International Scholarly Research Notices

International Scholarly Research Notices / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 752964 | 20 pages | https://doi.org/10.5402/2012/752964

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

Academic Editor: S. Blanes
Received30 May 2012
Accepted26 Jun 2012
Published09 Aug 2012

Abstract

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(π‘₯,𝑑), which is a function of position π‘₯ and time 𝑑. For instance, a subcase of the LLE is as follows: 𝑉𝑑=𝛼𝑉×𝐻eο¬€ξ€·βˆ’π›½π‘‰Γ—π‘‰Γ—π»eff,(1.1) where 𝑉=(𝑣1,𝑣2,𝑣3). 𝛼, 𝛽 are constants. Γ— denotes a vector cross-product. The effective magnetic field 𝐻eff is given by the derivative of the magnetic crystal energy 𝐸 with respect to the vector 𝑉: 𝐻eff=βˆ’πœ•πΈ,πœ•π‘‰(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)=𝐸𝑒π‘₯(𝑉)+πΈπ‘Žπ‘›(𝑉)+πΈπ‘šπ‘“(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): 𝑉𝑑=𝑉×Δ𝑉𝑅2,Δ𝑉𝑅2=πœ•2π‘‰πœ•π‘₯21+πœ•2π‘‰πœ•π‘₯22.(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 (𝑛β‰₯2). 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 limπ‘‘β†’π‘‡βˆ’supβ€–βˆ‡π‘‰(β‹…,𝑑)β€–βˆž=∞.(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: 𝑉𝑑=π‘‰Γ—π‘‰π‘Ÿπ‘Ÿ+1π‘Ÿπ‘‰Γ—π‘‰π‘Ÿ,(2.1) where ξ”π‘Ÿ=π‘₯21+π‘₯22, 𝑉=(𝑣1(𝑑,π‘Ÿ),𝑣2(𝑑,π‘Ÿ),𝑣3(π‘Ÿ)).

We will find the explicit solution of (2.1) in the form of 𝑣1𝑣(𝑑,π‘Ÿ)=cos(π‘š(𝑑,π‘Ÿ))𝑓(π‘Ÿ),2𝑣(𝑑,π‘Ÿ)=sin(π‘š(𝑑,π‘Ÿ))𝑓(π‘Ÿ),3(𝑑,π‘Ÿ)=𝛾𝑓(π‘Ÿ),(2.2) where π‘š(𝑑,π‘Ÿ) and 𝑓(π‘Ÿ) are functions to be determined, 𝛾 is constant.

