The solitary wave solutions to the Klein–Gordon–Zakharov equations by extended rational methods

Research on the exact traveling wave solutions to nonlinear partial differential equations (PDEs) and fractional differential, which portray physics phenomena, has become very important in the literature.^1,2 Because many natural phenomena can be modeled in the form of these equations, it is always important to find the exact solution to them. In particular, problems in physical phenomena, engineering, and basic sciences can be modeled in the form of the above equations. Thus, a proliferation of coalition techniques, including multiple scale method,³ new extended direct algebraic method,⁴ auxiliary equation method,^5,6 sine–cosine method,^7,8 sine-Gordon expansion method,⁹ Liu’s extended trial function method,¹⁰ Hirota bilinear method,¹¹ the first integral method,^12,13 similarity transformation,¹⁴ extended trial method,¹⁵ Jacobi’s elliptic function expansion method,¹⁶ and extended F-expansion and projective Ricatti equation methods,¹⁷ were involved in the study of specific solitary wave solutions of PDEs.^18–44 Recently, the Klein–Gordon–Zakharov (KGZ) equations have engrossed much consideration in different branches, such as biology, plasma physics, and optic fibers. In this paper, our purpose is to manufacture a new solitary wave solution by the extended rational sine–cosine and rational sinh–cosh methods to the KGZ equations. The residual part of this paper is formed as follows: in Sec. II, algorithms of extended rational sinh–cosh and sine–cosine methods are presented. In Sec. III, we represent these techniques with the KGZ equations. In addition, we consider the graphical representation of some acquired solutions in Sec. IV. Indeed, conclusion is provided in Sec. V.

Consider nonlinear PDE of the form as

where ψ = ψ(x, t) is a traveling wave solution of nonlinear PDE (1).

Using the following transformation:

where k is the wave speed, Eq. (1) can be converted into an ODE as follows:

where the prime denotes $∂∂ξ$⁠. Then integrate ODE (3) as long as all terms contain derivatives and neglect integration constants.⁴⁵ 

• Suppose that Eq. (3) has the solution in the form of

  $ψ(ξ)=a0⁡sin(ηξ)a2+a1⁡cos(ηξ),cos(ηξ)≠−a2a1$

  (4)

  or

  $ψ(ξ)=a0⁡cos(ηξ)a2+a1⁡sin(ηξ),sin(ηξ)≠−a2a1,$

  (5)

  where a_i, (i = 0, 1, 2) are parameters to be determined and η is the wave number.

• Unknown constants can be found by substituting Eq. (4) or Eq. (5) into Eq. (3), collecting all terms with the same powers of cos(ηξ)^m or sin(ηξ)^m and equating to zero all the coefficients of cos(ηξ)^m or sin(ηξ)^m yield a set of algebraic equations. By calculation using the Maple software, the solutions of the algebraic equations can be derived.

• By substituting the values of a₀, a₁, a₂, c, and η into Eq. (4) or Eq. (5), the solution of Eq. (3) can be found.

• Suppose that Eq. (3) has the solution in the form

  $ψ(ξ)=a0⁡sinh(ηξ)a2+a1⁡cosh(ηξ),cosh(ηξ)≠−a2a1$

  (6)

  or

  $ψ(ξ)=a0⁡cosh(ηξ)a2+a1⁡sinh(ηξ),sinh(ηξ)≠−a2a1,$

  (7)

  where a_i, (i = 0, 1, 2) are parameters to be determined and η is the wave number.

• Substituting Eq. (6) or Eq. (7) into Eq. (3), collecting all terms with the same powers of cosh(ηξ)^m or sinh(ηξ)^m and equating to zero all the coefficients of cosh(ηξ)^m or sinh(ηξ)^m yield a set of algebraic equations. The values of the unknowns are to be determined by solving the algebraic equations.

• The solution of Eq. (3) can be found by substituting the values of a₀, a₁, a₂, c, and η into Eq. (6) or Eq. (7).

