Table of Contents
ISRN Applied Mathematics
Volume 2012, Article ID 752964, 20 pages
Research Article

Some Blow-Up and Smooth Solutions about Landau-Lifshitz Equation and Their Spatial Curvature Behavior

Department of Applied Mathematics, Beijing University of Technology, Ping Le Yuan 100, Chaoyang District, Beijing 100124, China

Received 30 May 2012; Accepted 26 June 2012

Academic Editors: S. Blanes and X.-S. Yang

Copyright © 2012 Penghong Zhong et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We construct the exact solution of (2+1)-dimensional space-time Landau-Lifshitz equation (LLE) without the Gilbert term. Under suitable transformations, some exact solutions are obtained in the radially symmetric coordinates and nonsymmetric coordinates. The type of solutions cover the finite-time blow-up solution, smooth solution in time and vortex solution. At the end, some properties about these solutions and their spatial curvature are illustrated by the graphs.

1. Introduction

It is well known that Landau-Lifshitz equation (LLE) [1, 2] is one of the most important nonlinear equations in physics which bears a fundamental role in the understanding of nonequilibrium magnetism, just as the Navier-Stokes equation does in that of fluid dynamics. Under the principal assumption of the macroscopic theory of ferromagnetism, the state of a magnetic crystal is described by the magnetization vector 𝑉(𝑥,𝑡), which is a function of position 𝑥 and time 𝑡. For instance, a subcase of the LLE is as follows: 𝑉𝑡=𝛼𝑉×𝐻e𝛽𝑉×𝑉×𝐻e,(1.1) where 𝑉=(𝑣1,𝑣2,𝑣3). 𝛼, 𝛽 are constants. × denotes a vector cross-product. The effective magnetic field 𝐻e is given by the derivative of the magnetic crystal energy 𝐸 with respect to the vector 𝑉: 𝐻e=𝜕𝐸,𝜕𝑉(1.2) where 𝐸 is given by the sum of the exchange energy, the anisotropy energy, and the magnetic field energy; that is, 𝐸(𝑉)=𝐸𝑒𝑥(𝑉)+𝐸𝑎𝑛(𝑉)+𝐸𝑚𝑓(𝑉).(1.3)

Equation (1.1) exhibits a rich variety of dynamical properties of a spin vector in different backgrounds. In this letter, we discuss the following LLE which vanishes the Gilbert term (i.e., 𝛽=0), anisotropy energy and the magnetic field energy (here, we set 𝛼=1): 𝑉𝑡=𝑉×Δ𝑉𝑅2,Δ𝑉𝑅2=𝜕2𝑉𝜕𝑥21+𝜕2𝑉𝜕𝑥22.(1.4)

In the 1-dimensional motion LLE, the solition solutions have been studied by many physicists and mathematicians, see, for example, in 1976, Lakshmanan et al. constructed a class of solutions [3] of (1.4). Nakamura and Sasada constructed a solution [4] of the nonvanishing external magnetic field case in 1974. Tjon and Wright also found some solitons [5] for such case in 1977.

LLE is nonintegrable [6] in high dimensions (𝑛2). Furthermore, as far as we know, many famous direct methods such as Hirota bilinear method and auxiliary function method [7] are difficult for constructing the exact solution of (1.4). So, we can only find out some particular exact solutions [8, 9] by various direct methods which base on some special ansatz about the solution. In 2000 and 2001, Guo et al. constructed some exact blow-up solutions [10, 11] for 2-dimension radially symmetric LLE. Similarly, for 𝑛-dimensions LLE, some exact blow-up solutions were constructed in [12, 13]. In this situation, a solution 𝑉 blows up at time 𝑇>0 if lim𝑡𝑇sup𝑉(,𝑡)=.(1.5)

The present work is organized as follows. We firstly deduce a blow-up solution of (1.4) in Section 2. Then, we construct a vortex solution and some other periodic solutions about LLE in Section 3. In Section 4, some other exact solutions which different from blow-up solution and vortex solution are presented. In Section 6, some solutions and their spatial curvature are illustrated in a graphic way.

2. Blow-Up Solution

We deduce an exact blow-up solution in this section. Under radially symmetric coordinates, (1.4) takes the form: 𝑉𝑡=𝑉×𝑉𝑟𝑟+1𝑟𝑉×𝑉𝑟,(2.1) where 𝑟=𝑥21+𝑥22, 𝑉=(𝑣1(𝑡,𝑟),𝑣2(𝑡,𝑟),𝑣3(𝑟)).

