Optimal Homotopy Asymptotic Method to Nonlinear Damped Generalized Regularized Long-Wave Equation

Abstract

A new semianalytical technique optimal homology asymptotic method (OHAM) is introduced for deriving approximate solution of the homogeneous and nonhomogeneous nonlinear Damped Generalized Regularized Long-Wave (DGRLW) equation. We tested numerical examples designed to confine the features of the proposed scheme. We drew 3D and 2D images of the DGRLW equations and the results are compared with that of variational iteration method (VIM). Results reveal that OHAM is operative and very easy to use.

1. Introduction

Partial differential equations used in modeling different problems in physics, biology, chemical reactions, and engineering sciences problems are frequently too difficult to be solved exactly, and even if an exact solution is possible, the required calculations may be too difficult.

The DGRLW equation is a partial differential equation that describes the amplitude of long-wave, which takes the following form: 𝑢_𝑡−(𝜑(𝑥,𝑡)𝑢_𝑥𝑡)_𝑥−𝛼𝑢_𝑥𝑥+𝑢_𝑥+𝑢^𝑝𝑢_𝑥. (1) Here 𝛼>0, 𝑝≥1 is an integer, 𝜑(𝑥,𝑡) is known function, and 𝑢(𝑥,𝑡) is the amplitude of the long-wave at the position 𝑥 and time 𝑡. For 𝛼≠0, (1) features a balance between nonlinear and dispersive effects but also takes into account mechanisms of dissipation. In the physical sense, (1) with the dissipative term 𝛼𝑢_𝑥𝑥 is suggested if the good predictive power is preferred; such type of problem arises in the bore propagation as well as for water waves [1].

The Equal Width (EW) Wave, Regularized Long-Wave (RLW), and Generalized Regularized Long-Wave (GRLW) equations are special cases of the DGRLW equation [2–4].

The EW equation corresponds to 𝛼=0, 𝜑(𝑥,𝑡)=1, whereas the RLW or Benjamin-Bona-Mahony equation corresponds to 𝛼=0, 𝜑(𝑥,𝑡)=1, and 𝑝=1. On the other hand, the GRLW equation corresponds to 𝛼=0, 𝜑(𝑥,𝑡)=1, and 𝑝≥1.

Different methods have been used for numerical solutions of DGRLW equations. VIM was presented by Demir et al. for numerical solutions for the DGRLW equation [5]. Yousefi et al. used Bernstein Ritz-Galerkin Method for solving the DGRLW equation [6]. Achouri et al. worked on an article called “A fully Galerkin method for the damped generalized regularized long-wave (DGRLW) equation,” namely, in [7]. For the mathematical theory and physical significance of DGRLW equation see [8–16] and references therein.

Marinca and Herişanu proposed semianalytical technique OHAM for deriving approximate solution of nonlinear problems of thin film flow of a fourth grade fluid down a vertical cylinder [17–19]. The method has been used by many researchers for obtaining numerical approximations of linear and nonlinear differential equations [20–24]. The author has successfully applied the OHAM for deriving approximate solution of Equal Width Wave equation, Burger equations, and tenth order boundary value problems [25, 26]. The convergence criterion of proposed method is similar to that of homotopy analysis method (HAM) and homotopy perturbation method (HPM), but this method is more efficient and flexible. To improve the efficiency and accuracy of OHAM, Herişanu and Marinca introduced more generalized and new advances in OHAM which shows that the auxiliary function includes the functions of physical parameters in addition to the convergence control parameter [27, 28].

Here, we investigate the approximate solution of the DGRLW equation with a variable coefficient using OHAM. The whole paper is divided into 3 sections. Section 2 is devoted to the analysis of the proposed method. In Section 3, solution of homogeneous and non-homogenous (DGRLW) equations is presented by OHAM, and absolute errors are also compared with VIM. The 3D and 2D images of the approximate solution and exact solution are also drawn. In all cases, the proposed method yields very encouraging results.

2. Fundamental Theory of OHAM

Consider the partial differential equation of the following form: 𝐿(𝑢((𝑢(()𝐵𝑥,𝑡))+𝑁𝑥,𝑡))+𝑔𝑥,𝑡=0,𝑥∈Ω,𝑢,𝜕𝑢𝜕𝑡=0, (2) where 𝐿 is a linear operator and 𝑁 is nonlinear operator. 𝐵 is boundary operator, 𝑢(𝑥,𝑡) is an unknown function, 𝑥 and 𝑡 denote spatial and time variables, respectively, Ω is the problem domain, and 𝑔(𝑥,𝑡) is a known function.

