The Numerical Convergence of the Landau-Lifshitz Equations and Its Simulation

Abstract

A difference scheme of Landau-Lifshitz (LL for short) equations is studied. Their convergence and stability are proved. Furthermore, a new solution of LL equation is given for testing our scheme. At the end, three subcases of this LL equation are concerned about, and some properties about these equations are shown by a numeric simulation way.

1. Introduction and the Number Scheme

The LL equation [1] has aroused considerable interest among physicists and mathematicians. For the one-dimensional case, there have been many contributions to the study of the soliton solution, the interaction of solitary waves, and other properties of the solitary waves [2–4]. However, we point out here that it will be a more challenging task study the high-dimensional dynamics [5, 6] about LL equation. In the classical study of ferromagnetic chain, we often consider the following system (vector form): 𝜕𝑢𝜕𝑡=𝑢×Δ𝑢+𝑓(𝑥,𝑡,𝑢), (1.1) where “×” denotes a 3-dimensional cross product. The isotropic Heisenberg chain [2] is the special case of (1.1). Moreover, another special case of (1.1) is the LL equation with an easy plane 𝜕𝑢𝜕𝑡=𝑢×Δ𝑢+𝑢×(𝑃𝑢),𝑃=diag{𝑃₁,𝑃₂,𝑃₃},𝑃₁≤𝑃₂≤𝑃₃. (1.2)

Equation (1.2) has been studied by the inverse scattering method in [7]. Its integrability theories [8] are established. As far as we know, no explicit solutions of (1.2) are given by various direct methods such as Jacobi elliptical function method [9]. Therefore, the numerical method to study this equation and its more generalized form, which gives some of the visual characteristics of equations, is necessary. However, we do not intend to construct efficient (or optimal control) and high-precision algorithms [10–15] here. In this paper, we are concerned about the convergence of the discrete scheme. Furthermore, we hope that by means of numerical simulation, more intuitive understanding of the nature of the equation will be obtained.

For universality, we consider the following system with the Gilbert damping term which covers the above situations. The periodic condition is about the space variables 𝑥𝜕𝑢𝜕𝑡=−𝛼₁𝑢×(𝑢×Δ𝑢)+𝛼₂(𝑢×Δ𝑢)+𝑓(𝑥,𝑡,𝑢)(𝑥,𝑡)∈𝑅×𝐼,𝑢(𝑥+2𝜋,𝑡)=𝑢(𝑥,𝑡)(𝑥,𝑡)∈𝑅×𝐼,𝑢(𝑥,0)=𝑢₀(𝑥)𝑥∈𝑅, (1.3) where the spin vector 𝑢(𝑥,𝑡)=(𝑢¹,𝑢²,𝑢³)^𝑇 is a 3-dimensional vector-valued unknown function with respect to space variables 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛) and time 𝑡, and period about 𝑢(𝑥,𝑡) is 2𝜋. 𝑓(𝑥,𝑡,𝑢) is continuous with respect to 𝑥,𝑡, and 𝑢. 𝑢₀(𝑥) is the initial value function which is also a 3-dimensional function. 𝐼=[0,𝑇],(𝑇>0), 𝑅^𝑛 is the 𝑛-dimensional real value dominion. 𝛼₁≥0 is damping coefficient. 𝛼₂>0, Laplace operator Δ𝑢=(𝜕²𝑢¹/𝜕𝑥²,𝜕²𝑢²/𝜕𝑥²,𝜕²𝑢³/𝜕𝑥²)^𝑇.

Although the existence of the global attractor of (1.3) has been proved in [16], no exact solutions have been proposed as far as we know. In this paper, we give an exact solution of it, which can test the property of numerical scheme of (1.3). From a physical point of view, the value of 𝑢(𝑥,𝑡) is finite, just like the situation which is mentioned in [5, 16]. In this section, we just suppose that |𝑢(𝑥,𝑡)|=1 for convenience. So the first equation of problem (1.3) is equal to the following form: 𝜕𝑢𝜕𝑡=−𝛼₁Δ𝑢+𝛼₁|∇𝑢|²𝑢+𝛼₂𝐴(𝑢)Δ𝑢+𝑓(𝑥,𝑡,𝑢), (1.4) where ∇𝑢=(𝑢¹_𝑥,𝑢²_𝑥,𝑢³_𝑥)^𝑇, 𝐴(𝑢)=0−𝑢³𝑢²𝑢³0−𝑢¹−𝑢²𝑢¹0∈𝑅^3×3. (1.5) Furthermore, we consider a more general type of Landau-Lifshitz equation (as the variety of (1.4)): 𝜕𝑢𝜕𝑡=−𝛼₁Δ𝑢+𝛼₁𝑔(∇𝑢)𝑢+𝛼₂𝐴(𝑢)Δ𝑢+𝑓(𝑥,𝑡,𝑢)(𝑥,𝑡)∈𝑅^𝑛×𝐼,𝑢(𝑥+2𝜋,𝑡)=𝑢(𝑥,𝑡)(𝑥,𝑡)∈𝑅^𝑛×𝐼,𝑢(𝑥,0)=𝑢₀(𝑥)𝑥∈𝑅^𝑛, (1.6) where 𝑔(𝑣)∶𝑅³→𝑅 is a continuous differential function. When 𝑔(𝑠)=|𝑠|², (1.6) takes the form of (1.4).