We will find the explicit solution of (2.1) in the form of 𝑣1𝑣(𝑡,𝑟)=cos(𝑚(𝑡,𝑟))𝑓(𝑟),2𝑣(𝑡,𝑟)=sin(𝑚(𝑡,𝑟))𝑓(𝑟),3(𝑡,𝑟)=𝛾𝑓(𝑟),(2.2) where 𝑚(𝑡,𝑟) and 𝑓(𝑟) are functions to be determined, 𝛾 is constant.

Example 2.1. If 1𝑚(𝑡,𝑟)=43227𝐶31𝑟4+108𝐶21𝑟8/3𝐶2+144𝐶1𝑟4/3𝐶22+64𝐶32𝑡𝑇+𝐶3,(2.3)𝑓(𝑟)=81𝐶31𝑟4324𝐶21𝑟8/3𝐶2432𝐶1𝑟4/3𝐶22192𝐶32𝛾𝐶2116𝑟1/3𝐶22+24𝐶1𝑟5/3𝐶2+9𝐶21𝑟32,(2.4) where 𝛾,𝐶1,𝐶2,𝐶3 are any constants, 𝑇>0, then (2.2) is an exact blow-up solution of (2.1).

Proof. Substituting (2.2) into (2.1), we get 𝛾𝑟sin(𝑚)𝑚2𝑟𝑓𝛾𝑟cos(𝑚)𝑚𝑟𝑟𝑓2𝛾𝑟cos(𝑚)𝑚𝑟𝑓𝑟𝑓𝛾cos(𝑚)𝑚𝑟+sin(𝑚)𝑚𝑡𝑟=0,(2.5)𝛾𝑟cos(𝑚)𝑚2𝑟𝑓+𝛾𝑟sin(𝑚)𝑚𝑟𝑟𝑓+2𝛾𝑟sin(𝑚)𝑚𝑟𝑓𝑟+𝑓𝛾sin(𝑚)𝑚𝑟+cos(𝑚)𝑚𝑡𝑟=0,(2.6)𝑟𝑚𝑟𝑟𝑓+2𝑟𝑚𝑟𝑓𝑟+𝑚𝑟𝑓=0.(2.7) Furthermore, if we insert (2.7) into (2.5) and (2.6), we have 𝑟𝑚𝑟𝑟𝑓+2𝑟𝑚𝑟𝑓𝑟+𝑚𝑟𝑓=0,𝛾𝑚2𝑟𝑓+𝑚𝑡=0.(2.8)
If we settle down on the following ansatzs: 𝑚=𝑔(𝑟)𝑡𝑇+𝐶,𝑓=𝑔(𝑟)𝑗(𝑟),(2.9) where 𝑔(𝑟) and 𝑗(𝑟) are the functions about 𝑟 to be determined, 𝑇 and 𝐶 are constants.
From (2.8) and (2.9), we have 𝛾𝑔2𝑟𝑔𝑗𝑔=0,𝑟𝑔2𝑟𝑔𝑗+2𝑟𝑔𝑟𝑔𝑟𝑗+𝑔𝑗𝑟+𝑔𝑟𝑔𝑗=0.(2.10)
To construct the exact solution of the ordinary differential equations (ODEs for short), we can resort to the mathematic software. Concretely, we find out the exact solution of (2.10) as follows: 1𝑔=𝐶6431𝑟4+1𝐶1621𝑟8/3𝐶2+1𝐶121𝑟4/3𝐶22+1𝐶2732,𝑗=20736𝛾𝐶219𝐶21𝑟3+24𝐶1𝑟5/3𝐶2+16𝑟1/3𝐶222,(2.11) where 𝛾,𝐶1,𝐶2 are any constants.
This completes the proof of Example 2.1.

In the case of ODE, we mention here that deducing process of software tries to solve it using either classification methods or symmetry methods. In these process, the coefficient about 𝑟 in (2.11) will be extend to a general form. It is why the forms of (2.3) and (2.4) are so specific.

For this finite-time blow-up solution (2.2), it is easy to verify the following.

Remark 2.2. (i) It holds that ||||𝑉(𝑡,𝑟)2=9144𝐶1𝑟4/3𝐶22+108𝐶21𝑟8/3𝐶2+27𝐶31𝑟4+64𝐶3221+𝛾2𝛾2𝐶4116𝑟1/3𝐶22+24𝐶1𝑟5/3𝐶2+9𝐶21𝑟34(2.12) is independent of 𝑡, so 𝑟1𝑟0𝑟|𝑉|2𝑑𝑟(0<𝑟0𝑟1<) is a conservation quantity.
(ii) When 𝑟0𝑟𝑟1, 𝑇>0, it is not difficult to verify lim𝑡𝑇|𝑉𝑟|2=+. For example, in the case of 𝑇=1 and 𝛾=𝐶1=𝐶3=1, 𝐶2=0, we can verify that ||𝑉𝑟||2=1𝑟168256𝑡+128+128𝑡2𝑟6(𝑡1)2(2.13) is not bounded as 𝑡1.
(iii) For any first orthogonal matrix 𝐵 (the elements of 𝐵 are constants.), 𝑉𝐵 is also a blow up solution of (2.1).

