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Research ArticleSolitary Wave Solutions to the Multidimensional Landau–Lifshitz Equation
Ahmad Neirameh
Department of Mathematics, Faculty of Sciences, Gonbad Kavous University, Gonbad, Iran
Correspondence should be addressed to Ahmad Neirameh; [email protected]
Received 23 January 2021; Revised 20 February 2021; Accepted 19 March 2021; Published 31 March 2021
Academic Editor: F. Rabiei
Copyright © 2021 Ahmad Neirameh. 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.
In this paper, we study the different types of new soliton solutions to the Landau–Lifshitz equation with the aid of the auxiliaryequation method. Then, we get some special soliton solutions for this equation. Without the Gilbert damping term, we present atravelling wave solution with a finite energy in the initial time. The parameters of the soliton envelope are obtained as a functionof the dependent model coefficients.
1. Introduction
Nonlinear partial differential equations have different typesof equations; one of them is the Landau–Lifshitz equationthat is relevant to the classical and quantummechanics. Non-linear evolution equations (NEEs) which describe manyphysical phenomena are often illustrated by nonlinear partialdifferential equations. So, the exact solutions of NLPDE areexplored in detail in order to understand the physicalstructure of natural phenomena that are described by suchequations. A variety of powerful methods have been used tostudy the nonlinear evolution equations, for the analyticand numerical solutions. Some of these methods, theRiccati Equation method [1], Hirota’s bilinear operators[2], exponential rational function method [3], the Jacobielliptic function expansion [4], the homogeneous balancemethod [5], the tanh-function expansion [6], first integralmethod [7, 8], the subequation method [9], the exp-function method [10], the Backlund transformation, andsimilarity reduction [11–29], are used to obtain the exactsolutions of NLPDE.
In physics, the Landau–Lifshitz–Gilbert equation, namedfor Lev Landau, Evgeny Lifshitz, and T. L. Gilbert, is a nameused for a differential equation describing the processionalmotion of magnetization M in a solid. It is a modification
by Gilbert of the original equation of LLG equation (see[30]) can be written down as
∂∂t
S = αSΛΔS − βSΛ SΛΔSð Þ, ð1Þ
here, S = ðS1ðt, x!Þ, S2ðt, x!Þ, S3ðt, x!ÞÞ ∈ S2↪R3, α ≥ 0, α2 + β2
= 1, Λ denotes the cross product. The term multiplying withα represents the exchange interaction, while the β termdenotes the Gilbert damping term. Especially, two extremecases of (1) (β = 0 and α = 0, respectively) include as specialcases the well-known Schrödinger map equation and har-monic map heat flow, respectively. The well-posedness prob-lem of the LL(G) equation are intensively studied inmathematics, to list a few, in 1986, of the weak solution ofthe LL(G) equation. Under the small initial value, the globalexistence of the solution in different spaces [31–33] wasproved. The first progress on the existence of partially regularsolutions to the LLG equation was found [30, 33–36]. Evenfor the small initial data, the exact form of the solution is stillunknown. On the other hand, whether the LLG equationadmits a global solution will develop a finite time singularityfrom the large initial data is an open equation.