Using the basic idea of OHAM, the optimal homotopy 𝜓(𝑥,𝑡;𝑞)∶Ω×[0,1]→𝑅 is constructed which satisfies the following condition: 𝐿𝜓(𝑞𝐿𝜓𝜓(,1−𝑞𝑥,𝑡;𝑞+𝑔𝑥,𝑡)=𝐻𝑥,𝑡;𝑞+𝑁𝑥,𝑡;𝑞+𝑔𝑥,𝑡) (3) where 𝑞∈[0,1] is an embedding parameter and 𝐻(𝑞) is a nonzero auxiliary function for 𝑞≠0, 𝐻(0)=0. In such a case, (3) is called optimal homotopy equation. Clearly, we have 𝜓()𝜓()()𝜓()(1𝐿𝜓𝜓(𝑞=0⟹𝐻𝑥,𝑡;0,0=𝐿𝑥,𝑡;0+𝑔𝑥,𝑡=0,𝑞=1⟹𝐻𝑥,𝑡;1,1=𝐻)𝑥,𝑡;𝑞+𝑁𝑥,𝑡;𝑞+𝑔𝑥,𝑡)=0, (4)(5) when 𝑞=0 and 𝑞=1, then 𝜓(𝑥,𝑡;0)=𝑢₀(𝑥,𝑡) and 𝜓(𝑥,𝑡;1)=𝑢(𝑥,𝑡) hold. Thus, as 𝑞 varies from 0to1, the solution 𝜓(𝑥,𝑡;𝑞) approaches from 𝑢₀(𝑥,𝑡) to 𝑢(𝑥,𝑡), where 𝑢₀(𝑥,𝑡) is obtained from (3) for 𝑞=0: 𝐿𝑢₀(𝑥,𝑡)+𝑔(𝑥,𝑡)=0,𝐵𝑢₀,𝜕𝑢₀𝜕𝑡=0. (6) Next, we choose auxiliary function 𝐻(𝑞) of the following general form: 𝐻𝑞=𝑞𝑐1+𝑞²𝑐2+⋯. (7)

Here 𝑐1,𝑐2,… are constants to be determined later.

To get an approximate solution, we expand 𝜓(𝑥,𝑡;𝑞,𝑐𝑖) in Taylor’s series about 𝑞 in the following manner, 𝜓𝑥,𝑡;𝑞,𝑐𝑖=𝑢₀(𝑥,𝑡)+^∞_𝑘=1𝑢_𝑘(𝑥,𝑡;𝑐𝑖)𝑞^𝑘,𝑖=1,2,…. (8)

Substituting (8) into (3) and equating the coefficient of like powers of 𝑞, we obtain Zeroth order problem, given by (6); the first and second order problems are given by (9) and (10), respectively, and the general governing equations for 𝑢_𝑘(𝑥,𝑡) are given by (11) as follows: 𝐿𝑢₁(𝑥,𝑡)=𝑐1𝑁₀𝑢₀(𝑥,𝑡),𝐵𝑢₁,𝜕𝑢₁𝐿𝑢𝜕𝑡=0,₂(𝑥,𝑡)−𝐿𝑢₁(𝑥,𝑡)=𝑐2𝑁₀𝑢₀()𝐿𝑢𝑥,𝑡+𝑐1₁(𝑥,𝑡)+𝑁₁𝑢₀(𝑥,𝑡),𝑢₁(),𝐵𝑢𝑥,𝑡₂,𝜕𝑢₂𝐿𝑢𝜕𝑡=0_𝑘(𝑥,𝑡)−𝐿𝑢_𝑘−1(𝑥,𝑡)=𝑐𝑘𝑁₀𝑢₀(𝑥,𝑡)11+^𝑘−1_𝑖=1𝑐𝑖𝐿𝑢_𝑘−𝑖(𝑥,𝑡)+𝑁_𝑘−𝑖𝑢₀(𝑥,𝑡),𝑢₁(𝑥,𝑡),…,𝑢_𝑘−𝑖(),𝐵𝑢𝑥,𝑡_𝑘,𝜕𝑢_𝑘𝜕𝑡=0,𝑘=2,3,…, (9)(10)(11) where 𝑁_𝑘−𝑖(𝑢₀(𝑥,𝑡),𝑢₁(𝑥,𝑡),…,𝑢_𝑘−𝑖(𝑥,𝑡)) is the coefficient of 𝑞^𝑘−𝑖 in the expansion of 𝑁(𝜓(𝑥,𝑡;𝑞)) about the embedding parameter 𝑞: 𝑁𝜓𝑥,𝑡;𝑞,𝑐𝑖=𝑁₀𝑢₀(𝑥,𝑡)+_𝑘≥1𝑁_𝑘𝑢₀,𝑢₁,𝑢₂,…,𝑢_𝑘𝑞^𝑘. (12)

Here 𝑢_𝑘 for 𝑘≥0 are a set of linear equations with the linear boundary conditions, which can be easily solved.

The convergence of the series in (8) depends upon the auxiliary constants 𝑐1,𝑐2,… . If it is convergent at 𝑞=1, then 𝑢(𝑥,𝑡;𝑐𝑖)=𝑢₀(𝑥,𝑡)+_𝑘≥1𝑢_𝑘(𝑥,𝑡;𝑐𝑖). (13)

Substituting (13) into (1), it results with the following expression for residual: 𝑅(𝑥,𝑡;𝑐𝑖)=𝐿(𝑢(𝑥,𝑡;𝑐𝑖))+𝑔(𝑥,𝑡)+𝑁(𝑢(𝑥,𝑡;𝑐𝑖)). (14)

If 𝑅(𝑥,𝑡;𝐶𝑖)=0, then 𝑢(𝑥,𝑡;𝑐𝑖) will be the exact solution.