3. Vortex Solution and Periodic Solutions

In this section, we concentrate on the solution on the 𝑆2 (here, we mean that |𝑉|2=1). We firstly consider the vortex solution in the radially symmetric coordinates. Here, these time periodic solutions, which under the radially symmetric coordinates are called magnetic vortices or vortex solutions are of particular importance in manifesting interesting physical phenomena and topological structures of LLE. After that, some exact solutions which under the nonradially symmetric coordinates are proposed.

Under the radially symmetric coordinates, we set 𝑣1𝑣(𝑡,𝑟)=𝑓(𝑡)sin(𝑞(𝑟)),2(𝑡,𝑟)=1(𝑓(𝑡))2𝑣sin(𝑞(𝑟)),3(𝑡,𝑟)=cos(𝑞(𝑟)),(3.1)

Submitting (3.1) into (2.1), (2.1) can be transformed into 1(𝑓(𝑡))2𝑟2𝑑2𝑑𝑟2𝑞(𝑟)+1(𝑓(𝑡))2𝑑𝑑𝑑𝑟𝑞(𝑟)𝑟+𝑑𝑡𝑓(𝑡)sin(𝑞(𝑟))𝑟2=0.(3.2)

So we have 1(𝑓(𝑡))2+1𝐶𝑑𝑟𝑑𝑡𝑓(𝑡)=0,2𝑑2𝑑𝑟2𝑑𝑞(𝑟)+𝑑𝑟𝑞(𝑟)𝑟𝐶sin(𝑞(𝑟))𝑟2=0.(3.3) Solving (3.3), we have 𝑓(𝑡)=sin𝐶𝑡+𝐶1or𝑓(𝑡)=cos𝐶𝑡+𝐶1,1𝑓2(𝑡)=cos𝐶𝑡+𝐶1or1𝑓2(𝑡)=sin𝐶𝑡+𝐶1,𝑟𝑞(𝑟)=±4,arctan±𝐶𝜆,(3.4) where 𝐶>0, 𝐶1, 𝜆 are constants.

We point out there that (3.4) is vortex solution of (2.1). Similarly, if we set (𝐶 is a constant) 𝑣1𝑣(𝑡,𝑥,𝑦)=cos(𝑡+𝐶)sin(𝑄(𝑥,𝑦)),2𝑣(𝑡,𝑥,𝑦)=sin(𝑡+𝐶)sin(𝑄(𝑥,𝑦)),3(𝑡,𝑥,𝑦)=cos(𝑄(𝑥,𝑦)),(3.5)

Equation (1.4) can be transformed into 𝜕2𝑄𝜕𝑥2+𝜕2𝑄𝜕𝑦2=sin𝑄.(3.6)

Here, we propose the following two subcase solutions of (3.6).