For convenien, we discuss our scheme in one-dimensional case which the 𝑛-dimensional case can be discussed similarly. Firstly, we give some notion and symbol. Set Ω=[0,2𝜋],𝐽 is a positive integer. We divide the region Ω×𝐼 as a discrete mesh which ℎ=2𝜋/𝐽 in its space step and 𝑘 in its time step. 𝑥_𝑗=𝑗ℎ,𝑡_𝑛=𝑛𝑘(𝑗=0,1,…,𝐽;𝑛=0,1,…,[𝑇/𝑘]). Here (𝑢^𝑚)^𝑛_𝑗 denotes the approximate value of 𝑢^𝑚(𝑚=1,2,3) on (𝑥_𝑗,𝑡_𝑛). Let 𝑢^𝑛 denote the layer mesh function, that is, 𝑢^𝑛=(𝑢^𝑛₁,𝑢^𝑛₂,…,𝑢^𝑛_𝐽) where 𝑢^𝑛_𝑗=((𝑢¹)^𝑛_𝑗,(𝑢²)^𝑛_𝑗,(𝑢³)^𝑛_𝑗)^𝑇.

Define the discrete inner product and norm as follows: (𝑢^𝑛,𝑣^𝑛)=ℎ^𝐽_𝑗=1𝑢^𝑛_𝑗⋅𝑣^𝑛_𝑗=ℎ^𝐽3_𝑗=1𝑚=1(𝑢^𝑚)^𝑛_𝑗⋅(𝑣^𝑚)^𝑛_𝑗;‖𝑢^𝑛‖=(𝑢^𝑛,𝑢^𝑛)^1/2. (1.7)

We define the following finite difference approximations of derivatives along the space direction: 𝑢^𝑛_𝑗𝑥=𝑢^𝑛_𝑗+1−𝑢^𝑛_𝑗ℎ,𝑢^𝑛_𝑗𝑥=𝑢^𝑛_𝑗−𝑢^𝑛_𝑗−1ℎ,𝑢^𝑛_𝑗𝑥=𝑢^𝑛_𝑗+1−𝑢^𝑛_𝑗−12ℎ. (1.8) Similarly, we can define the derivatives along the time direction and the Laplace operator: 𝑢^𝑛_𝑗𝑡,𝑢^𝑛_𝑗𝑡,𝑢^𝑛_𝑗̂𝑡,and𝑢^𝑛_𝑗𝑥𝑥.

Under the definition and the symbol setting as above, we now define a finite difference approximation of (1.6) by 𝑢^𝑛_𝑗𝑡=𝛼₁𝑢^𝑛_𝑗𝑥𝑥+𝛼₁𝑔(𝑢^𝑛_𝑗𝑥)𝑢^𝑛_𝑗+𝛼₂𝐴(𝑢^𝑛_𝑗)𝑢^𝑛_𝑗𝑥𝑥+𝑓^𝑛_𝑗(𝑢^𝑛_𝑗𝑇)0≤𝑗≤𝐽,0≤𝑛≤[𝑘𝑢]−1,_{𝑛𝑗+𝑟𝐽}=𝑢^𝑛_𝑗𝑢,0≤𝑗≤𝐽,𝑟∈𝑍,_0𝑗=𝑢₀(𝑥_𝑗)0≤𝑗≤𝐽. (1.9)

Obviously, according to difference approximation (1.9), we can start from the zero layer to any layer about𝑢^𝑛(𝑛=1,2,…,[𝑇/𝑘]).

2. Existence of the Solution and the Stability of the Scheme

For studying the convergence and the stability, we first introduce several lemmas.

Lemma 2.1. Considering mesh-function 𝑢^𝑛, 𝑣^𝑛 defined on mesh-points, one has the following relationship:(1)(𝑢^𝑛_𝑥𝑥,𝑣^𝑛)=−(𝑢^𝑛_𝑥,𝑣^𝑛_𝑥),(2)(𝑢^𝑛_𝑡,𝑢^𝑛)≥(1/2)‖𝑢^𝑛‖²_𝑡−(𝑘/2)‖𝑢^𝑛_𝑡‖²,(3)‖𝑢𝑥^𝑛‖≤‖𝑢^𝑛_𝑥‖,(4)‖𝑢_𝑥‖≤(2/ℎ)‖𝑢‖.