For computing the optimal values of auxiliary constants, 𝑐𝑖,𝑖=1,2,…,𝑚, there are many methods available like Galerkin’s, Ritz, Least Squares, and Collocation method. One can apply the method of Least Squares as under the following: 𝐽𝐶_𝑖=^𝑡_0Ω𝑅²(𝑥,𝑡,𝑐𝑖)𝑑𝑥𝑑𝑡, (15) where 𝑅 is the residual, 𝑅𝑥,𝑡;𝐶_𝑖=𝐿𝑢𝑥,𝑡;𝐶_𝑖+𝑔(𝑥,𝑡)+𝑁𝑢𝑥,𝑡;𝐶_𝑖,𝜕𝐽=𝜕𝑐1𝜕𝐽𝜕𝑐2=⋯=𝜕𝐽𝜕𝑐3=0. (16)(17)

The constants 𝑐𝑖 can also be determined by another method as under: 𝑅ℎ₁;𝑐𝑖=𝑅ℎ₂;𝑐𝑖=⋯=𝑅ℎ_𝑚;𝑐𝑖=0,𝑖=1,2,…,𝑚 (18) at any time 𝑡, where ℎ_𝑖∈Ω. The convergence depends upon constants 𝑐1,𝑐2,…, which can be optimally identified and minimized by (18).

Example 1. Consider (1) with 𝛼=1, 𝑝=2, and 𝜑(𝑥,𝑡)=−(1/6)𝑒^{−(2𝑥+4𝑡)}𝑓(𝑥,𝑡)=0 which in the simplest form is given as 𝑢_𝑡+16𝑒^{−(2𝑥+4𝑡)}𝑢_𝑥𝑡𝑥−𝑢_𝑥𝑥+𝑢_𝑥+𝑢²𝑢_𝑥=0. (19) The initial condition is 𝑢(𝑥,0)=𝑒^−𝑥 and exact solution given by 𝑢(𝑥,𝑡)=𝑒^{(−𝑥+2𝑡)}. (20)

Zeroth Order Problem. Consider the following: 𝑢_0𝑡=0,𝑢₀(𝑥,0)=𝑒^−𝑥. (21) Its solution is given as 𝑑 under 𝑢₀(𝑥,𝑡)=𝑒^−𝑥. (22)

First Order Problem. Consider the following: −𝑢_0𝑡−𝑐𝑢_0𝑡+𝑢_1𝑡−𝑐1𝑢_0𝑥−𝑐1𝑢²₀𝑢_0𝑥+13𝑐1𝑒^{4𝑡−2𝑥}𝑢_0𝑥𝑡+𝑐1𝑢_0𝑥𝑥+𝑐1𝑒^{4𝑡−2𝑥}𝑢_{0𝑥𝑥𝑡}=0. (23) Its solution is as follows: 𝑢₁(𝑥,𝑡,𝑐1)=−𝑐1𝑒^−3𝑥1+2𝑒^2𝑥𝑡. (24)

Second Order Problem. Consider the following: −𝑐2𝑢_0𝑡−𝑢_1𝑡−𝑐1𝑢_1𝑡+𝑢_2𝑡−𝑐2𝑢_0𝑥−𝑐2𝑢₀²𝑢_0𝑥−2𝑐1𝑢₀𝑢₁𝑢_0𝑥−𝑐1𝑢_1𝑥−𝑐1𝑢₀²𝑢_1𝑥+13𝑐2𝑒^{4𝑡−2𝑥}𝑢_0𝑥,𝑡+13𝑐1𝑒^{4𝑡−2𝑥}𝑢_1𝑥,𝑡=0. (25) Its solution is under 𝑢₂()=1𝑥,𝑡,𝑐1,𝑐28𝑒^−5𝑥5𝑐1²−5𝑐1²𝑒^4𝑡+2𝑐1²𝑒^2𝑥−2𝑐1²𝑒^4𝑡+2𝑥−8𝑐1𝑒^2𝑥𝑡−8𝑐1²𝑒^2𝑥𝑡−8𝑐2𝑒^2𝑥𝑡−16𝑐1𝑒^4𝑥𝑡−16𝑐1²𝑒^4𝑥𝑡−16𝑐2𝑒^4𝑥𝑡+20𝑐1²𝑡²+72𝑐1²𝑒^2𝑥𝑡²+16𝑐1²𝑒^4𝑥𝑡². (26)

Third Order Problem. Consider the following: −𝑐3𝑢_0𝑡−𝑐2𝑢_1𝑡−𝑢_2𝑡−𝑐1𝑢_2𝑡+𝑢_3𝑡−𝑐3𝑢_0𝑥−𝑐3𝑢₀²𝑢_0𝑥−2𝑐2𝑢₀𝑢₁𝑢_0𝑥−𝑐1𝑢₁²𝑢_0𝑥−2𝑐1𝑢₀𝑢₂𝑢_0𝑥−𝑐2𝑢_1𝑥−𝑐2𝑢₀²𝑢_1𝑥−2𝑐1𝑢₀𝑢₁(𝑢₁)_𝑥−𝑐1𝑢_2𝑥−𝑐1𝑢₀²𝑢_2𝑥+13𝑐3𝑒^{4𝑡−2𝑥}𝑢_0𝑥,𝑡+13𝑐2𝑒^{4𝑡−2𝑥}𝑢_1𝑥,𝑡+13𝑐1𝑒^{4𝑡−2𝑥}𝑢_2𝑥,𝑡+𝑐3𝑢_2𝑥,𝑡+𝑐2𝑢_1𝑥,𝑥+𝑐1𝑢_2𝑥,𝑥−16𝑐3𝑒^{4𝑡−2𝑥}𝑢_{0𝑥,𝑥,𝑡}−16𝑐2𝑒^{4𝑡−2𝑥}𝑢_{1𝑥,𝑥,𝑡}−16𝑐1𝑒^{4𝑡−2𝑥}𝑢_{2𝑥,𝑥,𝑡}=0. (27)

