New Solutions for the Generalized BBM Equation in terms of Jacobi and Weierstrass Elliptic Functions

The Jacobi elliptic function method is applied to solve the generalized Benjamin-Bona-Mahony equation (BBM). Periodic and soliton solutions are formally derived in a general form. Some particular cases are considered. A power series method is also applied in some particular cases. Some solutions are expressed in terms of the Weierstrass elliptic function.


Introduction
The regularized long-wave (RLW) equation is a famous nonlinear wave equation which gives the phenomena of dispersion and weak nonlinearity, including magneto hydrodynamic wave in plasma, phonon packets in nonlinear crystals, and nonlinear transverse waves in shallow water or in ion acoustic. This equation is also called the BBM (Benjamin-Bona-Mahony) equation and reads It describes approximately the unidirectional propagation of long waves in certain nonlinear dispersive systems, and it was proposed by Benjamin et al. in 1972 [1] as a more satisfactory model than the KdV equation [2]: It is easy to see that Equation (1) can be derived from the equal width equation [3]: by means of the change of variable u ⟵ u + 1, that is, by replacing u with u + 1. This last equation is considered an equally valid and accurate model for the same wave phenomena simulated by (1) and (2). On the other hand, some researches analyzed the generalized KdV equation with variable coefficients u t + σu n u x + μu xxt = 0, ð4Þ because this model has important applications in several fields of science. Motivated by these facts, we will consider here the generalized equal width (RW) equation with constant coefficients u t + σu n u x − μu xxt = 0: Our aim is to present solutions different from those in [2,[4][5][6].
1.1. Trigonometric and Soliton Solutions. In order to obtain new solutions to Equation (5), we make the traveling wave transformation Inserting (6) into (5) gives Integrating once with respect to ξ to get where c 1 is the constant of integration. Now, we multiply Equation (8) by u ′ ðξÞ, and we integrate the resulting equation with respect to ξ: where c 2 is the constant of integration. If c 1 = c 2 = 0, Equation (9) takes the form Inserting Equation (11) into (10), we obtain From Equation (12) A sech solution to Equation (13) is It is clear that function is a solution to equation u t + σu n u x − μu xxt = 0 for any choice of the parameters λ and ξ 0 .
Finally, if μ < 0, then taking into account that we obtain from (15) the trigonometric solution

The Novel Solutions
In this section, we will give analytical solutions to Equation (5) for two special cases: n = 1 and n = 2: First of all, observe that Equation (5) is equivalent to Equation (9).
2.1. The Case n = 1. If n = 1, Equation (9) reads To solve this equation, we consider the reduced Duffing equation where the coefficient q does not depend on ξ. Observe that and integrating this equation w.r.t. ξ gives Suppose that y = yðξÞ is a solution to Equation (19) and let We have Taking into account Equations (18) and (23), we must have 2 Abstract and Applied Analysis Equating to zero the coefficients of y 6 , y 2 , and y 0 , we conclude that q = Aσ 6λμ , We have proved that the general solution to the nonlinear ODE may be written in the form uðξÞ = Ay 2 ðξÞ + λ/σ, where y = y ðξÞ is a solution to the Duffing equation According to [2], the general solution to Equation (29) may be expressed in terms of the Jacobian elliptic functions. More exactly, the general solution to Equation (19) is The value of ξ 0 is determined by solving equation y′ð0Þ = y 1 . In principle, we may consider that ξ 0 is any number as well as y 0 . Taking into account these facts, we have proved that the general solution to Equation (28) is Finally, an exact solution to the EW-equation Solution (32) involves four arbitrary constants: A, y 0 , λ, and ξ 0 (these do not depend on ξ). We choose A so that Aσ y 2 0 /6λμ > 0.

The Case n = 2.
This case corresponds to the so-called modified BBM or modified equal width equation. When n = 2, Equation (9) reads This equation is harder to solve. We seek solution to Equation (33) in the form Observe that function y = yðξÞ = rcnð ffiffiffi ffi w p ðξ + ξ 0 Þ, ffiffiffiffi m p Þ satisfies the nonlinear differential equation for any constants γ and ξ 0 . This equation is Duffing equation y ′ ′ + αy + βy 3 = 0 with coefficients α and β. The relationship between the frequency ω, the modulus m, and the coefficients is given by the system Solving this system for ω and m gives To solve the modified BBM equation, we will make use of the power series method (PSM). This is a promising method that also may be applicable to obtain an approximate solution for those ODE's that do not admit solution in closed form. In our case, we consider the function where The values of q and r are obtained from the initial conditions uðξ 0 Þ = 0 and uðξ 0 + Kð ffiffiffiffi m p Þ/ ffiffiffi ffi w p Þ = 0, that is, Abstract and Applied Analysis Thus, Inserting the ansatz (39) into (38), we obtain an expression for FðξÞ in terms of the unknowns w, m, p, and λ. Observe that function uðξÞ given by (39) is a solution to Equation (33), if and only if FðξÞ ≡ 0: The series method consists in coefficients of Taylor series of FðξÞ around some point. Usually, this is the origin. Since the calculations are enormous, we cannot realize this work by hand. Instead, we may use Mathematica 12 or Maple 16. The Mathematica command has the form Series ½F½ξ, fξ, ξ 0 , ng. This gives us the nth degree Taylor polynomial of FðξÞ around the point ξ = ξ 0 . In our case, we will use the command Series ½F½ξ, fξ , 0, ng for n ≥ 3, since we need at least four equations to determine the five unknowns w, m, p, and λ in terms of the coefficients of Equation (33). Of course, the series method is applicable to solve the BBM when n = 1. Our aim is to show two different methods. Let us proceed. We have that F ð2n+1Þ ðξ 0 Þ = 0 for all n = 0, 1, 2, 3, ⋯, and where F ξ 0 ð Þ = 1 12 12c 1 + 12 2 + 6λ + σ ð Þ = 0, ð44Þ Solving this last equation for m and taking into account (45), (47), and (49), we obtain after long algebraic calculations following solution: Direct calculations show that function u = uðξÞ defined by (39) with parameters given by (51)-(55) and (41) is a solution to Equation (41). Then, function u = uðx − λt − ξ 0 Þ is a solution to the modified BBM equation Observe that cnðx, 1Þ = sech ðxÞ: It may be verified that the following function is a solution to Equation (56): for the choice 4 Abstract and Applied Analysis We obtain soliton solutions for ð1 − γ 2 Þμ > 0 and trigonometric solutions for ð1 − γ 2 Þμ < 0:

Solution for a Generalized BBM Equation.
Let us consider the following generalized BBM equation: In order to obtain new solutions to Equation (59), we make the travelling wave transformation Inserting (60) into (59) gives Equation (61) is hard to integrate. We seek for an exact solution of the form Inserting ansatz (62) into (61) gives RðξÞ = 0, where being Equating to zero the coefficients of ℘ j ðξ ; g 2 , g 3 Þ gives a nonlinear algebraic system. Solving it with the aid of Wolfram Mathematica, we obtain Other solutions may be obtained choosing B from the condition where g 2 , g 3 , and λ are obtained from (65). The obtained solution is valid for We have solved the generalized BBM equation for any p such that pðp + 1Þðp + 2Þ ≠ 0: The exact traveling wave solution is given by The numbers A and ξ 0 are arbitrary. Particular cases are (1) p = 1 : (2) p = 2 : (3) p = 1/2 : After doing a traveling wave transformation, we get the linear ode whose general solution is given by Case 2 (ðp + 1Þðp + 2Þ = 0). Making the traveling wave transformation ξ = x − λt + ξ 0 and making use of the ansatz (62), we obtain that B = 0, which is a trivial solution. So, in the cases p = −1 and p = −2, we cannot get nontrivial solutions to the generalized BBM equation.
6 Abstract and Applied Analysis

Analysis and Discussion
We obtained the traveling wave solutions to BBM equation u t + σuu x − μu xxt = 0 and modified BBM equation u t + σu 2 u x − μu xxt = 0: Other solutions were obtained in [7] using the extended Jacobi elliptic function expansion method. However, in our approach, we got nonzero integration constants. Most authors get zero integration constants to simplify matters, but this may cause loss of solutions. We also may consider the following combined BBM equation: Making the traveling wave transformation u = vðξÞ, We now integrate once with respect to ξ, and denoting the constant of integration by c 1 , we obtain the ode This last equation is a Duffing-Helmholtz equation. The general form of the undamped and unforced Duffing-Helmholtz equation reads The general solution to Equation (79) may be expressed in the ansatz form Indeed, given the initial conditions the solution to the initial value problem (80)-(81) is given by (80), where 9A 4 pr 2 − 3A 4 q 2 r + 12A 3 pqr − 4A 3 q 3 + 18A 2 p 2 r À − 6A 2 pq 2 + 36Anpr − 12Anq 2 + 27n 2 r − 6npq + p 3 Á , The number A is a solution to the quartic 3rA 4 + 4qA 3 + 6pA 2 + 12nA − 12nv 0 + 6pv 2 0 + 4qv 3 In soliton theory, we are interested in sech or tanh solutions to Equation (79). The soliton solutions may occur only when In this case, we will have the following soliton solution to Equation (79): v ξ ð Þ = − q 3r + 3Ar + q 3r sech 3Ar + q 3 ffiffiffiffi ffi 2r p ξ , for 3Ar + q ≠ 0: Let us examine condition (85) for (78). We must have a 3 + 6aλσ + 12c 1 σ 2 = 0: Solving this equation for c 1 , we will have the following soliton solution to equation (18): u x, t ð Þ= − a 2σ + a + 2Aσ 2σ sech a + 2Aσ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi δ 4aAσ + 4A 2 σ 2 − 5a 2 ð Þ p Á x + 1 24 In (88), the constants A and ξ 0 are arbitrary. In the case when σ = 0, Equation (18) converts into usual BBM equation u t + au x − μu xxt = 0: In a more general fashion, we solved Equation (59). This equation was solved in [2,5,8] using the substitution u = v 1/p . In the present paper, we solved it by using the substitution u = v −2/p . Let us examine the odes obtained by means of these different substitutions. Using u = v 1/p gives On the other hand, letting u = v −2/p gives a different ode given by (61). So, our method gives other than that obtained in [2,5,8] solutions.
The authors in [5] claim that there are no periodic solutions for the parameter values of different form p = 1, p = 2,