Lemma 2.2. Let 𝐴(𝑢) stand for an antisymmetric matrix, 𝑣 is the three-dimension vector, then the following relationships hold:(1)𝐴(𝑢)𝑢=0,(2)(𝑤,𝐴(𝑢)𝑣)=−(𝐴(𝑢)𝑤,𝑣),(3)𝐴(𝑢)𝑣=−𝐴(𝑣)𝑢.

Let 𝑈(𝑥,𝑡) denote the solution of (1.6) (here 𝑈(𝑥,𝑡) will be regarded as a function with smoothness in some degree). 𝑈^𝑛_𝑗 denotes the value of 𝑈(𝑥,𝑡) on (𝑥_𝑗,𝑡_𝑛). 𝑈^𝑛=(𝑈^𝑛₁,𝑈^𝑛₂,…,𝑈^𝑛_𝑗)^𝑇 denotes the value of 𝑛-layer mesh about 𝑈(𝑥,𝑡).

Definition 2.3. One has 𝐶⁰_𝑏(𝑅³,𝑅^3×3)={𝐵(𝑢)∈𝑅^3×3∣‖𝐵(𝑢)‖<∞,∀𝑢∈𝑅³𝐶},_1𝑏(𝑅³)={𝑓(𝑢)∈𝑅∣‖𝑓(𝑢)‖_∞+‖𝑓^′(𝑢)‖_∞<∞,∀𝑢∈𝑅³}. (2.1)

Lemma 2.4. Assume that 𝐴(𝑢)∈𝐶⁰_𝑏(𝑅³;𝑅^3×3) and 𝑈∈𝐶⁰(𝐼;(𝐶³(0;2𝜋))³) are the solution of (1.6), then for any 𝜀>0𝛼₂(𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥,𝑒^𝑛)≤𝜀‖𝑒^𝑛_𝑥‖²+𝛼²₂𝑀²₁4𝜀‖𝑒^𝑛‖², (2.2) where 𝑀₁=sup_𝑢∈𝑅3‖𝐴(𝑢)‖,and𝑒^𝑛=𝑈^𝑛−𝑢^𝑛.

Proof. We have, by (1) of Lemma 2.2, −𝛼₂(𝑈^𝑛_𝑥𝑥,𝐴(𝑒^𝑛)𝑒^𝑛)=0. According to the antiproperty of antisymmetric matrix, we have 𝛼₂(𝑒^𝑛_𝑥,𝐴(𝑢^𝑛)𝑒^𝑛_𝑥)=0. Here, we set 𝑓∈𝐶¹_𝑏(𝑅,𝐼), hence (𝑓(𝑥,𝑡_𝑛)𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝑓(𝑥,𝑡_𝑛)𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥,𝑒^𝑛)=(𝐴(𝑒^𝑛)𝑈^𝑛_𝑥𝑥+𝐴(𝑢^𝑛)𝑒^𝑛_𝑥𝑥,𝑓(𝑥,𝑡_𝑛)𝑒^𝑛)=−(𝑈^𝑛_𝑥𝑥,𝑓(𝑥,𝑡_𝑛)𝐴(𝑒^𝑛)𝑒^𝑛)−(𝑒^𝑛_𝑥𝑥,𝑓(𝑥,𝑡_𝑛)𝐴(𝑢^𝑛)𝑒^𝑛)=(𝑒^𝑛_𝑥,(𝑓(𝑥,𝑡_𝑛)𝐴(𝑢^𝑛)𝑒^𝑛)_𝑥)=ℎ^𝐽−1_𝑗=0𝑒^𝑛_𝑗𝑥[𝑓(𝑥_𝑗+1,𝑡_𝑛)𝐴(𝑢^𝑛_𝑗+1)𝑒^𝑛_𝑗+1+𝑓(𝑥_𝑗,𝑡_𝑛)𝐴(𝑢^𝑛_𝑗)𝑒^𝑛_𝑗]ℎ=ℎ^𝐽−1_𝑗=0𝑒^𝑛_𝑗𝑥𝑓(𝑥_𝑗+1,𝑡_𝑛)𝐴(𝑢^𝑛_𝑗+1)𝑒^𝑛_𝑗𝑥+ℎ^𝐽−1_𝑗=0𝑒^𝑛_𝑗𝑥𝜕𝑓(𝑥_𝑗,𝑡_𝑛)𝜕𝑥𝐴(𝑢^𝑛_𝑗)𝑒^𝑛_𝑗+1. (2.3)
Let 𝑓(𝑥_𝑗,𝑡_𝑛)=1(𝑗=0,1,…,𝐽;𝑛=0,1,…,[𝑇/𝑘]). By 𝜀-inequality, we have𝛼₂(𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥,𝑒^𝑛)≤|𝛼₂|‖𝑒^𝑛_𝑥‖‖𝐴(𝑢^𝑛_𝑥)‖‖𝑒^𝑛‖≤𝜀‖𝑒^𝑛_𝑥‖²+𝛼²₂𝑀²₁4𝜀‖𝑒^𝑛‖². (2.4)