Its solution is given as follows: 𝑢₃()=1𝑥,𝑡,𝑐1,𝑐2,𝑐3𝑒96^−7𝑥245𝑐1³−70𝑐1³𝑒^4𝑡−175𝑐1³𝑒^8𝑡+120𝑐1²𝑒^2𝑥−60𝑐1³𝑒^2𝑥+120𝑐1𝑐2𝑒^2𝑥+48𝑐1²𝑒^4𝑥−12𝑐1³𝑒^4𝑥+48𝑐1𝑐2𝑒^4𝑥−120𝑐1²𝑒^4𝑡+2𝑥+90𝑐1³𝑒^4𝑡+2𝑥−120𝑐1𝑐2𝑒^4𝑡+2𝑥−30𝑐1³𝑒^8𝑡+2𝑥−48𝑐1²𝑒^4𝑡+4𝑥+12𝑐1³𝑒^4𝑡+4𝑥−48𝑐1𝑐2𝑒^4𝑡+4𝑥−420𝑐1³𝑡+700𝑐1³𝑒^4𝑡𝑡−1920𝑐1³𝑒^2𝑥𝑡−96𝑐1𝑒^4𝑥𝑡−192𝑐1²𝑒^4𝑥𝑡−384𝑐1³𝑒^4𝑥𝑡−96𝑐2𝑒^4𝑥𝑡−192𝑐1𝑐2𝑒^4𝑥𝑡−96𝑐3𝑒^4𝑥𝑡−192𝑐1𝑒^6𝑥𝑡−384𝑐1²𝑒^6𝑥𝑡−192𝑐1³𝑒^6𝑥𝑡−192𝑐2𝑒^6𝑥𝑡−384𝑐1𝑐2𝑒^6𝑥𝑡−192𝑐3𝑒^6𝑥𝑡+1080𝑐1³𝑒^4𝑡+2𝑥𝑡+48𝑐1³𝑒^4𝑡+4𝑥𝑡+480𝑐1²𝑒^2𝑥𝑡²+480𝑐1³𝑒^2𝑥𝑡²+480𝑐1𝑐2𝑒^2𝑥𝑡²+1728𝑐1²𝑒^4𝑥𝑡²+1728𝑐1³𝑒^4𝑥𝑡²+1728𝑐1𝑐2𝑒^4𝑥𝑡²+384𝑐1²𝑒^6𝑥𝑡²+384𝑐1³𝑒^6𝑥𝑡²+384𝑐1𝑐2𝑒^6𝑥𝑡²−784𝑐1³𝑡³−4480𝑐1³𝑒^2𝑥𝑡³−4032𝑐1³𝑒^4𝑥𝑡³−128𝑐1³𝑒^6𝑥𝑡³. (28) The third order approximate solution is given by the following equation: 𝑢(𝑥,𝑡,𝑐1,𝑐2,𝑐3)=𝑢₀(𝑥,𝑡)+𝑢₁(𝑥,𝑡,𝑐1)+𝑢₂(𝑥,𝑡,𝑐1,𝑐2)+𝑢₃(𝑥,𝑡,𝑐1,𝑐2,𝑐3),𝑢=𝑒^−𝑥−𝑐1𝑒^−3𝑥1+2𝑒^2𝑥𝑡+18𝑒^−5𝑥5𝑐1²−5𝑐1²𝑒^4𝑡+2𝑐1²𝑒^2𝑥−2𝑐1²𝑒^4𝑡+2𝑥−8𝑐1𝑒^2𝑥𝑡−8𝑐1²𝑒^2𝑥𝑡−8𝑐2𝑒^2𝑥𝑡−16𝑐1𝑒^4𝑥𝑡−16𝑐1²𝑒^4𝑥𝑡−16𝑐2𝑒^4𝑥𝑡+20𝑐1²𝑡²+72𝑐1²𝑒^2𝑥𝑡²+16𝑐1²𝑒^4𝑥𝑡²+1𝑒96^−7𝑥245𝑐1³−70𝑐1³𝑒^4𝑡−175𝑐1³𝑒^8𝑡+120𝑐1²𝑒^2𝑥−60𝑐1³𝑒^2𝑥+120𝑐1𝑐2𝑒^2𝑥+48𝑐1²𝑒^4𝑥−12𝑐1³𝑒^4𝑥+48𝑐1𝑐2𝑒^4𝑥−120𝑐1²𝑒^4𝑡+2𝑥+90𝑐1³𝑒^4𝑡+2𝑥−120𝑐1𝑐2𝑒^4𝑡+2𝑥−30𝑐1³𝑒^8𝑡+2𝑥−48𝑐1²𝑒^4𝑡+4𝑥+12𝑐1³𝑒^4𝑡+4𝑥−48𝑐1𝑐2𝑒^4𝑡+4𝑥−420𝑐1³𝑡+700𝑐1³𝑒^4𝑡𝑡−1920𝑐1³𝑒^2𝑥𝑡−96𝑐1𝑒^4𝑥𝑡−192𝑐1²𝑒^4𝑥𝑡−384𝑐1³𝑒^4𝑥𝑡−96𝑐2𝑒^4𝑥𝑡−192𝑐1𝑐2𝑒^4𝑥𝑡−96𝑐3𝑒^4𝑥𝑡−192𝑐1𝑒^6𝑥𝑡−384𝑐1²𝑒^6𝑥𝑡−192𝑐1³𝑒^6𝑥𝑡−192𝑐2𝑒^6𝑥𝑡−384𝑐1𝑐2𝑒^6𝑥𝑡−192𝑐3𝑒^6𝑥𝑡+1080𝑐1³𝑒^4𝑡+2𝑥𝑡+48𝑐1³𝑒^4𝑡+4𝑥𝑡+480𝑐1²𝑒^2𝑥𝑡²+480𝑐1³𝑒^2𝑥𝑡²+480𝑐1𝑐2𝑒^2𝑥𝑡²+1728𝑐1²𝑒^4𝑥𝑡²+1728𝑐1³𝑒^4𝑥𝑡²+1728𝑐1𝑐2𝑒^4𝑥𝑡²+384𝑐1²𝑒^6𝑥𝑡²+384𝑐1³𝑒^6𝑥𝑡²+384𝑐1𝑐2𝑒^6𝑥𝑡²−784𝑐1³𝑡³−4480𝑐1³𝑒^2𝑥𝑡³−4032𝑐1³𝑒^4𝑥𝑡³−128𝑐1³𝑒^6𝑥𝑡³. (29)

