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

Authors: Penghong Zhong; Shu Wang; Ke Wang; Yiping Bai
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## 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.


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## Body

## 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): (1.1)∂u∂t=u×Δu+f(x,t,u),
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 (1.2)∂u∂t=u×Δu+u×(Pu),P=diag{P1,P2,P3},P1≤P2≤P3.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 variablesx(1.3)∂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,
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,…,xn) 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), Rn 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: (1.4)∂u∂t=-α1Δu+α1|∇u|2u+α2A(u)Δu+f(x,t,u),
where ∇u=(ux1,ux2,ux3)T,
(1.5)A(u)=[0-u3u2u30-u1-u2u10]∈R3×3.
Furthermore, we consider a more general type of Landau-Lifshitz equation (as the variety of (1.4)):
(1.6)∂u∂t=-α1Δu+α1g(∇u)u+α2A(u)Δu+f(x,t,u)(x,t)∈Rn×I,u(x+2π,t)=u(x,t)(x,t)∈Rn×I,u(x,0)=u0(x)x∈Rn,
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 then-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. xj=jh,tn=nk(j=0,1,…,J;n=0,1,…,[T/k]). Here (um)jn denotes the approximate value of um(m=1,2,3) on (xj,tn). Let un denote the layer mesh function, that is, un=(u1n,u2n,…,uJn) where ujn=((u1)jn,(u2)jn,(u3)jn)T.Define the discrete inner product and norm as follows:(1.7)(un,vn)=h∑j=1Jujn·vjn=h∑j=1J∑m=13(um)jn·(vm)jn;∥un∥=(un,un)1/2.We define the following finite difference approximations of derivatives along the space direction:(1.8)ujxn=uj+1n-ujnh,ujx̅n=ujn-uj-1nh,ujx̂n=uj+1n-uj-1n2h.
Similarly, we can define the derivatives along the time direction and the Laplace operator: ujtn,ujt̅n,ujt̂n,andujxx̅n.Under the definition and the symbol setting as above, we now define a finite difference approximation of (1.6) by (1.9)ujtn=α1ujxx̅n+α1g(ujx̂n)ujn+α2A(ujn)ujxx̅n+fjn(ujn)0≤j≤J,0≤n≤[Tk]-1,uj+rJn=ujn,0≤j≤J,r∈Z,uj0=u0(xj)0≤j≤J.Obviously, according to difference approximation (1.9), we can start from the zero layer to any layer aboutun(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-functionun, vn defined on mesh-points, one has the following relationship:(1)
(uxx̅n,vn)=-(uxn,vxn),(2)
(utn,un)≥(1/2)∥un∥t2-(k/2)∥utn∥2,(3)
∥ux̂n∥≤∥uxn∥,(4)
∥ux∥≤(2/h)∥u∥.Lemma 2.2.
LetA(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.LetU(x,t) denote the solution of (1.6) (here U(x,t) will be regarded as a function with smoothness in some degree). Ujn denotes the value of U(x,t) on (xj,tn). Un=(U1n,U2n,…,Ujn)T denotes the value of n-layer mesh about U(x,t).Definition 2.3.
One has(2.1)Cb0(R3,R3×3)={B(u)∈R3×3∣∥B(u)∥<∞,∀u∈R3},Cb1(R3)={f(u)∈R∣∥f(u)∥∞+∥f′(u)∥∞<∞,∀u∈R3}.Lemma 2.4.
Assume  thatA(u)∈Cb0(R3;R3×3) and U∈C0(I;(C3(0;2π))3) are the solution of (1.6), then for any ε>0(2.2)α2(A(Un)Uxx̅n-A(un)uxx̅n,en)≤ε∥exn∥2+α22M124ε∥en∥2,
where M1=supu∈R3∥A(u)∥,anden=Un-un.Proof.
We have, by (1) of Lemma 2.2, -α2(Uxx̅n,A(en)en)=0. According to the antiproperty of antisymmetric matrix, we have α2(exn,A(un)exn)=0. Here, we set f∈Cb1(R,I), hence
(2.3)(f(x,tn)A(Un)Uxx̅n-f(x,tn)A(un)uxx̅n,en)=(A(en)Uxx̅n+A(un)exx̅n,f(x,tn)en)=-(Uxx̅n,f(x,tn)A(en)en)-(exx̅n,f(x,tn)A(un)en)=(exn,(f(x,tn)A(un)en)x)=h∑j=0J-1ejxn[f(xj+1,tn)A(uj+1n)ej+1n+f(xj,tn)A(ujn)ejn]h=h∑j=0J-1ejxnf(xj+1,tn)A(uj+1n)ejxn+h∑j=0J-1ejxn∂f(xj,tn)∂xA(ujn)ej+1n.