Lemma 2.5. Under the same condition of Lemma 2.4, one has ‖𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥‖≤𝑀₁(‖𝑒^𝑛2‖+ℎ‖𝑒^𝑛_𝑥‖). (2.5)

Proof. We first note that, by (3) of Lemma 2.2, we have, 𝐴(𝑒^𝑛)𝑈^𝑛_𝑥𝑥=−𝐴(𝑈^𝑛_𝑥𝑥)𝑒^𝑛; recalling (4) of Lemma 2.1, we have ‖𝑒^𝑛_𝑥𝑥‖≤(2/ℎ)‖𝑒^𝑛_𝑥‖, then we obtain ‖𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥‖=‖𝐴(𝑒^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑒^𝑛_𝑥𝑥‖≤‖𝐴(𝑒^𝑛)𝑈^𝑛_𝑥𝑥‖+‖𝐴(𝑢^𝑛)𝑒^𝑛_𝑥𝑥‖≤‖−𝐴(𝑈^𝑛_𝑥𝑥)𝑒^𝑛‖+‖𝐴(𝑢^𝑛)𝑒^𝑛_𝑥𝑥‖≤𝑀₁(‖𝑒^𝑛2‖+ℎ‖𝑒^𝑛_𝑥‖). (2.6)

Lemma 2.6. If 𝑔(𝑢)∈𝐶¹_𝑏(𝑅³) and 𝑈∈𝐶⁰(𝐼;(𝐶³(0,2𝜋)))³ are the solution of (1.6), then for any 𝜀>0𝛼₁(𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛,𝑒^𝑛)+(𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛),𝑒^𝑛)≤𝜀‖𝑒^𝑛_𝑥‖²𝛼+(²¹𝑀²₂𝑀²₃4𝜀+𝛼₁𝑀₄+𝑀₀𝛼₁)‖𝑒^𝑛‖², (2.7) where 𝑀₀=sup_{1≤𝑛≤[𝑇/𝐾]}‖𝜕𝑓^𝑛𝜕𝑢‖,𝑀₂=sup_𝑧∈𝑅3‖∇𝑔(𝑧)‖,𝑀₃=sup_{1≤𝑛≤[𝑇/𝐾]}‖𝑈^𝑛‖_∞,𝑀₄=sup_𝑧∈𝑅3‖𝑔(𝑧)‖,𝑒^𝑛=𝑈^𝑛−𝑢^𝑛. (2.8)

Proof. By (3) of Lemma 2.1, we have 𝛼₁(𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛,𝑒^𝑛)+(𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛),𝑒^𝑛)≤𝛼₁[([𝑔(𝑈𝑥^𝑛)−𝑔(𝑢^𝑛𝑥)]𝑈^𝑛,𝑒^𝑛)+(𝑔(𝑢𝑥^𝑛)𝑒^𝑛,𝑒^𝑛)]+𝑀₀‖𝑒^𝑛‖²≤𝛼₁[‖∇𝑔(𝑈𝑥^𝑛+𝜃𝑒𝑥^𝑛)‖‖𝑈^𝑛‖_∞‖𝑒𝑥^𝑛‖‖𝑒^𝑛‖+‖𝑔(𝑢𝑥^𝑛)‖‖𝑒^𝑛‖²]+𝑀₀‖𝑒^𝑛‖²≤𝜀‖𝑒^𝑛_𝑥‖²𝛼+(²¹𝑀²₂𝑀²₃4𝜀+𝛼₁𝑀₄+𝑀₀𝛼₁)‖𝑒^𝑛‖². (2.9)

Lemma 2.7. One has 𝛼₁‖𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛‖+‖𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛)‖≤𝛼₁(2𝑀₂𝑀₃ℎ+𝑀₄+𝑀₀𝛼₁)‖𝑒^𝑛‖. (2.10)

Proof. According to (1) and (2) of Lemma 2.1, we have 𝛼₁‖𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛‖+‖𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛)‖≤𝛼₁(‖∇𝑔(𝑈𝑥^𝑛+𝜃𝑒𝑥^𝑛)‖‖𝑈^𝑛‖_∞‖𝑒𝑥^𝑛‖‖𝑒^𝑛‖+‖𝑔(𝑢𝑥^𝑛)‖‖𝑒^𝑛‖²)+𝑀₀‖𝑒^𝑛‖. (2.11)

Definition 2.8. 𝑆_𝜀(𝑈)={𝑣∈𝐻¹_𝑝(Ω)∣‖𝑈−𝑣‖_𝐻¹_𝑝(Ω)≤𝜀}.

