({u } ) (Ω × I)) ≤ K1 |I|. (2) There is a constant K2 = K2 (E0 , 0 ) > 0 such that for any t > 0, set T t S ({u } ) := S({u } ) (Ω × {t}) has at most K2 points.

T

Proof. (1) By (4) of Lemma 4.5.15, we have for any z0 = (x0 , t0 ) ∈ S({u } ), any R > 0, and sufficiently small 0 <  ≤ (z0 ) 1 R2

Z

t0

Z

t0 −R2

1 g  (u ) > δ0 0 . Ω (x ) 4 BR 0

(4.5.45)

Fix a compact interval I ⊂ R+ and δ > 0. By compactness and Vitali’s Covering T theorem [7], any covering of S({u } ) (Ω × I) by parabolic cylinders QΩ R with T S 2 Ω 0 < R < δ and z ∈ S({u } ) (Ω × I) contains a finite covering j Q5Rj (zj ) ⊃ T S({u } ) (Ω × I) such that cylinders QΩ Since Rj (zj ) are pairwise disjoint. T S({u } ) (Ω × I) is a closed subset of the compact set Ω × I, it is compact. And therefore the countable subcovering stated in the Vitali theorem is indeed finite. By (4.5.45) and the energy estimate, we obtain for 0 <  ≤ min j {(zj )}, X j

ω2 (5Rj2 )

4ω2 ≤ 25 Σj δ 0 0

Z

tj tj −R2j

Z

Ω (x ) BR j

g (u (x, t))dxdt <

100ω2 (|I| + δ)E0 . δ 0 0

By letting δ & 0, and by the definition of the Hausdorff measure we find that H2 (S({u } )

\

(Ω × I)) ≤

100ω2 E0 |I|. δ 0 0

(2) Pick any (x1 , T ), . . . , (xk , T ) ∈ S({u }). By assumption we conclude that ∀ R, δ, γ > 0, ∃  ∈ (0, γ) such that sup T −δ
Z

Ω (x ) BR l

g (u (x, t))dx >

0 2

for 1 ≤ l ≤ k,

Landau–Lifshitz Equations and Harmonic Maps

197

Ω where we have used Lemma 4.5.15. We may choose R > 0 such that B2R (xl ) (1 ≤ l ≤ 2 R 0 k) are pairwise disjoint. Choose δ ∈ (0, 4CE0 ), where C is the constant from Lemma R 4.5.13 and  ∈ (0, γ) as above. Since t 7→ BRΩ (xl ) g (u (x, t))dx is continuous, we may R find tlδ ∈ (T − δ, T ) such that B Ω (xl ) g (u (x, tlδ ))dx ≥ 20 for 1 ≤ l ≤ k. The energy R estimates and the local energy inequality now imply

E0 ≥ ≥

k Z X l=1

Ω (x ) B2R l

k Z X l=1

Ω (x ) BR l

g (u (x, T − δ))dx

g (u (x, tlδ ))dx −

0 δ). Since δ < Thus E0 ≥ k( 20 − CE R2 is proved.

4.5.7

R 2 0 , 4CE0

C ZT Z Ω g (u , B2R (xl ))dt. Ω (x ) R2 T −δ B2R l

we finally get k ≤

4E0 0

:= K2 . Theorem 4.5.1

Passing to the Limits

In this section, we prove Theorem 4.5.2 Let u be a solution of (4.5.4) and (4.5.5) with u0 in H 1,2 (Ω; S 2 ) 3 1,2 (Ω × H 2 ,2 (∂Ω; S 2 ). Then there is at least one sequence {ui }i and u∗ ∈ Hloc T 1,2 R+ ; S 2 ) L∞ (R+ ; H 1,2 (Ω; S 2 )) such that ui * u∗ weakly in Hloc (Ω × R+ ; R3 ) and weak* in L∞ (R+ ; H 1,2 (Ω; R3 )). Moreover the following hold : (1) For any such sequence {ui }i , we have limi→∞ ui = u∗ and 12 (1 − |u |2 ) → T |∇u∗ |2 in C ∞ (Reg({ui }i ) (Ω × R+ )). T (2) u∗ is a smooth solution of (4.5.1) in Reg({ui }i ) (Ω × R+ ) and a global distriT 1,2 butional solution in Hloc (Ω × R+ ) L∞ (R+ ; H 1,2 (Ω; R3 )). Furthermore, u∗ satisfies the initial and boundary condition in the sense lim t&0 u∗ (·, t) = u0 in H 2,2 (Ω; R3 ) and u∗ |∂Ω = u0 |∂Ω as an H 2,2 (Ω; R3 )-trace for a.e. t > 0, respectively. R (3) If u∗ is regular at z0 = (x0 , t0 ) in the sense limR&0 supt0 −R2 ≤t≤t0 BR (x0 ) a(x)|∇u∗ |2 dx = 0, then z0 is parabolic isolated for {ui }i , i.e. BR (x0 ) × (t0 − R02 , t0 ) − {z0 } ⊂ Reg({ui }) for some R0 > 0. T

Proof. From the local energy estimates in Lemma 4.5.13, we see that {ui } is T 1,2 uniformly bounded in Hloc (Ω × R+ ) L∞ (R+ ; H 1,2 (Ω; R3 )). From the weak com1,2 pactness, and using the diagonal method, we can see that, there is u∗ ∈ Hloc (Ω × T 3 ∞ 1,2 3 R+ ; R ) L (R+ ; H (Ω; R )), and a subsequence {i }i such that ui * u∗ weakly 1,2 in Hloc (Ω × R+ ; R3 ) and weakly* in L∞ (R+ ; H 1,2 (Ω; R3 )). T (1) From Lemma 4.5.9, we can see that ∀ z0 ∈ Reg({ui }i ) (Ω × R+ ), C1 supPδR (z0 ) g (u ) ≤ (1−δ) 2 R2 . By Lemma 4.5.8, we obtain lim i→∞ ui = u∗ in PδR(z0 ) . 0 T Since point z0 is arbitrary, we get limi→∞ ui = u∗ in C ∞ (Reg({ui }i ) (Ω × R+ ); R3 ). (2) We multiply Eq. (4.5.4) by ui to get !

1 1 u i × ∂ t u i − u  i × u × ∂t ui − ui × ∇ · (a(x)∇ui ) = 0. 2 2 i

Landau–Lifshitz Equations

198

Using Lemma 4.5.8 that 12 (1 − |ui |2 ) is uniformly bounded in Reg({ui }i ) (Ω × i T R+ ), we get (1 − |ui |2 ) → 0 smoothly, i.e. |u∗ (x, t)| = 1 in Reg({ui }i ) (Ω × R+ ). Now from → − − − → − − → − − −c )→ − a ×( b ×→ c ) = (→ a ·→ b − (→ a · b )→ c T

and the fact that |u∗ (x, t)| = 1, we can let i to approach 0 to obtain 1 1 u∗ × ∂t u∗ − u∗ × (u∗ × ∂t u∗ ) − u∗ × ∇ · (a(x)∇u∗ ) = 0 2 2 in the sense of distribution. It is easy to get 1 1 ∂t u∗ − u∗ × ∂t u∗ − ∇ · (a(x)∇u∗ ) = a(x)|∇u∗ |2 u∗ . 2 2

(4.5.46)

From Eq. (4.5.4) 1 1 1 ∂t ui − ui × ∂t ui − ∇ · (a(x)∇ui ) = 2 a(x)(1 − |ui |2 )ui . 2 2 i We can see that the left side of the above equation converges to that of (4.5.46) and correspondingly we obtain \ 1 (1 − |ui |2 ) → |∇u∗ |2 in C ∞ (Reg({ui }i ) (Ω × R+ )). 2 i 1,2 In the sequel, we will rigorously prove that u∗ is the distributional Hloc solution of (4.5.1)

1 1 ∂t u∗ − u∗ × ∂t u∗ − ∇ · (a(x)∇u∗ ) = a(x)|∇u∗ |2 u∗ 2 2

T

L∞ (H 1,2 )

in Ω × R+ .

Note that sequence {ui }i converges weakly in H 1,2 (Ω × R+ ; S 2 ) and smoothly on T T Reg({ui }i ) (Ω × R+ ). Furthermore, since S t ({ui }i ) := S({ui }) (Ω × {t}) is finite for all t ≥ 0, we have both ui → u∗ pointwise a.e. in Ω × R+ and ui (·, t) → u∗ (·, t) pointwise a.e. in Ω for all t ∈ R+ . R R From the energy estimates in Lemma 4.5.12, we have 0∞ Ω |∂t ui |2 dxdt ≤ E0 . By R Fatou’s lemma, the complement of F := {t ≥ 0| lim inf i &0 Ω |∂t ui |2 (x, t)dx < ∞} has measure zero. For t0 ∈ F , there is a subsequence, still denoted by ui , such that ∂t ui (·, t0 ) * ∂t u∗ (·, t0 ) weakly in L2 (Ω; R3 ). By the local energy estimate, we may assume that, for the same subsequence, we also have ∂t ui (·, t0 ) * ∂t u∗ (·, t0 ) weakly in H 1,2 (Ω; S 2 ). By the uniqueness of the limit, the whole sequence converges. Hence, u∗ (·, t0 ) ∈ H 1,2 (Ω; S 2 ) and ∂t u∗ (·, t0 ) ∈ L2 (Ω; R3 ) for all t0 ∈ F . Moreover, since S t0 ({ui }i ) consists of finitely many points, it has zero 2-capacity in R2 , i.e. Cap2 (S t0 ({ui }i )) = 0. Therefore, from the definition of capacity, there exists a sequence {ηk }k = {ηk,q }k ⊂ Cc∞ (R2 ) such that ηk (x) = 1, ∀ x ∈ S t0 ({ui }i ) and kηk kH 1,2 (R2 ) → 0 as k → ∞.

Landau–Lifshitz Equations and Harmonic Maps

199

For φ ∈ Cc∞ (Ω), we multiply Eq. (4.5.4) by (1−ηk (t))φ(x), with its support contained in Reg({ui }i ). After passing to limit k → ∞ and using the property of the test function η, we see that for any t ∈ F Z

Ω

1 1 ∂t u∗ (x, t)φ(x)dx − u∗ (x, t) × ∂t u∗ (x, t)φ(x) 2 2 Z + a(x)∇u∗ (x, t)∇φ(x)dx =

Ω

a(x)|∇u∗ |2 u∗ (x, t)φ(x)dx.

The above equation holds for a.e. t ≥ 0. On the other hand, we have u∗ ∈ H 1,2 (Ω × [0, t]; S 2 ) for any t ≥ 0. Therefore both sides of the above equation are locally integrable on R+ . Multiplying the above equation by ψ ∈ Cc∞ [0, ∞) and integrating it over R+ , and noticing that linear combinations of Σk ak φk (x)ψk (t) with φk (x) ∈ Cc∞ (Ω) and ψk (t) ∈ Cc∞ ([0, ∞)) being dense in Cc∞ (Ω × [0, ∞)), we therefore get Z

∞ 0

Z

Ω

1 1 ∂t u∗ (x, t)φ(x, t)dx − u∗ (x, t) × ∂t u∗ (x, t)φ(x, t) 2 2

+ a(x)∇u∗ (x, t)∇φ(x, t)dxdt =

Z

∞ 0

Z

Ω

a(x)|∇u∗ |2 u∗ (x, t)φ(x, t)dxdt

for any φ(x, t) ∈ Cc∞ (Ω × [0, ∞)). Thus we have proved that u∗ is a distributional solution to (4.5.1). We still need to verify that u∗ satisfies the initial and boundary condition. Now the equation can be written as −a(x)4u∗ (·, t0 ) = a(x)|∇u∗ |2 u∗ (·, t0 ) + f, where 1 1 f = − ∂t u∗ (·, t0 ) + u∗ × ∂t u∗ (·, t0 ) + ∇a(x)∇u∗ (·, t0 ) ∈ L2 (Ω; R3 ). 2 2 By a regularity result due to Rivi`ere (see [126]), we have u∗ (·, t0 ) ∈ H 2,2 (Ω; S 2 ) if T 3 u0 ∈ H 2 ,2 (∂Ω; S 2 ) H 2,2 (Ω; S 2 ). This implies u∗ (·, t)|∂Ω = u0 |∂Ω as an H 2,2 -trace for any t ∈ F . As for the initial condition, we have lim u∗ (·, t) = 0 in H 1,2 (Ω; S 2 ). t&0

As a matter of fact, it follows from the following commutative diagram: u (x, t) −→ u∗ (x, t)

as  & 0 in C ∞ (Reg({u }) (Ω × R+ ))

u (x, 0) −→ u0 (x)

as  & 0 in C ∞ ,

↓t→0

T

↓t→0

3

where u0 ∈ H 1,2 H 2 ,2 (∂Ω) is the boundary value of u∗ ∈ H 2,2 in the trace sense. Thus we have proved statement (2) of the theorem. (3) From the regularity assumption, ∃ R > 0 such that T

sup t0 −R2 ≤t≤t0

Z

BR (x0 )

1 0 a(x)|∇u∗ (x, t)|2 dx < . 2 4

Landau–Lifshitz Equations

200

2

R 0 Set δ := min{R2 , 2CE } and s0 = t0 − 12 δ. We may also assume that for same R > 0, 0 Ω P2R (z0 )\{z0 } ⊂ Reg({ui }). Using Lemmas 4.5.8 and 4.5.11 and the above steps, we T have |ui | → |u∗ | = 1 in Reg({ui }) Ω and

lim

Z

i→∞ BR (x0 )

T

Ω

gi (ui (x,s0 ) )dx =

Z

BR (x0 )

From Lemma 4.5.13 we have Z

sup

s0 ≤t≤s0 +δ B 1 R (x0 )

≤

Z

2

BR (x0 )

T

Ω

T gi (ui (x, t))dx

0 1 a(x)|∇u∗ (x, s0 )|2 dx < . 2 4

Ω

T gi (ui (x, s0 ))dx + Ω

0 0 δCE0 < + = 0 2 R 2 2

for sufficiently large i. By construction, t0 ∈ (s0 , s0 + δ); hence the claim follows. Corollary 4.5.2 There is T0 > 0, such that lim&0 u = u∗ in C ∞ (Ω × (0, T0 ); S 2 ), where u∗ is the smooth solution of (4.5.1) with initial and boundary data u0 . Proof. From above discussions we know that there exists T0 > 0 such that Ω × [0, T0 ) ⊂ Reg({u }). Using Theorem 4.5.2, we obtain u (x, t) → u∗ (x, t),

4.5.8

in C ∞ (Reg({ui }i )).

Chen–Struwe Solution

The main purpose of this section is to show that the solution to (4.5.1) is indeed a Struwe solution. In Theorem 4.5.1 of Sec. 4.5.6, we have proved that for each t > 0, there are at most finitely many singularities; in particular, so is the maximal existence time T0 . By iterating the short time existence result, a short time smooth solution can be extended to a global weak solution which is smooth except for finitely many point singularities and has decreasing energies t → E(u∗ (t)). This extension solution is referred to as a Struwe solution. First, we give a lemma describing the energy of sublimits u∗ . Lemma 4.5.16 Let u be a solution of (4.5.4) and (4.5.5) for each fixed  > 0, and assume u∗ to be weak H 1,2 limits of ui for a sequence 0 < i & 0. If s < t and T S s ({ui }) := (Ω × {s}) S({ui }) = ∅ and S t ({ui }) 6= ∅, then Z

Ω

Z 1 1 a(x)|∇u∗ |2 (x, s)dx ≥ a(x)|∇u∗ |2 (x, τ )dx 2 Ω 2

∀ τ >s

and Z

Ω

1 a(x)|∇u∗ |2 (x, s)dx ≥ 2

Z

Ω

1 a(x)|∇u∗ |2 (x, t)dx + 0 2

where 0 is the constant from Lemma 4.5.11.

∀ t>s

(4.5.47)

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201

Proof. Set x := (x1 , x2 , . . . , xk ) if S t ({ui }) = (x1 , x2 , . . . , xk ) and BR (x) := j=1 (BR (xj )). Using the fact that u∗ (x, t) = 1, on Reg({u }), we have

Sk

E(u∗ (s), Ω) :=

Z

Ω

1 a(x)|∇u∗ |2 (x, s)dx = lim i→∞ 2 Z

≥ lim sup i→∞

≥

Z

Ω

Ω

Z

Ω

gi (ui )(x, s)dx

gi (ui )(x, τ )dx (∀ τ > s) (by Lemma 4.5.12)

1 a(x)|∇u∗ |2 (x, τ )dx (by weak lower semi-continuity). 2

Also E(u∗ (s), Ω) ≥ lim sup i→∞

Z

Ω−BR (x)

gi (ui ) (x, τ )dx +

Z

BR (x)

gi (ui ) (x, τ )dx

!

for all τ ∈ (s, t). Now from Lemma 4.5.15, for all δ ∈ (0, 1) and R > 0, there are sequences s < ti % t and 0 < δi & 0 such that Z

BR (x)

gi (ui ) (x, ti )dx =

sup

Z

t−δi <τ
gi (ui ) (x, τ )dx) ≥ δ0

and so E(u∗ (s)) = lim sup i→∞

Z

Ω−BR (x)

gi (ui ) (x, ti )dx + δ0 ≥

Z

Ω−BR (x)

gi (ui ) (x, τ )dx + δR0

∀ R > 0, and δ ∈ (0, 1). Since the last inequality holds for any R > 0 and δ ∈ (0, 1), the claim follows. Theorem 4.5.3 Let u0 ∈ H 1,2 (Ω; S 2 ).Then there is a global distributional solution T 1,2 u ∈ Hloc (Ω × (0, ∞); S 2 ) L∞ ((0, ∞); H 1,2(Ω; S 2 )) with ∂t u ∈ L2 (Ω × (0, ∞); R3 ) of (4.5.1) with initial and boundary data u0 . u is smooth on Ω × (0, ∞) except at finitely many points and has decreasing and right continuous energy. If in addition u 0 ∈ 3 H 2 ,2 (∂Ω; S 2 ), then u is unique among the solutions of (4.5.1) with initial and boundary data u0 . Also u is smooth except for isolated singular points, with lim t&s E(u(t)) < E(u(s)) + 0 for all s ≥ 0. Proof. By Theorem 4.5.2, the -approximation scheme provides a smooth short time solution u ∈ C ∞ (Ω × (0, T0 ); S 2 ) to (4.5.1) with boundary data u0 and limt%0 u(·, t) = u0 in H 1,2 (Ω; R3 ). Also, there are {x1 , . . . , xk } ⊂ Ω such that limt%T0 u(·, t) = u(·, T0 ) ∈ C ∞ (Ω\{x1 , . . . , xk }, R3 ) and k

q

a(x)∇u(·, T0 )k2L2 (Ω)

q

≤ lim inf k a(x)∇u(·, t)k2L2 (Ω) ≤ 2E0 . t%T0

We have used the maximum principle to prove that u(·, t) ≤ 1; therefore, u(·, T0 ) ∈ H 1,2 (Ω). If we now set u ˜0 := u(·, T0 ) and repeat the same procedure with u ˜0 instead of u0 , we obtain step by step a global solution with point singularities. To see that

Landau–Lifshitz Equations

202

∂t u ∈ L2 (Ω × R+ ; R3 ) we first use Lemma 4.5.12 to interval (tk , tk+1 ): for each t ∈ (tk , tk+1 ), 1 G (u (t) + 2

Z

t tk

Z

Ω

|∂t u |dxdt = G (u (tk )

Setting t % tk+1 , we have lim sup G (u (t) + t%tk+1

1 2

Z

tk+1 tk

Z

Ω

|∂t u |dxdt = G (u (tk ).

By Lemma 4.5.16 we get lim sup E(u(t)) ≥ E(u(tk+1 )) + 0 . t%tk+1

This implies that 21 0∞ Ω |∂t u|2 dxdt ≤ E0 − Σk 0 and also shows there can only be finitely many singular time tk . Now assume we have two solutions u1 and u2 of (4.5.1) with initial and boundary data u0 and both with finitely many singularities and limt&s E(u(t)) < E(u(s)) + 0 for all s ≥ 0. We have u1 = u2 on Ω × (0, T1 ), where T1 is the maximal common smooth existence time, i.e., either u1 or u2 has point singularities at T1 . However, if u1 admits a smooth extension up to T1 , so does u2 and vice versa. Moreover, since the criterion for the existence of a smooth extension is local, both the solutions have the same singularities x1 , . . . , xk at time T1 and u1 (·, T1 ) = u2 (·, T1 ) on Ω\{x1 , . . . , xk }. By Theorem 6 in [34], and the assumption on the energy, the extension of u1 and u2 after T1 is again unique for a short time, and an iteration of the previous argument leads to the claimed uniqueness. R

R

4.6

Smooth Solution and Decay Estimates to the L–L System with Small Initial Data in Higher Dimensions

4.6.1

Initial Value Problem to the L–L System in Higher Dimensions

1. Existence and uniqueness of the global smooth solution to the Cauchy problem Consider the existence and uniqueness of a smooth solution of the Cauchy problem to the Landau–Lifshitz system with the Gilbert term in higher dimensions: ~ t = λ1 Z ~ × ∆Z ~ − λ2 Z ~ × (Z ~ × ∆Z), ~ x ∈ Rn , t > 0, Z

(4.6.1)

~ t) = (Z1 (x, t), Z2 (x, t), Z3 (x, t)). For simplicity, we let λ1 = where n ≥ 2, and Z(x, ~ t)| = 1, system (4.6.1) is equivalent to the following λ2 = 1. Under condition |Z(x, system: ~ t = ∆Z ~ + |∇Z| ~ 2Z ~ +Z ~ × ∆Z, ~ x ∈ Rn , t > 0. Z (4.6.2)

Landau–Lifshitz Equations and Harmonic Maps

203

~ : |Z(x, ~ t)| = 1} onto the complex Taking the mirror mapping from surface B = {Z plane by Z1 (x, t) + iZ2 (x, t) w(x, t) = , (4.6.3) 1 + Z3 (x, t) we may rewrite (4.6.2) into a nonlinear Schr¨odinger equation as follows: 2(1 + i)w ∗ |∇w|2 = 0, 1 + |w|2

wt − (1 + i)∆w +

(4.6.4)

where w ∗ is the conjugate of w. Then our problem is changed to prove the existence and uniqueness of a smooth solution of the Cauchy problem with small initial data of nonlinear Schr¨odinger equation (4.6.4) with the following initial condition: x ∈ Rn ,

w(x, 0) = w0 (x),

n ≥ 2.

(4.6.5)

If this is proved, then making inverse transformation Z1 =

2<w(x, t) , 1 + |w|2

Z2 =

2=w(x, t) , 1 + |w|2

Z3 =

1 − |w|2 , 1 + |w|2

(4.6.6)

we may get the existence and uniqueness of a global smooth solution to the Cauchy problem ~ 0) = Z ~ 0 (x), x ∈ Rn , n ≥ 2 Z(x, (4.6.7) of Eq. (4.6.2), where <w and =w are the real and the image part of a complex number w. To do this we need to introduce some preliminary knowledge which can be found in [121] and [133]. 2. Heat conduct equation with complex coefficient For the Cauchy problem of complex heat conduct equation ut = a∆u,

0, x ∈ Rn , t > 0,

u|t=0 = ϕ(x),

x ∈ Rn , n ≥ 2,

(4.6.8)

we also have the following expression for the solution:

since

1 u(x, t) = S(t)ϕ = q √ ( 2a πt)n

Z

(

)

|x − ξ|2 exp − ϕ(ξ)dξ, 4at Rn

ku(·, t)k2 ≤ kϕk2 and

n

ku(·, t)k∞ ≤ Ct− 2 kϕkL1 (Rn )

(4.6.9)

(4.6.10) (4.6.11)

and the Riesz–Therin interpolation inequality n n

1

ku(·, t)kW N,q (Rn ) ≤ Ct− 2 ( p − 2 ) kϕkW N,p (Rn ) ,

(4.6.12)

Landau–Lifshitz Equations

204

where 1 < p, q < ∞,

1 1 + = 1. p q

If q = ∞, we have n

ku(·, t)kW N,∞(Rn ) ≤ Ct− 2 kϕkW N,l (Rn ) , ∀ l > 0,

(4.6.13)

ku(·, t)kW N,∞(Rn ) ≤ C(1 + t)

(4.6.14)

−n 2

kϕkW N +n+1,1 (Rn ) , ∀ l > 0.

3. Smoothness of vector-valued functions Lemma 4.6.1 Suppose that F = F (w) is smooth enough and F (0) = 0, where w = (w1 , w2 , . . . , wn ). For any integer s ≥ 0 if w = w(x) ∈ W s,p (Rn ), 1 ≤ p ≤ ∞, and kwkL∞ (Rn ) ≤ M, then F (w) ∈ W s,p (Rn ) and kF (w)kW s,p(Rn ) ≤ C(M )kwkW s,p (Rn ) .

(4.6.15)

Lemma 4.6.2 Suppose that F = F (w) is smooth enough, where w = (w1 , w2 , . . . , wn ). If for any |w| ≤ ν0 : F (w) = O(|w|1+α ), and if for any integer s ≥ 0 :

α ≥ 1 is integer,

kwkL∞ (Rn ) ≤ ν0 ,

and the integer appearing in the following formula makes sense, then kF (w)kW s,r (Rn ) ≤ CkwkW s,q (Rn ) kwkLp (Rn ) kwkα−1 L∞ (Rn ) , where 1 ≤ p, q, r ≤ ∞,

1 1 1 + = . p q r

Taking r = 2, q = 2, p = ∞; or r = 1, p = 2, q = 2; or r = 1, q = 1, p = ∞; respectively, we have

(4.6.16)

Landau–Lifshitz Equations and Harmonic Maps

205

Corollary 4.6.1 Under the condition of Lemma 4.6.2, kF (w)kH s (Rn ) ≤ CkwkH s (Rn ) kwkαL∞ (Rn ) ,

kF (w)kW s,1 (Rn ) ≤ CkwkW s,2 (Rn ) kwkαL2 (Rn ) kwkα−1 L∞ (Rn ) ,

kF (w)kW s,1 (Rn ) ≤ CkwkW s,1 (Rn ) kwkαL∞ (Rn )

(4.6.17) (4.6.18) (4.6.19)

hold. Lemma 4.6.3 Let 1 ≤ p, q, r ≤ ∞,

1 p

+

f ∈ W s,p (Rn ),

1 q

= 1r , for any given integer s ≥ 0, g ∈ W s,q (Rn ).

Then kD s (f g)kLr (Rn ) ≤ C(kf kLp (Rn ) kD s gkLq (Rn ) + kD s f kLp (Rn ) kgkLq (Rn ) ).

(4.6.20)

If s ≥ 1, then kD s (f g) − f (D s g)kLr (Rn )

≤ C(kDf kLp (Rn ) kD s−1 gkLq (Rn ) + kD s f kLp (Rn ) kDgkLs (Rn ) ).

(4.6.21)

If s ≥ 0, then kD s (f g)kLr (Rn ) ≤ C(kf kLp (Rn ) kgkW s,q (Rn ) + kf kW s,p (Rn ) kgkLp (Rn ) ). Let F (v) = Take space Xs,E as

2(1 + i)v ∗ |∇v|2 . 1 + |v|2

(4.6.22)

(4.6.23)

Xs,E = {u(x, t) : Ds (u) ≤ E},

(4.6.24)

where n

Ds (u) = sup(1 + t) 2 ku(·, t)kW s−n−3,∞(Rn ) + sup ku(·, t)kW s,l(Rn ) t≥0

 Z  +

t≥0

2

∞ 0

1

X

|k|≤2

kDxk u(·, t)k2H s (Rn ) 

,

(4.6.25)

where s ≥ n + 4. Construct a mapping Tˆ defined on Xs,E , Tˆ : ~v ∈ Xs,E → Tˆ v = w by (

wt − (1 + i)∆w + F (v) = 0, w|t=0 = ϕ(x).

(4.6.26)

Lemma 4.6.4 If n ≥ 2 and E is small, then mapping Tˆ maps Xs,E into itself.

Landau–Lifshitz Equations

206

Proof. From Poisson formula (4.6.9), we have w(x, t) = S(t)ϕ +

Z

t 0

S(t − τ )F (v(τ ))dτ.

It follows from (4.6.14) that Z

n

kwkS s−n−3,∞ (Rn ) ≤ C(1 + t)− 2 kϕkW s−2,1 (Rn ) +

t 0

n

(1 + t − τ )− 2 kF (v(τ ))kW s−2,1 (Rn ) dτ.

(i) It follows from Lemma 4.6.1 and Lemma 4.6.2 that kF (v)kW s,r (Rn )

" # ∗

v∗

v

≤C

k(∇v)2 kW s,q + k k s,q k(∇v)2 kLp 2 W

1 + |v|2 r 1 + |v| L h i

≤ C kvkLp k∇vkW s,q k∇vkL∞ + kvkW s,q k∇vkL∞ k∇vkLp h

i

≤ C kvkLp kvkW s+1,q + kvkW s,q kvkW 1,p k∇vkL∞ k∇vkL∞ .

Taking r = 1, p = ∞, we have

  kF (v)kW s,1 (Rn ) ≤ C[kvkL∞ k∇vkL∞ kvkW s+1,1 + kvkW s,1 k∇vk2L∞ ],  kF (v)k

W s,2 (Rn )

≤ CkvkW s−1,1 (Rn ) · kvk2W 1,∞ (Rn ) .

When s ≥ n + 4, it follows from the definition of Xs,E that

(4.6.27)

(4.6.28)

n

kvkW 1,∞ ≤ kv(·, τ )kW s−n−3,∞ ≤ E(1 + τ )− 2 ;

kvkW s−1,1 (Rn ) ≤ E. Hence, if

kϕkW s,1 + kϕkH s+1 , ≤ δE,

(4.6.29)

then from (4.6.26) and (4.6.28)

kw(·, t)kW s−n−3,∞ (Rn ) ≤ CkϕkW s−2,1 (1 + t) ≤ CδE(1 + t) That is

−n 2

−n 2

+ CE

Z

3

3

+ CE (1 + t)

t

0 −n 2

n

(1 + t − τ )− 2 (1 + τ )−n dτ

.

n

sup(1 + t) 2 kw(·, t)kW s−n−3,∞ (Rn ) ≤ C1 (δE + E 3 ).

(4.6.30)

t≥0

(ii) In (4.6.27), taking r = 1, p = q = 2, we have kF (v)kW s,1 (Rn ) ≤ Ckvk2H s+1 (Rn ) kvkW s−n−3,∞ (Rn ) . Therefore it follows from (4.6.26) that kw(·, t)kW s,1 (Rn ) ≤ kϕkW s,1 (Rn ) +

Z

t 0

≤ kϕkW s,1 (Rn ) + C

kF (v)kW s,1(Rn ) dt

Z

t 0

kvk2H s+1 (Rn ) kvkW 1,∞ (Rn ) dτ.

(4.6.31)

Landau–Lifshitz Equations and Harmonic Maps

207

Noting that kvk2H s+1 (Rn ) = kvk2H 1 +

X

|k|=2

kDxk v(·, τ )k2H s−1 (Rn )

≤ Ckv(·, τ )kW 1,∞ kv(·, τ )kW 1,1 + ≤ CE 2 (1 + τ )

− n2

X

+

|k|=2

X

|k|=2

kDxk v(·, τ )k2H s−1 (Rn )

kDxk v(·, τ )k2H s−1 (Rn ) ,

and substituting this inequality into (4.6.31), we have kw(·, t)kW s,1(Rn ) ≤ δE + +CE

3

Z

t 0

(1 + τ )−n dτ + CE 2 sup kvkW 1,∞ (Rn ) . t≥0

When s ≥ n + 4, we have sup kvkW 1,∞ (Rn ) ≤ C sup kvkW s,1 (Rn ) ≤ CE. t≥0

t≥0

Therefore kw(·, t)kW s,1(Rn ) ≤ δE + C2 E 3 .

(4.6.32)

(iii) For the complex heat conduct equation ut − a∆u = F (x, t)

(4.6.33)

u|t=0 = ϕ(x),

(4.6.34)

with initial condition where 0, we have the following theorem: Theorem 4.6.1 Let ϕ(x) ∈ H s+1 (Rn ), F ∈ L2 (0, T ; H s (Rn )), where s ≥ 0 is an integer. Then, the Cauchy problem (4.6.33) and (4.6.34) admits a unique solution u ∈ L2 (0, T ; H s+2(Rn )),

ut ∈ L2 (0, T ; H s(Rn )),

and there holds the estimate Z

t 0

X

|k|=2

kDxk u(·, τ )k2H s (Rn ) dτ ≤ C0 kϕk2H s+1 (Rn ) +

Z

t 0

!

kF (·, τ )k2H s (Rn ) dτ ,

where C0 is independent of T, and k is a multi-index : k = (k1 , k2 , . . . , kn ), |k| = k k1 + k2 + · · · + kn , Dxk = k1∂ kn . ∂x1 ···∂xn

Now we finish the proof of Lemma 4.6.4. It follows from (4.6.25) that Z

t

X

0 |k|=2

kDxk u(·, τ )k2H s (Rn ) dτ

≤C

kϕk2H s+1 (Rn )

+

Z

t 0

kF (v)k2H s (Rn ) dτ

!

.

(4.6.35)

Landau–Lifshitz Equations

208

Let r = q = 2, p = ∞ in (4.6.27). We have kF (v)kH s (Rn ) ≤ Ckv(·, t)kH s+2 kv(·, t)k2W 1,∞ . Note that kv(·, t)k2H s+2 = kvk2H 1 +

X

|k|=2

2

kDxk v(·, t)k2H s n

≤ CE (1 + τ )− 2 +

X

|k|=2

kDxk v(·, t)k2H s .

Substituting this into (4.6.35), we have Z

t

kDxk u(·, τ )k2H s (Rn ) dτ

X

0 |k|=2

"

≤ C δE +

Z

t 0

!1

2

kv(·, t)k2H s+2 kv(·, t)k4W 1,∞ dτ

!1 # 2

3

≤ C3 (δE + E ).

(4.6.36)

It follows from (4.6.30), (4.6.32) and (4.6.36) that Ds (w) ≤ max{1, C1 , C2 , C3 }(δE + E 3 ) = C(δE + E 3 ). Taking δ = δ0 =

√1 , 2C

when E ≤ E0 =

1 , 2C

(4.6.37)

we have

Ds (w) ≤ E(Cδ0 + CE 2 ) ≤ E.

(4.6.38)

This implies that Tˆ : v = Tˆ v maps Xs,E into itself. The lemma is proved. 5. Contracting mapping In the following we want to prove that map Tˆ is contracting. Lemma 4.6.5 If E is small enough, then Tˆ is contracting in the metric of Xs,E . Proof. For any v1 , v2 ∈ Xs,E , from Theorem 4.6.1, when E is small enough, we have wi = Tˆ vi ∈ Xs,E , i = 1, 2. Set v = v1 − v2 , w = w1 − w2 . It suffices to prove that if E is small, then there exists a constant 0 < η < 1 such that ρ(w1 , w2 ) ≤ ηρ(v1 , v2 ).

(4.6.39)

(i) From the definition of Tˆ we know (

wt − (1 + i)∆w = F (v1 ) − F (v2 ), w|t=0 = 0.

(4.6.40)

It follows from (4.6.26) that kw(·, t)kW s−n−3,∞(Rn ) ≤C

Z

t

0

n

(1 + t − τ )− 2 kF (v1 ) − F (v2 )kW s−2,1 (Rn ) dτ.

(4.6.41)

Landau–Lifshitz Equations and Harmonic Maps

209

Since "

v1∗ v2∗ 2 F (v1 ) − F (v2 ) = 2(1 + i) |∇v | − |∇v2 |2 1 1 + |v1 |2 1 + |v2 |2 v1∗ (|∇v1 |2 − |∇v2 |2 ) = 2(1 + i) 2 1 + |v1 | "

#

#

v1∗ v2∗ + 2(1 + i)|∇v2 | − 1 + |v1 |2 1 + |v2 |2 v1∗ = 2(1 + i) (∇v1 + ∇v2 )∇v 1 + |v1 |2 (1 + |v1 |2 )v ∗ + (|v1 | + |v2 |)2 v1∗ + 2(1 + i)|∇v2 |2 , (1 + |v1 |2 )(1 + |v2 |2 ) 2

we have kF (v1 ) − F (v2 )kW s,r (Rn ) v1∗ (∇v1 + ∇v2 )kLp (Rn ) k∇vkW s,q (Rn ) ≤k 1 + |v1 |2



v1∗

(∇v + ∇v ) k∇vkLp (Rn ) + 1 2

s,q n

1 + |v1 |2 W (R )

|∇v |2 (1 + |v |2 + (|v | + |v |)v ) 2 2 1 2 1

kvkW s,q (Rn )

p n (1 + |v1 |2 )(1 + |v2 |2 ) L (R )

|∇v |2 (1 + |v |2 + (|v | + |v |)v ) 2 2 1 2 1

+

kvkLp (Rn ) ,

s,q n (1 + |v1 |2 )(1 + |v2 |2 ) W

where



v1∗

(∇v + ∇v )

1 2

1 + |v1 |2

(R )

Lp (Rn )

≤ (k∇v1 kL∞ (Rn ) + k∇v2 kL∞ (Rn ) )kv1 kLp (Rn ) ;



v1∗

(∇v + ∇v )

1 2

1 + |v1 |2

W s,q (Rn )

≤ C[k∇v1 kLp0 (k∇v1 kW s,q0 + k∇v2 kW s,q0 ) + k∇v1 kW s,q0 (k∇v1 kLp0 + k∇v2 kLp0 )];



|∇v |2 (1 + |v |2 + (|v | + |v |)v ) 2 2 1 2 1



(1 + |v1 |2 )(1 + |v2 |2 )

Lp (Rn )

≤ Ck|∇v2 |2 kLp (Rn ) ≤ Ck∇v2 kL∞ k∇v2 kLp ;

(4.6.42)

Landau–Lifshitz Equations

210

and

|∇v |2 (1 + |v |2 + (|v | + |v |)|v |) 2 2 1 2 1



(1 + |v1 |2 )(1 + |v2 |2 )

W s,q

≤ Ck∇v2 kLp0 k∇v2 kW s,q0 ,

where 1q = p10 + q10 . Therefore, we have kF (v1 ) − F (v2 )kW s,r (Rn )

≤ C[(k∇v1 kL∞ + k∇v2 kL∞ )kv1 kLp kvkW s+1,q + {k∇v1 kLp0 (kv1 kW s+1,q0 + kv2 kW s+1,q0 )

+ kv1 kW s,q0 (k∇v1 kLp0 + k∇v2 kLp0 )}k∇vkLp + k∇v2 k2L∞ kv2 kLp0 kvkW s,q

+ k∇v2 kLp0 kv2 kW s+1,q0 kvkLp ].

(4.6.43)

In the above inequality, we take s as s − 2, r = 1, p = ∞, q = 1 to give kF (v1 ) − F (v2 )kW s−2,1 (Rn )

≤ C[(k∇v1 kL∞ + k∇v2 kL∞ )kv1 kL∞ kvkW s−1,1 + {kv1 kL∞ (kv1 kW s−1,1 + kv2 kW s−1,1 )

+ kv1 kW s,1 (k∇v1 kL∞ + k∇v2 kL∞ )}k∇vkL∞

+ kv2 k2L∞ kv1 kL∞ kvkW s−2,1

+ k∇v1 kL∞ kv2 kW s−1,1 kvkL∞ ]

≤ CkvkW s,1 (kv1 kW 1,∞ + kv2 kW 1,∞ )kv1 kW 1,∞

+ CkvkW s−n−3,∞ (kv1 kW s,1 + kv2 kW s,1 )kv1 kL∞

+ CkvkW s−n−3,∞ (k∇v1 kL∞ + k∇v2 kL∞ )}kv1 kW s,1 + CkvkW s,1 kv2 k2W s−n−3,∞ kv1 kW s−n−3,∞

+ CkvkW s−n−3,∞ kv1 kW s−n−3,∞ kv2 kW s,1 .

Since for i = 1, 2 n

kvi kW 1,∞ ≤ Ckvi kW s−n−3,∞ ≤ E(1 + τ )− 2 ,

k∇vi kL∞ ≤ kvi kW 1,∞ , we have

kF (v1 ) − F (v2 )kW s−2,1 ≤ CE 2 (1 + τ )−n Ds (v). On the other hand, since for n > 1 Z

t 0

n

n

(1 + t − τ )− 2 (1 + τ )−n dτ = C(1 + t)− 2 ,

Landau–Lifshitz Equations and Harmonic Maps

211

we have from (4.6.41) that n

kw(·, t)kW s−n−3,∞ ≤ CE 3 (1 + t)− 2 Ds (v). That is

n

sup(1 + t)− 2 kw(·, t)kW s−n−3,∞ ≤ CE 3 Ds (v). t>0

(ii) Similarly to (4.6.31) we have kw(·, t)kW s,1 ≤

Z

t 0

kF (v1 ) − F (v2 )kW s,1 dτ.

(4.6.44)

Taking r = 1, p = q = 2 in (4.6.43), we have kF (v1 ) − F (v2 )kW s,1 (Rn )

≤ CkvkH s+1 (k∇v1 kL∞ + k∇v2 kL∞ )kv1 kL2

+ Ck∇vkL∞ [kv1 kL∞ (kv1 kH s+1 + kv2 kH s+1 ) + kv1 kH s (k∇v1 kL∞ + k∇v2 kL∞ )] + CkvkH s k∇v1 k2L∞ kv1 kL2

+ CkvkL2 k∇v1 kL∞ kv2 kH s+1 .

(4.6.45)

However, kvk2H s+2 ≤ Ckv(·, τ )kW s−n−3,∞ kv(·, τ )kW s,1 + ≤ C(1 + τ )

−n 2

Ds2 (v) +

X

|k|=2

X

|k|=2

kDxk v(·, τ )k2H s

kDxk v(·, τ )k2H s

(4.6.46)

and kvk2H 1 ≤ Ckv(·, τ )kW 1,∞ kv(·, τ )kW 1,1 ≤ Ckv(·, τ )kW s−n−3,∞ kvkW s,1 n

≤ C(1 + τ )− 2 Ds2 (v),

(4.6.47)

we have         

kvkH s+2 (Rn ) ≤ C(1 + τ )

−n 4

Ds (v) +

∞ X

|k|=2

n

kDxk v(·, τ )k2H s (Rn )

!1

2

,

(4.6.48)

kvkH 1 (Rn ) ≤ C(1 + τ )− 4 Ds (v).

Similarly, kv1 kH s+2 + kv2 kH s+2 ≤ CE(1 + τ ) X

i=1,2

− n4

+

X

|k|=2,i=1,2 n

kvi (·, τ )kH 1 ≤ CE(1 + τ )− 4

kDxk vi (·, τ )k2H s

!1

2

, (4.6.49) (4.6.50)

Landau–Lifshitz Equations

212

and

n

X

i=1,2

kvi (·, τ )kW 1,∞ ≤ CE(1 + τ )− 2 .

(4.6.51)

Hence kF (v1 ) − F (v2 )kW s,1 (Rn ) "

≤ C (1 + τ )

−n 4

X

Ds (v) +

|k|=2

kDxk v(·, τ )k2H s

!1 # 2

n

n

E 2 (1 + τ )−( 2 + 4 )

+ C(1 + τ )−n Ds (v)E 2 "

+ C (1 + τ )

−n 4

X

Ds (v) +

|k|=2 2

× CDs (v)[E (1 + τ ) X

+C

|k|=2

−n

kDxk v(·, τ )k2H s

!1 # 2

3n

E 3 (1 + τ )−

5n 4

3n

+ E 3 (1 + τ )− 2 + E 2 (1 + τ )− 4 ]

kDxk v(·, τ )k2H s

!1 2

(E 3 (1 + τ )−

5n 4

3n

+ E 2 (1 + τ )− 4 ).

Applying this inequality to (4.6.44) and using the H¨older inequality, we have kw(·, t)kW s,1 ≤ CDs (v)(E 2 + E 3 ) + C(E 2 + E 3 ), Z

∞ 0

X

|k|=2

kDxk v(·, τ )k2H s dτ

!1 2

≤ 2CDs (v)(E 2 + E 3 ).

(4.6.52)

(iii) It follows from (4.6.35) that Z

t 0

X

|k|=2

kDxk v(·, τ )k2H s dτ

≤C

Z

t 0

kF (v1 ) − F (v2 )kH s dτ.

Taking r = q = 2, p = ∞ in (4.6.43), we have kF (v1 ) − F (v2 )kH s (Rn )

≤ CkvkH s+1 kv1 kL∞ (k∇v1 kL∞ + k∇v2 kL∞ )

+ Ck∇vkL∞ [kv1 kL∞ (kv1 kH s+1 + kv2 kH s+1 ) + kv1 kH s (k∇v1 kL∞ + k∇v2 kL∞ )]

+ CkvkH s k∇v1 k2L∞ kv2 kL∞

+ CkvkL∞ k∇v1 kL∞ kv2 kH s+1

≤ CkvkH s+2 kv1 kW 1,∞ (kv1 kW 1,∞ + k∇v2 kW 1,∞ )

+ CkvkW 1,∞ kv1 kH s+2 (kv1 kW 1,∞ + kv2 kW 1,∞ ) + CkvkH s+2 kv1 kH s+2 (kv1 kW 1,∞ + kv2 kW 1,∞

≤ CE 2 (1 + τ )

−n 2



 (1 + τ )

−n 4



Ds (v) + 

X

|k|=2

1  2 k 2   kDx v(·, τ )kH s 

(4.6.53)

Landau–Lifshitz Equations and Harmonic Maps 



n

+ C E(1 + τ )− 4 +  

3

X

|k|=2



1  2 k 2   kDx v1 (·, τ )kH s  E(1 + τ )−n 

n

+ CE 3 (1 + τ )− 2 n (1 + τ )− 4 Ds (v) +  



213

n

|k|=2



+ CEDs (v)(1 + τ )−n E(1 + τ )− 4 +  

X

X

|k|=2

1  2 k 2   kDx v(·, τ )kH s 

1  2 k 2   kDx v(·, τ )kH s  .

Substituting this into (4.6.53) and applying the H¨older inequality, we have Z

∞ 0

X

|k|=2

kDxk v(·, τ )k2H s dτ ≤ CDs2 (E 2 + E 3 )2 .

(4.6.54)

In summary, we get Ds (w) ≤ C0 Ds (E 2 + E 3 ), where C0 = C0 (s, n) is a positive constant. Choose E1 such that C0 (E12 + E13 ) = η < 1.

(4.6.55)

E ≤ min{E0 , E1 },

(4.6.56)

Letting we get Ds (w) ≤ ηDs (v), where E0 is given in Lemma 4.6.4. The lemma is proved.

4.6.2

Decay of Solution to the Higher-Dimensional L–L Equations

1. Decay of the Cauchy problem of nonlinear Schr¨ odinger equations From Lemmas 4.6.4 and 4.6.5, we have Theorem 4.6.2 Consider the Cauchy problem of the following nonlinear Schr¨ odinger system:  ∗  w − (1 + i)∆w + 2(1+i)w |∇w|2 = 0, t 1+|w|2 (4.6.57)  w(x, 0) = w (x), x ∈ Rn , n ≥ 2. 0

If for s ≥ n + 4 there exists small constants δ0 and E0 such that for δ ≤ δ0 , E ≤ E0 the initial data w0 (x) satisfies w0 (x) ∈ W s,1 (Rn ) ∩ H s+1 (Rn ),

(4.6.58)

kw0 (x)kW s,1 (Rn ) + kw0 (x)kH s+1 (Rn ) ≤ δE,

(4.6.59)

and

Landau–Lifshitz Equations

214

then problem (4.6.57) admits a unique smooth solution w(x, t) ∈ Xs,E , that is w(x, t) ∈ L2 (0, T ; H s+2 (Rn )) ∩ L∞ (0, T ; H s+1(Rn )),

wt (x, t) ∈ L2 (0, T ; H s(Rn )) ∩ L∞ (0, T ; H s−2(Rn )), and the decay estimate holds, i.e. n

kw(·, t)kW s−n−3,∞ (Rn ) ≤ C(1 + t)− 2 , ∀ t ≥ 0, where C is independent of t. Now we return to the Cauchy problem (4.6.2). Note that Z10 (x) + iZ20 (x) w0 (x) = . 1 + Z30 (x) If there exists a constant ε0 > 0 such that ε0 ≤ 1 + Z30 (x) < 2,

∀ x ∈ Rn ,

since (Z10 (x))2 + (Z20 (x))2 (1 + Z30 (x))2 1 − (Z30 (x))2 1 − Z30 (x) = = , (1 + Z30 (x))2 1 + Z30 (x)

|w0 (x)|2 =

and

(4.6.60)

~ ~ 0 (x) · ∂ Z0 (x) = 0, j = 1, 2, . . . , n, Z ∂xj

we have 3 ∂w (x) 2 X 0 = ∂xj i=1

∂ 0 0 ∂ 0 Zi (1 + Z30 ) − Z Z ∂xj ∂xj i 3

!2

/(1 + Z30 )4

!

∂ ∂ (Z10 )2 + (Z 0 )2 /(1 + Z30 )2 = ∂xj ∂xj 2 ∂ + (Z 0 )2 ((Z10 )2 + (Z20 )2 )/(1 + Z30 )4 ∂xj 3 !

∂ 0 0 ∂ 0 ∂ 0 −2 Z3 Z1 · Z10 + Z2 Z2 /(1 + Z30 )4 ∂xj ∂xj ∂xj =

3 X i=1

∂ 0 Z ∂xj i

Then |∇w0 |2 =

!2

/(1 + Z30 )2 .

~ 0 |2 |∇Z 1 ~ 0 |. , |∇w0 | ≤ |∇Z 0 2 (1 + Z3 ) ε0

(4.6.61)

Landau–Lifshitz Equations and Harmonic Maps

215

For k > 1 it is easily verified that ~0| + |∇k w0 | ≤ Ck (ε0 ) |∇k Z

k−1 X i=1

!

~ 0 ||∇k−i Z ~0| . |∇i Z

This implies k~

k

k∇ w0 kL1 (Rn ) ≤ Ck (ε0 ) k∇ Z0 kL1 (Rn ) +

k−1 X i=1

i~

k∇ Z0 kL2 (Rn ) k∇

k−i ~

Z0 kL2 (Rn )

~ 0 kL1 (Rn ) + k∇Z ~ 0 k2 k−1 n ). ≤ Ck (ε0 )(k∇k Z H (R )

!

(4.6.62)

It follows from (4.6.60) and (4.6.61) that ~ 0 kW s−1,1 (Rn ) + k∇Z ~ 0 k2H s−1 (Rn ) ). k∇w0 kW s−1,1 (Rn ) ≤ C(ε0 )(k∇Z It is easy to verify that ~ 0 kH s + k∇Z ~ 0 k2 s ). k∇w0 kH s ≤ C(ε0 , n, s)(k∇Z H In summary, we have for s ≥ n + 4, kw0 kW s,1 + kw0 kH s+1 "

!1

#

1 − Z30 + dx ≤ 1 + Z30 Rn ~ 0 kW s−1,1 + k∇Z ~ 0 kH s + k∇Z ~ 0 k2 s ). + C(ε0 , n, s)(k∇Z H Z

1 − Z30 1 + Z30

2

(4.6.63)

~ 0 = (Z 0 , Z 0 , Z 0 ) such that Hence we only need to choose small Z 1 2 3 "

!1

#

1 − Z30 + dx 1 + Z30 Rn ~ 0 kW s−1,1 + k∇Z ~ 0 kH s + k∇Z ~ 0 k2H s ) ≤ δE. + C(ε0 , n, s)(k∇Z

Z

1 − Z30 1 + Z30

2

2. Global smooth solution to the Cauchy problem We have from Theorem 4.6.2 that ~ 0 (x) = (Z 0 (x), Z 0 (x), Z 0 (x)) such that Theorem 4.6.3 Let Z 1 2 3 ~ 0 (x)| = 1, inf x∈Rn Z 0 (x) > −1, Z 0 (x) < 1; (i) |Z 3 3 ~ 0 (x) ∈ W s−1,1 (Rn ) ∩ H s (Rn ); (ii) ∇Z (iii) Z

Rn

"

1 − Z30 1 + Z30

!1

#

1 − Z30 + dx 1 + Z30 ~ 0 kW s−1,1 + k∇Z ~ 0 kH s + k∇Z ~ 0 k2 s ) ≤ d, + C(ε0 , n, s)(k∇Z H 2

where d = d(n, s) is a small positive number, s ≥ n + 4, n ≥ 2.

(4.6.64)

Landau–Lifshitz Equations

216

Then the Cauchy problem for the Landau–Lifshitz equations ~ t = ∆Z ~ + |∇Z| ~ 2Z ~ +Z ~ × ∆Z, ~ x ∈ Rn , t > 0, Z ~ t=0 = Z ~ 0 (x), x ∈ Rn , n ≥ 2, Z|

(4.6.65) (4.6.66)

~ t) such that has a unique global smooth solution Z(x, ~ t)| = 1, ∀ (x, t) ∈ Rn × R+ ; (i) |Z(x, ~ ∈ L2 (0, T ; H s(Rn )) ∩ L∞ (0, T ; H s−1(Rn )); (ii) ∇Z ~ t ∈ L2 (0, T ; H s (Rn )) ∩ L∞ (0, T ; H s−2 (Rn )); (iii) Z ~ t)kW s−n−4,∞ (Rn ) ≤ C(1 + t)− n2 , ∀ t ≥ 0. (iv) k∇Z(·, Proof. It follows from (4.6.63), (4.6.64) and Theorem 4.6.2 that the Cauchy ~ t) = (Z1 (x, t), problem (4.6.57) admits a unique solution w(x, t) ∈ Xs,E . Define Z(x, Z2 (x, t), Z3 (x, t)) by Z1 =

2<w , 1 + |w|2

Z1 =

2=w , 1 + |w|2

Z1 =

1 − |w|2 . 1 + |w|2

~ t)| = 1 for all (x, t) ∈ Rn × R+ , and It is easy to verify that |Z(x, 2<∇w 4w∇w<w − , 2 1 + |w| (1 + |w|2 )2 2=∇w 4w∇w=w ∇Z2 = − , 2 1 + |w| (1 + |w|2 )2 −4w∇w ∇Z3 = . (1 + |w|2 )2 ∇Z1 =

Since

n

1− n

kw(·, t)kL∞ ≤ CkD s wkLs 1 kwkL1 s ≤ CkwkW s,1 , we have sup kw(·, t)kL∞ ≤ C sup kw(·, t)kW s,1 ≤ CE. t≥0

t≥0

Applying Lemmas 4.6.1 and 4.6.2, we have

w

≤ CkwkW k,p(Rn ) ,

1 + |w|2 k,p n W (R )



1



≤ CkwkW k,p(Rn ) kwk∞ ≤ CkwkW k,p .

1 + |w|2 k,p n W

Hence

~ W k,p (Rn ) k∇Zk

(R )



1

≤ C[(k<∇wkW k,p + k=∇wkW k,p )

2 1 + |w| ∞



1

+ (k<∇wk∞ + k=∇wk∞)

1 + |w|2 k,p W

Landau–Lifshitz Equations and Harmonic Maps

217





w w=w



+

+

(1 + |w|2 )2 k,p

(1 + |w|2 )2 k,p W W

!

w<w

+

k∇wk∞

(1 + |w|2 )2 k,p W





w w=w



+

+

(1 + |w|2 )2

(1 + |w|2 )2 ∞ ∞

! #

w<w

+

k∇wkW k,p

(1 + |w|2 )2 ∞

≤ C1 [k<∇wkW k,p + k=∇wkW k,p

+ (k<∇wk∞ + k=∇wk∞) · kwkW k,p ]

+ C2 kwkW k,p k∇wk∞ + C3 k∇wkW k,p

≤ d1 k∇wkW k,p + d2 k∇wk∞ kwkW k,p .

(4.6.67)

Taking k = s − 1, p = 1 in (4.6.67), we have

~ t)kW s−1,1 (Rn ) ≤ d1 sup k∇wkW s−1,1 sup k∇Z(·, t≥0

t≥0

+ d2 sup kw(·, t)kW s−n−3,∞ sup kw(·, t)kW s−1,1 < ∞. t≥0

t≥0

Taking k = s + 1, p = 2 in (4.6.67), we have ~ t)kH s+1 ≤ d1 k∇wkH s+1 k∇Z(·, n

+d2 E(1 + t)− 2 kwkH s+1 ≤ dkwkW s+2 ,

and

Z

T 0

~ t)k2 s+1 dt ≤ d2 k∇Z(·, H

Z

T 0

kw(·, t)k2W s+2 dt ≤ d2 E 2 < ∞.

(4.6.68) (4.6.69)

Finally, we take k = s − n − 4, p = ∞ in (4.6.67) to give ~ t)kW s−n−4,∞ ≤ d1 kw(·, t)kW s−n−3,∞ k∇Z(·,

+ d2 kw(·, t)kW s−n−3,∞ kw(·, t)kW s−n−4,∞ n

≤ d1 E(1 + t)− 2 + d2 E 2 (1 + t)−n n

< (d1 E + d2 E 2 )(1 + t)− 2 , ∀ t > 0.

(4.6.70)

~ t) satisfies the equation and the initial conDirect computations show that Z(x, dition. The theorem is proved.

4.7

Radial Solution

The result of this section shows that the radial symmetric Landau–Lifshitz equations always admits smooth solution outside a cylinder centered at r = 0. In other words, the only possible blowing-up must occur at the axis r = 0 for the radial symmetric problem.

Landau–Lifshitz Equations

218

4.7.1

Two-Dimensional Radial Symmetric Landau–Lifshitz Equation

This section is devoted to the radial solutions of the Cauchy problem 1 ut = u × urr + u × ur , r

(4.7.1)

u(r, t = 0) = φ(r),

(4.7.2)

where u : Ω × R+ −→ S 2 , r = |x|, x ∈ Ω = {x ∈ R2 | |x| > R0 } and u satisfies the following boundary condition: ∂r u|r=R0 = 0, (4.7.3) where constant R0 > 0. The following results will be established. Theorem 4.7.1 Let initial data φ(r) ∈ S 2 and φr ∈ H m (Ω) (∀ m ≥ 1). Then for all time T > 0, there exists a solution of problem (4.7.1)–(4.7.3) such that u ∈ S 2 and urk1 tk2 (x, t) ∈ L∞ (0, T ; L2 (Ω)), where 1 ≤ 2k2 + k1 ≤ m + 1. Moreover, for all integer m ≥ 3, the solution is unique.

4.7.2

A Priori Estimates

To prove Theorem 4.7.1, we will construct the local (in time t) solutions of problem (4.7.1)–(4.7.3), and then extend it to large time t by using a priori estimates. Proposition 4.7.1 There exists a positive constant C such that the following inequality holds for all u ∈ H01 (R0 , ∞) and r α u, r α ur ∈ L2 (R0 , ∞) kr β ukLpr ≤ Ckr α ur kbL2r kr α uk1−b L2r

(4.7.4)

if and only if the following relations hold : 1 1 + β = + α − b, p 2 α−σ ≥0 α−σ ≤1

if b > 0, 1 1 if b > 0 and α − = + β, 2 p

where p > 0, 0 ≤ b ≤ 1, α > − 12 , β > − 1p and β = bσ + (1 − b)α. Lemma 4.7.1 For all solutions of problem (4.7.1)–(4.7.3), we have the following estimations for all time T > 0: kr 1/2 ur kL∞ (0,T ; L2r ) = kr 1/2 φr kL2r ,

(4.7.5)

kr 1/2 ut kL∞ (0,T ;L2r ) + kr 1/2 urr kL∞ (0,T ;L2r ) ≤ C(T, R0 , kr 1/2 φr kL2r , kr 1/2 φrr kL2r ), (4.7.6) 



kur kL∞ (0,T ;L∞ (R0 ,∞)) ≤ C T, R0 , kr 1/2 φr kL2r , kr 1/2 φrr kL2r , where k · kL2r = k · kL2 (R0 ,∞) .