HindawiAdvances in Mathematical PhysicsVolume 2021, Article ID 5538516, 7 pageshttps://doi.org/10.1155/2021/5538516
2. Auxiliary Equation Method
Let us consider a typical nonlinear PDE for q = qðx, tÞ,given by
U qx, qt ,⋯:ð Þ = 0: ð2Þ
Under the wave transformations of qðx, tÞ =QðξÞ andξ = σx − lt, Equation (2) becomes an ordinary differentialequation given by
U Q, σQ′,−lQ′,⋯:� �
= 0: ð3Þ
We assume that the solution ϕðξÞ of the nonlinear Equation(18) can be presented as
Q ξð Þ = 〠M
i=0Aiϒ
i ξð Þ, ð4Þ
in which Aiði = 0,⋯, nÞ are all real constants to be deter-mined, the balancing number M is a positive integer whichcan be determined by balancing the highest order derivativeterms with the highest power nonlinear terms in Equation(2) andϒ iðξÞ expresses the solutions of the following auxiliaryordinary differential equation:
dϒdξ
� �2= s1ϒ
2 ξð Þ + s2ϒ3 ξð Þ + s3ϒ
4 ξð Þ, ð5Þ
where s1, s2, and s3 are real parameters. Equations (6), (7), (8),(9), (10), (11), (12), (13), (14), (15), (16), (17), (18), and (19)with Δ = s22 − 4s1s3 give the following solutions:
Case 1. For s1 > 0,
−s1s2 sech2ffiffiffiffiffiffiffiffiffiffiffis1/2ð Þp
ξ� �
s22 − s1s3 1 + ξ tanhffiffiffiffiffiffiffiffiffiffiffis1/2ð Þp
ξ� �� � : ð6Þ
Case 2. For s1 > 0,
−s1s2 csch2ffiffiffiffiffiffiffiffiffiffiffis1/2ð Þp
ξ� �
s22 − s1s3 1 + ξ tanhffiffiffiffiffiffiffiffiffiffiffis1/2ð Þp
ξ� �� � : ð7Þ
Case 3. For s1 > 0, Δ > 0,
2s1 sechffiffiffiffis1
pξ
� �εffiffiffiffiΔ
p− s2 sech
ffiffiffiffis1
pξ
� � : ð8Þ
Case 4. For s1 < 0, Δ > 0,
2s1 secffiffiffiffiffiffiffi−s1
pξ
� �εffiffiffiffiΔ
p− s2 sec
ffiffiffiffiffiffiffi−s1p
ξ� � : ð9Þ
Case 5. For s1 > 0, Δ < 0,
2s1 cschffiffiffiffis1
pξ
� �εffiffiffiffiffiffi−Δ
p− s2 csch
ffiffiffiffis1
pξ
� � : ð10Þ
Case 6. For s1 < 0, Δ > 0,
2s1 cscffiffiffiffiffiffiffi−s1
pξ
� �εffiffiffiffiΔ
p− s2 csc
ffiffiffiffiffiffiffi−s1p
ξ� � : ð11Þ
Case 7. For s1 > 0, s3 > 0
−s1 sech2ffiffiffiffis1
p /2� �
ξ� �
s2 + 2ξ ffiffiffiffiffiffiffis1s3
p tanh ffiffiffiffis1
p /2� �
ξ� � : ð12Þ
Case 8. For s1 < 0, s3 > 0,
−s1 sec2ffiffiffiffiffiffiffi−s1
p /2� �
ξ� �
s2 + 2ξ ffiffiffiffiffiffiffiffiffiffi−s1s3p tan ffiffiffiffiffiffiffi−s1
p /2� �
ξ� � : ð13Þ
Case 9. For s1 > 0, s3 > 0
s1 csch2ffiffiffiffis1
p /2� �
ξ� �
s2 + 2ξ ffiffiffiffiffiffiffis1s3
p coth ffiffiffiffis1
p /2� �
ξ� � : ð14Þ
Case 10. For s1 < 0, s3 > 0,
−s1 csc2ffiffiffiffiffiffiffi−s1
p /2� �
ξ� �
s2 + 2ξ ffiffiffiffiffiffiffiffiffiffi−s1s3p cot ffiffiffiffiffiffiffi−s1
p /2� �
ξ� � : ð15Þ
Case 11. For s1 > 0, Δ = 0,
−s1s2
1 + ξ tanhffiffiffiffis1
p2 ξ
� �� �: ð16Þ
Case 12. For s1 > 0, Δ = 0,
−s1s2
1 + ξ cothffiffiffiffis1