(I) Functional separable solutions of (3.6)
let 𝑄𝑓𝜉(𝑥,𝑦)=±4arctan1𝑔𝜉2,(3.7) where 𝜉1=𝑎𝑥+𝑏, 𝜉2=𝑐𝑦+𝑑.
Substituting (3.7) into (3.6), we can deduce the following equations: 4𝑎2𝑓𝑔+4𝑎2𝑓𝑓2𝑔38𝑎2(𝑓)2𝑓𝑔3+4𝑐2𝑔𝑓+4𝑐2𝑔𝑓3𝑔28𝑐2(𝑔)2𝑓3𝑔4𝑓𝑔+4𝑓3g3=0.(3.8)
Introducing two auxiliary function about 𝑓, 𝑔, (𝑓)2=𝐴1𝑓4+𝐵1𝑓3+𝐶1𝑓2+𝐷1𝑓+𝐸1,((3.9)𝑔)2=𝐴2𝑔4+𝐵2𝑔3+𝐶2𝑔2+𝐷2𝑔+𝐸2.(3.10)
According to (3.9) and (3.10), we have 𝑓=12𝐷1+2𝐴1𝑓3+32𝐵1𝑓2+𝐶1𝑔𝑓,(3.11)=12𝐷2+2𝐴2𝑔3+32𝐵2𝑔2+𝐶2𝑔.(3.12)
Substituting (3.9)–(3.12) into (3.8), we have 2𝐵1𝑎2𝑓4𝑔42𝐵2𝑐2𝑓3𝑔4𝐶41𝑎2+𝐶2𝑐2𝑓13𝑔36𝐷2𝑐2𝑓3𝑔2𝐴+81𝑎2𝐸2𝑐2𝑓3𝑔6𝐷1𝑎2𝑓2𝑔3+6𝐵1𝑎2𝑓2𝑔𝐴+82𝑐2𝐸1𝑎2𝑓𝑔3+6𝐵2𝑐2𝑓𝑔2𝐶+41𝑎2+𝐶2𝑐21𝑓𝑔+2𝐷2𝑐2𝑓+2𝐷1𝑎2𝑔=0.(3.13)
Equating the coefficients of different powers of 𝑓𝑔 in (3.13), we obtain a set of algebraic equations: 𝐵1=𝐵2=𝐷1=𝐷2𝐶=0,1𝑎2+𝐶2𝑐2𝐴1=0,1𝑎2𝐸2𝑐2𝐴=0,2𝑐2𝐸1𝑎2=0.(3.14)
If we set 𝐴=𝐴1𝑎2,𝐵=𝐶1𝑎2,𝐶=𝐸1𝑎2,(3.15) according to (3.14), functions 𝑓=𝑓(𝜉1) and 𝑔=𝑔(𝜉2) are determined by the first-order autonomous ordinary differential equations as follows: 𝑓𝜉12=𝐴𝑓4+𝐵𝑓2𝑔+𝐶,𝜉22=C𝑔4+(1𝐵)𝑔2+𝐴,(3.16)
Here, we omit some details about the deducing and present some exact solutions of (3.16). (A) If 𝐴=𝐶=0, 0<𝐵<1, then ±𝑓=exp𝐵𝜉1±,𝑔=exp1𝐵𝜉2.(3.17)(B) If 𝐴=0, 𝐵=1, 𝐶>0, solving (3.16), we have 𝑓=𝜉𝐶sinh11,𝑔=±𝐶𝜉2.(3.18)(C) If 𝐴=0, 𝐵>1, 𝐶>0, then 𝑓=𝐶𝐵sinh𝐵𝜉1,𝑔=𝐵1𝐶sec𝐵1𝜉2.(3.19)(D) According to Table 3.1 which is proposed in [7], we can deduce a solution easily. Defining 𝑠𝑐(𝜉,𝑅)=𝑠𝑛(𝜉,𝑅)𝑐𝑛(𝜉,R),𝑑𝑠(𝜉,𝑅)=𝑑𝑛(𝜉,𝑅),𝑠𝑛(𝜉,𝑅)(3.20) where 𝑅 is the modulus of Jacobi elliptic function.
Suppose 𝑅=3/2, 𝐴=3/16, 𝐵=1/2, 𝐶=1, we have 𝜉𝑓=𝑠𝑑1,32𝜉,𝑔=𝑑𝑠2,32.(3.21)

(II)Another functional separable solution of (3.6)
𝑄(𝑥,𝑦)=4arctancot𝐶1cosh𝑓cosh𝑔,𝑓=cos𝐶11+𝐶22𝑥𝐶2𝑦,𝑔=sin𝐶11+𝐶22𝑦𝐶2𝑥,(3.22) where 𝐶1 and 𝐶2 are arbitrary constants.

According to the above derivation, we conclude our main result in the following example.

Example 3.1. (i) (3.1) and (3.4) are a vortex solution of (2.1);
(ii) (3.5), (3.7) (3.17)–(3.19), and (3.21) are exact periodic solutions about (1.4);
(iii) (3.5) and (3.22) are a periodic solution about (1.4).

Remark 3.2. (i) For the solution (3.1) and (3.4), ||𝑉𝑟(||𝑡,𝑟)2𝑟=162𝐶2𝐶𝜆2𝜆2+𝑟2𝐶2(3.23) is independent of 𝑡. Furthermore, 0|𝑉𝑟|2𝑟𝑑𝑟 is a conservation quantity. For example, if we set 𝐶=4 and 𝜆=1 in (3.23), we have 0|𝑉𝑟|2𝑟𝑑𝑟=160.
(ii) For the solutions proposed in (ii)-(iii) in Example 3.1, it is a tedious expression for |𝑉(𝑡,𝑥,𝑦)|2 ( is about the spacial directions) to each of them. Therefore, if we omit the details of 𝑄(𝑡,𝑥,𝑦), obviously |𝑉(𝑡,𝑥,𝑦)|2 of (ii) and (iii) will adopt the same form as follows: ||||𝑉(𝑡,𝑥,𝑦)2=𝜕𝜕𝑥𝑄(𝑥,𝑦)2+𝜕𝜕𝑦𝑄(𝑥,𝑦)2.(3.24) The graphic demonstration will be provided in Section 6.
(iii) For any first orthogonal matrix 𝐵 (the elements of 𝐵 are constants.), 𝑉𝐵 is also a blow up solution of (1.4) or (2.1).

