The (1 + 1)-dimensional nonlinear dispersive modified Benjamin-Bona
Mahony equation: In this section, we will apply the modified simple equation method to find the exact solutions and then the solitary wave solutions of (1 + 1)-dimensional nonlinear DMBBM equation,
where α is a nonzero constant. This equation was first derived to describe an approximation for surface long waves in nonlinear dispersive media. It can also characterize the hydro magnetic waves in cold plasma, acoustic waves in inharmonic crystals and acoustic gravity waves in compressible fluids (Yusufoglu 2008; Zayed and Al-Joudi 2010).
The traveling wave transformation is
Using traveling wave Eq. (3.2), Eq. (3.1) transforms into the following ODE
Integrating with respect to ξ choosing constant of integration as zero, we obtain the following ODE
Now balancing the highest order derivative u″ and non-linear term u3, we get
3N = N + 2, which gives N = 1
Now for N = 1, becomes
where β0 and β1 are constants to be determined such that β1 ≠ 0, while ψ(ξ) is an unknown function to be determined. It is easy to see that
Now substituting the values of u″, u, u3 in equation (3.3) and then equating the coefficients of Φ0, Φ- 1, Φ- 2, Φ- 3 to zero, we respectively obtain
Solving Eq. (3.8), we get
Solving Eq. (3.11), we get
and β1 ≠ 0
Case I: when β0 = 0 solving Eqs. (3.9), and (3.10) we get trivial solution. So this case is rejected.
Case II: when , Eqs. (3.9) and (3.10) yields
Integrating, Eq. (3.12) with respect to ξ, we obtain
From Eqs. (3.13) and (3.10), we obtain
Therefore, upon integration, we obtain
where c1 and c2 are arbitrary constants.
Substituting the values of Φ and Φ′ into Eq. (3.5), we obtain the following exact solution,
Putting the values of β0, β1, l and simplifying, we obtain
Since c1 and c2 are arbitrarily constants, consequently, if we set c1 = - 2 c2(1 - ω) and , Eq. (3.17) reduces to the following traveling wave solution:
Again setting c1 = 2 c2(1 - ω) and if , Eq. (3.17) reduces to the following singular traveling wave solutions:
If , Eqs. (3.18) and (3.19) yields the following periodic solutions:
Remark 1: From solutions (3.18)-(3.21) we conclude that ω ≠ 1.
The coupled Klein-Gordon equation
Now we will bring to bear the MSE method to find exact solutions and then the solitary wave solutions to the cKG Equation in the form,
The traveling wave Eq. (3.23) reduces Eqs. (3.22) into the following ODEs
By integrating Eq. (3.25) with respect to ξ, and neglecting the constant of integration we obtain
Substituting Eq. (3.26) into Eq. (3.24) we get,
Balancing the highest order derivative u″ and nonlinear term u3 from Eq. (3.27), we obtain N = 1
Now for N = 1, Eq. (2.4) becomes
where β0 and β1 are constants to be determined such that β1 ≠ 0, while Φ(ξ) is an unknown function to be determined. It is easy to see that
Now substituting the values of u″, u, u3 in Eq. (3.27) and then equating the coefficients of Φ0, Φ- 1, Φ- 2, Φ- 3 to zero, we respectively obtain
Solving Eq. (3.31), we get
Solving Eq. (3.34), we get
Case-I: When β0 = 0, Eq. (3.32) and (3.33) yields a trivial solution. So this case is rejected.
Case-II: When , Eqs. (3.32) and (3.33) yields,
Integrating, Eq. (3.35) with respect to ξ, we obtain
From Eqs. (3.36) and (3.32), we obtain
Therefore, integration Eq. (3.37), we obtain
where c1 and c2 are constants of integration.
Substituting the values of Φ and Φ′ into Eq. (3.28), we obtain the following exact solution,
Putting the values of β0, β1 into Eq. (3.39) and then simplifying, we obtain
We can freely choose the constants c1 and c2. Therefore, if we set , Eq. (3.40) reduces to:
Again, if we set , Eq. (3.40) reduces to:
Using hyperbolic identities, in trigonometric form Eqs. (3.41) and (3.42) can be written as follows:
Now applying Eqs. (3.41)-(3.44) into Eq. (3.26), we get
Remark 2: From solutions (3.41)-(3.48) we conclude that ω ≠ ± 1.