p2 ξ
� �� �: ð17Þ
Case 13. For s1 > 0,
4s1eεffiffiffis1
p ξ
eεffiffiffis1
p ξ − s2� �2 − 4s1s2
: ð18Þ
Case 14. For s1 > 0, s2 = 0,
±4s1εeεffiffiffis1
p ξ
1 − 4s1s3e2εffiffiffis1
p ξ: ð19Þ
Stage 2: substituting Equations (4) and (5) into Equation(3) and collecting all terms with the same order of ϒ j
together, we convert the left-hand side of Equation (3) intoa polynomial in ϒ j. Setting each coefficient of each polyno-mial to zero, we derive a set of algebraic equations for A0,
2 Advances in Mathematical Physics
A1, andA2. By solving these algebraic equations, we obtainseveral cases of variables solutions [15, 37]
Stage 3: by substituting the obtained solutions in stage 2into Equation (4) along with general solutions of Equations(6), (7), (8), (9), (10), (11), (12), (13), (14), (15), (16), (17),(18), and (19), it finally generates new exact solutions forthe nonlinear PDE (1)
3. Travelling Wave Solutions of the LL Equationvia Aem
In this section, we construct a travelling wave solution with-out the Gilbert term [30]. Under the condition of the auxil-iary equation method, we consider the following wavetransformations:
Wc,w t,�rð Þ = e−iwtϕ �r − ctð Þeiψ �r−ctð Þ, ð20Þ
where c and w are constants undetermined.Here, we assume −c2 + 4w > 0: Substitute (5) into (1)
[30], the separate real and the virtual parts, respectively, as
ϕ ξð Þ w − 2 ξð Þ2 + cψ′ ξð Þ − ψ′ ξð Þ2� �+ ϕ ξð Þ3 w + cψ′ ξð Þ + ψ′ ξð Þ2
� �+ ϕ ξð Þ2ϕ′′ ξð Þ = 0,
ð21Þ
ϕ′ ξð Þ −c + 2ψ′ ξð Þ� �
− ϕ ξð Þ2ϕ′ ξð Þ c + 2ψ′ ξð Þ� �
+ ϕ ξð Þψ′′ ξð Þ + ϕ ξð Þ3ψ′′ ξð Þ = 0,ð22Þ
where ξ =�r − ct. (21) and (22) are the nonlinear constantcoefficients of ordinary differential equation system withthe variable ξ. According to (22), we can obtain a relationshipbetween ψ and ϕ:
ψ′ ξð Þ =1 + ϕ ξð Þ2� �
−c + 2C1 + 2C1 ξð Þ2� �2ϕ ξð Þ2
, ð23Þ
where C1 is the arbitrary constant. If we set C1=0, we have
ψ′ ξð Þ = −c 1 + ϕ ξð Þ2� �2ϕ ξð Þ2
: ð24Þ
Substituting (24) into (21), we get
c2 + 3c2 ϕ ξð Þ2 + c2 − 4w� �
ϕ ξð Þ6 + ϕ ξð Þ2 3c2 − 4w + 8ϕ′ ξð Þ2� �
− 4ϕ ξð Þ3 1 + ϕ ξð Þ2� �
ϕ′′ ξð Þ = 0:
ð25Þ
4. Results
By the auxiliary equation method, the solution of (25) isassumed as
ϕ ξð Þ = A1ϒ ξð Þ + A0: ð26Þ
From (5), we have
ϒ ′� �2
= s1ϒ2 ξð Þ + s2ϒ
3 ξð Þ + s3ϒ4 ξð Þ,
ϒ″ = 12 s1ϒ ξð Þ + 3s2ϒ 2 ξð Þ + 4s3ϒ 3 ξð Þ� �
,ð27Þ
where A1 and A0 are constants. Substituting (26) into (25)along with (27) and comparing the coefficients of alikepowers of ϒðξÞ provides an algebraic system of equations,and solving this set of algebraic equations by used of theMaple, we obtain several case solutions. For example isas follows.