4. Non-Blow-Up Solutions on the Disc

In Section 1, we propose some blow-up solutions on the disc. However, whether or not there are non-blow-up solutions (here the non-blow-up solution contains discontinuous one) on the disc is not so clear as far as we know. By constructing some solutions, we give a positive answer about it.

According to results which provided in [13], some exact solutions can be constructed via an explicit transform between LLE with effective magnetic field and the one without external magnetic field. So, if we meet some difficulties in searching the solutions about (1.4), we can firstly try to find out some solutions about (1.1).

Similar to Section 2, we can find out an exact solution for (2.1) as follows: 𝑣1𝐶(𝑡,𝑟)=cos3𝐶𝑡+(3/4)𝑟1𝑟2/3+𝐶2𝐶3𝐶1𝑟4/3,𝑣2𝐶(𝑡,𝑟)=sin3𝐶𝑡+(3/4)𝑟1𝑟2/3+𝐶2𝐶3𝐶1𝑟4/3,𝑣3𝐶(𝑟)=3𝐶1𝑟4/3,(4.1) where 𝐶1,𝐶2,𝐶3 are any constants.

Although (4.1) is an exact solution, we can find out some other exact solutions about (2.1) by adding an effective magnetic field. This fact will be demonstrated in the following illustration which concentrates on Proposition 4.4.

Imposing a special anisotropy energy on (1.4), we have the following system: 𝑈𝑡=𝑈×(Δ𝑈+𝑈𝐽),(4.2) where 𝐽=diag{𝑎,𝑎,𝑏}, 𝑎,𝑏 are constants.

Under radially symmetric coordinates, (4.2) takes the form: 𝑈𝑡=𝑈×𝑈𝑟𝑟+1𝑟𝑈×𝑈𝑟+𝑈×𝑈𝐽,(4.3) where 𝑟=𝑥21+𝑥22, 𝑈=(𝑢1(𝑡,𝑟),𝑢2(𝑡,𝑟),𝑢3(𝑡,𝑟)).

We deduce a solution of (4.3) in the following example.

Example 4.1. If 𝑍(𝑟) is the root of 𝑍(𝑟)4𝐶1𝑢𝑟𝑍(𝑟)𝑎+𝑏=0,(4.4)1𝐶(𝑡,𝑟)=cos3𝑡+𝑍(𝑟)2𝑑𝑟+𝐶2𝐶3𝑍(𝑟)4,𝑢𝑏+𝑎2𝐶(𝑡,𝑟)=sin3𝑡+𝑍(𝑟)2𝑑𝑟+𝐶2𝐶3𝑍(𝑟)4,𝑢𝑏+𝑎3𝐶(𝑟)=3𝑍(𝑟)4,𝑏+𝑎(4.5) where 𝐶1,𝐶2,𝐶3 are any constants, then (4.4)-(4.5) is a solution of (4.3).

Proof. We settle down on the following setting: 𝑢1𝑢(𝑡,𝑟)=cos(𝑚(𝑡,𝑟))𝑓(𝑟),2𝑢(𝑡,𝑟)=sin(𝑚(𝑡,𝑟))𝑓(𝑟),3(𝑟)=𝑓(𝑟),(4.6) where 𝑚(𝑡,𝑟) and 𝑓(𝑟) are functions to be determined.
Similar to the deducing of Example 2.1, if we insert (4.6) into (4.3), we have 𝑟𝑚𝑟𝑟𝑓+2𝑟𝑚𝑟𝑓𝑟+𝑚𝑟𝑓=0,𝛾𝑚2𝑟𝑓+(𝑏𝑎)𝑓+𝑚𝑡=0.(4.7) Solving (4.7), we get 𝑚(𝑡,𝑟)=𝐶3𝑡+𝑍(𝑟)2𝑑𝑟+𝐶2,𝐶𝑓(𝑟)=3𝑍(𝑟)4,𝑏+𝑎(4.8) where 𝑍(𝑟) satisfies 𝑍(𝑟)4𝐶1𝑟𝑍(𝑟)𝑎+𝑏=0.
Therefore, we have completed the proof of Example 4.1.