The constants 𝑐1, 𝑐2, and 𝑐3 are calculated using the Least Squares, we have their optimal values as follows: 𝑐1=−0.8410486261349961,𝑐2=−0.022653709109618163,𝑐3=0.0031930834039587894. (30)

The 3rd order OHAM solution yields very encouraging results after comparing with 3rd order VIM solution [5]. Tables 1(a–c), and Figures 1, 2, 3, and 4 show the effectiveness of OHAM for 𝑥=1.2, 𝑥=1.4and 𝑥=1.6.

Table 1 

(a) Comparison of absolute errors of 3rd order OHAM solution and 3rd order VIM solution for Example 1 at 𝑥=1.2 and various values of t. (b) Comparison of absolute errors of 3rd order OHAM solution and 3rd order VIM solution for Example 1 at 𝑥=1.4 and various values of t. (c) Comparison of absolute errors of 3rd order OHAM solution and 3rd order VIM solution for Example 1 at 𝑥=1.6 and various values of t.

Figure 1 

Plot of of 3rd order approximate solution for the homogeneous DGRLW equation (2).

Figure 2 

Variation profile of exact solution for the homogeneous DGRLW equation (2).

Figure 3 

Approximate solution plot for 𝑡=0.04.

Figure 4 

Exact solution plot for 𝑡=0.04.

Example 2. Consider (1) with 𝛼=1, 𝑝=2 and 𝜑(𝑥,𝑡)=−(1/6)𝑒^{−(2𝑥+4𝑡)} which in the simplest form is given as 𝑢_𝑡+16𝑒^{−(2𝑥+4𝑡)}𝑢_𝑥𝑡𝑥−𝑢_𝑥𝑥+𝑢_𝑥+𝑢²𝑢_𝑥=0. (31)
The initial condition is 𝑢(𝑥,0)=sech[𝑥/4]² and exact solution given by 𝑢()𝑥𝑥,𝑡=sech4−𝑡3². (32)

Zeroth Order Problem. Consider the following: 𝑢_0𝑡=0,𝑢₀()𝑥𝑥,0=sech4². (33) Its solution is given by 𝑢₀()𝑥𝑥,𝑡=sech4². (34)

First Order Problem. Consider the following: −𝑢_0𝑡−𝑐1𝑢_0𝑡+𝑢_1𝑡−𝑐1𝑢_0𝑥−𝑐1𝑢²₀𝑢_0𝑥+13𝑐1𝑒^{4𝑡−2𝑥}𝑢_0𝑥𝑡+𝑐1𝑢_0𝑥𝑥+𝑐1𝑒^{4𝑡−2𝑥}𝑢_{0𝑥𝑥𝑡}=0. (35) Its solution is 𝑢₁()1𝑥,𝑡,𝑐1=−8𝑡𝑥−𝑐1sech4⁴𝑥+4𝑐1sech4²𝑥tanh4𝑥+4𝑐1sech4⁶𝑥tanh4𝑥+2𝑐1sech4²𝑥tanh4². (36)

Second Order Problem−𝑐2𝑢_0𝑡−𝑢_1𝑡−𝑐1𝑢_1𝑡+𝑢_2𝑡−𝑐2𝑢_0𝑥−𝑐2𝑢²₀𝑢_0𝑥−2𝑐1𝑢₀𝑢₁𝑢_0𝑥−𝑐1𝑢_1𝑥−𝑐1𝑢²₀𝑢_1𝑥+13𝑐2𝑒^{4𝑡−2𝑥}𝑢_0𝑥,𝑡+13𝑐1𝑒^{4𝑡−2𝑥}𝑢_1𝑥,𝑡+𝑐2𝑢_0𝑥,𝑥+𝑐1𝑢_1𝑥,𝑥−16𝑐2𝑒^{4𝑡−2𝑥}𝑢_{0𝑥,𝑥,𝑡}−16𝑐1𝑒^{4𝑡−2𝑥}𝑢_{1𝑥,𝑥,𝑡}=0. (37)