Example 2.1. If 1π‘š(𝑑,π‘Ÿ)=βˆ’43227𝐢31π‘Ÿ4+108𝐢21π‘Ÿ8/3𝐢2+144𝐢1π‘Ÿ4/3𝐢22+64𝐢32π‘‘βˆ’π‘‡+𝐢3,(2.3)𝑓(π‘Ÿ)=βˆ’81𝐢31π‘Ÿ4βˆ’324𝐢21π‘Ÿ8/3𝐢2βˆ’432𝐢1π‘Ÿ4/3𝐢22βˆ’192𝐢32𝛾𝐢21ξ€·16π‘Ÿ1/3𝐢22+24𝐢1π‘Ÿ5/3𝐢2+9𝐢21π‘Ÿ3ξ€Έ2,(2.4) where 𝛾,𝐢1,𝐢2,𝐢3 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 π›Ύπ‘Ÿsin(π‘š)π‘š2π‘Ÿπ‘“βˆ’π›Ύπ‘Ÿcos(π‘š)π‘šπ‘Ÿπ‘Ÿπ‘“βˆ’2π›Ύπ‘Ÿcos(π‘š)π‘šπ‘Ÿπ‘“π‘Ÿβˆ’π‘“π›Ύcos(π‘š)π‘šπ‘Ÿ+sin(π‘š)π‘šπ‘‘π‘Ÿ=0,(2.5)π›Ύπ‘Ÿcos(π‘š)π‘š2π‘Ÿπ‘“+π›Ύπ‘Ÿsin(π‘š)π‘šπ‘Ÿπ‘Ÿπ‘“+2π›Ύπ‘Ÿsin(π‘š)π‘šπ‘Ÿπ‘“π‘Ÿ+𝑓𝛾sin(π‘š)π‘šπ‘Ÿ+cos(π‘š)π‘šπ‘‘π‘Ÿ=0,(2.6)π‘Ÿπ‘šπ‘Ÿπ‘Ÿπ‘“+2π‘Ÿπ‘šπ‘Ÿπ‘“π‘Ÿ+π‘šπ‘Ÿπ‘“=0.(2.7) Furthermore, if we insert (2.7) into (2.5) and (2.6), we have π‘Ÿπ‘šπ‘Ÿπ‘Ÿπ‘“+2π‘Ÿπ‘šπ‘Ÿπ‘“π‘Ÿ+π‘šπ‘Ÿπ‘“=0,π›Ύπ‘š2π‘Ÿπ‘“+π‘šπ‘‘=0.(2.8)
If we settle down on the following ansatzs: π‘š=𝑔(π‘Ÿ)π‘‘βˆ’π‘‡+𝐢,𝑓=𝑔(π‘Ÿ)𝑗(π‘Ÿ),(2.9) where 𝑔(π‘Ÿ) and 𝑗(π‘Ÿ) are the functions about π‘Ÿ to be determined, 𝑇 and 𝐢 are constants.
From (2.8) and (2.9), we have 𝛾𝑔2π‘Ÿπ‘”π‘—βˆ’π‘”=0,π‘Ÿπ‘”2π‘Ÿπ‘”π‘—+2π‘Ÿπ‘”π‘Ÿξ€·π‘”π‘Ÿπ‘—+π‘”π‘—π‘Ÿξ€Έ+π‘”π‘Ÿπ‘”π‘—=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=𝐢6431π‘Ÿ4+1𝐢1621π‘Ÿ8/3𝐢2+1𝐢121π‘Ÿ4/3𝐢22+1𝐢2732,𝑗=20736𝛾𝐢21ξ€·9𝐢21π‘Ÿ3+24𝐢1π‘Ÿ5/3𝐢2+16π‘Ÿ1/3𝐢22ξ€Έ2,(2.11) where 𝛾,𝐢1,𝐢2 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 π‘Ÿ 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(𝑑,π‘Ÿ)2=9ξ€·144𝐢1π‘Ÿ4/3𝐢22+108𝐢21π‘Ÿ8/3𝐢2+27𝐢31π‘Ÿ4+64𝐢32ξ€Έ2ξ€·1+𝛾2𝛾2𝐢41ξ€·16π‘Ÿ1/3𝐢22+24𝐢1π‘Ÿ5/3𝐢2+9𝐢21π‘Ÿ3ξ€Έ4(2.12) is independent of 𝑑, so βˆ«π‘Ÿ1π‘Ÿ0π‘Ÿ|𝑉|2π‘‘π‘Ÿ(0<π‘Ÿ0β‰€π‘Ÿ1<∞) is a conservation quantity.
(ii) When π‘Ÿ0β‰€π‘Ÿβ‰€π‘Ÿ1, 𝑇>0, it is not difficult to verify lim𝑑→𝑇|π‘‰π‘Ÿ|2=+∞. For example, in the case of 𝑇=1 and 𝛾=𝐢1=𝐢3=1, 𝐢2=0, we can verify that ||π‘‰π‘Ÿ||2=1π‘Ÿ168βˆ’256𝑑+128+128𝑑2π‘Ÿ6(π‘‘βˆ’1)2(2.13) is not bounded as 𝑑→1.
(iii) For any first orthogonal matrix 𝐡 (the elements of 𝐡 are constants.), 𝑉𝐡 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(𝑑,π‘Ÿ)=𝑓(𝑑)sin(π‘ž(π‘Ÿ)),2(𝑑,π‘Ÿ)=1βˆ’(𝑓(𝑑))2𝑣sin(π‘ž(π‘Ÿ)),3(𝑑,π‘Ÿ)=cos(π‘ž(π‘Ÿ)),(3.1)

Submitting (3.1) into (2.1), (2.1) can be transformed into 1βˆ’(𝑓(𝑑))2π‘Ÿ2𝑑2π‘‘π‘Ÿ2ξ”π‘ž(π‘Ÿ)+1βˆ’(𝑓(𝑑))2ξ‚€π‘‘ξ‚ξ‚€π‘‘π‘‘π‘Ÿπ‘ž(π‘Ÿ)π‘Ÿ+𝑑𝑑𝑓(𝑑)sin(π‘ž(π‘Ÿ))π‘Ÿ2=0.(3.2)

So we have 1βˆ’(𝑓(𝑑))2+1πΆπ‘‘π‘Ÿπ‘‘π‘‘π‘“(𝑑)=0,2𝑑2π‘‘π‘Ÿ2ξ‚€π‘‘π‘ž(π‘Ÿ)+ξ‚π‘‘π‘Ÿπ‘ž(π‘Ÿ)π‘Ÿβˆ’πΆsin(π‘ž(π‘Ÿ))π‘Ÿ2=0.(3.3) Solving (3.3), we have 𝑓(𝑑)=sin𝐢𝑑+𝐢1ξ€Έξ€·or𝑓(𝑑)=cos𝐢𝑑+𝐢1ξ€Έ,1βˆ’π‘“2ξ€·(𝑑)=cos𝐢𝑑+𝐢1or1βˆ’π‘“2ξ€·(𝑑)=sin𝐢𝑑+𝐢1ξ€Έ,ξƒ©π‘Ÿπ‘ž(π‘Ÿ)=Β±4,arctanΒ±βˆšπΆπœ†ξƒͺ,(3.4) where 𝐢>0, 𝐢1, πœ† are constants.