(4.7.7)

Landau–Lifshitz Equations and Harmonic Maps

219

Proof. Taking the scalar product of rurr and Eq. (4.7.1), and then integrating the result over (R0 , ∞) × [0, T ], we have (4.7.5). To prove (4.7.6), we first differentiate (4.7.1) with respect to t, and get 1 1 utt = ut × urr + u × urrt + ut × ur + u × urt . r r

(4.7.8)

Taking the scalar product of rut and Eq. (4.7.8), and then integrating the result over (R0 , ∞), we get Z ∞ Z ∞ rut · utt dr = − rut · (ur × urt )dr. (4.7.9) R0

R0

We differentiate (4.7.1) with respect to r, and get 1 1 urt = ur × urr + u × urrr + u × urr − 2 u × ur . r r

(4.7.10)

Substituting (4.7.10) into (4.7.9), we have Z

∞ R0

rut · utt dr =

Z

∞ R0

(r(ur · ur )ut · urr − r(ur · urr )ur · ut ) dr

Z ∞ 1d Z∞ 2 ur · ur (u × ur ) · urr dr. (4.7.11) = r(ur · ur ) dr + 8 dt R0 R0

By using (4.7.1), one has Z

∞ R0

rut · ut dr =

Z

∞ R0



1 rurr · urr − r(ur · ur )2 + ur · ur dr. r 

(4.7.12)

Substituting (4.7.12) into (4.7.11), we get d dt

Z

∞ R0



1 5 rurr · urr + ur · ur − r(ur · ur )2 dr = 2 r 4 

Z

∞ R0

ur ·ur (u×ur )·urr dr. (4.7.13)

The right-hand side of (4.7.13) is bounded from above by 2

Z

∞ R0

2

r |urr | dr

!1/2

Z

∞ R0

1 |ur |6 dr r

!1/2

≤ Ckr 1/2 urr kL2 kr 1/2 ur k3L6 ≤ Ckr 1/2 urr k2L2 kr 1/2 ur k2L2 ,

(4.7.14)

where we have used Proposition 4.7.1. The third term on the left-hand side of (4.7.13) is bounded by 5 −1 1/2 4 1 R0 kr ur kL4 ≤ Ckr 1/2 urr kL2 kr 1/2 ur k3L2 ≤ kr 1/2 urr k2L2 + Ckr 1/2 ur k6L2 , (4.7.15) 4 2 where we have used Proposition 4.7.1. Putting (4.7.13)–(4.7.15) together and using Gronwall’s inequality, we get 



kr 1/2 urr kL∞ (0,T ;L2r ) ≤ C T, R0 , kr 1/2 φr kL2r , kr 1/2 φrr kL2r . Putting (4.7.12), (4.7.15) and (4.7.16) together, we have (4.7.6).

(4.7.16)

Landau–Lifshitz Equations

220

By using Proposition 4.7.1, we have −1/4

kur kL∞ (0,T ;L∞ (R0 ,∞)) ≤ R0

kr 1/4 ur kL∞ (0,T ;L∞ (R0 ,∞))

−1/4

≤ CR0

3/4

1/4

kr 1/2 urr kL2 kr 1/2 ur kL2 .

(4.7.17)

Putting (4.7.5), (4.7.6) and (4.7.17) together, we get (4.7.7). Lemma 4.7.2 For all solutions of problem (4.7.1)–(4.7.3), we have the following estimations for all time T > 0: kr 1/2 utt kL∞ (0,T ;L2r ) + kr 1/2 urrt kL∞ (0,T ;L2r ) + kr 1/2 ur4 kL∞ (0,T ;L2r ) 



≤ C T, R0 , kr 1/2 φr kL2r , . . . , kr 1/2 φr4 kL2r ,

kut kL∞ (ΩT ) + kurt kL∞ (ΩT ) + kurr kL∞ (ΩT ) + kur3 kL∞ (ΩT ) ≤ C,

(4.7.18) (4.7.19)

where k · kL2r = k · kL2 (R0 ,∞) and ΩT = [0, T ] × Ω. Proof. To prove (4.7.18), we first differentiate (4.7.8) with respect to t, and get uttt = utt × ∆u + 2ut × ∆ut + u × ∆utt ,

(4.7.20)

where ∆w = wrr + wrr . Taking the scalar product of rutt and Eq. (4.7.20), and then integrating the result over (R0 , ∞), we get Z

∞ R0

ruttt · utt dr =

Z

∞ R0

{rurtt · (ur × utt ) − 2rurtt · (ut × urt )} dr.

(4.7.21)

After substituting (4.7.8) and (4.7.10) in (4.7.21), the right-hand side of (4.7.21) becomes Z

∞ R0

rur · ut ∆u · urtt dr + +

Z

∞ R0

Z

∞ R0

rur · ∆u ut · urtt dr

r (ur · ∆ut − 2ut · (∆u)r ) u · urtt dr.

(4.7.22)

Using integration by parts and estimates (4.7.5)–(4.7.7), the first term of (4.7.22) is equal to Z ∞ − r {ur · urt ∆u · utt + ur · ut (∆u)r · utt } dr R0

and is bounded from above by 



C kr 1/3 urt kL3 kr 1/6 ∆ukL6 + kr 1/6 ut kL6 kr 1/3 (∆u)r kL3 kr 1/2 utt kL2 ; the second term of (4.7.22) is equal to −

Z

∞ R0

r {ur · ∆u urt · utt + ur · (∆u)r ut · utt } dr,

(4.7.23)

Landau–Lifshitz Equations and Harmonic Maps

221

and is also bounded from above by (4.7.23). Using the fact u · u = 1, we have u · urtt = −utt · ur − 2ut · urt .

(4.7.24)

Substituting (4.7.24) in (4.7.22), the third term of (4.7.22) is bounded from above by 

C kr 1/2 utt kL2 + kr 1/3 urt kL3 kr 1/6 ut kL6



kr 1/2 urrt kL2 + 1



+ Ckr 1/6 ut kL6 kr 1/3 (∆u)r kL3 kr 1/2 utt kL2 + Ckr 1/6 ut k2L6 kr 1/3 (∆u)r kL3 kr 1/3 urt kL3 ,

(4.7.25)

where we have used the following inequality: 

1/2



kr 1/2 urt kL2 ≤ C kr 1/2 urrt kL2 + 1 . In fact, we have kr 1/2 urt k2L2 = ≤

Z

∞

rurt · urt dr = −

R0

Z

(4.7.26)

∞ R0

(ut · urt + rut · urrt ) dr

1 1/2 kr urt k2L2 + Ckr 1/2 ut k2L2 + kr 1/2 urrt kL2 kr 1/2 ut kL2 . 2

Using Proposition 4.7.1 and (4.7.25), one has 1/3



2/3



2/3

kr 1/3 urt kL3 ≤ Ckr 1/2 urrt kL2 kr 1/2 urt kL2 ≤ C kr 1/2 urrt kL2 + 1 . By using (4.7.1), we get

ut · ut = ∆u · ∆u − (ur · ur )2 .

(4.7.27)

(4.7.28)

Using (4.7.5)–(4.7.7), (4.7.28) and Sobolev’s imbedding theorems, we have 







kr 1/6 ut kL6 ≤ C kr 1/6 ∆ukL6 + 1 ≤ C kr 1/6 urr kL6 + 1 .

(4.7.29)

Similarly, using (4.7.10), (4.7.5)–(4.7.7) and Sobolev’s imbedding theorems, one has 2 (∆u)r · (∆u)r = urt · urt + (ur · ∆u)2 + ur · ∆u − ur · ur r 



2

− ur · ur ∆u · ∆u,

kr 1/3 (∆u)r kL3 ≤ C kr 1/3 urt kL3 + kr 1/3 ∆ukL3 + 1 

2/3

1/2





≤ C kr 1/2 urrt kL2 + kr 1/6 urr kL6 + 1 . Using integration by parts, (4.7.27) and estimations (4.7.5)–(4.7.7), we get kr 1/6 urr k6L6 =

Z

∞ R0

=−

Z

r (urr · urr )3 dr ∞

R0

{ur · urr (urr · urr )2 + rur · urrr (urr · urr )2

+ 2rur · urr urr · urr urr · urrr }dr 1 ≤ kr 1/6 urr k6L6 + C + Ckr 1/6 urr k4L6 kr 1/3 urrr kL3 , 2

(4.7.30)

Landau–Lifshitz Equations

222



1/2

1/2

1/2

kr 1/6 urr kL6 ≤ C(kr 1/3 urrr kL3 + 1) ≤ C kr 1/3 urt kL3 + kr 1/3 urr kL3 + 1   1 1/6 1/3 kr urr kL6 + C kr 1/2 urrt kL2 + 1 . 2

≤



(4.7.31)

Substituting estimates (4.7.22) and (4.7.23), (4.7.25)–(4.7.27) and (4.7.29)–(4.7.31) into (4.7.21), we get d 2 dt

Z

∞ R0





rutt · utt dr ≤ C kr 1/2 utt k2L2 + kr 1/2 urrt k2L2 + 1 .

(4.7.32)

Using Eq. (4.7.8), one has utt · utt = ∆ut · ∆ut + 2ur · ur ut · ∆ut + ut · ut ∆u · ∆u − 4 (ur · urt )2 ,

(4.7.33)

where we have used the facts ∆u · ut = 0, u · u = 1, u · ∆u = −ur · ur and u · ∆ut = −2ur · urt . Substituting (4.7.33) into (4.7.32), the left-hand side of (4.7.32) reads d dt

Z

∞ R0

n

o

r |utt |2 + |∆ut |2 + 2 |ur |2 ut · ∆ut + |ut |2 |∆u|2 − 4 (ur · urt )2 dr.

(4.7.34)

Using (4.7.5)–(4.7.7) and (4.7.26), the second term of (4.7.34) is rewritten as Z

∞ R0

2

r |∆ut | dr =

Z

∞ R0



r |urrt |2 +

 1 |urt |2 dr ≥ kr 1/2 urrt k2L2 , r

(4.7.35)

and the third and fifth terms of (4.7.34) is bounded from above by 1 1/2 kr urrt k2L2 + C. 5

(4.7.36)

The fourth term of (4.7.34) is bounded from above by 





C kr 1/4 urr k4L4 + 1 ≤ C kr 1/6 urr k3L6 + 1



  1 ≤ C kr 1/2 urrt kL2 + 1 ≤ kr 1/2 urrt k2L2 + C, 5

(4.7.37)

where we have used (4.7.5)–(4.7.7), (4.7.28) and (4.7.31). Putting (4.7.32) and (4.7.34)–(4.7.37) together and using Gronwall’s inequality, we get kr 1/2 utt kL∞ (0,T ;L2r ) + kr 1/2 urrt kL∞ (0,T ;L2r ) ≤ C.

(4.7.38)

Using (4.7.10), (4.7.26) and (4.7.38), one has kr 1/2 urt kL∞ (0,T ;L2r ) + kr 1/2 (∆u)r kL∞ (0,T ;L2r ) ≤ C.

(4.7.39)

Landau–Lifshitz Equations and Harmonic Maps

223

Differentiating (4.7.10) with respect to r, we get 1 urrt = urr × ur + 2ur × (∆u)r + u × (∆u)rr = I1 + u × (∆u)rr , r (4.7.40)

(∆u)rr · (∆u)rr ≤ 2urrt · urrt + 2I1 · I1

+ {2ur · (urr + ∆u)r + urr · (2urr + ∆u)}2 ,

where we have used the fact u · (∆u)rr = −2ur · (urr + ∆u)r − urr · (2urr + ∆u). Putting together (4.7.38)–(4.7.40), we have (4.7.18). By using (4.7.18) and Proposition 4.7.1, we have kurt kL∞ (0,T ;L∞ (R0 ,∞)) + kurrr kL∞ (0,T ;L∞ (R0 ,∞)) ≤ C.

(4.7.41)

Using (4.7.18) and Sobolev’s imbedding theorems, one has kut kL∞ (0,T ;L∞ (R0 ,∞)) + kurr kL∞ (0,T ;L∞ (R0 ,∞)) ≤ C.

(4.7.42)

Putting (4.7.41) and (4.7.42) together, we get (4.7.19). Lemma 4.7.3 For all solutions of problem (4.7.1)–(4.7.3), we have the following estimations for all time T > 0 and all integer m ≥ 2: kr 1/2 utm kL∞ (0,T ;L2r ) + kr 1/2 urrtm−1 kL∞ (0,T ;L2r ) + kr 1/2 ur4 tm−2 kL∞ (0,T ;L2r ) 



≤ C T, R0 , kr 1/2 φr kL2r , . . . , kr 1/2 φr2m kL2r ,

(4.7.43)

kutm−1 kL∞ (ΩT ) + kurtm−1 kL∞ (ΩT ) + kurrtm−2 kL∞ (ΩT ) + kur3 tm−2 kL∞ (ΩT ) ≤ C, (4.7.44) where k · kL2r = k · kL2 (R0 ,∞) and ΩT is defined in Lemma 4.7.2. Proof. This lemma is proved by mathematical induction as follows. For m = 2, estimates (4.7.43) and (4.7.43) have been proved in Lemma 4.7.2. Suppose that (4.7.43) and (4.7.43) are valid for all 2 ≤ m ≤ M . Differentiating (4.7.1), we have utM +2 =

M +1 X k=0

!

M +1 utM +1−k × ∆utk , k

(4.7.45)

where ∆w = wrr + wrr . Taking the scalar product of rutM +1 and Eq. (4.7.45), and then integrating the result over (R0 , ∞), we get Z

∞ R0

rutM +2 · utM +1 dr = (M + 1) +

Z

+

Z

Z

M ∞X

R0 k=2 ∞ R0

∞

R0

rutM +1 · (utM × ∆ut ) dr !

M +1 rutM +1 · (utM +1−k × ∆utk ) dr k

rutM +1 · (u × ∆utM +1 ) dr = I1 + I2 + I3 . (4.7.46)

Landau–Lifshitz Equations

224

By the induction assumption and Sobolev’s imbedding theorems, we obtain I1 ≤ CkutM kL∞ (Ω) kr 1/2 ∆ut kL2 kr 1/2 utM +1 kL2 



≤ C 1 + kr 1/2 ∆utM kL2 kr 1/2 utM +1 kL2 ,

I2 ≤ C

M X

(4.7.47)

kutM +1−k kL∞ (Ω) kr 1/2 ∆utk kL2 kr 1/2 utM +1 kL2 ≤ Ckr 1/2 utM +1 kL2 .

k=2

(4.7.48)

Integrating by parts, one has Z

I3 =

∞ R0

r (ur × utM +1 ) · urtM +1 dr.

(4.7.49)

Differentiating (4.7.1), we have utM +1 =

M X

k=0

ur × utM +1 =

M X

k=0

M k

!

M k

!

utM −k × ∆utk ,

(4.7.50)

(ur · ∆utk utM −k − ur · utM −k ∆utk ) .

(4.7.51)

Using the fact that u · u = 1, one has u · urtM +1 = −

M X

k=0

!

M +1 utM +1−k · urtk . k

(4.7.52)

Substituting (4.7.51) and (4.7.52) into (4.7.49) and integrating by parts, −I3 becomes Z

−1 ∞M X

R0 k=0

M k

!

(ur · ∆utk utM −k + rurr · ∆utk utM −k + rur · ∆utk urtM −k

− ur · utM −k ∆utk − rurr · utM −k ∆utk − rur · urtM −k ∆utk ) · utM +1 dr + + +

Z

−2 ∞M X

R0 k=0 Z ∞ R0

Z

M k

!

(rur · (∆u)rtk utM −k − rur · utM −k (∆u)rtk ) · utM +1 dr

(rur · (∆u)rtM −1 ut − rur · ut (∆u)rtM −1 ) · utM +1 dr

M ∞X

R0 k=0

!

M +1 rur · ∆utM utM +1−k · urtk dr. k

(4.7.53)

By the induction assumption, the integrations corresponding to the first integral operator and the second integral operator of (4.7.53) is bounded from above by Ckr 1/2 utM +1 kL2 ;

(4.7.54)

and the integration corresponding to the fourth integral operator of (4.7.53) is bounded from above by 



Ckr 1/2 urrtM kL2 kr 1/2 utM +1 kL2 + kr 1/2 urtM kL2 + 1 .

(4.7.55)

Landau–Lifshitz Equations and Harmonic Maps

225

Differentiating (4.7.1), we have urtM = u × (∆u)rtM −1 + ur × ∆utM −1 +

M −2 X k=0

!

M −1 (utM −1−k × ∆utk )r , (4.7.56) k

(u × (∆u)rtM −1 ) · (u × (∆u)rtM −1 ) = |(∆u)rtM −1 |2 − (u · (∆u)rtM −1 )2 .

(4.7.57)

By the fact u · u = 1, we have M −2 X

!

M −1 u · (∆u)rtM −1 = − (2ur · urr + ur · ∆u)tM −1 − utM −1−k × (∆u)rtk . k k=0 (4.7.58) Putting (4.7.56)–(4.7.58) together and by the induction assumption, one has 



kr 1/2 (∆u)rtM −1 kL2 ≤ C 1 + kr 1/2 urtM kL2 . By the same argument as in the proof of (4.7.26), we obtain 

1/2

(4.7.59)



kr 1/2 urtM kL2 ≤ C 1 + kr 1/2 urrtM kL2 .

(4.7.60)

Using (4.7.59) and (4.7.60), the integration corresponding to the third integral operator of (4.7.53) is bounded from above by 

1/2



Ckr 1/2 utM +1 kL2 kr 1/2 (∆u)rtM −1 kL2 ≤ Ckr 1/2 utM +1 kL2 1 + kr 1/2 urrtM kL2 . (4.7.61) Substituting all estimates (4.7.47)–(4.7.49), (4.7.53)–(4.7.55), (4.7.60) and (4.7.61) in (4.7.46), we get d 2 dt

Z

∞ R0





rutM +1 · utM +1 dr ≤ C kr 1/2 utM +1 k2L2 + kr 1/2 urrtM k2L2 + 1 .

(4.7.62)

By the fact u · u = 1, we obtain (u × ∆utM ) · (u × ∆utM ) = ∆utM · ∆utM − (u · ∆utM )2 , u · ∆utM = −(ur · ur )tM −

M −1 X k=0

M k

!

(4.7.63)

(utM −k · ∆utk ) .

(4.7.64)

Using Eqs. (4.7.50), (4.7.63) and (4.7.64), and the induction assumption, one has 



kr 1/2 urrtM k2L2 ≤ C kr 1/2 utM +1 k2L2 + kr 1/2 urtM k2L2 + 1 .

(4.7.65)

By substituting (4.7.60) and (4.7.65) into (4.7.62), (4.7.62) can be rewritten as d dt

Z

∞ R0









r |utM +1 |2 + |urrtM |2 dr ≤ C kr 1/2 utM +1 k2L2 + kr 1/2 urrtM k2L2 + 1 . (4.7.66)

Using Gronwall’s inequality, we get 



kr 1/2 utM +1 k2L2 +kr 1/2 urrtM k2L2 ≤ C T, R0 , kr 1/2 φr kL2r , . . . , kr 1/2 φr2M +2 kL2r . (4.7.67)

Landau–Lifshitz Equations

226

Using (4.7.59) and (4.7.60), one has kr 1/2 urtM kL∞ (0,T ;L2r ) + kr 1/2 (∆u)rtM −1 kL∞ (0,T ;L2r ) ≤ C.

(4.7.68)

Differentiating (4.7.56) with respect to r, we get urrtM = u × (∆u)rrtM −1 + 2ur × (∆u)rtM −1 + urr × ∆utM −1 M −2 X

!

(4.7.69)

(u × (∆u)rrtM −1 ) · (u × (∆u)rrtM −1 ) = |(∆u)rrtM −1 |2 − (u · (∆u)rrtM −1 )2 .

(4.7.70)

+

k=0

M −1 (utM −1−k × ∆utk )rr , k

By the fact u · u = 1, we have

u · (∆u)rrtM −1 = − (2ur · urrr + 2urr · urr + 2ur · (∆u)r + urr · ∆u)tM −1 −

M −2 X k=0

!

M −1 utM −1−k × (∆u)rrtk . k

(4.7.71)

Putting (4.7.68)–(4.7.71) together and by the induction assumption, one has 



kr 1/2 (∆u)rrtM −1 kL2 ≤ C 1 + kr 1/2 urrtM kL2 ≤ C.

(4.7.72)

By using (4.7.67), (4.7.68), (4.7.72) and the same arguments in the proof of (4.7.41) and (4.7.42), we have kutM kL∞ (ΩT ) + kurtM kL∞ (ΩT ) + kurrtM −1 kL∞ (ΩT ) + kur3 tM −1 kL∞ (ΩT ) ≤ C.

(4.7.73)

Thus putting (4.7.67), (4.7.72) and (4.7.73) together and using the induction procedure, one finishes the proof. Corollary 4.7.1 For all solutions of problem (4.7.1)–(4.7.3), we have the following estimations for all time T > 0 and all integer m ≥ 0: 



kurk1 tk2 kL∞ (0,T ; L2 (Ω)) ≤ C R0 , T, kφr kH m (Ω) ,

(4.7.74)

where 1 ≤ 2k2 + k1 ≤ m + 1. By using the results of Lemmas 4.7.1–4.7.3, repeating the same procedure used in the proof of 4.7.73, and employing the method of induction, Corollary 4.7.1 can be established. We shall not reproduce the procedure here.

4.7.3

Existence of Local Solutions

In the above section we have done the a priori estimate. To obtain the global smooth solution, we only need to construct the local (in time t) solution u of problem (4.7.1)– (4.7.3). This local solution u is the limit of sequence {uh } when h tends to zero, where sequence {uh } satisfies the following ordinary differential-difference system D+ D− u j 1 D+ u j duj = uj × + uj × , 2 dt h rj h

j = 1, . . . ,

(4.7.75)

Landau–Lifshitz Equations and Harmonic Maps

227

with initial data uj (0) = φj = φ(rj ),

j = 0, 1, . . . ,

(4.7.76)

and boundary conditions D+ u0 = 0,

(4.7.77)

where h is step size, 0 < h < 1, rj = R0 + jh, uj = u(rj , t) (j = 0, 1, . . .), and D+ and D− denote the forward and backward difference operators, respectively. It is well-known that there exists a local smooth solution of problem (4.7.75)–(4.7.77), uh = {uj = u(rj , t)|j = 0, 1, . . .}. In order to verify the local existence of problem (4.7.1)–(4.7.3), it suffices to get the uniform a priori estimates of uh with respect to h. D u D u Let us define δuh = { +h j |j = 0, 1, . . .} = { −h j |j = 1, . . .}. Similarly, the discrete functions δ k uh (k ≥ 2) can be defined. The norms of discrete functions δ k uh (k ≥ 0) are defined as follows: p !1/p p !1/p ∞ ∞ k k−l l X X β D+ u j β D+ D− u j kr δ uh kp = rj h = r h , j hk hk j=0 j=l D k−l D l u Dk u j j β β kr β δ k uh k∞ = sup rj +k = sup rj + k − , h h j=0,1,... j≥l β k

where 1 ≤ p < ∞ and β ≥ 0. First we are going to derive the following discrete versions of Sobolev and Gagliardo–Nirenberg type inequalities. Lemma 4.7.4 For any discrete function vh = {v0 , v1 , . . .} with v0 = 0, the following inequality holds: 1

kr β vh kp ≤ Ckr β vh k22

+ p1



kr β vh k2 + kr β δvh k2

1−1 2

p

(4.7.78)

,

where 2 ≤ p ≤ ∞, β ≥ 0, constant C is independent of the discrete functions v h and the step size h. Proof. When β = 0, (4.7.78) is well known, which has been obtained in [6, 15, 16]. When β > 0, for any m ∈ {0, 1, . . .}, we have 2β rm vm · vm =

m−1 X j=0







β rj+1 vj+1 + rjβ vj D+ rjβ vj ≤ Ckr β vh k2 {kr β δvh k2 + kr β vh k2 }.

This completes the proof of (4.7.78) for p = ∞. For 2 ≤ p < ∞, it is clear that 1− 2

2

1

kr β vh kp ≤ kr β vh k∞ p kr β vh k2p ≤ Ckr β vh k22 The proof of this lemma is complete.

+ p1



kr β vh k2 + kr β δvh k2

1−1 2

p

.

Landau–Lifshitz Equations

228

Lemma 4.7.5 For any discrete function vh = {v0 , v1 , . . .} with kr β vh k2 < ∞, the following inequality holds: 1

kr β vh kp ≤ Ckr β vh k22

+ p1



kr β vh k2 + kr β δvh k2

1−1 2

p

,

(4.7.79)

where 2 ≤ p ≤ ∞, β ≥ 0, constant C is independent of the discrete functions v h and the step size h. Proof. Since kr β vh k2 < ∞, for any m ∈ {0, 1, . . .}, we have 2β rm vm · vm = −

∞  X

j=m







β rj+1 vj+1 + rjβ vj D+ rjβ vj ≤ Ckr β vh k2 {kr β δvh k2 + kr β vh k2 }.

Thus, repeating the same procedure in the proof of Lemma 4.7.4, this lemma is proved. Lemma 4.7.6 Let φ(x) = φ(|x|) ∈ S 2 , x ∈ Ω, ∇φ(x) ∈ L2 (Ω), and suppose that uh is the solution of problem (4.7.75)–(4.7.77). Then uj (t) ∈ S 2 for all t ≥ 0 and kr 1/2 δuh k2 ≤ C,

∀ t ≥ 0,

(4.7.80)

where constant C is independent of h. Proof. Taking the scalar product of (4.7.75) and uj , we obtain that uj · ujt = 0 (j = 1, . . .). Using that φ ∈ S 2 , we have uj ∈ S 2 (j = 0, 1, . . .). In addition, taking the scalar product of (4.7.75) and rj D+ hD2− uj , and then summing the result over j from 1 to ∞, we have ∞ X

∞ X D+ u j D+ D− u j D+ D− u j · u h = uj × h. · jt 2 h h h2 j=1

rj

j=1

(4.7.81)

By direct calculation, the left-hand side of (4.7.81) is equal to ∞ ∞ X 1d X D+ D− u j D+ u j D+ u j D+ u j − · rj+1 · h− uj × h. 2 dt j=1 h h h2 h j=1

(4.7.82)

Putting (4.7.81) and (4.7.82) together, we obtain (4.7.80). Lemma 4.7.7 Suppose that the conditions of Lemma 4.7.7 are satisfied and ∇φ(x) ∈ H 2 (Ω). Then there exists a constant T0 = T0 (k∇φkH 2 ) > 0 such that sup 0≤t≤T0

sup 0≤t≤T0

sup 0≤t≤T0







kr 1/2 uht k2 + kr 1/2 δ 2 uh k2 ≤ C,





kr 1/2 δuht k2 + kr 1/2 δ 3 uh k2 ≤ C, 

kr 1/2 δuh k∞ + kr 1/2 uht k∞ + kr 1/2 δ 2 uh k∞ ≤ C,

where constants C and T0 are independent of h.

(4.7.83) (4.7.84) (4.7.85)

Landau–Lifshitz Equations and Harmonic Maps

229

Proof. We first differentiate (4.7.75) with respect to t ujtt

!

!

D+ D− ujt D+ ujt D+ D− u j D+ u j = uj × + +ujt × + , j = 1, 2, . . . . (4.7.86) 2 h rj h h2 rj h

Taking the scalar product of (4.7.86) and rj ujt , and then summing the result over j from 1 to ∞, we have ∞ X

j=1

rj ujt · ujtt h =

∞ X

j=1

=−

r j uj ×

∞ X

rj+1

j=1

∞ X D+ D− ujt D+ ujt · u h + u × · ujt h jt j h2 h j=1

D+ uj D+ ujt × · ujt h ≤ Ckδuh k∞ kr 1/2 uht k2 kr 1/2 δuht k2 . h h (4.7.87)

Taking the scalar product of (4.7.86) and rj D+ Dh2− ujt , and then summing the result over j from 1 to ∞, we have ∞ X

j=1

D+ D− ujt h h2

rj ujtt · ∞ X

∞ X D+ D− uj D+ D− ujt D+ uj D+ D− ujt = rj ujt × · h+ ujt × · h 2 2 h h h h2 j=1 j=1

+

∞ X

j=1

uj ×

D+ ujt D+ D− ujt · h. h h2

(4.7.88)

Using (4.7.86), the left-hand side of (4.7.88) is equal to −

∞ X

rj+1

j=1

∞ X D+ ujt D+ ujt D+ ujtt · h− · ujtt h h h h j=1

∞ ∞ X 1d X D+ D− uj D+ ujt D+ ujt D+ ujt =− rj+1 · h− ujt × · h 2 dt j=1 h h h2 h j=1

+

∞ X

j=1

uj ×

∞ X D+ ujt D+ D− ujt 1 D+ uj D+ ujt · h − ujt × · h; 2 h h h h j=1 rj

(4.7.89)

the first term on the right-hand side of (4.7.88) is equal to −

∞ X

j=1

rj+1 uj+1 t ×

∞ 2 X D+ D− uj D+ ujt D+ D− uj D+ ujt · h − ujt × · h; 3 h h h2 h j=1

(4.7.90)

the second term on the right-hand side of (4.7.88) is equal to −

∞ X

j=1

ujt ×

2 D+ uj D+ ujt · h. h2 h

(4.7.91)

Landau–Lifshitz Equations

230

Substituting (4.7.89)–(4.7.91) into (4.7.88), we have ∞ 1d X D+ ujt D+ ujt rj+1 · h 2 dt j=1 h h

=

∞ X

j=1

−

rj+1 uj+1 t ×

∞ X 1

j=1

rj

ujt ×

∞ 2 2 X D+ D− uj D+ ujt D+ uj D+ ujt · h + u × · h jt 3 2 h h h h j=1

D+ uj D+ ujt · h. h h

(4.7.92)

The third term on the right-hand side of (4.7.92) is bounded from above by Ckδuh k∞ kr 1/2 uht k2 kr 1/2 δuht k2 .

(4.7.93)

Substituting (4.7.75) into (4.7.92), the second term on the right-hand side of (4.7.92) is bounded from above by 



C kδuh k∞ kr 1/2 δ 2 uh k2 + kr 1/2 δ 2 uh k24 kr 1/2 δuht k2 .

(4.7.94)

Using (4.7.75), one has D 2 D− u j D+ u j D+ D− u j D+ ujt + = uj+1 × + 3 × h h h h2 uj+1 D+ D− uj uj D+ u j + − × × rj+1 h2 rj rj+1 h 2 D+ D− u j + I1j , (4.7.95) h3 2 D 2 D− u j D+ D− u j = + 3 · uj+1 h h3 2 2 D 2 D− u j D+ D− u j D− u j D+ − uj+1 · + 3 + × I1j . (4.7.96) 3 3 h h h

= uj+1 × 2 D+ D− uj D+ ujt × h3 h

Using the fact uj ∈ S 2 , we get uj+1 ·

2 D+ D− u j D+ uj+1 + D− uj+1 D+ D− uj+1 3D+ uj + D− uj D+ D− uj =− · − · . 3 h 2h h2 2h h2 (4.7.97)

Substituting (4.7.75), (4.7.96), and (4.7.97) in (4.7.92), the first term on the righthand side of (4.7.92) is bounded from above by C (kδuh k∞ + 1) kr 1/2 δ 2 uh k24 kr 1/2 δ 3 uh k2 



+ C kδuh k2∞ + kδuh k∞ kr 1/2 δ 2 uh k2 kr 1/2 δ 3 uh k2

+ Ckδuh k∞ kr 1/2 uht k2 kr 1/2 δ 3 uh k2 .

(4.7.98)

Landau–Lifshitz Equations and Harmonic Maps

231

Using the fact uj ∈ S 2 , we get 1 D+ u j D+ u j D− u j D− u j D+ D− u j =− · + · . uj · 2 h 2 h h h h 



(4.7.99)

Using (4.7.75) and (4.7.99), we infer that ujt · ujt ≥

D+ uj 4 D u 4 C D+ uj 2 1 D+ D− u j D+ D− u j −C − j − . · − C 2 h2 h2 h h rj2 h











(4.7.100)

Similarly, using (4.7.95) and (4.7.97), we have

j+1 2 2 X D− u j D+ D− u j D+ ujt D+ ujt 1 D+ · ≥ · − C h h 2 h3 h3 k=j

C D+ D− uj 2 C − 2 − 2 2 2 rj+1 h rj rj+1



Using (4.7.77) and Lemma 4.7.5, we get 1/2

1/2

kδuh k∞ ≤ Ckδuh k2 kδ 2 uh k2

! D u 2 D+ D− u k 2 D+ u k 2 + − k 2

h

h

h

D+ uj 2 .

(4.7.101)

h

1/2

1/2

≤ Ckr 1/2 δuh k2 kr 1/2 δ 2 uh k2 .

(4.7.102)

By direct calculation, we have kr

1/2 2

δ

uh k22

=−

∞ X

D+

j=1



D+ D− u j rj h2

!

·

D+ u j h h 

≤ C kr 1/2 δ 2 uh k2 + kr 1/2 δ 3 uh k2 , kr 1/2 δ 2 uh k44 = − −

∞ X

D+ rj2

∞ X

rj2



j=1

j=1

D+ D− u j D+ D− u j · h2 h2

(4.7.103) 

D+ D− uj+1 D+ uj · h h2 h

2 D− u j D+ u j D+ D− u j D+ D− u j D+ · · h 2 2 h h h3 h



≤ Ckr 1/2 δ 2 uh k24 kr 1/2 δ 2 uh k2 + kr 1/2 δ 3 uh k2 ∞ X

j=1

rj

! D+ u j 2 D u 2 D+ D− u j 2 + − j 2

h

h

5/4

(4.7.104)

,

h



≤ Ckδuh k2∞ kr 1/2 δ 2 uh k22 ≤ C kr 1/2 δ 2 uh k2 + kr 1/2 δ 3 uh k2

3/2

,

(4.7.105)

where we have used (4.7.77), (4.7.80), (4.7.102) and (4.7.103). Putting estimates (4.7.87), (4.7.92)–(4.7.94), (4.7.98) and (4.7.100)–(4.7.105) together, for all t ≥ 0, we finally obtain kr 1/2 uht k22 + kr 1/2 δ 2 uh k22 + kr 1/2 δuht k22 + kr 1/2 δ 3 uh k22 ≤C +C

Z t 0

kr 1/2 uht k2 + kr 1/2 δ 2 uh k2 + kr 1/2 δuht k2 + kr 1/2 δ 3 uh k2

3

dτ.

Landau–Lifshitz Equations

232

Thus there exists a constant T0 = T0 (k∇φkH 2 ) > 0 such that sup 0≤t≤T0





kr 1/2 uht k2 + kr 1/2 δ 2 uh k2 + kr 1/2 δuht k2 + kr 1/2 δ 3 uh k2 ≤ C.

Moreover, estimates (4.7.83) and (4.7.84) are proved. Employing (4.7.80), (4.7.83), (4.7.84), and Lemma 4.7.5, (4.7.85) is proved. This completes the proof of the lemma. Lemma 4.7.8 Suppose that the conditions of Lemma 4.7.6 are satisfied and ∇φ(x) ∈ H 3 (Ω). Then sup 0≤t≤T0





kr 1/2 uhtt k2 + kr 1/2 δ 2 uht k2 + kr 1/2 δ 4 uh k2 ≤ C, sup 0≤t≤T0





kr 1/2 δuht k∞ + kr 1/2 δ 3 uh k∞ ≤ C,

(4.7.106) (4.7.107)

where T0 is defined in Lemma 4.7.7, and constant C is independent of h. Proof. In what follows, let 0 ≤ t ≤ T0 . We first differentiate (4.7.86) with respect to t D+ D− ujtt D+ D− ujt D+ D− u j + 2ujt × + ujtt × 2 2 h h h2 uj D+ ujtt ujt D+ ujt ujtt D+ uj + × +2 × + × , j = 1, 2, . . . . (4.7.108) rj h rj h rj h

ujttt = uj ×

Taking the scalar product of (4.7.108) and rj ujtt , and then summing the result over j from 1 to ∞, we have ∞ X

j=1

rj ujtt · ujttt h = 2

∞ X

j=1

−

rj ujt ×

∞ X

j=1

rj+1

∞ X D+ ujt D+ D− ujt · u h + 2 ujt × · ujtt h jtt 2 h h j=1

D+ uj D+ ujtt × · ujtt h. h h

(4.7.109)

Substituting (4.7.75) into (4.7.109) and using (4.7.80), (4.7.83)–(4.7.85), the first term on the right-hand side of (4.7.109) is bounded from above by kuht k∞ kr 1/2 uhtt k2 kr 1/2 δ 2 uht k2 ≤ Ckr 1/2 uhtt k2 kr 1/2 δ 2 uht k2 ,

(4.7.110)

and the second term on the right-hand side of (4.7.109) is bounded from above by kuht k∞ kr 1/2 δuht k2 kr 1/2 uhtt k2 ≤ Ckr 1/2 uhtt k2 .

(4.7.111)

Using (4.7.75), we have 2 2 D+ ujtt uj+1 + uj D+ D− ujt uj+1 t + ujt D+ D− u j j + + I2t ; = × × 3 h 2 h 2 h3

(4.7.112)

Landau–Lifshitz Equations and Harmonic Maps

I2j

D+ uj+1 D+ u j D+ D− (uj+1 + uj ) uj + = × + 2 h 2h rj rj+1 2rj+1 h

!

+

233

2 uj uj+1 + uj D+ × ; 2 2rj+1 h (4.7.113)

2 2 D− ujt uj+1 + uj D+ uj D+ D− uj uj+1 t + ujt D+ uj D+ ujtt D+ u j D+ × = · + · 3 h h h h 2 h h3 2 2 D+ uj uj+1 t + ujt D+ D− uj D+ uj j − · + × I2t . (4.7.114) 3 h 2 h h

Substituting (4.7.113) and (4.7.114) in (4.7.109) and using (4.7.80) and (4.7.83)– (4.7.85), the third term on the right-hand side of (4.7.109) is bounded from above by 2   D+ u j D+ D− ujt uj+1 + uj 1/2 2 1/2 kr δ u k + 1 . rj+1 · · u h + Ckr u k − ht 2 jtt htt 2 h h3 2 j=1 (4.7.115) By direct calculation, the first term of (4.7.115) is equal to ∞ X

∞ X

rj+1

j=1

+

2 ujt uj+1 + uj D+ ujtt D+ u j D+ · · h 2 h h 2 h ∞ X D+ uj+1

j=1

+

h

2 D+ ujt uj+1 + uj+2 · uj+1 tt h h2 2

∞ X

rj+1

2 ujt D+ (uj+1 + uj ) D+ uj+1 D+ · · uj+1 tt h 2 h h 2h

∞ X

rj+1

2 2 ujt uj + uj+1 D+ u j D+ · · uj+1 tt h. 2 2 h h 2

j=1

+

·

j=1

(4.7.116)

Substituting (4.7.112) and (4.7.113) into (4.7.116) and using (4.7.80), (4.7.83)– (4.7.85), the first term of (4.7.115) is bounded from above by 



Ckr 1/2 δ 2 uht k2 kr 1/2 uhtt k2 + kr 1/2 δ 2 uht k2 + 1 .

(4.7.117)

Using the fact uj ∈ S 2 , we get u j D+ u j D+ D− ujt uj · =− × 2 h rj h

!

!

D+ uj D+ ujt D− uj D− ujt D+ D− u j · − · + · . 2 h h h h h (4.7.118) Using (4.7.86) and (4.7.118), we infer that D+ D− u j 4 D+ uj 2 D+ ujt 2 1 D+ D− ujt D+ D− ujt ujtt · ujtt ≥ · −C −C 2 h2 h2 h2 h h D− uj 2 D− ujt 2 C D− ujt 2 C D− uj 4 C D+ uj 4 −C − 2 − 2 − 4 . (4.7.119) h h rj h rj h rj h









Landau–Lifshitz Equations

234

Putting (4.7.110), (4.7.111), (4.7.115), (4.7.117) and (4.7.119) together, and using (4.7.80), (4.7.83)–(4.7.85), we finally obtain kr

1/2

uhtt k22

+ kr

1/2 2

δ

uht k22

≤ C +C

Z

t 0





kr 1/2 uhtt k22 + kr 1/2 δ 2 uht k22 dτ.

Thus by Gronwall’s inequality, one has sup 0≤t≤T0





kr 1/2 uhtt k2 + kr 1/2 δ 2 uht k2 ≤ C.

(4.7.120)

Using the fact uj ∈ S 2 , we get 2 2 2 D+ D− uj D+ u j D+ D− uj D− I3j = − · − , h4 h h3 h

uj ·

(4.7.121)

D+ (uj+1 + uj ) D+ D− uj+1 D− (3uj+1 + uj ) D+ D− uj · + · . 2h h2 2h h2 Using (4.7.95), one has I3j =

(4.7.122)

2 2 2 D+ D− u j D+ u j D+ D− uj D− I1j D+ D− ujt = u × + + × . j h2 h4 h h3 h

(4.7.123)

By using (4.7.121)–(4.7.123), we have 2 2 2 2 D− uj D+ D− u j D+ · h4 h4

2

2

2 !

2 D Ij D Ij D− uj D+ D− ujt D+ D− ujt D+ uj 2 D+ − 3 − 1 ≤C · + + + 2 2 3 h h h h h h









. (4.7.124)

Using (4.7.80), (4.7.83)–(4.7.85), (4.7.120) and (4.7.124), we infer that sup kr 1/2 δ 4 uh k2 ≤ C.

(4.7.125)

0≤t≤T0

Putting (4.7.84), (4.7.120) and (4.7.125) together and using Lemma 4.7.5, this lemma is proved. Lemma 4.7.9 Suppose that the conditions of Lemma 4.7.6 are satisfied and ∇φ(x) ∈ H 2m−1 (Ω). Then for all integer m ≥ 2 sup 0≤t≤T0





kr 1/2 uhtm k2 + kr 1/2 δ 2 uhtm−1 k2 + kr 1/2 δ 4 uhtm−2 k2 ≤ C, sup 0≤t≤T0



(4.7.126)

kr 1/2 uhtm−1 k∞ + kr 1/2 δuhtm−1 k∞ 

+ kr 1/2 δ 2 uhtm−2 k∞ + kr 1/2 δ 3 uhtm−2 k∞ ≤ C, where T0 is defined in Lemma 4.7.7, and constant C is independent of h.

(4.7.127)

Landau–Lifshitz Equations and Harmonic Maps

235

Proof. In what follows, let 0 ≤ t ≤ T0 . This lemma is proved by mathematical induction as follows. For m = 2, estimates (4.7.126) and (4.7.127) have been proved in Lemma 4.7.7 and Lemma 4.7.8. Suppose that (4.7.126) and (4.7.127) are valid for all 2 ≤ m ≤ M − 1. Differentiating (4.7.75), we have ujtM +1 = uj × +

D+ D− ujtM uj D+ ujtM × + h2 rj h

M −1 X

M k

k=0

!

ujtM −k

!

D+ D− ujtk D+ ujtk , × + h2 hrj

j = 1, 2, . . . . (4.7.128)

Taking the scalar product of (4.7.128) and rj ujtM , and then summing the result over j from 1 to ∞, we have ∞ X

j=1

rj ujtM +1 · ujtM h

=

∞ X

j=1

+

r j uj ×

≤−

rj

∞ X

rj+1

k=1

j=1

+C ≤−

M −1 X

∞ X

j=1

∞ X D+ D− ujtM D+ ujtM M · u · ujtM h h + u × j jt h2 h j=1

M −1 X

k=1 ∞ X

!

ujtM −k

D+ D− ujtk D+ ujtk × + h hrj

!

· ujtM h

D+ uj D+ ujtM × · ujtM h h h 



kuhtM −k k∞ kr 1/2 δ 2 uhtk k2 + kr 1/2 δuhtk k2 kr 1/2 uhtM k2

rj+1

j=1

M k

D+ uj D+ ujtM × · ujtM h h h





+ C kuhtM −1 k∞ + kr 1/2 δuhtM −1 k2 + kr 1/2 δ 2 uhtM −1 k2 + 1 kr 1/2 uhtM k2 . (4.7.129) By direct calculation, we get kr

1/2

δuhtM −1 k22

=− ≤

∞ X

D−

j=1

D+ ujtM −1 rj h

!

· ujtM −1

  1 1/2 kr δuhtM −1 k22 + C kr 1/2 δ 2 uhtM −1 k2 + 1 . 2

(4.7.130)

Employing Lemma 4.7.5, one has 

kuhtM −1 k∞ ≤ C kr 1/2 δuhtM −1 k2 + 1

1/2



≤ C kr 1/2 δ 2 uhtM −1 k2 + 1

1/4

,

(4.7.131)

Landau–Lifshitz Equations

236

where we have used the induction assumption in (4.7.129)–(4.7.131). Using (4.7.75), we have 2 D+ ujtM uj+1 + uj D+ D− ujtM −1 = × h 2 h3 ! M −2 X M − 1 uj+1 tM −1−k + ujtM −1−k D 2 D− u k j + × + 3 jt + I2t M −1 , k 2 h k=0 (4.7.132) 2 D+ uj D+ ujtM D+ uj D+ D− ujtM −1 uj+1 + uj D+ uj j × = · + × I2t M −1 3 h h h h 2 h ! ! M −2 2 X uj+1 tM −1−k + ujtM −1−k D+ D− ujtk M − 1 D+ u j . + × × k h 2 h3 k=0 (4.7.133)

where I2j is defined in (4.7.113). Substituting (4.7.132) and (4.7.132) into (4.7.129) and using the induction assumption, the first term on the right-hand side of (4.7.129) is bounded from above by −

∞ X

j=1

rj+1

2 D+ u j D+ D− ujtM −1 uj+1 + uj · · ujtM h h h3 2





+ Ckr 1/2 uhtM k2 kr 1/2 δ 2 uhtM −1 k2 + kr 1/2 δuhtM −1 k2 + kr 1/2 δ 3 uhtM −2 k2 + 1 . (4.7.134) By direct calculation, the first term of (4.7.134) is equal to ∞ X

rj+1

j=1

+

2 ujtM −1 uj+1 + uj D+ ujtM D+ u j D+ · · h h h2 2 h ∞ X D+ uj+1

j=1

+

h

2 D+ ujtM −1 uj+1 + uj+2 · uj+1 tM h h2 2

∞ X

rj+1

2 ujtM −1 D+ (uj+1 + uj ) D+ uj+1 D+ · · uj+1 tM h h h2 2h

∞ X

rj+1

2 2 D+ u j D+ ujtM −1 uj + uj+1 · · uj+1 tM h. 2 h h2 2

j=1

+

·

j=1

(4.7.135)

Substituting (4.7.132) into (4.7.135) and using the induction assumption, the first term of (4.7.134) is bounded from above by 

Ckr 1/2 δ 2 uht k2 kr 1/2 uhtM k2 + kr 1/2 δ 2 uhtM −1 k2 

+ kr 1/2 δuhtM −1 k2 + kr 1/2 δ 3 uhtM −2 k2 + 1 .

(4.7.136)

Landau–Lifshitz Equations and Harmonic Maps

237

Using the fact uj ∈ S 2 , we get M −2 X D+ D− ujtk D+ D− ujtM −1 M −1 uj · = − u M −1−k · jt k h2 h2 k=0

!

−

1 D+ u j D+ u j D− u j D− u j · + · 2 h h h h 



tM −1

, (4.7.137)

2 D+ D− ujtM −2 h3 ! M −3 X D 2 D− u k M −2 =− uj+1 tM −2−k · + 3 jt k h k=0

uj+1 ·

D+ uj+1 + D− uj+1 D+ D− uj+1 3D+ uj + D− uj D+ D− uj − · + · 2h h2 2h h2 



. tM −2

(4.7.138)

Using (4.7.137) and (4.7.138), the induction assumption, and the following equations, ujtM = uj × +

D+ D− ujtM −1 uj D+ ujtM −1 + × h2 rj h

M −2 X k=0

!

!

D+ D− ujtk D+ ujtk M −1 ujtM −1−k × + , k h2 hrj

j = 1, 2, . . . , (4.7.139)

D 2 D− u M −2 D+ ujtM −1 = uj+1 × + 3 jt h h ! M −3 2 X M −2 D+ D− ujtk j ujtM −2−k × + + I1t M −2 , 3 k h k=0

j = 1, 2, . . . , (4.7.140)

we infer that 

kr 1/2 δ 2 uhtM −1 k22 ≤ C kr 1/2 uhtM k22 + kr 1/2 δuhtM −1 k22 + 1 





≤ C kr 1/2 uhtM k22 + kr 1/2 δ 2 uhtM −1 k2 + 1 , 





(4.7.141) 

kr 1/2 δ 3 uhtM −2 k22 ≤ C kr 1/2 δuhtM −1 k22 + 1 ≤ C kr 1/2 δ 2 uhtM −1 k2 + 1 , (4.7.142) where I1j is defined in (4.7.95) and (4.7.130) has been used. Putting (4.7.129)– (4.7.131), (4.7.134), (4.7.136), (4.7.141) and (4.7.142) together and using the induction assumption, we finally obtain kr 1/2 uhtM k22 + kr 1/2 δ 2 uhtM −1 k22 ≤ C + C Thus by Gronwall’s inequality, one has sup 0≤t≤T0



Z

t 0





kr 1/2 uhtM k22 + kr 1/2 δ 2 uhtM −1 k22 dτ. 

kr 1/2 uhtM k2 + kr 1/2 δ 2 uhtM −1 k2 ≤ C.

(4.7.143)

Landau–Lifshitz Equations

238

Putting (4.7.130), (4.7.142) and (4.7.143) together, one has sup 0≤t≤T0





kr 1/2 δuhtM −1 k2 + kr 1/2 δ 3 uhtM −2 k2 ≤ C.

(4.7.144)

Using the fact uj ∈ S 2 , we get M −3 X D 2 D 2 ujtk D 2 D 2 ujtM −2 M −2 uj · + − 4 =− ujtM −2−k · + − k h h4 k=0

!

2 D+ u j D+ D− u j − · h h3

!

tM −2

−

j D− I3t M −2 , h

(4.7.145)

where I3j is defined in (4.7.122). Employing (4.7.123), one has −3 2 2 2 2 D+ D− ujtk D+ D− ujtM −2 MX D+ D− ujtM −1 M −2 u M −2−k × = u × + j jt 2 4 4 k h h h k=0

!

2 D− u j D+ u j D+ + × h h3

!

+ tM −2

j D− I1t M −2 . h

(4.7.146)

Using (4.7.143)–(4.7.146) and the induction assumption, we infer that 



sup kr 1/2 δ 4 uhtM −2 k2 ≤ C kr 1/2 δ 2 uhtM −1 k2 + 1 ≤ C.

0≤t≤T0

(4.7.147)

Putting (4.7.143), (4.7.144) and (4.7.147) together, and from the induction assumption and Lemma 4.7.5, we obtain sup 0≤t≤T0





kr 1/2 uhtM −1 k∞ + kr 1/2 δuhtM −1 k∞ + kr 1/2 δ 2 uhtM −2 k∞ + kr 1/2 δ 3 uhtM −2 k∞ ≤ C.

Thus by the induction procedure, this lemma is proved. Using the results of Lemmas 4.7.5–4.7.9, one can achieve the local existence of solution for problem (4.7.1)–(4.7.3). Thus, we propose Proposition 4.7.2 For problem (4.7.1)–(4.7.3), suppose that the initial function φ(x) ∈ S 2 and ∇φ(x) ∈ H 2m−1 (Ω), m ≥ 1. Then there exists a constant T0 = T0 (k∇φkH 2m−1 ) > 0 and a solution such that ∂tm u(x, t), ∆∂tm−1 u(x, t), ∆2 ∂tm−2 u(x, t) ∈ L∞ (0, T0 ; L2 (Ω)). Using Lemmas 4.7.1–4.7.3 and Corollary 4.7.1, the local smooth solution u(x, t) constructed in Proposition 4.7.2 can be extended to all time t > 0. Thus theorem 4.7.1 is proved.

Landau–Lifshitz Equations and Harmonic Maps

4.8

239

Bibliography Comments

This chapter is mainly regarding the global solution theory for the higher-dimensional Landau–Lifshitz equation on the Riemannian manifold. The studies on higher-dimensional Landau–Lifshitz equations began in the early 1980s by Yulin Zhou, Boling Guo and Hesheng Sun (see [150–157]). They were mainly concerned with the global existence of weak solutions by the Galerkin method and the difference method. At the same time, the same problem was also discussed by Sulem and Sulem [132]. F. Alouger and A. Soyeur in 1992 [4] obtained a similar result to that by Zhou and Guo et al. [150–156]. After Zhou, Guo and Tan obtained smooth solution for the one-dimensional problem in 1991 [158], Guo and Hong, in 1993 [77], first discussed the two-dimensional problem. They discovered the relationships between the Landau–Lifshitz equations and the harmonic maps and established the partial regularity theory by using the method dealing with harmonic maps, that is, they proved that two-dimensional Landau–Lifshitz equation admits a global almost smooth solution which is now called the Chen–Struwe solution in the field of harmonic map heat flow. In [87], Guo and Wang proved the above results for the generalized Landau– Lifshitz systems which are from Rn → S 2 , n ≥ 2. Later, Guo and Ding applied the penalty method to the initial boundary value problems of saturated or unsaturated or the case with applied field Landau–Lifshitz systems in two dimensions. The existence theory of global solution to the L–L systems in higher dimensions can be found in [59]. In [29, 31], Chen, Guo and Ding proved that any weak solution with finite energy, decreasing in time, must be almost smooth for the two-dimensional problem. In 2003, Xiangao Liu [109] proved a partial regularity theorem for the stationary weak solution using the similar method applied to the harmonic map heat flow by Feldman [58]. And, Roger Moser [115] obtained a similar result for the lowerdimensional problem (n ≤ 4). In 2004, Ding and Guo [49] proved a similar result for the Landau–Lifshitz–Maxwell equations, see the next chapter. Recently, Melcher [113] proved that there exists a partially regular solution for the three-dimensional Cauchy problem of Landau–Lifshitz equations. This result extends the regularity theorem from a stationary weak solution to a weak solution. However, Melcher’s method does not apply to higher-dimensional problems. In 2006, Changyou Wang [143] extended Melcher’s result to three and four dimensions by the Pohozaev method. In 2007, Ding and Wang [52] proved that for some initial data, the finite time blowup for the solutions to the Dirichlet problem and the Neumann problem does occur. These results answer the long unsolved questions: Do Landau–Lifshitz equations really blow up at finite time? For the other problems, such as vortices and domain walls, as well as steady state problems, we refer to [18–24, 44–46, 116, 117, 120].

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Chapter 5 Landau–Lifshitz–Maxwell Equations 5.1

Global Weak Solution in Three Dimensions

5.1.1

The Periodic Initial Value Problem

In this chapter, we first study the following coupled Landau–Lifshitz–Maxwell equations: ~ t = α1 Z ~ × (∆Z ~ + H) ~ − α2 Z ~ × (Z ~ × (∆Z ~ + H)), ~ Z (5.1.1) ~ ~ ~ = ∂ E + σ E, (5.1.2) ∇×H ∂t ~ ~ ~ = − ∂H − β ∂Z , ∇×E (5.1.3) ∂t ∂t ~ + β∇ · Z ~ = 0, ~ = 0, ∇·H ∇·E (5.1.4) ~ = (H1 (x, t), H2 (x, t), H3 (x, t)) where α1 , α2 , σ and β are constants, α2 ≥ 0, σ ≥ 0, H ~ is the magnetic field, E(x, t) = (E1 (x, t), E2 (x, t), E3 (x, t)) is the electric field and e ~ = ∆Z ~ +H ~ is the effective magnetic field. H We study the existence of the global generalized solutions for the three-dimensional Landau–Lifshitz–Maxwell system (5.1.1)–(5.1.4) with the periodic initial value condition: ~ + 2D~ei , t) = Z(x, ~ t), H(x ~ + 2D~ei , t) = H(x, ~ Z(x t), ~ + 2D~ei , t) = E(x, ~ E(x t), (x ∈ Ω ⊂ R3 , t ≥ 0) (5.1.5) ~ 0) = Z ~ 0 (x), Z(x,

~ ~ 0 (x), H(x, 0) = H

~ ~ 0 (x), E(x, 0) = E

or with the initial value condition: ~ 0) = Z ~ 0 (x), Z(x,

~ ~ 0 (x), H(x, 0) = H

(x ∈ Ω ⊂ R3 )

~ ~ 0 (x), E(x, 0) = E

x ∈ R3 ,

(5.1.6)

(5.1.7)

where x + 2D~ei = (x1 , . . . , xi−1 , xi + 2D, xi+1 , . . . , xn ), (i = 1, 2, 3), D > 0, Ω ⊂ R3 represents the three-dimensional cube with width 2D along each direction, i.e. Ω = {x = (x1 , x2 , x3 ), |xi | ≤ D, i = 1, 2, 3}, QT = {(x, t), x ∈ Ω, 0 ≤ t ≤ T }. 241

Landau–Lifshitz Equations

242

5.1.2

Approximate Solutions and the A Priori Estimates for the Periodic Initial Value Problem

We use Galerkin method to establish the a priori estimate for the approximate solutions of (5.1.1)–(5.1.6). Let wn (x) (n = 1, 2, . . .) be the unit eigenfunctions satisfying the equation ∆wn + λn wn = 0, with periodicity wn (x − D~ei ) = wn (x + D~ei ) (i = 1, 2, 3), and λn (n = 1, 2, . . .) the corresponding eigenvalues different from each other. {wn (x)} consists of the orthonormal base in L2 . ~ N (x, t), Denote the approximate solution of the problem (5.1.1)–(5.1.6) by Z ~ N (x, t) and E ~ N (x, t) in the following form H N X ~ N (x, t) = α ~ sN (t)ws (x), Z  s=1   N  X  ~  HN (x, t) = β~sN (t)ws (x),   s=1   N X ~



EN (x, t) =

(5.1.8)

~γsN (t)ws (x),

s=1

where α ~ sN (t), β~sN (t), ~γsN (t), (t ∈ R+ ), (s = 1, 2, . . . , N; N = 1, 2, . . .) are the threedimensional vector-valued functions satisfying the following system of ordinary differential equations of first order: Z

Ω

~ N t ws (x)dx = α1 Z

Z

Ω

− α2

Z

Z Ω

Ω

~ N × (∆Z ~N + H ~ N ) · ws (x)dx Z Z

Ω

~ N × (Z ~ N × (∆Z ~N + H ~ N ))ws (x)dx, Z

~ Nt + βZ ~ N t )ws (x)dx = − (H

~ N t ws (x)dx + σ E

Z

Ω

~ N · ws (x)dx = E

Z

Z

Ω

Ω

~ N · ws (x)dx, ∇×E

~ N · ws (x)dx, ∇×H

(5.1.9) (5.1.10) (5.1.11)

and the initial condition Z

Z

Ω

Ω

Z

Ω

~ N (x, 0)ws (x)dx = Z ~ N (x, 0)ws (x)dx = H ~ N (x, 0)ws (x)dx = E

Z Z

Z

Ω

Ω

Ω

~ 0 (x)ws (x)dx, Z ~ 0 (x)ws (x)dx, H

(5.1.12)

~ 0 (x)ws (x)dx. E

Obviously there holds Z

Ω

~ N t ws (x)dx = α0 (t), Z sN

Z

Ω

~ N t ws (x)dx = β 0 (t), H sN

Z

Ω

~ N t ws (x)dx = γ 0 (t). E sN

For the sake of simplicity, denote k · kLp (Ω) = k · kp , p ≥ 2.

(5.1.13)

Landau–Lifshitz–Maxwell Equations

243

~ 0 (x), H ~ 0 (x), E ~ 0 (x)) ∈ (H 1 (Ω), L2 (Ω), L2 (Ω)), α2 ≥ 0. Lemma 5.1.1 Assume (Z Then for the solution of the initial value problem (5.1.9)–(5.1.12) we have the following estimates: h

i

~ N (·, t)k2 1 + kE ~ N (·, t)k2 2 + kH ~ N (·, t)k2 2 sup kZ H (Ω) L (Ω) L (Ω) ≤ K0 ,

0≤t≤T

~ N (·, t)k6 ≤ K0 , sup kZ

~ N (·, t) × ∇Z ~ N (·, t)k 3 ≤ K0 , sup kZ

0≤t≤T

(5.1.14)

2

0≤t≤T

where the constant K0 is independent of N, α2 and D. When α0 > 0, there is ~ N × (∆Z ~N + H ~ N )kL2 (0,T ;L2 (Ω)) ≤ K00 , kZ

(5.1.15)

where the constant K00 is independent of N and D. Proof. (1) Multiplying (5.1.9) by α ~ sN (t), and summing up the products for s = 1, 2, . . . , N, we get Z d ~ N (x, t)|2 dx = 0. |Z dt Ω Then ~ N (·, t)k22 ≤ kZ ~ N (·, 0)k22 ≤ kZ ~ 0 (x)k22 . kZ (5.1.16) (2) Making the scalar product of (−λs α ~ sN (t) + β~sN (t)) with (5.1.9), and summing up the resulting product for s = 1, 2, . . . , N, and noticing that ~N + H ~N = ∆Z

N X

(−λs α ~ sN + β~sN )ws (x),

s=1

we have Z

Ω

~ N t (∆Z ~N + H ~ N )dx = −α2 Z = α2

Z

Z

Ω

Ω

~ N × (Z ~ N × (∆Z ~N + H ~ N )) · (∆Z ~N + H ~ N )dx Z

~ N × (∆Z ~N + H ~ N )|2 dx, |Z

i.e. Z 1d Z ~ N (x, t)|2 dx + α2 kZ ~ N × (∆Z ~N + H ~ N )k2 − Z ~N t · H ~ N dx = 0. |∇Z 2 2 dt Ω Ω

(5.1.17)

Making the scalar product of β~sN (t) with (5.1.10), and ~γsN (t) with (5.1.11), respectively, and adding the two equalities obtained, and then summing up the resulting products for s = 1, 2, . . . , N, we get 1d 2 dt

Z

Ω

~ N (x, t)|2 + |H ~ N (x, t)|2 )dx + σkE ~ N (·, t)k2 + β (|E 2

Z

Ω

~N t · H ~ N dx = 0. (5.1.18) Z

From (5.1.17) and (5.1.18) (multiply (5.1.18) by δ0 ) it follows that d dt

Z " Ω

#

1 ~ 2 δ0 ~ 2 ~ N k2 ~ N |2 ) dx + σδ0 kE |∇ZN | + (|EN | + |H 2 2 2

~ N × (∆Z ~N + H ~ N )k2 + (βδ0 − 1) + α 2 kZ 2

Z

Ω

~N t · H ~ N dx = 0. (5.1.19) Z

Landau–Lifshitz Equations

244

Multiplying (5.1.10) by (βδ0 − 1)~ αsN (t), and summing up the products for s = 1, 2, . . . , N, we obtain ~ Nt · Z ~ N dx + 1 β(βδ0 − 1) d H 2 dt Ω Z ~N · Z ~ N dx = 0. + (βδ0 − 1) ∇ × E

(βδ0 − 1)

Z

Z

Ω

~ N |2 dx |Z (5.1.20)

Ω

Adding (5.1.19) and (5.1.20) gives i 1d Z h ~ 2 1 d Z ~ 2 2 2 ~ ~ |∇ZN | + δ0 (|EN | + |HN | ) dx + β(βδ0 − 1) |ZN | dx 2 dt Ω 2 dt Ω Z d ~N · H ~ N dx + α2 kZ ~ N × (∆Z ~N + H ~ N )k2 + (βδ0 − 1) Z 2 dt Ω Z ~ N (·, t)k2 + (1 − βδ0 ) ∇ × E ~N · Z ~ N dx = −σδ0 kE 2

Ω

~ N (·, t)k2 + (1 − βδ0 ) = −σδ0 kE 2

Z

Ω

~N · E ~ N dx ∇×Z

2 ~ N (·, t)k2 + ε0 kE ~ N (·, t)k2 + (1 − βδ0 ) k∇Z ~ N (·, t)k2 , ≤ −σδ0 kE 2 2 2 4ε0

(ε0 > 0).

