AA note on the Sine-Gordon expansion method and itsapplications
Nizhum Rahman
Email: [email protected] of Mathematics and Physics, The University of Queensland, St Lucia, 4072, AustraliaFaculty of Science and Information Technology, Daffodil International University Dhaka,1207,BangladeshJanuary 15, 2021
Abstract
The sine-Gordon expansion method; which is a transformation of the sine-Gordon equation has beenapplied to the potential-YTSF equation of dimension (3+1) and the reaction-diffusion equation. Weobtain new solitons of this equation in the form hyperbolic, complex and trigonometric function byusing this method. We plot 2D and 3D graphics of these solutions using symbolic software.
Keywords:
Sine-Gordon expansion method; the potential-YTSF equation of dimension (3+1); thereaction-diffusion equation; trigonometric function solutions; hyperbolic function solutions; complexsolutions.
In recent decades, the nonlinear evolution equations (NLEEs) have become one of the vital topics ofinterest among the researchers from different field including physics, mathematics, engineering andbiology [1–7]. The analytic solutions in the form of a travelling wave of the NLEEs play a crucial rolefor the development of scientific fields. A variety of effective methods have been constructed from theend of the 20 th century by the researchers. Many of these methods have been used successfully indifferent NLEEs to obtain new solutions [8–32].In this article we will apply the sine-Gordon expansion method [33] to the potential-YTSF equationof dimension (3+1) and the reaction-diffusion equation. Recently this method has been applied toonly a few research articles [34–37]. The remaining part of the article is described as follows. Thesine-Gordon expansion method has been stated in section 2. The next section is the applications ofthis method. Finally, we describe the necessity of this method briefly in the last section. The sine-Gordon equation (SGE) [33] ∂ u∂x − ∂ u∂t = h sin ( u ) (1)arises in many physical science applications [38, 39], where h is a constant. By considering the movingcoordinate as u ( x, t ) = U ( η ) where η = k ( x − ct ), after simplifying this equation, we get d Udη = h k (1 − c ) sin ( U ) (2)1 a r X i v : . [ m a t h - ph ] J a n here η and c represent the width and velocity of the travelling waves respectively. After multiplying dUdη on both sides of Eq. (2) and integrating, we get, (cid:104)(cid:16) U (cid:17) (cid:48) (cid:105) = h k (1 − c ) sin (cid:16) U (cid:17) , (3)where the constant of integration is considered to zero. Substituting ω ( η ) = U and h k (1 − c ) = 1 intoEq.(3), we obtain ω (cid:48) = sin ( ω ) . (4)This is a modified formation of the SGE. We can write the solution of Eq. (4) as of the form sin [ ω ( η )] = 2 r exp ( η ) r exp (2 η ) + 1 (cid:12)(cid:12)(cid:12)(cid:12) r =1 = sech ( η ) , (5)or cos [ ω ( η )] = r exp (2 η ) − r exp (2 η ) + 1 (cid:12)(cid:12)(cid:12)(cid:12) r =1 = tanh ( η ) , (6)with r is a non-zero constant of integration.The travelling wave solution U ( η ) of the NLEEs of the form Q ( u, u x , u t , u xx , u tx , u tt , ............. ) = 0 , (7)can be written as U ( η ) = n (cid:88) j =1 tanh j − ( η ) [ B j sech ( η ) + A j tanh ( η )] + A . (8)Using Eq. (5) and Eq. (6), the solutions of the Eq. (8) take the following form U ( ω ) = n (cid:88) j =1 cos j − ( ω ) [ B j sin ( ω ) + A j cos ( ω )] + A . (9)We calculate the value of unknown n by apply the homogeneous balance assumption. After letting thecollection of the coefficients of sin i ( ω ) cos j ( ω ) of equal power to be all 0 (zero), we get an algebraicsystem of equations. By find out the unknowns of the system and using Eq.(9), we get the solutionsto Eq. (7) is the form of (8). We can implement the sine-Gordon expansion method in many NLEEs. To demonstrate our method,we examine the method to the potential-YTSF equation of dimension (3+1) and the reaction-diffusionequation.