Remark 4.2. (i) For (4.4), we can find out the exact solution of it. Let 𝐺=108𝐶21𝑟+12768𝑎32304𝑎2𝑏+2304𝑎𝑏2768𝑏3+81𝐶41𝑟2,𝑄=𝐺2/348𝑎+48𝑏𝐺1/3,(4.9)
Then 𝑍(𝑟)=±6112𝑄±126𝑄𝐺2/3288𝑄𝑎+288𝑄𝑏726𝐶1𝑟𝐺1/3𝐺1/3𝑄.(4.10)
(ii) it holds that ||||𝑈(𝑡,𝑟)2=2𝐶23𝐶21𝑟𝑍(𝑟)2(4.11) is independent of 𝑡, so 𝑟1𝑟0𝑟|𝑈|2𝑑𝑟(0<𝑟0𝑟1<) is a conservation quantity.
(iii) When 0<𝑟𝑟1, it will not a difficult work to verify |𝑉𝑟|2𝐶 (𝐶 is a positive constant). For instant, in the case of 𝑏=0 and 𝑎=𝐶1=𝐶3=1, we can verify that ||𝑈𝑟||2=𝑟1/2𝑍(𝑟)+18+16𝑟2+9𝑟3𝑍(𝑟)2+8𝑟1/2𝑍(𝑟)+24𝑟5/2𝑍(𝑟)𝑟324𝑟1/2𝑍(𝑟)3+16𝑍(𝑟)2+9𝑟3/2𝑍(𝑟)+9𝑟(4.12) is bounded as 0<𝑟𝑟1. In fact, |𝑈𝑟|2 decay quickly as 𝑟+. We can see this property in Figure 6.
(iv) For any first orthogonal matrix 𝐵 (the elements of 𝐵 are constants.), 𝑈𝐵 is also a solution of (4.3).
In order to get the exact solution of (2.1), we firstly propose an important Lemma which borrow from [13] as follows.

Lemma 4.3. Let 𝐵=𝑏22+𝑏23𝑏21+𝑏22+𝑏23𝑏1𝑏2𝑏22+𝑏23𝑏21+𝑏22+𝑏23𝑏1𝑏3𝑏22+𝑏23𝑏21+𝑏22+𝑏230𝑏3𝑏22+𝑏23𝑏2𝑏22+𝑏23𝑏1𝑏21+𝑏22+𝑏23𝑏2𝑏21+𝑏22+𝑏23𝑏3𝑏21+𝑏22+𝑏23,𝐵=𝑏22+𝑏23𝑏1𝑏2𝑏22+𝑏23𝑏1𝑏3𝑏22+𝑏230𝑏3𝑏22+𝑏23𝑏2𝑏22+𝑏23𝑏1𝑏2𝑏3,𝐴=22+32cos(̃𝜌)||𝐻||12cos(̃𝜌)||𝐻||22+323sin(̃𝜌)22+3213cos(̃𝜌)||𝐻||22+32+2sin(̃𝜌)22+3222+32sin(̃𝜌)||𝐻||12sin(̃𝜌)||𝐻||22+32+3cos(̃𝜌)22+3213sin(̃𝜌)||𝐻||22+322cos(̃𝜌)22+321||𝐻||2||𝐻||3||𝐻||,(4.13) where 𝐻=(1,2,3)=(1,2,3)𝐵𝑇, |𝐻|=21+22+23, and ̃𝜌=|𝐻|𝑡0𝜇(𝑠)𝑑𝑠.
Considering the following two LLEs (1) and (2): 1𝜕𝑉𝜕𝑡=𝑉×(Δ𝑉+𝜇(𝑡)𝐻+𝐻(𝑉)),in𝑅𝑛×(0,),𝑉𝑡=0=𝑉0,in𝑅𝑛,(4.14) where 𝑡0𝜇(𝑠)𝑑𝑠𝐶[0,) and 𝜇(𝑡)𝐿2(0,), 𝐻(𝑉)=3𝑖,𝑗=1𝑏𝑖𝑗𝑢𝑖(𝑏1,𝑏2,𝑏3), 𝑏30, 𝑏𝑖𝑗=𝛼+(𝛽𝛼)𝑏𝑖𝑏𝑗(,𝑖=𝑗,𝛽𝛼)𝑏𝑖𝑏𝑗,𝑖𝑗.(4.15)2𝜕𝑈𝜕𝑡=𝑈×Δ𝑈+𝐻1(𝑈),in𝑅𝑛×(0,),𝑈𝑡=0=𝑈0,in𝑅𝑛,(4.16) where 𝐻1(𝑈)=(0,0,(𝛽𝛼)(𝑏21+𝑏22+𝑏23)𝑢3).
By the transform 𝐵𝑉=𝑈𝐴𝐵, (4.14) is equivalent to (4.16).

Combining Example 4.1 with Lemma 4.3, one has the following Proposition, in which one also adopts the notations of Lemma 4.3.