According to the lemmas mentioned above, we now come to discuss the convergence of difference equation  (1.9).

Theorem 2.9. Let 𝑈∈𝐶²(𝐼;(𝐻⁴_𝑝(0,2𝜋))³) be the solution of (1.6), 𝑢^𝑛 is the solution of (1.9), 𝑔(𝑢)∈𝐶¹(𝑅³), 𝑓(𝑢)∈𝐶(𝑅³). For any positive integer 𝜎, if (𝑘/ℎ²)≤(𝛼₁−𝜎)/6(𝛼₁+|𝛼₂|𝑀₁)², then there exists a positive constant 𝐶_𝑖(𝑖=1,2,3,4) which independent of ℎ and 𝑘. If ℎ≤𝐶₁, 𝑘≤𝐶₂, one has sup_{0≤𝑛≤[𝑇/𝑘]}‖𝑈^𝑛−𝑢^𝑛‖+‖|𝑈^𝑛_𝑥−𝑢^𝑛_𝑥|‖≤𝐶₃(𝑘+ℎ²), (2.12) furthermore, ‖𝑈^𝑛−𝑢^𝑛‖_∞≤𝐶₄ℎ, (2.13) where 𝑀₁=𝑚sup_{𝑢∈𝑆𝜀(𝑈)}‖𝐴(𝑢)‖(𝑚>1),‖|⋅|‖=(𝑘∑^[𝑇/𝑘]_𝑛=1‖⋅‖²)^1/2,‖𝑣‖_∞=max_{1≤𝑗≤𝐽}|𝑣_𝑗|.