We point out there that (3.4) is vortex solution of (2.1). Similarly, if we set (𝐢 is a constant) 𝑣1𝑣(𝑑,π‘₯,𝑦)=cos(𝑑+𝐢)sin(𝑄(π‘₯,𝑦)),2𝑣(𝑑,π‘₯,𝑦)=sin(𝑑+𝐢)sin(𝑄(π‘₯,𝑦)),3(𝑑,π‘₯,𝑦)=cos(𝑄(π‘₯,𝑦)),(3.5)

Equation (1.4) can be transformed into πœ•2π‘„πœ•π‘₯2+πœ•2π‘„πœ•π‘¦2=sin𝑄.(3.6)

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

(I) Functional separable solutions of (3.6)
let π‘„ξ€Ίπ‘“ξ€·πœ‰(π‘₯,𝑦)=Β±4arctan1ξ€Έπ‘”ξ€·πœ‰2ξ€Έξ€»,(3.7) where πœ‰1=π‘Žπ‘₯+𝑏, πœ‰2=𝑐𝑦+𝑑.
Substituting (3.7) into (3.6), we can deduce the following equations: 4π‘Ž2π‘“ξ…žξ…žπ‘”+4π‘Ž2π‘“ξ…žξ…žπ‘“2𝑔3βˆ’8π‘Ž2(𝑓′)2𝑓𝑔3+4𝑐2π‘”ξ…žξ…žπ‘“+4𝑐2π‘”ξ…žξ…žπ‘“3𝑔2βˆ’8𝑐2(𝑔′)2𝑓3π‘”βˆ’4𝑓𝑔+4𝑓3g3=0.(3.8)
Introducing two auxiliary function about 𝑓′, π‘”ξ…ž, (𝑓′)2=𝐴1𝑓4+𝐡1𝑓3+𝐢1𝑓2+𝐷1𝑓+𝐸1,((3.9)𝑔′)2=𝐴2𝑔4+𝐡2𝑔3+𝐢2𝑔2+𝐷2𝑔+𝐸2.(3.10)
According to (3.9) and (3.10), we have π‘“ξ…žξ…ž=12𝐷1+2𝐴1𝑓3+32𝐡1𝑓2+𝐢1𝑔𝑓,(3.11)ξ…žξ…ž=12𝐷2+2𝐴2𝑔3+32𝐡2𝑔2+𝐢2𝑔.(3.12)
Substituting (3.9)–(3.12) into (3.8), we have βˆ’2𝐡1π‘Ž2𝑓4𝑔4βˆ’2𝐡2𝑐2𝑓3𝑔4ξ€·πΆβˆ’41π‘Ž2+𝐢2𝑐2ξ€Έπ‘“βˆ’13𝑔3βˆ’6𝐷2𝑐2𝑓3𝑔2𝐴+81π‘Ž2βˆ’πΈ2𝑐2𝑓3π‘”βˆ’6𝐷1π‘Ž2𝑓2𝑔3+6𝐡1π‘Ž2𝑓2𝑔𝐴+82𝑐2βˆ’πΈ1π‘Ž2𝑓𝑔3+6𝐡2𝑐2𝑓𝑔2𝐢+41π‘Ž2+𝐢2𝑐2ξ€Έβˆ’1𝑓𝑔+2𝐷2𝑐2𝑓+2𝐷1π‘Ž2𝑔=0.(3.13)
Equating the coefficients of different powers of 𝑓𝑔 in (3.13), we obtain a set of algebraic equations: 𝐡1=𝐡2=𝐷1=𝐷2𝐢=0,1π‘Ž2+𝐢2𝑐2π΄βˆ’1=0,1π‘Ž2βˆ’πΈ2𝑐2𝐴=0,2𝑐2βˆ’πΈ1π‘Ž2=0.(3.14)
If we set 𝐴=𝐴1π‘Ž2,𝐡=𝐢1π‘Ž2,𝐢=𝐸1π‘Ž2,(3.15) according to (3.14), functions 𝑓=𝑓(πœ‰1) and 𝑔=𝑔(πœ‰2) are determined by the first-order autonomous ordinary differential equations as follows: ξ€·π‘“πœ‰1ξ€Έ2=𝐴𝑓4+𝐡𝑓2𝑔+𝐢,πœ‰2ξ€Έ2=C𝑔4+(1βˆ’π΅)𝑔2+𝐴,(3.16)
Here, we omit some details about the deducing and present some exact solutions of (3.16). (A) If 𝐴=𝐢=0, 0<𝐡<1, then ξ‚€Β±βˆšπ‘“=expπ΅πœ‰1ξ‚ξ‚€Β±βˆš,𝑔=exp1βˆ’π΅πœ‰2.(3.17)(B) If 𝐴=0, 𝐡=1, 𝐢>0, solving (3.16), we have βˆšπ‘“=ξ€·πœ‰πΆsinh1ξ€Έ1,𝑔=Β±βˆšπΆπœ‰2.(3.18)(C) If 𝐴=0, 𝐡>1, 𝐢>0, then 𝑓=πΆπ΅ξ‚€βˆšsinhπ΅πœ‰1,𝑔=π΅βˆ’1πΆξ‚€βˆšsecπ΅βˆ’1πœ‰2.(3.19)(D) According to Table 3.1 which is proposed in [7], we can deduce a solution easily. Defining 𝑠𝑐(πœ‰,𝑅)=𝑠𝑛(πœ‰,𝑅)𝑐𝑛(πœ‰,R),𝑑𝑠(πœ‰,𝑅)=𝑑𝑛(πœ‰,𝑅),𝑠𝑛(πœ‰,𝑅)(3.20) where 𝑅 is the modulus of Jacobi elliptic function.
Suppose βˆšπ‘…=3/2, 𝐴=βˆ’3/16, 𝐡=1/2, 𝐢=1, we have ξƒ©πœ‰π‘“=𝑠𝑑1,√32ξƒͺξƒ©πœ‰,𝑔=𝑑𝑠2,√32ξƒͺ.(3.21)

