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A note on the Sine-Gordon expansion method and its applications

Nizhum Rahman
Email: nizhum.rahman@uq.edu.au
School of Mathematics and Physics, The University of Queensland, St Lucia, 4072, Australia

Abstract

The sine-Gordon expansion method; which is a transformation of the sine-Gordon equation has been applied to the potential-YTSF equation of dimension (3+1) and the reaction-diffusion equation. We obtain new solitons of this equation in the form hyperbolic, complex and trigonometric function by using 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); the reaction-diffusion equation; trigonometric function solutions; hyperbolic function solutions; complex solutions.

1 Introduction

In recent decades, the nonlinear evolution equations (NLEEs) have become one of the vital topics of interest among the researchers from different field including physics, mathematics, engineering and biology [1, 2, 3, 4, 5, 6, 7]. The analytic solutions in the form of a travelling wave of the NLEEs play a crucial role for the development of scientific fields. A variety of effective methods have been constructed from the end of the 20th20^{th} century by the researchers. Many of these methods have been used successfully in different NLEEs to obtain new solutions [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32].
In this article we will apply the sine-Gordon expansion method [33] to the potential-YTSF equation of dimension (3+1) and the reaction-diffusion equation. Recently this method has been applied to only a few research articles [34, 35, 36, 37]. The remaining part of the article is described as follows. The sine-Gordon expansion method has been stated in section 2. The next section is the applications of this method. Finally, we describe the necessity of this method briefly in the last section.

2 Method

The sine-Gordon equation (SGE) [33]

2ux22ut2=h2sin(u)\frac{\partial^{2}u}{\partial x^{2}}-\frac{\partial^{2}u}{\partial t^{2}}=h^{2}sin(u) (1)

arises in many physical science applications [38, 39], where hh is a constant. By considering the moving coordinate as u(x,t)=U(η)u(x,t)=U(\eta) where η=k(xct)\eta=k(x-ct), after simplifying this equation, we get

d2Udη2=h2k2(1c2)sin(U)\frac{d^{2}U}{d\eta^{2}}=\frac{h^{2}}{k^{2}(1-c^{2})}sin(U) (2)

where η\eta and cc represent the width and velocity of the travelling waves respectively. After multiplying dUdη\frac{dU}{d\eta} on both sides of Eq. (2) and integrating, we get,

[(U2)]2=h2k2(1c2)sin2(U2),\Big{[}\Big{(}\frac{U}{2}\Big{)}^{\prime}\Big{]}^{2}=\frac{h^{2}}{k^{2}(1-c^{2})}sin^{2}\Big{(}\frac{U}{2}\Big{)}, (3)

where the constant of integration is considered to zero. Substituting ω(η)=U2\omega(\eta)=\frac{U}{2} and h2k2(1c2)=1\frac{h^{2}}{k^{2}(1-c^{2})}=1 into Eq.(3), we obtain

ω=sin(ω).\omega^{\prime}=sin\,(\omega). (4)

This is a modified formation of the SGE. We can write the solution of Eq. (4) as of the form

sin[ω(η)]=2rexp(η)r2exp(2η)+1|r=1=sech(η),sin\,[\omega(\eta)]=\frac{2r\,exp(\eta)}{r^{2}\,exp(2\eta)+1}\bigg{|}_{r=1}=sech(\eta), (5)

or

cos[ω(η)]=r2exp(2η)1r2exp(2η)+1|r=1=tanh(η),cos\,[\omega(\eta)]=\frac{r^{2}\,exp(2\eta)-1}{r^{2}\,exp(2\eta)+1}\bigg{|}_{r=1}=tanh(\eta), (6)

with rr is a non-zero constant of integration.

The travelling wave solution U(η)U(\eta) of the NLEEs of the form

Q(u,ux,ut,uxx,utx,utt,.)=0,Q(u,u_{x},u_{t},u_{xx},u_{tx},u_{tt},.............)=0, (7)

can be written as

U(η)=j=1ntanhj1(η)[Bjsech(η)+Ajtanh(η)]+A0.U(\eta)=\sum_{j=1}^{n}tanh^{j-1}(\eta)\,[B_{j}\,sech(\eta)+A_{j}\,tanh(\eta)]+A_{0}. (8)

Using Eq. (5) and Eq. (6), the solutions of the Eq. (8) take the following form

U(ω)=j=1ncosj1(ω)[Bjsin(ω)+Ajcos(ω)]+A0.U(\omega)=\sum_{j=1}^{n}cos^{j-1}(\omega)\,[B_{j}\,sin(\omega)+A_{j}\,cos(\omega)]+A_{0}. (9)

We calculate the value of unknown nn by apply the homogeneous balance assumption. After letting the collection of the coefficients of sini(ω)cosj(ω)sin^{i}\,(\omega)\,cos^{j}\,(\omega) of equal power to be all 0 (zero), we get an algebraic system of equations. By find out the unknowns of the system and using Eq.(9), we get the solutions to Eq. (7) is the form of (8).