Its approximate solution 𝑢₂(𝑥,𝑡,𝑐1,𝑐2) is obtained in similar manner. The second order approximate solution is given by 𝑢(𝑥,𝑡,𝑐1,𝑐2)=𝑢₀(𝑥,𝑡)+𝑢₁(𝑥,𝑡,𝑐1)+𝑢₂()𝑥𝑥,𝑡,𝑐1,𝑐2=sech4²−18𝑡𝑥−𝑐1sech4⁴𝑥+4𝑐1sech4²𝑥tanh4𝑥+4𝑐1sech4⁶𝑥tanh4𝑥+2𝑐1sech4²𝑥tanh4²+𝑢₂(𝑥,𝑡,𝑐1,𝑐2). (38)

Using method of Least Squares, the optimal values of constants 𝑐1and𝑐2 are computed and are given as under 𝑐1=−0.8122876966282708,𝑐2=−0.011502183336221636. (39)

Table 2(a) shows the effectiveness of OHAM for 𝑥=15, 𝑥=20, and 𝑥=25, while Table 2(b), and Figures 5, 6, 7, and 8 shows the effectiveness of OHAM for various values of 𝑥 and 𝑡.

Table 2 

(a) Absolute error of the solution of Example 2 by optimal homotopy asymptotic method (OHAM) at 𝑥=15, 𝑥=20, and 𝑥=25 and various values of 𝑡. (b) Absolute error of the solution of Example 2 by optimal homotopy asymptotic method (OHAM) at various values of 𝑥 and 𝑡.

Figure 5 

Surface 2nd order approximate solution for (4.12).

Figure 6 

Exact solution for (4.12).

Figure 7 

Approximate solution plot for 𝑡=0.05.

Figure 8 

Exact solution plot for 𝑡=0.05.

Example 3. Consider (1) with 𝛼=1, 𝑝=1, and 𝜑(𝑥,𝑡)=1−(3cosh[𝑥/4−𝑡/3]sinh[𝑥/4 − 𝑡/3]/(−3+2cosh[𝑡/3−𝑥/4]²)) which in the simplest form is given as 𝑢_𝑡+[][]1−3cosh𝑥/4−𝑡/3sinh𝑥/4−𝑡/3−3+2cosh[𝑡/3−𝑥/4]²𝑢_𝑥𝑡𝑥−𝑢_𝑥𝑥+𝑢_𝑥+𝑢𝑢_𝑥=0. (40) The initial condition is 𝑢(𝑥,0)=sech[𝑥/4]² and exact solution given by 𝑢()𝑥𝑥,𝑡=sech4−𝑡3². (41)

Zeroth Order Problem. Consider the following: 𝑢_0𝑡=0,𝑢₀()𝑥𝑥,0=sech4². (42) Its solution is as follows: 𝑢₀()𝑥𝑥,𝑡=sech4². (43)

First Order Problem. Consider the following: −𝑢_0𝑡−𝑐1𝑢_0𝑡+𝑢_1𝑡−𝑐1𝑢_0𝑥−𝑐1𝑢₀𝑢_0𝑥[]−𝑐13cosh𝑡/3−𝑥/4²4−3+2cosh[𝑡/3−𝑥/4]²−3cosh[𝑡/3−𝑥/4]²sinh[𝑡/3−𝑥/4]²−3+2cosh[𝑡/3−𝑥/4]²²+3sinh[𝑡/3−𝑥/4]²4−3+2cosh[𝑡/3−𝑥/4]²𝑢_0𝑥,𝑡+𝑐1𝑢_0𝑥,𝑥[][]−𝑐1−1−3cosh𝑡/3−𝑥/4sinh𝑡/3−𝑥/4−3+2cosh[𝑡/3−𝑥/4]²×𝑢_{0𝑥,𝑥,𝑡}=0. (44)

Its solution is 𝑢₁𝑥,𝑡,𝑐₁1=−8𝑡−𝑐₁𝑥sech4⁴+4𝑐₁𝑥sech4²𝑥tanh4+4𝑐₁𝑥sech4⁴𝑥tanh4+2𝑐₁𝑥sech4²𝑥tanh4². (45)

The first order approximate solution is given by 𝑢𝑥,𝑡,𝑐₁=𝑢₀(𝑥,𝑡)+𝑢₁𝑥,𝑡,𝑐₁𝑢𝑥,𝑡,𝑐₁𝑥=sech4²−18𝑡−𝑐₁𝑥sech4⁴+4𝑐₁𝑥sech4²𝑥tanh4+4𝑐₁𝑥sech4⁴𝑥tanh4+2𝑐₁𝑥sech4²𝑥tanh4². (46)(47) The constants 𝑐1 is calculated using the Least Squares that we have its optimal values as follows: 𝑐₁=−0.802767563787412. (48) The first order optimum solution using OHAM is as follows: 𝑢()𝑥𝑥,𝑡=sech4²−18𝑡𝑥0.802767563787412sech4⁴𝑥−3.211070255149648sech4²𝑥tanh4𝑥−3.211070255149648sech4⁴𝑥tanh4𝑥−1.605535127574824sech4²𝑥tanh4². (49)

The first order OHAM solution yields very encouraging results after comparing with 2nd order VIM solution [5]. Tables 3(a–d), and Figures 9, 10, 11, and 12 show the effectiveness of OHAM for 𝑥=−2.5, 𝑥=0, 𝑥=2.5 and 𝑥=5.