Suppose the KGZ equations defined by Eqs. (8) and (9),¹⁸ 

where ϕ(x, t) stands for a complex function, while v(x, t) is a real function, t shows the time, and x is the distance along the direction of publication. Therefore, α and β are nonzero parameters.

To construct the soliton wave solution to the KGZ equations [Eqs. (8) and (9)], we adopt the transformation in the form of

where ψ(x, t) = ψ(ξ) is a real function, while k and ω are constants to be determined, and θ stands for an arbitrary constant.

Inserting (10) into (8) and (9) gives

Using (10), it is obtained from (11) that

Next, to surmise v(x, t) = W(ξ) and insert (14) in (13), after integrating twice with a nonzero constant value, it is recovered that

where C is an integration constant. Inserting (15) in (11) leads to

where Δ = αC + k² − ω² + 1. Now, instead of solving the main equation, we solve the last equation [Eq. (16)], and with the method mentioned in the second part and using the Maple software, we obtain the equation coefficients. After solving the system of equations, the following set of solutions is obtained for numerical coefficients.⁴⁶ 

Suppose that Eq. (16) has solutions in the form of

Substituting Eq. (17) into Eq. (16) and collecting all terms with the same powers of cos(ηξ)^m and equating to zero all the coefficients of cos(ηξ)^m, the following algebraic equations are obtained:

After solving these algebraic equations, the following solutions are obtained:

• Taking set 1 into account, the solutions of (16) can be obtained as

  $ψ1(ξ)=±Δk2−ω2αβtan12−2Δk2−ω2ξω.$

  (18)

  Combining Eqs. (10) and (18), we get

  $ϕ1(x,t)=±Δk2−ω2αβtan12−2Δk2−ω2kt+ωxeikx+ωt+θω,$

  (19)

  $W1(x,t)=Δtan−Δ2k2−2ω2kt+ωx2α+C.$

  (20)
• Similarly, for set 2, the solutions of (16) can be obtained as

  $ψ2(ξ)=±Δk2−ω2αβsin−2Δk2−ω2ξωcos−2Δk2−ω2ξ−1.$

  (21)

  Combining Eqs. (10) and (21), we get

  $ϕ2(x,t)=±Δk2−ω2αβsin−2Δk2−ω2kt+ωxeikx+tω+θωcos−2Δk2−ω2kt+ωx−1,$

  (22)

  $W2(x,t)=Δsin−2Δk2−ω2kt+ωx2αcos−2Δk2−ω2kt+ωx−12+C.$

  (23)
• Similarly, for set 2, the solutions of (16) can be obtained as

  $ψ3(ξ)=±Δk2−ω2αβsin−2Δk2−ω2ξωcos−2Δk2−ω2ξ+1.$

  (24)

  Combining Eqs. (10) and (24), we get

  $ϕ3(x,t)=±Δk2−ω2αβsin−2Δk2−ω2kt+ωxeikx+tω+θωcos−2Δk2−ω2kt+ωx+1,$

  (25)

  $W3(x,t)=Δsin−2Δk2−ω2kt+ωx2αcos−2Δk2−ω2kt+ωx+12+C.$

  (26)

  OR
  Suppose that Eq. (16) has solutions in the form of

  $ψ(ξ)=a0⁡cos(ηξ)a2+a1⁡sin(ηξ).$

  (27)

  Substituting Eq. (27) into Eq. (16) and collecting all terms with the same powers of sin(ηξ)^m and equating to zero all the coefficients of sin(ηξ)^m, the following algebraic equations are obtained:

  $sin(ηξ)2:−Cαk2a12+Cαω2a12+αβω2a02−k4a12+2k2ω2a12−ω4a12−k2a12+ω2a12a0,sin(ηξ)1:−η2a2a1k4+2η2a2a1k2ω2−η2a2a1ω4−2Cαk2a2a1+2Cαω2a2a1−2a1a2k4+4a1a2k2ω2−2a1a2ω4−2k2a2a1+2ω2a2a1a0,sin(ηξ)0:−a0k−ωk+ωη2−1ω2+−η2+1k2+Cα+1a22+2η2ω4a12+−4η2k2a12+αβa02ω2+2η2k4a12.$