3 Applications

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-diffusion equation.

3.1 The potential-YTSF equation of dimension (3+1)

We consider the potential-YTSF equation of dimension (3+1) [40, 41, 42] in the form

4uxt+uxxxz+4uxuxz+2uxxuz+3uyy=0,-4\,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, 44, 45]. The moving coordinate

u(x,y,z,t)=U(η),η=x+y+zct,u(x,y,z,t)=U(\eta)\,,\,\,\eta=x+y+z-ct, (11)

allows to convert Eq.(10) in to an ODE

U′′′′+6UU′′+(4c+3)U′′=0,U^{\prime\prime\prime\prime}+6U^{\prime}U^{\prime\prime}+(4c+3)U^{\prime\prime}=0,

integrating it with respect to η\eta, we acquire

U′′′+3(U)2+(4c+3)U=0,U^{\prime\prime\prime}+3(U^{\prime})^{2}+(4c+3)U^{\prime}=0, (12)

where integrating constant is set to zero. Setting U=VU^{\prime}=V, we have

V′′+3V2+(4c+3)V=0.V^{\prime\prime}+3V^{2}+(4c+3)V=0. (13)

Applying homogeneous balance principle between V′′V^{\prime\prime} and V2V^{2} in Eq. (13) based on Eq. (9) we obtain n+2=n+nn+2=n+n, which implies n=2n=2. Now we can write Eq. (9) as

V(ω)=B1sin(ω)+A1cos(ω)+B2sin(ω)cos(ω)+A2cos2(ω)+A0V(\omega)=B_{1}\,sin(\omega)+A_{1}\,cos(\omega)+B_{2}\,sin(\omega)\,cos(\omega)+A_{2}\,cos^{2}(\omega)+A_{0} (14)

and {IEEEeqnarray}rCl V”&=-B_1 sin^3(ω)+B_1 cos^2(ω) sin(ω)-2A_1 sin^2(ω) cos(ω)-5 B_2 cos(ω) sin^3(ω)+B_2 cos^3(ω) sin(ω)
+ 2 A_2 sin^4(ω)-4 A_2cos^2(ω) sin^2(ω). By substituting Eq. (14) and Eq. (14) into Eq. (13) and equaling all polynomials with same degree to zero, we get a system of equation as below.

constant:               4vA0+3A02+3B12+3A0+2A2=04\,v\,A_{{0}}+3\,{A_{{0}}}^{2}+3\,{B_{{1}}}^{2}+3\,A_{{0}}+2\,A_{{2}}=0
cos(ω):                   4vA1+6A0A1+6B1B2+A1=0cos(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,v\,A_{{1}}+6\,A_{{0}}A_{{1}}+6\,B_{{1}}B_{{2}}+A_{{1}}=0
sin(ω):                   4vB1+6A0B1+2B1=0sin(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,v\,B_{{1}}+6\,A_{{0}}B_{{1}}+2\,B_{{1}}=0
cos2(ω):                 4vA2+6A0A2+3A123B12+3B225A2=0cos^{2}(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,v\,A_{{2}}+6\,A_{{0}}A_{{2}}+3\,{A_{{1}}}^{2}-3\,{B_{{1}}}^{2}+3\,{B_{{2}}}^{2}-5\,A_{{2}}=0
sin(ω)cos(ω):  4vB2+6A0B2+6A1B12B20=0sin(\omega)\,cos(\omega):\,\,4\,v\,B_{{2}}+6\,A_{{0}}B_{{2}}+6\,A_{{1}}B_{{1}}-2\,B_{{2}}0=0
cos3(ω):                  6A1A26B1B2+2A1=0cos^{3}(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,A_{{1}}A_{{2}}-6\,B_{{1}}B_{{2}}+2\,A_{{1}}=0
sin(ω)cos2(ω):6A1B2+6A2B1+2B1=0sin(\omega)\,cos^{2}(\omega):6\,A_{{1}}B_{{2}}+6\,A_{{2}}B_{{1}}+2\,B_{{1}}=0
cos4(ω):                  3A223B22+6A2=0cos^{4}(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,{A_{{2}}}^{2}-3\,{B_{{2}}}^{2}+6\,A_{{2}}=0
sin(ω)cos3(ω):6A2B2+6B2=0sin(\omega)\,cos^{3}(\omega):6\,A_{{2}}B_{{2}}+6\,B_{{2}}=0

After solving the above system , we find the traveling wave solution U(η)U(\eta) to Eq. (10) in the form of (8).