Set 1:
A1 = −203 s3
3s1 − 12s2240s3 − 4w
� �,
A0 =3s1 − 12s2240s3 − 4w ,
s2 = 0,
c = 130
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi25s3 3s1ð Þ2 + −3120w − 10800s3ð Þs3 3s1ð Þ + 144 w + 90s3ð Þ2
q+ 150s1 − 10800ð Þs3 + 480w
r !: ð28Þ
3Advances in Mathematical Physics
From set 1 and Eq. (26), we have
ϕ ξð Þ = −203 s3
3s1240s3 − 4w
� �ϒ ξð Þ + 3s1
240s3 − 4w : ð29Þ
So, solutions of Equation (5) are obtained in (6), (7), (8),(9), (10), (11), (12), (13), (14), (15), (16), (17), (18), and (19),we have final solutions of Equation (1) and Equation (25) asfollows:
ϕ1 �r, tð Þ = −203 s3
3s1240s3 − 4w
� �
× −s1 csc2ffiffiffiffiffiffiffi−s1
p /2� �
�r − ctð Þ� �2ε ffiffiffiffiffiffiffiffiffiffi−s1s3p cot ffiffiffiffiffiffiffi−s1
p /2� �
�r − ctð Þ� � + 3s1 − 12s2240s3 − 4w :
ð30Þ
In Figure 1, the graphical behavior of solutions ϕ1 fors1 = −1, s3 = 1, andw = 1 in different values of �r and t isillustrated.
ϕ2 �r, tð Þ = −203 s3
3s1240s3 − 4w
� �× 2s1 sec
ffiffiffiffiffiffiffi−s1p
�r − ctð Þ� �εffiffiffiffiffiffiffiffiffiffiffiffi−4s1s3
p
+ 3s1 − 12s2240s3 − 4w :
ð31Þ
In Figure 2, the graphical behavior of solutions ϕ2 fors1 = −1, s3 = 1, andw = 1 in different values of �r and t isillustrated.
ϕ3 �r, tð Þ = −203 s3
3s1240s3 − 4w
� �
× s1 csch2ffiffiffiffis1
p /2� �
�r − ctð Þ� �2ε ffiffiffiffiffiffiffi
s1s3p coth ffiffiffiffi
s1p /2� �
�r − ctð Þ� � + 3s1240s3 − 4w :
ð32Þ
150
100
50
0
0 0
–50
–100
–150
–π
–π
π
π
2–
π
2π
2
π
2–
t r
π
Figure 1: Graphical behavior for ϕ1ð�r, tÞ, for �r, t = −π::π, s1 = −1,s3 = 1, andw = 1.
–100
–200
–300
0
0 0–π
–π
π
π
2– π
2–π
2π
2t r
Figure 2: Graphical behavior for ϕ2ð�r, tÞ, for �r, t = −π::π, s1 = −1,s3 = 1, andw = 1.
π
2–
π
4–
3π
4–
π
π
2π
2π
4
3π
4
00
–10–8–6–4–2
–π
π
02468
t
r
π
2 0
Figure 3: Graphical behavior for ϕ3ð�r, tÞ, for �r, t = −π::π, s1 = 1,s3 = 1, andw = 1.
4 Advances in Mathematical Physics
In Figure 3, the graphical behavior of solutions ϕ3 for s1 =1, s3 = 1, andw = 1 in different values of �r and t is illustrated.
Set 2:
A1 =−s12 s1
2 + 225s22� �
−90s2 + s1ð Þ12w 2s14 − 75s13s2 + 118125s24ð Þ ,
A0 = −115
s1s2,
s3 = 0,
c =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi415
s1s2
+ 4� �
w/ 6 − 115
s1s2
� �s:
ð35Þ
From set 2 and Eq. (26), we have
ϕ ξð Þ = −s12 s12 + 225s22
� �−90s2 + s1ð Þ
12w 2s14 − 75s13s2 + 118125s24ð Þϒ ξð Þ − 115
s1s2: ð36Þ
So, solutions of Equation (5) are obtained in (6), (7), (8),(9), (10), (11), (12), (13), (14), (15), (16), (17), (18), and (19),we have final solutions of Equation (1) and Equation (25) asfollows:
ϕ4 �r, tð Þ = −203 s3
3s1240s3 − 4w
� �× 4s1eε
ffiffiffis1
p �r−ctð Þ
eεffiffiffis1
p �r−ctð Þ� �2 + 3s1240s3 − 4w ,
ϕ5 �r, tð Þ = −203 s3
3s1240s3 − 4w
� �× ±4s1εeε