Letf(xj,tn)=1(j=0,1,…,J;n=0,1,…,[T/k]). By ε-inequality, we have(2.4)α2(A(Un)Uxx̅n-A(un)uxx̅n,en)≤|α2|∥exn∥∥A(uxn)∥∥en∥≤ε∥exn∥2+α22M124ε∥en∥2.Lemma 2.5.
Under the same condition of Lemma2.4, one has
(2.5)∥A(Un)Uxx̅n-A(un)uxx̅n∥≤M1(∥en∥+2h∥exn∥).Proof.
We  first note that, by (3) of Lemma 2.2, we have, A(en)Uxx̅n=-A(Uxx̅n)en; recalling (4) of Lemma 2.1, we have ∥exx̅n∥≤(2/h)∥exn∥, then we obtain
(2.6)∥A(Un)Uxx̅n-A(un)uxx̅n∥=∥A(en)Uxx̅n-A(un)exx̅n∥≤∥A(en)Uxx̅n∥+∥A(un)exx̅n∥≤∥-A(Uxx̅n)en∥+∥A(un)exx̅n∥≤M1(∥en∥+2h∥exn∥).Lemma 2.6.
Ifg(u)∈Cb1(R3) and U∈C0(I;(C3(0,2π)))3 are the solution of (1.6), then for any ε>0(2.7)α1(g(Ux̂n)Un-g(ux̂n)un,en)+(f(Un)-f(un),en)≤ε∥exn∥2+(α12M22M324ε+α1M4+M0α1)∥en∥2,
where
(2.8)M0=sup1≤n≤[T/K]∥∂fn∂u∥,M2=supz∈R3∥∇g(z)∥,M3=sup1≤n≤[T/K]∥Un∥∞,M4=supz∈R3∥g(z)∥,en=Un-un.Proof.
By (3) of Lemma 2.1, we have
(2.9)α1(g(Ux̂n)Un-g(ux̂n)un,en)+(f(Un)-f(un),en)≤α1[([g(Ux̂n)-g(ux̂n)]Un,en)+(g(ux̂n)en,en)]+M0∥en∥2≤α1[∥∇g(Ux̂n+θex̂n)∥∥Un∥∞∥ex̂n∥∥en∥+∥g(ux̂n)∥∥en∥2]+M0∥en∥2≤ε∥exn∥2+(α12M22M324ε+α1M4+M0α1)∥en∥2.Lemma 2.7.
One has(2.10)α1∥g(Ux̂n)Un-g(ux̂n)un∥+∥f(Un)-f(un)∥≤α1(2M2M3h+M4+M0α1)∥en∥.Proof.
According to (1) and (2) of Lemma 2.1, we have
(2.11)α1∥g(Ux̂n)Un-g(ux̂n)un∥+∥f(Un)-f(un)∥≤α1(∥∇g(Ux̂n+θex̂n)∥∥Un∥∞∥ex̂n∥∥en∥+∥g(ux̂n)∥∥en∥2)+M0∥en∥.Definition 2.8.
Sε(U)={v∈Hp1(Ω)∣∥U-v∥Hp1(Ω)≤ε}.According to the lemmas mentioned above, we now come to discuss the convergence of difference equation  (1.9).Theorem 2.9.
LetU∈C2(I;(Hp4(0,2π))3) be the solution of (1.6), un 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 Ci(i=1,2,3,4) which independent of h and k. If h≤C1, k≤C2, one has
(2.12)sup0≤n≤[T/k]∥Un-un∥+∥|Uxn-uxn|∥≤C3(k+h2),
furthermore,
(2.13)∥Un-un∥∞≤C4h,
where M1=msupu∈Sε(U)∥A(u)∥(m>1),∥|·|∥=(k∑n=1[T/k]∥·∥2)1/2,∥v∥∞=max1≤j≤J|vj|.Proof.
According to the condition, the solution of (1.6) U∈C2(I;(Hp4(0,2π))3), substituting the according part of (1.9) with U(x,t), we have
(2.14)Utn=α1Uxx̅n+α2A(Un)Uxx̅n+α1g(Ux̂n)Un+rn,
where rn is the error estimate of (1.9). According to (2.14) and (1.9), denote en=Un-un, we have
(2.15)etn=α1exx̅n+α2(A(Un)Uxx̅n-A(un)uxx̅n)+α1(g(Ux̂n)Un-g(ux̂n)un)+f(Un)-f(un)+rn.
Take the inner product of (2.15) and en, combine them with (1) and (2) of Lemma 2.1, we have(2.16)12∥en∥t2-k2∥etn∥2+α1∥exn∥2=α2(A(Un)Uxx̅n-A(unuxx̅n),en)+α1(g(Ux̂n)Un-g(ux̂n)un,en)+(f(Un)-f(un),en)+(rn,en).
Just like the method mentioned in [17], we can assume  that A*(u)∈C0b(R3,R3×3) and  g*(u)∈C1b(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:(2.17)12∥en∥t2-k2∥etn∥2+(α1-2ε)∥exn∥2≤M5∥en∥2+12∥rn∥2+M0∥en∥2≤M5∥en∥2+C2(k+h2)2+M0∥en∥2,
where
(2.18)M0=max1≤n≤[T/k]∥∂fn∂u∥,M5=α12M22M324ε+α1M4+α22M124ε+12.