Table 3 

(a) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at 𝑥=−2.5 and various values of t. (b) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at 𝑥=0 and various values of t. (c) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at 𝑥=−2.5 and various values of t. (d) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 3 at 𝑥=−2.5 and various values of t.

Figure 9 

1st order approximate solution for (4.12).

Figure 10 

Exact solution plot for (4.12).

Figure 11 

Approximate solution plot for 𝑡=0.1.

Figure 12 

Exact solution plot for 𝑡=0.1.

Example 4. Let us consider the inhomogeneous DGRLW equation: 𝑢_𝑡−𝜑(𝑥,𝑡)𝑢_𝑥𝑡𝑥−𝑢_𝑥𝑥+𝑢_𝑥+𝑢𝑢_𝑥=𝑓(𝑥,𝑡), (50) where 𝜑(𝑥,𝑡)=𝑥𝑡 and 𝑓(𝑥,𝑡)=(−𝑡cos[𝑥]+𝑥𝑡sin[𝑥]+cos[𝑥]+sin[𝑥]cos[𝑥]𝑒^−𝑡)𝑒^−𝑡.
The initial condition is 𝑢(𝑥,0)=sin[𝑥] (51) and exact solution given by 𝑢(𝑥,𝑡)=sin[𝑥]𝑒^−𝑡. (52)
According to OHAM scheme presented in Section 2.

Zeroth Order Problem. Consider the following: 𝑢_0𝑡=0,𝑢₀(𝑥,0)=sin[𝑥]. (53) Its solution is 𝑢₀(𝑥,𝑡)=sin[𝑥]. (54)

First Order Problem. Consider the following: 𝑐1𝑒^−𝑡cos[𝑥]−𝑡cos[𝑥]+𝑡𝑥sin[𝑥]+𝑒^−𝑡[𝑥][𝑥]−𝑢cossin_0𝑡−𝑐1𝑢_0𝑡+𝑢_1𝑡−𝑐1𝑢_0𝑥−𝑐1𝑢₀𝑢_0𝑥+𝑐1𝑡𝑢_0𝑥,𝑡+𝑐1𝑢_0𝑥,𝑥+𝑐1𝑡𝑥𝑢_{0𝑥,𝑥,𝑡}=0. (55) Its solution is as follows: 𝑢₁()1𝑥,𝑡,𝑐1=−2𝑐1𝑒^−2𝑡2𝑒^𝑡𝑡cos[𝑥]−2𝑒^2𝑡𝑡cos[𝑥]−2𝑒^2𝑡𝑡sin[𝑥]−2𝑒^𝑡𝑥sin[𝑥]+2𝑒^2𝑡𝑥sin[𝑥]−2𝑒^𝑡[𝑥][𝑥][𝑥]𝑡𝑥sin−cossin+𝑒^2𝑡cos[𝑥]sin[𝑥]−2𝑒^2𝑡𝑡cos[𝑥]sin[𝑥]. (56)

Second Order Problem. Consider the following: 𝑐2𝑒^−𝑡[𝑥][𝑥][𝑥]cos−𝑡cos+𝑡𝑥sin+𝑒^−𝑡[𝑥][𝑥]𝑢cossin−𝑐2_0𝑡−𝑢_1𝑡−𝑐1𝑢_1𝑡+𝑢_2𝑡−𝑐2𝑢_0𝑥−𝑐2𝑢₀𝑢_0𝑥−𝑐1𝑢₁𝑢_0𝑥−𝑐1𝑢_1𝑥−𝑐1𝑢₀𝑢_1𝑥+𝑐2𝑡𝑢_0𝑥,𝑡+𝑐1𝑡𝑢_1𝑥,𝑥+𝑐2𝑢_0𝑥,𝑥+𝑐1𝑢_1𝑥,𝑥+𝑐2𝑡𝑥𝑢_{0𝑥,𝑥,𝑡}+𝑐1𝑡𝑥𝑢_{0𝑥,𝑥,𝑡}=0. (57)