Case-1

c=7/4,A0=2,A1=0,B1=0,A2=2,B2=0,c=-7/4\,,\,A_{0}=2\,,\,A_{1}=0\,,\,B_{1}=0\,,\,A_{2}=-2\,,\,B_{2}=0\,,

which gives:

u(x,y,z,t)=U(η)= 2tanh(η),withη=x+y+z+74t.u(x,y,z,t)=\,U(\eta)=\,2\,tanh\,(\eta),\text{with}\,\,\eta=x+y+z+\frac{7}{4}\,t. (15)
Refer to caption
Figure 1: The graphical representation of Eq. (15) (3D on the left side and 2D on the right side).

Case-2

c=1/4,A0=2/3,A1=0,B1=0,A2=2,B2=0,c=1/4\,,\,A_{0}=2/3\,,\,A_{1}=0\,,\,B_{1}=0\,,\,A_{2}=-2\,,\,B_{2}=0\,,

which gives:

u(x,y,z,t)=U(η)= 2tanh(η)43η,withη=x+y+z14t.u(x,y,z,t)\,=U(\eta)=\,2\,tanh\,(\eta)-\frac{4}{3}\,\eta,\,\text{with}\,\,\eta=\,x+y+z-\frac{1}{4}\,t. (16)
Refer to caption
Figure 2: The graphical representation of Eq. (16) (3D on the left side and 2D on the right side).

Case-3

c=1,A0=1,A1=0,B1=0,A2=1,B2=±i,c=-1\,,\,A_{0}=1\,,\,A_{1}=0\,,\,B_{1}=0\,,\,A_{2}=-1\,,\,B_{2}=\pm i\,,

which gives:

u(x,y,z,t)=U(η)=tanh(η)±isech(η),withη=x+y+z+t.u(x,y,z,t)=\,U(\eta)=\,tanh\,(\eta)\pm\,i\,sech\,(\eta),\,\,\text{with}\,\,\eta=\,x+y+z+t. (17)
Refer to caption
Figure 3: The graphical representation of Eq. (17) (3D on the left side and 2D on the right side).

Case-4

c=1/2,A0=2/3,A1=0,B1=0,A2=1,B2=±i,c=-1/2\,,\,A_{0}=2/3\,,\,A_{1}=0\,,\,B_{1}=0\,,\,A_{2}=-1\,,\,B_{2}=\pm i\,,

which gives:

u(x,y,z,t)=U(η)=tanh(η)±isech(η)13η,withη=x+y+z+12t.u(x,y,z,t)\,=U(\eta)\,=tanh\,(\eta)\pm\,i\,sech\,(\eta)-\frac{1}{3}\,\eta,\,\,\text{with}\,\,\eta=x+y+z+\frac{1}{2}t. (18)
Refer to caption
Figure 4: The graphical representation of Eq. (18) (3D on the left side and 2D on the right side).

3.2 The Reaction-Diffusion Equation

We consider the following form of the reaction-diffusion equation [46],

utt+αuxx+βu+γu3=0u_{tt}+\alpha\,u_{xx}+\beta\,u+\gamma\,u^{3}=0\, (19)

where α\alpha, β\beta and γ\gamma are constants (nonzero). The moving coordinate u(x,t)=U(η)u(x,t)=U(\eta), where η=xct\eta=x-ct leads the Eq. (19) in an ordinary differential equation (ODE)

(α+c2)U′′+βU+γU3=0.(\alpha+c^{2})\,U^{\prime\prime}+\beta\,U+\gamma\,U^{3}=0. (20)

Considering homogeneous balance between U′′U^{\prime\prime} and U3U^{3} in Eq. (20) based on Eq. (9) we obtain n+2=n+n+nn+2=n+n+n, which implies n=1n=1. Now we can write Eq. (9) as

U(ω)=B1sin(ω)+A1cos(ω)+A0.U(\omega)=B_{1}\,sin(\omega)+A_{1}\,cos(\omega)+A_{0}. (21)

Using Eq. (21) into Eq. (20), and equaling all polynomials with same degree to get, we have a system of equation as below.