(II)Another functional separable solution of (3.6)
𝑄(π‘₯,𝑦)=4arctancot𝐢1cosh𝑓cosh𝑔,𝑓=cos𝐢11+𝐢22ξ€·π‘₯βˆ’πΆ2𝑦,𝑔=sin𝐢11+𝐢22ξ€·π‘¦βˆ’πΆ2π‘₯ξ€Έ,(3.22) where 𝐢1 and 𝐢2 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), ||π‘‰π‘Ÿ(||𝑑,π‘Ÿ)2π‘Ÿ=162βˆšπΆβˆ’2πΆπœ†2ξ‚€πœ†2+π‘Ÿ2βˆšπΆξ‚2(3.23) is independent of 𝑑. Furthermore, ∫∞0|π‘‰π‘Ÿ|2π‘Ÿπ‘‘π‘Ÿ is a conservation quantity. For example, if we set 𝐢=4 and πœ†=1 in (3.23), we have ∫∞0|π‘‰π‘Ÿ|2π‘Ÿπ‘‘π‘Ÿ=16≫0.
(ii) For the solutions proposed in (ii)-(iii) in Example 3.1, it is a tedious expression for |βˆ‡π‘‰(𝑑,π‘₯,𝑦)|2 (βˆ‡ is about the spacial directions) to each of them. Therefore, if we omit the details of 𝑄(𝑑,π‘₯,𝑦), obviously |βˆ‡π‘‰(𝑑,π‘₯,𝑦)|2 of (ii) and (iii) will adopt the same form as follows: ||||βˆ‡π‘‰(𝑑,π‘₯,𝑦)2=ξ‚€πœ•ξ‚πœ•π‘₯𝑄(π‘₯,𝑦)2+ξ‚΅πœ•ξ‚Άπœ•π‘¦π‘„(π‘₯,𝑦)2.(3.24) The graphic demonstration will be provided in Section 6.
(iii) For any first orthogonal matrix 𝐡 (the elements of 𝐡 are constants.), 𝑉𝐡 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: 𝑣1𝐢(𝑑,π‘Ÿ)=βˆ’cos3𝐢𝑑+(3/4)π‘Ÿ1βˆšπ‘Ÿξ‚2/3+𝐢2𝐢3𝐢1βˆšπ‘Ÿξ‚4/3,𝑣2𝐢(𝑑,π‘Ÿ)=βˆ’sin3𝐢𝑑+(3/4)π‘Ÿ1βˆšπ‘Ÿξ‚2/3+𝐢2𝐢3𝐢1βˆšπ‘Ÿξ‚4/3,𝑣3𝐢(π‘Ÿ)=βˆ’3𝐢1βˆšπ‘Ÿξ‚4/3,(4.1) where 𝐢1,𝐢2,𝐢3 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: π‘ˆπ‘‘=π‘ˆΓ—(Ξ”π‘ˆ+π‘ˆπ½),(4.2) where 𝐽=diag{π‘Ž,π‘Ž,𝑏}, π‘Ž,𝑏 are constants.