Integrating the above inequality with respect to t we get 1 ~ N (·, t)k2 + 1 δ0 (kE ~ N (·, t)k2 + kH ~ N (·, t)k2 ) k∇Z 2 2 2 2 2 Z 1 ~ N (·, t)k2 + (βδ0 − 1) H ~ N (x, t) · Z ~ N (x, t)dx + β(βδ0 − 1)kZ 2 2 Ω + α2

Z

t

0

~ N × (∆Z ~N + H ~ N )k2 dt kZ 2

1 ~ 0 k2 + 1 δ0 (kE ~ 0 k2 + k H ~ 0 k2 ) + 1 |β(βδ0 − 1)|kZ ~ 0 k2 ≤ k∇Z 2 2 2 2 2 2 2 ~ 0 k2 kZ ~ 0 k2 + (ε0 − σδ0 ) + |βδ0 − 1|kH +

(1 − βδ0 )2 4ε0

Z

t 0

Z

t 0

~ N (·, t)k2 dt kE 2

~ N (·, t)k2 dt. k∇Z 2

Denote c0 =

1 ~ 0 k2 + 1 δ0 (kE ~ 0 k2 + k H ~ 0 k2 ) k∇Z 2 2 2 2 2 1 ~ 0 k2 + |βδ0 − 1|kH ~ 0 k2 kZ ~ 0 k2 . + |β(βδ0 − 1)|kZ 2 2

(5.1.21)

Landau–Lifshitz–Maxwell Equations

245

Then, by (5.1.21) and (5.1.16) Z t 1 1 2 2 2 ~ N × (∆Z ~N + H ~ N )k22 dx ~ ~ ~ kZ k∇ZN (·, t)k2 + δ0 (kEN (·, t)k2 + kHN (·, t)k2 ) + α2 2 2 0 |β(βδ0 − 1)| ~ ~ N (·, t)k2 kZ ~ N (·, t)k2 ≤ c0 + kZN (·, t)k22 + |βδ0 − 1|kH 2 Z Z t (1 − βδ0 )2 t 2 ~ N (·, t)k2 dt ~ k∇Z + (ε0 − σδ0 ) kEN (·, t)k2 dt + 2 4ε0 0 0 !

δ0 ~ |β(βδ0 − 1)| (βδ0 − 1)2 2 ~ 0 k22 ≤ c 0 + kH kZ (·, t)k + + N 2 4 2 δ0 + (ε0 − σδ0 )

Z

t 0

(1 − βδ0 )2 Z t ~ N k22 dt. k∇Z 4ε0 0

~ N k22 dt + kE

Therefore, 1 ~ N (·, t)k2 + δ0 kH ~ N (·, t)k2 + δ0 kE ~ N (·, t)k2 k∇Z 2 2 2 2 4 2 + α2

Z

t 0

~ N × (∆Z ~N + H ~ N )k2 dt kZ 2

≤ d0 + (ε0 − σδ0 ) where

Z

t 0

~ N k2 + kE 2

(1 − βδ0 )2 4ε0

Z

t 0

~ N k2 dt, k∇Z 2

(5.1.22)

!

|β(βδ0 − 1)| (βδ0 − 1)2 ~ 0 k22 . d0 = c 0 + + kZ 2 δ0

When β > 0, σ ≥ 0, we let δ0 = β1 , ε0 = σδ0 + 41 δ0 . Then, from (5.1.22) it follows that 1 ~ N (·, t)k2 + 1 (kE ~ N (·, t)k2 + kH ~ N (·, t)k2 ) k∇Z 2 2 2 2 4β + α2

Z

t 0

~ N × (∆Z ~N + H ~ N )k22 dt ≤ d0 , kZ

∀ t ≥ 0.

(5.1.23)

When σ and β are arbitrary constants, we take ε0 = δ0 = 1. By (5.1.22) there is Z t 1 ~ N (·, t)k22 + 1 kH ~ N (·, t)k22 + 1 kE ~ N (·, t)k22 + α2 ~ N × (∆Z ~N + H ~ N )k22 dt k∇Z kZ 2 4 2 0 Z t 2 Z t (1 − β) ~ N (·, t)k2 dt + ~ N (·, t)k2 dt k∇Z ≤ d0 + |1 − σ| kE 2 2 4 0 0

≤ d0 + e0 where

Z

t

0

~ N k22 + k∇Z ~ N k22 )dt, (kE (

(5.1.24)

)

(1 − β)2 . e0 = max |1 − σ|, 4

Landau–Lifshitz Equations

246

Applying Gronwall inequality for (5.1.24) yields ~ N (·, t)k2 + kH ~ N (·, t)k2 + kE ~ N (·, t)k2 ) ≤ K0 , sup (k∇Z 2 2 2

(5.1.25)

t∈[0,T ]

where K0 is independent of N , α2 , D. By Sobolev imbedding theorem and H¨older inequality, we have ~ N (·, t)k2 ≤ K0 , sup kZ 6

~ N (·, t) × ∇Z ~ N (·, t)k 3 ≤ K0 . sup kZ

0≤t≤T

2

0≤t≤T

(5.1.26)

When α2 > 0, using (5.1.23), (5.1.24) and (5.1.25), we get ~ N × (∆Z ~N + H ~ N )kL2 (0,T ;L2 (Ω)) ≤ K 0 , kZ 0

(5.1.27)

where K00 is independent of N . The proof of the lemma is complete. ~ N (x, t), Lemma 5.1.2 Under the conditions of Lemma 5.1.1, for the solution (Z ~ N (x, t), E ~ N (x, t)) of the initial value problem (5.1.9)–(5.1.12) there are H (1) when α2 = 0, h

i

~ N t kH −2 (Ω) + kE ~ N t kH −2 (Ω) + kH ~ N t kH −2 (Ω) ≤ K1 ; sup kZ

t∈[0,T ]

(5.1.28)

(2) when α2 > 0, ~N tk 3 kZ L 2 (Q

T)

~ N t kL2 (0,T ;H −1 (Ω)) + kH ~ N t kL2 (0,T ;H −1 (Ω)) ≤ K2 , + kE

(5.1.29)

where the constants K1 and K2 are independent of N and D. Proof. (1) When α2 = 0, for any ϕ ∈ H02 , ϕ can be represented by ϕ = ϕN + ϕ¯N , where ϕN =

N X

βs ws (x),

ϕ¯N =

s=1

For s ≥ N + 1,

∞ X

βs ws (x).

s=N +1

Z

Ω

~ N t · ws (x)dx = 0. Z

Then, by Lemma 5.1.1 there are Z

Ω

~ N t ϕdx = Z

Z

Ω

= α1

~ N t ϕN (x)dx = α1 Z Z

Ω

Z

Ω

~ N × (∆Z ~N + H ~ N )ϕN (x)dx Z

~N × Z ~ N · ∇ϕN dx + α1 ∇Z

Z

Ω

~N × H ~ N · ϕN dx Z

~ N k2 kZ ~ N k2 + k Z ~ N k2 kH ~ N k2 )(k∇ϕN k2 + kϕN k∞ ) ≤ |α1 |(k∇Z ≤ CkϕN kH 2 (Ω) ≤ CkϕkH 2 (Ω) ,

Landau–Lifshitz–Maxwell Equations Z Z

Ω

247

~ N t ϕdx ≤ C1 (kE ~ N k2 + k H ~ N k2 )(k∇ϕk2 + kϕk2 ) ≤ CkϕkH 2 (Ω) , E Z

~ N t ϕdx ≤ |β| H Ω

Ω



~ N t ϕdx + Z

Z

Ω

~ N |dx |∇ϕ||E

Z ~ ~ N k2 ≤ |β| ZN t ϕdx + k∇ϕk2 kE Ω

≤ CkϕkH 2 (Ω) + C2 k∇ϕk2 ≤ C3 kϕkH 2 . Then there holds ~ N t kH −2 (Ω) + kE ~ N t kH −2 (Ω) + kH ~ N t kH −2 (Ω) ≤ K1 . kZ (2) For α2 > 0, Z Z ZZ T ~ ~ ~ ~ (ZN t · ϕ)dxdt ≤ |α1 | ZN × (∆ZN + HN )ϕdxdt 0 Ω QT ZZ ~ ~ ~ ~ + α2 ZN × (ZN × (∆ZN + HN ))ϕdxdt QT

~ N × (∆Z ~N + H ~ N )kL2 (Q ) kϕkL2 (Q ) ≤ |α1 |kZ T T

~ N kL6 (Q ) kZ ~ N × (∆Z ~N + H ~ N )kL2 (Q ) kϕkL3 (Q ) + α 2 kZ T T T

≤ C4 kϕkL3 (QT ) ,

Z (Z ~ N t · ϕ)dx ≤ C5 kϕkH 1 (Ω) , Ω ZZ Z T Z ~ ~ EN t ϕdxdt ≤ |σ| kEN k2 kϕk2 dt + QT

0

T 0

~ N k2 k∇ϕN k2 dt kH

1

ZZ

QT



≤ C6 T 2 kϕkL2 (0,T ;H 1 (Ω)) ,

~ N t ϕdxdt = −β H

ZZ

QT

~ N t ϕN (x, t)dxdt − Z

≤ C7 kϕN kL2 (0,T ;H 1 (Ω)) + C8

Z

T 0

ZZ

QT



~ N ϕN (x, t)dxdt ∇×E

~ N k2 k∇ϕN kdt kE

1

≤ C7 kϕkL2 (0,T ;H 1 (Ω)) + C8 T 2 kϕN kL2 (0,T ;H 1 (Ω)) ≤ C9 kϕkL2 (0,T ;H 1 (Ω)) . The lemma is proved. ~ N (x, t), Lemma 5.1.3 Under the conditions of Lemma 5.1.1, for the solution (Z ~ ~ HN (x, t), EN (x, t)) of the initial value problem (5.1.9)–(5.1.12) there are (1) when α2 = 0, ~ N (·, t1 ) − Z ~ N (·, t2 )k2 + kH ~ N (·, t1 ) − H ~ N (·, t2 )k2 + kE ~ N (·, t1 ) − E ~ N (·, t2 )k2 kZ 1

≤ K3 |t1 − t2 | 3 ,

(5.1.30)

248

Landau–Lifshitz Equations

where the constant K3 is independent of N, D. (2) when α2 > 0, ~ N (·, t1 ) − Z ~ N (·, t2 )k3 ≤ K4 |t1 − t2 | 23 , kZ

(5.1.31) 1

~ N (·, t1 )− H ~ N (·, t2 )kH −1 (Ω) +kE ~ N (·, t1 )− E ~ N (·, t2 )kH −1 (Ω) ≤ K5 |t1 −t2 | 2 , kH

(5.1.32)

where the constant K4 is independent of N and D. Proof. (1) When α2 = 0, by the Sobolev interpolation inequality of negative order, there is 2

1

~ N (·, t1 ) − Z ~ N (·, t2 )k 3 1 ~ N (·, t1 ) − Z ~ N (·, t2 )k2 ≤ CkZ ~ N (·, t1 ) − Z ~ N (·, t2 )k 3 −2 kZ kZ H (Ω) H (Ω)

1

Z t ~ N

3 2 ∂Z

≤ C 0

dt

t1 ∂t

−2 H

1

(Ω)

≤ C 00 |t2 − t1 | 3 .

~ N (·, t1 ) − H ~ N (·, t2 )k2 , and kE ~ N (·, t1 ) − Similarly, above inequality holds for kH ~ EN (·, t2 )k2 . (2) When α2 > 0,

Z t ~ N

2 ∂Z ~ N (·, t1 ) − Z ~ N (·, t2 )k3 = dt

kZ

t1 ∂t 2 3



≤ |t2 − t1 |  

3

ZZ

QT

2

≤ K4 |t2 − t1 | 3 , ~ N (·, t1 ) − H ~ N (·, t2 )kH −1 (Ω) kH



Z t ~ N

2 ∂H =

dt

∂t

t1

1 2



 ≤ |t2 − t1 | 

Z

1

H −1 (Ω)

T

0

≤ K5 |t2 − t1 | 2 .

1 3 ~ 3 ∂ ZN  dxdt ∂t



~ 2

∂ HN



∂t

H −1 (Ω)

1

2

 dt

~ N (x, t1 ) − E ~ N (x, t2 ) similar inequality holds. The lemma is proved. For E By using the above integral estimates of the approximate solution, we have Lemma 5.1.4 Under the conditions of Lemma 5.1.1, the initial value problem for the system of the ordinary differential equation (5.1.9)–(5.1.12) has at least one continuously differentiable and global solution α ~ sN (t), β~sN (t), ~γsN (t) (s = 1, 2, . . . , N, t ∈ [0, T ]).

Landau–Lifshitz–Maxwell Equations

5.1.3

249

Existence of Generalized Solution

~ t) ∈ Definition 5.1.5 A triple of three-dimensional vector functions Z(x, ~ ~ L∞ (0, T ; H 1 (Ω)), H(x, t) ∈ L∞ (0, T ; L2 (Ω)), E(x, t) ∈ L∞ (0, T ; L2 (Ω)) is called the generalized solution for the periodic initial value problem (5.1.5) and (5.1.6) of the ~ t), and test function system (5.1.1)– (5.1.4), if for any vector-valued test function ψ(x, ~ T ) = 0, ψ(x ~ + Dei , t) = ψ(x ~ − Dei , t), i = 1, 2, 3, the ζ(x, t) ∈ C 1 (QT ) with ψ(x, following equalities hold ZZ

QT

~ ~ t · ψdxdt Z − α1 ZZ

− α2

QT

ZZ

QT

~ ~ × (∆Z ~ + H) ~ · ψdxdt Z

~ ~ × (∆Z ~ + H) ~ ·Z ~ × ψdxdt Z = 0,

(α2 > 0)

(5.1.33)

or ZZ

~ ~ t · ψdxdt Z + α1

QT

−α1 ZZ

QT

Z

Ω

QT

QT

QT

~ ~ × ∇Z ~ · ∇ψdxdt Z

~ ~ × H) ~ · ψdxdt (Z = 0,

~t eσt dxdt + ~ ·ψ E +

ZZ

ZZ

ZZ

ZZ

QT

~ · H(x, ~ eσt ∇ × ψ t)dxdt

~ 0)dx = 0, ~ 0 (x) · ψ(x, E

~t dxdt − ~ ~ t))ψ (H(x, t) + β Z(x, +

Z

Ω

(α2 = 0),

ZZ

(5.1.34)

QT

~ · E(x, ~ ∇×ψ t)dxdt

~ 0)dx = 0, ~ 0 (x) + β Z ~ 0 (x)) · ψ(x, (H ZZ

QT

~ + β Z)dxdt ~ ∇ζ · (H = 0,

ZZ

QT

~ ∇ζ · Edxdt = 0,

~ 0) = Z ~ 0 (x), Z(x,

a.e. x ∈ Ω.

(5.1.35) (5.1.36) (5.1.37) (5.1.38)

~ 0 (x), H ~ 0 (x) satisfy the condition Lemma 5.1.5 Let the initial value vector functions E Z

Ω

~ 0 dx = 0, ∇ξ · E

Z

Ω

~ 0 (x) + β Z ~ 0 (x))dx = 0, ∇ξ · (H

(5.1.39)

for all ξ(x) ∈ C 1 (Ω). Then for any ζ(x, t) ∈ C 1 (QT ) with ζ(x, T ) = 0, ζ0 = ζ(x, 0), we have, from (5.1.36), (5.1.37) that ZZ

QT

~ ∇ζ · E(x, t)dxdt = 0,

ZZ

QT

~ ~ t))dxdt = 0. ∇ζ · (H(x, t) + β Z(x,

Landau–Lifshitz Equations

250

Proof. Take ~t (x, t) = ψ

Z

t 0

e−στ ∇ζ(x, τ )dτ −

Z

T 0

e−στ ∇ζ(x, τ )dτ,

ζ ∈ C 1 (QT ).

By (5.1.34) we get ZZ

QT

Since

~ · ∇ζdxdt + E Z

then

Ω

Z

T

e

−στ

dτ

0

Z

Ω

~ 0 (x)dx = 0. ∇ζ(x, τ ) · E

~ 0 (x)dx = 0, ∇ζ(x, τ ) · E ZZ

QT

~ · ∇ζdxdt = 0. E

~ = R t ∇ζdt − R T ∇ζdt, we have, from (5.1.35) By letting ψ 0 0 ZZ

From it follows that

QT

~ + β Z)∇ζdxdt ~ (H − Z

Ω

ZZ

Z

T

0

Z

Ω

~ 0 + βZ ~ 0 (x)) · ∇ζdxdτ = 0. (H

~ 0 + βZ ~ 0 (x))∇ζdx = 0, (H

QT

~ + β Z) ~ · ∇ζdxdt = 0. (H

~ 0 (x) ∈ H 1 (Ω), H ~ 0 (x) ∈ L2 (Ω), E ~ 0 (x) ∈ L2 (Ω), and they Theorem 5.1.1 Assume Z are periodic functions with periodicity D, and satisfy (5.1.36). Then the periodic initial value problem (5.1.5) and (5.1.6) for the system of Landau–Lifshitz–Maxwell ~ t), H(x, ~ ~ (5.1.1)–(5.1.4) has at least one generalized solution Z(x, t), E(x, t). When α2 > 0, ~ t) ∈ L∞ (0, T ; H 1 (Ω)) T C (0, 32 ) (0, T ; L3 (Ω)), Z(x,

(5.1.40)

~ t) ∈ L∞ (0, T ; H 1 (Ω)) T C (0, 12 ) (0, T ; L2 (Ω)), Z(x,

(5.1.41)

T 1 ~ ~ E(x, t), H(x, t) ∈ L∞ (0, T ; L2 (Ω)) C (0, 2 ) (0, T ; H −1 (Ω)).

When α2 = 0,

T 1 ~ ~ E(x, t), H(x, t) ∈ L∞ (0, T ; L2 (Ω)) C (0, 3 ) (0, T ; L2 (Ω)).

~ t) ∈ C 1 (QT ), ψ(x, ~ T ) = 0, we Proof. For any vector-valued test function ψ(x, define an approximate sequence ~N (x, t) = ψ

N X

n=1

β~n (t)wn (x),

Landau–Lifshitz–Maxwell Equations

251

R ~ t)wn (x)dx. We know that ψ ~N is uniformly convergent to ψ(x, ~ t) where β~n (t) = Ω ψ(x, 1 in C (QT ). n ~ N (x, t), H ~ N (x, t), From the uniform estimate for the approximate solution Z o ~ N (x, t) in Sec. 5.1.2, it follows that E

~ N (x, t) → Z(x, ~ t) Z ~ N (x, t) → Z(x, ~ t) Z ~ N (x, t) → Z(x, ~ t) Z

weakly in L6 (QT ), strongly in L6−η (QT ),

(η > 0),

strongly in L∞ (0, T ; H 1 (Ω)),

~ N (x, t) → H(x, ~ H t) ~ N (x, t) → E(x, ~ E t) ~ N t (x, t) → Z ~ t (x, t) Z ~ N t (x, t) → Z ~ t (x, t) Z

(5.1.42)

weakly in L∞ (0, T ; L2 (Ω)), weakly in L∞ (0, T ; L2 (Ω)), 3 weakly in L 2 (QT ), (α2 > 0), weakly in L∞ (0, T ; H −2(Ω)), (α2 = 0).

Making the scalar products of β~s (t) with (5.1.9), and eσt β~s with (5.1.10), and summing up the products for s = 1, 2, . . . , we get ZZ

QT

~N dxdt − α1 ~N t · ψ Z − α2

ZZ

QT

ZZ

QT

ZZ

QT

~N dxdt ~ N × (∆Z ~N + H ~N) · ψ (Z

~N )dxdt = 0, ~ N × (∆Z ~N + H ~ N ) · (Z ~N × ψ Z

~N (x, t)dxdt = − ~ Nt + βZ ~ N t) · ψ (H

ZZ

QT

d σt ~ ~N (x, t)dxdt = (e EN ) · ψ dt

ZZ

ZZ

QT

QT

~N (x)dxdt, ~N · ψ ∇×E

~N (x)dxdt. ~N · ψ eσt ∇ × H

(5.1.43)

(5.1.44) (5.1.45)

Rewriting (5.1.44), we get ZZ

QT

~N t dxdt + ~ N + βZ ~ N (x, t))ψ (H −

ZZ

QT

Z

Ω

~N (x, 0)dxdt ~ N (x, 0) + β Z ~ N (x, 0)) · ψ (H

~N · E ~ N (x, t)dxdt = 0. ∇×ψ

(5.1.46)

Rewriting (5.1.45), we get ZZ

QT

~N t eσt )dxdt + ~ N · (ψ E +

Z

Ω

ZZ

QT

~N · H ~ N (x, t)dxdt eσt ∇ × ψ

~N (·, 0)dx = 0. ~ N (·, 0) · ψ E

(5.1.47)

(1) By Lemmas 5.1.1 and 5.1.2, we get ZZ

QT

~N dxdt = ~N t · ψ Z

ZZ

QT

~ ~ N t · ψdxdt Z →

ZZ

QT

~ ~ t ψdxdt, Z

Landau–Lifshitz Equations

252

Z

ZZ Ω

QT

~N dxdt → ~ Nt + βZ ~ N t )ψ (H

ZZ

~N (x, 0)dx → ~ N (x, 0) + β Z ~ N (x, 0))ψ (H

(2) (i) ZZ

(ii)

ZZ

In fact ZZ

~N · H ~ N dxdt → e ∇×ψ σt

QT

Z

~N · E ~ N (x, t)dxdt → ∇×ψ

QT

~ ~ t + βZ ~ t ) · ψdxdt, (H

QT

ZZ

Ω

~ 0)dx. ~ 0 (x) + β Z ~ 0 (x))ψ(x, (H

ZZ

QT

QT

~ · Edxdt, ~ ∇×ψ

~ · Hdxdt. ~ eσt ∇ × ψ

~N · E ~ N dxdt ∇×ψ

QT

=

ZZ

=

ZZ

+

QT

~N − ψ) ~ ·E ~ N dxdt + ∇ × (ψ

ZZ

QT

~N − ψ) ~ ·E ~ N dxdt + ∇ × (ψ

ZZ

ZZ

QT

QT

~ ·E ~ N dxdt (∇ × ψ)

QT

~ · Edxdt ~ (∇ × ψ)

~ E ~ N − E)dxdt. ~ (∇ × ψ)(

As ZZ

QT



~N − ψ) ~ ·E ~ N dxdt ≤ ∇ × (ψ ZZ

QT

ZZ

QT

~N − ψ)| ~ 2 dxdt |∇(ψ

1 2

~ N kL2 (Q ) → 0, kE T



~ · (E ~ N − E)dxdt ~ → 0, (∇ × ψ)

(i) is proved. Similarly (ii) can be proved. ~ N (x, t)} such that, as N → +∞ (3) There exists a subsequence {Z (i) ~ ~ ~ N × ∂ ZN → Z ~ × ∂Z Z ∂xi ∂xi

weakly-∗ in

3

L∞ (0, T ; L 2 (Ω)),

(i = 1, 2, 3).

(5.1.48)

(ii) 







~ ~ ~ × ∂Z  ~ N × ∂ ZN  → Z Z ∂xi ∂xi xi

weakly-∗ in

L2 (QT ),

(i = 1, 2, 3). (5.1.49)

xi

~ t) ∈ C 1 (QT ), there is In fact, for any test function ψ(x, ZZ

QT

~N × (Z =

ZZ

+

~N ~ ∂Z ~ ~ × ∂ Z ) · ψdxdt −Z ∂xi ∂xi

QT

ZZ





~ ~ ~ N − Z) ~ × ∂ ZN  · ψdxdt (Z ∂xi QT







~ ~ ~ ×  ∂ ZN − ∂ Z  dxdt. Z ∂xi ∂xi

Landau–Lifshitz–Maxwell Equations

253

~ N (x, t) is strongly convergent to Z(x, ~ t) in L2 (QT ) and ∂ Z~ N is weakly conSince Z ∂xi ~N ~ ∂Z ∂Z 2 ~ in L (Q ), and Z and are uniformly bounded in L2 (QT ) with vergent to ∂x T N ∂xi i respect to N , it is easy to prove that the right-hand side of the above equality is convergent to zero as N → +∞. ~ N ×(∆Z ~N + H ~ N ) is In the following we prove (ii). By Lemma 5.1.1, when α2 > 0, Z 2 uniformly bounded in L (QT ) with respect to N . Then there exists a vector function ~ t) ∈ C 1 (QT ), there holds ~ (x, t) ∈ L2 (QT ) such that for any test function ψ(x, V ZZ

QT

~ ~ N × (∆Z ~N + H ~ N ) · ψ)dxdt (Z

−

ZZ

QT



~ t)dxdt → 0. ~ (x, t) + Z ~N × H ~ N ) · ψ(x, (V

On the other hand, as N → +∞, ZZ

~ ~ N × (∆Z ~N + H ~ N )) · ψdxdt (Z

QT

=−

ZZ

→−

QT

ZZ

~+Z ~ ~ N × ∇Z ~ N · ∇ψ ~N × H ~ N · ψ)dxdt (Z

QT

~ ~ × ∇Z) ~ · ∇ψdxdt (Z +

ZZ

QT

~ ~ × H) ~ · ψdxdt, (Z

where we have used (i) and the fact that ZZ

QT



~ ~N × H ~N − Z ~ × H) ~ · ψdxdt (Z

ZZ ≤

≤

Z

T

0

→ 0,

QT



ZZ

~ ~ N − Z) ~ ×H ~ N · ψdxdt + (Z

~N kZ

ZZ ~ ~ ~ − Zk5 kHN k2 kψk 10 dt + 3

(N → +∞).

QT

QT



~ ~ × (H ~ N − H) ~ · ψdxdt Z

Then ZZ

QT

~ ~ · ψdxdt V =−

ZZ

QT

~ ~ × ∇Z) ~ · ∇ψdxdt, (Z

~ × ∆Z ~ =V ~ ∈ L2 (QT ). Z To prove the existence of the generalized solution it remains to prove ZZ

QT

~N )dxdt ~ N × (∆Z ~N + H ~ N ) · (Z ~N × ψ Z →

ZZ

QT



~ ~ × (H ~ N − H) ~ · ψdxdt Z

~ ~ × (∆Z ~ + H) ~ · (Z ~ × ψ)dxdt. Z

(5.1.50)

Landau–Lifshitz Equations

254

In fact ZZ

QT

~N )dxdt ~ N × (∆Z ~N + H ~ N ) · (Z ~N × ψ Z ZZ

− =

ZZ

QT

+

ZZ

+

ZZ

+

ZZ

QT

h

~ ~ × (∆Z ~ + H) ~ · (Z ~ × ψ)dxdt Z i

~ ~ N × ∆Z ~ N ) − (Z ~ × ∆Z) ~ ·Z ~ × ψdxdt (Z

QT

h

i

~N − Z ~ dxdt ~ N × ∆Z ~ N ) · (Z ~N × ψ ~ × ψ) (Z

QT

~ ~N × H ~N − Z ~ × H) ~ ·Z ~ × ψdxdt (Z

QT

~N − Z ~ ~N × H ~ N · (Z ~N × ψ ~ × ψ)dxdt Z

= I1N + I2N + I3N + I4N . By (5.1.49), I1N → 0 as N → ∞. Moreover, ~ N × ∆Z ~ N kL2 (Q ) |I2N | ≤ kZ T ≤C

ZZ

QT

ZZ

QT

~−Z ~ 2 dxdt ~N × ψ ~ × ψ| |Z

1 2

~N − ψ) ~ + (Z ~ 2 dxdt → 0. ~ N × (ψ ~ N − Z) ~ × ψ| |Z

I3N → 0 (N → ∞), which is similar to (5.1.50). ~N − Z ~ L4 (Q ) → 0. ~ N kL2 (Q ) kZ ~ N kL4 (Q ) kZ ~N × ψ ~ × ψk |I4N | ≤ kH T T T Therefore, by taking N → ∞ in (5.1.43), (5.1.44) and (5.1.45), we obtain the limit ~ t), H(x, ~ ~ function Z(x, t), E(x, t) satisfying the integral equality (5.1.33)–(5.1.35). Obviously (5.1.36)–(5.1.38) hold. The generalized solution of the periodic initial value problem (5.1.1)–(5.1.6) is obtained.

5.1.4

Existence of Solution for the Initial Problem

Note that the above a priori estimates are independent of D. By using the diagonal method and letting D → ∞, we can get Theorem 5.1.2 Assume that the conditions of Theorem 5.1.1 hold in R 3 . Then for the initial value problem (5.1.1)–(5.1.4) there exists one generalized solution satisfying ~ t) ∈ L∞ (0, T ; H 1 (R3 )) Z(x, ~ ~ H(x, t), E(x, t) ∈ L∞ (0, T ; L2 (R3 ))

Cloc 3 (0, T ; L3 (R3 )),

\

Cloc 2 (0, T ; H −1(R3 )),

(0, 1 )

\

Cloc 3 (0, T ; H −1 (R3 )),

\

Cloc 3 (0, T ; L2 (R3 )).

~ t) ∈ L∞ (0, T ; H 1 (R3 )) Z(x, ~ ~ H(x, t), E(x, t) ∈ L∞ (0, T ; L2 (R3 ))

(0, 2 )

\

(0, 1 )

(0, 1 )

(α2 > 0),

Landau–Lifshitz–Maxwell Equations

5.2 5.2.1

255

Global Smooth Solution in One or Two Dimensions with Small Initial Data The Problem

In the above subsection, we have obtained the global weak solution to the periodic initial boundary value problem (5.1.1)–(5.1.6), or Cauchy problem (5.1.1)–(5.1.5) and (5.1.7) for Landau–Lifshitz–Maxwell equations. When n ≤ 2, in this subsection, we prove the global existence of smooth solution to the periodic problem (5.1.1)–(5.1.6), or to the initial (5.1.1)–(5.1.5) and (5.1.7) with small initial data. For this aim we make the a priori estimates.

5.2.2

A Priori Estimates

~ 0 (x)| = 1. Then for the smooth solution of the periodic initial Lemma 5.2.1 Let |Z value problem of the system (5.1.1)–(5.1.6) ~ t)| = 1, |Z(x,

x ∈ Ω,

t ≥ 0.

(5.2.1)

~ with (5.1.1), we get Proof. Making the scalar product of Z ~ ·Z ~ t = 0, Z

~ t)|2 = 0. |Z(x, t

Then ~ t)| = |Z ~ 0 (x)| = 1. |Z(x, ~ 0 (x) ∈ L2 (Ω), E ~ 0 (x) ∈ L2 (Ω), Lemma 5.2.2 Assume β ≥ 0, α2 > 0, σ ≥ 0, ∇Z ~ H0 (x) ∈ L2 (Ω). Then for the smooth solution of the periodic initial value problem of the system (5.1.1)–(5.1.6) h

i

~ t)k2L (Ω) + kE(·, ~ t)k2L (Ω) + kH(·, ~ t)k2L (Ω) ≤ K1 , sup k∇Z(·, 2 2 2

0≤t≤T

Z

T 0

~ × (∆Z ~ + H)k ~ 2 dt ≤ K2 , kZ L2

(5.2.2)

~ 0 (x)kL , kE ~ 0 (x)kL and where the constants K1 and K2 depend only on k∇Z 2 2 ~ kH0 (x)kL2 . ~ with (5.1.2), and Proof. First assume β > 0. Making the scalar product of E ~ making the scalar product of −H with (5.1.3), and then adding these two equalities obtained, we obtain ~ ~ ~ ~ + ∂H · H ~ + β ∂Z · H ~ + σ|E| ~ 2. ~ ·E ~ − (∇ × E) ~ ·H ~ = ∂E · E (∇ × H) ∂t ∂t ∂t

(5.2.3)

By using the formula ~ ·E ~ − (∇ × E) ~ ·H ~ = ∇ · (H ~ × E), ~ (∇ × H)

(5.2.4)

Landau–Lifshitz Equations

256

and integrating (5.2.3) over x ∈ Ω, we get  1d  ~ 2 ~ 2 + σkEk ~ 2 +β kEkL2 + kHk L2 L2 2 dt

Z

Ω

~ ∂Z ~ · Hdx = 0. ∂t

(5.2.5)

~ + H) ~ with (5.1.1), we get Making the scalar product of (∆Z ~ ~ + H) ~ · ∂ Z = −α2 (∆Z ~ + H) ~ · [Z ~ × (Z ~ × (∆Z ~ + H))], ~ (∆Z ∂t

(5.2.6)

~ + H) ~ · [Z ~ × (Z ~ × (∆Z ~ + H))] ~ ~ × (∆Z ~ + H)| ~ 2. −(∆Z = |Z

(5.2.7)

where Integrating (5.2.6) over x ∈ Ω, one gets Z

Ω

~ + H) ~ · (∆Z

~ ∂Z dx = α2 ∂t

Z

Ω

~ × (∆Z ~ + H)| ~ 2 dx, |Z

and then Z

Ω

Z Z ~ ~ ∂Z ∂Z ~ ~ ~ × (∆Z ~ + H)| ~ 2 dx · Hdx = − ∆Z · dx + α2 |Z ∂t ∂t Ω Ω Z 1d 2 ~ ~ × (∆Z ~ + H)| ~ 2 dx. k∇ZkL2 + α2 |Z = 2 dt Ω

(5.2.8)

Adding (5.2.5) and (5.2.8), we obtain 1d ~ 2 ~ 2 ) + σkEk ~ 2 (kEkL2 + kHk L2 L2 2 dt β d ~ 2 + βα2 kZ ~ × (∆Z ~ + H)k ~ 2 = 0. k∇Zk + L2 L2 2 dt

(5.2.9)

Integrating the above equality with respect to t ∈ [0, T ], we have t 1 ~ ~ t)k2 ) + σ ~ t)k2 dt E(t) ≡ (kE(·, t)k2L2 + kH(·, kE(·, L2 L2 2 0 Z t β ~ t)k2 + βα2 ~ × (∆Z ~ + H)k ~ 2 dt + k∇Z(·, kZ L2 L2 2 0 1 ~ β 2 2 ~ ~ 0 (x)k2 . = E(0) = (kE k∇Z 0 (x)kL2 + kH0 (x)kL2 ) + L2 2 2

Z

(5.2.10)

It follows that the estimate (5.2.2) holds. Now assume β = 0, then we get 1d ~ 2 ~ 2 ) + σkEk ~ 2 = 0. (kEkL2 + kHk L2 L2 2 dt Therefore, ~ t)k2 + kH(·, ~ t)k2 ≤ kE ~ 0 (x)k2 + kH ~ 0 (x)k2 . kE(·, L2 L2 L2 L2

(5.2.11)

Landau–Lifshitz–Maxwell Equations

Note that

Z

Ω

~ ·H ~ t dx = − Z

Adding (5.2.8) to (5.2.12) yields d dt

Z

Z

Ω

~ · Zdx ~ =− ∇×E

1d ~ · Hdx ~ ~ 2 + α2 Z = k∇Zk L2 2 dt Ω

Z

Ω

Z

Ω

257

~ · Edx. ~ ∇×Z

~ × (∆Z ~ + H)| ~ 2 dx − |Z

Z

Ω

(5.2.12)

~ E)dx. ~ (∇ × Z,

So, ~ t)k2 + 2α2 k∇Z(·, L2 ≤2

Z

Ω

Z

+

Z

t 0

~ × (∆Z ~ + H)k ~ 2 dt kZ L2

~ t) · H(x, ~ Z(x, t)dx − 2 t

~ t)k2 dt + k∇Z(·, L2

0

1 ~ t)k2 + C ≤ k∇Z(·, 2 2 ≤C

Z

t 0

Z

t 0

Z

t 0

Z

Ω

~ 0 (x) · H ~ 0 (x)dx Z

~ t)k2 dt kE(·, L2

~ 2 dt + C1 kHk 2

~ 22 dt + C2 . k∇Zk

From the Gronwall inequality (5.2.2) is proved. Lemma 5.2.3 (Gagliardo–Nirenberg Inequality) Assume that u ∈ Lq (Ω), D m u ∈ Lr (Ω), Ω ⊂ Rn , 1 ≤ q, r ≤ ∞, 0 ≤ j ≤ m. Then kD j ukLp (Ω) ≤ CkukaWrm (Ω) kuk1−a Lq (Ω) ,

(5.2.13)

where C is a positive constant, and !

1 j 1 1 m = +a − + (1 − a) , p n r n q

j ≤ a ≤ 1. m

Lemma 5.2.4 Assume that the conditions of Lemmas 5.2.1 and 5.2.2 hold, and ~ 0 (x) ∈ H 2 (Ω), E ~ 0 (x), H ~ 0 (x) ∈ H 1 (Ω), Ω ⊂ Rn , n = 1, 2. When n = 2, β > 0, Z there is a small δ > 0 such that ~ 0 (x)k2 ≤ δ, ~ 0 (x)k2 + kH ~ 0 k2 + k E k∇Z 2 2 2

(5.2.14)

then h

~ t)k22 + k∇H(·, ~ t)k22 + k∇E(·, ~ t)k22 sup k∆Z(·,

0≤t≤T

+

Z

T 0

~ 2 dt ≤ K3 k∇∆Zk 2

i

(5.2.15)

~ 0 (x)kH 2 (Ω) , kH ~ 0 (x)kH 1 (Ω) and kE ~ 0 (x)kH 1 (Ω) . holds, where K3 depends only on kZ ~ 0 (x) ∈ H 2 (Ω), H ~ 0 (x) ∈ H 2 (Ω), there are When β = 0, E h

i

~ t)k22 + k∆H(·, ~ t)k22 ≤ K30 , sup k∆E(·,

0≤t≤T

~ 0 k2 and k∆H ~ 0 k2 . where K30 also depends on k∆E

(5.2.16)

Landau–Lifshitz Equations

258

~ 0 (x)| = 1, Eq. (5.1.1) in the classic Proof. We know that under the condition |Z sense is equivalent to the equation ~ t = α 2 ∆Z ~ + α2 |∇Z| ~ 2Z ~ + α1 Z ~ × (∆Z ~ + H) ~ − α2 Z ~ × (Z ~ × H). ~ Z

(5.2.17)

~ + H) ~ with (5.2.17), and integrating the resulting Making the scalar product of ∆(∆Z equality with respect to x ∈ Ω, we have Z 1∂ Z ~ 2 dx + α2 k∇∆Z(·, ~ t)k22 + Z ~ t · ∆Hdx ~ |∆Z| 2 ∂t Ω Ω

= −α1

Z

Ω

− α2

Z

+ α2

Z

= −α1

Z

~ × (∆Z ~ + H) ~ · (∇∆Z ~ + ∇H)dx ~ ∇Z

Ω

~ 2 ∇Z ~ + 2(∇Z ~ · ∇2 Z) ~ Z) ~ · (∇∆Z ~ + ∇H)dx ~ (|∇Z|

Ω

~ × (Z ~ × H)) ~ · (∇∆Z ~ + ∇H)dx ~ ∇(Z + α2

Ω

+ α1

Z

− α2

Z

Z

Ω

~ · ∆Hdx ~ ∆Z

~ × ∆Z) ~ · ∇∆Z ~ + (∇Z ~ × H) ~ · ∇∆Z ~ + (∇Z ~ × H) ~ · ∇H)dx ~ ((∇Z

Ω

Ω

~ × ∇∆Z ~ · H)dx ~ (∇Z ~ 2 ∇Z ~ · ∇∆Z ~ + 2(∇Z ~ · ∇2 Z) ~ Z ~ · ∇∆Z ~ (|∇Z|

~ 2 ∇Z ~ · ∇H ~ + 2(∇Z ~ · ∆Z) ~ Z ~ · ∇H)dx ~ + |∇Z| + α2

Z

Ω

~ × (Z ~ × H))(∇∆ ~ ~ + ∇H)dx ~ ∇(Z Z − α2

Z

Ω

~ · ∇Hdx ~ ∇∆Z

~ ∞ k∆Zk ~ 2 k∇∆Zk ~ 2 + 2|α1 |k∇Zk ~ ∞ kHk ~ 2 k∇∆Zk ~ 2 ≤ (2α2 + |α1 |)k∇Zk ~ 2 k∇Hk ~ 2 k∇Zk ~ ∞ + α2 k∇Zk ~ 3 k∇∆Zk ~ 2 + α2 k∇Zk ~ 3 k∇Hk ~ 2 + α1 kHk L6 L6 ~ ∞ k∆Zk ~ 2 k∇Hk ~ 2 + 2α2 kHk ~ 2 k∇Zk ~ ∞ k∇∆Zk ~ 2 + 2α2 k∇Zk ~ 2 k∇Zk ~ ∞ k∇Hk ~ 2 + α2 k∇Hk ~ 2 + α2 k∇Hk ~ 2 k∇∆Zk ~ 2. + 2α2 kHk 2

(5.2.18)

By using Lemma 5.2.3 and interpolation formulas and Poincar´e inequality ~ 6L ≤ Ck∇∆Zk ~ n2 k∆Zk ~ 6−n k∇Zk , 2 6 ~ 2 ≤ Ck∇∆Zk ~ 2 k∇Zk ~ 2, k∆Zk 2 n 4

1− n 4

~ ∞ ≤ Ck∇∆Zk ~ 2 k∇Zk ~ 2 k∇Zk

where Z

Ω

~ = 0, ∇Zdx

(5.2.19) ,

Landau–Lifshitz–Maxwell Equations

259

and substituting (5.2.19) into (5.2.18), we get 1∂ 2 ∂t

Z

Ω

~ 2 dx + α2 k∇∆Zk ~ 2+ |∆Z| 2 3

Z

Ω

n

~ t · ∆Hdx ~ Z n

3

n

n

2 ~ 22 − 4 k∇∆Zk ~ 22 + 4 + Cα2 k∇Zk ~ 23− 2 k∇∆Zk ~ 1+ ≤ (2α2 + |α1 |)k∇Zk 2 n

n

n

n

2 ~ 21− 4 k∇∆Zk ~ 21+ 4 + Cα2 k∇Zk ~ 3− ~ 22 k∇Hk ~ 2 + Cα1 k∇Zk k∇∆Zk 2 n

n

~ 2 k∇∆Zk ~ 24 k∇Zk ~ 21− 4 + C|α1 |k∇Hk n

n

n

~ 21− 4 k∇∆Zk ~ 21+ 4 + Cα2 k∇Hk ~ 2 k∇∆Zk ~ 24 + Cα2 k∇Zk ~ 2 + α2 k∇Hk ~ 2 k∇∆Zk ~ 2 + α2 k∇Hk 2

≤

 7α 2 ~ 2 + C1 k∇Hk ~ 2 + d1 ,  k∇∆Zk  2 2   hM    

n=1 i

~ 2 ~ 2 + Cα2 k∇Zk ~ 2 k∇∆Zk (2α2 + |α1 |)k∇Zk 2 2 5α2 2 2 ~ ~ + k∇∆Zk + C2 k∇Hk + d2 , 2

M

2

(5.2.20) n = 2,

where Ci and di (i = 1, 2) are the constants independent of t, D and M is a constant to be chosen later. Here the inequality (5.2.20) will be discussed for n = 1 and 2, respectively. (1) For n = 1, letting M = 14, then (5.2.20) yields Z 1d Z α2 Z 2 2 ~ ~ ~ t · Hdx ~ ~ 22 + d. |∆Z| dx + |∇∆Z| dx + Z ≤ Ck∇Hk 2 dt Ω 2 Ω Ω

(5.2.21)

~ 0 k2 + k E ~ 0 (x)k2 + kH ~ 0 (x)k2 is small enough, (2) For n = 2, take M = 10. If k∇Z then by Lemma 5.2.2 there holds ~ 2 + Cα2 k∇Zk ~ 2 ≤ α2 . (2α2 + |α1 |)k∇Zk 2 2

(5.2.22)

So, (5.2.20) is still true. ~ with (5.1.2) and ∆H ~ with (5.1.3), respectively, Next making the products of ∆E and summing the two equalities, and then integrating the resulting equality with respect to x ∈ Ω, we obtain (β > 0) 1d 2 dt

Z

~ 2 + |∇H| ~ 2 )dx + σ (|∇E| β Ω

Z

Ω

~ 2 dx − |∇E|

Z

Ω

~ t · ∆Hdx ~ Z = 0,

(5.2.23)

where Z

Ω

~ · ∆E ~ −∇×E ~ · ∆H)dx ~ (∇ × H =−

Z

Ω

~ · ∇E ~ − ∇ × ∇E ~ · ∇H)dx ~ (∇ × ∇H = 0.

(5.2.24)

Landau–Lifshitz Equations

260

Combining (5.2.21) with (5.2.23) we have d dt

Z

~ 2 dx + 1 d |∆Z| β dt Ω + α2

Z

Ω

Z

Ω

~ 2 + |∇H| ~ 2 )dx (|∇Z|

~ 2 dx ≤ Ck∇Hk ~ 2 + d. |∇∆Z| 2

(5.2.25)

From the Gronwall inequality it follows that (5.2.15) is proved. ~ − ∆E ~ with (5.1.2) and ∆2 H ~ − ∆H ~ When β = 0, by making the products of ∆2 E with (5.1.3), respectively, and then integrating the resulting equalities with respect to x ∈ Ω, we obtain Z

Ω

~ t (∆2 E ~ − ∆E)dx ~ E +σ Z

Ω

Z

Ω

~ 2E ~ − ∆E)dx ~ E(∆ =

~ t (∆2 H ~ − ∆H)dx ~ H =−

Z

Ω

Z

Ω

~ 2E ~ − ∆E)dx, ~ ∇ × H(∆

~ 2H ~ − ∆H)dx. ~ ∇ × E(∆

It is easy to see that 1d 2 dt

Z

Ω

~ 2 + |∇E| ~ 2 + |∆H| ~ 2 + |∇H| ~ 2 )dx ≤ 0, (|∆E|

i.e. there are h

~ t)k2 + k∇E(·, ~ t)k2 + k∆H(·, ~ t)k2 k∇H(·, ~ t)k2 sup k∆E(·, 2 2 2 2

t∈[0,T ]

~ 0 k2 + k∇E ~ 0 k2 + k∆H ~ 0 k2 + k∇H ~ 0 k2 ≤ K 0 . ≤ k∆E 2 2 2 2

i

(5.2.26)

Note that ~ t |2 = α 2 |Z ~ × (∆Z ~ + H)| ~ 2 + α 2 |Z ~ × (Z ~ × (∆Z ~ + H))| ~ 2 |Z 1 2

~ + H| ~ 2 ≤ 2(α1 + α2 )(|∆Z| ~ 2 + |H| ~ 2 ), ≤ (α12 + α22 )|∆Z 2

Z Z ~ t · ∆Hdx ~ ≤ (α12 + α22 )(k∆Zk ~ 22 + kHk ~ 22 ) + 2k∆Hk ~ 22 . Ω

From (5.2.20) it follows that

1d Z ~ 2 dx + α2 k∇∆Zk ~ 22 ≤ C3 k∇∆Zk ~ 22 + d3 . |∆Z| 2 dt Ω 2 By the Gronwall inequality and (5.2.25), we get (5.2.16).

(5.2.27)

Landau–Lifshitz–Maxwell Equations

261

~ t), H(x, ~ ~ Lemma 5.2.5 Assume that (Z(x, t), E(x, t)) is the smooth solution of the ~ 0 (x), H ~ 0 (x), problem (5.1.1)–(5.1.6). If the conditions of Lemma 5.2.4 hold and ( Z ~ 0 (x)) ∈ (H m (Ω), H m−1 (Ω), H m−1 (Ω)) (m ≥ 2), then for any T > 0, E h

i

~ t)k2 m + kH(·, ~ t)k2 m−1 + kE(·, ~ t)k2 m−1 sup kZ(·, H (Ω) H (Ω) H (Ω) ≤ Km ,

t∈[0,T ]

(5.2.28)

where the constant Km is independent of D, Ω ⊂ Rn , n ≤ 2. Proof. The lemma will be proved by the induction for m. Obviously, when m = 2, (5.2.28) is true. Now suppose that (5.2.28) hold for m = K(≥ 2), i.e. 

sup 

t∈[0,T ]

X

l≤K−2

~ 2 k∇l Zk ∞

+

X

l≤K

~ 2 k∇l Zk 2

+

~ 2 (k∇l Ek 2

X

l≤K−1

+ k∇

l



~ 2 ) Hk 2

≤ El . (5.2.29)

We shall prove (5.2.28) holds for m = K + 1. ~ + H) ~ with (5.2.17), and integrating the (1) Making the scalar product of ∆K (∆Z resulting equality with respect to x ∈ Ω, we have ~ 2 dx ~ t , ∆K H) ~ + 1 (−1)K+1 d |∇K+1 Z| (Z 2 dt Ω Z ~ 2 dx = (−1)K α2 |∇K+2 Z| Z

Ω

+ (−1)K α1 + α2 (−1)K − α2 (−1)

K

Z

Ω

Z

Ω

Z

Ω

K+2 ~ ~ × (∆Z ~ + H))(∇ ~ ~ ∇K ( Z Z + ∇K H)dx

~ 2 Z) ~ · (∇K+2 Z ~ + ∇K H)dx ~ ∇K (|∇Z| ~ × (Z ~ × H)) ~ · (∇K+2 Z ~ + ∇K H)dx. ~ ∇K ( Z

Then 1d 2 dt

Z

Ω

≤

~ 2 dx + α2 k∇K+2 Zk ~ 2 − (−1)K |∇K+1 Z| 2

Z

Ω

~ t · ∇K Hdx ~ Z

α2 K+2 ~ 2 ~ 22 k∇ Zk2 + Ck∇K Hk 4 Z ~ × (∆Z ~ + H)) ~ −Z ~ × ∇K (∆Z ~ + H)| ~ 2 dx + α2 |∇K (Z 1

+ α2

Ω

Z

Ω

~ 2 Z)| ~ 2 dx + α2 |∇K (|∇Z|

Z

Ω

~ × (Z ~ × H))| ~ 2 dx, |∇K (Z

where for the third term on the right-hand side of the inequality (5.2.30) ~ × (∆Z ~ + H)) ~ −Z ~ × ∇K (∆Z ~ + H) ~ ∇K ( Z =

K−1 X i=0

i ~ × (∇i+2 Z ~ + ∇i H) ~ CK ∇K−i Z

(5.2.30)

Landau–Lifshitz Equations

262

~ × (∇K+1 Z ~ + ∇K−1 H) ~ + ∇K Z ~ × (∆Z ~ + H) ~ = K∇Z +

K−2 X i=1

i ~ × (∇i+2 Z ~ + ∇i H) ~ CK ∇K−i Z

K+1 ~ K−1 ~ ~ ~ ~ Z| ~ ≤ K|∇Z||∇ Z| + K|∇Z||∇ H| + |∇K Z||∆

~ H| ~ +C + |∇K Z||

K−2 X i=1

i+2 ~ ~ |∇K−i Z||∇ Z| + C

K−2 X i=1

i~ ~ |∇K−i Z||∇ H|,

~ × (∆Z ~ + H)) ~ −Z ~ × ∇K (∆Z ~ + H)k ~ 2 k∇K ((Z 2 2 K+1 2 2 ~ k∇ ~ + C2 k∇Zk ~ ~ 2 ≤ C1 k∇Zk Zk + C3 k∇K Zk 2

∞

+ C3

K−2 X i=1

~ 2∞ k∇i+2 Zk ~ 22 + C4 k∇K−i Zk

≤

n ~ 21+ K+1 C5 k∇K+2 Zk

≤

α2 K+2 ~ 2 k∇ Zk2 + C7 , M



∞

n K+1

~ 2 + k∇K+2 Zk

∞

K−2 X i=1

~ 2∞ k∇i Hk ~ 22 k∇K−i Zk n

1+ n 3

~ 22 + k∇K+2 Zk ~ 2 + k∇K+2 Zk

(M ≥ 3),

n

n

2(K+2−l) ~ 2 ≤ Ck∇K+2 Zk ~ 22(K+2−l) k∇l Zk ~ 1− k∇l Zk , 2 ∞

(l ≤ K),

~ 2 ≤ Ck∇K+2 Zk ~ 2 k∇K Zk ~ 2. k∇K+1 Zk 2 For the fourth term on the right-hand side of the inequality (5.2.30) ~ 2 Z) ~ = ∇ (|∇Z| =

K X

i ~ K−i (∇Z ~ · ∇Z) ~ CK ∇i Z∇

i=0

K K−i X X i=0 j=0

=

K X

j=0

+

j i ~ K+1−i−j Z ~ · ∇j+1 Z) ~ CK CK−i ∇i Z(∇

j ~ ~ ∇j+1 Z) ~ CK Z(∇K+1−j Z,

K K−i X X i=1 j=0

j i ~ K+1−i−j · ∇j+1 Z) ~ CK CK−i ∇i Z(∇

~ K+1 Z ~ · ∇Z) ~ + = 2Z(∇ +

K K−i X X i=1 j=0

+ C6

(5.2.31)

where the Sobolev interpolation inequality in Lemma 5.2.3 have been used

K



K−1 X j=1

j ~ · ∇j+1 Z) ~ Z ~ CK (∇K+1−j Z

j i ~ K+1−i−j Z ~ · ∇j+1 Z). ~ CK CK−i ∇i Z(∇

Landau–Lifshitz–Maxwell Equations

263

Then ~ 2 Z)k ~ 2 ≤ 2k∇Zk ~ 2 k∇K+1 Zk ~ 2 +C k∇K (|∇Z| 2 ∞ ∞ +C

K K−i X X i=1 j=0

j=1

~ 2 k∇j+1 Zk ~ 2 k∇K+1−j Zk ∞ 2

~ 2 k∇j+1 Zk ~ 2 k∇K+1−i−j Zk ~ 2 k∇i Zk ∞ ∞ 2 n 1+ K+1

~ 2 ≤ C1 k∇K+2 Zk + C3

K−1 X

K K−i X X i=1 j=0

~ 2 + k∇K−1 Zk ~ 2 ) + C2 (k∇K Zk ∞ ∞ n

~ 2K+2−i k∇K+2 Zk

n + K+1−j

.

(5.2.32)

Notice that n (i) when K ≥ 2, and n ≤ 2, there is 1 + K+1 < 2; (ii) when 1 ≤ i ≤ K and 0 ≤ j ≤ K − i, there is

1 1 1 1 + < + < 2; K +2−i K +1−j 2 i+1 n

n

n

2 ~ 2∞ ≤ Ck∇K Zk ~ 2− ~ 22 ≤ Ck∇K+1 Zk ~ 22 ; (iii) k∇K Zk k∇K+2 Zk 2 n ~ 2 ≤ Ck∇K+2 Zk ~ 23 . (iv) k∇K−1 Zk ∞ From (5.2.32) we obtain ~ 22 + C. ~ 2 Z)k ~ 22 ≤ α2 k∇K+2 Zk (5.2.33) k∇K (|∇Z| M By the same way we can prove for the last term on the right-hand side of the inequality (5.2.30) ~ 2 + Ck∇K Hk ~ 2 + C 0. ~ × (Z ~ × H))k ~ 2 ≤ α2 k∇K+2 Zk (5.2.34) k∇K (Z 2 2 2 M Combining (5.2.30), (5.2.31), (5.2.33), and (5.2.34), and taking M ≥ 3, we get

Z 1d Z K+1 ~ 2 K ~ t · ∆K Hdx ~ ~ 22 + C 00 . |∇ Z| dx − (−1) Z ≤ Ck∇K Hk (5.2.35) 2 dt Ω Ω ~ with (5.1.2) and ∆K H ~ with (5.1.3), re(2) Making the scalar products of ∆K E spectively, and integrating the resulting equalities with respect to x ∈ Ω, we have

1 d ~ 2 dx + (−1)K σk∇K Ek ~ 2= ~ · ∆K Edx, ~ (−1)K |∇K E| ∇×H 2 2 dt Ω Ω Z Z Z 1 K d K~ 2 K~ ~ ~ · ∆K Hdx. ~ (−1) |∇ H| dx + β Zt · ∆ Hdx = − ∇ × E 2 dt Ω Ω Ω Summing up the two equalities above gives Z

Z

Z 1d Z K~ 2 K~ 2 K~ 2 K ~ t · ∆K Hdx ~ (|∇ Z| + |∇ H| )dx + σk∇ Ek2 + (−1) β Z 2 dt Ω Ω Z ~ · ∇K E ~ − ∇ × (∇K E) ~ · ∇K H)dx ~ = (∇ × ∇K H) Ω

=

Z

Ω

~ × ∇K E)dx ~ ∇ · (∇K H = 0.

(5.2.36)

Landau–Lifshitz Equations

264

From (5.2.35) and (5.2.36) it follows that 1d 2 dt

"Z

Ω

|∇

1 Z| dx + β

K+1 ~ 2

Z

~ + |∇ H| ~ )dx (|∇ Z| K

Ω

2

K

2

#

~ 2 + C 0. ≤ Ck∇K Hk 2 Then (5.2.28) is true for m = K + 1. The proof of the lemma is complete.