We consider the potential-YTSF equation of dimension (3+1) [40–42] in the form − u xt + u xxxz + 4 u x u xz + 2 u xx u z + 3 u yy = 0 , (10)which arise in many physical problems and have been solved in different ways by the researchers[43–45].The moving coordinate u ( x, y, z, t ) = U ( η ) , η = x + y + z − ct, (11)allows to convert Eq.(10) in to an ODE U (cid:48)(cid:48)(cid:48)(cid:48) + 6 U (cid:48) U (cid:48)(cid:48) + (4 c + 3) U (cid:48)(cid:48) = 0 , η , we acquire U (cid:48)(cid:48)(cid:48) + 3( U (cid:48) ) + (4 c + 3) U (cid:48) = 0 , (12)where integrating constant is set to zero. Setting U (cid:48) = V , we have V (cid:48)(cid:48) + 3 V + (4 c + 3) V = 0 . (13)Applying homogeneous balance principle between V (cid:48)(cid:48) and V in Eq. (13) based on Eq. (9) we obtain n + 2 = n + n , which implies n = 2. Now we can write Eq. (9) as V ( ω ) = B sin ( ω ) + A cos ( ω ) + B sin ( ω ) cos ( ω ) + A cos ( ω ) + A (14)and V (cid:48)(cid:48) = − B sin ( ω ) + B cos ( ω ) sin ( ω ) − A sin ( ω ) cos ( ω ) − B cos ( ω ) sin ( ω ) + B cos ( ω ) sin ( ω )+ 2 A sin ( ω ) − A cos ( ω ) sin ( ω ) . (15)By substituting Eq. (14) and Eq. (15) into Eq. (13) and equaling all polynomials with same degreeto zero, we get a system of equation as below.constant: 4 v A + 3 A + 3 B + 3 A + 2 A = 0 cos ( ω ) : 4 v A + 6 A A + 6 B B + A = 0 sin ( ω ) : 4 v B + 6 A B + 2 B = 0 cos ( ω ) : 4 v A + 6 A A + 3 A − B + 3 B − A = 0 sin ( ω ) cos ( ω ) : 4 v B + 6 A B + 6 A B − B cos ( ω ) : 6 A A − B B + 2 A = 0 sin ( ω ) cos ( ω ) : 6 A B + 6 A B + 2 B = 0 cos ( ω ) : 3 A − B + 6 A = 0 sin ( ω ) cos ( ω ) : 6 A B + 6 B = 0After solving the above system , we find the traveling wave solution U ( η ) to Eq. (10) in the formof (8).Case-1 c = − / , A = 2 , A = 0 , B = 0 , A = − , B = 0 , which gives: u ( x, y, z, t ) = U ( η ) = 2 tanh ( η ) , with η = x + y + z + 74 t. (16)Figure 1: The graphical representation of Eq. (16) (3D on the left side and 2D on the right side).3ase-2 c = 1 / , A = 2 / , A = 0 , B = 0 , A = − , B = 0 , which gives: u ( x, y, z, t ) = U ( η ) = 2 tanh ( η ) − η, with η = x + y + z − t. (17)Figure 2: The graphical representation of Eq. (17) (3D on the left side and 2D on the right side).Case-3 c = − , A = 1 , A = 0 , B = 0 , A = − , B = ± i , which gives: u ( x, y, z, t ) = U ( η ) = tanh ( η ) ± i sech ( η ) , with η = x + y + z + t. (18)Figure 3: The graphical representation of Eq. (18) (3D on the left side and 2D on the right side).Case-4 c = − / , A = 2 / , A = 0 , B = 0 , A = − , B = ± i , u ( x, y, z, t ) = U ( η ) = tanh ( η ) ± i sech ( η ) − η, with η = x + y + z + 12 t. (19)Figure 4: The graphical representation of Eq. (19) (3D on the left side and 2D on the right side). We consider the following form of the reaction-diffusion equation [46], u tt + α u xx + β u + γ u = 0 (20)where α , β and γ are constants (nonzero). The moving coordinate u ( x, t ) = U ( η ), where η = x − ct leads the Eq. (20) in an ordinary differential equation (ODE)( α + c ) U (cid:48)(cid:48) + β U + γ U = 0 . (21)Considering homogeneous balance between U (cid:48)(cid:48) and U in Eq. (21) based on Eq. (9) we obtain n + 2 = n + n + n , which implies n = 1. Now we can write Eq. (9) as U ( ω ) = B sin ( ω ) + A cos ( ω ) + A . (22)Using Eq. (22) into Eq. (21), and equaling all polynomials with same degree to get, we have a systemof equation as below.constant: γ A + 3 γ A B + β A = 0 cos ( ω ) : 3 γ A A + 3 γ A B − v A − α A + β A = 0 sin ( ω ) : 3 γ A B + γ B − v B − α B + β B = 0 cos ( ω ) : 3 γ A A − γ A B = 0 sin ( ω ) cos ( ω ) : 6 γ A A B = 0 5 os ( ω ) : γ A − γ A B + 2 v A + 2 α A = 0 sin ( ω ) cos ( ω ) : 3 γ A B − γ B + 2 v B + 2 α B = 0After solving the above system , we find the traveling wave solution u ( x, y, z, t ) to Eq. (20) in theform of (8).Case-1 c = ± (cid:114) β − α , A = 0 , A = ± (cid:115) − βγ , B = 0 , α = α , β = β , γ = γ, which gives: u ( x, t ) = U ( η ) = ± (cid:115) − βγ tanh ( η ) , with η = x ± (cid:114) β − α t. (23)Figure 5: The graphical representation of Eq. (23) (3D on the left side and 2D on the right side).Case-2 c = ± (cid:112) − β − α , A = 0 , B = ± (cid:115) − βγ , A = 0 , α = α , β = β , γ = γ, which gives: u ( x, t ) = U ( η ) = ± (cid:115) − βγ sech ( η ) , with η = x ± (cid:112) − β − αt. (24)Figure 6: The graphical representation of Eq. (24) (3D on the left side and 2D on the right side).6ase-3 c = ± (cid:112) β − α , A = 0 , A = ± (cid:115) − βγ , B = ± (cid:115) βγ , α = α , β = β , γ = γ, which gives: u ( x, t ) = U ( η ) = ± (cid:115) − βγ tanh ( η ) ± (cid:115) βγ sech ( η ) , with η = x ± (cid:112) β − αt. (25)Figure 7: The graphical representation of Eq. (25) (3D on the left side and 2D on the right side). In this article, we have applied the Sine-Gordon expansion method for calculating new travelling wavesolutions to the potential-YTSF equation of dimension (3+1) and the reaction-diffusion equation. Wehave found these solutions of the equation in the trigonometric, complex and hyperbolic functionforms. This method is powerful and very efficient to finding travelling wave solutions to the NLEEs.We can solve various NLEEs by this method using any symbolic software.
Abbreviations
NLEEs, SGE and ODE.
DeclarationsAvailability of data and material : Not applicable.
Competing interests : The author has no any financial or non-financial conflict of inter-ests. 7 unding : There was no funding for this research.
Authors’ contributions : The Author did the research (methodology, examples and writing )by himself and approved the final manuscript.
Acknowledgements : The author is very grateful to the editorial team and reviewers for theirvaluable comments and suggestions towards improving this article.
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