Under radially symmetric coordinates, (4.2) takes the form: π‘ˆπ‘‘=π‘ˆΓ—π‘ˆπ‘Ÿπ‘Ÿ+1π‘Ÿπ‘ˆΓ—π‘ˆπ‘Ÿ+π‘ˆΓ—π‘ˆπ½,(4.3) where ξ”π‘Ÿ=π‘₯21+π‘₯22, π‘ˆ=(𝑒1(𝑑,π‘Ÿ),𝑒2(𝑑,π‘Ÿ),𝑒3(𝑑,π‘Ÿ)).

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

Example 4.1. If 𝑍(π‘Ÿ) is the root of 𝑍(π‘Ÿ)4βˆ’πΆ1βˆšπ‘’π‘Ÿπ‘(π‘Ÿ)βˆ’π‘Ž+𝑏=0,(4.4)1𝐢(𝑑,π‘Ÿ)=cos3βˆ«π‘‘+𝑍(π‘Ÿ)2π‘‘π‘Ÿ+𝐢2𝐢3βˆ’π‘(π‘Ÿ)4,π‘’βˆ’π‘+π‘Ž2𝐢(𝑑,π‘Ÿ)=sin3βˆ«π‘‘+𝑍(π‘Ÿ)2π‘‘π‘Ÿ+𝐢2𝐢3βˆ’π‘(π‘Ÿ)4,π‘’βˆ’π‘+π‘Ž3𝐢(π‘Ÿ)=3βˆ’π‘(π‘Ÿ)4,βˆ’π‘+π‘Ž(4.5) where 𝐢1,𝐢2,𝐢3 are any constants, then (4.4)-(4.5) is a solution of (4.3).