5.2.3

Existence of Global Smooth Solution

By using the following result about the local existence of the smooth solution for (5.1.1)–(5.1.6) and the a priori estimates of the smooth solution, we can obtain the global existence of the smooth solution for the problem (5.1.1)–(5.1.6) from the extension principle. By the Galerkin method, we can prove Theorem 5.2.1 (Local existence of the smooth solution) Assume the constants αh 2 i> ~ 0 (x), H ~ 0 (x), E ~ 0 (x)) ∈ (H k (Ω), H k−1 (Ω), H k−1 (Ω)) k ≥ 3 + n , 0, β ≥ 0, σ ≥ 0, (Z 3 ~ 0 + βZ ~ 0 ) = 0, ∇ · E ~ 0 = 0. Then the periodic initial value problem n ≤ 2, ∇ · (H (5.1.1)–(5.1.6) has a local smooth solution k 2] s ~ t) ∈ ∩[s=0 Z(x, W∞ (0, T0 ; H k−2s (Ω)),

s k−1−s ~ ~ H(x, t), E(x, t) ∈ ∩k−1 (Ω)). s=0 W∞ (0, T0 ; H

Remark. From the relations ~ 0 + β∇ · Z ~ 0 = 0, ∇·H

~ 0 = 0, ∇·E

we can deduce that ~ + β∇ · Z ~ = 0, ∇·H

~ = 0. ∇·E

In fact, by applying the divergence operator ∇· for (5.1.2) and (5.1.3), we get ∂ ~ + σ∇ · E ~ = 0, ∇·E ∂t

∂ ~ + β Z) ~ = 0, ∇ · (H ∂t

and so the assertions above hold. From the Theorem 5.2.1 and a priori estimates in Sec. 5.2.2 it follows that Theorem 5.2.2 Assume that the constants α2 > 0, β ≥ 0, σ ≥ 0, the hinitial value i n k k−1 k−1 ~ ~ ~ functions (Z0 (x), H0 (x), E0 (x)) ∈ (H (Ω), H (Ω), H (Ω)), k ≥ 3 + 2 , Ω ⊂ Rn ~ 0 (x)| = 1, ∇ · E ~ 0 = 0, ∇ · (H ~ 0 + βZ ~ 0 ) = 0. (n ≤ 2) is a bounded domain, and |Z Moreover, when n = 2, there is ~ 0 k2 + k E ~ 0 k2 + k H ~ 0 k2 ≤ δ, k∇Z

Landau–Lifshitz–Maxwell Equations

265

where δ is a suitable small constant. Then the periodic problem (5.1.1)–(5.1.6) has ~ t)| = 1, x ∈ Ω, t ∈ R+ , and one global smooth solution such that |Z(z, k 2] s ~ t) ∈ ∩[s=0 Z(x, W∞ (0, T ; H k−2s(Ω)),

s k−1−s ~ H(x, t) ∈ ∩k−1 (Ω)), s=0 W∞ (0, T ; H s k−1−s ~ E(x, t) ∈ ∩k−1 (Ω)). s=0 W∞ (0, T ; H

Noticing that all the constants K of the estimates are independent of the periodic D, and letting D → ∞, we can obtain the solution of the Cauchy problem. Theorem 5.2.3 Under the conditions of Theorem 5.2.2, the Cauchy problem ~ t), H(x, ~ ~ (5.1.1)–(5.1.5) and (5.1.7) has one global smooth solution Z(x, t), E(x, t) such that ~ t)| = 1, |Z(x, ∀ (x, t) ∈ Rn × R+ , (n ≤ 2), [ k2 ] s ~ t) ∈ ∩s=0 W∞ (0, T ; H k−1−2s(Ω)), ∇Z(x, s k−1−s ~ H(x, t) ∈ ∩k−1 (Ω)), s=0 W∞ (0, T ; H s k−1−s ~ E(x, t) ∈ ∩k−1 (Ω)), s=0 W∞ (0, T ; H

where Ω = R2 . Theorem 5.2.4 Assume that the conditions of Theorem 5.2.2 are satisfied. Then the global smooth solution for the periodic initial value problem (the Cauchy problem) for Landau–Lifshitz–Maxwell system is unique. ~i, H ~ i, E ~ i ) (i = 1, 2) be the smooth solutions for the problem Proof. Let (Z (5.1.1)–(5.2.6), or (5.1.1)–(5.1.5) and (5.1.7). Denote ~ t) = Z ~ 1 (x, t) − Z ~ 2 (x, t), Z(x, ~ ~ 1 (x, t) − H ~ 2 (x, t), H(x, t) = H ~ ~ 1 (x, t) − E ~ 2 (x, t). E(x, t) = E

(5.2.37)

~ t), H(x, ~ ~ Then Z(x, t) and E(x, t) satisfy the following system: ~ t = α1 Z ~ × ∆Z ~ 2 + α1 Z ~ 1 × ∆Z ~ + α1 Z ~ ×H ~ 2 + α1 Z ~1 × H ~ + α 2 ∆Z ~ Z ~ 2 |2 Z ~ + α2 (∇Z ~ · ∇(Z ~1 + Z ~ 2 ))Z ~ 1 + α2 H ~ + α2 |∇Z ~2 · H ~ 2 )Z ~ − α 2 (Z ~2 · H ~ +H ~ 1 · Z) ~ Z ~1, − α 2 (Z ~ t + σE ~ = ∇ × H, ~ E ~ t + βZ ~ t = −∇ × E, ~ H

~ 0) = H(x, ~ ~ Z(x, 0) = E(x, 0) = 0.

(5.2.38) (5.2.39) (5.2.40) (5.2.41)

Landau–Lifshitz Equations

266

~ with (5.2.38) and integrating the equality (1) Making the scalar product of Z obtained with respect to x ∈ Ω, we get 1d 2 dt

Z

Ω

~ 2 dx + α2 k∇Zk ~ 2 |Z| 2

~ 2 + kZk ~ 2 + kHk ~ 2 ). ≤ C(|α1 | + α2 )(k∇Zk 2 2 2

(5.2.42)

~ with (5.2.38) and integrating the equality (2) Making the scalar product of ∆Z obtained with respect to x ∈ Ω, we get 1d 2 dt

Z

Ω

~ 2 dx + α2 k∆Zk ~ 2 |∇Z| 2

= −α1

Z

− α1 − α2 + α2

Z

~ ×H ~ 2 · ∆Zdx ~ Z

Ω

~ × ∆Z ~ 2 · ∆Zdx ~ − α1 Z

Z

Ω

~1 × H ~ · ∆Zdx ~ − α2 Z

Ω

~ · (∇(Z ~1 + Z ~ 2 ))Z ~ 1 · ∆Zdx ~ + α2 (∇Z

Ω

~2 · H ~ 2 )Z ~ · ∆Zdx ~ + α2 (Z

Z

Z

Z

Ω

Ω

~ 2 |2 Z ~ · ∆Zdx ~ |∇Z

Z

Ω

Z

Ω

~ · ∆Zdx ~ H

~2 · H ~ +H ~ 1 · Z) ~ Z ~ 1 · ∆Zdx. ~ (Z

(5.2.43)

Now, we estimate each term on the right-hand side of (5.2.43). Z ~ ~ ~ −α1 Z × ∆Z2 · ∆Zdx = Ω

Z ~ ~ ~ α1 Z × ∇∆Z2 · ∇Zdx Ω

~ ∞ kZk ~ 2 k∇Zk ~ 2 ≤ C|α1 |(kZk ~ 2 + k∇Zk ~ 2 ), ≤ |α1 |k∇∆Zk 2 2

Z Z ~ ~ ~ ~ ~ ~ −α1 Z × H2 · ∆Zdx = α1 Z × ∇H2 · ∇Zdx Ω

Ω

~ 2 k∞ kZk ~ 2 k∇Zk ~ 2 ≤ C|α1 |(kZk ~ 2 + k∇Zk ~ 2 ), ≤ |α1 |k∇H 2 2

Z 2~ ~ ~ ~ 2 k2 kZk ~ 2 k∆Zk ~ 2 −α2 |∇Z2 | Z · ∆Zdx ≤ α2 k∇Z ∞ Ω

≤

α2 ~ 2 + C(K)kZk ~ 2, k∇Zk 2 2 K

(K ≥ 6),

Z ~ ~ ~ ~ ~ ~1 + Z ~ 2 )k∞ k∇Zk ~ 2 k∆Zk ~ 2 −α2 ∇Z · (∇(Z1 + Z2 ))Z1 · ∆Zdx ≤ α2 k∇(Z Ω

α2 ~ 2 + C(K)α2 k∇Zk ~ 2, k∆Zk 2 2 K Z α 2 ~ · ∆Zdx ~ ≤ α2 kHk ~ 2 k∆Zk ~ 2 H Ω α2 ~ 2 + C(K)α2 kHk ~ 2, ≤ k∆Zk 2 2 K Z ~2 · H ~ 2 )Z ~ · ∆Zdx ~ ≤ α2 kH ~ 2 k∞ kZk ~ 2 k∆Zk ~ 2 α 2 ( Z Ω α2 ~ 2 + C(K)α2 kZk ~ 2, ≤ k∆Zk 2 2 K ≤

Landau–Lifshitz–Maxwell Equations

267

Z ~2 · H ~ +H ~ 1 · Z) ~ · ∆Zdx ~ ≤ α2 (kHk ~ 2 k∆Zk ~ 2 + kH ~ 1 k∞ kZk ~ 2 k∆Zk ~ 2) α 2 ( Z Ω

≤

α2 ~ 2 + C(K)α2 (kHk ~ 2 + kZk ~ 2 ). k∆Zk 2 2 2 K

Then (5.2.43) can be rewritten as 1d 2 dt

Z

Ω

~ ~ 2 + α1 |∇Z|dx + α2 k∆Zk 2

Z

Ω

~1 × H ~ · ∆Zdx ~ Z

5 ~ 2 + C(K)α2 (k∇Zk ~ 2 + kZk ~ 2 + kHk ~ 2) k∆Zk 2 2 2 2 K ~ 2 + k∇Zk ~ 2 ). + C|α1 |(kZk

≤

2

2

(5.2.44)

~ with (5.2.39) and H ~ with (5.2.40), respec(3) Making the scalar products of E tively, and integrating the equalities obtained with respect to x ∈ Ω, and then summing up the two equations, we get 1 d 2β dt

Z

Ω

~ 2 + |H| ~ 2 )dx + (|E|

σ ~ 2 kEk2 = − β

Z

Ω

~ t · Hdx. ~ Z

(5.2.45)

From (5.2.42), (5.2.44) and (5.2.45) it follows that 1d 2 dt

Z " Ω

#

~ 2 + |H| ~ 2 + 1 (|E| ~ 2 + |H| ~ 2 ) dx + α2 k∆Zk ~ 2 |E| 2 β

5 ~ 22 ~t · H ~ + α1 Z ~ ×H ~ · ∆Z)dx ~ (Z + α2 k∆Zk K Ω 2 2 2 ~ + kZk ~ + kHk ~ ) + C|α1 |(k∇Zk

≤−

Z

2

+

~ 22 C(K)α2 (k∇Zk

2

+

~ 22 kZk

2

~ 22 ). + kHk

(5.2.46)

~ with (5.2.38), and integrating the equality (4) Making the scalar product of H obtained with respect to x ∈ Ω, we get Z (Z ~t · H ~ + α1 Z ~1 × H ~ · ∆Z)dx ~ Ω Z Z ~ ~ ~ ~ ~ ~ ~ ≤ |α1 | (Z × ∆Z2 + Z × H2 ) · Hdx + α2 ∆Z · Hdx Ω Ω Z Z 2~ ~ ~ ~ ~ ~ ~ ~ + α2 |∇Z2 | Z · Hdx + α2 (∇Z · ∇(Z1 + Z2 )Z1 · Hdx Ω Ω Z 2 ~ ~ ~ ~ ~ + α2 kHk2 + α2 (Z2 · H2 )Z · Hdx Ω Z Z ~ ~ ~ ~ ~ ~ ~ ~ + α2 (Z2 · H)(Z1 · H)dx + α2 (H1 · Z)Z1 · Hdx Ω

Ω

α2 ~ 22 + (C(K)α2 + C|α1 |)(k∇Zk ~ 22 + kZk ~ 22 + kHk ~ 22 ). ≤ k∆Zk K

(5.2.47)

Landau–Lifshitz Equations

268

Substituting (5.2.47) into (5.2.46), and taking K ≥ 6, we get "

#

d Z ~ 2 + |Z| ~ 2 + 1 (|E| ~ 2 + |H| ~ 2 ) dx |∇Z| dt Ω β ~ 22 + kZk ~ 22 + kHk ~ 22 ). ≤ C(α2 + |α1 |)(k∇Zk ~ t) = H(x, ~ ~ From the Gronwall inequality it follows that Z(x, t) = E(x, t) ≡ 0, ∀ (x, t) ∈ Ω × (0, T ). When β = 0, by (5.2.45) we obtain d ~ 2 ~ 2 ) + σkEk ~ 2 = 0. (kEk2 + kHk 2 2 dt ~ ~ ~ t) ≡ 0. Then E(x, t) ≡ H(x, t) ≡ 0. From (5.2.42) and (5.2.44) it follows that Z(x, The proof of the theorem is complete. 2

5.3

Global Smooth Solution to One-Dimensional L–L–M with Large Data

In Sec. 5.1, we have obtained the global weak solutions to three-dimensional Landau– Lifshitz–Maxwell equations and, in Sec. 5.2, we proved the existence and uniqueness of global smooth solution to this system with small initial data. Now we are going to derive the existence and uniqueness of global smooth solution to this system with large initial data in one dimension. Especially, we deal with the equations without Gilbert damping, that is, α2 = 0 in (5.1.1). So it states as follows: ~ t = α1 Z ~ × (∆Z ~ + H), ~ Z

(5.3.1)

~ ~ = ∂ E + σ E, ~ ∇×H ∂t ~ ~ ~ = − ∂H − β ∂Z , ∇×E ∂t ∂t ~ + β∇ · Z ~ = 0, ∇ · E ~ = 0, ∇·H

(5.3.2) (5.3.3) (5.3.4)

with the periodic initial value condition: ~ + 2D, t) = Z(x, ~ t), Z(x

~ + 2D, t) = H(x, ~ H(x t),

~ + 2D, t) = E(x, ~ E(x t), ~ 0) = Z ~ 0 (x), Z(x,

~ ~ 0 (x), H(x, 0) = H

(x ∈ R, t ≥ 0)

~ ~ 0 (x), E(x, 0) = E

(5.3.5) (x ∈ R),

(5.3.6)

x ∈ R.

(5.3.7)

where D > 0; or with the initial value condition: ~ 0) = Z ~ 0 (x), Z(x,

~ ~ 0 (x), H(x, 0) = H

~ ~ 0 (x), E(x, 0) = E

Landau–Lifshitz–Maxwell Equations

5.3.1

269

Viscosity Vanishing Method

In one space dimension we shall prove the existence of the global smooth solution of the periodic problem (5.3.1)–(5.3.6) or the initial problem (5.3.1)–(5.3.4) and (5.3.7) by the viscosity vanishing method. It means that we approximate (5.3.1) by an equation with viscosity term of small parameter, i.e. the following periodic initial value problem is considered: ~ t = α1 Z ~ × (∆Z ~ + H) ~ − εZ ~ × (Z ~ × (∆Z ~ + H)), ~ Z ~ ~ = ∂ E + σ E, ~ ∇×H ∂t ~ ~ ~ = − ∂H − β ∂Z , ∇×E ∂t ∂t ~ + β∇ · Z ~ = 0, ∇ · E ~ = 0, ∇·H

~ + 2D, t) = Z(x, ~ t), Z(x

~ ~ 0 (x), H(x, 0) = H

(5.3.9) (5.3.10) (5.3.11)

~ + 2D, t) = H(x, ~ H(x t),

~ + 2D, t) = E(x, ~ E(x t), ~ 0) = Z ~ 0 (x), Z(x,

(5.3.8)

(x ∈ R, t ≥ 0)

~ ~ 0 (x), E(x, 0) = E

(5.3.12) (x ∈ R)

(5.3.13)

and the a priori estimate independent of ε is proved, then the solution of the problem (5.3.1)–(5.3.6) is obtained as ε → 0. ~ 0 (x)| = 1. Then for the smooth solution of the periodic initial Lemma 5.3.1 Let |Z value problem (5.3.8)–(5.3.13) there holds ~ t)| = 1, |Z(x,

x ∈ Ω ≡ [0, 2D],

t ≥ 0.

(5.3.14)

~ with (5.3.8), we get Proof. Making the scalar product of Z ∂ ~ |Z(x, t)|2 = 0. ∂t Then the conclusion of the lemma is proved. ~ 0 (x) ∈ L2 (Ω), E ~ 0 (x) ∈ L2 (Ω), Lemma 5.3.2 Assume β ≥ 0, σ ≥ 0, ∇Z ~ 0 (x) ∈ L2 (Ω). Then for the smooth solution of the periodic initial value problem H (5.3.8)–(5.3.13) h

i

~ t)k2 + kE(·, ~ t)k2 + kH(·, ~ t)k2 ≤ K1 , sup k∇Z(·, 2 2 2

0≤t≤T

ε

Z

0

T

~ × (∆Z ~ + H)k ~ 2 dt ≤ K2 , kZ 2

(5.3.15)

~ 0 (x)k2 , kE ~ 0 (x)k2 and kH ~ 0 (x)k2 , where the constants K1 and K2 depend only on k∇Z but not on ε and D.

Landau–Lifshitz Equations

270

~ with (5.3.9), and Proof. First assume β > 0. Making the scalar product of E ~ making the scalar product of −H with (5.3.10), and then adding these two equalities obtained, we have ~ ·E ~ − (∇ × E) ~ ·H ~ = (∇ × H)

~ ~ ~ ∂E ~ + σ|E| ~ 2 + ∂H · H ~ + β ∂ Z · H. ~ ·E ∂t ∂t ∂t

(5.3.16)

By using the formula ~ ·E ~ − (∇ × E) ~ ·H ~ = ∇ · (H ~ × E), ~ (∇ × H)

(5.3.17)

and integrating (5.3.16) over x ∈ Ω, we get

Z  ~ 1d  ~ 2 ∂Z 2 2 ~ ~ ~ kEk2 + kHk2 + σkEk2 + β · Hdx = 0. 2 dt Ω ∂t ~ + H) ~ with (5.3.8), we get Making the scalar product of (∆Z

(5.3.18)

~ ~ + H) ~ · ∂ Z = −ε(∆Z ~ + H) ~ · [Z ~ × (Z ~ × (∆Z ~ + H))], ~ (∆Z ∂t

(5.3.19)

~ + H) ~ · [Z ~ × (Z ~ × (∆Z ~ + H))] ~ ~ × (∆Z ~ + H)| ~ 2. −(∆Z = |Z

(5.3.20)

where

From integrating (5.3.19) over x ∈ Ω it follows that Z

Z ~ ~ + H) ~ · ∂ Z dx = ε |Z ~ × (∆Z ~ + H)| ~ 2 dx, (∆Z ∂t Ω Ω

and then ~ ~ ∂Z ~ ~ · ∂ Z dx + ε |Z ~ × (∆Z ~ + H)| ~ 2 dx · Hdx = − ∆Z ∂t Ω Ω Ω ∂t Z 1d ~ 22 + ε |Z ~ × (∆Z ~ + H)| ~ 2 dx. k∇Zk = 2 dt Ω By adding (5.3.18) and (5.3.21) we obtain Z

Z

Z

1d ~ 2 ~ 2 ) + σkEk ~ 2 (kEk2 + kHk 2 2 2 dt β d ~ 2 + βεkZ ~ × (∆Z ~ + H)k ~ 2 = 0. + k∇Zk 2 2 2 dt Integrating the above equality with respect to t ∈ [0, T ], we have

1 ~ ~ t)k2 ) + σkE(·, ~ t)k2 (kE(·, t)k22 + kH(·, 2 2 2 Z t β ~ t)k2 + βε kZ ~ × (∆Z ~ + H)k ~ 2 dt + k∇Z(·, 2 2 2 0 1 ~ 2 2 ~ = E(0) = (kE 0 (x)k2 + kH0 (x)k2 ) 2 ~ 0 (x)k22 + β k∇Z ~ 0 (x)k22 . + σkE 2

(5.3.21)

(5.3.22)

E(t) ≡

(5.3.23)

Landau–Lifshitz–Maxwell Equations

271

It follows that the estimate (5.3.15) holds. Now assume β = 0. Then Eq. (5.3.10) becomes ~ ~ = − ∂H . ∇×E ∂t Combining the equality above with the equation (5.3.9), we get 1d ~ 2 ~ 2 ) + σkEk ~ 2 = 0. (kEk2 + kHk 2 2 2 dt Therefore, ~ t)k2 + kH(·, ~ t)k2 ≤ kE ~ 0 (x)k2 + kH ~ 0 (x)k2 . kE(·, 2 2 2 2 Note that

Z

Ω

~ ·H ~ t dx = − Z

Z

~ · Zdx ~ =− ∇×E

Ω

Z

Ω

~ · Edx. ~ ∇×Z

(5.3.24) (5.3.25)

Adding (5.3.21) and (5.3.25) yields d dt

Z

1d ~ · Hdx ~ ~ 2 + εkZ ~ × (∆Z ~ + H)k ~ 2− Z = k∇Zk 2 2 2 dt Ω

Z

Ω

~ · Edx. ~ (∇ × Z)

So, ~ t)k2 + 2ε k∇Z(·, 2 ≤2

Z

+

Ω

Z

Z

0

t

~ × (∆Z ~ + H)k ~ 2 dt kZ 2

~ t) · H(x, ~ Z(x, t)dx − 2 t

0

~ t)k2 dt + k∇Z(·, 2

1 ~ t)k2 + C ≤ k∇Z(·, 2 2 ≤C

Z

t 0

Z

t 0

Z

t 0

Z

Ω

~ 0 (x) · H ~ 0 (x)dx Z

~ t)k2 dt kE(·, 2

~ t)k2 dt + C1 k∇Z(·, 2

~ t)k22 dt + C2 . k∇Z(·,

From the Gronwall inequality (5.3.15) is proved. Lemma 5.3.3 Assume that the conditions of Lemmas 5.3.1 and 5.3.2 hold, and ~ 0 (x) ∈ H 2 (Ω), E ~ 0 (x), H ~ 0 (x) ∈ H 1 (Ω). Then there holds Z h

i

(5.3.26)

i

(5.3.27)

~ t)k22 + k∇H(·, ~ t)k22 + k∇E(·, ~ t)k22 ≤ K3 , sup k∆Z(·,

0≤t≤T

h

~ t (·, t)k2 + kH ~ t (·, t)k2 + kE ~ t (·, t)k2 ≤ K4 , sup kZ 2 2 2

0≤t≤T

~ 0 (x)kH 2 (Ω) , kH ~ 0 (x)kH 1 (Ω) and kE ~ 0 (x)kH 1 (Ω) , but where K3 and K4 depend only on kZ not on ε and D.

Landau–Lifshitz Equations

272

Proof. The proof of this lemma is longer than others. In the following the constants Ci , dj , C, d are the constants independent of ε and D. First, by the argument ~ 0 (x)| = 1, Eq. (5.3.8) in the used in [20], we can prove that under the condition |Z classic sense is equivalent to the equation ~ t = α1 Z ~ × (∆Z ~ + H) ~ + ε∆Z ~ + ε|∇Z| ~ 2Z ~ − ε(Z ~ · H) ~ Z ~ + εH. ~ Z

(5.3.28)

Differentiating (5.3.28) with respect to t we get ~ tt = α1 Z ~ t × (∆ + H) ~ + α1 Z ~ × (∆Z ~t + H ~ t ) + ε∆Z ~ t + ε|∇Z| ~ 2Z ~ Z t ~ 2Z ~ t − ε(Z ~ · H) ~ tZ ~ − ε(Z ~ · H) ~ Z ~ t + εH ~ t. + ε|∇Z|

(5.3.29)

~ t with the above equality, and integrating the (I) Making the scalar product of Z resulting equality with respect to x ∈ Ω, we have 1d Z ~ 2 ~ t k22 |Zt | dx + εk∇Z 2 dt Ω = α1

Z

+ε

Ω

~ × ∆Z ~t · Z ~ t dx + α1 Z

Z

~ 2 |Z ~ t |2 dx − ε |∇Z|

Ω

Z

Z

Ω

Ω

~ ×H ~t · Z ~ t dx + ε Z

Z

Ω

~t · Z ~ t dx H

~ · H)| ~ Z ~ t |2 dx. (Z

(5.3.30)

By using the Sobolev inequality in Lemma 5.3.3, we estimate each term at the righthand side of (5.3.30) except the first term. (i) Z α 1 ~ ~ ~ Z × Ht · Zt dx = Ω

Z −α1 ~ ~ ~ Z × (∇ × E) · Zt dx Ω  

~ 2 + kZ ~ t k2 , ≤ C k∇Ek 2 2

(ii)

Z Z Z 2 ~ ~ ~ ~ ~ ε Ht · Zt dx = −εβ |Zt | dx − ε ∇ × E · Zt dx Ω Ω Ω  

~ 2 + kZ ~ t k2 , ≤ C k∇Ek 2 2

(iii) ε

Z

Ω

~ 2 |Z ~ t |2 dx ≤ εk∇Zk ~ 2 kZ ~ t k2 ≤ Cεk∇Z ~ t k2 kZ ~ t k2 |∇Z| 2 ∞ ≤

ε ~ t k2 + CkZ ~ t k2 , k∇Z 2 2 K

K is a constant to be chosen later,

(iv) Z 3 1 −ε (Z ~ · H)| ~ Z ~ t |2 dx ≤ εkHk ~ 2 kZ ~ t k24 ≤ CεkHk ~ 2 kZ ~ t k22 k∇Z ~ t k22 Ω

≤

ε ~ t k22 + CkZ ~ t k22 . k∇Z K

(5.3.31)

Landau–Lifshitz–Maxwell Equations

273

(II) To estimate the first term on the right-hand side of (5.3.30), we apply the operator ∆ to (5.3.28) and obtain ~ t = α 1 (Z ~ × ∆2 Z ~ + 2∇Z ~ × ∇∆Z ~ + ∆Z ~ ×H ~ + 2∇Z ~ × ∇H ~ +Z ~ × ∆H) ~ ∆Z 

~ + |∇Z| ~ 2 ∆Z ~ + 2|∆Z| ~ 2Z ~ + ε ∆2 Z

~ · ∇∆Z) ~ Z ~ + 4(∇Z ~ · ∆Z)∇ ~ Z ~ + ∆H ~ + 2(∇Z



~ · H)∆ ~ Z ~ − 2(∇Z ~ · H)∇ ~ Z ~ − 2(Z ~ · ∇H)∇ ~ Z ~ . − (Z

(5.3.32)

~ with the equality above, and making the scalar Making the cross product of Z ~ t with the resulting equality, and then using Z ~ ·Z ~ t = 0, ∇Z ~ ·Z ~ = 0, and products of Z 2 ~ ~ ~ Z · ∆Z = −|∇Z| , we obtain ~ × ∆Z ~t · Z ~ t = α2 (−∆2 Z ~ ·Z ~ t + 2(Z ~ · ∇3 Z)(∇ ~ ~ ·Z ~t) α1 Z Z 1 ~ × (∆Z ~ + H) ~ ·Z ~ t − ∆H ~ ·Z ~ t + 2(Z ~ · ∇H)(∇ ~ ~ ·Z ~t) +Z Z h

~ × ∆2 Z ~ ·Z ~ t + |∇Z| ~ 2Z ~ × ∆Z ~ ·Z ~t + α1 ε Z

~ · ∇2 Z) ~ Z ~ × ∆Z ~ ·Z ~t + Z ~ × ∆H ~ ·Z ~ t − (Z ~ · H) ~ Z ~ × ∆Z ~ ·Z ~t + 4(∇Z i

~ · H) ~ Z ~ × ∇Z ~ ·Z ~ t − 2(Z ~ · ∇H) ~ Z ~ × ∇Z ~ ·Z ~t . − 2(∇Z

(5.3.33)

We estimate each term on the right-hand side of (5.3.32) except the fourth term, ~ ·Z ~t. −∆H (1) Z Z 2 ~ 2 dx. ~ ·Z ~ t dx = − α1 d |∆Z| −α12 ∆2 Z 2 dt Ω Ω ~ 2 ), we have ~ · ∇3 Z ~ = − 3 ∇(|∇Z| (2) Since Z 2 2α12

Z

Ω

~ · ∇3 Z)(∇ ~ ~ ·Z ~ t )dx (Z Z

= −3α12 =

3α12

3 = α12 4

Z

Z

Ω

Ω

~ 2 )(∇Z ~ ·Z ~ t )dx ∇(|∇Z|

~ 2 (∇Z ~ · ∇Z ~ t )dx + 3α2 |∇Z| 1

Z

Ω

~ 4 dx |∇Z|

!

+ 3α12 t

Z

Ω

Z

Ω

~ 2 ∆Z ~ ·Z ~ t dx |∇Z|

~ 2 ∆Z ~ · [ε|∇Z| ~ 2Z ~ |∇Z|

~ + α1 Z ~ ×H ~ − εZ ~ × (Z ~ × H)]dx ~ + ε∆Z Z  Z 3 ~ 4 dx − 3α2 ε |∇Z| ~ 6 dx ≤ α12 |∇Z| 1 4 Ω Ω t 1

1

2 2 ~ 2+ ~ 1+ ~ 2 kHk ~ 2 k∆Zk ~ 2 + 3Cα12 εk∇Zk k∇∆Zk + Ck∇Zk 2 2 2 Z  Z 3 ~ 4 dx − 3α2 ε |∇Z| ~ 6 dx ≤ α12 |∇Z| 1 4 Ω Ω t ε ~ 22 + C3 k∆Zk ~ 22 + d. + k∇∆Zk K

Landau–Lifshitz Equations

274

(3) α12

Z

Ω

~ × (∆Z ~ + H) ~ ·Z ~ t dx Z Z

=

α12

=

α12

ZΩ

+

α12

≤

Ω

~ · H)(∆ ~ ~ ·Z ~ t )dx + α2 (Z Z 1

Z

Ω

~ 2Z ~ t · Hdx ~ |∇Z|

~ · H)∆ ~ Z ~ · (α1 Z ~ ×H ~ + ε∆Z ~ + ε|∇Z| ~ 2Z ~ − εZ ~ × (Z ~ × H))dx ~ (Z Z

~ 2Z ~ t · Hdx ~ |∇Z|

Ω 3 ~ ~ 24 + εα12 kHk ~ 2 (k∆Zk ~ 24 + k∇Zk ~ 48 ) α1 k∆Zk2 kHk ~ 2 kHk ~ 2 + α2 kHk ~ 2 kZ ~ t k2 k∇Zk ~ 2 + α12 εk∆Zk 4 1 ∞ 10

1

3

~ 28 + k∇∆Zk ~ 24 ) + CkZ ~ t k2 k∆Zk ~ 2 ~ 2 k∇Hk ~ 22 + Cε(k∇∆Zk ≤ Ck∆Zk ε ~ 2 + C1 k∆Zk ~ 2 + C 2 kZ ~ t k2 + C3 k∇Hk ~ 2 + d1 ≤ k∇∆Zk 2 2 2 2 K ε ~ 2 + C4 k∆Zk ~ 2 + C3 k∇Hk ~ 2 + d2 . ≤ k∇∆Zk 2 2 2 K (4) 2α12

Z

Ω

~ · ∇H)(∆ ~ ~ ·Z ~ t )dx (Z Z

= =

2α12

Ω

−2α12 

−α12

Z

Ω

Z

~ · H)∇ ~ Z ~ · ∆Z ~ t dx − 2α2 (Z 1

Ω

Ω

Ω

~ · H)∆ ~ Z ~ ·Z ~ t dx (Z

~ · H)∇ ~ Z ~ ·Z ~ t dx (∇Z

~ · H)|∇ ~ ~ 2 dx (Z Z|

Z h Ω

Z



+ t

α12

Z

Ω

~t · H ~ +Z ~ ·H ~ t )|∇Z| ~ 2 dx (Z i

~ · H)∆ ~ Z ~ ·Z ~ t + (∇Z ~ · H)(∇ ~ ~ ·Z ~ t ) dx. (Z Z

~ ·H ~ t = −β Z ~ ·Z ~t − Z ~ ·∇×E ~ = −Z ~ · ∇ × E, ~ Z

it follows that 2α12

Ω

Z

− 2α12 From (3) and

~ · H) ~ − ∇Z ~ · H)(∆ ~ ~ ·Z ~ t )dx (∇ · (Z Z

Z

2α12

− =

Z

~ · ∇H)(∆ ~ ~ ·Z ~ t )dx (Z Z

≤



−α12

Z

Ω

~ · H)|∇ ~ ~ 2 dx (Z Z|



t

~ 2 kHk ~ 2 kZ ~ t k2 + k∇Zk ~ 2 k∇Zk ~ 2) + α12 (k∇Zk ∞ 4

ε ~ 2 + C3 k∇Zk ~ 2 + C4 k∇Hk ~ 2 + d2 k∇∆Zk 2 2 2 K   Z ~ · H)|∇ ~ ~ 2 dx ≤ −α2 (Z Z| +

1

Ω

t

ε ~ 2 + C(k∆Zk ~ 2 + k∇Hk ~ 2 + k∇Ek ~ 2 ) + d, + k∇∆Zk 2 2 2 2 K where the constants C and d are independent of ε and D.

Landau–Lifshitz–Maxwell Equations

275

(5) α1 ε

Z

Ω

~ × ∆2 Z ~ ·Z ~ t dx Z

= α1 ε

Z

3~

Ω

~ × ∇Z ~ t dx + α1 ε ∇ Z ·Z

Z

Ω

~ · ∇Z ~ ×Z ~ t dx. ∇3 Z

Note that ~ t = ε∇∆Z ~ + ε|∇Z| ~ 2 ∇Z ~ + 2ε(∇Z ~ · ∆Z) ~ Z ~ + α 1 ∇Z ~ × ∆Z ~ ∇Z ~ × ∇∆Z ~ + α1 ∇(Z ~ × H) ~ − ε∇(Z ~ × (Z ~ × H)). ~ + α1 Z

(5.3.34)

Then α1 ε

Z

Ω

~ × ∆2 Z ~ ·Z ~ t dx Z

= α1 ε

Z

Ω

h

~ ×Z ~ · ε∇3 Z ~ + ε|∇Z| ~ 2 ∇Z ~ − ε∇(Z ~ × (Z ~ × H)) ~ ∇3 Z i

~ × H) ~ + α1 Z ~ × ∇3 Z ~ + α 1 ∇Z ~ × ∆Z ~ dx + α1 ε + α1 ∇(Z

Z

Ω

~ · ∇Z ~ ×Z ~ t dx ∇3 Z

~ × Zk ~ 2 + C1 εk∇∆Zk ~ 2 (k∇Zk ~ 3 + k∇Zk ~ ∞ + k∇Hk ~ 2) ≤ −α12 εk∇3 Z 2 6 ~ 2 k∇Zk ~ ∞ (k∆Zk ~ 2 + kZ ~ t k2 ) + C1 εk∇∆Zk 1

5

~ × Zk ~ 2 + C2 ε(k∇∆Zk ~ 21+ 2 + k∇∆Zk ~ 24 ≤ −α12 εk∇3 Z 2 7

~ 24 + k∇∆Zk ~ 2 k∇Hk ~ 2) + k∇∆Zk ~ 2 + Ck∇Hk ~ 2 + d. ~ × Zk ~ 2 + ε k∇∆Zk ≤ −α12 εk∇3 Z 2 2 2 K (6) Z 2~ α1 ε ~ ~ ~ |∇Z| Z × ∆Z · Zt dx Ω

~ 2 k∆Zk ~ 2 kZ ~ t k2 ≤ |α1 |εk∇Zk ∞ ~ 2 k∆Zk ~ 2 + C1 εk∇Zk ~ 2 k∆Zk ~ 2 kHk ~ 2 ≤ C1 εk∇Zk ∞ 2 ∞ 3

~ 2) ~ 22 + k∇∆Zk ≤ C2 ε(k∇∆Zk ε ~ 2 + d. ≤ k∇∆Zk 2 K (7) By (6) we have Z ~ · ∇2 Z) ~ Z ~ × ∆Z ~ ·Z ~ t dx 4α1 ε (∇Z Ω

~ 2 k∆Zk ~ 2 kZ ~ t k2 ≤ 4|α1 |εk∇Zk ∞ ε ~ 22 + d. ≤ k∇∆Zk K

Landau–Lifshitz Equations

276

(8) Z α 1 ε ~ ~ ~ Z × ∆H · Zt dx Ω Z ~ ~ ~ ~ ~ = α1 ε ∇H · (∇Z × Zt + Z × ∇Zt )dx Ω

~ 2 kZ ~ t k2 k∇Zk ~ ∞ + k∇Hk ~ 2 k∇Z ~ t k2 ) ≤ |α1 |ε(k∇Hk h

~ 2 k∆Zk ~ 2 k∇Zk ~ ∞ + k∇Hk ~ 2 (k∇Zk ~ ∞ + k∇Z ~ t k2 ) ≤ C1 ε k∇Hk 1

3

~ 24 + k∇Z ~ t k2 ) ~ 2 (k∇∆Zk ~ 24 + k∇∆Zk ≤ C2 εk∇Hk ε ~ 2 + ε k∇Z ~ t k2 + Ck∇Hk ~ 2 + d. ≤ k∇∆Zk 2 2 2 K K (9) Z −α1 ε (Z ~ · H) ~ Z ~ × ∆Z ~ ·Z ~ t dx Ω

~ ∞ k∆Zk ~ 2 kZ ~ t k2 ≤ |α1 |εkHk 1

~ 22 (k∆Zk ~ 2 + k∆Zk ~ 2) ≤ C1 εk∇Hk 2 1

1

~ 2 + k∇∆Zk ~ 22 ) ~ 22 (k∇∆Zk ≤ C2 εk∇Hk ε ~ 2 + Ck∇Hk ~ 2 + d. ≤ k∇∆Zk 2 2 K (10) Z −2α1 ε (∇Z ~ · H) ~ Z ~ × ∇Z ~ ·Z ~ t dx Ω

~ 2 kHk ~ 2 kZ ~ t k2 ≤ 2|α1 |εk∇Zk ∞ ~ 2 kZ ~ t k2 ≤ C1 εk∆Zk

~ 2 + k∆Zk ~ 2 ) ≤ Ck∆Zk ~ 2 + d. ≤ C2 ε(k∆Zk 2 2

(11) Z −2α1 ε (Z ~ · ∇H) ~ Z ~ × ∇Z ~ ·Z ~ t dx Ω

~ ∞ k∇Hk ~ 2 kZ ~ t k2 ≤ 2|α1 |εk∇Zk 1

~ 2 k∆Zk ~ 22 (k∆Zk ~ 2 + kHk ~ 2) ≤ C1 εk∇Hk 1

3

~ 2 k∆Zk ~ 22 + C3 εk∇Hk ~ 2 k∇∆Zk ~ 24 ≤ C2 εk∇Hk ε ~ 2 + C(k∇Hk ~ 2 + k∆Zk ~ 2 ) + d. ≤ k∇∆Zk 2 2 2 K

i

Landau–Lifshitz–Maxwell Equations

277

Combining the results of (1)–(11) and using (5.3.31) we have α1

Z

Ω

~ × ∆Z ~t · Z ~ t dx Z

≤

−α12

Z

α12 ~ ~ Zt · ∆Hdx − 2 Ω

3 2 + α 4 1 

−

Z

Ω

Z

~ 4 dx − α2 |∇Z| 1

~ 6 3α12 εk∇Zk 6

−

Z

~ α12 εk∇3 Z



Ω

~ 2 dx |∆Z|

Ω

~ · H)|∇ ~ ~ 2 dx (Z Z|

×

t



t

~ 2 Zk 2

9ε ~ 22 + ε k∇Z ~ t k22 k∇∆Zk K K 2 ~ ~ 2 + k∇Ek ~ 2 ) + d. + C(k∆Zk + k∇Hk

+

2

2

(5.3.35)

2

Substituting (5.3.35) into (5.3.30) we get 1 2

Z

Ω

~ t |2 + α12 |∆Z| ~ 2 )dx (|Z

~ t k22 + α12 + εk∇Z ≤

Z

Ω



+ t

Z

Ω

~ (2α12 (Z

~ t · Hdx ~ Z + α12 ε

Z

Ω

~ 4 )dx ~ ~ 2 − 3 α12 |∇Z| · H)|∇ Z| 2



t

~ 6 + |∇3 Z ~ × Z| ~ 2 )dx (3|∇Z|

9ε ~ 22 + 3ε k∇Z ~ t k22 + C0 (k∆Zk ~ 22 + k∇Hk ~ 22 + k∇Ek ~ 22 ) + d0 . k∇∆Zk K K

(5.3.36)

Moreover, from (5.3.28) ~ × ∇Z ~ t = εZ ~ × ∇∆Z ~ + ε|∇Z| ~ 2Z ~ × ∇Z ~ + α1 Z ~ × (∇Z ~ × ∆Z) ~ − α1 ∇∆Z ~ Z ~ · ∇3 Z) ~ Z ~ + α1 Z ~ × ∇(Z ~ × H) ~ − εZ ~ × ∇(Z ~ × (Z ~ × H)). ~ + α 1 (Z ~ · ∇3 Z ~ = −3∇Z ~ · ∆Z ~ and the equality above, we have Then, by Z ~ ≤ |∇Z ~ t | + ε|∇∆Z| ~ + ε|∇Z| ~ 3 + 4|α1 ||∇Z||∆ ~ Z| ~ |α1 ||∇∆Z| ~ H| ~ + |∇H|). ~ + (|α1 | + ε)(|∇Z|| Take ε <

|α1 | . 2

The above inequality yields ~ ≤ C(|∇Z ~ t | + |∇Z||∆ ~ Z| ~ + |∇Z|| ~ H| ~ + |∇H|). ~ |α1 ||∇∆Z|

Therefore, ~ 2 ≤ Ck∇Z ~ t k2 + Ck∇Zk ~ ∞ (k∆Zk ~ 2 + kHk ~ 2 ) + Ck∇Hk ~ 2 |α1 |k∇∆Zk 3

1

~ t k2 + C1 (k∇∆Zk ~ 24 + k∇∆Zk ~ 24 ) + Ck∇Hk ~ 2 ≤ Ck∇Z ≤

|α1 | ~ 2 + C(k∇Z ~ t k2 + k∇Hk ~ 2 ) + d, k∇∆Zk 2

and it follows that ~ 2 ≤ C0 (k∇Z ~ t k2 + k∇Hk ~ 2 ) + d0 . k∇∆Zk

(5.3.37)

Landau–Lifshitz Equations

278

Substituting (5.3.37) into (5.3.36) and choosing K ≥ 9C02 + 3, we obtain d dt

Z

Ω

~ t (·, t)|2 + α2 |∆Z(·, ~ t)|2 )dx (|Z 1 d + dt

Z " Ω

~ 2α12 (Z

#

~ ~ 2 − 3 α2 |∇Z| ~ 4 dx + 2α2 · H)|∇ Z| 1 2 1

Z

Ω

~ t · Hdx ~ Z

~ 22 + k∇Hk ~ 22 + k∇Ek ~ 22 ) + d. ≤ C(k∆Zk

(5.3.38)

~ with (5.3.9) and (III) In the case of β > 0, we make the scalar products of ∆E ~ ∆H with (5.3.10), respectively, and then get 1 d 2β dt

Z

Ω

~ 2 + |∇H| ~ 2 )dx + (|∇E|

σ ~ 2= k∇Ek 2 β

Z

Ω

~ t · Hdx. ~ Z

(5.3.39)

In the case of β = 0, from (5.3.9) and (5.3.10) it follows that ~ t)k2 1 + kE(·, ~ t)k2 1 ≤ kH ~ 0 (x)k2 1 + kE ~ 0 (x)k2 1 . kH(·, H (Ω) H (Ω) H (Ω) H (Ω)

(5.3.40)

(1) Substituting (5.3.39) into (5.3.38) we get d dt

Z " Ω

~t| + |Z 2

~ 2 α12 |∆Z|

#

α2 ~ 2 + |∇H| ~ 2 ) dx + 1 (|∇E| β

~ ~ 2 − 3 α2 |∇Z| ~ 4 dx · H)|∇ Z| 2 1 Ω ~ 2 + k∇Hk ~ 2 + k∇Ek ~ 2 ) + d. ≤ C(k∆Zk d + dt

Z 



~ 2α12 (Z

2

2

2

(5.3.41)

Denote α12 2 2 ~ ~ ~ t)k22 + k∇H(·, ~ t)k22 ) A(t) = kZt (·, t)k2 + k∆Z(·, t)k2 + (k∇E(·, β ~ × (∆Z ~ + H)k ~ 2 + ε 2 kZ ~ × (Z ~ × (∆Z ~ + H))k ~ 2 = α 2 kZ 1

2

~ 2+ + k∆Zk 2 B(t) =

Z  Ω

~ 2α12 (Z

α12 β

2

~ 2 + k∇Hk ~ 2 ), (k∇Ek 2 2

~ ~ 2 − 3 α2 |∇Z| ~ 4 dx. · H)|∇ Z| 1 2 

Integrating (5.3.41) over (0, t) (0 < t ≤ T ), we get A(t) + B(t) ≤ A(0) + B(0) + C

Z

t 0

~ t)k2 + k∇H(·, ~ t)k2 + k∇E(·, ~ t)k2 )dt (k∆Z(·, 2 2 2

and A(t) ≤ A(0) + B(0) + |B(t)| +

Z

t 0

~ 2 + k∇Hk ~ 2 + k∇Ek ~ 2 )dt. (k∆Zk 2 2 2

(5.3.42)

Landau–Lifshitz–Maxwell Equations

279

Note that ~ ~ 2 dx + 3 α2 |B(t)| ≤ |H||∇ Z| 2 1 Ω ~ ∞ + C2 k∇Zk ~ 4 ≤ C1 kHk 2α12

Z

Z

Ω

~ 4 dx |∇Z|

4

2

≤

α12

|α1 | ~ 2+ ~ 2 + d1 . k∆Zk k∇Hk 2 2 2 2β

(5.3.43)

Substituting (5.3.43) into (5.3.42) we obtain "

α12 ~ t)k2 + 1 (k∇H(·, ~ t)k2 + k∇E(·, ~ t)k2 k∆Z(·, 2 2 2 2 β ≤ A0 + B0 + d1 + dT +

Z

t 0

#

~ 22 + k∇Hk ~ 22 + k∇Ek ~ 22 )dt. (k∆Zk

Applying the Gronwall inequality for the above inequality gives sup 0≤t≤T





~ t)k2 + k∇H(·, ~ t)k2 + k∇E(·, ~ t)k2 ≤ C, k∆Z(·, 2 2 2

(5.3.44)

where the constant C is independent of D and ε. (2) When β = 0, it follows that d dt

Z

~ t |2 + α2 |∆Z| ~ 2 )dx + d B(t) ≤ Ck∆Z(·, ~ t)k2 + d. (|Z 1 2 dt Ω

Similar argument as (1) gives ~ t)k2 ≤ C, sup k∆Z(·, 2

t∈[0,T ]

where the constant C is independent of D and ε. By (5.3.39) and the above inequality we conclude that (5.3.26) and (5.3.27) are true still for β = 0. The proof of the lemma is complete. By using the induction and the analogous derivation as that of Lemma 5.3.3, after complicated calculation we can get the following result. ~ t), H(x, ~ ~ Lemma 5.3.4 Assume that (Z(x, t), E(x, t)) is the smooth solution of the ~ 0 (x), H ~ 0 (x), periodic value problem (5.3.8)–(5.3.13), and |α1 | > 0, σ ≥ 0, β ≥ 0. If (Z k k−1 k−1 ~ 0 (x)) ∈ (H (Ω), H (Ω), H (Ω)) (k ≥ 2), then there are estimates E 

X

sup 

t∈[0,T ]

[ k2 ]

s=0

~ t)k2 + kDts Dxk−2s Z(·,

k−1 X s=0

  ~ t)k2  ~ t)k2 + kDts Dxk−1−s E(·, kDts Dxk−1−s H(·,  ≤ C,

(5.3.45)

where the constant C is independent of D and ε, and k and s are nonnegative integers.

Landau–Lifshitz Equations

280

5.3.2

Global Existence of Smooth Solution

From the a priori estimates independent of ε for the smooth solution of the periodic value problem (5.3.8)–(5.3.13), we prove the global existence of the smooth solution. Theorem 5.3.1 [42] Assume the constants ε > 0, β ≥ 0, σ ≥ 0, the periodic value h i n k k−1 k−1 ~ ~ ~ functions (Z0 (x), H0 (x), E0 (x)) ∈ (H (Ω), H (Ω), H (Ω)) k ≥ 3 + 3 , Ω ⊂ Rn ~ 0 (x)| = 1, ∇·(H ~ 0 +β Z ~ 0 ) = 0, ∇· E ~ 0 = 0. Moreover, (n ≤ 2) is bounded domain, and |Z

when n = 2, there is

~ 0 k2 + k E ~ 0 k2 + k H ~ 0 k2 ≤ δ, k∇Z where δ is a suitable small constant. Then the periodic initial value problem (5.3.8)–(5.3.13) of the Landau–Lifshitz–Maxwell system has a unique global smooth ~ t)| = 1, x ∈ Ω, t ∈ R+ , and solution such that |Z(z, s 2] ~ t) ∈ ∩[s=0 Z(x, W∞ (0, T ; H k−2s(Ω)), s k−1−s ~ H(x, t) ∈ ∩k−1 (Ω)), s=0 W∞ (0, T ; H ~ E(x, t) ∈ ∩k−1 W s (0, T ; H k−1−s(Ω)). k

s=0

∞

By the standard principle of compactness and letting ε → 0, we obtain Theorem 5.3.2 Under the conditions of Theorem 5.3.1, the one-dimensional periodic initial value problem (5.3.1)–(5.3.6) of the strong degenerate Landau–Lifshitz– Maxwell system admits a unique global smooth solution. Remark. The following assumptions are imposed : ~ 0 + β∇ · Z ~ 0 = 0, ∇·H

~ 0 = 0, ∇·E

to assure that ~ + β∇ · Z ~ = 0, ∇·H

~ = 0. ∇·E

Since all the a priori estimates are independent of the periodicity D, we let D → ∞ and then obtain the following result. Theorem 5.3.3 Under the conditions of Theorem 5.3.1, the one-dimensional initial value problem (5.3.1)–(5.3.4) and (5.3.7) for strong degenerate Landau–Lifshitz– ~ t), H(x, ~ ~ Maxwell system has a unique global smooth solution (Z(x, t), E(x, t)) satisfying ~ t)| = 1, |Z(x, ∀ (x, t) ∈ Rn × R+ , k 2] s ~ t) ∈ ∩[s=0 ∇Z(x, W∞ (0, T ; H k−1−2s(R)), s k−1−s ~ H(x, t) ∈ ∩k−1 (R)), s=0 W∞ (0, T ; H ~ E(x, t) ∈ ∩k−1 W s (0, T ; H k−1−s(R)).

s=0

∞

Landau–Lifshitz–Maxwell Equations

5.4 5.4.1

281

Global Weak Solution to L–L–M System on Riemannian Manifold L–L–M System on Riemannian Manifold

Assume that (M, g) is an n-dimensional compact Riemannian manifold without boundary with metric g. In local coordinates x = (x1 , . . . , xn ), the Laplace–Beltrami operator and |∇u(x)| read as ∆M

1 ∂ √ ∂ g αβ g α =√ β g ∂x ∂x |∇u(x)|2 =

!

= g αβ

XX αβ

j

g αβ

∂2 ∂ − Γkαβ k , α β ∂x ∂x ∂x

∂uj ∂uj , ∂xα ∂xβ

where (g αβ ) is the inverse of (gαβ ). The L–L–M system on Riemannian manifold M reads as ~ t = α1 Z ~ × (∆M Z ~ + H) ~ − α2 Z ~ × (Z ~ × (∆M Z ~ + H)), ~ Z ∇j Flj = 0, ∇j Flj = σF0l ,

(5.4.1)

l = 0,

l 6= 0,

(5.4.2)

∇l (Fjk + βZjk ) + ∇j (Fkl + βZkl ) + ∇k (Flj + βZlj ) = 0,

ljk 6= 0,

∇l (Fjk + βZjk ) + ∇j (Fkl + βZkl ) + ∇k (Flj + βZlj ) = −β

∂Zlj , ∂t

one of l, j, k = 0,

(5.4.3)

where Flj = g jk Fkl ,

~ = (Z1 , Z2 , Z3 ) = (Z23 , Z31 , Z12 ). Z

Fkl is second order asymmetric covariant tensor energy: 0 F01 F02 F03 F 0 F12 F13  (Fkl ) =  10  F20 F21 0 F23 F30 F31 F32 0 

    

0 E1 E2 E3  −E 0 H3 −H2    1 = ,  −E2 −H3 0 H1  −E3 H2 −H1 0 



(5.4.4)

~ = (E1 , E2 , E3 ), H ~ = (H1 , H2 , H3 ), (g jk ) is the metric of the manifold where E M (x0 , x1 , . . . , xn ). In fact we have jk

(g ) =

!

1 0 , 0 −g jk

(5.4.5)

Landau–Lifshitz Equations

282

where (g jk ) is the metric of the manifold M (x1 , . . . , xn ), Dj is the covariant derivative. Since ∇j g¯ij = 0,

(5.4.6)

(5.4.2) is equivalent to ik

g ∇j Fkl =

(

0, l = 0, −σF0l , l = 6 0.

(5.4.7)

In particular, taking g 00 = 1, g 11 = g 22 = g 33 = −1 and x0 = −t, we may get the L–L–M system on Euclid space Rn . Now we calculate the electromagnetic field energy Tij of (5.4.1)–(5.4.3). It follows from the definition that 1 Tij = g il Fkl F kl − g kl Fil Flj , 4

(5.4.8)

F kl = g ki g ls Fis .

(5.4.9)

where

From ∇l g 0k = 0,

det(g jk ) 6= 0,

we have ∇j g ij = 0, ∇j T ij = = = g ik (∇j Fkl )F jl = = 1 ij kl g F ∇j Fkl = 2

1 ∇j (g ij Fkl F kl ) − ∇j (F il Flj ) 4 1 ij 1 g ∇Fkl · F kl + g ij Fkl ∇j F kl − (∇j F il )Flj − Fjl ∇j Flj 4 4 1 ij kl g F ∇j Fkl − g ij (∇j Fkl )F jl − σF il F0l ; 2 1 ik g ∇j Fkl (F jl − F lj ) 2 1 − g ik F lj (∇j Fkl − ∇l Fkj ), 2 1 ik lj g F ∇k Flj , 2

therefore it follows from (5.5.8) that 1 ik lj g F ∇j Fkl − g ik ∇j Fkl F jl + σF il F0l 2 1 = g ik Flj (∇k Flj + ∇j Fkl + ∇l Fjk ) + σF il F0l , 2

∇j T ij =

Landau–Lifshitz–Maxwell Equations

283

β ∂Zlj ∇j T 0j = − g 00 F lj − σF 0l F0l , 2 ∂t F lj

(5.4.10)

∂Zlj ∂Z =H . ∂t ∂t

Let n be the unit vector on M × {t}. When i > 0, ni = 0. Integrating (5.4.10) over M × [0, t], we have Z

M ×{t}

T 00 −

Z

M ×{0}

T 00 = −σ

Z

M ×[0,t]

F 0l F0l +

~ β Z t Z 00 ~ ∂ Z dMdt. g H 2 0 M ∂t

It follows from (5.4.1) that ~ + H) ~ ·Z ~ t = −α2 (∆M Z ~ + H)[ ~ Z ~ × (Z ~ × (∆M Z ~ + H))] ~ (∆Z ~ × (∆Z ~ + H)| ~ 2, = α 2 |Z and hence Z

M

~ ·Z ~ t dM = − H

Z

M

+ α2 =−

Z

M

~ ·Z ~ t dM ∆M Z Z

M

~ × (∆Z ~ + H)| ~ 2 dM − |Z

~ 1 ∂ αβ √ ∂ Z g g √ g ∂xβ ∂xα

!

·

Z

M

~√ ∂Z gdM ∂t

~ ~ ∂2Z √ ∂Z g αβ g α · β dM ∂x ∂x ∂t M ! Z 1 2 ~ = |∇Z| dM , 2 M t =

Z

1 00 g Fkl F kl − F0l F 0l 4 X 1 1 = g 00 Fij F ij + g 00 F0l F 0l 4 2 i>0,j>0

T 00 =

=

1 00 ~ 2 ~ 2 ), g (|H| + |E| 2

where ~ 2= |H|

~ ·Z ~ t dM ∆M Z

1 X Fij F ij , 2 i,j>0

~ 2 = −g 00 g lj F0l F0j ≥ 0. |E|

(5.4.11)

Landau–Lifshitz Equations

284

It follows from (5.4.11) that 1 ~ 2 + |H| ~ 2 )dM + σ g 00 (|E| F 0l F0l dMdt 2 M ×{t} M ×[0,t] Z Z 1 ~ 2 dM + ~ × (∆Z ~ + H)| ~ 2 dM |∇Z| α 2 |Z = 2 M ×{t} M ×[0,t] Z 1Z ~ 2 dM. ~ 2 + |H| ~ 2 )dM + 1 + |∇Z| g 00 (|E| 2 M ×{0} 2 M ×{0} Z

Z

(5.4.12)

In order to prove the existence of weak solution to the initial value problem of L–L–M equations, we derive the L–L–M equations on Riemannian manifold by much more simple method and get its energy conservation law. Now we give the definitions of gradient operator and spin operator on Riemannian manifold. For a scalar field, its covariant derivative is defined by ∇j ϕ =

∂ϕ ∂xj

which is the first order covariant energy. gij ∇j ϕ or ∇j ϕ is the second order inverse variant energy. The gradient is defined by gradϕ = g ij ∇j ϕ~ei ,

(5.4.13)

where ~ei is the covariant local coordinate. ~ = ai~e i = ai~ei . Its covariant derivatives ∇i aj Suppose that there is a vector A form a covariant tensor ∇i a j − ∇ j a i ,

(5.4.14)

which is the second order antisymmetry covariant energy. In three-dimensional space, there are only three independent elements of the second order antisymmetric tensor. We may construct a vector whose inverse variant elements are 1 ξ i = ijk (Dj ak − Dk aj ) = ijk Dj ak , 2

(5.4.15)

ijk = ei · (ej × ek ).

(5.4.16)

where

Landau–Lifshitz–Maxwell Equations

285

~ Vector ξ~ is the rotation of vector A,   ξ1 =    

ξ2 =

     ξ3 =

√1 g √1 g √1 g

  

∂a3 ∂x2 ∂a1 ∂x3 ∂a2 ∂x1

−

∂a2 ∂x3

−

∂a3 ∂x1

−

∂a1 ∂x2



,



,



(5.4.17)

,

~ = ξ i~e i . rot A

(5.4.18)

Therefore, we may write L–L–M equations on manifold as           

∂E 1 ∂t ∂E 2 ∂t H3 ∂t

−

√1 g

−

√1 g

−

√1 g

 



∂H 3 ∂x2 ∂H 1 ∂x3

∂H 2 ∂x1

∂H 2 ∂x3

−

∂H 3 ∂x1

− −

 

 1

∂H ∂x2

+ σE 1 = 0, + σE 2 = 0,

+ σE 3 = 0,

~ t = α1 Z ~ × (∆M Z ~ + H) ~ − α2 Z ~ × (Z ~ × (∆Z ~ + H)), ~ Z           

∂H 1 ∂t ∂H 2 ∂t ∂H 3 ∂t

1

+ β ∂Z + ∂t 2

+ β ∂Z + ∂t +β

∂Z 3 ∂t

+

√1 g √1 g √1 g



∂E 3 ∂x2



∂E 2



∂E 1 ∂x3 ∂x1

(5.4.19)

 2

= 0,



= 0.

−

∂E ∂x3

−

∂E ∂x1

−

∂x2

 3

∂E 1

= 0,

(5.4.20)

(5.4.21)

As for the divergence equation we have ~ + β Z) ~ = 0, ∇ · (H

~ = 0, ∇·E

(5.4.22)

where ~ = (Z 1 , Z 2 , Z 3 ), Z

~ = (H 1 , H 2 , H 3 ), H

~ = (E 1 , E 2 , E 3 ). E

We first give the definition of the divergence. The divergence of a vector means the invariant quantity after contraction for the index of covariant derivative of its inverse variant components: ~ = ∇ i ai , div A ~ = ai~ei . From where A √ ∂ ln g ∂ai i k k k ∇i a = i + Γij a , Γij = , ∂x ∂xk one has √ ∂ai ∂ ln g i i ∇i a = + a ∂xi ∂xi √ ! 1 √ ∂ai ∂ g i = √ g i+ a g ∂x ∂xi √ 1 ∂ gai = √ . g ∂xi

Landau–Lifshitz Equations

286

Therefore, the divergence of a vector is √ 1 ∂( gai ) ~ div A = √ . g ∂xi

(5.4.23)

On Riemannian manifold M , Eq. (5.4.22) read as √ √ β ∂( gZ i ) 1 ∂( gH i ) +√ = 0, √ g ∂xi g ∂xi √ 1 ∂( gE i ) = 0. √ g ∂xi

(5.4.24) (5.4.25)

Now we prove the existence of Eqs. (5.4.19)–(5.4.21), (5.4.24) and (5.4.25) with the initial conditions E i |t=0 = E0i ,

H i |t=0 = H0i ,

Z i |t=0 = Z0i ,

(5.4.26)

where i = 1, 2, 3 and ~ 0 = div(H ~ 0 + βZ ~ 0 ) = 0. div E

(5.4.27)

We will use Galerkin method. Let {λj } be the eigenvalues of the operator −∆M , that is −∆M ϕj = λj ϕj ,

j = 1, 2, . . . ,

(5.4.28)

and the eigenfunctions {ϕj } form the normal orthogonal basis of H 1 (M ; R3 ). Suppose that the approximate solutions of problem (5.4.19)–(5.4.21), (5.4.28) and the initial conditions (5.4.26) and (5.4.27) are  m X    E = αjm (t)ϕj (x),  m    j=1     m  X

Hm =

βjm (t)ϕj (x),

  j=1     m  X    Z = γjm (t)ϕj (x),  m  j=1

where 1 2 3 Em = (Em , Em , Em ), 1 2 3 Hm = (Hm , Hm , Hm ), 1 2 3 Zm = (Zm , Zm , Zm ),

(5.4.29)

Landau–Lifshitz–Maxwell Equations

287

and αm (t), βm (t), γm (t) are three-dimensional vectors in t ∈ R+ which solve the following ordinary differential systems: (Zmt − α1 Zm × (∆M Zm + Hm ) + α2 Zm × (Zm × (∆M Zm + Hm )), ϕj ) = 0, 1 3 2 ∂Em 1 ∂Hm ∂Hm −√ − ∂t g ∂x2 ∂x3

!

