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Traveling wave solutions of Zakharov–Kuznetsov-modified equal-width and Burger’s equations via $$ {\text{exp}}( - \varphi \left( \eta \right)) $$ ...

Traveling wave solutions of Zakharov–Kuznetsov-modified equal-width and Burger’s equations via $$... syedtauseefs@hotmail.com Department of Mathematics, In this article, a technique is proposed for obtaining better and accurate results for Faculty of Sciences, HITEC nonlinear PDEs. We constructed abundant exact solutions via exp(−ϕ(η))‑ expansion University Taxila, Taxila, method for the Zakharov–Kuznetsov‑ modified equal ‑ width (ZK‑ MEW ) equation and Pakistan the (2 + 1)‑ dimensional Burgers equation. The traveling wave solutions are found through the hyperbolic functions, the trigonometric functions and the rational func‑ tions. The specified idea is very pragmatic for PDEs, and could be extended to engi‑ neering problems. Keywords: Exp(−ϕ(η))‑ expansion method, Nonlinear evolution equation, (ZK‑ MEW ) equation, Burger’s equation, Solitary wave solutions Background Over the past few decades, researchers have shown keen interest in the solutions of nonlinear partial differential equations (PDEs).In the study of nonlinear physical phe - nomena, the investigation of solitary wave solutions [1–44] of nonlinear wave equa- tions shows an important role. Scientific problems arise nonlinearly in numerous fields of mathematical physics, such as fluid mechanics, plasma physics, solid-state physics and geochemistry. Due to exact interpretation of nonlinear phenomena, these problems have gained much importance. However, in recent years, a variety of effective analyti - cal methods has been developed to study soliton solutions of nonlinear equations, such as Backlund transformation method [1], tanh method [2–6], extended tanh method [7– 12], pseudo-spectral method [13], trial function [14], sine–cosine method [15], Hirota method [16], exp function method [17–25], (G /G)-expansion method [26–30], homo- geneous balance method [31, 32], F-expansion method [33–35] and Jacobi elliptic func- tion expansion method [36–38]. Ma et al. [39–44] established the complexiton solutions for Toda lattice equation. The theme of the method is that the exact solutions of nonlin - ear evolution equations can be articulated by exp(−ϕ(η)), where ϕ(η) gratifies the ordi - nary differential equation (ODE): ϕ (η) = exp(−ϕ(η)) + μ exp(ϕ(η)) + (1) where η = x − Vt . © 2015 Mohyud‑ Din et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 2 of 10 (−ϕ(η)) Explanation of exp ‑expansion method Now, the exp(−ϕ(η))-expansion method will be explained for constructing traveling wave solutions. Consider the general nonlinear partial differential equation for u(x, t) is given by, φ(u, u , u , u , u , u , . . .) = 0, t x tt xx xxx (2) where u(η) = u(x, t), φ is a polynomial of u and its derivatives. Solving (2), the following steps are as. Step 1 We Combine the variables by η, u = u(η), η = x − Vt , (3) where V is the speed of wave. Using Eqs. (3, 2) reduced to the following ODE for u = u(η) ′ ′′ ′′′ ′′′′ G u, u , u , u , u , . . . = 0, (4) Step 2 The solution of Eq. (4) can be articulated as u(η) = a (exp(−ϕ(η))) , n (5) n=0 where a 0 ≤ n ≤ M are constants such that a �= 0 and ϕ(η) satisfies Eq. (1). Our solu - n n tions now depend on the parameters involved in (1). Family 1: When  − 4μ> 0, we have       � �  �  − 4μ � �  2    ϕ(η) = ln −  − 4μ tanh (η + c ) −  . (6)  2μ 2  Family 2: When  − 4μ< 0, we have      � � � 2   − 4μ � �     ϕ(η) = ln  − 4μ tan (η + c ) −  . (7)  2μ 2  Family 3: When  − 4 μ> 0 μ = 0 and  �= 0, ϕ(η) = −ln . (8) exp ((η + k)) − 1 Family 4: When  − 4 μ = 0,  �= 0, and μ �= 0, 2((η + k) + 2) ϕ(η) = ln . (9) (η + k) Family 5: When  − 4 μ = 0,  = 0, and μ = 0, ϕ(η) = ln(η + k) (10) Step 3 By considering the homogenous principal, in Eq.  (4). Considering Eqs.  (1, 4, 5), Mϕ(η) we have e . We get algebraic equations with a , V , ,μ, after comparing the same n Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 3 of 10 ϕ(η) powers of e to zero. We put the above values in Eq. (5) and with Eq. (1), we get some valuable traveling wave solutions of Eq. (2). Solution procedure Zakharov–Kuznetsov‑modified equal‑width equation Consider the equation, u + α u + βu + δu = 0, t xt yy (11) x x where α, β and δ are some nonzero parameters. We use u = u(η), η = x + y − Vt , we can convert Eq. (11) into an ODE. ′ ′′′ ′′′ ′ −Vu − βVu + δu + 2αuu = 0, (12) where the dash denotes the derivative w. r. t. η. Now integrating Eq. (12), we have, ′′ ′′ 2 −Vu − βVu + δu + αu + C = 0, (13) ′′ 2 Using homogenous principle, balancing u and u , we have 2M = M + 2, M = 2. The trial solution of Eq. (12) can be stated as, u(η) = a (exp(−ϕ(η))) + a (exp(−ϕ(η))) + a , (14) 2 1 0 where a �= 0, a and a are constants, while ,μ are any constants. 2 1 0 ′ ′′ 2 Putting u, u , u , u in Eq. (13) and comparing, we get, 2 2 2 αa + δa μ + C − 2βVa μ − βVa μ + 2δa μ − Va = 0, 1 2 1 2 0 2 2 2αa a + δa  + 2δa μ + −2βVa μ − 6βV μ − βVa  + 6δa μ − Va = 0, 0 1 1 1 1 1 2 1 2αa a + 10δa  + 2δa + −2βVa − 10βVa  = 0, 2 1 2 1 1 2 2αa a + 10a  + 2a + −2βVa − 10βVa  = 0, 2 1 2 1 1 2 (15) αa + 6δa − 6βVa = 0, 2 2 By solving the algebraic equations, the required solution is given below. 1 αa + 6δ 1 1 V = ,  = 0, a = a , a = 0, μ = 2 βα 6Cβ + 6αβa − αa a − 6a δ , 0 0 1 0 2 0 6 β 2 βαa where  and μ are any constants. Now putting the values in Eq. (14), we obtain −2ϕ(η) u = a + a e , (16) 0 2 where η = x − Vt . By putting (6–10) in (16), we obtain the solutions which are given below. Case 1 When  − 4μ> 0 and μ �= 0, we have, 4a μ u (η) = a + , 1 0 −4μ −  − 4μtanh (η + c ) − 2 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 4 of 10 1 αa +6δ where and where c is any constant. η = x − t 6 β Case 2 When  − 4μ< 0 and μ �= 0, we have, 4a μ u (η) = a + , 2 0 − +4μ − + 4μtan (η + c ) − αa +6δ 1 2 where η = x − t and where c is any constant. 6 β Case 3 When μ = 0 and  �= 0, we have, u (η) = a + , 3 0 exp (η + c ) − 1 1 αa +6δ where η = x − t and where c is any constant. 6 β Case 4 When  − 4μ = 0,  �= 0, and μ �= 0, we obtain, 2 4 a (η + c ) 2 1 u (η) = a + , 4 0 2(η + c ) + 2 1 αa +6δ where η = x − t and where c is any constant. 6 β a αa +6δ 2 1 2 Case 5 When  = 0, and μ = 0, we have, u η = a + , where η = x − t ( ) 5 0 6 β (η+c ) and where c is any constant. Graphical demonstration The graphs are given in Figs. 1, 2, 3, 4 and 5. (2 + 1)‑dimensional Burger’s equation Consider the equation, u − uu − u − u = 0, t x xx yy (17) Fig. 1 Kink wave solution of u when a = 1, a = 2, y = 0,  = 3, μ = 2, c = 1 1 2 0 1 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 5 of 10 Fig. 2 Singular kink wave solution u when a = 10, a = 8, y = 0,  = 7, μ = 5, c = −10 2 2 0 1 Fig. 3 Singular kink wave solution u when a = 1, a = 2, y = 0,  = 1, c =−1 3 2 0 1 Fig. 4 Singular kink wave solution u when a = 3, a = 2, y = 0,  = 5,μ = 4, c = −2 4 2 0 1 where α, β and δ are some nonzero parameters. We have, u = u(η), η = x + y − Vt , we can convert Eq. (17) into an ODE. ′ ′′ ′ −Vu − 2u − uu = 0, (18) where dash denotes the derivative w. r. t.η. Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 6 of 10 Fig. 5 Singular kink wave solution u when a = 0.5, a = 0.2, y = 0,  = 0.1, c = −0.1 5 2 0 1 Integrating Eq. (18), we have, ′ 2 −Vu − 2u − u + C = 0, (19) ′ 2 u u , M = 1 Using homogenous principle, balancing the and we have, . The trial solution of Eq. (18) can be stated as, u(η) = a (exp(−ϕ(η))) + a , 1 0 (20) ′ ′′ 2 where a �= 0, a is a constant, while ,μ are any constants. By putting u, u , u , u in 1 0 Eq. (19) and comparing, we get − a + 2a μ + C − Va = 0, 1 0 − a a + 2a  − Va = 0, 0 1 1 1 (21) − a + 2a = 0, By solving the algebraic equations, the required solution is given below. 