  After solving these algebraic equations, the following solutions are obtained:

  $Set 3)η=±−Δ2k2−2ω2,a0=±−Δ2k2−2ω2,a1=a1,a2=0,Set 4)η=±−2Δk2−ω2,a0=±a2ω−Δ(k2−ω2)αβ,a1=±a2,a2=a2.$
• Taking set 3 into account, the solutions of (16) can be obtained as

  $ψ4(ξ)=±Δk2−ω2αβcot12−2Δk2−ω2ξω.$

  (28)

  Combining Eqs. (10) and (28), we get

  $ϕ4(x,t)=±Δk2−ω2αβcot12−2Δk2−ω2kt+ωxeikx+ωt+θω,$

  (29)

  $W4(x,t)=Δcot−Δ2k2−2ω2kt+ωx2α+C.$

  (30)
• Similarly, for set 4, the solutions of (16) can be obtained as

  $ψ5(ξ)=±Δk2−ω2αβcos−2Δk2−ω2ξωsin−2Δk2−ω2ξ+1.$

  (31)

  Combining Eqs. (10) and (31), we get

  $ϕ5(x,t)=±Δk2−ω2αβcos−2Δk2−ω2kt+ωxeikx+tω+θωsin−2Δk2−ω2kt+ωx+1,$

  (32)

  $W5(x,t)=−Δcos−2Δk2−ω2kt+ωx2αcos−2Δk2−ω2kt+ωx2−2⁡sin−2Δk2−ω2kt+ωx−2+C.$

  (33)
• Similarly, for set 4, the solutions of (16) can be obtained as

  $ψ6(ξ)=±Δk2−ω2αβcos−2Δk2−ω2ξωsin−2Δk2−ω2ξ+1.$

  (34)

  Combining Eqs. (10) and (34), we get

  $ϕ6(x,t)=±Δk2−ω2αβcos−2Δk2−ω2kt+ωxeikx+tω+θωsin−2Δk2−ω2kt+ωx−1,$

  (35)
  $W6(x,t)=−Δcos−2Δk2−ω2kt+ωx2αcos−2Δk2−ω2kt+ωx2−2⁡sin−2Δk2−ω2kt+ωx−2+C.$

  (36)

Suppose that Eq. (16) has solutions in the form of

Substituting Eq. (37) into Eq. (16) and collecting all terms with the same powers of cosh(ηξ)^m and equating to zero all the coefficients of cosh(ηξ)^m, the following algebraic equations are obtained:

After solving these algebraic equations, the following solutions are obtained:

• Taking set 5 into account, the solutions of (16) can be obtained as

  $ψ7(ξ)=±Δk2−ω2αβtanh12−2Δk2−ω2ξω.$

  (38)

  Combining Eqs. (10) and (38), we get

  $ϕ7(x,t)=±Δk2−ω2αβtanh12−2Δk2−ω2kt+ωxeikx+ωt+θω,$

  (39)

  $W7(x,t)=Δtanh−Δ2k2−2ω2kt+ωx2α+C.$

  (40)
• Similarly, for set 6, the solutions of (16) can be obtained as

  $ψ8(ξ)=±Δk2−ω2αβsinh−2Δk2−ω2ξωcosh−2Δk2−ω2ξ−1.$

  (41)

  Combining Eqs. (10) and (41), we get

  $ϕ8(x,t)=±Δk2−ω2αβsinh−2Δk2−ω2kt+ωxeikx+tω+θωcosh−2Δk2−ω2kt+ωx−1,$

  (42)