2 1 3 ∂Em 1 ∂Hm ∂Hm −√ − ∂t g ∂x3 ∂x1

!

3 2 1 Hm 1 ∂Hm ∂Hm −√ − ∂t g ∂x1 ∂x2

!

(5.4.30)

+

1 σEm , ϕj

!

= 0,

(5.4.31)

+

2 σEm , ϕj

!

= 0,

(5.4.32)

+

3 σEm , ϕj

!

= 0,

(5.4.33)

!

!

= 0,

(5.4.34)

!

!

= 0,

(5.4.35)

!

!

= 0,

(5.4.36)

1 2 3 ∂Hm ∂Z 1 1 ∂Em ∂Em +β m + √ − , ϕj ∂t ∂t g ∂x2 ∂x3 2 3 1 ∂Hm ∂Em ∂Z 2 1 ∂Em − , ϕj +β m + √ ∂t ∂t g ∂x3 ∂x1 3 1 2 ∂Hm ∂Em ∂Z 3 1 ∂Em − , ϕj +β m + √ ∂t ∂t g ∂x1 ∂x2

with initial conditions Z Z   Zm (x, 0)ϕj (x)dM = Z0 (x)ϕj (x)dM,    M   ZM  Z 

H (x, 0)ϕ (x)dM =

H (x)ϕ (x)dM,

m j 0 j  M M    Z Z      Em (x, 0)ϕj (x)dM = E0 (x)ϕj (x)dM, M

(5.4.37)

M

j = 1, 2, 3, . . . . From the following a priori estimates, we know that this problem admits a global solution on [0, T ]. In the following we give the a priori estimate for the approximate solutions. 1 2 3 1 2 3 Multiplying (5.4.30)–(5.4.36) by γm , αm , αm , αm , βm , βm and βm , respectively, and summing from j = 1 to m, we have (Zmt − α1 Zm × (∆M Zm + Hm )

+α2 Zm × (Zm × (∆M Zm + Hm )), Zm ) = 0,

3 2 1 ∂Em 1 ∂Hm ∂Hm −√ − ∂t g ∂x2 ∂x3

!

+

1 1 σEm , Em

!

= 0,

(5.4.38)

(5.4.39)

Landau–Lifshitz Equations

288

2 1 3 ∂Em 1 ∂Hm ∂Hm −√ − ∂t g ∂x3 ∂x1

!

3 2 1 Hm 1 ∂Hm ∂Hm −√ − ∂t g ∂x1 ∂x2

!

2 2 + σEm , Em

!

= 0,

(5.4.40)

3 3 σEm , Em

!

= 0,

(5.4.41)

!

!

= 0,

(5.4.42)

!

!

= 0,

(5.4.43)

!

!

= 0.

(5.4.44)

+

1 3 2 ∂Hm ∂Z 1 1 ∂Em ∂Em 1 , Hm +β m + √ − ∂t ∂t g ∂x2 ∂x3 2 1 3 ∂Hm ∂Z 2 1 ∂Em ∂Em 2 , Hm +β m + √ − ∂t ∂t g ∂x3 ∂x1 3 3 2 1 ∂Hm ∂Zm 1 ∂Em ∂Em 3 , Hm +β +√ − 1 2 ∂t ∂t g ∂x ∂x

Adding (5.4.39)–(5.4.44), we have 1∂ (kEm k22 + kHm k22 ) 2 ∂t + It follows from (5.4.38) that

σkEm k22 d kZm k22 dt

+β

Z

M

∂Zm · Hm dM = 0. ∂t

(5.4.45)

= 0, then

kZm (·, t)k2 = kZm (·, 0)k2 = kZ0 k2 .

(5.4.46)

On the other hand, we have from (5.4.30) that (Zmt − α1 Zm × (∆M Zm + Hm )

+ α2 Zm × (Zm × (∆M Zm + Hm )), ∆m Zm + Hm ) = 0,

that is Z

M

Zmt (∆M Zm + Hm )dM − α2

Z

M

|Zm × (∆M Zm + Hm )|2 dM = 0.

Since !

∂ √ ∂Zm − Zmt ∆Zm dM = − Zmt β g αβ g α dM ∂x ∂x M M Z

Z

√ ∂Zmt ∂Zm dM g αβ g ∂xβ ∂xα M Z ∂Zmt ∂Zm = g αβ dM ∂xβ ∂xα M ! Z 1 = |∇Zm |2 dM , 2 M t =

Z

(5.4.47)

Landau–Lifshitz–Maxwell Equations

289

we have from (5.4.47) that 1d k∇Zm k22 + α2 kZm × (∆M Zm + Hm )k22 = 2 dt

Z

M

Zmt · Hm dM.

(5.4.48)

From (5.4.45) and (5.4.48) and the method in Sec. 5.1, we have Lemma 5.4.1 Let α2 ≥ 0 and (Z0 (x), E0 (x), H0 (x)) ∈ (H 1 (M ), L2 (M ), L2 (M )). Then for the approximate solutions of problem (5.4.30)–(5.4.37) there holds (i)

sup (kZm (·, t)k2H 1 + kEm (·, t)k22 + kHm (·, t)k22 ) ≤ C1 ,

(5.4.49)

0≤t≤T

where C1 is independent of α2 and m. When α2 > 0, there holds (ii) kZm × (∆M Zm + Hm )kL2 (0,T ;L2 (M )) ≤ C2 , where C2 is independent of m.

5.4.2

Existence of Generalized Solution

It follows from Lemma 5.4.1 that we may choose a subsequence such that Zm (x, t) → Z(x, t) in L∞ (0, T ; H 1(M )) weakly-∗, as m → ∞; Em (x, t) → E(x, t) in L∞ (0, T ; L2 (M )) weakly-∗, as m → ∞; Hm (x, t) → H(x, t) in L∞ (0, T ; L2 (M )) weakly-∗, as m → ∞. If α2 > 0, Zm × (∆Zm + Hm ) weakly converge in L2 (0, T ; L2 (M )). Definition 5.4.2 Three-dimensional vectors Z(x, t) ∈ L∞ (0, T ; H 1 (M )), E(x, t) ∈ L∞ (0, T ; L2 (M )), H(x, t) ∈ L∞ (0, T ; L2 (M )) are called global weak solutions to problem (5.4.19)–(5.4.21), and (5.4.24) and (5.4.25) if for any test function v(x, t) ∈ C 1 (QT ) with v(x, T ) = 0 there holds Z

T 0

Z

M

Zvt dM dt + α1 T

+ α1

Z

+ α2

Z

+ α2

Z

0 T 0 T 0

Z

Z

Z Z

T 0

M

M

M

Z

Z

T 0

Z

M

∇Z × Z · ∇vdMdt

Z × H · vdMdt + α2

Z

T 0

Z

M

(Z × H) × Z · ∇vdMdt

∇Z × (Z × ∇Z) · vdMdt (∇Z × Z) × Z · ∇vdMdt +

M

Evt dM dt +

−σ

Z

T 0

Z

M

Z

T 0

Z

M

Z

M

Z0 v0 dM = 0,

(5.4.50)

∇ × v · HdMdt

E · vdMdt +

Z

M

E0 v0 dM = 0,

(5.4.51)

Landau–Lifshitz Equations

290

Z

T 0

Z

M

Z

(H + βZ)vt dMdt −

+

Z

T 0

Z

M

T 0

Z

M

∇ × v · Et dMdt

(H0 + βZ0 )dM = 0,

(5.4.52)

and

Z

T 0

Z Z

M

T

Z

∇ξ · EdMdt = 0,

(5.4.53)

∇η · (H + βZ)dMdt = 0,

(5.4.54)

0

M

for any test function ξ, η ∈ L2 (0, T ; C0∞ (M )). Theorem 5.4.1 Let α2 ≥ 0 and the initial data (Z0 , E0 , H0 ) ∈ (H 1 (M ), L2 (M ), L2 (M )), such that ∇ · E0 ,

∇ · (H0 + βZ0 ) = 0

(5.4.55)

in the sense of distribution. Then problem (5.4.19)–(5.4.21), and (5.4.24) and (5.4.25) admits at least one global weak solution (Z(x, t), H(x, t), E(x, t)). Proof. Similar to Sec. 5.1 we may verify that (Z, H, E) satisfy integral equalities (5.4.50)–(5.4.52). So we only need to prove that conditions (5.4.53) and (5.4.54) hold. (i) For any ξ ∈ L2 (0, T ; C0∞(M )), take test function v(x, t) as follows: Z

v(x, t) =

t

0

∇ξ(x, τ )dτ −

Z

T

0

∇ξ(x, τ )dτ.

It is easy to verify that v(x, t) ∈ C 1 (QT ), v(x, T ) = 0, and vt (x, t) = ∇ξ(x, t),

v(x, 0) = −

Z

T

0

∇ξ(x, τ )dτ.

Substituting them into (5.4.52) we have Z

T 0

Z

M

Z

(H + βZ)∇ξdMdt =

T 0

Z

M

(H0 + βZ0 )∇ξdMdt = 0.

(5.4.56)

(ii) For any ξ ∈ L2 (0, T ; C0∞ (M )), denote ξ1 = ξe−σt . Take test function v(x, t) as follows ! Z t Z T ∇ξ1 (x, τ )dτ − ∇ξ1 (x, τ )dτ eσt . v(x, t) = 0

0

It is easy to verify that v(x, t) ∈ C 1 (QT ), v(x, T ) = 0, and vt − σv = ∇ξ1 (x, t)e

σt

= ∇ξ(x, t),

v(x, 0) = −

Z

T 0

∇ξ1 (x, τ )dτ.

Substituting them into (5.4.51) we have Z

T 0

Z

M

E · ∇ξdMdt =

Z

T 0

Z

M

E0 · ∇ξ1 dMdt = 0.

(5.4.57)

conditions (5.4.53) and (5.4.54) follow from (5.4.56) and (5.4.57). The theorem is proved.

Landau–Lifshitz–Maxwell Equations

5.5

291

Partial Regularity for Stationary Solutions to L–L–M Equations

In this section, we discuss the equations as follows: 1 1 ut − (u × ut ) = ∆u + u|∇u|2 + H − hH, uiu, 2 2

in B 3 × R+ ,

(5.5.1)

where H(u) is the nonlocal term that satisfies the following quasi-steady state Maxwell equations curl H = 0, div(H + u) = 0,

in D 0 (R3 ),

in D 0 (R3 ).

(5.5.2) (5.5.3)

We impose to the equations (5.5.1)–(5.5.3) the initial condition u(x, 0) = u0 (x)

(5.5.4)

and boundary condition

∂u =0 ∂ν ∂B 3

(5.5.5)

in (5.5.4), |u0 (x)| ≡ 1, in (5.5.5), ν is the unit outer normal to the boundary of B 3 , u is the zero extension of u from B 3 to R3 , u = (u1 , u2 , u3 ) is the spin vector, “×” denotes the vector cross product in R3 . We should notice that this extension guarantees u ∈ L∞ (R3 × R+ ) ∩ L∞ (0, ∞; W −1,∞(R3 )).

5.5.1

Quasi-Static Maxwell Equations

Lemma 5.5.1 Let u ∈ H 1 (B 3 , S 2 ). Let H = ∇Φ ∈ L2 (R3 , R3 ) be the solution of curl H = 0,

div(H + u ˜) = 0

(5.5.6)

in D 0 (R3 ) where u ˜ is equal to u in B 3 (|u| = 1 a.e. in B 3 ) and zero outside B 3 . Then H ∈ ∩1≤p<∞ Lp (R3 )

(5.5.7)

and for all p ∈ (1, ∞) there exists a constant Kp > 0 such that kHkLp (R3 ) ≤ Kp kukLp (B 3 ) .

(5.5.8)

Proof. This is because that |˜ u| = |u| = 1 a.e in B 3 and u ˜ = 0 outside of B 3 , then u ˜ ∈ L∞ (R3 ) and div u ∈ W −1,∞ . Consider Φ ∈ H 1 (R3 ) such that H = ∇Φ and ∆Φ = −div u in R3 . We have that Φ ∈ ∩1≤p<∞ W 1,p and ∀ 1 < p < ∞, k∇ΦkLp (R3 ) = kHkLp (R3 ) ≤ Kp kukLp (B 3 ) . The lemma is proved.

(5.5.9)

Landau–Lifshitz Equations

292

Remark 5.5.1 For n = 3, taking p = 3∗ = 6, we get from (5.5.9) k∇ΦkL6 (R3 ) = kHkL6 (R3 ) ≤ KkukL6 (B 3 ) ≤ CkukH 1 = Ck∇ukL2 .

(5.5.10)

1

Therefore, Φ ∈ W 1,6 (R3 ) with embedding W 1,6 (R3 ) ⊂ C 0, 2 (R3 ) (note that n = 3 < 6) kΦkL∞ ≤ Ck∇ukL2

(5.5.11)

and kΦkLp ≤ Ck∇ukL2 ,

∀ 1 ≤ p < ∞.

(5.5.12)

Remark 5.5.2 We have −

Z

R3

u · ∇Φ =

Z

R3

Φ div u = −

Z

R3

Φ∆Φ =

Z

R3

|∇Φ|2 =

Z

R3

|H|2 .

(5.5.13)

Remark 5.5.3 Next we recall some results concerning the singular integral transforms, Riesz transform. Let G(r) = Cr be the Laplace kernel in R3 , u be as above. Then the solution of the Poisson equation with parameter t ∆Φ(x, t) = −div u, in D 0 (R3 ) can be expressed by Φ(x, t) = −

Z

B3

G(|x − y|)div u(y, t)dy +

Z

∂B 3

G(|x − y|)hu(y, t), n(y)idσ(y).

On the other hand, since div u ∈ L∞ (0, T ; L2 (B 3 )), from the Lp theory of Riesz transforms we have ∂2Φ = −Ri Rj (−div u), ∂xi ∂xj

i, j = 1, 2, 3,

where Ri are the Riesz transforms Γ((3 + 1)/2) Ri (−div u) = π (3+2)/2

Z

R2

(xi − yi )(−div u(y, t)) dy |x − y|(3+1)

and ϕ ∈ W 2,2 (R3 ) with kΦkW 2,2 (R3 ) (t) ≤ Ckdiv ukL2 (R3 ) (t). This result can also be deduced from Coifman–Fefferman [36] for general singular integral operators. We should also note that (5.5.9)–(5.5.12) hold for H = ∇Φ and Φ. Remark 5.5.4 If Lε (x, t) =

Z

y∈B 3 \B(x,ε)

x−y 0 G (|x − y|)div u(y, t)dy |x − y|

(5.5.14)

Landau–Lifshitz–Maxwell Equations

293

and L(x, t) = limε→0 Lε (x, t) = −H + v, then we have from the theory of Riesz R dσ(y) transform that H(x, t) ∈ L∞ (0, ∞; Lp(B 3 )), |v(x, t)| ≤ K ∂B 3 |x−y| 2 ≤ K ln(1 − |x|) ∈ ∞ p 3 L (0, ∞; L (B )) for any p > 1 and kL(x, t)kL2 ≤ Ck∇ukL2 .

(5.5.15)

Remark 5.5.5 (Definition of weak solution) A function u ∈ L∞ (0, T ; H 1 (B 3 )) with ut ∈ L2 (0, T ; L2 (B 3 )) is said a weak solution if (5.5.1)–(5.5.3) hold in the sense of distribution.

5.5.2

Definition of Stationary Solution

Definition 5.5.1 A weak solution u of (5.5.1) is said a stationary solution if for any ξ(x, t) ∈ C01 (B 3 ×R+ , R3 ), θ(x, t) ∈ C01 (B 3 ×R+ , R) with ξ(x, t), θ(x, t), ∇(x,t) ξ, ∇(x,t) θ bounded on B 3 × R+ and ξ, θ ≡ 0 for t = 0 and t ≥ t∗ > 0 such that x + τ ξ|∂B 3 = Id, t + τ θ|∂B 3 = Id, there holds Z

+∞ 0

Z

1 1 ut − u × u t 2 2

B3

!

∂uτ ∂τ

!

+

∂τ+

τ =0

Z

+∞ 0

Z

B3

e(uτ ) + |H(uτ )|2 dxdt ≤ 0, (5.5.16)

where uτ (x, t) = u(x + τ ξ(x, t), t + τ θ(x, t)), e(u) = 21 |∇u(x, t)|2 .

We want to make the definition applicable in the following. To this aim we first compute the right derivative in the definition. For simplicity, we simply compute the derivative at τ = 0 and without loss of generality, we simply assume that u is smooth. It follows from (5.5.2) and (5.5.3) that Hτ = H(uτ ) satisfies in D 0 (R3 ),

curl Hτ = 0, div(Hτ + uτ ) = 0,

in D 0 (R3 ).

(5.5.17) (5.5.18)

Let the potential ατ (x, t) be such that Hτ = ∇ατ , then (5.5.17) and (5.5.18) become ∆ατ = −div uτ ,

in D 0 (R3 ).

(5.5.19)

Therefore, ατ = −

Z

τ

B3

G(|x − y|)div u dy +

Z

∂B 3

G(|x − y|)huτ , ~n(y)idσ(y),

(5.5.20)

where G is the fundamental solution in three dimensions. Denote (xτ , tτ ) = (x + R R R R τ ξ(x, t), t + τ θ(x, t)), Aτ = 21 0+∞ B 3 |∇uτ (x, t)|2 dxdt, Bτ = − 0+∞ B 3 |Hτ |2 dxdt. It is simple to compute div uτ (x, t) = div u(xτ , tτ ) + τ uixj (xτ , tτ )ξxj i + τ uit (xτ , tτ )θxi .

(5.5.21)

We have

dAτ = dτ τ =0

Z

+∞ 0

Z

B

(uixj uixk ξxkj 3

+

uit uixj θxj )

1 − 2

Z

+∞ 0

Z

B3

|∇u|2 (div ξ + θt ).

(5.5.22)

Landau–Lifshitz Equations

294

On the other hand Z

Bτ =

+∞ 0

=−

Z

=−

Z

Z

=

Z

+∞

Z

0 +∞

0 +∞

Z

0

|Hτ |2 dxdt

B3

Z

B3

B3

Hτ · uτ dxdt

B3

∇ατ · uτ dxdt

(div uτ ) · ατ dxdt −

Z

+∞ 0

Z

∂B 3

(uτ · ~n)ατ dσ.

(5.5.23)

It follows from (5.5.20), (5.5.21) and (5.5.23) that dB1 dB2 dB3 dB4 dBτ = + + + , dτ dτ dτ dτ dτ

(5.5.24)

where −B1 =

Z

+∞

dt

0

h

Z

B3

Z

B3

[div u(xτ , tτ ) + τ uixj (xτ , tτ )ξxj i + τ uit (xτ , tτ )θxi ] i

× div u(yτ , tτ ) + τ uiyj (yτ , tτ )ξyji + τ uit (yτ , tτ )θyi G(|x − y|). (5.5.25) Hence we have dB1 − = dτ τ =0

Z

+∞

dt

0

+

Z

+∞

Z

B3

dt 0

Z

Z

B3

Z

B3

[uixi xk ξ k + uixi t θ + uixj ξxj i + uit θxi ][div u(y, t)]G(|x − y|)dxdy B3

[uiyi yk ξ k + uiyi t θ + uiyj ξyji + uit θyi ][div u(x, t)]G(|x − y|).

(5.5.26)

It follows from Carbou’s result [21] that Z

B3

Z

B3

[uixi xk ξ k + uixj ξxj i ][div u(y, t)]G(|x − y|)dxdy +

Z

= −2 +

B3

Z

Z

Z

B3

B3

B3

Z

Z

[uiyi yk ξ k + uiyj ξyji ][div u(x, t)]G(|x − y|)dxdy

B3

[div u div ξ − uixj ξxj i ][div u(y, t)]G(|x − y|)dxdy +

*

x−y , ξ(x, t) − ξ(y, t) div u(x, t)div u(y, t). (5.5.27) G (|x − y|) |x − y| B3 0

Therefore,

dB1 − = −2 dτ τ =0 +

Z

Z

+∞

dt

0 +∞

0

dt

Z

Z

B3

B3

Z Z

B3

B3

[div u div ξ − uixj ξxj i ][div u(y, t)]G(|x − y|)dxdy G0 (|x − y|)

Landau–Lifshitz–Maxwell Equations

×

*

+

Z

+

Z

295

+

x−y , ξ(x, t) − ξ(y, t) div u(x, t)div u(y, t)dxdy |x − y| +∞

dt

Z

Z

Z

Z

B3

0 +∞

dt

B3

0

B3

(uixi t θ + uit θxi )div u(y, t)G(|x − y|)dxdy

B3

(uiyi t θ + uit θyi )div u(x, t)G(|x − y|)dxdy.

(5.5.28)

The last two terms can be estimated as follows: Z

(uixi t θ + uit θxi )div u(y, t)G(|x − y|)dxdy

B3

=− Z

Z

uit (x, t)θ(x, t)div u(y, t)

B3

=−

Z

uit (y, t)θ(y, t)div u(x, t)

B3

So, we have from (5.5.28)–(5.5.30) that

Z

−

×

+∞

Z

dt

B3

0

Z

+∞

dt

0

*

+2

Z

Z

B3

B3

Z

xi − y i 0 G (|x − y|)dxdy. |x − y|

B3

G0 (|x − y|) +

x−y , ξ(x, t) − ξ(y, t) div u(x, t)div u(y, t)dxdy |x − y|

Z

+∞

dt

Z

B3

0

B2 =

uit (x, t)θ(x, t)div u(y, t)

Z

+∞

dt

Z

x∈B 3

0

Z

y∈∂B 3

xi − y i 0 G (|x − y|)dxdy. (5.5.31) |x − y|

[div u(xτ , tτ )

+ τ uixj ξxj i + τ uit θxi ](u(y, t), ~n(y))G(|x − y|)dxdσ(y), B3 =

Z

+∞

dt

Z

x∈∂B 3

0

Z

y∈B 3

τ =0

(5.5.32)

[div u(yτ , tτ )

+ τ uiyj ξyji + τ uit θyi ](u(x, t), ~n(x))G(|x − y|)dydσ(x),

(5.5.33)

+ uixi t θ + uixj ξxj i + uit θxi ](u(y, t), ~n(y))G(|x − y|)dxdσ(y),

(5.5.34)

we have Z +∞ Z Z dB2 = dt [uixi xk ξ k 3 3 dτ τ =0 0 x∈B y∈∂B =

(5.5.30)

[div u div ξ − uixj ξxj i ][div u(y, t)]G(|x − y|)dxdy

Moreover, since B2 and B3 in (5.5.24) are

dB3 dτ

(5.5.29)

(uiyi t θ + uit θyi )div u(x, t)G(|x − y|)dxdy

B3

dB1 =2 dτ τ =0

xi − y i 0 G (|x − y|)dxdy, |x − y|

Z

+∞ 0

dt

Z

x∈∂B 3

Z

y∈B 3

[uiyi yk ξ k

+ uiyi t θ + uiyj ξyji + uit θyi ](u(x, t), ~n(x))G(|x − y|)dydσ(x),

(5.5.35)

Landau–Lifshitz Equations

296

and similar to the above we have



dB2 dB3 + dτ τ =0 dτ τ =0 =2

Z

+∞

dt

0

Z

x∈B 3

Z

y∈∂B 3

[−div u(x, t)div ξ(x, t)

+ uixj ξxj i + ui xi t + uit θxi ](u(y, t), ~n(y))G(|x − y|)dxdσ(y). (5.5.36) We note that the term B4 is determined only by the boundary data which is 4 | = 0. We finally get independent of τ , then dB dτ τ =0

dBτ = −2 dτ τ =0

Z

+2 +

Z

+∞

dt

B3

0

Z

Z

+∞

0 +∞

dt

dt

0

*

Z

Z

div u(x, t)div ξ(x, t)Φ(x, t)dxdt Φ(x, t)uixj (x, t)ξxj i (x, t)

B3

B3

Z

B3

+

x−y × , ξ(x, t) − ξ(y, t) G0 (|x − y|)div u(x, t)div u(y, t)dxdy |x − y| −2 *

Z

+∞

dt

0

Z

B3

Z

B3

+

x−y , ut (x, t) θ(x, t)div u(y, t)G0 (|x − y|)dxdy, × |x − y|

(5.5.37)

where Φ(x, t) = −

Z

B3

G(|x − y|)div u(y, t)dy +

Z

B3

G(|x − y|)hu(y, t), n(y)idσ(y)

which satisfies (5.5.12). Combining (5.5.22) and (5.5.37), we may now simplify the definition of stationary weak solution for our problem (5.5.1)–(5.5.5) as follows. Remark 5.5.6 A stationary solution u of (5.5.1) satisfies for any ξ(x, t) ∈ C01 (B 3 × R+ , R3 ) and θ(x, t) ∈ C01 (B 3 × R+ , R) as above, the following inequality Z

+∞ 0

−

Z

B3

!

1 1 ut − u × ut (∇u · ξ + ut θ) + 2 2

Z

+∞

dt

0

1 Z +∞ Z dt |∇u|2 (div ξ + θt )dx 3 2 0 B

−2

Z

+2

Z

+∞

dt

0

0

+∞

dt

Z Z

B3

B3

div u(x, t)div ξ(x, t)Φ(x, t)dxdt Φ(x, t)uixj (x, t)ξxj i (x, t)

Z

B3

(uixj uixk ξxkj + uit uixj θxj )dx

Landau–Lifshitz–Maxwell Equations

+

Z

+∞

dt

0

−2

Z

+∞

Z

B3

dt

0

Z

Z

B3

B3

Z

*

+

x−y , ξ(x, t) − ξ(y, t) G0 (|x − y|)div u(x, t)div u(y, t)dxdy |x − y|

*

B3

297

+

x−y , ut (x, t) θ(x, t)div u(y, t)G0 (|x − y|)dxdy ≤ 0. |x − y| (5.5.38)

In what follows we want to derive some more applicable inequalities and equalities from (5.5.38) to be used in the future. Now let Mε = B 3 × B 3 \ {|x − y| < ε} for ε > 0 be small enough. Since ξ(·, t) is in (5.5.37) is uniformly bounded. smooth, ξ(x,t)−ξ(y,t) |x−y| Z

+∞

dt

0

Z

Z

B3

*

B3

+

x−y , ξ(x, t) − ξ(y, t) G0 (|x − y|)div u(x, t)div u(y, t)dxdy |x − y| +

*

x−y , ξ(x, t) − ξ(y, t) G0 (|x − y|)div u(x, t)div u(y, t)dxdy = lim dt ε→0 0 |x − y| Mε Z +∞ Z x−y 0 ξ(x, t) = lim 2 dt G (|x − y|)div u(x, t)div u(y, t)dxdy ε→0 |x − y| 0 Mε Z +∞ Z Z x−y 0 = lim 2 dt ξ(x, t)div u(x, t) G (|x − y|)div u(y, t). 3 3 ε→0 0 x∈B y∈B \B(x,ε) |x − y| (5.5.39) +∞

Z

Z

Let Lε (x, t) = 2

Z

y∈B 3 \B(x,ε)

We have

x−y 0 G (|x − y|)div u(y, t)dy. |x − y|

lim Lε (x, t) = L(x, t) = −H(x, t) + v(x, t)

ε→0

with H(x, t) ∈ L∞ (0, ∞; Lp (B 3 )), |v(x, t)| ≤ K Lp (B 3 )) for any p > 1 and there holds

R

dσ(y) ∂B 3 |x−y|2

(5.5.40)

≤ K ln(1 − |x|) ∈ L∞ (0, ∞;

kL(x, t)kL2 ≤ Ck∇ukL2 . Then (5.5.39) can be rewritten as Z

+∞

dt

0

=

Z

Z

B3 ∞

0

Z

+

B3

*

B3

ξ(x, t) · L(x, t)div u(x, t)dxdt.

Z

x−y , ξ(x, t) − ξ(y, t) G0 (|x − y|)div u(x, t)div u(y, t)dxdy |x − y| (5.5.41)

Similarly we deduce that −

Z

+∞

dt

0

=−

Z

Z

B3

0

B3

*

B3

ut · L(x, t)θ(x, t)dxdt.

Z

∞Z

+

x−y , ut (x, t) θ(x, t)div u(y, t)G0 (|x − y|)dxdy |x − y| (5.5.42)

Landau–Lifshitz Equations

298

Finally, inequality (5.5.38) can be rewritten as Assumption (S) Z

+∞ 0

Z

!

1 1 ut − u × ut (∇u · ξ + ut θ) 2 2

B3

+

Z

+∞

dt

0

Z

B3

(uixj uixk ξxkj + uit uixj θxj )dx

1 Z +∞ Z dt |∇u|2 (div ξ + θt )dx − 2 0 B3 −2

Z

+2

Z

+

Z

+∞

+∞

0 ∞

0

dt

0

Z

dt

B3

Z Z

B3

B3

div u(x, t)div ξ(x, t)Φ(x, t)dxdt Φ(x, t)uixj (x, t)ξxj i (x, t)

L(x, t)[ξdiv u(x, t) − ut θ]dxdt ≤ 0.

(5.5.43)

Formula (5.5.43), Assumption (S), is just the starting point of all the following discussions. From Assumption (S), one can easily derive the following lemma as in [58] which will be used to get the generalized monotonicity inequality in the following section. Lemma 5.5.2 Let u be a stationary weak solution of (5.5.1)–(5.5.5) and ξ, θ as before. Then we have Z

B 3 ×{t}

(ut − u × ut )∇u · ξ − |∇u|2 div ξ + 2uxj uxk ξxkj

− 2 div u(x, t)div ξ(x, t)Φ(x, t) + 2Φ(x, t)uixj ξxj i + 2 div u(x, t)(ξ · L(x, t))dxdt = 0 (5.5.44) and Z

B 3 ×{t2 }

≤

5.5.3

Z

−

t2 t1

Z

Z

B 3 ×{t1 }

B3

!

|∇u|2 θdx

|∇u|2 θt − |ut |2 θ − 2(∇u · ut )∇θ + 2(ut · L(x, t))θdxdt.

(5.5.45)

Estimates for Local Energy

In this section, we use inequalities (5.5.44) and (5.5.45) to derive the generalized monotonicity inequality which will be used to deduce the energy decay in the next section. Denote Bρ = Bρ (0), Pρ (z) = Bρ (x) × (t − ρ2 , t + ρ2 ) for z = (x, t). In the following, we always denote by u the stationary solution of (5.5.1)–(5.5.5).

Landau–Lifshitz–Maxwell Equations

299

Lemma 5.5.3 For any Bs (x0 ) ⊂ Br/2 (0) we have for the stationary solution u that 1 s

Z

1 |∇u| ≤ r Bs (x0 ) 2

Z

2

Br

|∇u| + Kr

Z

Br

|ut |2 .

(5.5.46)

Proof. Taking ξ(x) = φ(|x|)x in inequality (5.5.44) where for any fixed 0 < r, r + h < 1, h > 0   if s ≤ r  1, r−s φ(s) =  1 + h , if r ≤ s ≤ r + h  0, if r + h ≤ s < 1

and noting that

ξxkj (x) = δjk φ(|x|) + φ0 (|x|)

xj xk , |x|

div ξ = 3φ(|x|) + φ0 (|x|)|x|

(5.5.47)

where φ0 (s) = 0 if s ≤ r, φ0 (s) = − h1 if r < s < r + h and φ0 (s) = 0 if r + h < s < 1, we obtain from (5.5.44) that Z

Br+h

(ut − u × ut )(∇u · x)φ(|x|) − |∇u|2 (3φ(|x|) + φ0 (|x|)|x|) + 2uxj uxk

xj xk δjk φ(|x|) + φ (|x|) |x| 0

!

− 2Φ(x, t)div u(x, t)(3φ(|x|) + φ0 (|x|)|x|) +

2Φ(x, t)ujxk

xj xk δjk φ(|x|) + φ (|x|) |x| 0

!

+ 2 div u(x, t)φ(|x|)(x · L(x, t))dx = 0.

(5.5.48)

Sending h → 0 we may estimate every term in (5.5.48) as follows. lim

Z

− lim

Z

h→0 Br+h

(ut − u × ut )(∇u · x)φ(|x|) = 2

h→0 Br+h

Z

Br

0

|∇u| (3φ(|x|) + φ (|x|)|x|) = −3

(ut − u × ut )(∇u · x), Z

2

Br

|∇u| + r

xj xk lim 2uxj uxk δjk φ(|x|) + φ (|x|) h→0 Br+h |x| Z Z 2 =2 |∇u|2 − |x · ∇u|2 dσ, r ∂Br Br Z

− lim

0

Z

h→0 Br+h

= −6

Z

Z

∂Br

(5.5.49)

|∇u|2 dσ, (5.5.50)

!

(5.5.51)

2Φ(x, t)div u(x, t)(3φ(|x|) + φ0 (|x|)|x|)

Br

Φ(x, t)div u(x, t) + 2r

Z

∂Br

Φ div u(x, t)dσ,

(5.5.52)

Landau–Lifshitz Equations

300

!

xj xk − lim 2Φ(x, t)ujxk δjk φ(|x|) + φ0 (|x|) h→0 Br+h |x| Z Z 2 Φ(x · ∇u) · xdσ, = −2 Φ(x, t)div u(x, t) − r ∂Br Br Z

lim

Z

h→0 Br+h

=2

(5.5.53)

2 div u(x, t)φ(|x|)(x · L(x, t))

Z

Br

(x · L(x, t))div u(x, t).

(5.5.54)

Substituting (5.5.49)–(5.5.54) into (5.5.48), we get Z

Br

(ut − u × ut )(∇u · x) −

Z

2

Br

|∇u| + r

Z

∂Br

|∇u|2 dσ

Z Z 2Z 2 − |x · ∇u| dσ − 8 Φ(x, t)div u(x, t) + 2r Φ div u(x, t)dσ r ∂Br Br ∂Br Z Z 2 (x · L(x, t))div u(x, t) = 0. (5.5.55) − Φ(x · ∇u) · xdσ + 2 r ∂Br Br

On the other hand, since (

d 1 dr r

Z

Br

h

|∇u|2 − (ut − u × ut )(∇u · x)

+ 8Φ(x, t)div u(x, t) − 2(x · L(x, t))div u(x, t) =−

1 r2

Z

Br

h

1 r

Z

∂Br

)

|∇u|2 − (ut − u × ut )(∇u · x)

+ 8Φ(x, t)div u(x, t) − 2(x · L(x, t))div u(x, t) +

i

h

i

|∇u|2 − (ut − u × ut )(∇u · x)

+ 8Φ(x, t)div u − 2(x · L(x, t))div u

i

(5.5.56)

we get from (5.5.55) and (5.5.56) that (

d 1 dr r

Z

Br

h

|∇u|2 − (ut − u × ut )(∇u · x)

+ 8Φ(x, t)div u(x, t) − 2(x · L(x, t))div u(x, t)

i

1Z = [−(ut − u × ut )(∇u · x) r ∂Br + 8Φ(x, t)div u(x, t) − 2(x · L(x, t))div u(x, t)] +

Z

∂B 3

"

)

2 2 2 |x · ∇u|2 − Φ div u(x, t) + 3 Φ(x · ∇u) · x 3 r r r

#

Landau–Lifshitz–Maxwell Equations

=2

Z

∂B 3

"

|x · ∇u|2 (x · ∇u)(ut − u × ut ) − |x|3 2|x|

301

#

3Φ div u (x · L)div u Φ(x · ∇u) · x . + − + |x| |x| |x|3

(5.5.57)

Denote 1 ψ(ρ, t) = ρ

Z

Bρ

h

i

|∇u|2 − (ut − u × ut )(∇u · x) + 8Φ(x, t)div u − 2(x · L)div u . (5.5.58)

Then by integrating (5.5.57) from s to ρ (s < ρ), we obtain ψ(ρ, t) − ψ(s, t) = 2

Z

Bρ \Bs

"

|x · ∇u|2 (x · ∇u)(ut − u × ut ) − |x|3 2|x|

3Φ div u (x · L)div u Φ(x · ∇u) · x + − + |x| |x| |x|3

or

ψ(s, t) = ψ(ρ, t) − 2

Z

Bρ \Bs

"

#

(5.5.59)

|x · ∇u|2 (x · ∇u)(ut − u × ut ) − |x|3 2|x| #

3Φ div u (x · L)div u Φ(x · ∇u) · x + − + , |x| |x| |x|3

(5.5.60)

that is 1 s

Z

Bs

|∇u|2 =

1 [(ut − u × ut )(∇u · x) s Bs − 8Φ(x, t)div u(x, t) + 2(x · L(x, t))div u(x, t)] Z

+ ψ(ρ, t) − 2

Z

Bρ \Bs

"

|x · ∇u|2 (x · ∇u)(ut − u × ut ) − |x|3 2|x| #

3Φ div u (x · L)div u Φ(x · ∇u) · x + − + . |x| |x| |x|3

(5.5.61)

By hole-filling method and note that |x| ≤ s if x ∈ Bs , we have "

Z 1Z |(x · ∇u)(ut − u × ut )| 4|Φ div u| 2 + |∇u| ≤ ψ(ρ, t) + 2 s Bs 2|x| |x| Bρ #

|(x · L)div u| |Φ||x · ∇u| + + . |x| |x|2

(5.5.62)

Now we can estimate all the terms on the right-hand side of (5.5.62) and get the generalized monotonicity inequality. It follows from (5.5.62) and H¨older inequality that Z 1Z 1Z |∇u|2 ≤ |∇u|2 + Kρ [|ut |2 + |Φ|2 + |Φ|8 + |L|2 ]. (5.5.63) s Bs ρ Bρ Bρ

Landau–Lifshitz Equations

302

In this inequality, we should note that Φ(x, t) ∈ L∞ (0, T ; W 1,p(Ω)), L(x, t) ∈ L∞ (0, T ; Lp (Ω)) for any p > 1. Then it follows from Lemma 5.5.1, (5.5.12) and (5.5.15) that 1 s

Z

1 |∇u| ≤ K ρ Bs 2

Z

2

Bρ

|∇u| + Kρ

Z

Bρ

|ut |2

(5.5.64)

with K independent of t. It is not difficult to get, for any Bs (x0 ) ⊂ Br/2 (0)

Z 1Z 1Z |∇u|2 ≤ |∇u|2 + Kr |ut |2 . s Bs (x0 ) r Br Br

(5.5.65)

Lemma 5.5.3 is proved. In the following we want to prove Lemma 5.5.4 (Generalized monotonicity inequality) There exists a constant C > 0 such that for any θ ∈ (0, 1/16] there is ε0 > 0, if u ∈ H 1 (Pr (z0 ), S 2 ) is a stationary solution, we have Z

r −3

|∇u|2 dz ≤ δε2 + C1 r −5

Pr/8 (0)

under the condition r −3

Z

Pr (0)

Z

Pr (0)

|u − (u)Pr (0) |2 dz

|∇u|2 dz ≤ ε2 ≤ ε20 .

Proof. It follows from (5.5.45) that Z

B 3 ×{t

≤

2}

Z

−

t2 t1

Z

Z

B 3 ×{t

B3

1}

!

|∇u|2 θdx

|∇u|2 θt − |ut |2 θ − 2(∇u · ut )∇θ + 2(ut · L(x, t))θdxdt.

(5.5.66)

Therefore, we have Z

B 3 ×{t2 }

−

Z

B 3 ×{t1 }

!

2

|∇u| θdx ≤ C

Z

t2 t1

Z

B3

|∇u|2 (θt + |∇θ|2 ) + |L(x, t)|2 θ.

Hence, if taking θ(x, t) such that |∇θ|2 , |θt | ≤ rC2 , θ(x, t1 ) = 0 and θ ≡ 1 for x ∈ Br/2 , t ∈ ((− 2r )2 , ( 2r )2 ), we obtain from Lemma 5.5.1 for almost every t = t2 ∈ (−( 2r )2 , ( 2r )2 ) 1 r

Z

Br/2 (0)×{t}

|∇u|2 ≤ K1

1 r3

Z

Pr (0)

!

|∇u|2 dxdt .

(5.5.67)

Claim. There exists a constant C > 0 such that for any given 0 < λ < 1, there exists a set Λ ⊂ (−r 2 /2, r 2 /2) with |Λ| ≤ λ satisfying 1Z Cε2 |ut |2 dx ≤ r B r2 (0)×{t} λ

for almost every t 6∈ Λ where

1 r3

R

Pr

|∇u|2 dz ≤ ε2 ≤ ε20 .

(5.5.68)

Landau–Lifshitz–Maxwell Equations

303

Proof. Taking smooth test function θ(x, t) such that θ ≡ 1 in Pr/2 (0) but θ ≡ 0 outside Pr (0), t1 = −r 2 , t2 = r 2 , and noticing that L(x, t) ∈ L∞ (0, ∞; Lp (B 3 )) for any p ≥ 1, we get from Lemma 5.5.1 that Z

2

|ut | dz ≤ C0

Pr/2 (0)

1 ≤ C0 r 3 r

Z

Z

Pr (0)

|∇u|2 (θt + |∇θ|2 ) + θ|L(x, t)|2 dz !

|∇u|2 ≤ C0 rε2 .

Pr (0)

(5.5.69)

If the Claim is false, then for any C > 0, there exists 0 < λ < 1 such that for some set Λ ⊂ (−r 2 /2, r 2 /2) with |Λ| > λ there holds: if t ∈ Λ, then 1 r

Z

Pr/2 (0)

|ut |2 dz ≥

1 r

Z Z Λ

Br/2 (0)

|ut |2 dz ≥ Cε2 .

This contradicts (5.5.69) by taking C > C0 . The Claim follows. We can even prove by (5.5.65) (Lemma 5.5.3) that if t 6∈ Λ then 1 s

sup Bs (x0 )⊂Br/4 (x0 )

Z

2

Bs (x0 )×{t}

!

|∇u| dx ≤

Cε2 . λ

(5.5.70)

In the sequence, we mean t 6∈ Λ. In order to prove the lemma, we take ξ ∈ C0∞ (Br/4 (0)) such that 0 ≤ ξ ≤ 1, ξ ≡ 1 in Br/8 (0), |∇ξ| ≤ 16 , to compute r Z

2

Br (0)×{t}

ξ(x)|∇u| dx =

Z

Br (0)×{t}

=−

Z

−

ξ(x)∇u · ∇(u − (u)Pr (0) )dx

Br (0)×{t}

Z

(u − (u)Pr (0) )∇ξ · ∇u

Br (0)×{t}

ξ(x)(u − (u)Pr (0) )∆u.

Using the equation, we have Z

Br (0)×{t}

ξ(x)|∇u|2 dx = −

Z

Br (0)×{t}

(u − (u)Pr (0) )∇ξ · ∇u

1 1 − ξ(x)(u − (u)Pr (0) ) ut − u × ut 2 2 Br (0)×{t} Z

+

Z

−

Z

Br (0)×{t}

Br (0)×{t}

!

ξ(x)(u − (u)Pr (0) )u|∇u|2 ξ(x)(u − (u)Pr (0) )uH(u).

(5.5.71)

In the following we estimate every term on the right-hand side of (5.5.71).

Landau–Lifshitz Equations

304

Estimate of the first term: Z

Br (0)×{t}

(u − (u)Pr (0) )∇ξ · ∇u ≤

16 k∇ukL2 (Br ) ku − (u)Pr (0) kL2 (Br ) r

16 ≤ √ r

!1/2

1 |∇u|2 dx ku − (u)Pr (0) kL2 (Br ) r Br (0) 1 ≤ C0 ε √ ku − (u)Pr (0) kL2 (Br ) (by (5.5.67)) r δλr C 1 ≤ Cε2 + ku − (u)Pr (0) k2L2 (Br ) . (5.5.72) 2 2 δλ r Z

Estimate of the second term: Since t 6∈ Λ, we get Z

1 1 ξ(x)(u − (u)Pr (0) ) ut − u × ut 2 2 Br (0)×{t}

!

≤ Ckut kL2 (Br ) ku − (u)Pr (0) kL2 (Br ) √

!1/2

1 ≤C r |ut |2 dx ku − (u)Pr (0) kL2 (Br ) r Br (0) √ Cε r ≤ √ ku − (u)Pr (0) kL2 (Br ) (by (5.5.68)) λ δr C ≤ Cε2 + ku − (u)Pr (0) k2L2 (Br ) . 2 δλ Z

(5.5.73)

Estimate of the third term: To estimate this term, we use H´elein method to decompose it. Since |u| = 1 a.e. we have ui |∇u|2 =

3 X

j=1

∇uj (ui ∇uj − uj ∇ui )

and then Z

Br (0)×{t}

=

ξ(x)(u − (u)Pr (0) )u|∇u|2

3 X 3 Z X

i=1 j=1 Br (0)×{t}

(ui − (ui )Pr (0) )∇uj (ξ(x)(ui ∇uj − uj ∇ui )).

On the other hand, div(ui ∇uj − uj ∇ui ) = ui ∆uj − uj ∆ui = ui w j − uj w i ,

Landau–Lifshitz–Maxwell Equations

305

where w = 21 ut + 12 u × ut , we obtain kdiv(ξ(x)(ui ∇uj − uj ∇ui ))kL2 (Br )

≤ k∇ξ(x) · (ui ∇uj − uj ∇ui )kL2 (Br ) + kξ(x)(ui w j − uj w i )kL2 (Br )

32 k∇ukL2 (Br/4 (0)) + 2kut kL2 (Br/4 (0)) r Cε ≤√ (by (5.5.67) and (5.5.68)). λr

≤

(5.5.74)

To continue the proof, we recall a lemma by Feldman [58]. Lemma 5.5.5 ([58]) Let f, h ∈ H 1 (Rn ) and g ∈ L2 (Rn , Rn ) with div g ∈ L2 (Rn ) in the distribution sense, and sup

x0

Then

Z

Rn

∈Rn ,

r r>0

2−n

Z

Br (x0 )

!

|∇h| dx = A2 < ∞. 2



f g · ∇h ≤ CA(k∇f kL2 kgkL2 + kf kL2 kdiv gkL2 )

for some universal constant C.

Now we apply Lemma 5.5.5 to f = ui − (ui )Pr (0) , h = uj and g = ξ(x)(ui ∇uj − uj ∇ui ) by extending them properly to the whole space R3 . By (5.5.67) and (5.5.74) Z

Br (0)×{t}

=

Z

ξ(x)(u − (u)Pr (0) )u|∇u|2

Br (0)×{t}

(ui − (ui )Pr (0) )∇uj (ξ(x)(ui ∇uj − uj ∇ui ))

Cε ≤ √ [k∇f kL2 kgkL2 + kf kL2 kdiv gkL2 ] λ " # Cε Cε 2 ≤ √ Cε r + √ ku − (u)Pr (0) kL2 λ λr ≤

Cε3 r Cε2 + 2 ku − (u)Pr (0) k2L2 . λ λr

(5.5.75)

Finally, we estimate the last term on the right-hand side of (5.5.71) as follows. Since H(u) = ∇Φ and ∆Φ = −div u, div u ∈ L∞ (0, ∞; H −1 ), we know kH(u)k2L2 (Br ) ≤ C0 k∇uk2L2 (Br ) ≤ Crε2 . Hence we derive Z

Br (0)×{t}

ξ(x)(u − (u)Pr (0) )uH(u)

√ ≤ C rεku − (u)Pr (0) kL2 (Br (x0 )) ≤ Cε2

δλr C + ku − (u)Pr (0) k2L2 (Br (x0 )) . 2 δλ

(5.5.76)

Landau–Lifshitz Equations

306

Combining (5.5.71)–(5.5.76), we have obtained for t 6∈ Λ (|Λ| < λ) Z

!

Cε δ 2 C + ε r+ ku − (u)Pr (0) k2L2 (Br (0)) . (5.5.77) 2 λ 2 δλr

2

Br/8 (0)×{t}

|∇u| dx ≤

Integrating over (−r 2 /2, r 2 /2) and using (5.5.67), we get Z

r 2 /2 −r 2 /2

Z

!

Cε δ 2 3 C + Cλ + εr + ku − (u)Pr (0) k2L2 (Pr (0)) . λ 2 δλr 2 (5.5.78)

2

Br/8 (0)×{t}

|∇u| dx ≤

Hence, choosing C, λ and ε0 properly we obtain r

−3

Z

2

Pr/8 (0)

2

|∇u| dz ≤ δε + C1 r

under the condition r −3

Z

Pr (0)

−5

Z

Pr (0)

|u − (u)Pr (0) |2 dz

(5.5.79)

|∇u|2 dz ≤ ε2 ≤ ε20 .

Lemma 5.5.4 is proved.

5.5.4

Energy Decay and Partial Regularity

In this section we derive the energy decay which will be used to prove the regularity. Lemma 5.5.6 There exists a constant C > 0 such that for any θ ∈ (0, 12 ], there is a number ε0 > 0 such that for any stationary solution u ∈ H 1 (Pr (z0 ); S 2 ) of (5.5.1)–(5.5.5) satisfying r

−3

Z

Pr (z0 )

|∇u|2 dz ≤ ε2 ≤ ε20 ,

we have (θr)

−5

Z

Pθr (z0 )

|u − (u)Pθr (z0 ) |2 dz ≤ Cθ 2 ε2 .

(5.5.80)

Proof. Since the integrals are invariant under the transformation (x, t) → (rx + x0 , r 2 t + t0 ), we may assume Pr (z0 ) = P1 (0). We argue by contradiction: If the conclusion is untrue, then for any C > 0 we may find θ ∈ (0, 1/2] and stationary weak solution uk ∈ H 1 (P1 (0); S 2 ) of the considered problem such that Z

P1 (0)

|∇uk |2 dz = ε2k → 0,

k → ∞,

(5.5.81)

but Z

Pθ (0)

|uk − (uk )Pθ (0) |2 > Cθ 3+4 ε2k .

(5.5.82)

Landau–Lifshitz–Maxwell Equations

307

It follows from these assumptions and the lemmas in the above sections that the sequence {vk } = { ε1k (uk − (uk )Pθ (0) )} is bounded in H 1 (P1/2 (0)) which allows us to assume that there is a map v ∈ H 1 (P1/2 (0); R3 ) such that vk → v, weakly in H 1 (P1/2 (0); R3 ); with

Z

Pθ (0)

vdz = 0;

vk → v, strongly in L2 (P1/2 (0); R3 ) Z

P1/2 (0)

|∇v|2 dz ≤ 1.

It is obvious that we can assume that uk → p strongly in L2 (P1/2 (0)) for some constant map p ∈ S 2 . Note that uk solves the following equation: 1 1 ukt − (uk × ukt ) = ∆uk + uk |∇uk |2 + H(uk ) − H(uk )uk , in B 3 × R+ , (5.5.83) 2 2 where curl H(uk ) = 0, div(H(uk ) + uk ) = 0,

in D 0 (R3 ),

in D 0 (R3 ).

(5.5.84) (5.5.85)

Let H(uk ) = ∇Φk . Then ∆Φk = −div uk . Note that uk = εk vk + (uk )Pθ (0) . Then for any φ ∈ C0∞ (P1/2 (0), R3 ) we have by multiplying (5.5.83) by φ Z

!

1 1 ukt − (uk × ukt ) φ 2 2

P1/2 (0)

=

Z

=

Z

P1/2 (0)

P1/2 (0)

(∆uk + uk |∇uk |2 + H(uk ) − H(uk )uk )φ (−∇uk · ∇φ + uk |∇uk |2 φ + φ∇Φk − uk φ∇Φk ).

(5.5.86)

Note that uk = εk vk + (uk )Pθ (0) . We have from (5.5.86) that ε

Z

!

1 1 vkt − (uk × vkt ) φ 2 2

P1/2 (0)

=

Z

P1/2 (0)

(−εk ∇vk · ∇φ + uk |∇uk |2 φ + φ∇Φk − uk φ∇Φk )

(5.5.87)

with ∆Φk = −εk div vk . Divide both sides of (5.5.87) by εk and send k → ∞ to give Z

P1/2 (0)

= since limk→∞ ε1k

R

!

1 1 vt − (p × vt ) φ 2 2

P1/2 (0)

Z

P1/2 (0)

(−∇v · ∇φ + φ∇Φ∞ − pφ∇Φ∞ )

(5.5.88)

|∇uk |2 uk φ = 0 from (5.5.81), where ∆Φ∞ = −div v. Since

v ∈ H 1 (P1/2 (0)) we know div v ∈ L2 (P1/2 (0)), then Φ∞ ∈ W22,2 . Denote H∞ = ∇Φ∞ .

Landau–Lifshitz Equations

308

It follows from (5.5.88) that v satisfies 1 1 vt − (p × vt ) = ∆v + H∞ − pH∞ . (5.5.89) 2 2 By standard estimates and the boot-strapping method, we know that v is smooth and there holds Z

Pθ (0)

|v|2 dz ≤

Z

Pθ (0)

|v − vPθ (0) |2 dz ≤ C0 θ 3+4

(5.5.90)

by Poincar´e inequality. Inequality (5.5.90) contradicts (5.5.82) from the strong L 2 convergence of vk to v if one chooses C > C0 . Combining these lemmas we get by iteration method Proposition 5.5.1 There is a constant C > 0 such that for any θ ∈ (0, 1/16], there exists a number ε0 > 0 such that if u ∈ H 1 (Pr (z0 ), S 2 ) is a stationary solution of (5.5.1)–(5.5.5) satisfying the small energy condition r

−3

Z

Pr (z0 )

|∇u|2 dz ≤ ε2 ≤ ε20 ,

(5.5.91)

then (θr)−3

Z

Pθr (z0 )

|∇u|2 dz ≤ Cθ 2 ε2 .

(5.5.92)

1 1 Proof. Given 0 < θ < 1/16, taking k such that 8k+2 ≤ θ ≤ 8k+1 . Denote R R −3 2 −5 2 Φ(r) = r Pr (z0 ) |∇u| dz, Ψ(r) = r Pr (z0 ) |u − (u)Pr (z0 ) | . It follows from (5.5.91) and Lemma 5.5.4 that for any δ > 0

r Φ 8

!

≤ δε2 + C1 Ψ(r).

(5.5.93)

Similarly we have r Φ 2 8

!

!

r r ≤ δΦ + C1 Ψ 8 8

!

!

r ≤ δ(δε + C1 Ψ(r)) + C1 Ψ . 8 2

(5.5.94)

Iterating, applying Lemma 5.5.6 and choosing δ properly, we get Φ

r 8k+1

!

≤δ

k X

r ε + C1 δ Ψ j 8 j=1

k+1 2

j

!

r +Ψ k 8

!

≤ Cδ

1 ε +C k 8

k+1 2

!2

ε2 ≤ Cθ 2 ε2 , (5.5.95)

where we have used Lemma 5.5.6 again which shows Ψ( 8rj ) ≤ C( 81j )2 ε2 . Then we have Φ(θr) = (θr)

≤ (8

−3

k+1

Z

θ)

2

Pθr

−3

|∇u| dz ≤ θ r 8k+1

!−3 Z

−3

r 8k+1

!−3

2

P

r 8k+1

(8

k+1 −3

)

3

|∇u| dz ≤ 8 Φ

Z

P

r 8k+1

r 8k+1

!

|∇u|2 dz ≤ Cθ 2 ε2

(by (5.5.95)).

Landau–Lifshitz–Maxwell Equations

309

Remark 5.5.1 By virtue of (5.5.70), Proposition 5.5.1 also holds if replacing Pθr (z0 ) by Ps (z1 ) for any z1 ∈ P15r/16 (z0 ) and s ∈ (0, r/16). Theorem 5.5.1 There exist constants ε0 > 0 and Ckl > 0 such that any stationary solution u ∈ H 1 (Pr (z0 )) of (5.5.1)–(5.5.4) satisfying the small energy condition (5.5.89) is smooth in Pr/2 (z0 ) and k∂tl ∇k ukL∞ (Pr/2 (z0 )) ≤ Ckl r −k−2l ε,

k, l = 0, 1, 2, . . . .

(5.5.96)

Proof. For any given Pr (z0 ) ⊂ B 3 × (0, T ), Proposition 5.5.1 shows that for any λ ∈ (0, 1), if ε0 is small enough, we have Z

Ps (z1 )

(|∇u|2 + s2 |ut |2 )dz ≤ C1 sn+2λ

(5.5.97)

for any z1 ∈ P15r/16 (z0 ) and s ∈ (0, r/16) with C1 only depending on λ. In fact, let θr = s, then θ = rs . Substituting this θ into (5.5.92), we have s−3

Z

Ps (z1 )

|∇u|2 dz ≤

s r

!2

ε20 .

(5.5.98)

On the other hand, we may control Ps (z1 ) s2 |ut |2 dz as in Feldman [58]. (5.5.97) is proved. By standard method we deduce that u is smooth in P15r/16 (z0 ) by Morrey’s Lemma which was done by Feldman [58]. The estimates (5.5.96) can be obtained by a scaling argument. By the standard method as in Giaquinta’s book [64], it is easy to conclude. R

Theorem 5.5.2 Let u ∈ H 1 (Ω×(0, T ); S 2 ) be a stationary solution of (5.5.1)–(5.5.4). There is an open set Q ⊂ Ω × (0, T ) such that u is smooth in Q and H3 (Ω × (0, T ) \ Q) = 0

(5.5.99)

where Ω × (0, T ) \ Q = {z = (x, t)| lim inf r −3 r→0

5.6 5.6.1

Z

Pr (z)

|∇u|2 dz ≥ ε0 }.

(5.5.100)

Weak Solutions to Landau–Lifshitz–Maxwell Equations with Polarization The Problem and Physics Background

In this section, we study the three-dimensional Landau–Lifshitz–Maxwell equations coupling with polarization as follows 







~ t = α1 Z ~ × 4Z ~ +H ~ − α2 Z ~× Z ~ × 4Z ~ +H ~ Z



,

(5.6.1)

310

Landau–Lifshitz Equations

~ = ∇×H

~ + P~ ) ∂(E ~ + σ E, ∂t

(5.6.2)

~ ~ ~ = − ∂H − β ∂Z , ∇×E ∂t ∂t

(5.6.3)

  ∂ 2 P~ ∂ P~ 2 ~ 2 ~ − 2P~ Φ0 (|P~ |2 ) , + λ curl P + µ = ν E ∂t2 ∂t

(5.6.4)

b P ~) = where P~ (x, t) = (P1 (x, t), P2 (x, t), P3 (x, t)) denotes the electric polarization, E( 2 0 2 2P~ Φ (|P~ | ) the equilibrium electric field, curl P~ = curl(curl P~ ) = ∇ × (∇ × P~ ), α2 ≥ 0 is the Gilbert damping coefficient, λ > 0 denotes the speed of light for the internal fields, σ ≥ 0 denotes the constant conductivity, constant β can be viewed as the magnetic permeability of free space. The physical meanings of parameters µ, ν can be found in [69]. We assume that Φ : R+ → R is a C 2 convex function such that

|Φ0 (r)| ≤ C0 ,

rΦ00 (r) ≤ C1

(5.6.5)

for all r ≥ 0. We also assume that function Φ(r 2 ) has unique minimum at some point r02 . These assumptions guarantee that rΦ0 (r 2 ) ≤ C2 for all r ≥ 0, where C2 = C0 +2C1 . Therefore, we have ~ 0 ~ 2 ~ Φ0 (|Y ~ |2 ) ≤ C2 X ~ −Y ~ XΦ (|X| ) − Y

~ Y ~ ∈ R3 . for all X,

(5.6.6)

Much more about the equilibrium relation of Φ may be found in Landau and Lifshitz [104], pp. 84–91. System (5.6.1)-(5.6.4) models the dynamics of magnetization, magnetic field, electric field. and electric polarization for the ferromagnetic–ferroelectric materials, which includes a new equation for polarization P~ . As we know that, some ferromagnetic substances, such as ferrites, are not only ferromagnetic materials, but also ferroelectric ones (such as LiFePO4 ), we call them the ferromagnetic–ferroelectrics [148]. If an electric field is applied to a medium (such as a dielectric one) made up of a large number of atoms or molecules, the charges bound in each molecule will respond to the applied field and will execute perturbed motions: the molecular charge density will be distorted. The multipole moments of each molecule will be different from what they were in the absence of the field. In simple substances, when there is no applied field the multipole moments are all zero, at least when they averaged over many molecules. The dominant molecular multipole with the applied fields is the dipole. There is thus produced in the medium an electric polarization P~ (the dipole moment per unit volume). A dielectric in which P~ differs from zero is said to be polarized. The vector P~ determines not only the volume charge density but also the density of the charge on the surface of the polarized dielectric [93]. One can learn more about polarization in [17, 43, 62, 104].