Proof. According to the condition, the solution of (1.6) 𝑈∈𝐶²(𝐼;(𝐻⁴_𝑝(0,2𝜋))³), substituting the according part of (1.9) with 𝑈(𝑥,𝑡), we have 𝑈^𝑛_𝑡=𝛼₁𝑈^𝑛_𝑥𝑥+𝛼₂𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥+𝛼₁𝑔(𝑈𝑥^𝑛)𝑈^𝑛+𝑟^𝑛, (2.14) where 𝑟^𝑛 is the error estimate of (1.9). According to (2.14) and (1.9), denote 𝑒^𝑛=𝑈^𝑛−𝑢^𝑛, we have 𝑒^𝑛_𝑡=𝛼₁𝑒^𝑛_𝑥𝑥+𝛼₂(𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥)+𝛼₁(𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛)+𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛)+𝑟^𝑛. (2.15)
Take the inner product of (2.15) and 𝑒^𝑛, combine them with (1) and (2) of Lemma 2.1, we have12‖𝑒^𝑛‖²_𝑡−𝑘2‖𝑒^𝑛_𝑡‖²+𝛼₁‖𝑒^𝑛_𝑥‖²=𝛼₂(𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛𝑢^𝑛_𝑥𝑥),𝑒^𝑛)+𝛼₁(𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛,𝑒^𝑛)+(𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛),𝑒^𝑛)+(𝑟^𝑛,𝑒^𝑛). (2.16)
Just like the method mentioned in [17], we can assume that 𝐴^∗(𝑢)∈𝐶^𝑏₀(𝑅³,𝑅^3×3) and 𝑔^∗(𝑢)∈𝐶^𝑏₁(𝑅³). Conveniently, we can denote 𝐴^∗(𝑢) as 𝐴(𝑢) and 𝑔^∗(𝑢) as 𝑔(𝑢). By Lemmas 2.4 and 2.6, (1.9) can change into the following form:12‖𝑒^𝑛‖²_𝑡−𝑘2‖𝑒^𝑛_𝑡‖²+(𝛼₁−2𝜀)‖𝑒^𝑛_𝑥‖²≤𝑀₅‖𝑒^𝑛‖²+12‖𝑟^𝑛‖²+𝑀₀‖𝑒^𝑛‖²≤𝑀₅‖𝑒^𝑛‖²+𝐶²(𝑘+ℎ²)²+𝑀₀‖𝑒^𝑛‖², (2.17) where 𝑀₀=max_{1≤𝑛≤[𝑇/𝑘]}‖𝜕𝑓^𝑛𝜕𝑢‖,𝑀₅=𝛼²₁𝑀²₂𝑀²₃4𝜀+𝛼₁𝑀₄+𝛼²₂𝑀²₁+14𝜀2. (2.18) In the following part, we propose the estimation about ‖𝑒^𝑛_𝑡‖ of (2.17).
By Lemmas 2.5 and 2.7 and (2.15), we have𝑘‖𝑒^𝑛_𝑡‖²≤|𝛼₂|‖𝐴(𝑈^𝑛)𝑈^𝑛_𝑥𝑥−𝐴(𝑢^𝑛)𝑢^𝑛_𝑥𝑥‖+𝛼₁‖𝑔(𝑈𝑥^𝑛)𝑈^𝑛−𝑔(𝑢𝑥^𝑛)𝑢^𝑛‖+‖𝑓(𝑈^𝑛)−𝑓(𝑢^𝑛)‖+‖𝑟^𝑛‖≤2𝛼₁ℎ‖𝑒^𝑛_𝑥‖+𝛼₁(2𝑀₂𝑀₃ℎ+𝑀₄)‖𝑒^𝑛‖+|𝛼₂|𝑀₁(‖𝑒^𝑛2‖+ℎ‖𝑒^𝑛_𝑥‖)+𝑀₀‖𝑒^𝑛‖+𝑟^𝑛≤𝑀₆ℎ‖𝑒^𝑛_𝑥𝑀‖+(⁷ℎ+𝑀₈+𝑀₀)‖𝑒^𝑛‖+𝐶(𝑘+ℎ²), (2.19) where 𝑀₆=2(𝛼₁+|𝛼₂𝑀₁|), 𝑀₇=2𝛼₁𝑀₂𝑀₃, and 𝑀₈=𝛼₁𝑀₄+|𝛼₂|𝑀₁.
So according to the inequality given above, we have𝑘‖𝑒^𝑛_𝑡‖²≤3𝑀²₆𝑘ℎ²‖𝑒^𝑛_𝑥‖²+(6𝑀²₇𝑘ℎ²+6𝑀²₈𝑘+3𝑘𝑀²₀)‖𝑒^𝑛‖²+3𝑘𝐶²(𝑘+ℎ²)². (2.20)
Substitute (2.16) into (2.15), let 𝜀=𝛼₁/4, according to the condition given in Theorem 2.9, we have12‖𝑒^𝑛‖²_𝑡𝛼+(¹2−3𝑀²₆𝑘𝑘²)‖𝑒^𝑛_𝑥‖²≤𝑀₅‖𝑒^𝑛‖²+𝐶(𝑘+ℎ²)², (2.21) where 𝑀₉=𝑀₀+𝑀₅+6𝑀²₈𝑇+𝑀²₇((𝛼₁−𝜎)/(𝛼₁+|𝛼₂|𝑀₁)²).
According to (2.17) and the condition given in Theorem 2.9,‖𝑒^𝑛_𝑡‖²+𝜎‖𝑒^𝑛_𝑥‖²≤2𝑀₉‖𝑒^𝑛‖²+𝐶(𝑘+ℎ²)² (2.22)
By ‖𝑒⁰‖=0 and Gronwall inequality, we have‖𝑒^𝑛‖≤𝐶(𝑘+ℎ²)²exp(𝑀₉𝑇)≤𝐶₃(𝑘+ℎ²)². (2.23)
By (2.22), 𝜎𝑘^𝑚_𝑛=1‖𝑒^𝑛_𝑥‖²≤𝐶𝑇(𝑘+ℎ²)²+2𝑀₉𝑘^𝑚_𝑛=1‖𝑒^𝑛‖². (2.24)
So we have𝑘^[𝑇/𝑘]_𝑛=1‖𝑒^𝑛_𝑥‖²≤𝑀²₁₀(𝑘+ℎ²)², (2.25) where 𝑀²₁₀=(𝐶𝑇+2𝑀₉𝑇𝐶²₃)/𝜎 by (2.25), hence (𝑘^𝑚_𝑛=1‖𝑒^𝑛_𝑥‖²)^1/2≤𝑀₁₀(𝑘+ℎ²)². (2.26)
So by (2.23) and (2.26), when 𝐴(𝑢)∈𝐶_𝑏(𝑅³,𝑅^3×3), 𝑔∈𝐶¹_𝑏(𝑅³), and 𝑓∈𝐶_𝑏(𝑅³), Theorem 2.9 is right. Similarly to [17], these presuppositions can be omitted according to the finite extensive method of nonlinear function. In fact, by (2.23) and ‖𝑒^𝑛_𝑥‖≤(2/ℎ)‖𝑒^𝑛‖≤𝐶ℎ, when 𝑘,ℎ→0, 𝑢^𝑛∈𝑆_𝜀(𝑈), according to the definition about 𝐴^∗(𝑢) and 𝑔^∗(𝑣) mentioned in [17] (𝑓^∗(𝑢) can be defined accordingly), we have 𝐴^∗(𝑢)=𝐴(𝑢) and 𝑔^∗(𝑢𝑥^𝑛)=𝑔(𝑢𝑥^𝑛) and 𝑓^∗(𝑢^𝑛)=𝑓(𝑢^𝑛). So this theorem and also hold when 𝐴(𝑢)∈𝐶⁰(𝑅³,𝑅^3×3), 𝑔(𝑢)∈𝐶¹(𝑅³), 𝑓(𝑢)∈𝐶(𝑅³). Here, we finish the proof of the first inequality of Theorem 2.9.