Its approximate solution is under 𝑢₂()=1𝑥,𝑡,𝑐1,𝑐2𝑒16^−2𝑡−16𝑐1²𝑒^𝑡+16𝑐1²𝑒^2𝑡−8𝑐1²𝑒^𝑡𝑡−8𝑐1²𝑒^2𝑡𝑡+80𝑐1²𝑒^𝑡cos[𝑥]−80𝑐1²𝑒^2𝑡cos[𝑥]−16𝑐1𝑒^𝑡𝑡cos[𝑥]+32𝑐1²𝑒^𝑡𝑡cos[𝑥]−16𝑐2𝑒^𝑡𝑡cos[𝑥]+16𝑐1𝑒^2𝑡𝑡cos[𝑥]+48𝑐1²𝑒^2𝑡𝑡cos[𝑥]+16𝑐2𝑒^2𝑡𝑡cos[𝑥]+8𝑐1²𝑒^2𝑡𝑡²cos[𝑥]−144𝑐1²𝑒^𝑡𝑥cos[𝑥]+144𝑐1²𝑒^2𝑡𝑥cos[𝑥]−128𝑐1²𝑒^𝑡𝑡𝑥cos[𝑥]−16𝑐1²𝑒^2𝑡𝑡𝑥cos[𝑥]−64𝑐1²𝑒^𝑡𝑡²𝑥cos[𝑥]+8𝑐1²𝑒^2𝑡𝑡²𝑥cos[𝑥]−8𝑐1²cos[2𝑥]+32𝑐1²𝑒^𝑡cos[2𝑥]−24𝑐1²𝑒^2𝑡cos[2𝑥]−8𝑐1²𝑡cos[2𝑥]+24𝑐1²𝑒^𝑡𝑡cos[2𝑥]+8𝑐1²𝑒^2𝑡𝑡²cos[2𝑥]+𝑐1²sin[𝑥]−96𝑐1²𝑒^𝑡sin[𝑥]+95𝑐1²𝑒^2𝑡sin[𝑥]−80𝑐1²𝑒^𝑡𝑡sin[𝑥]+16𝑐1𝑒^2𝑡𝑡sin[𝑥]+2𝑐1²𝑒^2𝑡𝑡sin[𝑥]+16𝑐2𝑒^2𝑡𝑡sin[𝑥]−32𝑐1²𝑒^𝑡𝑡²sin[𝑥]+6𝑐1²𝑒^2𝑡𝑡²sin[𝑥]+16𝑐1𝑒^𝑡𝑥sin[𝑥]−16𝑐1²𝑒^𝑡𝑥sin[𝑥]+16𝑐2𝑒^𝑡𝑥sin[𝑥]−16𝑐1𝑒^2𝑡𝑥sin[𝑥]+16𝑐1²𝑒^2𝑡𝑥sin[𝑥]−16𝑐2𝑒^2𝑡𝑥sin[𝑥]+16𝑐1𝑒^𝑡𝑡𝑥sin[𝑥]+16𝑐2𝑒^𝑡𝑡𝑥sin[𝑥]−16𝑐1²𝑒^2𝑡𝑡𝑥sin[𝑥]+8𝑐1²𝑒^2𝑡𝑡²𝑥sin[𝑥]+32𝑐1²𝑒^𝑡𝑥²sin[𝑥]−32𝑐1²𝑒^2𝑡𝑥²sin[𝑥]+32𝑐1²𝑒^𝑡𝑡𝑥²sin[𝑥]+16𝑐1²𝑒^𝑡𝑡²𝑥²sin[𝑥]+4𝑐1sin[2𝑥]−4𝑐1²sin[2𝑥]+4𝑐2sin[2𝑥]−4𝑐1𝑒^2𝑡sin[2𝑥]+4𝑐1²𝑒^2𝑡sin[2𝑥]−4𝑐2𝑒^2𝑡sin[2𝑥]+8𝑐1𝑒^2𝑡𝑡sin[2𝑥]−8𝑐1²𝑒^2𝑡𝑡sin[2𝑥]+8𝑐2𝑒^2𝑡𝑡sin[2𝑥]+24𝑐1²𝑒^2𝑡𝑡²sin[2𝑥]+8𝑐1²𝑥sin[2𝑥]−32𝑐1²𝑒^𝑡𝑥sin[2𝑥]+24𝑐1²𝑒^2𝑡𝑥sin[2𝑥]+16𝑐1²𝑡𝑥sin[2𝑥]−16𝑐1²𝑒^𝑡𝑡𝑥sin[2𝑥]−16𝑐1²𝑒^2𝑡𝑡𝑥sin[2𝑥]+16𝑐1²𝑒^2𝑡𝑡²𝑥sin[2𝑥]−3𝑐1²sin[3𝑥]+3𝑐1²𝑒^2𝑡sin[3𝑥]−6𝑐1²𝑒^2𝑡𝑡sin[3𝑥]+6𝑐1²𝑒^2𝑡𝑡²sin[3𝑥]. (58) The second order approximate solution is given by the following equation, 𝑢(𝑥,𝑡,𝑐1,𝑐2)=𝑢₀(𝑥,𝑡)+𝑢₁(𝑥,𝑡,𝑐1)+𝑢₂(𝑥,𝑡,𝑐1,𝑐2). (59) Using the method of Least Squares the optimum values of 𝑐1 and 𝑐2 are computed which are as follows: 𝑐1=−1.0433989069917953,𝑐2=−0.0018859250375958814. (60)

The 2nd order OHAM solution yields very encouraging results after comparing with 2nd order VIM solution [5]. Tables 4(a–d), and Figures 13, 14, 15, and 16 show the effectiveness of OHAM for 𝑥=0.2, 𝑥=0.4, 𝑥=0.6, and 𝑥=1.

Table 4 

(a) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at 𝑥=0.2 and various values of t. (b) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at 𝑥=0.4 and various values of t. (c) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at 𝑥=0.6 and various values of t. (d) Comparison of absolute errors of 1st order OHAM solution and 1st order VIM solution for Example 4 at 𝑥=1 and various values of t.

Figure 13 

2nd order approximate solution for (4.12).

Figure 14 

The surface shows Exact solution for (4.12).

Figure 15 

Approximate solution plot for 𝑡=0.1 for (4.12).

Figure 16 

Exact solution plot for 𝑡=0.1 for (4.12).

3. Conclusion

In this paper, the OHAM has been successfully implemented for the approximate solution of solutions of the Nonlinear Damped Generalized Regularized Long-Wave (DGRLW) equations. The results obtained by OHAM are very consistent in comparison with VIM.