Landau–Lifshitz–Maxwell Equations

311

The coupling of this classical Landau–Lifshitz–Maxwell system with P~ and Eq. (5.6.4) for P~ can be derived from the full Maxwell system as follows: ~ ∂B ~ = −curl E ∂t

~ ∂D ~ = curl H, ~ + σE ∂t

and

(5.6.7)

~ and H ~ are the electric and magnetic fields, σ ≥ 0 is the conductivity, D ~ and where E ~ the electric and magnetic displacements defined by B ~ = 0 E ~ + P~ , B ~ = µ 0 (H ~ + Z), ~ D where 0 is the permittivity of free space, µ0 is the magnetic permeability of free ~ is the magnetization, and P~ is the electric polarization. Substituting these space, Z ~ E, ~ H ~ and P~ by systems like (5.6.1)–(5.6.3). definitions into (5.6.7), one may couple Z, For the derivation of (5.6.4), we refer to [69]. For the system (5.6.1)–(5.6.4), we impose the following periodic initial conditions ~ 0), H(x, ~ ~ (Z(x, 0), E(x, 0), P~ (x, 0), P~t (x, 0)) ~ 0 (x), H ~ 0 (x), E ~ 0 (x), P~0 (x), P~1 (x)), = (Z

(5.6.8)

where we call a function f (x) is 2D-periodic if f (x + 2Dei ) = f (x), (i = 1, 2, 3), where (e1 , e2 , e3 ) forms the unit orthogonal basis of R3 , D > 0 is a constant. And, we always assume that Z0 (x), H0 (x), E0 (x), P0 (x) and P1 (x) are 2D-periodic. We denote by Ω ⊂ R3 the three-dimensional cube with width 2D along each direction, i.e. Ω = {x = (x1 , x2 , x3 )| |xi | < D; (i = 1, 2, 3)} and QT = {(x, t)| x ∈ Ω, 0 < t ≤ T }. ~ t), E(x, ~ ~ Definition 5.6.1 A 2D-periodic vector (Z(x, t), H(x, t), P~ (x, t)) ∈ (L∞ (0, T ; T H 1 (Ω)), L∞ (0, T ; L2 (Ω)), L∞ (0, T ; L2 (Ω)), W 1,∞ (0, T ; L2 (Ω)) L∞ (0, T ; H 1(Ω)) is called a weak solution to problem (5.6.1)–(5.6.4) and (5.6.8), if for any 2D-periodic ~ ~ vector-valued test function Ψ(x, t) ∈ C 1 (QT ) such that Ψ(x, T ) = 0, the following equalities hold ZZ

ZZ

QT

QT

+ α2

ZZ

− α2

ZZ



ZZ

QT

QT



QT

QT



QT

~ × ∇Z) ~ · ∇Ψ ~ − α1 (Z 

ZZ







~ t − λ2 P~t · Ψ ZZ

ZZ

QT



QT

~ ·H ~ + eσt ∇ × Ψ

~ + βZ ~ ·Ψ ~t− H

− 2ν





~ ×H ~ ·∇ Z ~ ×Ψ ~ + Z



QT



~ × ∇Z ~ ·∇ Z ~ ×Ψ ~ Z

~ + P~ · Ψ ~ t eσt + σ E

+ ZZ

ZZ

~ ·Ψ ~ t + α1 Z

ZZ

QT

ZZ

QT

Z

Ω

ZZ

QT

~ × H) ~ ·Ψ ~ (Z

~ 0 · Ψ(x, ~ Z 0) = 0,

(5.6.9)

~ eσt P~ · Ψ

Z  Ω



~ 0 + P~0 · Ψ(x, 0) = 0, E





~ ·E ~+ ∇×Ψ

~ −µ curl P~ · curl Ψ

~ + Φ0 (|P~ |2 )P~ · Ψ

Z

Ω

Z 

ZZ

Ω

QT

(5.6.10) 

~ 0 + βZ ~ 0 · Ψ(x, ~ H 0) = 0, (5.6.11)

~ +ν P~t · Ψ

~ P~1 · Ψ(x, 0) = 0.

ZZ

QT

~ ·Ψ ~ E (5.6.12)

Landau–Lifshitz Equations

312

5.6.2

Viscosity Approximation

Since Eq. (5.6.1) is strongly coupled, it is not easy to obtain weak solutions by use of the theory of semigroups as Habib Ammari in [6]. We are going to use Galerkin method here. However, it is easy to see that Eq. (5.6.4) lacks the compactness which we need to get the estimate of H 1 -norm for P~ . In fact, from this equation, we will only be able to obtain the L∞ (0, T ; L2 (Ω)) estimates for curl P~ , but do not have the L∞ (0, T ; L2 (Ω)) estimates for div P~ . To overcome this difficulty, we firstly apply the viscosity vanishing argument to get the weak solution for the viscosity problem and then, we will set the viscosity constant to zero. But the lack of compactness reappears in the limit procedure. Therefore, we secondly consider more regular (than energy ones) class weak solution and then obtain the additional a priori estimates for the div-component of Maxwell fields (cf. Lemma 5.6.9). Finally we obtain the desired weak solution to the original problem. Replacing (5.6.4) by the following viscosity approximation ∂ 2 P~ ∂ P~ 2 2~ ~ − 2P~ Φ0 (|P~ |2 )), + λ curl P + µ − 4P~ = ν(E 2 ∂t ∂t

(5.6.13)

we get a viscosity system (5.6.1)–(5.6.3) and (5.6.13) with the 2D-periodic initial conditions (5.6.8).

5.6.3

Solutions to the Viscosity Problem

Definition 5.6.2 The space HP (curl , Ω) is defined by ~ ∈ L2 (Ω); V ~ is 2D-periodic and curl V ~ ∈ L2 (Ω)}, HP (curl , Ω) = {V and is provided with the norm ~ kH (curl ,Ω) = {kV ~ k2 2 + kcurl V ~ k2 2 }1/2 . kV L (Ω) L (Ω) P The space HP (div, Ω) is defined by ~ ∈ L2 (Ω); V ~ is 2D-periodic and div V ~ ∈ L2 (Ω)}, HP (div, Ω) = {V and is provided with the norm ~ kH (div,Ω) = {kV ~ k2 2 + kdiv V ~ k2 2 }1/2 . kV L (Ω) L (Ω) P Finally, we set XP (Ω) = HP (curl , Ω) ∩ HP (div, Ω) with the norm ~ kX (Ω) = {kV ~ k2 2 + kcurl V ~ k2 2 + kdiv V~ k2 2 }1/2 . kV L (Ω) L (Ω) L (Ω) P

Landau–Lifshitz–Maxwell Equations

313

Lemma 5.6.1 Assume X ⊂ E ⊂ Y are Banach spaces and X ,→,→ E. Then the following imbedding are compact, if 1 ≤ q ≤ ∞, or 1 < r ≤ ∞ q

T

∞

T

(i) L (0, T ; X) (ii) L (0, T ; X)

(

)

(5.6.14)

(

)

(5.6.15)

∂ϕ ϕ: ∈ L1 (0, T ; Y ) ,→,→ Lq (0, T ; E), ∂t

∂ϕ ϕ: ∈ Lr (0, T ; Y ) ,→,→ C([0, T ]; E). ∂t

Let ωn (x), (n = 1, 2, 3, . . .) be the unit eigenfunctions satisfying the equation −4ωn = λn ωn , with periodicity ωn (x − Dei ) = ωn (x + Dei ) and λn , (n = 1, 2, 3, . . .) the corresponding eigenvalues different from each other. Denote the approximate ~  (x, t), H ~  (x, t), solutions of the problem (5.6.1)–(5.6.3), (5.6.13) and (5.6.8) by Z N N  ~N E (x, t), P~N (x, t) in the following form: ~  (x, t) = Z N

N X

 α ~ sN (t)ωs (x),

~  (x, t) = H N

s=1

~  (x, t) = E N

N X

N X

 β~sN (t)ωs (x),

s=1  ~γsN (t)ωs (x),

P~N (x, t) =

s=1

N X

~δ (t)ωs (x), sN

s=1

    where α ~ sN (t), β~sN (t), ~γsN (t), ~δsN (t), (t ∈ R+ ), (s = 1, 2, . . . , N; N = 1, 2, . . .) are three-dimensional vector-valued functions satisfying the following system of ordinary differential equations:

Z

Ω

~  ωs = α 1 Z Nt

Z

Ω

− α2 Z 

Ω

=

Ω

P~N tt ωs + λ2 =ν

Z

Ω

Ω







Z 

Z

2 ~ PN ωs (x)

Z 

~  ωs (x) − 2ν E N

Z

Ω

+µ

ωs ,



Ω

Z

Ω

(5.6.16)

(5.6.17)



~  + P~  ωs (x) E N N

~  ωs (x) + σ ∇×H N

curl Ω





 ~N ∇×E ωs (x),

Ω

~  + P~  ωs (x) + σ E Nt Nt Z 



~ N × Z ~ N × 4Z ~ N + H ~ N Z



Z 

Ω

Z

~ N t + β Z ~ N t ωs = − H

Ω

Z



~  × 4Z ~ + H ~  ωs Z N N N

Z

Ω

P~N ωs (x),

P~N t ωs (x) − 

P~N Φ0 (|P~N |2 )ωs (x)

Z

(5.6.18)

Ω

4P~N ωs (x) (5.6.19)

with initial conditions Z

Ω

~  (x, 0)ωs (x) = Z N

Z

Ω

~ 0 (x)ωs (x), Z

(5.6.20)

Landau–Lifshitz Equations

314 Z

Z

Z

Ω

Z

Ω

Ω

Ω

~  (x, 0)ωs (x) = H N  ~N E (x, 0)ωs (x) =

P~N (x, 0)ωs (x)dx =

P~N t (x, 0)ωs (x)dx =

Z Z Z Z

Ω

Ω

Ω

Ω

~ 0 (x)ωs (x), H

(5.6.21)

~ 0 (x)ωs (x), E

(5.6.22)

P~0 (x)ωs (x)dx,

(5.6.23)

P~1 (x)ωs (x)dx.

(5.6.24)

It follows from the standard theory on nonlinear ordinary differential equations that the problem (5.6.16)–(5.6.24) admits unique local solution. The following a priori estimates enable us to take the limit N → ∞ in (5.6.16)–(5.6.24) to obtain the global solution to the viscosity problem. For the sake of simplicity, we denote k · kLp (Ω) = k · kp , p ≥ 2. ~ 0 (x), H ~ 0 (x), E ~ 0 (x), P~0 (x), P~1 (x)) ∈ (H 1 (Ω), L2 (Ω), L2 (Ω), Lemma 5.6.2 Assume (Z H 1 (Ω), L2 (Ω)). Then for the solution of the initial value problem (5.6.16)–(5.6.24), we have the following estimates: ~  (·, t)k2 1 + kE ~  (·, t)k2 + kH ~  (·, t)k2 + kP~  (·, t)k2 sup [kZ N H (Ω) N 2 N 2 N 2

0≤t≤T

+ kcurl P~N (·, t)k22 + k∇P~N (·, t)k22 + kP~N t (·, t)k22 ] ≤ C3 , ~ N (·, t)k26 ≤ C4 , sup kZ

~ N (·, t) × ∇Z ~ N (·, t)k 3 sup kZ ≤ C5 , L 2 (Ω)

0≤t≤T

(5.6.25) (5.6.26)

0≤t≤T

where the constants C3 , C4 and C5 are independent of N, α2 and D. When α2 > 0, there is ~  × (4Z ~ + H ~  )kL2 (0,T ;L2 (Ω)) ≤ C6 , kZ N N N

(5.6.27)

where the constant C6 is independent of N and D.  Proof. 1. Multiplying (5.6.16) by α ~ sN (t), summing up the products for s = 1, 2, . . . , N, we get

d dt

Z

Ω

~  (·, t)|2 dx = 0. |Z N

Then we have ~  (·, t)k2 = kZ ~  (·, 0)k2 ≤ kZ ~ 0 (x)k2 , kZ N 2 N 2 2 

∀ t ≥ 0.

(5.6.28)



  2. Making the scalar product of −λs α ~ sN (t) + β~sN (t) with (5.6.16), summing up the resulting product for s = 1, 2, . . . , N and then integrating by parts, we have

Z 2

  1 d Z ~ 

~   2 ~ ~ ~ · H ~  dx = 0. ∇ZN (·, t) dx + α2 ZN × 4ZN + HN − Z Nt N 2 2 dt Ω Ω

(5.6.29)

Landau–Lifshitz–Maxwell Equations

315

  Multiplying (5.6.17) by β~sN (t) and (5.6.18) by ~γsN (t), summing up, and integrating by parts, we obtain

 2 2  1 d Z ~  ~ HN (·, t) + EN (·, t) 2 dt Ω

+σ

Z Z Z ~  2 ~ + β Z ~ · H ~  = 0. E N + P~N t · E N Nt N Ω

Ω

(5.6.30)

Ω

  Multiplying (5.6.19) by (~γsN (t) + ~δsN (t)), summing up, and integrating by parts, we get

1d 2 dt

Z Z 2 ~ ~  ~ ~  2 dx EN (·, t) + PN (·, t) dx + σ EN + P N Ω

Ω

=σ

Z





 ~N P~N · E + P~N dx +

Ω

Z  Ω

 



 ~ N · E ~N ∇×H + P~N dx. (5.6.31)

Putting these equalities together, we have  2  1 d Z ~  ~  ~  ~ EN (·, t) + PN (·, t) + 2 HN (·, t) + EN (·, t) dx 2 dt Ω

+σ

Z Z ~ ~  2 dx + σ E ~  2 dx EN + P N N Ω

+ 2β =σ

Z

Ω

Z

Ω

P~N

Ω

~ · H ~  dx + Z Nt N

·



 ~N E

+

P~N



Z

Ω

~  dx P~N t · E N

dx +

Z  Ω



~ N · P~N dx. ∇×H

(5.6.32)

 0 Multiplying (5.6.19) by ~δsN (t), summing up the product for s = 1, 2, . . . , N and integrating by parts, one has, by noticing that P~N is periodic,

1d 2 dt

Z Z 2 λ2 d ~ ~  (·, t) 2 dx PN t (·, t) dx + ∇ × P N

2 dt Ω Z Z 2  d ~ ~  2 + ∇PN (·, t) dx + µ PN t dx 2 dt Ω Ω Z Z   0  ~ · P~ dx − 2ν Φ (|P~ |2 )P~  · P~  dx. =ν E N Nt N N Nt Ω

Ω

(5.6.33)

Ω

From (5.6.29) and (5.6.32) (multiplying (5.6.32) by δ0 , chosen later), it follows that 1d 2 dt

Z  2 2  ~ ~ ~ ~  ~ ∇ZN (·, t) + δ0 EN (·, t) + PN (·, t) + 2δ0 HN (·, t) + δ0 EN (·, t) Ω

+ δ0 σ

Z Z

  ~  2

~ ~ + H ~ 

2 + (2βδ0 − 1) Z ~ · H ~  dx E N + α 2 Z × 4 Z N N N Nt N 2

Ω

Z Z ~  2 ~ ~  dx + δ0 σ EN + PN + δ0 P~N t · E N Ω

= δ0 σ

Z

Ω

Ω

Ω





~  + P~  + δ0 P~N · E N N

Z  Ω



~  · P~  . ∇×H N N

(5.6.34)

Landau–Lifshitz Equations

316

R  ~  ·H ~  dx, we multiply (5.6.17) by (2βδ0 −1)~ In order to deal with the term Ω Z αsN Nt N and sum up the product for s = 1, 2, . . . , N to obtain Z

(2βδ0 − 1)

Ω

~ ·Z ~  + (2βδ0 − 1) H Nt N

Z  Ω



~ · Z ~  = 0. ∇×E N N

(5.6.35)

Adding (5.6.34) and (5.6.35), one gets 1d 2 dt

Z  2 2  ~ ~ ~ ~  ~ ∇ZN (·, t) + δ0 EN (·, t) + PN (·, t) + 2δ0 HN (·, t) + δ0 EN (·, t) dx Ω

Z

  d

~   2 ~ ~ ~ · H ~  dx + α2 ZN × 4ZN + HN + (2βδ0 − 1) Z N N

dt

2

Ω

Z Z Z ~ ~  2   2 ~N ~ dx = −δ0 σ EN + PN dx − δ0 σ EN dx − δ0 P~N t · E Ω

Z

+ δ0 σ

+ δ0



~  + P~  dx − (2βδ0 − 1) P~N · E N N

Ω

Z  Ω

Ω

Ω





~  · P~  dx. ∇×H N N

Z  Ω



~ · Z ~  dx ∇×E N N (5.6.36)

Putting (5.6.33) and (5.6.36) together, we have 2 2 1 d Z h ~  ~ ~  ~ ∇ZN (·, t) + δ0 E (·, t) + P (·, t) + 2δ H (·, t) 0 N N N 2 dt Ω i ~  (·, t) + P~  (·, t) 2 + λ2 ∇ × P~  (·, t) 2 +  ∇P~  (·, t) 2 + δ0 E N

Nt

N

N

  d ~ N · H ~ N + α2

Z ~ N × 4Z ~ N + H ~ N

2 Z 2 dt Ω ! !

2 σ 2 δ02 (δ0 − ν)2

~ 

2

2 2

PN t + 8ν C0 + + δ0 σ P~N ≤ µ+2+ 2 2 4 4 Z

+ (2βδ0 − 1)

~ 

2 + δ0 σ

E ~  + P~ 

2 +

H ~ 

2 + |3 − σδ0 |

E N N N N

2



2



2

(1 − 2βδ0 )

~ 

2 δ02

~ 

2 , +

∇Z N +

curl P N

4

4

2

2

2

where C0 is given by (5.6.5). Integrating the above inequality with respect to t, we obtain

2

1

~  δ0

~ 

~  (·, t)

2 + δ0

H ~  (·, t)

2 + δ0

∇ZN (·, t) +

EN (·, t) + P N N 2 2 2 2 2 2 2



1 λ

2 ~  (·, t)

2 + 

∇P~  (·, t)

2

curl P + P~N t (·, t) + N N 2 2 2 2 2 2

+ (2βδ0 − 1)

Z

Ω

~ · H ~  dx + α2 Z N N

2

2

2

0

2

2

1



2

~ ~ + P~1 + |2βδ0 − 1| H 0 Z 0

2

2

2

2

Z t  

~ ~ + H ~ 

2 dt

Z N × 4 Z N N

2

2 δ0

~

2 1

~

2 δ0

~

~ ~ + ≤ ∇Z

+

E + P

+ δ

H

E 0 0 0 0 0 0

2

2

~

EN (·, t)

2

2

2

Landau–Lifshitz–Maxwell Equations

317

! 2 Z t

2

2

λ2

 (δ − ν) 0



~  2 ~0 + ∇P~0 + µ + 2 + +

curl P

PN t dt 2 2 2 2 2 4 0

+ 8ν

2

σ 2 δ02 + + δ0 σ 4

C02

!Z

t



~  2

PN dt 2

0

Z t Z t

~  2

~ ~N

2 dt + |3 − σδ0 | EN dt + δ0 σ

E N + P 2

0

+

Z

t

(1 − 2βδ0 )2

~  2

HN dt +

4

2

0

2

0

Z

t

0

Z

δ02 t

~  2 ~ 

2 dt.

∇ZN dt +

curl P N

4

2

2

0

Therefore we have

2



~

~ ~  (·, t)

2 + 2δ0

H ~  (·, t)

2

∇ZN (·, t) + δ0 E (·, t) + P N N N 2

2

2

2

2

~

~

2 ~  (·, t)

2 + δ 0 E (·, t)

+

P (·, t)

+ λ

curl P N Nt N 2 2 2 Z t

2   2

~ ~ N + H ~ N

dt +  ∇P~N (·, t) + 2α2

Z N × 4Z 2 2 0 ! 2 Z t

(δ0 − ν)

~  2 ≤ C7 + 2 µ + 2 +

PN t dt

+ 16ν 2 C02 +

4

0

σ 2 δ02 + 8|δ0 σ| 2

!Z

2

t



~  2

PN dt 2

0

Z t Z t

~  2

~  2 + (|6 − 2σδ0 | + 4|δ0 σ|) EN dt + 2 H N dt 2

0

+

(1 − 2βδ0 ) 2

where

2

Z

0

2

0

t δ02 Z t

~  2 ~ 

2 dt,

∇ZN dt +

curl P N

2

2

(5.6.37)

2

0

~ 2

~

~ 2

~ 2

~ 2 ~ 2 C7 = ∇Z 0 + δ0 E0 + P0 + 2δ0 H0 + δ0 E0 + P1



2



2

2





~ ~ 2 ~ 2 + 2|2βδ0 − 1| H 0 Z0 + λ curl P0 2

2

2



2

2

2 (1 − 2βδ0 )2

~

2

Z 0 . +  ∇P~0 dx +

δ0

2

2

On the other hand, we have

2

2

2

~

~

 ~N

PN (·, t) − 2 E (·, t) ≤ 2 P~N (·, t) + E (·, t) N 2 2 2

2

~  (·, t)

. ≤ 3 P~N (·, t) + E N

(5.6.38)

2

Taking δ0 = 3, we get from (5.6.37) and (5.6.38) that

2

2

2

2

2

~

~

~

~

~

∇ZN (·, t) + P (·, t)

+ 6

H (·, t)

+

E (·, t)

+ 

∇ P (·, t)

N N N N 2 2 2 2 2 Z t

2

2  



~ ~ N + H ~ N

2 dt + P~N t (·, t) + λ2 curl P~N (·, t) + 2α2

Z N × 4 Z 2

≤ C8

Z

t

0

2

2

0











~  2 ~  2 ~  2 ~  2 ~  2 ~ 

2 dt + C7 ,

∇ZN + PN + EN + HN + PN t + curl P N 2

2

2

2

2

2

(5.6.39)

Landau–Lifshitz Equations

318

where C8 = max

(

!

9 9σ 2 (1 − 6β)2 , 16ν 2 C02 + + 24σ , (|6 − 6σ| + 12σ), 2[µ + 2 + (3 − ν)2 ], 2 2 2

)

.

For λ2 > 0 (in fact, λ > 0 denotes the speed of light for the internal field), (5.6.39) combined with Gronwall’s inequality yields (5.6.25). Step 3. By Sobolev imbedding theorem and H¨older inequality, we have (5.6.26). Combining (5.6.36) and (5.6.25), we obtain (5.6.27) if α2 > 0. Lemma 5.6.2 is proved. ~ N , H ~ N , Lemma 5.6.3 Under the condition of Lemma 5.6.2, for the solution (Z ~  , P~  ) of the problem (5.6.16)–(5.6.24), there exist C9 > 0 and C10 > 0, both indeE N N pendent of N, D and , such that (i) when α2 = 0, 



~ sup Z Nt

0≤t≤T

H −2 (Ω)

(ii) when α2 > 0,

~

Z N t





~ + H Nt

3 L 2 (QT )

H −2 (Ω)



~  + H N t



+ P~N tt





~ + E Nt

H −2 (Ω)

L2 (0,T ;H −1 (Ω))

L2 (0,T ;H −2 (Ω))

≤ C10 .





+ P~N tt



~ + E N t

H −2 (Ω)



≤ C9 . (5.6.40)

L2 (0,T ;H −1 (Ω))

(5.6.41)

Remark 5.6.1 This lemma shows that if α2 > 0, then we may get better estimate like the above lemma. Proof. (i) When α2 = 0, for any periodic function ϕ ∈ H02 (Ω), ϕ can be represented by ϕ = ϕ N + ϕN ,

ϕN =

N X

ηs ωs (x),

ϕN =

s=1

For s ≥ N + 1, we have

Z

Ω

∞ X

ηs ωs (x).

(5.6.42)

s=N +1

~  ωs (x)dx = 0 Z Nt

Then, by Lemma 5.6.2, there holds Z Z Z   Z ~  ϕ(x)dx = ~  ϕN (x)dx = α1 Z ~  × 4Z ~ + H ~  ϕN (x)dx Z Nt Nt N N N Ω Ω Ω  

~ N k2 kZ ~ N k6 + kZ ~ N k2 kH ~ N k2 (k∇ϕN k3 + kϕN k∞ ) ≤ |α1 | k∇Z ≤ C11 kϕkH 2 (Ω) ,

where we have used Gagliardo–Nirenberg inequalities 1

3

kϕN k∞ ≤ CkϕN k24 k4ϕN k24 ,

1

1

k∇ϕN k3 ≤ Ck∇ϕN k22 k4ϕN k22 .

Landau–Lifshitz–Maxwell Equations

319

In the similar manner, we have Z H ~ N t ϕ(x)dx ≤ C12 kϕkH 2 (Ω) , ZΩ E ~  ϕ(x)dx ≤ C13 kϕkH 2 (Ω) , Nt Ω Z ~  ϕdx ≤ C14 kϕkH 2 (Ω) , P N tt Ω

where C11 , C12 , C13 and C14 are independent of N , D and . (5.6.40) follows. (ii) Now we assume α2 > 0. For any periodic function ϕ ∈ L3 (QT ), we have ZZ

QT



ZZ

~  ϕdxdt ≤ |α1 | Z N







~  × 4Z ~ + H ~  ϕdxdt Z N N N

Q

ZZT    ~ N × Z ~ N × 4Z ~ N + H ~ N ϕdxdt + α2 Z QT

 

~ ~ ~ ≤ |α1 | Z N × 4ZN + HN L2 (Q ) kϕkL2 (QT ) T

 

~   ~ ~ ~ 

+ α2 kZN kL6 (QT ) ZN × 4ZN + H kϕkL3 (QT ) N 2 L (QT )

≤ C15 kϕkL3 (QT ) .

Similarly, for any periodic function ϕ ∈ L2 (0, T ; H01 (Ω)), using (5.6.42) and Lemma 5.6.2, we get Z Z ~  ϕdx ≤ C16 kϕkH 1 (Ω) ZZ

QT

ZZ

QT





~  ϕdxdt = − H Nt

ZZ



QT



Nt

Ω





~  ϕdxdt − β ∇×E N

≤ C17 kϕkL2 (0,T ;H 1 (Ω)) , ZZ

~  ϕdxdt = E Nt

QT





~  ϕdxdt − ∇×H N

≤ C18 kϕkL2 (0,T ;H 1 (Ω)) .

ZZ

ZZ

QT



~  ϕN dxdt Z Nt

QT

P~N t ϕt dxdt − σ

ZZ

QT



~  ϕN dxdt E N

For any periodic function ϕ ∈ L2 (0, T ; H 2 (Ω)), using (5.6.42) again, we obtain ZZ ZZ ZZ  ~  ϕN dxdt +  ~ E 4P~N ϕN dxdt PN tt ϕdxdt = ν N QT QT QT ZZ

− 2ν −µ

QT

ZZ

QT

Φ0 (|P~N |2 )P~N ϕN dxdt

P~N t ϕN dxdt

−λ

≤ C19 kϕN kL2 (0,T ;H 1 (Ω)) +  + λ2

Z

T 0

Z

2

ZZ

T 0

QT

curl



2 ~ PN ϕN dxdt

kP~N k2 k4ϕN k2 dt

kcurl P~N k2 k∇ϕN k2 dt

≤ C20 kϕkL2 (0,T ;H 2 (Ω)) , where C15 –C20 are independent of N , D and . The proof of Lemma 5.6.3 is complete.

Landau–Lifshitz Equations

320

~ , H ~ ,E ~  , P~  ) Lemma 5.6.4 Under the condition of Lemma 2.1, for the solution (Z N N N N of the problem (5.6.16)–(5.6.24), there exist constants C21 > 0, C22 > 0, C23 > 0, C24 > 0 and C25 > 0, independent of N, D, and , such that (i) When α2 = 0, ~  (·, t1 ) − Z ~  (·, t2 )k2 ≤ C21 |t1 − t2 | 21 , kZ N N    ~ ,E ~ , P~ , P~  ∈ C([0, T ]; H −1 (Ω)). H N

N

N

(5.6.43) (5.6.44)

Nt

(ii) When α2 > 0, 2

~  (·, t1 ) − Z ~  (·, t2 )k3 ≤ C22 |t1 − t2 | 3 , kZ N N

(5.6.45)

~  (·, t1 ) − H ~  (·, t2 )kH −1 (Ω) + kE ~  (·, t1 ) − E ~  (·, t2 )kH −1 (Ω) ≤ C23 |t1 − t2 | 12 , kH N N N N

(5.6.46)

Proof. holds

1 kP~N t (·, t1 ) − P~N t (·, t2 )kH −2 (Ω) ≤ C24 |t1 − t2 | 2 ,

(5.6.47)

1 kP~N (·, t1 ) − P~N (·, t2 )k2 ≤ C25 |t1 − t2 | 2 .

(5.6.48)

(i) When α2 = 0, by the Sobolev interpolation of negative order, there 1

2

~  (·, t1 ) − Z ~  (·, t2 )k2 ≤ C26 kZ ~  (·, t1 ) − Z ~  (·, t2 )k 3 −2 kZ ~  (·, t1 ) − Z ~  (·, t2 )k 3 1 kZ N N N N N N H H

1

Z t ~ 

3

2 ∂Z 1 N ≤ C27

dt

≤ C22 |t1 − t2 | 3 .

t1 ∂t

−2 H

On the other hand, it follows from Lemma 5.6.1 and

L2 (Ω) ,→,→ H −1 (Ω) ,→ H −2 (Ω); ~  ∈ L∞ (0, T ; L2 (Ω)) H N

\

(

∂Ψ Ψ: ∈ L∞ (0, T ; H −2(Ω)) ∂t

that ~  ∈ C([0, T ]; H −1 (Ω)). H N Similarly, we also have ~  , P~  , P~  ∈ C([0, T ]; H −1 (Ω)). E N N Nt

)

Landau–Lifshitz–Maxwell Equations

321

(ii) When α2 > 0, we have

Z t

~ 

2 ∂Z

~

N ~  (·, t2 ) = dt

ZN (·, t1 ) − Z N

3

t1 ∂t 3  ZZ 2 ≤ |t1 − t2 | 3 

QT

2



~ ~  (·, t2 )

HN (·, t1 ) − H N

H −1 (Ω)

≤ C22 |t1 − t2 | 3 ,

Z t ~ N

2 ∂H =

dt

∂t

t1

1 2



 ≤ |t1 − t2 | 

Z

1 ~ 3 ∂ ZN dxdt ∂t

H −1

T

0



~  2

∂ HN



∂t

H −1 (Ω)

1 2

≤ C23 |t1 − t2 | .

1

2

 dt

~  (·, t1 ) − E ~  (·, t2 )|, a similar inequality holds. For |E N N At the same time, we have

~ ~N t (·, t2 )

PN t (·, t1 ) − P

H −2 (Ω)



Z t 2 ~ 

2 ∂ PN dt

=

t1 ∂t2 1 2



≤ |t1 − t2 |  

Z

H −2 (Ω)

T

0

H −2 (Ω)

1 2

≤ C24 |t1 − t2 | ,

Z t

~ 

2 ∂P

~

N ~  (·, t2 ) =

PN (·, t1 ) − P dt

N

2

t1 ∂t



2 ~  2

∂ PN



∂t2

1

2

dt 

2

1



≤ |t1 − t2 | 2  

Z 1 2

0

≤ C25 |t1 − t2 | .

T

1

2

~  2

∂ PN 

dt



∂t 2

Lemma 5.6.4 follows. In fact, it follows from (5.6.25)–(5.6.26) that the solution of ODE (5.6.16)–(5.6.24) does not blow up at any finite time. Hence, from ODE theory, Lemmas 5.6.2–5.6.4, we have the following lemma: Lemma 5.6.5 Under the conditions of Lemma 5.6.2, the initial value problem for the system of the ordinary differential equation (5.6.16)–(5.6.24) admits at least one continuously differentiable global solution  α ~ sN (t),

 β~sN (t),

 ~γsN (t),

~δ (t), sN

(s = 1, 2, . . . , N; t ∈ [0, T ]).

Landau–Lifshitz Equations

322

5.6.4

Existence of Weak Solution for the Viscosity Problem

First of all, similar to Definition 5.6.1, we may define the weak solution for the viscosity problem (5.6.1)–(5.6.3), and (5.6.13) and (5.6.8). In the proof of the following theorem, we must use the following lemma which is well known to all. Lemma 5.6.6 If un → u strongly in L2 (QT ) and vn → v weakly in L2 (QT ), then un vn → uv weakly in L1 (QT ) and in the sense of distribution. ~ 0 (x), H ~ 0 (x), E ~ 0 (x), P~0 (x), Theorem 5.6.1 Assume the 2D-periodic initial data (Z 1 2 2 1 2 ~ P1 (x)) ∈ (H (Ω), L (Ω), L (Ω), H (Ω), L (Ω)). Then the periodic initial value prob~  (x, t), lem (5.6.1)–(5.6.3), (5.6.13), (5.6.8) admits at least one global weak solution Z ~  (x, t), E ~  (x, t), P~  (x, t) such that H (i) When α2 = 0, there hold ~  (x, t) ∈ L∞ (0, T ; H 1(Ω)) C (0, 13 ) (0, T ; L2 (Ω)); Z \ ~  (x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1(Ω)); H \ ~  (x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1(Ω)); E \

P~  (x, t) ∈ L∞ (0, T ; H 1(Ω)) C(0, T ; H −1 (Ω)); \ P~t (x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1(Ω)). \

(5.6.49) (5.6.50) (5.6.51) (5.6.52) (5.6.53)

(ii) When α2 > 0, we have

~  (x, t) ∈ L∞ (0, T ; H 1 (Ω)) Z

~  (x, t) ∈ L∞ (0, T ; L2 (Ω)) H

~  (x, t) ∈ L∞ (0, T ; L2 (Ω)) E

2

C (0, 3 ) (0, T ; L3 (Ω));

(5.6.54)

C

(0, 12 )

(0, T ; H −1(Ω));

(5.6.55)

\

C

(0, 12 )

(0, T ; H −1(Ω));

(5.6.56)

\

C (0, 2 ) (0, T ; H −2(Ω)).

\

P~  (x, t) ∈ L∞ (0, T ; H 1 (Ω))

P~t (x, t) ∈ L∞ (0, T ; L2 (Ω))

\

\

C

(0, 21 )

(0, T ; L2 (Ω));

1

(5.6.57) (5.6.58)

~  (x, t), H ~  (x, t), Proof. The uniform estimates for the approximate solution Z N N ~  (x, t), P~  (x, t) in the above section yield that there is a subsequence of Z ~  (x, t), E N N N ~  (x, t), E ~  (x, t), P~  (x, t), still denoted by Z ~  (x, t), H ~  (x, t), E ~  (x, t), P~  (x, t), and H N N N N N N N ~  (x, t), H ~  (x, t), E ~  (x, t), P~  (x, t), such that Z ~ N (x, t) → Z ~  (x, t), Z

weakly-∗ in L6 (QT );

(5.6.59)

~ N (x, t) → Z ~  (x, t), Z

strongly in L6−% (QT ), (% > 0);

(5.6.60)

~ N (x, t) → Z ~  (x, t), Z

weakly-∗ in L∞ (0, T ; H 1 (Ω));

(5.6.61)

~ N (x, t) → H ~  (x, t), H

weakly-∗ in L∞ (0, T ; L2 (Ω));

(5.6.62)

weakly-∗ in L∞ (0, T ; L2 (Ω));

(5.6.63)

~  (x, t) → E ~  (x, t), E N

Landau–Lifshitz–Maxwell Equations

P~N (x, t) → P~  (x, t),

P~N t (x, t) → P~t (x, t),

curl P~N (x, t) → curl P~  (x, t),

323

weakly-∗ in L∞ (0, T ; H 1 (Ω));

(5.6.64)

weakly-∗ in L∞ (0, T ; L2 (Ω));

(5.6.65)

weakly-∗ in L∞ (0, T ; L2 (Ω));

(5.6.66)

3

~  (x, t) → Z ~  (x, t), Z Nt t

weakly-∗ in L 2 (QT ), (α2 > 0);

(5.6.67)

~  (x, t) → Z ~  (x, t), Z Nt t

weakly-∗ in L∞ (0, T ; H −2(Ω)),

(α2 = 0). (5.6.68)

From Lemma 5.6.1(ii), we get that L∞ (0, T ; H 1(Ω))

\

{ϕ :

∂ϕ ∈ L∞ (0, T ; L2 (Ω))} ∂t

,→,→ C([0, T ]; L2 (Ω)) ⊂ L2 (0, T ; L2 (Ω)). Since P~N is bounded uniformly in L∞ (0, T ; H 1 (Ω)) and ∂t P~N is bounded uniformly in L∞ (0, T ; L2 (Ω)), we deduce that there exists a subsequence of {P~N }, still denoted by {P~N }, such that as N → ∞ P~N (x, t) → P~  (x, t),

strongly in L∞ (0, T ; L2 (Ω)).

(5.6.69)

~ ~ For any vector-valued periodic test function Ψ(x, t) ∈ C 1 (QT ), Ψ(x, T ) = 0, we define an approximate sequence ~ N (x, t) = Ψ

N X

~ηs (t)ωs (x),

s=1

where ~ηs (t) =

R

Ω

~ Ψ(x, t)ωs (x)dx, then

~ N (x, t) → Ψ(x, ~ Ψ t) in C 1 (QT ) and in Lp (QT ), ∀ p > 1.

(5.6.70)

Making the scalar product of ~ηs (t) with (5.6.16), (5.6.17) and eσt ~ηs (t) with (5.6.18), ~ηs (t) with (5.6.19), and summing up the products for s = 1, 2, . . . , N, we get from integration by parts ZZ

QT

~ · Ψ ~ Nt − Z N − α1

ZZ

− α2

ZZ

ZZ

QT

QT

QT

Z

Ω

~  (x, 0) · Ψ(x, ~ Z 0)dx + α1 N

~ × H ~ )·Ψ ~ N + α2 (Z N N 

ZZ

 



QT

~  × ∇Z ~  ) · ∇Ψ ~N (Z N N 



~  × ∇Z ~ · ∇ Z ~ × Ψ ~N Z N N N



~ × H ~ · Z ~ × Ψ ~ N = 0, Z N N N





~  + βZ ~ · Ψ ~ N t dxdt − H N N

+

QT

ZZ

Z  Ω

ZZ

QT







(5.6.71)



~N ·E ~  dxdt ∇×Ψ N

~  (x, 0) + β Z ~  (x, 0) · Ψ ~ N (x, 0)dx = 0, H N N

(5.6.72)

Landau–Lifshitz Equations

324 ZZ

QT

~  + P~  ) · (eσt Ψ ~ N t )dxdt + (E N N +σ

ZZ

~ N dxdt + eσt P~N · Ψ

QT

ZZ

QT

~ N t − λ2 P~N t · Ψ +ν

ZZ

− 2ν

QT

ZZ

ZZ

ZZ

Z

Ω

QT

QT

~  (·, 0) + P~  (·, 0)) · Ψ ~ N (·, 0)dx = 0, (5.6.73) (E N N ~N −µ curl P~N · curl Ψ

 ~N ~N − E ·Ψ

QT

~  dxdt eσt ∇ × ΨN · H N

ZZ

QT

ZZ

QT

~N P~N t · Ψ

~N ∇P~N · ∇Ψ

~N + Φ0 (|P~N |2 )P~N · Ψ

Z

Ω

~ N (·, 0) = 0. P~N t (·, 0) · Ψ

(5.6.74)

~  (x, t), H ~  (x, t), E ~  (x, t), P~  (x, t)) is a Now we are in the position to prove that (Z weak solution of (5.6.1)–(5.6.3), (5.6.13) and (5.6.8). To this aim, one should set N to ∞ in (5.6.71)–(5.6.74). From (5.6.59)–(5.6.70) and Lemma 5.6.6, it suffices to deal with the nonlinear terms in (5.6.71)–(5.6.74). First of all, we are able to prove ZZ

QT

~ N dxdt → Φ0 (|P~N |2 )P~N · Ψ

ZZ

QT

~ Φ0 (|P~  |2 )P~  · Ψdxdt.

(5.6.75)

In fact, using the Lipschitz condition (5.6.6), we get ZZ

QT

~N − Φ0 (|P~N |2 )P~N · Ψ

ZZ ≤

≤ C∗

QT

ZZ

h

ZZ

QT



~ Φ0 (|P~  |2 )P~  · Ψ i

~N + Φ (|P~N |2 )P~N − Φ0 (|P~  |2 )P~  · Ψ

QT

0

ZZ

~ N | + CkP~ kL∞ (0,T ;L2 (Ω)) |P~N − P~  ||Ψ



  ~N −Ψ ~ Φ (|P~  |2 )P~  · Ψ 0

QT

Z

T 0

~ N − Ψk ~ L2 (Ω) kΨ

~ N kL2 (0,T ;L2 ) kP~  − P~  kL2 (0,T ;L2 ) + CkP~  kL∞ (0,T ;L2 ) ≤ C ∗ kΨ N

Z

T 0

~ N − Ψk ~ 2 kΨ

→ 0 (as N → +∞), where we have used (5.6.69). ~  , still denoted by Z ~  , such Secondly, we claim that there exist subsequences of Z N N that, as N → +∞, for i = 1, 2, 3, (1) (2)

~ ~ ~  × ∂Z ~ N × ∂ ZN → Z Z ∂xi ∂xi ~ Z N

~ ∂Z N × ∂xi

!

~ ∂Z → Z × ∂xi ~

xi

3

!

xi

weakly-∗ in L∞ (0, T ; L 2 (Ω)),

(5.6.76)

weakly in L2 (QT ),

(5.6.77)

(α2 > 0).

Landau–Lifshitz–Maxwell Equations

325

~ In fact, for any periodic test function Ψ(x, t) ∈ C 1 (QT ), we obtain ZZ





~ ~ ~  × ∂ Z  · Ψdxdt ~ ~  × ∂ ZN − Z Z N ∂xi ∂xi

QT

=

ZZ



QT

ZZ

+



~ ~ − Z ~  ) × ∂ ZN  · Ψdxdt ~  (Z N ∂xi

QT







~ ~ ∂Z ∂Z ~  · Ψdxdt Z × N − ∂xi ∂xi

~

~ ∂Z ~ ~ − Z ~  kL2 (Q ) ∞ ≤ kΨkL (QT )k N kL2 (QT ) kZ N T ∂xi ZZ

+

QT







~ ~ ~  ×  ∂ ZN − ∂ Z  · Ψdxdt ~ Z → 0, ∂xi ∂xi

(as N → +∞).

Therefore, (5.6.76) is proved. ~  ×(4Z ~  +H ~  )) Now we turn to prove (5.6.77). By Lemma 5.6.2, when α2 > 0, (Z N N N is bounded in L2 (QT ) uniformly with respect to N . Then there exist a subse~  × (4Z ~ + H ~  )), still denoted by (Z ~  × (4Z ~ + H ~  )), and a vector quence of (Z N N N N N N  2 1 ~ (x, t) ∈ L (QT ), such that for any test function Ψ(x, ~ U t) ∈ C (QT ), there holds that as N → +∞ ZZ ZZ   ~  × 4Z ~ + H ~ ·Ψ ~ → ~  · Ψ. ~ Z U N N N QT

QT

On the other hand, as N → +∞, ZZ

QT





~  × 4Z ~ + H ~ ·Ψ ~ Z N N N =−

ZZ

→−

QT

ZZ

QT





~  × ∇Z ~  · ∇Ψ ~ + Z N N





~  × ∇Z ~  · ∇Ψ ~ + Z

ZZ

ZZ

QT

QT





~ × H ~ ·Ψ ~ Z N N 



~ ~ × H ~  · Ψ, Z

where we have used (5.6.76) and the fact that, as N → ∞ ZZ

QT



ZZ

≤ ≤





~ N × H ~ N − Z ~ × H ~ ·Ψ ~ Z

Z

T

0

QT







ZZ

~ − Z ~ × H ~ ·Ψ ~ + Z N N

QT

ZZ

~ − Z ~  k5 kH ~  k2 kΨk ~ 10 dt + kZ N N 3



QT

Then we have ZZ

QT

~ ·Ψ ~ − U

ZZ

QT









~ × H ~ −H ~ ·Ψ ~ Z N

~ × H ~ ·Ψ ~ =− Z



~  × (H ~ −H ~ ) · Ψ ~ → 0. (5.6.78) Z N ZZ

QT





~  × ∇Z ~  · ∇Ψ. ~ Z

Landau–Lifshitz Equations

326

Therefore, one gets in the sense of distribution ~  × 4Z ~  = (U ~  − (Z ~ × H ~  )) ∈ L2 (QT ). Z So (5.6.77) is proved. It remains to prove that ZZ

QT









~  × ∇Z ~ · ∇ Z ~ × Ψ ~N − Z N N N

→

ZZ

QT





ZZ



QT



ZZ

ZZ





~  × ∇Z ~ · ∇ Z ~ × Ψ ~ − Z

 

~ × H ~ · Z ~ × Ψ ~N Z N N N

QT





 



~ × H ~ · Z ~ × Ψ ~ . Z

In fact, we have ZZ

QT









~  × ∇Z ~ · ∇ Z ~ × Ψ ~N − Z N N N −

=

ZZ

− =

ZZ

ZZ

ZZ

+

ZZ

+

ZZ

+

ZZ

QT





QT



~ × Ψ ~ + ~  × ∇Z ~ · ∇ Z Z 

ZZ

 

~  × 4Z ~ + H ~ · Z ~ × Ψ ~N Z N N N N

QT

QT



~ × Ψ ~ ~  × 4Z ~ + H ~ · Z Z

QT

h QT

QT

QT

 





~ × Ψ ~N − Z ~ × Ψ ~ Z N



 h

~ × H ~ · Z N N

. = IN + JN + KN + LN .

~ × Ψ ~ · Z N









 h

~ N × H ~ N − Z ~ × H ~ Z



 



~ × Ψ ~ ~ × H ~ · Z Z

i 

~  × 4Z ~ · Z N N

h

QT







~  × 4Z ~ − Z ~  × 4Z ~ Z N N 

 

~ × H ~ · Z ~ × Ψ ~N Z N N N





i

i 





i

~ N × Ψ ~ · Z 

~ × Ψ ~N − Z ~ × Ψ ~ Z N

From (5.6.77), we get IN → 0 as N → +∞. At the same time, as N → +∞, we have |JN |

~  × 4Z ~  kL2 (Q ) ≤ kZ N N T ≤C

ZZ

QT

ZZ

QT

~ × Ψ ~N −Z ~  × Ψ| ~ 2 |Z N 



~ N × (Ψ ~ N − Ψ) ~ + Z ~ N − Z ~ × Ψ| ~ 2 |Z

1

2

1 2

→ 0.

Similarly, one gets that KN → 0, as N → +∞ and ~  kL2 (Q ) kZ ~  kL4 (Q ) kZ ~ × Ψ ~N −Z ~  × Ψk ~ L4 (Q ) → 0. |LN | ≤ kH N N N T T T Finally, from the above arguments, one may take N → +∞ in (5.6.71)–(5.6.74) ~  (x, t), H ~  (x, t), E ~  (x, t), P~  (x, t)) is a global weak solution of the to obtain that (Z viscosity problem (5.6.1)–(5.6.3), (5.6.13) and (5.6.8). This completes the proof.

Landau–Lifshitz–Maxwell Equations

327

Note that the above a priori estimates are independent of D. By using the diagonal method and letting D → +∞, we can obtain the global existence of weak solution to the Cauchy problem of system (5.6.1)–(5.6.3) and (5.6.13). For simplicity, we do not state the theorem here.

5.6.5

A Priori Estimates Uniform in ε

In Sec. 5, we have obtained a global weak solution for viscosity problem (5.6.1)– (5.6.3), (5.6.13) and (5.6.8) for fixed ε > 0. In this section we will derive the a priori estimates uniform in ε for solutions to viscosity problem. These uniform estimates enable us to pass to the limit ε → 0 and then get the global weak solution to the problem (5.6.1)–(5.6.4) and (5.6.8). We need the following lemmas ~ ∈ Xp (Ω). Lemma 5.6.7 Assume Ω = {x = (x1 , x2 , x3 ); |xi | < D, i = 1, 2, 3}, Q ~ ∈ H 1 (Ω) and there holds Then Q ~ 2 1 = kQk ~ 2 kQk H (Ω) Xp (Ω) . Proof. It follows from the relation ~ = ∇(∇ · Q) ~ − ∇ × (∇ × Q) ~ ∆Q that

Z

~ Q ~ = Q∆ Ω

Z

Ω

~ ~ − Q∇(∇ · Q)

~ implies The periodicity of Q Z

Ω

~ 2= |∇Q|

Z

Ω

Z

Ω

~ 2+ |∇ · Q|

Z

~ × (∇ × Q). ~ Q∇

Ω

~ 2, |∇ × Q|

and therefore we obtain the conclusion of the lemma. From the above estimates and convergence, one easily gets the following. ~ 0 (x), H ~ 0 (x), E ~ 0 (x), P~0 (x), P~1 (x)) ∈ (H 1 (Ω), L2 (Ω), L2 (Ω), Lemma 5.6.8 Assume (Z 1 2 H (Ω), L (Ω)). Then for the solution of the initial value problem (5.6.1)–(5.6.3), (5.6.13) and (5.6.8), there hold following estimates: ~  (·, t)k2 1 + kE ~  (·, t)k2 + kH ~  (·, t)k2 sup [kZ H (Ω) 2 2

0≤t≤T

+ kP~  (·, t)k22 + kcurl P~  (·, t)k22 + kP~t (·, t)k22 ] ≤ M1 , ~  (·, t)k2 ≤ M1 , sup kZ 6

0≤t≤T

~  (·, t) × ∇Z ~  (·, t)k 3 sup kZ ≤ M1 , L 2 (Ω)

(5.6.79) (5.6.80)

0≤t≤T

where the constant M1 is independent of α2 , D, and . When α2 > 0, there is ~  × (4Z ~ + H ~  )kL2 (0,T ;L2 (Ω)) ≤ M2 , kZ where the constant M2 is independent of  and D.

(5.6.81)

Landau–Lifshitz Equations

328

In the following we will prove that ∇P~  is uniformly bounded in L∞ (0, T ; L2 (Ω)). We shall consider the compatibility conditions associated with the viscosity problem given by the following set of equations that hold in the sense of distributions ∂(e + p ) + σe = 0, ∂t ~ ) ∂(h + β∇ · Z = 0, ∂t

(5.6.82) (5.6.83)

 ∂ 2 p ∂p   0 ~ 2  (2) ~  2   ∂Pj , +µ − 4p − νe + 2νΦ (|P | )p = −4νΦ (|P | )Pi Pj ∂t2 ∂t ∂xi

(5.6.84)

where ~  , e = div E ~  , p = div P~  h = div H and Pi is the i-th component of P~  and the relation: 

∂Pj div(P~  Φ0 (|P~  |2 )) = Φ0 (|P~ |2 )p + 2Φ(2) (|P~  |2 )Pi Pj . ∂xi In order to obtain the L2 (Ω) estimate of ∇P~  (·, t), we shall assume that ~ 0, div H

~ 0, div E

div P~0 ,

div P~1 ∈ L2 (Ω).

(5.6.85)

We have the following lemma. Lemma 5.6.9 Under the conditions of Lemma 5.6.8 and assuming that the hypotheses (5.6.85) hold, then for the solutions of the viscosity problem, we have sup k∇P~ (·, t)kL2 (Ω) ≤ M3 ,

(5.6.86)

0≤t≤T

where M3 is independent of D and . Proof. For simplicity we present the proof for the case that ∇(div P~0 ) ∈ L2 (Ω), since the general case of div P~0 ∈ L2 (Ω) can be handled by the modifying technique or the proper approximation of the div P~0 . Multiplying (5.6.82) by 3e and 2(e + p ), we have 3d 2 dt

Z

Z Z d Z  |e + p |2 dx + 2σ |e + p |2 dx − 2σ (e + p )p dx = 0. dt Ω Ω Ω

(5.6.88)

∂p , ∂t

Ω

e

∂p dx + 3σ ∂t

(5.6.87)

Ω

|e |2 dx + 3

Z

|e |2 dx = 0,

Multiplying (5.6.84) by 1d 2 dt

Z

Ω

one gets

Z  2 Z  2 Z Z  d ∂p ∂p ∂p  2 dx + µ dx + |∇p | dx − ν e dx 2 dt Ω ∂t Ω ∂t Ω ∂t Ω Z Z   

+ 2ν

∂p Φ0 (|P~  |2 )p dx + 4ν ∂t Ω

∂Pj ∂p Φ(2) (|P~ |2 )Pi Pj dx = 0. ∂xi ∂t Ω

(5.6.89)

Landau–Lifshitz–Maxwell Equations

329

Combining (5.6.87)–(5.6.89), we obtain 1d 2 dt

Z



 Z Z  2 ∂p 2 ∂p   2  2  2   2 2|e + p | + 3|e | + + |∇p |  + 2σ |e + p | + µ ∂t Ω Ω Ω ∂t Z

= −3σ

 2

|e | dx + 2σ

Z







(p + e )p dx + (ν − 3)

Z

 ∂p



dx ∂t Z Z ∂P  ∂p ∂p − 2ν Φ0 (|P~  |2 )p dx − 4ν Φ(2) (|P~  |2 )Pi Pj j dx ∂t ∂xi ∂t Ω Ω

≤ M3

Z

Ω



Ω 

e

Ω

 Z ∂p 2  |e | + |p | + |∇P~  |2 dx. dx + M4 + M5 ∂t Ω Ω  2

 2

Therefore, we get that 1d 2 dt

Z

 ∂p 2  2 2|e + p |2 + 3|e |2 + + |∇p |  dx ∂t Ω 

≤ M3

Z



 Z ∂p 2  |e |2 + |p |2 + dx + M4 + M5 |∇P~ |2 dx. ∂t Ω Ω

Integrating with respect to t, we have

∂p (·, t)k22 + k∇p (·, t)k22 2k(e + p + 3ke + ∂t   Z tZ Z tZ ∂p 2  |e |2 + |p |2 + ≤ M6 + 2M5 , |∇P~  |2 + 2M3 ∂t 0 Ω 0 Ω 



)(·, t)k22



(·, t)k22

~ 0 + div P~0 k2 + 3kdiv E ~ 0 k2 + kdiv P~1 k2 + k∇(div P~0 )k2 + 2M4 is a where M6 = 2kdiv E 2 2 2 2 constant from hypotheses (5.6.85). On the other hand,we get kp (·, t)k22 = kp (·, t) + e (·, t) − e (·, t)k22

≤ 2kp (·, t) + e (·, t)k22 + 2ke (·, t)k22 .

We obtain that ∂p kp (·, t)k22 + ke (·, t)k22 + (·, t)k22 + k∇p (·, t)k22 ∂t   Z tZ Z tZ ∂p 2  |e |2 + |p |2 + ≤ M6 + 2M5 |∇P~  |2 + 2M3 . ∂t 0 Ω 0 Ω

Landau–Lifshitz Equations

330

By Gronwall inequality we get kp



(·, t)k22

+ ke



∂p + (·, t)k22 ∂t  Z tZ

(·, t)k22



≤ M6 + M 5 ≤ M7 + M8

Z

0

0 Ω tZ Ω

|∇P~  |2 (1 + M3 teM3 t )

|∇P~ |2 .

Therefore, we obtain that kp



(·, t)k22

≤ M7 + M8

Z

t 0

Z

Ω

|∇P~ |2 .

(5.6.90)

Using Lemma 5.6.7 for P~  (x, t), we get k∇P~  (·, t)k22 ≤ M9 (kcurl P~  (·, t)k22 + kdiv P~  (·, t)k22 + kP~  (·, t)k22 ) ≤ M10 + M11

Z tZ 0

Ω

|∇P~  |2 .

By Gronwall’s inequality one gets k∇P~  (·, t)k22 ≤ M12 , where M12 is independent of . Lemma 5.6.9 is proved. Remark 5.6.2 Lemma 5.6.8 and Lemma 5.6.9 show that {P~  } is bounded in L∞ (0, T ; H 1 (Ω)).

5.6.6

Global Existence of Weak Solutions

By a priori estimates uniform in ε obtained in Sec. 5.6.4 for the viscosity problem and passing to the limit ε → 0 in Eqs. (5.6.1)–(5.6.3) and (5.6.13), we can get the global weak solution of problem (5.6.1)–(5.6.4) and (5.6.8). ~ 0 (x), H ~ 0 (x), E ~ 0 (x), P~0 (x), P~1 (x)) ∈ Theorem 5.6.2 Assume the periodic functions (Z 1 2 2 1 2 (H (Ω), L (Ω), L (Ω), H (Ω), L (Ω)) and satisfying (5.6.85). Then the periodic initial value problem (5.6.1)–(5.6.4) and (5.6.8) admits at least one global 2D-periodic weak ~ t), H(x, ~ ~ solution Z(x, t), E(x, t), P~ (x, t) such that (i) When α2 = 0, ~ t) ∈ L∞ (0, T ; H 1 (Ω)) T C (0, 31 ) (0, T ; L2 (Ω)); Z(x, T ~ H(x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1 (Ω)); T ~ E(x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1 (Ω));

T P~ (x, t) ∈ L∞ (0, T ; H 1 (Ω)) C(0, T ; H −1 (Ω)); T P~t (x, t) ∈ L∞ (0, T ; L2 (Ω)) C(0, T ; H −1 (Ω)).

(5.6.91)

Landau–Lifshitz–Maxwell Equations

331

(ii) When α2 > 0, ~ t) ∈ L∞ (0, T ; H 1 (Ω)) T C (0, 23 ) (0, T ; L3 (Ω)); Z(x, T 1 ~ H(x, t) ∈ L∞ (0, T ; L2 (Ω)) C (0, 2 ) (0, T ; H −1 (Ω)); T 1 ~ E(x, t) ∈ L∞ (0, T ; L2 (Ω)) C (0, 2 ) (0, T ; H −1 (Ω));

(5.6.92)

T 1 P~ (x, t) ∈ L∞ (0, T ; H 1 (Ω)) C (0, 2 ) (0, T ; L2 (Ω)); T 1 P~t (x, t) ∈ L∞ (0, T ; L2 (Ω)) C (0, 2 ) (0, T ; H −2 (Ω)).

5.7

Bibliography Comments

In this chapter we mainly introduce the readers to the main results of Landau– Lifshitz–Maxwell equations. Guo and Su [80] first proved the existence of global weak solution to three-dimensional L–L–M equations. They also showed the existence of global smooth solution to two-dimensional L–L–M equations [81, 82]. Meanwhile, Guo and Su obtained the existence and uniqueness of global smooth solution to L–L– M equations without Gilbert damping term and small initial data. Guo and Ding [74] gave the existence of global weak solution of Neumann boundary value problem to L–L–M equations. In [84], Guo and Su studied the existence of global weak solution to L–L–M equations on Riemannian manifold. The Landau–Lifshitz–Maxwell equations coupling with polarizations was first proposed by Ding, Guo, Lin, and Zeng in 2006 which will be published in DCDS 2007.

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Chapter 6 Long Time Behavior of Solutions to the System of Ferromagnetic Spin Chain 6.1

Existence and Stability of Steady State Solutions

6.1.1

One-Dimensional Landau–Lifshitz Equations

Consider the existence and stability of a steady state solution to the one-dimensional Landau–Lifshitz equations  ∂Z  × ∆Z  − λZ  × (Z  × ∆Z).  =Z ∂t

(6.1.1)

The steady state solution solves  × ∆Z  = λZ  × (Z  × ∆Z),  Z

(6.1.2)

 =Z  xx . where ∆Z  × ∆Z,  we have Taking the inner product (6.1.2) by Z  × ∆Z|  2 = 0. |Z

(6.1.3)

 = (Z1 , Z2 , Z3) is a steady state solution of (6.1.1) if and only Thus, it follows that Z  is a solution of Z  × ∆Z  = 0 and ∆Z  is parallel to Z.  It is clear that ∆Z  is if Z  parallel to Z if and only if there exists a real function l(r) such that  = l(r)Z.  ∆Z

(6.1.4)

 and noting |Z|  = 1, we get Multiplying the two sides of (6.1.4) by Z  · ∆Z  = l(r). Z 333

(6.1.5)

334

Landau–Lifshitz Equations

 = 1, i.e. Z  ·Z  = 1, it follows that Z  · ∇Z  = 0. Thus (6.1.5) reads Since |Z|  · ∆Z  = −|∇Z|  2 = l(r). Z

(6.1.6)

 2 into (6.1.4), we have It is clear that l(r) ≤ 0. Substituting l(r) = −|∇Z|   = −|∇Z|  2 Z. ∆Z

(6.1.7)

 one gets Multiplying (6.1.7) by ∇Z, 1 ∂  2 = −|∇Z|  2 = 0.  2 1 ∂ |Z| |∇Z| 2 ∂r 2 ∂r  2 = −l2 < 0 independent of r. Thus (6.1.4) implies This implies that l(r) = −|∇Z|   = −l2 Z. ∆Z

(6.1.8)

Solving (6.1.8), we know that the steady state solutions can be expressed by  Z(r) = C1 exp(ilr) + C2 exp(−ilr),

(6.1.9)

 is a where Cj (j = 1, 2) are two three-dimensional complex vectors. Because Z  can be rewritten as three-dimensional real vector function, Z  2 sin(lr).  =R  1 cos(lr) + R Z  2 | = 1 and R 1 ⊥ R  2.  = 1, we have that |R  1 | = |R Since |Z|  = (Z1 , Z2, Z3 ) is the steady state solution of (6.1.1) if and only if Theorem 6.1.1 Z  1 cos(lr) + R  2 sin(lr),  l (r) = R Z

∀ l ∈ R,

(6.1.10)

 2 | = 1 and R 1 ⊥ R  2.  1 | = |R where |R Remark 6.1.1 By Theorem 6.1.1, we know that the inhomogeneous steady-state solutions of (6.1.1) are independent of λ, i.e., steady state solutions of (6.1.1) with the damping term are absolutely the same as those of (6.1.1) without the damping term.