ffiffiffis1
p �r−ctð Þ
1 − 4s1s3e2εffiffiffis1
p �r−ctð Þ +3s1
240s3 − 4w , ð33Þ
Wc,w t,�rð Þ = e−iwtϕ �r − ctð Þeiψ �r−ctð Þ,
ψ′ ξð Þ = −c 1 + ϕ ξð Þ2� �2ϕ ξð Þ2
soψ ξð Þ = −ð c 1 + ϕ ξð Þ2� �2ϕ ξð Þ2
dξ,
c = 130
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi25s3 3s1ð Þ2 + −3120w − 10800s3ð Þs3 3s1ð Þ + 144 w + 90s3ð Þ2
q+ 150s1 − 10800ð Þs3 + 480w
r !: ð34Þ
π
2–
π
2–π
4–
3π
4–
π
2π
2π
4
–π
–π
–0.16
–0.15
–0.14
–0.13
–0.12
–0.11
–0.10
–0.09
–0.08
–0.07
00
x
3π
4π
t
π
2–
π
2–π
3π
4––π
π
00
Figure 5: Graphical behavior for ϕ7ð�r, tÞ, for �r, t = −π::π, s1 = 1,and s2 = 1
π
2–
π
2–π
4–
3π
4–
π
2π
2π
43π
4
–π
–π
00
π
–0.0675
–0.0674
–0.0673
–0.0672
–0.0671
–0.0670
–0.0669
–0.0668
–0.0667
π
2–
π
2–π
3π
4–
–
π
00
t x
Figure 4: Graphical behavior for ϕ6ð�r, tÞ, for �r, t = −π::π, s1 = 1,and s2 = 1.
5Advances in Mathematical Physics
ϕ6 �r, tð Þ = −s12 s12 + 225s22
� �−90s2 + s1ð Þ
12w 2s14 − 75s13s2 + 118125s24ð Þ
× −s1s2 sech2ffiffiffiffiffiffiffiffis1/2
p�r − ctð Þ� �
s22−
115
s1s2:
ð37Þ
In Figure 4, the graphical behavior of solutions ϕ6 fors1 = 1, s2 = 1 in different values of �r and t is illustrated.
ϕ7 �r, tð Þ = −s12 s12 + 225s22
� �−90s2 + s1ð Þ
12w 2s14 − 75s13s2 + 118125s24ð Þ
× −s1s2 csch2ffiffiffiffiffiffiffiffis1/2
p�r − ctð Þ� �
s22−
115
s1s2:
ð38Þ
In Figure 5, the graphical behavior of solutions ϕ7 fors1 = 1, s3 = 1 in different values of �r and t is illustrated.
ϕ8 �r, tð Þ = −s12 s12 + 225s22
� �−90s2 + s1ð Þ
12w 2s14 − 75s13s2 + 118125s24ð Þ
× 2s1 sechffiffiffiffis1
p�r − ctð Þ� �
εs2 − s2 sechffiffiffiffis1
p�r − ctð Þ� � − 1
15s1s2,
ϕ9 �r, tð Þ = −s12 s12 + 225s22
� �−90s2 + s1ð Þ
12w 2s14 − 75s13s2 + 118125s24ð Þ
× 4s1eεffiffiffis1
p �r−ctð Þ
eεffiffiffis1
p �r−ctð Þ − s2� �2 − 4s1s2
−115
s1s2,
ð39Þ
Wc,w t,�rð Þ = e−iwtϕ �r − ctð Þeiψ �r−ctð Þ,
ψ′ ξð Þ = −c 1 + ϕ ξð Þ2� �2ϕ ξð Þ2
soψ ξð Þ = −ð c 1 + ϕ ξð Þ2� �2ϕ ξð Þ2
dξ,
c =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi415
s1s2
+ 4� �
w/ 6 − 115
s1s2
� �s:
ð40ÞFor more convenience and understanding, the graphical
behavior of the answers is considered (see Figures 1–5).
5. Conclusions
This paper derived new optical soliton solutions of theLandau–Lifshitz equation, which describe the propagationof ultrashort pulses in nonlinear optical fibers by usingthe auxiliary equation method. We boldly say that thework here is valuable and may be beneficial for studyingin other nonlinear science. The exact solutions obtainedfrom the model equations provide important insight intothe dynamics of solitary waves. The solutions obtained inthis paper have not been reported in the old research.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
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