  $W8(x,t)=Δsinh−2Δk2−ω2kt+ωx2αcosh−2Δk2−ω2kt+ωx−12+C.$

  (43)
• Similarly, for set 6, the solutions of (16) can be obtained as

  $ψ9(ξ)=±Δk2−ω2αβsinh−2Δk2−ω2ξωcosh−2Δk2−ω2ξ+1.$

  (44)

  Combining Eqs. (10) and (44), we get

  $ϕ9(x,t)=±Δk2−ω2αβsinh−2Δk2−ω2kt+ωxeikx+tω+θωcosh−2Δk2−ω2kt+ωx+1,$

  (45)

  $W9(x,t)=Δsinh−2Δk2−ω2kt+ωx2αcosh−2Δk2−ω2kt+ωx+12+C.$

  (46)

  OR
  Suppose that Eq. (16) has solutions in the form of

  $ψ(ξ)=a0⁡cosh(ηξ)a2+a1⁡sinh(ηξ).$

  (47)

  Substituting Eq. (47) into Eq. (16) and collecting all terms with the same powers of sinh(ηξ)^m and equating to zero all the coefficients of sinh(ηξ)^m, the following algebraic equations are obtained:

  $sinh(ηξ)2:Cαk2a12−Cαω2a12−αβω2a02+k4a12−2k2ω2a12+ω4a12+k2a12−ω2a12a0,$

  $sinh(ηξ)1:−η2a2a1k4+2η2a2a1k2ω2−η2a2a1ω4−2aCk2a2a1+2aCω2a2a1−2a2a1k4+4a2a1k2ω2−2a2a1ω4−2a2a1k2+2ω2a1a2a0,sinh(ηξ)0:−a0k+ωk−ωη2−1ω2+−η2+1k2+αC+1a22+2η2ω4a12+−4η2k2a12+αβa02ω2+2η2k4a12.$
  After solving these algebraic equations, the following solutions are obtained:

  $Set 7)η=±−Δ2k2−2ω2,a0=±a1ω−Δ(k2−ω2)αβ,a1=a1,a2=0,Set 8)±η=−2Δk2−ω2,a0=±a2ω−Δ(k2−ω2)αβ,a1=±a2,a2=a2.$
• Taking set 7 into account, the solutions of (16) can be obtained as

  $ψ10(ξ)=±Δk2−ω2αβcoth12−2Δk2−ω2ξω.$

  (48)

  Combining Eqs. (10) and (48), we get

  $ϕ10(x,t)=±Δk2−ω2αβcoth12−2Δk2−ω2kt+ωxeikx+ωt+θω,$

  (49)

  $W10(x,t)=Δcoth−Δ2k2−2ω2kt+ωx2α+C.$

  (50)
• Similarly, for set 8, the solutions of (16) can be obtained as

  $ψ11(ξ)=±Δk2−ω2αβcosh−2Δk2−ω2ξωsinh−2Δk2−ω2ξ+1.$

  (51)

  Combining Eqs. (10) and (51), we get

  $ϕ11(x,t)=±Δk2−ω2αβcosh−2Δk2−ω2kt+ωxeikx+tω+θωsinh−2Δk2−ω2kt+ωx+1,$

  (52)

  $W11(x,t)=Δcosh−2Δk2−ω2kt+ωx2αcosh−2Δk2−ω2kt+ωx2−2⁡sinh−2Δk2−ω2kt+ωx+2+C.$

  (53)
• Similarly, for set 8, the solutions of (16) can be obtained as

  $ψ12(ξ)=±Δk2−ω2αβcosh−2Δk2−ω2ξωsinh−2Δk2−ω2ξ−1.$

  (54)

  Combining Eqs. (10) and (54), we get

  $ϕ12(x,t)=±Δk2−ω2αβcosh−2Δk2−ω2kt+ωxeikx+tω+θωsinh−2Δk2−ω2kt+ωx−1,$

  (55)

  $W12(x,t)=Δcosh−2Δk2−ω2kt+ωx2αcosh−2Δk2−ω2kt+ωx2−2⁡sinh−2Δk2−ω2kt+ωx−2+C.$

  (56)