6.1.2

Stability of the Steady State Solutions

Now we study the linear stability of the steady state solutions of Landau–Lifshitz equation (6.1.1). We know that Eq. (6.1.1) is equivalent to the following equation in the classical sense:  ∂Z  × ∆Z  + λ∆Z  + λ|∇Z|  2 Z.  =Z ∂t

(6.1.11)

335

Long Time Behavior

l = R  1 cos(lr) + R  2 sin(lr). Let Z  = Z  l + V, where Linearize (6.1.11) about Z V is initially, at least, a small perturbation (i.e., V(r, 0) is arbitrarily small). Then  into (6.1.11), and retaining only linear terms in V, we have substituting Z ∂V  l × ∆V + V × ∆Z  l + λ∆V + λ|∇Z  l |2 V + 2λ(∇Z  l · ∇V)Z  l , (6.1.12) =Z ∂t with the condition V(r + (6.1.12). Then we have

2π ) l

= V(r). Let V = d exp(ilmr + Ω(t)) (|d|  1) in

d  l · V)Z l.  l × V + λ(1 − m2 )l2 V + 2ilmλ(∇Z ΩV = (1 − m2 )l2 Z dt

(6.1.13)

In the following, we calculate the last term on right-hand side of (6.1.13). By the  = 1 and Z  =Z  l + V, we have facts |Z|  l · V) = −|V|2 . 2(Z

(6.1.14)

 l (r − π ). Then one has  l = −lZ On the other hand, ∇Z 2l


l r −  l (r) · V(r) = −lZ ∇Z













π π π V r− exp im . 2l 2l 2

(6.1.15)

By (6.1.14) and (6.1.15), we have 











 l · V)Z  l = −il2 mλ|V|2 exp im 2ilmλ(∇Z

π  Zl . 2

2  l (r) · V(r) = − l V r − 2π  exp im π . ∇Z   2 l 2

Thus






(6.1.16)

Substituting (6.1.16) into (6.1.13) and then multiplying it by V and noting (6.1.14), we have




1 π d Ω|V|2 = λ(1 − m2 )l2 |V|2 + iλl2 m|V|4 exp im . dt 2 2

(6.1.17)

Set Ω(t) = Ωr (t) + iΩi (t) and note |V| = |d| exp(Ωr (t)). (6.1.17) implies   d Ωr 

dt d Ω dt i

= λ(1 − m2 )l2 − 12 λl2 m sin( mπ )|d|2 exp(Ωr (t)), 2 = 12 λl2 m cos( mπ )|d|2 exp(Ωr (t)). 2

(6.1.18)

By the first equation of (6.1.18), we have   d Ωr 

dt d Ω dt r

= am ,

m ≡ 0 mod(2),

= am + bm exp(Ωr (t)), m ≡ 1 mod(2),

(6.1.19)

where am = λ(1 − m2 )l2 and bm = − 12 λl2 m(−1)[m/2] |d|2 . Here [m/2] denotes the maximum integer which is less than or equal to m/2.

336

Landau–Lifshitz Equations

In the following, we solve Eq. (6.1.19). First if m = 0, then am = λl2 > 0; thus  l is linearly unstable. Z Second, if m ≡ 0 mod(2) and m = 0, then am = λ(1 − m2 )l2 < 0 and eΩr (t) = am t+Ωr (0)  l is linear ; it is clear that limt→∞ eΩr (t) = 0. Thus the steady state solution Z e stable. Third, if m ≡ 1 mod(2), by the second equation of (6.1.19), we have that dΩr (t) = dt; am + bm e2Ωr (t) integrating the identity given above implies 

 t 1 − ln am + bm e2Ωr (t) am 2am

t 0

= t;

solving the above identity yields eΩr (t) =



1/2 1 (am + bm e2Ωr (0) )e2(1−am )t − am . bm

(6.1.20)

 l is By (6.1.20), it follows that limt→∞ eΩr (t) = ∞; thus the steady state solution Z linearly unstable.  l (r), (l = 0), is hyperTheorem 6.1.2 The inhomogeneous steady state solutions Z bolic. Moreover, we have u  (Zl ) = span{d exp(ilmr + Ω(t)); m ≡ 1 mod(2) and m = 0}; Wloc s  Wloc (Zl ) = span{d exp(ilmr + Ω(t)); m ≡ 0 mod(2) but m = 0}.

6.2 6.2.1

Asymptotic Behavior of L–L Equations Estimates for Energy

First, we introduce an energy function  E(Z(t)) =

1 2

 Ω

 t)|2 dr, |∇Z(r,

where Ω = [0, 2π]. In the following we give the properties of the energy function.  t) is a solution of (6.1.1), then Proposition 6.2.1 If Z(r,   dE(Z(t))  × ∆Z|  2 dr ≤ 0; (i) = −λ Ω |Z dt  Z(t))  is the steady state Z  l (r); = 0 if and only if Z (ii) dE(dt 2  l (r)) = πl ; (iii) E(Z   = R (R is a constant (iv) the energy function E(Z(t)) arrives minimum at Z vector with |R| = 1).

337

Long Time Behavior

 and integrating it over Ω, we get Proof. (i) Multiplying (6.1.1) by ∆Z −

 1d   2 dr = −λ ∆Z  · [Z  × (Z  × ∆Z)]dr.  |∇Z| 2 dt Ω Ω

 · [Z  × (Z  × ∆Z)]  = (∆Z  × Z)  · (Z  × ∆Z)  = −|Z  × ∆Z|  2 , (i) holds. Since ∆Z   = Z  l (r) and (i), then |Z  × ∆Z|  = 0, and dE(Z(t)) (ii) If Z = 0. On the other dt  dE(Z(t))  t1 ) = Z  l(t1 ) (r) and then by the periodicity of the hand, if dt = 0, we can get Z(r, solution, it follows that l(t1 ) is an integer. Thus the item is proved.  l (r)) = l2 π. Thus (iv) is also proved. (iii) It is clear that E(Z  t) is a solution of (6.1.1), then Lemma 6.2.1 If Z(r,  t) ≤ ∇Z(·,  0) .

∇Z(·,  Lemma 6.2.2 If Z(r) is a 2π-periodic function, then  2.  2 ≤ 2π 2 ∆Z

∇Z

Proof. Note the following identity:  y 

2 2     + |∇Z(y)| − 2∇Z(x) · ∇Z(y) =  |∇Z(x)|

x

2 

 ≤ 2π  ∆Z(r)dr 

 Ω

2  |∆Z(r)| dr.

Integrating twice over Ω, we get  2−2 4π ∇Z


 Ω

2

 ∇Z(x)dx

 2. ≤ (2π)3 ∆Z

   By the periodicity of Z(r), it follows that Ω ∇Z(x)dx = 0; thus the lemma holds. 1 (Ω)) and Lemma 6.2.3 If u(r, t) ∈ C 0 (R+ ; Hper



lim

t→∞ Ω

|∇u(r, t)|2dr = 0,

then ∀ r ∈ Ω lim u(r, t) = V,

t→∞

where V is a constant. Proof. For any x, y ∈ Ω, it is clear that  x

u(x, t) = u(y, t) +

y

∇u(r, t)dr.

Taking the absolute value above, we have |u(x, t)| ≤ |u(y, t)| +

 Ω

|∇u(r, t)|dr ≤ |u(y, t)| +

√



2π

Ω

|∇u(r, t)|2 dr.

338

Landau–Lifshitz Equations

By the hypothesis of the lemma, lim |u(x, t)| ≤ lim |u(y, t)|.

t→∞

(6.2.1)

t→∞

Similar to the proof as above, we have lim |u(y, t)| ≤ lim |u(x, t)|.

t→∞

(6.2.2)

t→∞

Thus by (6.2.1) and (6.2.2) and the continuity of u(r, ·) about t, the proof of the lemma is complete.  0) satisfies Theorem 6.2.1 Assume that the initial data Z(r, 1−δ √ . K1 K22 2π

 0) ≤

∇Z(·,

(6.2.3)

Then there exists a constant vector R such that  t) = R, lim Z(r,

(6.2.4)

t→∞

where δ ∈ (0, 1), |R| = 1 and K1 , K2 are the constants in the Sobolev inequalities, respectively. Proof. By (i) of Proposition 6.2.1, we have 1d  × ∆Z

 2 = −λ  2 = −λ Z

∇Z

2 dt

 Ω

 2 dr + λ |∆Z|

 Ω

 · ∆Z)  2 dr. (Z

(6.2.5)

 · ∆Z  = −|∇Z|  2 ; thus It is clear that Z 

λ

Ω

 · ∇Z)  2 dr = λ (Z

 Ω

 4 dr. |∇Z|

 = 1, we have By Sobolev’s inequality and |Z|

√   3 ≤ K1 K 2 Z

∇    2 = K1 K 2 2π ∇Z

∆   2.  4 4 ≤ K1 ∆Z

∇ Z

Z

∆ Z

Z

∇Z

L 2 2 (6.2.6)

Substituting (6.2.6) into (6.2.5), we have √ 1d  2 + λ(1 − K1 K 2 2π ∇Z ) ∆   2 ≤ 0.

∇Z

Z

2 2 dt

(6.2.7)

By the hypothesis of this theorem and Lemma 6.2.2, it follows that d  2 + 2λδ ∆Z

 2 ≤ 0.

∇Z

dt

(6.2.8)

d  2 + λδ ∇Z

 2 ≤ 0.

∇Z

dt π2

(6.2.9)

By Lemma 6.2.2, it yields

339

Long Time Behavior

Then by Gronwall’s inequality, it yields 2 2   ≤ ∇Z(0)

exp(−λδt/π 2 );

∇Z(t)

(6.2.10)

 2  = 0, i.e., limt→∞ Ω |∇Zj |2 dr = 0 (j = 1, 2, 3). By Lemma thus limt→∞ ∇Z(t)

6.2.3, it yields  t) = R. lim Z(r, t→∞

This completes the proof of the theorem. Theorem 6.2.2 Under the hypothesis of Theorem 6.2.1, then R=

1 2π



 0)dr + 1 r, Z(r, 2π Ω

(6.2.11)

 2 Zdrdτ  t) and r = λ  ∞  |∇Z|  where R = limt→∞ Z(r, satisfies Ω 0 |r| ≤

(1 − δ)2 ; 4πK12 K24 δ

(6.2.12)

here δ ∈ (0, 1) is defined in Theorem 6.2.1. Proof. Integrating (6.1.1) over Ω, we have d dt

 Ω

 t)dr = Z(r,

 Ω

 × ∆Z)dr  (Z +λ



 +λ ∆Zdr

Ω

 Ω

 2 Zdr.  |∇Z|

 × ∆Z  = ∇(Z  × ∇Z),  Since Z  Ω

then

 × ∆Z)dr  (Z = 0,



λ

Ω

 = 0; ∆Zdr

 d    2 Zdr.  Z(r, t)dr = λ |∇Z| dt Ω Ω

It follows that  Ω

 t)dr = Z(r,

 Ω

 0)dr + λ Z(r,

 t 0

Ω

 2 Zdrdτ  |∇Z|

(6.2.13)

and |r| ≤ λ lim

 t

t→∞ 0

Ω

2  2 drdτ ≤ λ ∇Z(0)

 |∇Z| lim

 t

t→∞ 0

exp(−2λδτ )dτ =

1 2 

∇Z(0)

. 2δ

From the hypothesis of the theorem, we observe that |r| satisfies the condition. This completes the proof of the theorem.

340

6.2.2

Landau–Lifshitz Equations

Bifurcation and Chaos

1. Magnetization equations, chaos and bifurcation Recently, Gibson and Jeffries [69] completed a series of experiments on ferromagnetic resonance in which the signal power exceeds the second order Sullo unstable threshold. When the signal power increases there appear limit cycles, double periodic bifurcations, chaos, and periodic windows. In earlier time, Nahwmuha et al. [121] considered the chaos formed from two inhomogeneous spin modes, taking the parallel pump into account. For the transverse pump, if the signal power increases, there are also spin wave double bifurcations and chaos with periodicity 3 and 5 as observed in the GJ experiment. The magnetization equation is dm   − αm  = −γ m  ×H  × (m  × H) dt

(6.2.14)

 consists of three in which the Gilbert damping term has been considered. Field H parts: applied field, diplar field, and exchange field. Writing Eq. (6.2.14) in the back coefficient form and retaining the long terms, we have  dB  dτk  dB0 dτ

∗ = iB02 B−k + iΓ|B0 |2 Bk − ηBk ,

(6.2.15)

= iBk2 B0∗ − iΩs − ηB0 ,

where B0 , Bk are the oscillations of the spin wave, η is the damping constant, dτ = 0.5Ωm dt, and Γ is a constant proportional to Ωm . Ωm is the magnetization, and Ωs is the strength of the driving field. Suppose Bk = B−k , η = 0.005, and Γ = 0.5. Equations (6.2.15) and its conjugate equations form 4 ordinary equations whose fixed points are 

  |B0 | 

1

|B0 | = Ωs /η, |Bk | = 0.

(6.2.16)

1

= η 2 /(1 − Γ2 ) 4 , 1

2

1

1

1

|Bk | = [−η(1 − Γ2 ) 2 + { Ωηs (1 − Γ2 ) 2 − η 2 Γ2 } 2 ] 2 . 1

(6.2.17) 1

When Ωs < η1 (η/Ωm ) 2 (Sullo threshold), (6.2.17) is stable; when Ωs > η1 (η/Ωm ) 2 (Sullo threshold), (6.2.11) is unstable, but (6.2.17) is stable. When |Bk |  |B0 |, the condition of stability for fixed point (6.2.17) is (λ + η)3 − (λ + η)[(1 − Γ2 )|B0 |4 + 4Γ|Bk |2 |B0 |2 − 8|Bk |2 |B0 |2 − 4Γ2 η|Bk |2 |B0 |2 ] = 0,

(6.2.18)

where λ is the Lyapunov index. When |Bk |2 ≤ 0.1|B0 |2 , λ must be negative. If |Bk | ∼ |B0 |, the computation is very complicated.

Long Time Behavior

341

Figure 6.2.1. (a) Projection of the first limit cycle on the plan Re Be − Im Be , ws = 1.518 × 10−3 , f = 1.087 × 10−3 wm ; (b) twisting and expansion when ws = 0.0022, f = 1.470 × 10−3 wm ; (c) double period, ws = 0.0045, f = 0.018 × 10−3 wm ; (d) period 4, ws = 0.0055, f = 0.018 × 10−3 wm ; (e) chaos, ws = 0.005.

342

Landau–Lifshitz Equations

Figure 6.2.2. Period 3 to period 6 (a) window of period 3, ws = 0.00625, f = 0.085 × 10−3 wm ; (b) bifurcation of period 6, ws = 0.0065, f = 0.442 × 10−3 wm ; (c) window of period 5, ws = 0.0090, f = 0.636 × 10−3 wm . Ignore the choas between window of period 5 and window of period 6.

Long Time Behavior

343

Figure 6.2.3. Limit cycle (2) (a) projection of limit cycle on plan n0 − nh , ws = 0.00155, f = 1.068 × 10−3 wm ; (b) twisting and expansion when ws = 0.0022, f = 1.240 × 10−3 wm ; (c) double period ws = 0.0260.

344

Landau–Lifshitz Equations

2. Number of magnetic waves It follows from (6.2.14) that d (|B0 |2 + |Bk |2 ) = −αΩs Im B0 − 2η(|B0|2 + |Bk |2 ). (6.2.19) dt |Bk |2 and |B0 |2 can be explained as the number of magnetic wave without damping and driving. The number of magnetic wave is conservative which can be taken as the Hamiltonian system. If (6.2.17) hold, the right-hand side of (6.2.19) becomes zero; If the parcels of the magnetic wave increase (or decrease) such that driving increases (or decrease), the steady state becomes unstable. From this point of view and the fact Re B0 ∼ 0, we see that when Ωs > 4η|Im B0 | the fixed point (6.2.17) becomes unstable. Thus we have Ωs,critical = 6.1518 × 10−3 . At this time, the fixed point is the bifurcation indicated by Jeffries. The frequency of the limit cycle is, from numerical computation, f = 6.1087 × 10−3 Ωm . In the GJ experiment, 4πΩs = 300G. Hence, f = 5543 ×106 Hz. Figures 6.2.1–6.2.3 give out the limit cycle, double periodic bifurcation, and periodicities 3, 4, 5, and 6 with different values of f and Ωs .

6.3 6.3.1

Approximate Inertial Manifold for One-Dimensional L–L Equations One-Dimensional Landau–Lifshitz Equations

Consider the following one-dimensional Landau–Lifshitz equations: ∂u = −αu × (u × uxx ) + βu × uxx , Ω × R+ ∂t with the initial data and periodic boundary conditions as follows: u(x, 0) = u0 (x), x ∈ Ω = (−D, D),

(6.3.1)

|u(x0 )| = 1, 0 < D < ∞,

u(x − D, t) = u(x + D, t),

t ∈ R+ .

x ∈ R,

(6.3.2) (6.3.3)

Here u = (u1 , u2 , u3) : R × R+ → R3 is the spin vector, α, β are constants, and α > 0 denotes the Gilbert damping constant. As we know, for u(x0 ) ∈ H 2 (Ω), |u(x0 )| = 1, problem (6.3.1)–(6.3.3) admits a unique solution u(t) such that u(x, t) ∈ L∞ (R+ ; H 2(Ω)). Lemma 6.3.1 Suppose |u(x0 )| = 1, x ∈ R. Then for the solution u(t) to problem (6.3.1)–(6.3.3); |u(x, t)|2 = 1, 2

(x, t) ∈ R × R+ , 2

ux (t) ≤ ux (0) , hold.

t∈R

+

(6.3.4) (6.3.5)

345

Long Time Behavior

Proof. Multiplying (6.3.1) by u, we have ∂ |u(x, t)|2 = 0, (x, t) ∈ R × R+ . ∂t From this, (6.3.4) follows. Taking the inner product of (6.3.1) with uxx , we arrive at 1d −

ux 2 = −α 2 dt = −α





=α

Ω



Ω

Ω

(u × (u × uxx )) · uxx dx (uxx × u) · (u × uxx )dx

|u × uxx |2 dx.

Hence

d

ux 2 ≤ 0. dt And (6.3.5) follows. Thus the proof is complete. We now introduce subset Hρ of H = L2 (Ω) Hρ = {u ∈ H : |u(x)| = 1, ux ≤ ρ}. From Lemma 6.3.1 we assert that Hρ is an invariant subset of H. Hereafter, we assume that u0 ∈ Hρ . Lemma 6.3.2 Suppose that the assumptions of Lemma 6.3.1 hold. Then u(x, t) is a classical solution to problem (6.3.1)–(6.3.3) if and only if u(x, t) is a solution to the following problem: ∂u = αuxx + βu × uxx + α|ux|2 u, ∂t u(x, 0) = u0 (x), |u0 (x)| = 1, x ∈ Ω, u(x − D, t) = u(x + D, t),

x ∈ R,

(6.3.6) (6.3.7) +

t∈R .

(6.3.8)

Lemma 6.3.3 Suppose that u0 ∈ H 2 ∩ Hρ . Then for solution u(t) to problem (6.3.6)–(6.3.8), the following estimate holds:

uxx (t) ≤ C1 ,

 t+1 t

uxxx (t) 2 dt ≤ C1 , t ≥ t1 .

(6.3.9)

Here constant C1 depends on (α, β, ρ, Ω), and t1 depends on (α, β, ρ, Ω) and R,

u0 H 2 ≤ R. Proof. Since |u(x, t)| = 1, we assert that if |ux | = 0, then u, ux , and u × ux form the orthogonal base of R3 . Let uxx = α1 u + α2 ux + α3 u × ux . It is easy to see that α1 = −|ux |2 ,

α2 =

ux · uxx , |ux |2

α3 =

(u × ux ) · uxx . |ux |2

346

Landau–Lifshitz Equations

Differentiating (6.3.6) with respect to x twice, and taking inner product with uxx , we have 1d

uxx 2 = 2 dt



uxx · (αuxx + βu × uxx + α|ux |2 u)xx dx

Ω

2

= −α uxxx − β −α

 Ω

 Ω

(ux × uxx ) · uxxx dx

uxxx · (|ux |2 ux + 2u(ux · uxx ))dx.

(6.3.10)

Differentiate (6.3.4) with respect to x to get 3 u · uxxx = − (|ux |2 )x . 2

(6.3.11)

Using this we can handle the right-hand side of (6.3.10) as follows: 

Ω

 Ω

uxxx · (|ux |2 ux )dx = −

uxxx · (u(ux · uxx ))dx =



 Ω

Ω

Ω

(ux × uxx ) · uxxx dx = =

Ω

Ω

+ 

=

Ω

− 

=



  

Ω



Ω

|ux · uxx |2 dx,

(6.3.12)

(uxxx · u)(ux · uxx )dx

= −3 

|ux |2 |uxx |2 dx − 2

Ω

|ux · uxx |2 dx, 

(u × ux ) · uxx ux × − |ux | u + (u × ux ) |ux |2

(6.3.13) 

2

· uxxxdx

|ux |2 (u × ux ) · uxxx dx  Ω

(u × ux ) · uxx (ux × (u × ux )) · uxxx dx |ux |2

|ux |2 (u × ux ) · uxxx dx 3 2

 Ω

(|ux |2 )x (u × ux ) · uxx dx

|ux |2 (u × ux ) · uxxx dx

3 |ux |2 (u × ux )x · uxx dx + 2 Ω 3 |ux |2 (u × ux ) · uxxxdx + 2 Ω  5 |ux |2 (u × ux ) · uxxx dx. = 2 Ω

(6.3.14)

It follows from (6.3.10) and (6.3.12)–(6.3.14) that   1d

uxx 2 + α uxxx 2 = α |ux |2 |uxx |2 dx + 8α |ux · uxx |2 dx 2 dt Ω Ω  5 − β |ux |2 (u × ux ) × uxxx dx. 2 Ω

(6.3.15)

347

Long Time Behavior

Moreover, 1d 4 dt

 Ω



|ux |4 dx =



=

Ω Ω

|ux |2 ux · (αuxx + βu × uxx + α|ux |2 u)x dx |ux |2 ux · (αuxxx + βux × uxx + βu × uxxx

+ α|ux |2 ux dx + 2αu(ux · uxx ))dx 

= −α

Ω

|ux |2 |uxx |2 dx − 2α



+α and





6

|ux | dx − β

Ω

1d −β |ux | (u × ux ) · uxxx dx = 4 dt Ω 2



+ 2α



Ω 2

Ω

|ux | (u × ux ) · uxxx dx

4

Ω

|ux | dx + α



Ω

|ux · uxx |2 dx

 Ω

|ux |2 |uxx |2 dx 

2

|ux · uxx | − α

Ω

|ux |6 dx.

(6.3.16)

Inserting (6.3.16) into (6.3.15), one has d 5d

uxx 2 − dt 4 dt 

= 7α Since

4

Ω

Ω



2

|ux | |uxx | dx + 26α

Ω

|ux |2 |uxx |2 dx + 26α 

≤ 33α ≤ 5α ≤ 5α ≤ 5α



2

|ux | dx + 2α uxxx + 5α

Ω 2



7α



Ω



Ω

Ω

|ux · uxx |2 dx.

Ω Ω

(6.3.17)

|ux · uxx |2 dx

|ux | dx + C

 Ω

|uxx |3 dx 7



|ux |6 dx

|ux |2 |uxx |2 dx 6





Ω

5

|ux |6 dx + C uxxx 4 ux 4 |ux |6 dx +

α

uxxx 2 + C. 2

We can rewrite (6.3.17) as follows: d 5d

uxx 2 − dt 4 dt



3 |ux |4 dx + α uxxx 2 ≤ C. 2 Ω

(6.3.18)

By the Poincar´e inequality, we have α

uxxx ≥ K uxx 2 , 2 since



Ω



uxx dx = 0. We infer from (6.3.18) that

5 d

uxx 2 − dt 4







5 |ux | dx + K uxx − 4 Ω 4

2

 Ω

 4

|ux | dx + α uxx 2 ≤ C. (6.3.19)

348

Landau–Lifshitz Equations

Whence, 

d 5

uxx 2 − dt 4







5 |ux | dx + K uxx − 4 Ω 4

2

 Ω

 4

|ux | dx ≤ C.

Using the Gronwall inequality,

uxx (t) 2 − 

5 4

 Ω

|ux (t)|4 dx

5 ≤ uxx (0) − 4 2

2 −Kt

≤ uxx (0) e 2 −Kt

where t∗ =

1 K

≤R e

+ C,

≤ 2C,

t ≥ t∗ ,

 Ω

 4

|ux (0)| dx e−Kt +

C K

+C t≥0 (6.3.20)

2

ln RC . By the interpolation technique, one has 3

1

ux 4 ≤ C ux 4 uxx 4 . Hence, 5

ux 44 ≤ C ux 3 uxx

4 ≤ C uxx

1 ≤ uxx 2 + C. 2

(6.3.21)

From (6.3.20) and (6.3.21), it follows that

uxx (t) ≤ C,

t ≥ t∗ .

(6.3.22)

Using (6.3.22), and integrating (6.3.19) with respect to t over (t, t + 1), we arrive at  t+1 t

uxx 2 dt ≤ C,

∀ t ≥ t∗ .

Thus the proof of the lemma is complete. Lemma 6.3.4 Assume that u0 ∈ H k+1 ∩ Hρ , k ≥ 1. Then

 t+1 t

Dxk+1u(t) 2 ≤ Ck ,

t ≥ tk ,

Dxk+2u(t) 2 dt ≤ Ck ,

t ≥ tk ,

where constant Ck depends only on the initial data and k, and tk depends only on the initial data, k and R; u0 H 2 ≤ R.

Long Time Behavior

349

Proof. We apply the induction argument to prove this lemma. (i) For k = 1, invoking Lemma 6.3.3, we assert that the results of Lemma 6.3.4 are valid. (ii) Assume Lemma 6.3.4 is valid for k − 1. We intend to prove that Lemma 6.3.4 is valid for k ≥ 2. Differentiating (6.3.6) with respect to x and (k + 1) times, taking the inner product with Dxk+1 u, we have 1d

Dk+1 u 2 + α Dxk+2u 2 2 dt x =β

k+1 



i Ck+1

Ω

i=0



k+1 

+α

(Dxk+1−i u × Dxi uxx ) · Dxk+1udx

i Ck+1

Ω

i=0

(Dxi |ux |2 (Dxk+1−iu · Dxk+1 u)dx.

From a simple calculation, one gets β

k+1 

 i Ck+1

i=0

Ω

(Dxk+1−iu × Dxi uxx ) · Dxk+1 udx

k = −βCk+1

+β

k−1 

 Ω



k + βCk+1

−β

Ω

Ω

Ω

(Dxk+1−iu × Dxi uxx ) · Dxk+1 udx

(ux × Dxk+2u) · Dxk+2udx

(ux × Dxk+2u) · Dxk+1udx

≤ C Dxk u ∞ +C



i Ck+1

i=2



(Dxk u × uxx ) · Dxk+2udx

k−1 



Ω

|uxx ||Dxk+2u|dx

Dxk+1−iu ∞

i=2

+ C ux ∞

 Ω

 Ω

|Dxi+2 u||Dxk+1u|dx

|Dxk+2u||Dxk+1u|dx

≤ C Dxk u ∞ uxx

Dxk+2 u

+C

k−1 

Dxk+1−iu ∞ Dxi+2 u

Dxk+1u

i=2

+ C ux ∞ Dxk+2 u

Dxk+1u

≤ C Dxk+1 u

uxx

Dxk+2u

+C

k−1 

Dxk+2−iu

Dxi+2u

Dxk+1u

i=2

+ C uxx

Dxk+2u

Dxk+1u

(6.3.23)

350

Landau–Lifshitz Equations

≤ C Dxk+1 u

Dxk+2u + C

k−1 

Dxi+2u

Dxk+2u

i=2

+ C Dxk+1u 2

≤

C Dxk+1 u

Dxk+2u

≤

α k+2 2

D u + C Dxk+1u 2 . 4 x

(6.3.24)

By the Leibniz formula, we have k 

Dxk |ux |2 =

Ckj Dxk−j ux · Dxj ux

j=0 k−1 

=

Ckj Dxk−j+1u · Dxj+1 ux + 2Dxk+1u · Dx u,

(6.3.25)

j=1 k+1 

Dxk+1 |ux |2 =

j Ck+1 Dxk+1−j ux · Dxj ux

j=0 k−1 

=

j Ck+1 Dxk−j+2u · Dxj+1 u

j=2 1 Dxk+1u · Dx2 u. + 2Dxk+2 · Dx u + 2Ck+1

Then α

k+1 

 i Ck+1

i=0



=α +

Ω

Ω

Dxi |ux |2 (Dxk+1−i u · Dxk+1 u)dx

|ux |2 (Dxk+1u · Dxk+1u)dx

+α

k−1 



k + αCk+1



+α

Ω

i Ck+1

i=2

≤



1 2αCk+1

Ω

(ux · uxx )(Dxk u · Dxk+1u)dx  Ω

Dxi |ux |2 (Dxk+1−i u · Dxk+1 u)dx

Dxk |ux |2 (Dx u · Dxk+1 u)dx

Dxk+1|ux |2 (u

Ω C ux 2∞ Dxk+1u 2

· Dxk+1u)dx

+ C ux ∞ Dxk u ∞ uxx

Dxk+1u

+C

k−1 

Dxi |ux |2

Dxk+1−iu ∞ Dxk+1 u

i=2

+ C ux 2∞ Dxk+1 u 2 +C

k−1  j=1

ux ∞ Dxj+1u ∞ Dxk−j+1u

Dxk+1u

(6.3.26)

Long Time Behavior

+C

k−1 

351

Dxj+1 u ∞ Dxk−j+2u

Dxk+1u

j=2

+ C Dx u ∞ Dxk+2u

Dxk+1u

+ C Dxk+1 u

Dx2 u

Dxk+1u

≤ C Dx2 u 2 Dxk+1 u 2 +C

i k−1 

C j Dxi−j+1u ∞ Dxj+1u

Dxj+2−iu

Dxk+1u

i=2 j=0

+C

k−1 

uxx

Dxj+2u

Dxk−j+1u

Dxk+1u

j=1

+C

k−1 

Dxj+1 u

Dxk−j+2u

Dxk+1u

j=2

+ C uxx

Dxk+3 u

Dxk+1u

≤ Dxk+1u 2 + C Dxk+1u 2 Dxk u 2 + C Dxk+1 u

Dxk+2u

α ≤ Dxk+2u 2 + C Dxk+1u 2 . 4 It follows from (6.3.23), (6.3.24) and (6.3.27) that d

Dxk+1 u 2 + α Dxk+2u 2 ≤ C Dxk+1u 2 . dt

(6.3.27)

(6.3.28)

This implies d (6.3.29)

Dxk+1 u 2 ≤ C Dxk+1u 2 , t ≥ t∗ , dt where t∗ depends on the initial data and k. On the other hand, it follows from the induction assumption that  t+1 t

Dxk+1 u 2dt ≤ C,

t ≥ t∗ .

(6.3.30)

From (6.3.29) and (6.3.30) and the Gronwall inequality, we have

Dxk+1u 2 ≤ C,

t ≥ t∗ + 1.

(6.3.31)

For t ∈ (t, t + 1), integrating (6.3.28), we have  t+1 t

Dxk+2u 2 dt ≤ C,

t ≥ t∗ + 1.

(6.3.32)

The conclusion of this lemma follows from (6.3.31) and (6.3.32). Lemma 6.3.5 Let u0 ∈ H k+2 ∩ Hρ , k ≥ 0. There exists a constant Ck depending on the initial data and k such that

Dxk ut ≤ Ck ,

t ≥ tk ,

where tk depends on the initial data, k, and R with u0 H 2 ≤ R.

352

Landau–Lifshitz Equations

Proof. This lemma can be obtained by differentiating (6.3.6) k times with respect to x and applying Lemma 6.3.4. In order to study the approximate inertial manifold of the Landau–Lifshitz equations, rewrite (6.3.6)–(6.3.8) as the operators on H: du + αAu + B(u, u) + R(u) = 0, dt

(6.3.33)

where A = −∂xx is an unbounded self-conjugate operator defined on D(A) = {u : u ∈ H 2 , u satisfies (6.3.8)}. B(u, 0) = −βu × νxx is a bilinear form; R(u) = −α|ux |2 u is a nonlinear operator: D(A) → H. It is known that H has an orthogonal base consisting of the eigenfunctions of operator A: {Ωj }∞ j=1 , 0 = α1 < λ2 < · · · < λj → ∞,

AΩj = λj Ωj ,

j → ∞.

For a given m, let P = Pm : H → Span{Ω1 , Ω2 , . . . , Ωm } be a project operator. Q = Qm = P − Pm . Acting (6.3.33) by Pm and Qm , we have     

dp + αAp + Pm B(p + q, p + q) + Pm R(p + q) = 0, dt   dq   + αAq + Qm B(p + q, p + q) + Qm R(p + q) = 0, dt

(6.3.34)

where p = Pm u, q = Qm u. It follows from Lemmas 6.3.1, 6.3.4 and 6.3.5 that

u ,

Ak u ,

Ak+1 u ,

3

A 2 u ≤ C, 3

Ak+ 2 u ,

t ≥ t∗ ,

     k ∂u  A   ∂t 

(6.3.35)

≤ Ck ,

t ≤ tk ,

(6.3.36)

where tk , Ck are as before. Note  1 

A 2 u

      

= ∂u

, ∂x

Pm u ≤ u ,

u ∈ H 1,

Qm u ≤ u ,

A2 u ≥ λαm+1 u ,

It follows from (6.3.36) that

q ,

Aq ,

   3   2  A q ,  

α > 0,   d     q  dt 

u ∈ H,

(6.3.37)

u ∈ Qm D(A2 ).

≤ Ck λ−k m+1 ,

t ≥ tk .

(6.3.38)

Define a map Φ : Pm H → Qm H such that for any p ∈ Pm H, Φ(p) = Ψ is given by αAΨ + Qm B(p, p) + Qm R(p) = 0. 

Let = graph(Φ). It can be verified that of (6.3.6)–(6.3.8). Then



(6.3.39)

is the approximate inertial manifold

353

Long Time Behavior

Theorem 6.3.1 Let u0 ∈ H k+2 ∩ Hρ . Then for any integer k, there exists a constant Ck depending on the initial data and k such that any solution u(t) of (6.3.6)–(6.3.8)  when t > tk , where tk depends is within a H 2 -distance no large than Ck λk−1 m+1 from 2 on the initial data, k and R with u0 H ≤ R. Proof. Subtracting (6.3.34) from (6.3.39), we have dq + Qm B(p + q, p + q) − Qm B(p, q) dt + Qm R(p + q) − Qm R(p) dq + Qm B(q, p + q) + Qm B(p, q) = dt + Qm R(p + q) − Qm R(p).

αAΨ − αAq =

(6.3.40)

Since

B(q, p + q) + B(p, q) ≤ β q × Au + β p × Aq

≤ β q

Au ∞ + β p

Aq ∞ 3

3

≤ β q

A 2 u + β p

A 2 q

≤ Ck λ−k m+1 .

(6.3.41)

And

R(p + q) − R(p) = α |ux |2 u − |px |2 p

≤ α (|ux |2 − |px |2 )u

+ α (|px |2 (u − p)

≤ α (|ux | + |px |)(|ux| − |px |)u

+ α px 2∞ q

≤ 2α Au

Aq

u + α Au

q

≤ Ck λ−k m+1 .

(6.3.42)

It follows from (6.3.40)–(6.3.42) that

AΨ − Aq ≤ Ck λ−k m+1 ,

t ≥ tk .

Therefore we get 

distH 2 u(t),



≤ u(t) − (p(t) + Φ(p(t))) H 2 ≤ Φ(p(t)) − q(t) H 2 ≤ AΨ − Aq

≤ Ck λ−k m+1 ,

t ≥ tk .

354

6.4 6.4.1

Landau–Lifshitz Equations

Attractor of Landau–Lifshitz Equations on Riemannian Manifold Landau–Lifshitz Equations on Riemannian Manifold

Assume that (M, γ), (N, g) be two compact Riemannian manifolds. M has no boundary, and N is S 2 . Consider the following Landau–Lifshitz equations for u = (u1 , u2, u3 ) : M → S 2 , ∂t u = −α1 u × (u × ∆M u) + α2 u × ∆M u,

α1 > 0,

(6.4.1)

with initial condition u(x, 0) = u0 (x),

(6.4.2)

where |u0(x)| = 1, x = (x1 , x2 , . . . , xn ) ∈ M, ∆M is the Laplace–Beltrami operator ∂2 ∂ 1 ∂ αβ √γ ∂ αβ γ = γ − Γkαβ k . ∆M = √ β α α β γ ∂x ∂x ∂x ∂x ∂x In the classical sense, Eq. (6.4.1) is equivalent to 

√ ∂u 1 ∂ γ αβ γ α ∂t u = α1 √ β γ ∂x ∂x





+ α2 |∇u|2 u



√ ∂u 1 ∂ + α2 u × √ γ αβ γ α β γ ∂x ∂x

6.4.2



.

(6.4.3)

The A Priori Estimates

Our aim is to prove the existence of a global attractor of problem (6.4.1) and (6.4.2) and give the Hausdorff and fractal dimensions. To this aim we first give the a priori estimates. Let ∇ be the Riemannian connection corresponding to metric γ (or covariant derivative): |∇u(x)|2 =

 α,β

i

γ αβ

∂ui ∂ui . ∂xα ∂xβ

(6.4.4)

For real function ϕ ∈ C k (M) (k ≥ 0 is integer), we define |∇k ϕ|2 = ∇α1 ∇α2 · · · ∇αk ∇α1 ∇α2 · · · ∇αk ϕ. In particular, |∇1 ϕ| = |∇ϕ|, |∇1 ϕ|2 = |∇ϕ|2 = ∇ν ϕ∇ν ϕ, ∇k ϕ is kth order covariant derivative of ϕ. Consider the vector space Lpk = {ϕ : ϕ ∈ C ∞ , |∇l ϕ| ∈ Lp (M), 0 ≤ l ≤ k}; l, k are integers and p ≥ 1. Sobolev space Wpk (M) is the closure of Lpk in the norm 

ϕ Wkp = ki=0 ∇i ϕ Lp . In particular, W2k (M) = H k (M), · 2 = · .

355

Long Time Behavior

Lemma 6.4.1 Let |u0(x)|2 = 1. Then for the smooth solution of (6.4.1) and (6.4.2) there holds |u(x, t)|2 = 1,

∀ (x, t) ∈ M × (0, ∞).

(6.4.5)

Proof. Multiplying (6.4.1) by u we have u · ∂t u = 0,

∀ (x, t) ∈ M × (0, ∞);

the conclusion follows from (6.4.2) and (6.4.5). Lemma 6.4.2 Let |u0(x)|2 = 1 and ∇u0 < ∞. Then for the smooth solution of (6.4.1) and (6.4.2),   ∇u(·, t) 2 t 

2α1

≤ ∇u0 2 ,

u × ∆M u 2 dt ≤ ∇u0 2 ,

0

∀ 0≤t<∞

(6.4.6)

hold. Proof. Multiplying (6.4.1) by ∆M u we have ∆M u · ut = −α1 ∆M u · (u × (u × ∆M u)) = −α1 (u × ∆M u) · (∆M u × u) = α1 |u × ∆M u|2 . On the other hand,  M

∆M u · ut dM = −

1d

∇u(·, t) 2. 2 dt

Therefore

d

∇u(·, t) 2 ≤ 0. dt The lemma follows from this and the assumptions on the initial data u0 .

Lemma 6.4.3 (Sobolev interpolation inequality on Riemannian manifold) Let M be a compact Riemannian manifold with smooth boundary, q, r be real numbers, 1 ≤ q, r ≤ ∞, j, m be integers, 0 ≤ j < m. Then there exists a constant C depending only on n, m, j, q, r and α, M such that for any f ∈ Wrm (M) ∩ Lq (M), there holds

∇j f p ≤ C f αWrm f 1−α Lq , where





j 1 1 m 1 = +α − + (1 − α) , p n r n q for any α :

j m

≤ α ≤ 1 and nonnegative integer p.

(6.4.7)

356

Landau–Lifshitz Equations

Proof. It follows from the Aubin theorem ([8, Theorem 6.3.70]) that ,

∇j F p ≤ C ∇m F αr f 1−α q where

(6.4.8)



1 f dM, vol(M) M
 1 m j 1 1 = +α − + (1 − α) . p n r n q

F = f − f,

f=

(i) i > 0. It follows from (6.4.8) that

∇j F p ≤ C ∇m F αr ( f q + f q )1−α ≤ C1 ∇m F αr f 1−α q ≤ C1 F αWrm(M ) f 1−α . q (ii) i = 0, it follows from H¨older inequality that 

|f | dx =



|f |α |f |β dx

p

M

≤

M



1   l

|f | dx αl

M

M

where α + β = p, Then

lα = r,

1 = l





q−p r , q−r p

l β = q,

βl


 1


|f | dx

l

,

1 1 +  = 1. l l

1 1 =1− .  p l

(6.4.7) follows from (i) and (ii). Remark. Replacing space Lpk (M), T. Aubin in [8] introduced space Vkp which is the closure of space Skp in the norm

ϕ Vkp (M ) =



0≤l≤ k2

∆lM ϕ p +



0≤l≤ k−1 2

∇∆lM ϕ p ,

where Skp is a vector space of elements ϕ : ϕ ∈ C ∞ (M), ∆lM ϕ ∈ Lp (M), 0 ≤ l ≤ k2 , . and ∇∆lM ϕ ∈ Lp (M), 0 ≤ l ≤ k−1 2 Lemma 6.4.4 Let the conditions of Lemma 6.4.2 hold, and let

∇u0 ≤ λ,

n = 2,

(6.4.9)

where constant λ is small enough. Then for the smooth solutions of problem (6.4.1) and (6.4.2), E1 , ∀ x ∈ M, t > 0, 1 ≤ n ≤ 2, t holds, where constant E1 depends only on ∇u0 H 1 (M ) and 0 < t ≤ T .

∆M u(·, t) ≤

(6.4.10)

357

Long Time Behavior

Proof. Using operator ∆M on (6.4.7) and taking the inner product with t∆M u, we have 



t∆M u, ∆M ut − α1 ∆2M u − α1 ∆M (|∇u|2 u) − α2 ∆M (u × ∆M u) = 0,

where t(∆M u, ∆M ut ) = (∆M u, α1 ∆2M u)



1 ∂ √ ∂∆M u √ ∆M u. √ γ αβ γ γdx β γ ∂x ∂xα

= α1 = −α1

   

1d 1 (t ∆M u 2 ) − ∆M u 2 , 2 dt 2 



(6.4.11)



√ ∂∆M u ∂∆M u γ αβ γ dx = −α1 ∇∆M u 2 , α β ∂x ∂x 

 √ ∆M u.∆M (u × ∆M u) γdx

= = =

              







 ∂ √ ∂  ∆M u. β γ αβ γ α (u × ∆M u) dx  ∂x ∂x

∂∆M u γ αβ ∂xβ ∂∆M u γ αβ ∂xβ



 ∂(u × ∆M u) √   γdx α  ∂x 





 ∂u ∂∆M u √   × ∆ u + u × γdx M α α  ∂x ∂x

≤ C1 ∇∆M u

∇u ∞ ∆M u .

(6.4.12)

In (6.4.12), we have used the fact 

∂∆M u γ αβ . ∂xβ





∂∆M u √ u× γdx = 0. ∂xα

(6.4.13)

Lemma 6.4.3 implies that  

1 

1

 

1 

1

∇u ∞ ≤ C2 ∇3 u 2 ∇u 2 + C3

(n = 2),

∆M u ≤ C4 ∇3 u 2 ∇u 2 + C5 ,

(6.4.14)

where constants C3 and C5 depend on ∇u0 . Substituting (6.4.14) into (6.4.12), we get    

M

Here

 

 

2 

∆M u.∆M (u × ∆M u)dM  ≤ 2C1 C2 C4 C6 ∇u ∇3 u + C6 .  

 

∇∆M u ≤ C6 ∇3 u

and constant C6 depends on ∇u0 .

(6.4.15)

358

Landau–Lifshitz Equations

Now we are going to estimate term (∆M (|∇u|u), ∆M u) in (6.4.11).     2 (∆M u, ∆M (|∇u| u))

= = = =

≤

   ∂ αβ √ ∂|∇u|2 u    ∆ u. γ γ dx  M M ∂xβ  ∂xα     ∂ ∂   2 αβ √   γ γ |∇u| u. ∆ udx M α β  M  ∂x ∂x        ∂ ∂   2 2 ∂u αβ √   γ γ |∇u| u + |∇u| ∆ udx M  M  ∂xα ∂xα ∂xβ    ∂ 2 u ∂u  αβ lδ  γ 2γ (x) α l δ u  ∂x ∂x ∂x     2 ∂u ∂∆M u lδ  ∂u ∂u  dM + γ (x) l δ u + |∇u|  ∂x ∂x ∂xα ∂xβ     C7 ∇u ∞ u ∞ ∇2 u ∇∆M u + ∇u 2 u ∞ ∇∆M u

+ ∇u 2∞ ∇u

∇∆M u







≤ C7 2C2 C4 ∇u + (2C22 + 1) ∇u 2 ∇3 u + C8 ,

(6.4.16)

where constants C7 and C8 depend on ∇u0 and supx∈M (|γ αβ (x)|, |(γ αβ (x)) |). Hence from (6.4.11), (6.4.15) and (6.4.16), we have 1d 1 t ∆M u 2 − ∆M u 2 + α1 t ∇∆M u 2 2 dt 2 ≤ 2t [|α2 |C1 C2 C4 C6 + α1 C7 C2 C4

 

 

+ α1 (C22 + 1)C7 ∇u ∇u ∇3 u + C9 .

(6.4.17)

Now we estimate the lower bound for ∇∆M u 2 . By the Ricci formula ∆(∇ f ) = ∇ (∆f ) + k

k

k−1 

Ski (∇k−if ),

(6.4.18)

i=0

where Ski is a linear functional depending on the covariant derivative ∇i R of the curvature tensor, we get

∇∆M u 2 = ∆M (∇u) − S10 (∇u)

≥ ∆M (∇u) − S10 ∇u .

(6.4.19)

359

Long Time Behavior

Since 2

∆M (∇u) = =

  

 M

j





=− =−





 M



 M

j



ujii

i

M

j

=−



 j

ujii

i

M

j







ujkk dM

k





ujkk dM

k





uji

i





uji

i





uji

i





ujkki dM

k



l l (ujkik + ujl Rkki + ulk Rjki )dM

k



l (ujikk + ul Rjkik

k

l l i + ulk Rjki + ujl Rkki + ulk Rjki )dM

≥



M

   3 2 ∇ u dM

− C11



M

|∇u|2 dM.

(6.4.20)

Hence from (6.4.17), (6.4.19) and (6.4.20), we have d t ∆M u 2 − ∆M u 2 + 2t[α1 − (2α2 C1 C2 C4 C6 + 2α1 C7 C2 C4 dt  

2 

+ 2α1 (C22 + 1)C7 ∇u + C12 ∇u )] ∇u ∇3 u ≤ C13 .

(6.4.21)

Choose ∇u0 suitably small such that α1 − (2|α2 |C1 C2 C4 C6 + 2α1 C7 C2 C4 + 2α1 (C22 + 1)C7 ∇u0

α + C12 ∇u0 ) ∇u0 > > 0. 2 From (6.4.21), we get 2

t ∆M u −

 t 0

1  t  3 2

∆M u dt + α t∆M u dt ≤ C13 t. 2 0 2

(6.4.22)



In order to estimate term 0t ∆M u 2 dt on the left-hand side of inequality (6.4.22), we need the following lemma. Lemma 6.4.5 Under the conditions of Lemma 6.4.2, we have   t  ∂u 2     dt   ∂t 0  t 0

≤

(α12 + α22 )

∇u0 2 , α1

∆M u 2 dt ≤ C14 ,

where constant C14 depends on ∇u0 and T .

∀ t ∈ R+ ,

0 ≤ t ≤ T,

(6.4.23) (6.4.24)

360

Landau–Lifshitz Equations

Proof. Multiplying (6.4.3) by ∂t u and integrating with respect to x, t ∈ M × [0, t), we have  t 0

M

|∂t u|2 dMdt +

− α2

 t 0

M

α1 2

 t 0

d

∇u 2 dt dt

∂t u.(u × ∆M u)dMdt = 0.

(6.4.25)

By the cross product of Eq. (6.4.3) with u, we get u × ∂t u = α1 (u × (∆M u) + α2 (u × (u × ∆M u)) = α1 (u × ∆M u) − α2 ∆M u − α2 |∇u|2 u. Because

−∆M u − |∇u|2 u = −

This gives

1 α2 ∂t u + (u × ∆M u). α1 α1





α2 α2 u × ∂t u + ∂t u = α1 + 2 (u × ∆M u). α1 α1

Multiplying the above equation with ∂t u, we have ∂t u.(u × ∆M u) = α2 (α12 + α22 )−1 |∂t u|2 . Then we have  t

α2

0

M

∂t u.(u × ∆M u)dMdt =

α22 α12 + α22

 t 0

|∂t u|2 dMdt.

M

From (6.4.25), it follows that  t

α12 α12 + α22

0

M

|∂t u|2 dMdt +

α1 ( ∇u(., t) 2 − ∇u0 2 ) = 0, 2

i.e.,  t 0

M

|∂t u|2 dMdt ≤

α12 + α22

∇u0 2 , 2α1

∀ t ∈ R+ .

(6.4.26)

From (6.4.3), we have  t 0

∆M u 2 dt ≤ C15


 t 0

ut 2 dt +

 t 0

u × ∆M u 2 dt +

 t 0

M



|∇u|4 dMdt . (6.4.27)

By Sobolev’s inequality (6.4.7), it follows that  M

 

2 

|∇u|4 dM ≤ C16 ∇2 u ∇u 2 + C17 ,

where constant C17 depends on ∇u0 and 0 ≤ t ≤ T .

0 ≤ t ≤ T,

(6.4.28)

361

Long Time Behavior

The definition of Laplace–Beltrami operator implies  

2 

 

2 

C18 ∇2 u + C19 ∇u 2 ≥ ∆M u 2 ≥ C18 ∇2 u − C19 ∇u 2 , where constants C18 and C19 depend on tively. Therefore, from (6.4.27), we get

    supx∈M γ αβ (x)

and

    supx∈M Γkαβ ,

(6.4.29) respec-

 t  t    2 2  2 2 2 ∇ u dt − C15 C16 ∇ u dt. ∇u0

C18 0

≤

0



   t  t  ∂u 2   2 C15    dt +

u × ∆M u dt + C17  + C19 ∇u0 2 t.   ∂t 0 0

(6.4.30)

Choose ∇u0 suitably small such that C18 . 2

(6.4.31)

0 ≤ t ≤ T.

(6.4.32)

C18 − C15 C16 ∇u0 2 ≥ From (6.4.27)–(6.4.30), we have  t 0

∆M u 2 dt ≤ C14 ,

The lemma is proved. From (6.4.22) and (6.4.32), we get E1 , ∀x = (x1 , x2 , . . . , xn ), n ≤ 2, t > 0, t where constant E1 depends only on ∇u0 and T . Now we prove that inequality (6.4.10) holds in the case n = 1 without the restriction ∇u0 ≤ λ. In fact, from Lemma 6.4.3, we have

∆M u 2 ≤

 

1 

3

∇u ∞ ≤ C2 ∇3 u 4 ∇u 4 + C3 ,  

1 

1

∆M u ≤ C4 ∇3 u 2 ∇u 2 + C5 

M

|∇u|4 dM ≤

    C16 ∇2 u ∇u 3

(n = 1), + C17 .

The conclusion of inequality (6.4.10) can be easily obtained. Using the estimates in [77], we have Theorem 6.4.1 Let M be a compact manifold without boundary, and suppose that the following conditions are satisfied : (1) u0 (x) ∈ H m (M), m ≥ 2, |u0 (x)|2 = 1, x = (x1 , . . . , xn ) ∈ M, (2) In the case of n = 2,

1 ≤ n ≤ 2,

∇u0 (x) ≤ λ,

where constant λ is suitably small. Then there exists a unique global smooth solution u(x, t) : M × [0, ∞) → S 2 of the initial value problem (6.4.1) and (6.4.2) u(x, t) ∈ L∞ (0, ∞; H m(M)).

362

Landau–Lifshitz Equations

6.4.3

The Global Attractor

Theorem 6.4.2 Let M be an m-dimensional compact Riemannian manifold without boundary (n ≤ 2). If conditions (i) α1 > 0, |Z0 (x)| = 1, Z0 (x) ∈ H 1 (M), x = (x1 , . . . , xn ) ∈ M, n ≤ 2; (ii) ∇Z0 (x) ≤ λ, x ∈ M, n = 2, where constant λ is suitably small, are satisfied, then the initial value problem (6.4.1) and (6.4.2) of the Landau–Lifshitz equation has an attractor A which is compact in H 1 (M), where 

A = ω(B 1 ) =

 s≥0

t≥s

S(t)B1 .

(6.4.33)

Here B1 = {Z ∈ H 1 (M), |Z| = 1, Z H 1 (M ) ≤ ρ1 }

(6.4.34)

is a bounded absorbing set for S(t) in H 1 (M), and ρ1 is a positive constant. Proof. From Theorem 6.4.1, there exists a unique global smooth solution u(x, t) ∈ L∞ (0, ∞; H m(M)) of the initial value problem (6.4.1) and (6.4.2), which means that problem (6.4.1) and (6.4.2) generates a semigroup S(t) : Z(x, t) = S(t)Z0 . From Lemma 6.4.1 and Lemma 6.4.2, we know B1 = {Z ∈ H 1 (M), |Z| = 1, Z H 1 (M ) ≤ ∇Z0 (x) + Vol M = ρ1 } is a bounded absorbing set for S(t) in H 1 (M). From Lemma 6.4.4, we have

Z(., t) H 2 (M ) ≤

E1 , t

t > 0,

where constant E1 depends only on ∇Z0 (x) H 1 (M ) . This implies that semigroup operator S(t) is a completely continuous operator in H 1 (M) for t > 0. From Theorem 6.2, the semigroup operator S(t) has a compact attractor A=

6.4.4





s≥0

t≥s

S(t)B1 = ω(B1 ).

The Estimates of the Upper Bounds of Hausdorff and Fractal Dimensions for the Attractors

In the following, we give the estimates of the upper bounds of the Hausdorff and fractal dimensions for the attractors of the mentioned problems in the above section. We consider the linear variation corresponding to problem (6.4.1) and (6.4.2) v(t) + L(u(t))v = 0,

(6.4.35)

v(0) = v0 (x),

(6.4.36)

363

Long Time Behavior

where L(u(t))v = −α1 ∆M v − α1 |∇u|2 v − 2α1 ∇uu∇v − α2 A (u)∆M uv − α2 A(u)∆M v.

(6.4.37)

Since solution u of problem (6.4.1) and (6.4.2) is suitably smooth, we can prove that the linear problem (6.4.35) and (6.4.36) has a global solution v(x, t) ∈ L∞ (0, ∞; H 2(M)) as long as the initial data v0 (x) is mildly smooth. In fact, for Eq. (6.4.35), the principle part is given by m 

Dα (aαβ (x, t, u)Dβ u) =

α,β=1

m 

√ Dα γ(γ αβ Dβ u + u × γ αβ Dβ u).

α,β=1

The corresponding coefficients aij α,β are given by aij α,β = 

Here

√ αβ γγ gij .

α1  g = (gij ) =  α2 u3 −α2 u2

−α2 u3 α1 α2 u1



α2 u2  −α2 u1  , α1

u = (u1 , u2 , u3). Therefore, the Landau–Lifshitz equation (6.4.1) is strongly parabolic by the following relation: m 3  

α β aij α,β (x, t, η)ξ ξ ζi ζj = α1 |ζ|

i,j=1 α,β=1

m 

√ γ αβ γξ α ξ β > 0

α,β=1

for all (x, t, η) ∈ M × [0, T ] × R3 , for all ξ = (ξ 1 , . . . , ξ m ) ∈ Rm \{0}, and for all ζ = (ζ1 , ζ2 , ζ3 ) ∈ R3 \{0}. Let Gt be a solution operator of (6.4.1) and (6.4.2) such that v(t) = Gt v0 . It can be verified that the semigroup operator St u0 is differentiated in L2 (M), namely the Frechet derivative St u0 exists, and Gt v0 = St u0 . In fact, we set W (t) = St (u0 + v0 ) − St (u0 ) − Gt (u0 )v0 = w1 (t) − u(t) − v(t). Thus we have ∂t W (t) = L1 (u1 (t)) − L1 (u(t)) + L(u(t))v(t) = L1 (u(t) + v(t) + w(t)) − L1 (u(t)) + L(u(t))v(t), w(0) = 0, where ut = L1 (u) is the operator form of Eq. (6.4.3).

(6.4.38)

364

Landau–Lifshitz Equations

Therefore, (6.4.38) can be rewritten in the form ∂t w(t) + L(u(t))w(t) = Λ0 (u, v, w),

(6.4.39)

where Λ0 (u, v, w) = L1 (u(t) + v(t) + w(t)) − L1 (u(t)) + L(u(t))(v + w).

(6.4.40)

By applying the theorem of linear parabolic partial differential equations, we have the following L2 estimates:

w(t) 2 ≤ C v0 2 .

(6.4.41)

This implies that the semigroup operator S(t) is differentiated in L2 (M). Denote by V1 (t), V2 (t), . . . , VJ (t) the solution of linear equation (6.4.36) corresponding, respectively, to the initial data V1 (0) = ξ1 , . . . , VJ (0) = ξJ ; here ξ = (ξ1 , ξ2 , . . . , ξJ ) ∈ L2 (M), and by the simple computation [15], we can deduce that d

V1 (t) ∧ V2 (t) ∧ · · · ∧ VJ (t) 2 + 2 Tr(L(u(t)) · QJ (t)) dt · V1 (t) ∧ V2 (t) ∧ · · · ∧ VJ (t) 2 = 0,

(6.4.42)

where L(u(t)) = L(S(t)u0 ) is a linear map: v → L(u(t))v, “∧” denotes the exterior product, “Tr,” the trace of operator, and QJ (t) is the orthogonal projection of space L2 (M) to the spanning subspace generated by V1 (t), V2 (t), . . . , VJ (t). Therefore, from (6.4.42), we can obtain the change of volume ∧Jj=0 ξ of the J-dimensional cube by WJ (t) = sup

sup

u0 ∈A ξj ∈L2 (M ),|ξj |≤1

≤ sup exp

V1 (t) ∧ V2 (t) ∧ · · · ∧ VJ (t) 2∧j L2


 t

u0 ∈A

0



inf(Tr(L(S(τ )u0 ))(QJ (τ ))dτ .

(6.4.43)

Lemma 6.4.6 Let (M, g) be a Riemannian manifold of dimension n. For every p with   n n ,
ρ

p p−1

 2m(p−1)

dM

n

≤ k(M)

N   j=1 M

|∇m ϕj |2 dM



+ χ(M) Here ρ =

N j=1

M

ρdM.

(6.4.44)

|ϕj (x)|2 , and constants k(M) and χ(M) depend on m, n, p and (M, g).

365

Long Time Behavior

Lemma 6.4.7 Let M be a compact Riemannian manifold without (or with) boundary, and {λj } is the spectrum of M with respect to Laplace–Beltrami operator. Then the following inequalities hold δ λk ≥ e



k Vol(M)

2

n

,

where n = dim M and δ is the Sobolev constant of M, that is, for any u ∈ C ∞ (M) 

2

|∇u| dM ≥ δ

M




|u|

M

2n n−2

 n−2 n

.