By discrete Sobolev's inequality [18] and the first inequality of Theorem 2.9, the second inequality in this theorem can be proved.

Similarly, we have the stability theorem about the difference equation.

Theorem 2.10. Let 𝑢^𝑛 be the solution of (1.9), 𝑣^𝑛 is the solution of (1.9) under the disturbance 𝑢₀(𝑥)+𝛿(𝑥). 𝑔(𝑢)∈𝐶¹(𝑅³). For any positive constant 𝛿, if (𝑘/ℎ²)≤(𝛼₁−𝜎)/6(𝛼₁+|𝛼₂|𝑀₁)², then there exist constant 𝐶_𝑖(𝑖=5,6,7) which is independent on ℎ and 𝑘. When ℎ≤𝐶₅, 𝑘≤𝐶₆, one has sup_{0≤𝑛≤[𝑇/𝑘]}(‖𝑢^𝑛−𝑣^𝑛‖+‖|𝑢^𝑛_𝑥−𝑣^𝑛_𝑥|‖)≤𝐶₇‖𝛿₀‖, (2.27) where 𝑀₁=𝑚sup_{𝑢∈𝑆𝜀(𝑈)}‖𝐴(𝑢)‖(𝑚>1).

3. Numerical Experiment and Its Error Analysis

In this section, we propose the numerical examples and the error analysis of the solutions. Three subcases of (1.3) mentioned in Section 1 will be performed in our simulation respectively.

Conveniently for computation, first we rewrite (1.9) as the following form: 𝑢_𝑗^𝑛+1=𝑢^𝑛_𝑗+𝑘[(𝛼₁+𝛼₂𝐴(𝑢^𝑛_𝑗))𝑢^𝑛_𝑗𝑥𝑥+𝛼₁|𝑢^𝑛_𝑗𝑥|²𝑢^𝑛_𝑗+𝑓(𝑢^𝑛_𝑗𝑇)],𝑗=1,…,𝐽;𝑛=0,…,[𝑘𝑢]−1,_0𝑗=𝑢₀(𝑥_𝑗𝑢),𝑗=1,…,𝐽,_𝑛0=𝑢^𝑛_𝐽;𝑢^𝑛_𝐽+1=𝑢^𝑛₁𝑇,𝑛=1,…,[𝑘]. (3.1)

According to (3.1), in the first step, we can get the value of 𝑢⁰ on 𝑢⁰_𝑗. Second, according to second equation of (3.1), we can also get the solution 𝑢^𝑛+1 step by step. In each step of computation, the third equation of (3.1) will be used repeatedly.

(i) Setting 𝛼₁=𝛼₂=1 and 𝑓(𝑥,𝑡,𝑢)=𝑢×(0,0,1)^𝑇, we consider the following spin wave of 𝑢=𝑢(𝑥−𝑐𝑡): 𝑢=(1−𝑠²₀cos𝜉,1−𝑠²₀sin𝜉,𝑠₀)^𝑇, (3.2) where 𝜉=𝑥−𝑡,𝑠₀=0. In fact, we found that (3.2) is the solution of (1.3), where 𝜉=𝑎𝑥−𝑏𝑡+𝑐,𝑠₀=0. As far as we know, exact solutions of this case were still not constructed. For the simplicity, we have omitted the details of our constructing.

In accordance with (3.2), the initial and boundary conditions of this equation can be proposed. Let us first consider a domain 𝑥∈[0,6𝜋] with the Dirichlet boundary condition on the spin vector. We have implemented (3.1) where 𝑘=0.0001,ℎ=1/5 in numerical resolution. Figures 1(a) and 1(b) show the numerical solutions in time 𝑡=0.1, and 𝑡=0.5 respectively.

(a)

(b)

Figure 1 

The solution 𝑢 at (a) 𝑡=0.1, and (b) 𝑡=0.5; 𝑘=0.0001,ℎ=1/5.