Proof. We settle down on the following setting: 𝑒1𝑒(𝑑,π‘Ÿ)=cos(π‘š(𝑑,π‘Ÿ))𝑓(π‘Ÿ),2𝑒(𝑑,π‘Ÿ)=sin(π‘š(𝑑,π‘Ÿ))𝑓(π‘Ÿ),3(π‘Ÿ)=𝑓(π‘Ÿ),(4.6) where π‘š(𝑑,π‘Ÿ) and 𝑓(π‘Ÿ) are functions to be determined.
Similar to the deducing of Example 2.1, if we insert (4.6) into (4.3), we have π‘Ÿπ‘šπ‘Ÿπ‘Ÿπ‘“+2π‘Ÿπ‘šπ‘Ÿπ‘“π‘Ÿ+π‘šπ‘Ÿπ‘“=0,π›Ύπ‘š2π‘Ÿπ‘“+(π‘βˆ’π‘Ž)𝑓+π‘šπ‘‘=0.(4.7) Solving (4.7), we get π‘š(𝑑,π‘Ÿ)=𝐢3ξ€œπ‘‘+𝑍(π‘Ÿ)2π‘‘π‘Ÿ+𝐢2,𝐢𝑓(π‘Ÿ)=3βˆ’π‘(π‘Ÿ)4,βˆ’π‘+π‘Ž(4.8) where 𝑍(π‘Ÿ) satisfies 𝑍(π‘Ÿ)4βˆ’πΆ1βˆšπ‘Ÿπ‘(π‘Ÿ)βˆ’π‘Ž+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=108𝐢21ξ”π‘Ÿ+12768π‘Ž3βˆ’2304π‘Ž2𝑏+2304π‘Žπ‘2βˆ’768𝑏3+81𝐢41π‘Ÿ2,ξƒŽπ‘„=𝐺2/3βˆ’48π‘Ž+48𝑏𝐺1/3,(4.9)
Then βˆšπ‘(π‘Ÿ)=Β±6112π‘„Β±ξ„Άξ„΅ξ„΅βŽ·12βˆ’6𝑄𝐺2/3βˆšβˆ’288π‘„π‘Ž+288π‘„π‘βˆ’726𝐢1βˆšπ‘ŸπΊ1/3𝐺1/3𝑄.(4.10)
(ii) it holds that ||||π‘ˆ(𝑑,π‘Ÿ)2=2𝐢23𝐢21π‘Ÿπ‘(π‘Ÿ)2(4.11) is independent of 𝑑, so βˆ«π‘Ÿ1π‘Ÿ0π‘Ÿ|π‘ˆ|2π‘‘π‘Ÿ(0<π‘Ÿ0β‰€π‘Ÿ1<∞) is a conservation quantity.
(iii) When 0<π‘Ÿβ‰€π‘Ÿ1, it will not a difficult work to verify |π‘‰π‘Ÿ|2≀𝐢 (𝐢 is a positive constant). For instant, in the case of 𝑏=0 and π‘Ž=𝐢1=𝐢3=1, we can verify that ||π‘ˆπ‘Ÿ||2=ξ€·π‘Ÿ1/2𝑍(π‘Ÿ)+1ξ€Έξ€·8+16π‘Ÿ2+9π‘Ÿ3𝑍(π‘Ÿ)2+8π‘Ÿ1/2𝑍(π‘Ÿ)+24π‘Ÿ5/2𝑍(π‘Ÿ)π‘Ÿ3ξ€·24π‘Ÿ1/2𝑍(π‘Ÿ)3+16𝑍(π‘Ÿ)2+9π‘Ÿ3/2𝑍(π‘Ÿ)+9π‘Ÿ(4.12) is bounded as 0<π‘Ÿβ‰€π‘Ÿ1. In fact, |π‘ˆπ‘Ÿ|2 decay quickly as π‘Ÿβ†’+∞. We can see this property in Figure 6.
(iv) For any first orthogonal matrix 𝐡 (the elements of 𝐡 are constants.), π‘ˆπ΅ 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 βŽ‘βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ£ξ”π΅=𝑏22+𝑏23𝑏21+𝑏22+𝑏23βˆ’π‘1𝑏2𝑏22+𝑏23𝑏21+𝑏22+𝑏23βˆ’π‘1𝑏3𝑏22+𝑏23𝑏21+𝑏22+𝑏230𝑏3𝑏22+𝑏23βˆ’π‘2𝑏22+𝑏23𝑏1𝑏21+𝑏22+𝑏23𝑏2𝑏21+𝑏22+𝑏23𝑏3𝑏21+𝑏22+𝑏23⎀βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯⎦,ξ‚βŽ‘βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ£ξ”π΅=𝑏22+𝑏23βˆ’π‘1𝑏2𝑏22+𝑏23βˆ’π‘1𝑏3𝑏22+𝑏230𝑏3𝑏22+𝑏23βˆ’π‘2𝑏22+𝑏23𝑏1𝑏2𝑏3⎀βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯⎦,ξ‚βŽ‘βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ£ξ”π΄=ξ‚β„Ž22+ξ‚β„Ž32cos(ΜƒπœŒ)||𝐻||βˆ’ξ‚β„Ž1ξ‚β„Ž2cos(ΜƒπœŒ)||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž3sin(ΜƒπœŒ)ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž1ξ‚β„Ž3cos(ΜƒπœŒ)||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž32+ξ‚β„Ž2sin(ΜƒπœŒ)ξ”ξ‚β„Ž22+ξ‚β„Ž32ξ”ξ‚β„Ž22+ξ‚β„Ž32sin(ΜƒπœŒ)||𝐻||βˆ’ξ‚β„Ž1ξ‚β„Ž2sin(ΜƒπœŒ)||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž32+ξ‚β„Ž3cos(ΜƒπœŒ)ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž1ξ‚β„Ž3sin(ΜƒπœŒ)||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž2cos(ΜƒπœŒ)ξ”ξ‚β„Ž22+ξ‚β„Ž32ξ‚β„Ž1||𝐻||ξ‚β„Ž2||𝐻||ξ‚β„Ž3||𝐻||⎀βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯⎦,(4.13) where ξ‚ξ‚β„Žπ»=(1,ξ‚β„Ž2,ξ‚β„Ž3)=(β„Ž1,β„Ž2,β„Ž3)𝐡𝑇, |𝐻|=ξ‚β„Ž21+ξ‚β„Ž22+ξ‚β„Ž23, and ξ‚βˆ«ΜƒπœŒ=|𝐻|𝑑0πœ‡(𝑠)𝑑𝑠.
Considering the following two LLEs (β„™1) and (β„™2): ξ€·β„™1ξ€Έξƒ―πœ•π‘‰πœ•π‘‘=𝑉×(Δ𝑉+πœ‡(𝑑)𝐻+𝐻(𝑉)),in𝑅𝑛×(0,∞),π‘‰βˆ£π‘‘=0=𝑉0,in𝑅𝑛,(4.14) where βˆ«π‘‘0πœ‡(𝑠)π‘‘π‘ βˆˆπΆ[0,∞) and πœ‡(𝑑)∈𝐿2(0,∞), βˆ‘π»(𝑉)=3𝑖,𝑗=1𝑏𝑖𝑗𝑒𝑖(𝑏1,𝑏2,𝑏3), 𝑏3β‰ 0, 𝑏𝑖𝑗=𝛼+(π›½βˆ’π›Ό)𝑏𝑖𝑏𝑗(,𝑖=𝑗,π›½βˆ’π›Ό)𝑏𝑖𝑏𝑗ℙ,𝑖≠𝑗.(4.15)2ξ€Έξƒ―πœ•π‘ˆξ€·πœ•π‘‘=π‘ˆΓ—Ξ”π‘ˆ+𝐻1ξ€Έ(π‘ˆ),in𝑅𝑛×(0,∞),π‘ˆβˆ£π‘‘=0=π‘ˆ0,in𝑅𝑛,(4.16) where 𝐻1(π‘ˆ)=(0,0,(π›½βˆ’π›Ό)(𝑏21+𝑏22+𝑏23)𝑒3).
By the transform 𝐡𝑉=π‘ˆπ΄π΅, (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 𝐡𝑉=π‘ˆβ‰ˆπ΄π΅(4.17) be a transform between the initial problem of the isotropic Landau-Lifshitz equation: ξ€·β„š1ξ€Έξƒ―πœ•π‘‰πœ•π‘‘=𝑉×(Δ𝑉),in𝑅𝑛×(0,∞),π‘‰βˆ£π‘‘=0=𝑉0,in𝑅𝑛,(4.18) and the initial problem of the LLE with an anisotropic magnetic field: ξ€·β„š2ξ€Έξƒ―πœ•π‘ˆξ€·πœ•π‘‘=π‘ˆΓ—Ξ”π‘ˆ+𝐻1ξ€Έ(π‘ˆ),in𝑅𝑛×(0,∞),π‘ˆ|𝑑=0=π‘ˆ0,in𝑅𝑛,(4.19) where 𝐻1(π‘ˆ)=(0,0,(π›½βˆ’π›Ό)(𝑏21+𝑏22+𝑏23)𝑒3), ξ‚ξ‚βŽ‘βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ’βŽ£ξ”π΄=ξ‚β„Ž22+ξ‚β„Ž32||𝐻||βˆ’ξ‚β„Ž1ξ‚β„Ž2||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž1ξ‚β„Ž3||𝐻||ξ”ξ‚β„Ž22+ξ‚β„Ž320ξ‚β„Ž3ξ”ξ‚β„Ž22+ξ‚β„Ž32βˆ’ξ‚β„Ž2ξ”ξ‚β„Ž22+ξ‚β„Ž32ξ‚β„Ž1||𝐻||ξ‚β„Ž2||𝐻||ξ‚β„Ž3||𝐻||⎀βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯βŽ₯⎦.(4.20)
If 𝛼𝛽=βˆ’1+𝑏23𝑏23,𝑏1=𝑏2=𝑏3ξƒ°ξ€½or𝛼=0,𝑏1=βˆ’π‘2βˆ’b3ξ€Ύ,(4.21) then 𝑉 is a solution of (β„š1) if and only if π‘ˆ is a solution of (β„š2).