Proposition 4.4. Let 𝐵𝑉=𝑈𝐴𝐵(4.17) be a transform between the initial problem of the isotropic Landau-Lifshitz equation: 1𝜕𝑉𝜕𝑡=𝑉×(Δ𝑉),in𝑅𝑛×(0,),𝑉𝑡=0=𝑉0,in𝑅𝑛,(4.18) and the initial problem of the LLE with an anisotropic magnetic field: 2𝜕𝑈𝜕𝑡=𝑈×Δ𝑈+𝐻1(𝑈),in𝑅𝑛×(0,),𝑈|𝑡=0=𝑈0,in𝑅𝑛,(4.19) where 𝐻1(𝑈)=(0,0,(𝛽𝛼)(𝑏21+𝑏22+𝑏23)𝑢3), 𝐴=22+32||𝐻||12||𝐻||22+3213||𝐻||22+320322+32222+321||𝐻||2||𝐻||3||𝐻||.(4.20)
If 𝛼𝛽=1+𝑏23𝑏23,𝑏1=𝑏2=𝑏3or𝛼=0,𝑏1=𝑏2b3,(4.21) then 𝑉 is a solution of (1) if and only if 𝑈 is a solution of (2).

Proof. According to Lemma 4.3, to prove Proposition 4.4, we just need to set 𝜇(𝑡)𝐻=0 and 𝐻(𝑉)=3𝑖,𝑗=1𝑏𝑖𝑗𝑢𝑖(𝑏1,𝑏2,𝑏3)=0 in (4.19).
𝐻(𝑉)=0 if and only if 𝑏3𝛼+(𝛽𝛼)21+𝑏1𝑏2+𝑏1𝑏3𝑏=0,3𝛼+(𝛽𝛼)1𝑏2+𝑏22+𝑏2𝑏3𝑏=0,3𝛼+(𝛽𝛼)1𝑏3+𝑏2𝑏3+𝑏23=0.(4.22) Solving (4.22), we have {𝛽=𝛼(1+𝑏23)/𝑏23,𝑏1=𝑏2=𝑏3} or {𝛼=0,𝑏1=𝑏2𝑏3}.
On the other hand, if 𝜇(𝑡)=0, 𝐴 will degenerate as 𝐴.

Remark 4.5. According to Proposition 4.4, we can construct a smooth solution about (2.1) on the disc from (4.17), (4.21), and (4.4)-(4.5).

5. Discussions

In Sections 2, and 4, we construct some solutions in which |𝑉| is not a constant. In the original physical model [1, 2] about LLE, LLE is under the constrain of 𝑉𝑆2 (here, we means that |𝑉|=𝐶>0.). However, many papers [1013] do some investigations about the case that |𝑉| is not a constant. If |𝑉| is not a constant, is (1.4) still a physical significance model? To answer this question, we firstly mention another LLE. Exactly speaking, under radially symmetric coordinates, (2+1)-dimensional inhomogeneous LLE [14] takes the following form: 𝑈𝑡𝑈=q(𝑟)𝑈×𝑟𝑟+1𝑟𝑈𝑟+𝑞(𝑟)𝑟𝑈×𝑈𝑟,(5.1) where 𝑞(𝑟) is the inhomogeneous term.

Suppose 𝑉=(𝑣1(𝑡,𝑟),𝑣2(𝑡,𝑟),𝑣3(𝑡,𝑟)) is a solution of (2.1), 𝑍=(𝑧1(𝑡,𝑟),𝑧2(𝑡,𝑟),𝑧3(𝑡,𝑟)),𝑧21(𝑡,𝑟)+𝑧22(𝑡,𝑟)+𝑧23(𝑡,𝑟)=𝐶>0. Furthermore, the relationship between 𝑉 and 𝑍 is in the form of 𝑣1(𝑡,𝑟)=𝑧1𝑣(𝑡,𝑟)𝐺(𝛼𝑟),2(𝑡,𝑟)=𝑧2𝑣(𝑡,𝑟)𝐺(𝛼𝑟),3(𝑡,𝑟)=𝑧3(𝑡,𝑟)𝐺(𝛼𝑟),(5.2) where 𝐺(𝛼𝑟) is a functions, 𝛼 is constant.

Substituting (5.2) into (2.1), we get 𝑍𝑡𝑍=𝐺(𝛼𝑟)𝑍×𝑟𝑟+1𝑟𝑍𝑟+2𝛼𝐺(𝛼𝑟)𝑍×𝑍𝑟,(5.3) where “” is a derivative about 𝛼𝑟.

Furthermore, suppose 𝛼=1/2, (5.3) can be rearranged to give (for simplicity, we write 𝐺 for 𝐺() here. 𝑍𝑡𝑍=𝐺𝑍×𝑟𝑟+1𝑟𝑍𝑟+𝐺𝑍×𝑍𝑟.(5.4) So we can regard (5.4) as a modification version of (5.1). According to the above deducing, the following theorem holds

Theorem 5.1. Under the constrain (5.2), 𝑉=(𝑣1(𝑡,𝑟),𝑣2(𝑡,𝑟),𝑣3(𝑡,𝑟)) is a solution of (2.1) if and only if 𝑍=(𝑧1(𝑡,𝑟),𝑧2(𝑡,𝑟),𝑧3(𝑡,𝑟))(𝐺()=𝐺(𝑟/2)) is a solution of (5.4) where 𝑧21(𝑡,𝑟)+𝑧22(𝑡,𝑟)+𝑧23(𝑡,𝑟)=𝐶>0.