Theorem 6.4.3 Under the conditions of Theorem 6.4.6, the Hausdorff and fractal dimensions of the global attractor A of the problem (6.4.1) and (6.4.2) are finite, and dH (A) ≤ J0 ,

dF (A) ≤ 2J0 ,

(6.4.45)

where J0 is the smallest integer, such that −4m

J0 ≥ C0 α1 (4−m)(2+m) ,

(1 ≤ m ≤ 2),

(6.4.46)

where C0 is a constant that depends on M and on the norms u ∞ , ∇u ∞ and

∇2 u 2 . Proof. Suppose that {ϕ1 (x), . . . , ϕJ (x)} is an orthogonal basis of the subspace QJ L2 , we have Trace(L(u(t)) · QJ ) =

∞ 

(L(u(t)) · QJ (τ )ϕj , ϕj )

j=1

=

J 

(L(u(t))ϕj (τ ), ϕj (τ )

j=1

=

J 

{−α1 (∆M ϕj , ϕj ) − α1 (|∇u|2 ϕj , ϕj )

j=1

− 2α1 (∇u · u · ∇ϕj , ϕj ) − α2 (A (u)∇uϕj , ϕj ) − α2 (A(u) · ∆M ϕj , ϕj )} ≥ α1

J 

 

 

λj − α1 |∇u|2  ρ(x) 2 2

j=1

− 2α1 ∇(∇u · u) 2 ρ(x) 2 − α2 A (u) ∞ ∇u 2 ρ(x) 2 ≥ α1

J  j=1

λj − 2α1 ∇u ∞ ∇u 2 ρ(x) 2  

 

− 2α1 u ∞ ∇2 u ρ(x) 2 − α2 A (u) ∞ ∇u 2 ρ(x) 2 2

366

Landau–Lifshitz Equations

≥ α1

J 

 

 

λj − (2α1 ∇u ∞ ∇u 2 + 2α1 u ∞ ∇2 u

j=1

+ α2 A (u) ∞ ∇u 2 ) ρ(x) 2 ≥ α1

J 



λj − H k(M)

j=1



≥ α1 

J 

≥

α1 

J 

m 4

λj + χ(M)

j=1

m 4

λj 

 1− m 4 J    · λ  j

j=1



J 

2



C1  − H  − C2 H α1

j=1

m 4

λj 

− C2 H > 0,

j=1

where

 

 

H = 2α1 ∇u ∞ ∇u ∞ ∇u 2 + 2α1 u ∞ ∇2 u + α2 A (u) ∞ ∇u 2 ). 2

As 

J ≥ Max

C3



2 m



m 2+m



C1 · H +1 α1





4m (4−m)(2+m)

,

C3



2 m



m 2+m



C2 H α1

2+ 4 ' m

.

Here we have used the following inequality [49]: J 

j

2 m

≥ C3




j=1



2 2 J m +1 , m

and C1 , C2 are constants which depend on the manifold M. Now we consider the initial-boundary value problem for the following Landau– Lifshitz equation Zt = −α1 (Z × (Z × ∆M Z)) + Z × JZ, Z|t=0 = Z0 (x),

x ∈ M,

(6.4.47) (6.4.48)



∂Z  = 0, ∂ν ∂M

(6.4.49)

where J = diag(J1 , J2 , 0), α1 > 0, and ν is the outer unit normal vector to ∂M. Theorem 6.4.4 Assume that A is the global attractor of problem (6.4.47)–(6.4.49), and J1 J2 < 0. Then we have n

dim A ≥ Cα1 − 2 ,

(6.4.50)

where dim A is the Hausdorff or fractal dimension of A and C is a positive constant.

367

Long Time Behavior

Proof. It is easy to find out that Z = (0, 0, 1) is a fixed point of semigroup S(t) generated by the problem (6.4.47)–(6.4.49), i.e., Z(0, 0, 1) satisfies the equation −A1 (Z) = −α1 (Z × (Z × ∆M Z)) + Z × JZ = 0.

(6.4.51)

The linear variation equation −A1 (Z) for Eq. (6.4.51) is as follows: −A1 (Z)v = α1 ∆M v + Z × Jv = 0.

(6.4.52)

Let ζ denote the eigenvalues associated with the matrix B(Z)v = Z × Jv = ζv, where   0 −J2 0   0 0. B(Z) =  J1 0 0 0 Therefore, (

ζ1 =

ζ 2 + J1 J2 = 0,

−J1 J2 > 0,

(

ζ2 = − −J1 J2 .

Let λk , k ∈ N denote the eigenvalues associated with Neumann problem 

∂ψk  = 0, ∂ν ∂M

−∆M ψk = λk ψk , 0 < α1 ≤ λ2 ≤ · · · ,

λk → ∞,

(6.4.53)

k → ∞.

Let µk , k ∈ N denote the sequence of eigenvalues of linearized operator −A (Z)wk = α1 ∆M wk + B(Z)wk = µk wk .

(6.4.54)

Here wk (x) = ψk (x)pk , pk ∈ C 3 , that is, (α1 λk + B(Z))pk = µk pk .

(6.4.55)

If µk is roots of equation det(α1 λk + B(Z) − µk I) = 0,

Reµk > 0,

(6.4.56)

then there exists the nonzero solution pk . Under√ the assumptions of the theorem, when α1 = 0, there exists at least one root ζ1 = −J1 J2 > 0. Thus we have a root µk of Eq. (6.4.56), Re µk > 0, as α1 λk < δ, where δ is a suitable small constant. From 2 the asymptotic behavior of eigenvalues λk ∼ Ck n , we get the inequality n

−n 2

1 ≤ k ≤ C1 δ 2 α1 Therefore, we have

−n

= C2 α1 2 . −n

dim A ≥ Cα1 2 .

368

Landau–Lifshitz Equations

6.5 6.5.1

The Attractors for Landau–Lifshitz–Maxwell Equations Periodic Initial Value Problem to L–L–M Systems

Consider the following periodic initial value problem for L–L–M system  t = α1 Z  × (∆Z  + H)  − α2 Z  × (Z  × (∆Z  + H)),  Z  = ∇×H

 ∂E ∂t

 + σ E,

(6.5.1) (6.5.2)

 = − ∂ H − β ∂ Z , ∇×E ∂t ∂t

(6.5.3)

 + β∇ · Z  = 0, ∇ · E  = 0, ∇·H  + 2Dei , t) = Z(x,  t), H(x  + 2Dei, t) = H(x,  Z(x t),  + 2Dei , t) = E(x,  E(x t), (x ∈ Ω, t ≥ 0, i = 1, 2, . . . , n),

(6.5.4)

 0) = Z  0 (x), Z(x,

  0 (x), H(x, 0) = H

  0 (x), E(x, 0) = E

(x ∈ Ω),

(6.5.5) (6.5.6)

where α2 > 0, x+2Dei = (x1 , . . . , xi−1 , xi +2D, xi+1 , . . . , xn ), (i = 1, 2, . . . , n), D > 0, Ω ⊂ Rn is n-dimensional cube with width 2D. Our purpose of this section is to study the existence of the attractor.

6.5.2

A Priori Estimates

For the sake of simplicity denote · Lp = · p , p ≥ 2.  0 (x)| = 1. Then for the smooth solution of the periodic Lemma 6.5.1 Assume |Z initial value problem (6.5.1)–(6.5.6), there are  t)| = 1, |Z(x,

x ∈ Ω,

t ≥ 0.

(6.5.7)

 with (6.5.1), we get Proof. Making the scalar product of Z ∂  |Z(x, t)|2 = 0. ∂t Then the conclusion of the lemma is proved.  0 (x) ∈ L2 (Ω), E  0 (x) ∈ L2 (Ω), Lemma 6.5.2 Assume α2 > 0, β > 0, σ ≥ 0, ∇Z  0 (x) ∈ L2 (Ω). Then for the smooth solution of the periodic initial value problem H (6.5.1)–(6.5.6)

sup

0≤t<∞

 t) 2 + E(·,  t) 2 + H(·,  t) 2 ≤ K1 ,

∇Z(·, 2 2 2  ∞ 0

 × (∆Z  + H)

 2 dt ≤ K2 ,

Z 2

(6.5.8)

(6.5.9)

 0 (x) 2 , E  0 (x) 2 , and H  0 (x) 2 . where the constants K1 and K2 depend only on ∇Z

Long Time Behavior

369

 with (6.5.2), and making the scalar product Proof. Making the scalar product of E  of −H with (6.5.3), and then adding these two equalities obtained, we have     + σ|E|  2 + ∂H · H  + β ∂ Z · H.   ·E  − (∇ × E)  ·H  = ∂E · E (∇ × H) ∂t ∂t ∂t By using the formula  ·E  − (∇ × E)  ·H  = ∇ · (H  × E),  (∇ × H)

(6.5.10)

(6.5.11)

and integrating (6.5.10) over x ∈ Ω, we get   1d   2 ∂Z 2 2    · Hdx = 0.

E 2 + H 2 + σ E 2 + β 2 dt Ω ∂t  + H)  with (6.5.1), we get Making the scalar product of (∆Z

(6.5.12)

  + H)  · [Z  × (Z  × (∆Z  + H))],   + H)  · ∂ Z = −ε(∆Z (∆Z ∂t

(6.5.13)

 + H)  · [Z  × (Z  × (∆Z  + H))]   × (∆Z  + H)|  2. −(∆Z = |Z

(6.5.14)

where

From integrating (6.5.13) over x ∈ Ω, it follows that  Ω

 + H)  · (∆Z

 ∂Z dx = ε ∂t

 Ω

 × (∆Z  + H)|  2 dx, |Z

and then  Ω

    ∂Z ∂Z    × (∆Z  + H)|  2 dx · Hdx = − ∆Z · dx + ε |Z ∂t ∂t Ω Ω  1d 2   × (∆Z  + H)|  2 dx.

∇Z 2 + ε |Z = 2 dt Ω

(6.5.15)

By adding (6.5.12) and (6.5.15) we obtain 1d  2  2 ) + σ E

 2 ( E 2 + H

2 2 2 dt β d  2 + βε Z  × (∆Z  + H)

 2 = 0.

∇Z

+ 2 2 2 dt Integrating the above equality with respect to t ∈ [0, T ], we have 1   t) 2 ) + σ E(·,  t) 2 ( E(·, t) 22 + H(·, 2 2 2  t β  × (∆Z  + H)

 2 dt  t) 2 + βε Z + ∇Z(·, 2 2 2 0 1  2 2  = E(0) = ( E 0 (x) 2 + H0 (x) 2 ) 2  0 (x) 22 .  0 (x) 22 + β ∇Z + σ E 2

E(t) ≡

(6.5.16)

370

Landau–Lifshitz Equations

It follows that the estimates (6.5.8) and (6.5.9) hold. In order to get the uniform  H,  E)  ∈ (H 2 (Ω), H 1 (Ω), H 1(Ω)), we first rewrite estimates in t for the solution (Z, (6.5.2) and (6.5.3) as the following equivalent second order nonlinear wave equations:  + σ∇ × E,   = ∂ ∇×E ∇ × (∇ × H) ∂t  = −∂ ∇×H  − β ∂ ∇ × Z.  ∇ × (∇ × E) ∂t ∂t From the formula  = ∇(∇ · H)  − ∆H  ∇ × (∇ × H)  − ∆H,  = −β∇(∇ · Z)  = ∇(∇ · E)  − ∆E  = −∆E,  ∇ × (∇ × E) we have  − ∆H  = ∂ ∇×E  + σ∇ × E  −β∇(∇ · Z) ∂t  tt − β Z  tt − σ H  t − σβ Z  t, = −H ∂  − β ∂ (∇ × Z)  (∇ × H) ∂t ∂t  tt − σ E  − β(∇ × Z)  t, = NE

 =− −∆E

 tt − ∆H  + βZ  tt + σ H  t + σβ Z  t − β∇(∇ · Z)  = 0, H  tt − ∆E  + σE  t + β(∇ × Z)  t = 0. E

(6.5.17) (6.5.18)

 E,  H)  ∈ (H 2 , H 1, H 1 ) It is difficult to derive the a priori estimate uniform in t for (Z, from (6.5.11), (6.5.17) and (6.5.18). We prove instead by Lyapunov functional containing small parameter method. We define the Lyapunov functional as follows: e(t) =

1   t) 2 + E  t (·, t) 2 + ∇E(·,  t) 2 + ∆Z(·,  t) 2

Ht (·, t) 22 + ∇H(·, 2 2 2 2 2  H  t ) + η2 (E,  E  t ), + η1 (H, (6.5.19)

where η1 and η2 are constants to be determined. We want to prove that e(t) satisfies the following differential inequality: de(t) + ae(t) ≤ K, dt

(6.5.20)

where a > 0 is a constant, K is independent of t. Then, the a priori estimates can be derived.

371

Long Time Behavior

In fact, it follows from (6.5.19) that de(t)  t, H  tt ) + (∇H,  ∇H  t ) + (E  t, E  tt ) = (H dt  t , ∇E  t ) + (∆Z,  ∆Z  t ) + η1 (H  t, H  t) + (∇E  H  tt ) + η2 (E  t, E  t ) + η2 (E,  E  tt ), + η1 (H, in which  t, H  tt ) = (H  t , ∆H  − βZ  tt − σ H  t − σβ Z  t + β∇(∇ · Z))  (H  t , ∆H)  − β(H  t, H  tt ) − σ(H  t, H  t) = (H  t, Z  t ) + β(H  t , ∇(∇ · Z)),  −σβ(H

 H  t ),  ∇H  t ) = −(∆H, (∇H,  t, E  tt ) = (E  t , ∆E  − σE  t − β(∇ × Z)  t) (E  t , ∆E)  − σ(E  t, H  t ) − β(E  t , (∇ × Z)  t ), = (E  ∇E  t ) = −(∆E,  E  t ), (∇E,  ∆Z  t ) = α1 (∆(Z  × ∆Z),  ∆Z)  + α1 (∆(Z  × H),  ∆Z)  (∆Z,  2 Z,  ∆Z)  + α2 (∆2 Z,  ∆Z)  + α2 (∆(|∇Z|  ∆Z)  − α2 (∆2 (Z  · H)  Z,  ∆Z),  + α2 (∆H,  H  tt ) = η1 (H,  ∆H  − βZ  tt − σ H  t − σβ Z  t + β∇(∇ · Z))  η1 (H,  2 − βη1 (H,  Z  tt ) = −η1 ∇H

2

 H  t ) − σβη1 (H,  Z  t ) + η1 β(H,  ∇(∇ · Z)),  − ση1 (H,  E  tt ) = η2 (E,  ∆E  − σE  t − β(∇ × Z)  t) η2 (E,  2 − ση2 (E,  E  t ) − βη2 (E,  (∇ × Z)  t ). = η2 ∇E

2

Hence, we have de(t)  t 2 − σβ(H  t, Z  t ) − β(H  t, Z  tt ) = −σ H 2 dt  t 2 − β(E  t , (∇ × Z)  t) − σ E 2

 × ∆Z),  ∆Z)  + α1 (∆(Z  × H),  ∆Z)  + α1 (∆(Z  2 Z),  2  ∆Z)  − α2 ∇∆Z

+ α2 (∆(|∇Z| 2

(6.5.21)

372

Landau–Lifshitz Equations

 ∆H)  − α2 (∆(Z  · H),  ∆Z)  + α2 (∆Z,  t , ∇(∇ · Z))  + η1 β(H,  ∇(∇ · Z))  + β(H  t 2 − η1 ∇H

 2 + η1 H 2 2  Z  tt ) − ση1 (H,  H  t) − βη1 (H,  Z  t ) + η2 E  t 2 − σβη1 (H, 2 2     (∇ × Z)  t) − η2 Z − ση2 (Z, Et ) − η2 β(E, 2

 t 22 − (ση2 ) E  t 22 = −(σ − η1 ) H  2 − η2 ∇E

 2 − α2 ∇∆Z

 2 − η1 ∇H

2

2

2

 Z  tt ) − ση1 (H,  H  t ) − σβη1 (H,  Z  t) − βη1 (H,  E  t ) − βη2 (E,  (∇ × Z)  t) − ση2 (E,  t , ∇(∇ · Z))  + βη1 (H  t , ∇(∇ · Z))  − β(H  t, Z  tt ) + β(H  t, Z  t ) − β(E  t , (∇ × Z)  t) − σβ(H  × ∆Z),  ∆Z)  + α1 (∆(Z  × H),  ∆Z)  + α1 (∆(Z  2 Z),  ∆H)   ∆Z)  + α2 (∆Z, + α2 (∆(|∇Z|  · H)  Z,  ∆Z),  − α2 (∆(Z in which  Z  tt ) = (H,  Z  t )t − (H  t, Z  t ), (H,  t, Z  tt ) = (H  t, Z  t )t − (H  tt , Z  t) (H  t, Z  t )t − (Z  t , ∆H  − βZ  tt − σ H  t − σβ Z  t + β∇(∇ × Z))  = (H t, Z  t, Z  t )t + β (Z  t )t − (Z  t , ∆H)  + σ(Z  t, H  t) = (H 2  t 22 − β(Z  t , ∇(∇ · Z)),  + σβ Z  t ) = β(Z  t , ∇ × Z)  t − β(∆E  − σE  t − β(∇ × Z)  t , ∇ × Z)   t , (∇ · Z) β(Z β2  ∇ × Z)  t (∇ × Z, 2  t , ∇ × Z)  + β(∇Z,  ∇(∇ × Z)),  + σβ(Z

 t , ∇ × Z)  t+ = β(Z

 (∇ × Z)  t ) = η2 β(E,  ∇ × Z)  t − η2 β(E  t , ∇ × Z).  η2 β(E, Set e1 (t) = 12 G(t) + R(t), where  t 22 + ∇E

 22 + ∇H

 22 + ∆Z

 22  t 22 + H G(t) = E  H  t ) − 2η2 (E,  E  t ), = 2e(t) − 2η1 (H,

(6.5.22)

373

Long Time Behavior 2 2  t 2 + β(E  2  t, H  t ) + β Z  t , ∇ × Z)  + β ∇ × Z

R(t) = β(Z 2 2 2 2  ∇ × Z)  + 1 ση1 H

 2 + 1 ση2 E

 2 = −η2 β(E, 2 2 2 2  ∇ × Z)  + η2 (E,  E  t ) + η1 (H,  H  t ). + η2 β(E,

(6.5.23)

de1 (t)  t 2 + (σ − η2 ) E  t 2 + (σ − η1 ) H 2 2 dt  2 + η2 E

 2 + α2 ∇∆Z

 2 + σβ 2 Z  t 2 + η1 ∇H

2

2

2

2

 t, H  t ) + β(Z  t , ∆H)  − (σ − η2 )β(E  t , ∇ × Z)  = −(2σβ − η1 β)(Z  ∇(∇ × Z))  + α1 (∆(Z  × ∆), ∆Z)  − β(∇E,  × H),  ∆Z)  + α2 (∆Z,  ∆H)  + α1 (∆(Z  2 Z),  · H)  Z,  ∆Z)  − σβη1 (H,  Z  t)  ∆Z)  − α2 (∆(Z + α2 (∆(|∇Z|  t , ∇(∇ · Z))  + βη1 (H,  ∇(∇ · Z))  + β 2 (Z  t , ∇(∇ × Z)).  + β(H

(6.5.24)

In the sequel we estimate every term in (6.5.24). (1) Estimate of the first term. It follows from (6.5.1) that  × (∆Z  + H)|  2 + α 2 |Z  × (Z  × (∆Z  + H))|  2  t |2 = α 2 |Z |Z 1 2  + H|  2 2(α2 + α2 )(|∆Z|  2 + |H|  2 ). ≤ (α2 + α2 )|∆Z 1

2

1

2

Using Sobolev inequality and H¨older inequality we have  t, H  t )| | − (2σβ + η1 β)(Z  t 2 + C(ε1 , σ, β, η1 ) Z  t 2 ≤ ε 1 H 2

2

 t 22 + 2C(α12 + α22 ) ∆Z

 22 + d1 ≤ ε 1 H  t 2 + α2 ∇∆Z

 2 + d2 , ≤ ε 1 H 2 2 l where ε1 and l are constants to be determined. (2) Estimate of the second term. Acting ∇ on (6.5.1), we get  × (∆Z  + H)  + α1 Z  × (Z  × (∆Z  + H))   × ∇∆Z  − α2 ∇Z  t = α1 ∇Z ∇Z  × (∇Z  × (∆Z  + H))  − α2 Z  × (Z  × ∆Z)  − α2 Z  × (Z  × ∇H).  − α2 Z

374

Landau–Lifshitz Equations

Hence,  t , ∆H)  = −β(∇Z  t , ∇H)  β(Z  × (∆Z  + H),  ∇H)  = −βα1 (∇Z  × ∇∆Z,  ∇H)  + βα2 (∇Z  × (Z  × ∆Z),  ∇H)  − βα1 (Z  × (Z  × H),  ∇H)  + βα2 (Z  × (∇Z  × ∆Z),  ∇H)  + βα2 (∇Z  × (∇Z  × H),  ∇H)  + βα2 (Z  × (Z  × ∇∆Z),  ∇H)  − βα2 Z  × ∇H

 2 + βα2 (Z 2

× ≤ −βα2 Z

 2 ∇H

2

 ∞ ∆Z

 2 ∇H

 2 + (|α1 | + 2α2 )β ∇Z

 ∞ H

 2 ∇H

 2 + (|α1 | + 2α2 )β ∇Z

 2 ∇H

 2 + (|α1 | + α2 )β ∇∆Z

3

n

1

n

 22 − 4 ∇∆Z

 22 + 4 ∇H

 2 ≤ C1 (|α1 | + 2α2 )β ∇Z

n

 24 ∇H

 2 ∇H

 2 + C2 (|α1 | + 2α2 )β ∇∆Z

+ ≤

≤

|α1 | + α2  2  2 + (|α1 | + α2 )β 2 ∇H

∇∆Z

2 2 4

n n C02 1+ n 2 2 2  3−   (|α1 | + 2α2 )2 ∇Z

∇∆ Z

+ C

∇∆ Z

3 2 2 2 α2 |α1 | + α2  2 + 2α2 (|α1 | + α2 )β 2 ∇H

 2

∇∆Z

+ 2 2 4  α2   2  + |α1 |+α2 ∇∆Z



     

2

4

l

 2 + C, + 2(|α1 | + α2 ) ∇H

2



n = 1;

 |α1 |+α2 α2 2   2 ∇∆Z

 2  + + + (|α | + 2α )

∇Z

1 2  2 2 l 4       2 + C, n = 2. + 2(|α | + α ) ∇H

C02 α2

1

2

2

(3) Estimate of the third term of (6.5.24).   −(σ

 

 t , ∇ × Z)   ≤ ε 2 E  t 2 + C1 ∇Z

 2 + η2 )β(E 2 2  t 2 + C2 . ≤ ε 2 E 2

(4) Estimate of the fourth term of (6.5.24).     ∇(∇ × Z))   −β(∇E,





    = β(E, ∆(∇ × Z)) α2  22 + C1 E

 22

∇∆Z

≤ l α2  2 + C2 .

∇∆Z

≤ 2 l

375

Long Time Behavior

(5) Estimate of the fifth term of (6.5.24).    α1 (∆(Z







 ∆Z)   ≤ |α1 | (∇Z  × ∆Z,  ∇∆Z)   × ∆Z),  ∞ ∆Z

 2 ∇∆Z

 2 ≤ |α1 | ∇Z

2 3

n

3

n

 22 − 4 ∇∆Z

 22 + 4 ≤ C0 |α1 | ∇Z

≤

  2  α2 ∇∆Z

2 l



+ C,

n = 1;

 2 ∇∆Z

 2 , n = 2. C0 |α1 | ∇Z

2

(6) Estimate of the sixth term of (6.5.24).    α1 (∆(Z







 ∆Z)   = α1 (∆(Z  × H),  ∆Z)   × H),  ∞ H

 2 ∇∆Z

 2 + |α1 | ∇H

 2 ∇∆Z

 2 ≤ |α1 | ∇Z

2 α2  22 + C.  22 + 2α1 ∇H

∇∆Z

4 α2

≤

(7) Estimate of the seventh term of (6.5.24).     2 Z),  ∆Z)   α2 (∆(|Z|





  2 Z),  ∇∆Z)   = −α2 (∇(|Z|

 36 ∇∆Z

 2 + C|α2 | ∇Z

 ∞ ∆Z

 2 ∇∆Z

 2 ≤ |α2 | ∇Z




3− n 2

 2 ≤ C|α2 | ∇Z

≤

  2  α2 ∇∆Z

2 l



1+ n 2

 2

∇∆Z

3

n

3

n

 22 − 4 ∇∆Z

 22 + 4 + ∇Z

+ C,

n = 1;

 2 + ∇Z

 2 ) ∇∆Z

 2 , n = 2. Cα2 (|∆Z

2 2 2

(8) Estimate of the eighth term of (6.5.24).     ∆H)   α2 (∆Z,

≤

α2  22 + α2 ∇H

 22 .

∇∆Z

4

(9) Estimate of the ninth term of (6.5.24).    α2 (∆(Z







 Z,  ∆Z)   = α2 (∇(Z  × H),  ∇∆Z)   · H) 

 2 ∇Z

 ∞ ∇∆Z

 2 + ∇H

 2 ∇∆Z

 2 ≤ α2 2 H

α2  2 + 3α2 ∇H

 2 + C.

∇∆Z

≤ 2 2 8 (10) Estimate of the tenth term of (6.5.24).     Z  t ) −σβη1 (H,

 2 Z  t 2 ≤ C1 H

 2 + d1 ≤ C2 ∆Z

α2  22 + d2 .

∇∆Z

≤ l





376

Landau–Lifshitz Equations

(11) Estimate of the eleventh term of (6.5.24).       β(Ht , ∇(∇ · Z))

 t 2 ∆Z

 2 ≤ β H  2 + C.  t 2 + α2 ∇∆Z

≤ ε 3 H 2 2 l

(12) Estimate of the twelfth term of (6.5.24).     ∇(∇ · Z))   η1 β(H,

 2 ∆Z

 2 ≤ η1 β H

α2  2 + C. ≤

∇∆Z

2 l

Combining (1)–(12) and using (6.5.24) we have (i) If n = 1, de1 (t)  t 22 + (σ − η1 − ε1 − ε2 ) H dt  t 2 + η2 ∇E

 2 + (σ − η2 − ε2 ) E 2



+ η1 −


+ α2

2α12

+ 

4α22

2

2



+ 2β α2 (|α1 | + α2 )  2

∇H

2 α2

1 7  22 ≤ C1 . −

∇∆Z

8 l

(6.5.25)

(ii) If n = 2, de1 (t)  t 2 + (σ − η1 − ε1 − ε3 ) H 2 dt  t 22 + η2 E  t 22 + (σ − η2 − ε2 ) E 



2α2 + 4α22 + 2β 2 α2 (|α1 | + α2 )  2 + η1 − 1

∇H

2 α2 


1 6  2 − + α2 − C(|α1 | + α2 ) ∇Z

8 l C2 2  2 ] ∇∆Z

 2 ≤ C2 , − (α + (|α1 | + 2α2 )2 ) ∇Z

2 2 α2 2

(6.5.26)

where C1 , C2 are independent of t, positive numbers ε1 , ε2 , ε3 , η1 , η2 and l are to be determined. In (6.5.25) and (6.5.26) we let 2α12 + 4α22 + 2β 2α2 (|α1 | + α2 ) < σ, α2

(6.5.27)

2α12 + 4α22 + 2β 2 α2 (|α1 | + α2 ) < η1 < σ, α2

(6.5.28)

377

Long Time Behavior

η2 = ε2 = l > 56,

n = 1;

σ , 4

l = 60,

(6.5.29) n = 2.

(6.5.30)

 0 2 , ∇E  0 2 , ∇H  0 2 are small. It follows from Lemma 6.4.2 Suppose that ∇Z that α12 + (|α1 | + 2α2 )2  2

∇Z

2 α2

√ 1    0 2 + ∇Z  0 2 ≤ C(|α1 | + α2 ) √ E 2σ E 0 2 + H 0 2 + β

 t) 2 + C C(|α1 | + α2 ) ∇Z(·,



1 C + α22 + (|α1 | + 2α2 )2 α2 β





σ  0 2 + H  0 2 + β ∇Z  0 2 1+

E 2 β

'

α2 . 40 (6.5.31)

≤

Therefore, there exists a constant a > 0 such that  de1 (t)  t 22 + E  t 22 + ∇H

 22 + ∇E

 22 + ∇∆Z

 22 ≤ C, + a H dt

(6.5.32)

where C is independent of t.  = 0. From Poincar´e inequality we have  t) is periodic in x,  2D ∆Zdx Since ∆Z(x, 0  22 ≤ δ ∇∆Z

 22 .

∆Z

Choosing δ0 = min(a, δa0 ), we have from (6.5.32) that  de1 (t)  t 2 + E  t 2 + ∇H

 2 + ∇E

 2 + ∆Z

 2 ≤ C. + δ0 H 2 2 2 2 2 dt

(6.5.33)

Equation (6.5.33) can be rewritten as de1 (t) + 2δ0 e1 (t) ≤ C + 2δ0 R(t) + C + 2δ0 sup |R(t)|. dt t Hence, we have




C ≤ e1 (0) + + sup |R(t)| (e2δ0 − 1), e1 (t)e 2δ0 t C e1 (t) ≤ e1 (0) + + sup |R(t)|, 2δ0 t 2δ0 t

that is






G(t) ≤ 2C0 + 2 sup |R(t)| − R(t) t

≤ 2C0 + 4 sup |R(t)|. t

(6.5.34)

378

Landau–Lifshitz Equations

It follows from (6.5.23) that β2  2

Z t 2 2  t 2 ∇Z

 2 + η1 H

 2 ∇E

 2 + β Z

 t 2 +  t 2 H R(t) ≤ β Z

β2    2 + η2 E 2 Et 2 + ∇Z

2 2 1  2 + 1 ση2 E

 2 + βη2 E

 2 ∇Z

 2 + ση1 H

2 2 2 β + β2  2 β  2  t 2 + ∇E

 2 ) + C1

Zt 2 + ( Ht 2 + E ≤ 2 2 2 2  22 ≤ (β + β 2 )(α12 + α22 ) ∆Z

β  2  2  2 + ( H t 2 + Et 2 + ∇E 2 ) + C2 . 2 Taking β <

1 2

and (β + β 2 )(α12 + α22 ) < 14 , we have 

(6.5.35)

'

β , (β + β 2 )(α12 + α22 ) < 1. a0 = 4 max 2 It follows from (6.5.36) |R(t)| ≤

1  t 2 + E  t 2 + ∇E

 2 + ∆Z

 2 ) ≤ 1 a0 G(t). a0 ( H 2 2 2 2 4 4

(6.5.36)

Substituting (6.5.37) into (6.5.35), we have G(t) ≤ 2C0 + a0 sup G(t), t

that is sup G(t) ≤ t

2C0 = d0 . 1 − a0

This proves the lemma.  t), H(x,   Lemma 6.5.3 Assume that Z(x, t), E(x, t) are smooth solutions of (6.4.1) 2 1    and (6.4.2). (Z0 (x), H0 (x), E0 (x)) ∈ (H (Ω), H (Ω), H 1 (Ω)), Ω ⊂ Rn , 1 ≤ n ≤ 2.  0 (x)| = 1, and assume |Z (i) 4α2 + 2α12 + 2α2 β 2 (|α1 | + α2 ) ; α2 > 0, σ > 2 α2 (ii) 1 1 0 < β < , (β + β 2 )(α12 + 2α22 ) < ; 2 4 (iii) When n = 2,  0 (x) 2 + E  0 (x) 2 ≤ λ,  0 (x) 2 + H

∇Z 2 2 2

Long Time Behavior

379

where λ = λ(α1 , α2 , β) is a small constant, then  t) 2 2 + H(·,  t) 2 1 + E(·,  t) 2 1 ] ≤ K, sup [ Z(·, H (Ω) H (Ω) H (Ω)

t∈[0,∞)

where the constant K depends on 2  0 (x) 2H 1 (Ω) + E(x)

  0 (x) 2H 2 (Ω) + H

Z H 1 (Ω) .

 0 (x), H  0 (x), Theorem 6.5.1 Assume α2 > 0, β ≥ 0, σ > 0 and the initial data (Z n k k−1 k−1 n   E0 (x)) ∈ (H (Ω), H (Ω), H (Ω)), k ≥ 1 + [ 2 ], Ω ⊂ R , 1 ≤ n ≤ 2, |Z0 (x)| = 1,  0 = 0, ∇(H  0 + βZ  0 ) = 0. When n = 2, assume and ∇ · E  0 (x) 2 + H  0 (x) 2 + E  0 (x) 2 < δ,

∇Z where δ is a small constant; then the periodic initial value problem (6.5.1)–(6.5.2) admits a unique global smooth solution:  t)| = 1, |Z(x,

x ∈ Ω,

t ∈ R+ ,

k

] s 2  t) ∈ ∩[s=0 Z(x, W∞ (0, T ; H k−2s(Ω)), s k−s−1  H(x, t) ∈ ∩k−1 (Ω)), s=0 W∞ (0, T ; H  E(x, t) ∈ ∩k−1 W s (0, T ; H k−s−1(Ω)). s=0

∞

Theorem 6.5.2 Assume that the conditions of Theorem 6.5.1 hold. Then the periodic initial value problem (6.5.1)–(6.5.2) admits an attractor A (i) A is weakly compact in H 2 (Ω) × H 1 (Ω) × H 1 (Ω); (ii) S(t)A = A; (iii) limt→∞ dist(S(t)B, A) = 0 for any bounded set B ∈ D ⊂ H 2 (Ω) × H 1 (Ω) × H 1 (Ω), where  E,  H)  ∈ H 2 (Ω) × H 1 (Ω) × H 1(Ω), D = {(Z,  t)| = 1, ∇ · E  = 0, ∇ · (H  + β Z)  = 0}, |Z(x,  0, H  0, E  0 ) is the semigroup formed by probdist(x, y) = supx∈X inf y∈Y x − y . S(t)(Z lem (6.5.1)–(6.5.2). Proof. It follows from Theorem 6.5.1 that the periodic initial value problem  0, E  0 ) and it is continuous by [70].  0, H (6.5.1)–(6.5.2) has a semigroup operator S(t)(Z Taking the subset D, it follows from Lemma 6.5.2 that the operator S(t) : E → E is bounded and  t)| = 1, ∇ · E  = 0, ∇ · (H  + β Z)  = 0, A = {|Z(x,  t) ∈ H 2 (Ω), H(·,  t) ∈ H 1 (Ω), E(·,  t) ∈ H 1 (Ω), Z(·,  t) 2 2 + H(·,  t) 2 1 + E(·,  t) 2 1 ≤ ε0 + δ0 }

Z(·, H H H

380

Landau–Lifshitz Equations

is a bounded absorbing set in D, then A = ΩA is a weakly compact attractor of the periodic initial value problem. Now we estimate the Hausdorff dimension and Fractal dimension of A. To this aim we consider the linear variational problem of problem (6.5.1)–(6.5.2):  Z  + α2 |∇Z|  2 z + α1 Z  × ∆z zt = α2 ∆z + 2α2 (∇z · ∇Z)  + H)  × z + α2 h − α2 (z · H)  Z   × h − α1 (∆Z + α1 Z  · h)Z  − α2 (Z  · H)z,  − α2 (Z

(6.5.37)

et = ∇ × h − σe,

(6.5.38)

ht = −∇ × e − βzt ,

(6.5.39)

z(0) = z0 ,

h(0) = h0 ,

e(0) = e0 .

(6.5.40)

Simply write (6.5.37)–(6.5.40) as operator form: vt = −L(u)v,

v(0) = v0 ,

(6.5.41)

 E,  H  are all three E,  H),  v0 = (z0 , e0 , h0 ), z, e, h, Z, where v = (z, e, h), u = (Z, dimensional vector-valued functions. Since the periodic initial value problem (6.5.1)–(6.5.2) has smooth solution, the coefficients of the linear system (6.5.37)–(6.5.39) are smooth. If v0 is properly smooth, then problem (6.5.37)–(6.5.40) admits global smooth solution. Denote its solution operator by G(t), that is v(t) = G(t)v0 . Moreover, we can prove that the semigroup operator S(t)u0 is differentiable in L2 (Ω) and the Frechet differential S  (t)u0 = G(t)v0 . We first prove  t), H(x,   Lemma 6.5.4 The smooth solution (Z(x, t), E(x, t)) of problem (6.5.1)– (6.5.2) is continuously dependent on the initial data.  i(x, t), E  i (x, t)) (i = 1, 2), be smooth solutions of (6.5.1)  i (x, t), H Proof. Let (Z  0i (x), H  i (x, 0) = H  0i (x), E  i (x, 0) = E  0i (x)  and (6.5.2) with initial data Zi (x, 0) = Z  2 (x, t), H(x,   2 (x, t) − Z  1 (x, t), E(x,   t) = Z  2 (x, t) − Z t) = H t) = (i = 1, 2). Let Z(x,      E2 (x, t) − Z1 (x, t); then (Z(x, t), H(x, t), E(x, t)) satisfies  + α1 Z  × ∆Z  2 + α1 Z  1 × ∆Z  ×H 2  t = α2 Z Z  + α2 ∆Z  + α2 |∇Z  2 |2 Z 1 × H  + α1 Z  ∇(Z 1 + Z  2 ))Z  1 + α2 H  + α2 (∇Z, 2 · H  2 )Z  − α2 (Z 2 · H  +H  1 · Z)  Z  1, − α2 (Z t = ∇ × H  − σ E,  E  t = −∇ × E  − βZ  t, H  + β Z)  = 0, ∇ · (H

 = 0, ∇·E

(6.5.42) (6.5.43) (6.5.44) (6.5.45)

381

Long Time Behavior

 + D, t) = Z(x  − D, t), Z(x  + D, t) = H(x  − D, t), H(x  + D, t) = E(x  − D, t), E(x  0) = Z  0 (x), Z(x,

  0 (x), H(x, 0) = Z

  0 (x), E(x, 0) = E

(6.5.46)

 0 (x)| = 1, |Z  0 = 0. ∇·E

 0) = 0,  0 + βZ ∇ · (H

(6.5.47)

We may establish the following inequality  t) 2 1 + H(·,  t) 2 2 + E(·,  t) 2 2 ] sup [ ∇Z(·, H L L

0≤t≤T

 0 (x) 2 1 + H  0 (x) 2 2 + E  0 (x) 2 2 ), ≤ C( ∇Z H L L

(6.5.48)

where C is an absolute constant. It is clear that if (6.5.48) holds, then the conclusion of Lemma 6.5.6 is proved.  we have In fact, taking the inner product of (6.5.42) with ∆Z, 1∂ 2 ∂t

 Ω

 2 dx + α2 ∇Z

 2 ≤ C1 [ ∇Z

 2 + H

 2 + E

 2 ]. |∇Z| 2 2 2 2

(6.5.49)

 we have Taking the inner product of (6.5.42) with ∆Z, 1∂ 2 ∂t

 Ω

 2 dx + α2 ∆Z

 2 |∇Z| 2

≤ −α1 − α1 − α2

  

+ α2 

+ α2 in which

   −α1 



 − α1  × ∆Z  2 · ∆Zdx Z

 · ∆Zdx  − α2 1 × H Z





  ×H  2 · ∆Zdx Z

 2Z  · ∆Zdx  |∇Z|

 · ∇(Z 1 + Z  2 ))Z  1 · ∆Zdx  (∇Z  · ∆Zdx  + α2 H



2 · H  2 )Z  · ∆Zdx  (Z

2 · H  +H  1 · Z)  Z  1 · ∆Zdx,  (Z 



  = α1  × ∆Z  2 · ∆Zdx Z  



(6.5.50) 

   × ∇∆Z  2 · ∇Zdx Z 

 2 ∞ Z

 2 ∇Z

 2 ≤ |α1 | ∇∆Z  2 + ∇Z

 2 ), ≤ C|α1 |( Z

2

2

382

Landau–Lifshitz Equations    −α1 

   −α2 





  = α1  ×H  2 · ∆Zdx Z  

 2 |2 Z  |∇Z

   · ∆Zdx





   × ∇H  2 · ∇Zdx Z 

 2 ∞ Z

 2 ∇Z

 2 ≤ |α1 | ∇H  2 + ∇Z

 2 ), ≤ C|α1 |( Z

2

2

 2 2 Z

 2 ∆Z

 2 ≤ |α2 | ∇Z ∞ ≤

α2  2  2 + C(K)α2 Z

∆Z

2 2 K

where K is a constant to be determined.    −α2 



 · ∇(Z 1 + Z  2 ))Z  1 · ∆Zdx   (∇Z 

 1 + ∇Z  2 ∞ ∇Z

 2 ∆Z

 2 ≤ α2 ∇Z α2  2,  2 + C(K)α2 Z

≤ ∆Z

2 2 K    α2 



 2 ∆Z

 2  · ∆Zdx   ≤ α2 H

H  ≤

   α2 



α2  2 + C(K)α2 H

 2,

∆Z

2 2 K

2 · H  2 )Z  · ∆Zdx   ≤ α2 H  2 ∞ Z  2 ∞ Z

 2 ∆Z

 2 (Z  ≤    α2 

α2  2 + C(K)α2 Z

 2,

∆Z

2 2 K 

2 · H  +H  1 · Z)  Z  1 · ∆Zdx   (Z   2 ∆Z

 2 + H  1 ∞ Z

 2 ∆Z

 2) ≤ α2 ( H

α2  2 + C(K)α2 ( Z

 2 + H

 2 ). ≤ ∆Z

2 2 2 K

Therefore, it follows from (6.5.50) that 

1d  2 + α1 Z  · ∆Zdx  1 × H  2 + α2 ∇Z

∇Z

2 2 2 dt α2  22 + C(K)( ∇Z

 22 + Z

 22 + H

 22 ). ≤ 5 ∆Z

K

(6.5.51)

 and multiplying (6.5.44) by H  and integrating over Ω Multiplying (6.5.43) by E (β > 0), we have 1 d 2β dt

 Ω

 2 + |H|  2 )dx + (|E|

σ  2

Z 2 = − β

 Ω

  t · Hdx. Z

(6.5.52)

383

Long Time Behavior

It follows from (6.5.51) and (6.5.52) that 1d   22 + |Z

 22 + 1 (|Z|  2 ))dx + α2 ∆Z

 22  2 + |H| (|∇Z| 2 dt Ω β  α2  2 t · H  + α1 Z  · ∆Z)dx  1 × H ≤ − (Z + 5 ∆Z

2 K Ω  2 + Z

 2 + H

 2 ). + C(K)( ∇Z

2

2

(6.5.53)

2

 we have Multiplying (6.5.42) by H,    

 

Ω

t · H  + α1 Z  · ∆Z)dx  1 × H  (Z  ≤

          |α1 |  (Z × ∆Z2 + Z × H2 ) · Hdx Ω          · Hdx   + α2  |∇Z|  2Z  · Hdx   + α2  ∆Z    Ω Ω      · (∇(Z 1 + Z  + 2))Z  · Hdx   + α2  |∇Z  Ω    2 + α2  (Z 2 · H  2 )Z  · Hdx   + α2 H

  2 Ω        2 · H)(  Z  1 · H)dx   1 · Z)(  Z 1  + α2  (H + α2  (Z   Ω

Ω

α2  2 + C(K)( ∇Z

 2 + Z

 2 + H

 2 ). ≤ ∆Z

2 2 2 2 K

 

  · H)dx  (6.5.54)

It follows from (6.5.53) and (6.5.54) that 1d 2 dt

  Ω



 2 + |H|  2 + |Z

 2 + 1 (|Z|  2 ) dx |∇Z| 2 2 β

 2 + Z

 2 + H

 2 ]. ≤ C(K, α1 , α2 )[ ∇Z

2 2 2 The lemma is proved. In order to prove that the operator semigroup S(t) is Frechet differentiable, we consider a linear variational problem of (6.5.1)–(6.5.2) as follows: 1 + H  1 ) + α1 Z  1 × (∆w + I) wt = α1 w × (∇Z  1 × (Z 1 + H  1 )) − α2 w × (Z 1 + H  1 ))  1 × (w × (∆Z − α2 Z  1 × (∆w + I)),  1 × (Z − α2 Z

(6.5.55)

where ∇ × I = F + σ F , ∇ × F = −It − βwt ,

(6.5.56)

384

Landau–Lifshitz Equations

 + β Z)  = 0, ∇ · (H

 = 0, ∇·E

(6.5.57)

 0, H  0, E  0 ), (w(t), I(t), F (t))|t=0 = (Z (6.5.58) 1, H  1, E  1 ) = S(t)(Z  01 , H  01 , E  01 ) is a solution of (6.5.1) with initial data where (Z  01 , E  01 ). Set  01 , H (Z ˜ H, ˜ E) ˜ = (Z,  H,  E)  − (w, I, F ) (Z,  01 , H  01 , E  01 ) − S(t)(Z  0, H  0, E  0) = S(t)(Z  01 , H  01 , E  01 )(Z  0, H  0, E  0 ). − (DS(t))(Z

(6.5.59)

Hence,  × (∆Z 2 + H  2 ) − w × (∆Z 1 + H  1 )] Z˜t = α1 [Z  1 × (∆Z  + H)  −Z  1 × (∆w + I)] + α1 [Z  × (Z  2 × (∆Z 2 + H  2 )) − Z  × (Z  1 × (∆Z 1 + H  1 ))] − α2 [Z  1 × (Z  × (∆Z 2 + H  2 )) − Z  × (w × (∆Z  + H))]  − α2 [Z  1 × (Z  1 × (∆Z  + H))  −Z  × (Z  1 × (∆w + I))], − α2 [Z

(6.5.60)

˜ = E˜t + σ E, ˜ ∇×H

(6.5.61)

˜ = −H ˜ t − β Z˜t , ∇×E

(6.5.62)

˜ + β Z) ˜ = 0, ∇ · (H

∇ · E˜ = 0,

˜ H, ˜ E)| ˜ t=0 = 0. (Z,

(6.5.63) (6.5.64)

Rewrite (6.5.60) as follows: 1 + H  1) + Z  × (∆Z  + H)]  Z˜t = α1 [Z˜ × (∆Z ˜ + α1 [Z˜1 × (∆Z˜ + H)]  × (Z  × (∆Z 2 + H  2 )) + Z  × (Z  1 × (∆Z  + H))  − α2 [Z  1 × (∆Z 1 + H  1 ))] + Z˜ × (Z  × (Z˜ × (∆Z 1 + H  1 )) + Z  1 × (Z  × (∆Z  + H))]  − α2 [Z  1 × (Z  1 × (∆Z˜ + H))]. ˜ − α2 [Z

(6.5.65)

It follows from (6.5.65) that 1d ˜ 2 ˜ 2 + C2 ( Z

 2 + H

 2 + E

 2 )4 .

Z 2 ≤ C1 Z

2 2 2 2 2 dt This implies 2 2 C1 t ˜ ˜ +

Z(t)

2 ≤ Z(0) 2 e

 t

=

0

 t 0

2 2 2 2    eC1 (t−s) C2 ( Z(s)

2 + H(s) 2 + E(s) 2 ) ds

4    eC1 (t−s) C2 ( Z(s)

2 + H(s) 2 + E(s) 2 ) ds.

385

Long Time Behavior

It follows from Lemma 6.5.4 that for 0 ≤ t ≤ T ,  0 2 + E  0 2 )2 . ˜  0 2 + H

Z(t)

≤ C(T, K)( Z ˜ ˜ Similarly, we can estimate H(t)

2 and E(t) 2 . We have  0, H  0, Lemma 6.5.5 If solutions of (6.5.1)–(6.5.2) are properly smooth, then S(t) : (Z        E0 ) → (Z(t), H(t), E(t)) is uniformly differentiable. Its differential at (Z0 , H0 , E0 ) belongs to A and  0, E  0 ) = (w(t), I(t), F (t))  0, H DS(t)(Z is a solution of (6.5.55)–(6.5.58).

6.5.3

Dimension Estimate of Attractor

Now we estimate the Hausdorff dimension and fractal dimension of A. Rewrite (6.5.37)–(6.5.39) as  z , ∇z, ∆z , H;  Z,  ∇Z,  ∆Z,  H)  = 0, zt + f( et + σe − ∇ × h = 0, ht + βzt + ∇e = 0,

(6.5.66) (6.5.67) (6.5.68)

where  Z,  ∇Z,  ∆Z,  H)  f(z, ∇z , ∆z , H;  Z  − α2 |∇Z|  2z = −α2 ∆z − 2α2 (∇z · ∇Z)  × ∆z − α1 Z  × h − α1 Z  + H)  × z − α2h + α2 (Z  · H)  z + α1 (∆Z  · h)Z  + α2 (z · H)  Z.  + α2 (Z

(6.5.69)

Choosing periodic orthogonal function basis (ϕj (x), ej (x), hj (x)) such that (i) ∆ϕj = −λ2j ϕj ; (ii) we have

ϕj 2 = ej 2 = hj 2 = 1,

∇ϕj 2 = |λj |,

∆ϕj 2 = λ2j .

From the definition we have Trac{L(u(t))QJ (t)} =

J 

 j ; Z,  ∇Z,  ∆Z,  H),  ϕj ) (f(ϕj , ∇ϕj , ∆ϕj , H

j=1

+ σ(ej , ej ) − (∇ × hj , ej ) + (∇ × ej , hj )

 j , ∇ϕj , ∆ϕj , H  j ; Z,  ∇Z,  ∆Z,  H),  hj ) . − β(f(ϕ

(6.5.70)

386

Landau–Lifshitz Equations

Since −(∇ × hj , ej ) + (∇ × ej , hj ) =

 Ω

∇ · (ej × hj )dx = 0,

we only have in (6.5.70) the following two terms:  j ; Z,  ∇Z,  ∆Z,  H),  ϕj ) (f(ϕj , ∇ϕj , ∆ϕj , H and

 j ; Z,  ∇Z,  ∆Z,  H),  hj ). (f(ϕj , ∇ϕj , ∆ϕj , H

It follows from (6.5.69) that  j , ∇ϕj , ∆ϕj , H  j ; Z,  ∇Z,  ∆Z,  H),  ϕj ) (f(ϕ  · ∇ϕj )Z,  ϕj ) = −α2 (∆ϕj , ϕj ) − 2α2 ((∇Z  2 ϕj , ϕj ) − α1 (Z  × ∆ϕj , ϕj ) − α2 (|∇Z|  ×H  j , ϕj ) − α2 (hj , ϕj ) + α2 ((ϕj · H)  Z,  ϕj ) − α1 (Z  · hj )Z,  ϕj ) + α2 ((Z  · H)ϕ  j , ϕj ), + α2 ((Z in which

−α2 (∆ϕj , ϕj ) = α2 λ2j ,    −2α2 ((∇Z



 ϕj ) ≤ 2α2 ∇ϕj 2 ϕj 2 ∇Z

 ∞ · ∇ϕj )Z,

    2 ϕj , ϕj ) −α2 (|∇Z|    −α1 (Z



 ∞, = 2α2 |λj | ∇Z

 2 ϕj 2 ≤ α2 ∇Z

∞ 2  2, = α2 ∇Z

∞ 



   × ϕj ) × ∆ϕj , ϕj ) = α1 (∇ϕj , ∇Z

   −α1 (Z



 ∞ ≤ |α1 | ∇ϕj 2 ϕj 2 ∇Z

 ∞, = | − α1 λj | ∇Z

 j , ϕj ) ≤ |α1 | hj 2 ϕj 2 = |α1 |, ×H

|−α2 (hj , ϕj )| ≤ α2 ,

  α2 ((ϕj



 Z,  ϕj ) ≤ α2 H

 ∞ ϕj 2 · H) 2

    · hj )Z,  ϕj ) α2 ((Z     · H)ϕ  j , ϕj ) α2 ((Z

 ∞, = α2 H

≤ α2 hj 2 ϕj 2 = α2 ,  ∞. ≤ α1 H

Hence,  j , ∇ϕj , ∆ϕj , H  j ; Z,  ∇Z,  ∆Z,  H),  ϕj ) (f(ϕ  ∞ − α2 ∇Z

 2. ≥ α2 λ2j − (2α2 + |α1 |)λj ∇Z

∞

(6.5.71)

387

Long Time Behavior

Similarly, it follows from (6.5.69) that  j , ∇ϕj , ∆ϕj , hj ; Z,  ∇Z,  ∆Z,  H),  hj ) −β(f(ϕ  · ∇ϕj )Z,  , hj ) = α2 β(∆ϕj , hj ) + 2α2 β((∇Z  2 ϕj , hj ) + α1 β(Z  × ∆ϕj , hj ) + α2 β(|∇Z|  + H)  × ϕj , hj ) + α2 β(hj , hj ) − α1 β((∆Z  Z,  hj ) − α2 β((Z  · hj )Z,  hj ) + α2 β((ϕj · H)  · hj )Z,  hj ) − α2 β((Z  · H)ϕ  j , hj ), − α2 β((Z in which |α2 β(∆ϕj , hj )| ≤ α2 β ∆ϕj 2 hj 2 = α2 λ2j β,

   2α2 β((∇Z



 hj ) ≤ 2α2 β ∇ϕj 2 hj 2 ∇Z

 ∞ · ∇ϕj )Z,

    2 ϕj , hj ) α2 β(|∇Z|    α2 β(Z    −α1 β((∆Z

 

 ∞, = 2α2 β|λj | ∇Z

 2 hj 2 ϕj 2 ≤ α2 β ∇Z

∞  2, = α2 β ∇Z

∞

× ∆ϕj , hj ) ≤ α2 β ∆ϕj 2 hj 2 = α2 λ2j β,  

 × ϕj , hj ) ≤ |α1 |β ∆Z  + H

 ∞ ϕj 2 hj 2 + H)  + H

 ∞, = |α1 |β ∆Z |α2 β(hj , hj )| = α2 β hj 22 = α2 β,

    Z,  hj ) −α2 β((ϕj · H)     · hj )Z,  hj ) −α2 β((Z     · H)ϕ  j , hj ) −α2 β((Z

 ∞, ≤ α2 β H

≤ α2 β,  ∞. ≤ α2 β H

Hence  j ; Z,  ∇Z,  ∆Z,  H),  hj ) −β(f(ϕj , ∇ϕj , ∆ϕj , H  ∞ ≥ −|α1 |βλ2j − 2α2 βλ2j − 2α2 β|λj | ∇Z



 2 + |α1 |β ∆Z  + H

 ∞ + 2α2 β H

 ∞ . − α2 β ∇Z

∞

(6.5.72)

388

Landau–Lifshitz Equations

Substituting (6.5.71) and (6.5.72) into (6.5.70), we have Trac{L(u(t))QJ (t)} ≥ (α2 − (α2 + |α1 |β))

J 

λ2j

j=1

− (2α2 + 2α2 β + |α1 |)

J 

 ∞ λj ∇Z

j=1

 2 − 2α2 (1 + β) H

 ∞ + σ − α2 (1 + β) ∇Z

∞  + H

 ∞. − |α1 |β ∆Z (6.5.73) Choosing β small such that α2 > (α2 + |α1 |)β, we have 0<β<

α2 . α2 + |α1 |

(6.5.74)

Setting  δ            X                 

= α2 − (α2 + |α1 |)β, 

=



J 

1 2

λ2j 

,

j=1

 ∞, a = (2α2 + 2α2 β + |α1 |) ∇Z

 2 − 2α2 (1 + β) H

 ∞ b = σ − α2 (1 + β) ∇Z

(6.5.75)

∞

 + H

 ∞. − |α1 |β ∆Z

Noting that J 



|λj | ≤



j=1

J 

1 2

λ2j 

J,

j=1

we can rewrite (6.5.73) as 1

Trac{L(u(t))QJ (t)} ≥ δX 2 − aJ 2 X + bJ 

1

1 2

aJ 2  4δb − a2 J. + = δ X − 2δ 4δ

(6.5.76)

It follows from (6.5.76) that for 4δb − a2 ≤ 0 Trac{L(u(t))QJ (t)} √ √    a + a2 − 4δb 1 a − a2 − 4δb 1 J2 J2 . X− ≥δ X− 2δ 2δ Take J such that X≥

a+

√ a2 − 4δb 1 J 2, 2δ

(6.5.77)

389

Long Time Behavior

and estimate λj as in [138] as follows: 

2

1

2 1 (j − 1) n 1 − 1 = (j − 1) n − (j − 1) n + 1, λ2j ≥  2 4

that is λ2j ≥

1 (j

− 1)2 − j + 2 = 14 j 2 − 32 j + 94 , 1 1 − 1) − (j − 1) 2 + 1 = 14 j + 34 − (j − 1) 2 ,

4 1 (j 4

(i) n = 1. J 

λ2j ≥

j=1

J J 1 3 9 λ2j − λj + J 4 j=1 2 j=1 4

3 9 1 (J + 1)(2J + 1)J − (J + 1)J + J 24 4 4 1 3 8 2 37 J − J + J. = 12 5 24

=

In order to have (6.5.77), choose J0 such that 

1 3 8 2 37 (a + J0 − J0 + − 12 5 24

√  a2 − 4δb)2 J0 > 0, 4δ 2

that is √

a2 − 4δb)2 > 0, δ2 √
 15 2 71 3(a + a2 − 4δb)2 − J0 − + > 0. 4 16 δ2 2J02

When

− 15J0 + 37 −




15 J0 − 4

2

71 3(a + + − 16

that is

J0 >

√ a2 − 4δb)2 > 0, δ2

(a +

√ a2 − 4δb)2 < 2δ,

(a +

√ a2 − 4δb)2 ≥ 2δ,

we may choose J0 = 1. When

we may choose

6(a +

) ( *  * 3 a + (a2 +

δ2

− 4δb)

2

−

71 15 + . 16 4

n = 1, n = 2.

390

Landau–Lifshitz Equations

(ii) n = 2. J 

λ2j ≥

j=1

=

J J  1 1 3 λj + J − (j − 1) 2 4 j=1 4 j=1 J−1  1 1 3 J2 (J + 1)J + J − 8 4 j=1

J 2 + 7J − ≥ 8 ≥

 J−1  1 2√ 

j

J −1

j=1

J 2 + 7J 1 3 − √ J 2. 8 2

In order to have (6.5.77), choose J0 such that √ J0 + 7 1 12 (a + a2 − 4δb)2 − √ J0 > , 8 4δ 2 2 that is





√ a+ (J0 − 2 2)2 − 1 + 2 1 2

  



√ 2  a2 − 4δb  > 0, δ 

√ 2  12 2  2 − 4δb √ a + a  J0 >  2 2 +  1 + 2 > 0.  δ   

We have from the above results the following: Theorem 6.5.3 Let Ω ⊂ Rn (1 ≤ n ≤ 2) be a bounded set and assume (i) 4α2 + 2β 2α2 (|α1 | + α2 ) + 2α12 . α2 > 0, σ > 2 α2 (ii)



'

1 α2 , . 0 < β < min 2 α2 + |α1 | (iii) 1 (β + β 2 )(α12 + α22 ) < . 4 (iv) When n = 2

 0 2 + E  0 2 ≤ ν,  0 2 + H

∇Z

where ν = ν(α1 , α2 , β) is a small constant. Then the periodic initial value problem (6.5.1)–(6.5.2) has a attractor A = Ω(A) and  H,  E)  ∈ (H 2 (Ω), H 1 (Ω), H 1 (Ω)) Z

 H 2 + H

 H 1 + E

 H 1 ≤ ε0 + δ0 } A = {(Z,

391

Long Time Behavior

is a bounded absorbing set. The Hausdorff dimension and Fractal dimension of A are finite and satisfy (1) If a2 − 4δb < 0, then dH (A) ≤ 1,

dF (A) ≤ 2;

(2) If a2 − 4δb ≥ 0, then √ 2 (a) when n = 1 and √ a + a − 4δb < 2δ, then dH (A) ≤ 1, dF (A) ≤ 2; when n = 1 and a + a2 − 4δb > 2δ, then dH (A) ≤ J1 , dF (A) ≤ 2J1 where J1 is the smallest integer subject to

J0 >

)  * √ * 3 a + a2 +

2

− 4δb

δ2

−

71 15 + . 16 4

(b) When n = 2, then dH (A) ≤ J2 , dF (A) ≤ 2J2 , where J2 is the smallest integer subject to √    2 '2 √ a + a2 − 4δb , J0 > 2 2 + 1 + 2 δ in which δ = α2 − (α2 + |α1 |)β,  ∞, a = (2α2 + 2α2 β + |α1 |) ∇Z

 2 b = σ − α2 (1 + β) ∇Z

∞

 ∞ + |α1 |β ∆Z  + H

 ∞. − 2α2 (1 + β) H

6.6

Bibliography Comments

In this chapter we have studied the asymptotic properties for the system of ferromagnetic spin chain. Because the systems are weak degenerate, we do not have the expressions of the linearized equation. This makes the studies of properties when t → ∞ very difficult. Guo and Lu [79] studied the stationary solution and its nonstability and the asymptotic behavior as t → ∞ under some assumptions. Guo and Wang [85, 86] got the existence of approximate inertial manifold for the onedimensional L–L equations. Guo et al. [78] proved the existence of attractor and the estimates of upper and lower bounds of Hausdorff and Fractal dimensions for such attractor for the L–L system on Riemannian manifold. Guo and Su [83] proved the existence of attractor of L–L–M system by the method of multi-parameter Lyapunov functional and have given the estimate in three dimensions. It is worth mentioning that in 1957, Suhl [131] proposed the chaotic property for the attractor of L–L system with Gilbert damping term. The numerical results were given in [147].

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