Proof. According to Lemma 4.3, to prove Proposition 4.4, we just need to set πœ‡(𝑑)𝐻=0 and βˆ‘π»(𝑉)=3𝑖,𝑗=1𝑏𝑖𝑗𝑒𝑖(𝑏1,𝑏2,𝑏3)=0 in (4.19).
𝐻(𝑉)=0 if and only if 𝑏3𝛼+(π›½βˆ’π›Ό)21+𝑏1𝑏2+𝑏1𝑏3𝑏=0,3𝛼+(π›½βˆ’π›Ό)1𝑏2+𝑏22+𝑏2𝑏3𝑏=0,3𝛼+(π›½βˆ’π›Ό)1𝑏3+𝑏2𝑏3+𝑏23ξ€Έ=0.(4.22) Solving (4.22), we have {𝛽=𝛼(βˆ’1+𝑏23)/𝑏23,𝑏1=𝑏2=𝑏3} or {𝛼=0,𝑏1=βˆ’π‘2βˆ’π‘3}.
On the other hand, if πœ‡(𝑑)=0, 𝐴 will degenerate as β‰ˆπ΄.

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 π‘‰βˆˆπ‘†2 (here, we means that |𝑉|=𝐢>0.). However, many papers [10–13] 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: π‘ˆπ‘‘ξ‚ƒπ‘ˆ=q(π‘Ÿ)π‘ˆΓ—π‘Ÿπ‘Ÿ+1π‘Ÿπ‘ˆπ‘Ÿξ‚„+π‘ž(π‘Ÿ)π‘Ÿξ€Ίπ‘ˆΓ—π‘ˆπ‘Ÿξ€»,(5.1) where π‘ž(π‘Ÿ) is the inhomogeneous term.