2 2 = V + 2C + 16μ, a =−V + V + 2C + 16μ, a = 4, 0 1 where  and μ are any constants. Now putting the values in Eq. (20), we obtain, −ϕ(η) u =−V + V + 2C + 16μ + 4e , (22) where η = x − Vt . Now putting (6–10) in (22), we obtain the solutions as. Case 1 When  − 4μ> 0 and μ �= 0, we have, 8μ u (η) =−1 + 1 + 2C + 16μ + √ , −4μ −  − 4μtanh (η + c ) − 2 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 7 of 10 where η = x − Vt and where c is any constant. Case 2 When  − 4μ< 0 and μ �= 0, we obtain, 8μ u (η) == −1+ 1 + 2C + 16μ+ , − +4μ + − + 4μtanh (η + c ) − where η = x − Vt and where c is any constant. Case 3 When μ = 0 and  �= 0, we have, u (η) === −1 + 1 + 2C + 16μ +   , η + c − 1 ( ) where η = x − Vt and where c is any constant. Case 4 When  − 4μ = 0,  �= 0, and μ �= 0, we obtain, 4(η + c ) u (η) =−1 + 1 + 2C + 16μ + , 2(η + c ) + 2 where η = x − Vt and where c is any constant. Case 5 When  = 0, and μ = 0, we have, u (η) =−1 + 1 + 2C + 16μ + , (η + c ) where η = x − Vt and where c is any constant. Graphical illustration The graphs are given in Figs. 6, 7, 8, 9 and 10. Fig. 6 Kink wave solution u when C = 1, a = 1, y = 0,  = 3,μ = 1, c = 1 6 0 1 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 8 of 10 Fig. 7 Periodic solution u (η) when a = 2, C = 1, y = 0,  = 1,μ = 2, c = −1 7 2 1 Fig. 8 Singular kink wave solution u when μ = 1, C = 1, y = 0,  = 3, c =−1 8 1 Fig. 9 Singular kink wave solution u when a = 1, C = 1, y = 0,  = 13,μ = 1, c = −1 9 2 1 Conclusions The exp (−ϕ(η))-expansion method has been successfully applied to find the exact solu - tions of (ZK-MEW) equation and the Burger’s equation. The attained results show that Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 9 of 10 Fig. 10 Singular kink wave solution u when C = 15, y = 0, μ = 12, c = −1 the proposed technique is effective and capable for solving nonlinear partial differential equations. In this study, some exact solitary wave solutions, mostly solitons and kink solutions, are obtained through the hyperbolic and rational functions. This study shows that the proposed method is quite proficient and practically well organized in finding exact solutions of other physical problems. Authors’ contributions The work was carried out in cooperation among all the authors (STM‑D, AA and MAI). All authors have a good involve ‑ ment to plan the paper, and to execute the analysis of this research work together. All authors read and approved the final manuscript. Compliance with ethical guidelines Competing interests The authors declare that they have no competing interests. Received: 1 June 2015 Accepted: 20 September 2015 References 1. Ablowitz MJ, Clarkson PA (1991) Solitons, nonlinear evolution equations and inverse scattering. Cambridge Univer‑ sity Press, New York 2. Wazwaz AM (2004) The tanh‑method for traveling wave solutions of nonlinear equations. Appl Math Comput 154:713–723 3. Malfliet W, Hereman W (1996) The tanh method: exact solutions of nonlinear evolution and wave equations. Phys Scr 54:563–568 4. Wazwaz AM (2007) The tanh‑method for traveling wave solutions of nonlinear wave equations. Appl Math Comput 187:1131–1142 5. 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Traveling wave solutions of Zakharov–Kuznetsov-modified equal-width and Burger’s equations via $$ {\text{exp}}( - \varphi \left( \eta \right)) $$ ...

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Springer Journals
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Copyright © 2015 by Mohyud-Din et al.