One mentions here that if 𝛼=1, 𝐺(𝛼𝑟)=𝐺(𝑟), (5.3) will adopt the following form: 𝑈𝑡𝑍=𝐺(𝑟)𝑍×𝑟𝑟+1𝑟𝑍𝑟+2𝐺(𝑟)𝑟𝑈×𝑈𝑟.(5.5)

So, (5.5) will not equal to (5.1) because of the coefficient 2 before the 𝐺(𝑟)𝑟[𝑍×𝑍𝑟]. And one cannot deduce a the solution of (5.1) from (2.1) according to Theorem 5.1. One has to choose another way to construct the exact solution of (5.1) which is now beyond the topic of this paper.

6. Geometric Significance

In this section, we investigate the evolution of solutions and the spacial gradient about them. Five different kinds of solutions are demonstrated in the following notations.(1)Figure 1 display part shape of solutions on solution (2.2)–(2.4) where 𝑟[1,3], 𝑡[0,9/10], 𝑇=1, 𝛾=1, 𝐶1=1, 𝐶2=0, 𝐶3=1 and 𝐶4=1. Accordingly, from Figure 2, we can descry the blow up behavior of (2.13) near 𝑇=1. Different from Figure 1, here we set 𝑟[3/5,10] in Figure 2.(2)For vortex solution (3.1) and (3.4), as shown in Figure 3, we can see the decaying behavior of square spatial curvature (3.23). Here, we set 𝐶=4 and 𝜆=0,1,2,3,4, respectively. In Figure 3, begin from 𝜆=1, we can see that the shape of (3.23) has only a declining shift of the peak when 𝜆 increases.(3)From Figures 45 we can apperceive the periodic behavior of (3.24) which is determined by the periodic function: 𝑄(𝑥,𝑦)=4arctan𝑠𝑑𝑥+1,32𝑑𝑠𝑦+1,32𝑄,(6.1)(𝑥,𝑦)=4arctancot(1)cosh2/2cos(1)(𝑥𝑦)cosh2/2sin(1)(𝑦𝑥),(6.2)     respectively. A singularity behavior of (3.24) can be seen in Figure 4. (4) Considering the solution (4.1), and setting 𝐶1=1, 𝐶2=1 and 𝐶3=1, we draw the picture of |𝑈𝑟|2=(1/9)(9𝑟8/3+8/𝑟10/3) where 𝑟[1/2,25] in Figure 6. From Figure 6, we can find out that the energy cannot be bounded near 𝑡=0. Setting 𝑡[0,3] and 𝑟[1/10,3], we draw the picture of (4.1) in Figure 7. Amazedly, we point out here that (2.2)–(2.4) will perform a blow-up behavior while (4.1) will not under the infinite energy initial condition.

Figure 1: (2.2)–(2.4) with 𝑟[1,3], 𝑡[0,9/10].
Figure 2: Part shape of (2.13) with 𝑇=1.
Figure 3: (3.23) as 𝜆=0,1,2,3,4, respectively.
Figure 4: Part shape of periodic function (3.24) and (6.1).
Figure 5: Part shape of periodic function (3.24) and (6.2).
Figure 6: (4.1) with 𝑟[1/2,25] near 𝑡=0.
Figure 7: Part shape of (4.1) as 𝑡[0,3] and 𝑟[1/10,3].

7. Conclusions

In the present work, we have obtained the exact solutions of the LLE by using some ansatzs about the solutions. By using the functional separable technique and general Jacobi elliptic-function method, we have obtained the blow-up solution, vortex solution, various periodic solutions, and non-blow-up solutions in the radially symmetric coordinates or not. For a better understanding about the property of the solutions, some solutions and their spatial curvature which cover five different types of solutions have been also illustrated in a graphic way. From Sections 2 and 4, we can see that the solution of Section 2 will perform a blow-up behavior while the solutions in Section 4 will not under the infinite energy initial condition. Similarly, from Section 3, we can find out the same conclusion that the vortex solution will preserve the finite energy while the corresponding periodic solutions will not.


This work is supported by NSFC (Grant no. 11071009), Funding Project for Academic Human Resources Development in Institutions of Higher Leading Under the Jurisdiction of Beijing Municipality (PHR-IHLB 200906103), Subjects and Education Program for Graduate Student-Project for Innovative Human (Ph.D.), and Innovation Fund of Beijing University of Technology-YB201211 (held by P. Zhong).


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