Suppose 𝑉=(𝑣1(𝑑,π‘Ÿ),𝑣2(𝑑,π‘Ÿ),𝑣3(𝑑,π‘Ÿ)) is a solution of (2.1), 𝑍=(𝑧1(𝑑,π‘Ÿ),𝑧2(𝑑,π‘Ÿ),𝑧3(𝑑,π‘Ÿ)),𝑧21(𝑑,π‘Ÿ)+𝑧22(𝑑,π‘Ÿ)+𝑧23(𝑑,π‘Ÿ)=𝐢>0. Furthermore, the relationship between 𝑉 and 𝑍 is in the form of 𝑣1(𝑑,π‘Ÿ)=𝑧1𝑣(𝑑,π‘Ÿ)𝐺(π›Όπ‘Ÿ),2(𝑑,π‘Ÿ)=𝑧2𝑣(𝑑,π‘Ÿ)𝐺(π›Όπ‘Ÿ),3(𝑑,π‘Ÿ)=𝑧3(𝑑,π‘Ÿ)𝐺(π›Όπ‘Ÿ),(5.2) where 𝐺(π›Όπ‘Ÿ) is a functions, 𝛼 is constant.

Substituting (5.2) into (2.1), we get 𝑍𝑑𝑍=𝐺(π›Όπ‘Ÿ)π‘Γ—π‘Ÿπ‘Ÿ+1π‘Ÿπ‘π‘Ÿξ‚„+2𝛼𝐺(π›Όπ‘Ÿ)ξ…žξ€Ίπ‘Γ—π‘π‘Ÿξ€»,(5.3) where β€œβ€²β€ is a derivative about π›Όπ‘Ÿ.

Furthermore, suppose 𝛼=1/2, (5.3) can be rearranged to give (for simplicity, we write 𝐺 for 𝐺(β‹…) here. 𝑍𝑑𝑍=πΊπ‘Γ—π‘Ÿπ‘Ÿ+1π‘Ÿπ‘π‘Ÿξ‚„+πΊξ…žξ€Ίπ‘Γ—π‘π‘Ÿξ€».(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(𝑑,π‘Ÿ),𝑣2(𝑑,π‘Ÿ),𝑣3(𝑑,π‘Ÿ)) is a solution of (2.1) if and only if 𝑍=(𝑧1(𝑑,π‘Ÿ),𝑧2(𝑑,π‘Ÿ),𝑧3(𝑑,π‘Ÿ))(𝐺(β‹…)=𝐺(π‘Ÿ/2)) is a solution of (5.4) where 𝑧21(𝑑,π‘Ÿ)+𝑧22(𝑑,π‘Ÿ)+𝑧23(𝑑,π‘Ÿ)=𝐢>0.

One mentions here that if 𝛼=1, 𝐺(π›Όπ‘Ÿ)=𝐺(π‘Ÿ), (5.3) will adopt the following form: π‘ˆπ‘‘ξ‚ƒπ‘=𝐺(π‘Ÿ)π‘Γ—π‘Ÿπ‘Ÿ+1π‘Ÿπ‘π‘Ÿξ‚„+2𝐺(π‘Ÿ)π‘Ÿξ€Ίπ‘ˆΓ—π‘ˆπ‘Ÿξ€».(5.5)

So, (5.5) will not equal to (5.1) because of the coefficient 2 before the 𝐺(π‘Ÿ)π‘Ÿ[π‘Γ—π‘π‘Ÿ]. 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 π‘Ÿβˆˆ[1,3], π‘‘βˆˆ[0,9/10], 𝑇=1, 𝛾=1, 𝐢1=1, 𝐢2=0, 𝐢3=1 and 𝐢4=βˆ’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 π‘Ÿβˆˆ[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 𝐢=4 and πœ†=0,1,2,3,4, respectively. In Figure 3, begin from πœ†=1, we can see that the shape of (3.23) has only a declining shift of the peak when πœ† increases.(3)From Figures 4–5 we can apperceive the periodic behavior of (3.24) which is determined by the periodic function: ξƒ©ξƒ©βˆšπ‘„(π‘₯,𝑦)=4arctan𝑠𝑑π‘₯+1,32ξƒͺξƒ©βˆšπ‘‘π‘ π‘¦+1,32π‘„βŽ›βŽœβŽœβŽβˆšξƒͺξƒͺ,(6.1)(π‘₯,𝑦)=4arctancot(1)cosh2/2cos(1)(π‘₯βˆ’π‘¦)√coshξ‚€ξ‚€ξ‚ξ‚βŽžβŽŸβŽŸβŽ 2/2sin(1)(π‘¦βˆ’π‘₯),(6.2)     respectively. A singularity behavior of (3.24) can be seen in Figure 4. (4) Considering the solution (4.1), and setting 𝐢1=1, 𝐢2=1 and 𝐢3=1, we draw the picture of |π‘ˆπ‘Ÿ|2=(1/9)(9π‘Ÿ8/3+8/π‘Ÿ10/3) where π‘Ÿβˆˆ[1/2,25] in Figure 6. From Figure 6, we can find out that the energy cannot be bounded near 𝑑=0. Setting π‘‘βˆˆ[0,3] and π‘Ÿβˆˆ[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.

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.

Acknowledgments

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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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.

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