In the following part, we propose the estimation about ∥etn∥ of (2.17).
By Lemmas2.5 and 2.7 and (2.15), we have(2.19)k∥etn∥2≤|α2|∥A(Un)Uxx̂n-A(un)uxx̂n∥+α1∥g(Ux̂n)Un-g(ux̂n)un∥+∥f(Un)-f(un)∥+∥rn∥≤2α1h∥exn∥+α1(2M2M3h+M4)∥en∥+|α2|M1(∥en∥+2h∥exn∥)+M0∥en∥+rn≤M6h∥exn∥+(M7h+M8+M0)∥en∥+C(k+h2),
where M6=2(α1+|α2M1|), M7=2α1M2M3, and M8=α1M4+|α2|M1.
So according to the inequality given above, we have(2.20)k∥etn∥2≤3M62kh2∥exn∥2+(6M72kh2+6M82k+3kM02)∥en∥2+3kC2(k+h2)2.
Substitute (2.16) into (2.15), let ε=α1/4, according to the condition given in Theorem 2.9, we have(2.21)12∥en∥t2+(α12-3M62kk2)∥exn∥2≤M5∥en∥2+C(k+h2)2,
where M9=M0+M5+6M82T+M72((α1-σ)/(α1+|α2|M1)2).
According to (2.17) and the condition given in Theorem 2.9,(2.22)∥etn∥2+σ∥exn∥2≤2M9∥en∥2+C(k+h2)2
By∥e0∥=0 and Gronwall inequality, we have(2.23)∥en∥≤C(k+h2)2exp(M9T)≤C3(k+h2)2.
By (2.22), (2.24)σk∑n=1m∥exn∥2≤CT(k+h2)2+2M9k∑n=1m∥en∥2.
So we have(2.25)k∑n=1[T/k]∥exn∥2≤M102(k+h2)2,
where M102=(CT+2M9TC32)/σ by (2.25), hence
(2.26)(k∑n=1m∥exn∥2)1/2≤M10(k+h2)2.
So by (2.23) and (2.26), when A(u)∈Cb(R3,R3×3), g∈Cb1(R3), and f∈Cb(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 ∥exn∥≤(2/h)∥en∥≤Ch, when k,h→0, un∈Sε(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*(ux̂n)=g(ux̂n) and f*(un)=f(un). So this theorem and also hold when A(u)∈C0(R3,R3×3), 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.
Letun be the solution of (1.9), vn 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 Ci(i=5,6,7) which is independent on h and k. When h≤C5, k≤C6, one has
(2.27)sup0≤n≤[T/k](∥un-vn∥+∥|uxn-vxn|∥)≤C7∥δ0∥,
where M1=msupu∈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: (3.1)ujn+1=ujn+k[(α1+α2A(ujn))ujxx̂n+α1|ujx̂n|2ujn+f(ujn)],j=1,…,J;n=0,…,[Tk]-1,uj0=u0(xj),j=1,…,J,u0n=uJn;uJ+1n=u1n,n=1,…,[Tk].According to (3.1), in the first step, we can get the value of u0 on uj0. Second, according to second equation of (3.1), we can also get the solution un+1 step by step. In each step of computation, the third equation of (3.1) will be used repeatedly.(i) Settingα1=α2=1 and f(x,t,u)=u×(0,0,1)T, we consider the following spin wave of u=u(x-ct): (3.2)u=(1-s02cosξ,1-s02sinξ,s0)T,
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.The solutionu at (a) t=0.1, and (b) t=0.5; k=0.0001,h=1/5.
(a)(b)From Figure1, 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.The error ofu at (a) t=0.1 and (b) t=0.5; k=0.0001,h=1/5.
(a)(b)Observed from Figures1 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.The error ofu at (a) t=0.1 and (b) t=0.5 at space step 50; k=0.0001,h=1/5.
(a)(b)(ii) Settingα1=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]. uj(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: (3.3)u1=sech[(3x4)(t+δ)-1/2]{tanh[(3x4)(t+δ)-1/2]sin[(3x4)(t+δ)-1/2]-cos[(3x4)(t+δ)-1/2]},u2=-sech[(3x4)(t+δ)-1/2]{tanh[(3x4)(t+δ)-1/2]cos[(3x4)(t+δ)-1/2]+sin[(3x4)(t+δ)-1/2]},u3=tanh2[(3x4)(t+δ)-1/2].(a) Solution ofu at t=0.3, (b) error of u at t=0.3; k=0.0001,h=1/5.
(a)(b)(iii) Letα1=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).Whenα1=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.The solutionu at (a) t=0.1 and (b) t=0.5; k=0.0001,h=1/5.
(a)(b)



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