constant:               γA03+3γA0B12+βA0=0\gamma\,{A_{{0}}}^{3}+3\,\gamma\,A_{{0}}{B_{{1}}}^{2}+\beta\,A_{{0}}=0
cos(ω):                   3γA02A1+3γA1B122v2A12αA1+βA1=0cos(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\gamma\,{A_{{0}}}^{2}A_{{1}}+3\,\gamma\,A_{{1}}{B_{{1}}}^{2}-2\,{v}^{2}A_{{1}}-2\,\alpha\,A_{{1}}+\beta\,A_{{1}}=0
sin(ω):                   3γA02B1+γB13v2B1αB1+βB1=0sin(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\gamma\,{A_{{0}}}^{2}B_{{1}}+\gamma\,{B_{{1}}}^{3}-{v}^{2}B_{{1}}-\alpha\,B_{{1}}+\beta\,B_{{1}}=0
cos2(ω):                 3γA0A123γA0B12=0cos^{2}(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\gamma\,A_{{0}}{A_{{1}}}^{2}-3\,\gamma\,A_{{0}}{B_{{1}}}^{2}=0
sin(ω)cos(ω):  6γA0A1B1=0sin(\omega)\,cos(\omega):\,\,6\,\gamma\,A_{{0}}A_{{1}}B_{{1}}=0
cos3(ω):γA133γA1B12+2v2A1+2αA1=0cos^{3}(\omega):\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\gamma\,{A_{{1}}}^{3}-3\,\gamma\,A_{{1}}{B_{{1}}}^{2}+2\,{v}^{2}A_{{1}}+2\,\alpha\,A_{{1}}=0
sin(ω)cos2(ω):3γA12B1γB13+2v2B1+2αB1=0sin(\omega)\,cos^{2}(\omega):3\,\gamma\,{A_{{1}}}^{2}B_{{1}}-\gamma\,{B_{{1}}}^{3}+2\,{v}^{2}B_{{1}}+2\,\alpha\,B_{{1}}=0
After solving the above system , we find the traveling wave solution u(x,y,z,t)u(x,y,z,t) to Eq. (19) in the form of (8).

Case-1

c=±β2α2,A0=0,A1=±βγ,B1=0,α=α,β=β,γ=γ,c=\pm\,\sqrt{\frac{\beta-2\,\alpha}{2}}\,,\,A_{0}=0\,,\,A_{1}=\pm\,\sqrt{\frac{-\beta}{\gamma}}\,,\,B_{1}=0\,,\alpha=\alpha\,,\,\beta=\beta\,,\,\gamma=\gamma,

which gives:

u(x,t)=U(η)=±βγtanh(η),withη=x±β2α2t.u(x,t)=U(\eta)=\pm\,\sqrt{\frac{-\beta}{\gamma}}\,tanh\,(\eta),\,\text{with}\,\,\eta=x\pm\,\sqrt{\frac{\beta-2\,\alpha}{2}}t. (22)
Refer to caption
Figure 5: The graphical representation of Eq. (22) (3D on the left side and 2D on the right side).

Case-2

c=±βα,A0=0,B1=±2βγ,A1=0,α=α,β=β,γ=γ,c=\pm\,\sqrt{-\beta-\alpha}\,,\,A_{0}=0\,,\,B_{1}=\pm\,\sqrt{\frac{-2\,\beta}{\gamma}}\,,\,A_{1}=0\,,\alpha=\alpha\,,\,\beta=\beta\,,\,\gamma=\gamma,

which gives:

u(x,t)=U(η)=±2βγsech(η),withη=x±βαt.u(x,t)=U(\eta)=\pm\,\sqrt{\frac{-2\,\beta}{\gamma}}\,sech\,(\eta),\,\,\text{with}\,\,\eta=x\pm\,\sqrt{-\beta-\alpha}t. (23)
Refer to caption
Figure 6: The graphical representation of Eq. (23) (3D on the left side and 2D on the right side).

Case-3

c=±2βα,A0=0,A1=±βγ,B1=±βγ,α=α,β=β,γ=γ,c=\pm\,\sqrt{2\,\beta-\alpha}\,,\,A_{0}=0\,,\,A_{1}=\pm\,\sqrt{\frac{-\beta}{\gamma}}\,,\,B_{1}=\pm\,\sqrt{\frac{\beta}{\gamma}}\,,\alpha=\alpha\,,\,\beta=\beta\,,\,\gamma=\gamma,

which gives:

u(x,t)=U(η)=±βγtanh(η)±βγsech(η),withη=x±2βαt.u(x,t)=U(\eta)=\pm\,\sqrt{\frac{-\beta}{\gamma}}\,tanh\,(\eta)\pm\,\sqrt{\frac{\beta}{\gamma}}\,sech\,(\eta),\,\,\text{with}\,\,\eta=x\pm\,\sqrt{2\,\beta-\alpha}t. (24)
Refer to caption
Figure 7: The graphical representation of Eq. (24) (3D on the left side and 2D on the right side).

4 Conclusion

In this article, we have applied the Sine-Gordon expansion method for calculating new travelling wave solutions to the potential-YTSF equation of dimension (3+1) and the reaction-diffusion equation. We have found these solutions of the equation in the trigonometric, complex and hyperbolic function forms. 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.

Declarations

Availability of data and material
: Not applicable.

Competing interests: The author has no any financial or non-financial conflict of interests.

Funding : 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 their valuable comments and suggestions towards improving this article.

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