From Figure 1, we observe that the numerical solutions exhibit an irregular changing at the beginning of the space steps. This will happen in the range of probably 1-10 space steps in Figure 1(a) as well as mainly in 1-20 steps in Figure 1(b). This can be seen more clearly in Figure 2 which exhibit the error about the solution in Figure 1 accordingly.

(a)

(b)

Figure 2 

The error of 𝑢 at (a) 𝑡=0.1 and (b) 𝑡=0.5; 𝑘=0.0001,ℎ=1/5.

Observed from Figures 1 and 2, probably from the 10 or 20 space steps, the numerical solution is credible. But from then on, the creditable region about the solution is gradually shrinking when the time increase. At space step 50, by amplifying solution error image Figure 2, we found that the magnitude of error is approximately 10⁻⁴ which is consist with our Theorem 2.9. These details can be seen in Figure 3.

(a)

(b)

Figure 3 

The error of 𝑢 at (a) 𝑡=0.1 and (b) 𝑡=0.5 at space step 50; 𝑘=0.0001,ℎ=1/5.

(ii) Setting 𝛼₁=𝛼₂=1, and 𝑓(𝑥,𝑡,𝑢)=0 in (1.3), the solitary solution which proposed in [19] can be written as (3.3) (here we can set 𝛿=2). For this case, according to (3.3), the initial boundary conditions can be given in our numerical scheme (3.1). Furthermore, we mention here at (3.3) is not a periodic solution, but we can extend the problem into a periodic one. In fact, we take the truncated domain as Ω=[0,4𝜋]. 𝑢^𝑗(4𝜋,𝑡)≈0 and the smoothness of the solution ensure the extend can be done. Thus the finite difference scheme (3.1) still can be used for computation. Figure 4 shows the numerical solution in the 𝑡=0.3 and the corresponding error image. Seen in Figure 4, we find that solution error is quite large. Therefore, we suspect that (3.3) which is proposed in [19] is not correct. Moreover, a simple symbol computation by Maple confirms our speculation: 𝑢¹√=sech[(3𝑥4)(𝑡+𝛿)^−1/2√]{tanh[(3𝑥4)(𝑡+𝛿)^−1/2√]sin[(3𝑥4)(𝑡+𝛿)^−1/2]√−cos[(3𝑥4)(𝑡+𝛿)^−1/2𝑢]},₂√=−sech[(3𝑥4)(𝑡+𝛿)^−1/2√]{tanh[(3𝑥4)(𝑡+𝛿)^−1/2√]cos[(3𝑥4)(𝑡+𝛿)^−1/2]√+sin[(3𝑥4)(𝑡+𝛿)^−1/2𝑢]},₃=tanh²√[(3𝑥4)(𝑡+𝛿)^−1/2]. (3.3)

(a)

(b)

Figure 4 

(a) Solution of 𝑢 at 𝑡=0.3, (b) error of 𝑢 at 𝑡=0.3; 𝑘=0.0001,ℎ=1/5.

(iii) Let 𝛼₁=0, 𝛼₂=1, and 𝑓(𝑥,𝑡,𝑢)=(𝑢¹,𝑢²,𝑢³)^𝑇×(diag(1,2,3)⋅(𝑢¹,𝑢²,𝑢³)^𝑇) in (1.3). This is special case of LL equation (1.2) with an easy plane. Under situation (3.2), we can easily offer the initial-boundary condition about (3.1).

When 𝛼₁=0, Theorem 2.9 obviously can not give estimate about 𝑘/ℎ². But when we follow (3.2) to set initial boundary value, in addition to numerical viscosity began in the outside of space step, solution is very irregular. Therefore, we believe that the numerical solution is convergent. Nevertheless, for a better estimation, we should try to use other methods to estimate convergence rate of the discrete form solution. From Figure 5, we found that the evolution of 𝑢¹ and 𝑢² is similar to subcase (i) mentioned above; compared to subcase (i), the evolution of 𝑢³ exhibits a larger undulate behavior.

(a)

(b)

Figure 5 

The solution 𝑢 at (a) 𝑡=0.1 and (b) 𝑡=0.5; 𝑘=0.0001,ℎ=1/5.

Acknowledgments

This work is supported by NSFC (Grant No. 10771009 and No. 10861014), BSFC (Grant No. 1085001) of China, Funding Project for Academic Human Resources Development in Institutions of Higher Leading Under the Jurisdiction of Beijing Municipality (PHR-IHLB 200906103), Beijing Education Committee Funds and Ph.D. Innovation Project of Beijing University of Technology (Grant No. ykj-2010-3418 and No.ykj-2010-4151).