Subject
Engineering; Theoretical and Applied Mechanics; Computational Science and Engineering; Classical Continuum Physics; Mathematical Applications in the Physical Sciences
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Abstract

syedtauseefs@hotmail.com Department of Mathematics, In this article, a technique is proposed for obtaining better and accurate results for Faculty of Sciences, HITEC nonlinear PDEs. We constructed abundant exact solutions via exp(−ϕ(η))‑ expansion University Taxila, Taxila, method for the Zakharov–Kuznetsov‑ modified equal ‑ width (ZK‑ MEW ) equation and Pakistan the (2 + 1)‑ dimensional Burgers equation. The traveling wave solutions are found through the hyperbolic functions, the trigonometric functions and the rational func‑ tions. The specified idea is very pragmatic for PDEs, and could be extended to engi‑ neering problems. Keywords: Exp(−ϕ(η))‑ expansion method, Nonlinear evolution equation, (ZK‑ MEW ) equation, Burger’s equation, Solitary wave solutions Background Over the past few decades, researchers have shown keen interest in the solutions of nonlinear partial differential equations (PDEs).In the study of nonlinear physical phe - nomena, the investigation of solitary wave solutions [1–44] of nonlinear wave equa- tions shows an important role. Scientific problems arise nonlinearly in numerous fields of mathematical physics, such as fluid mechanics, plasma physics, solid-state physics and geochemistry. Due to exact interpretation of nonlinear phenomena, these problems have gained much importance. However, in recent years, a variety of effective analyti - cal methods has been developed to study soliton solutions of nonlinear equations, such as Backlund transformation method [1], tanh method [2–6], extended tanh method [7– 12], pseudo-spectral method [13], trial function [14], sine–cosine method [15], Hirota method [16], exp function method [17–25], (G /G)-expansion method [26–30], homo- geneous balance method [31, 32], F-expansion method [33–35] and Jacobi elliptic func- tion expansion method [36–38]. Ma et al. [39–44] established the complexiton solutions for Toda lattice equation. The theme of the method is that the exact solutions of nonlin - ear evolution equations can be articulated by exp(−ϕ(η)), where ϕ(η) gratifies the ordi - nary differential equation (ODE): ϕ (η) = exp(−ϕ(η)) + μ exp(ϕ(η)) + (1) where η = x − Vt . © 2015 Mohyud‑ Din et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 2 of 10 (−ϕ(η)) Explanation of exp ‑expansion method Now, the exp(−ϕ(η))-expansion method will be explained for constructing traveling wave solutions. Consider the general nonlinear partial differential equation for u(x, t) is given by, φ(u, u , u , u , u , u , . . .) = 0, t x tt xx xxx (2) where u(η) = u(x, t), φ is a polynomial of u and its derivatives. Solving (2), the following steps are as. Step 1 We Combine the variables by η, u = u(η), η = x − Vt , (3) where V is the speed of wave. Using Eqs. (3, 2) reduced to the following ODE for u = u(η) ′ ′′ ′′′ ′′′′ G u, u , u , u , u , . . . = 0, (4) Step 2 The solution of Eq. (4) can be articulated as u(η) = a (exp(−ϕ(η))) , n (5) n=0 where a 0 ≤ n ≤ M are constants such that a �= 0 and ϕ(η) satisfies Eq. (1). Our solu - n n tions now depend on the parameters involved in (1). Family 1: When  − 4μ> 0, we have       � �  �  − 4μ � �  2    ϕ(η) = ln −  − 4μ tanh (η + c ) −  . (6)  2μ 2  Family 2: When  − 4μ< 0, we have      � � � 2   − 4μ � �     ϕ(η) = ln  − 4μ tan (η + c ) −  . (7)  2μ 2  Family 3: When  − 4 μ> 0 μ = 0 and  �= 0, ϕ(η) = −ln . (8) exp ((η + k)) − 1 Family 4: When  − 4 μ = 0,  �= 0, and μ �= 0, 2((η + k) + 2) ϕ(η) = ln . (9) (η + k) Family 5: When  − 4 μ = 0,  = 0, and μ = 0, ϕ(η) = ln(η + k) (10) Step 3 By considering the homogenous principal, in Eq.  (4). Considering Eqs.  (1, 4, 5), Mϕ(η) we have e . We get algebraic equations with a , V , ,μ, after comparing the same n Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 3 of 10 ϕ(η) powers of e to zero. We put the above values in Eq. (5) and with Eq. (1), we get some valuable traveling wave solutions of Eq. (2). Solution procedure Zakharov–Kuznetsov‑modified equal‑width equation Consider the equation, u + α u + βu + δu = 0, t xt yy (11) x x where α, β and δ are some nonzero parameters. We use u = u(η), η = x + y − Vt , we can convert Eq. (11) into an ODE. ′ ′′′ ′′′ ′ −Vu − βVu + δu + 2αuu = 0, (12) where the dash denotes the derivative w. r. t. η. Now integrating Eq. (12), we have, ′′ ′′ 2 −Vu − βVu + δu + αu + C = 0, (13) ′′ 2 Using homogenous principle, balancing u and u , we have 2M = M + 2, M = 2. The trial solution of Eq. (12) can be stated as, u(η) = a (exp(−ϕ(η))) + a (exp(−ϕ(η))) + a , (14) 2 1 0 where a �= 0, a and a are constants, while ,μ are any constants. 2 1 0 ′ ′′ 2 Putting u, u , u , u in Eq. (13) and comparing, we get, 2 2 2 αa + δa μ + C − 2βVa μ − βVa μ + 2δa μ − Va = 0, 1 2 1 2 0 2 2 2αa a + δa  + 2δa μ + −2βVa μ − 6βV μ − βVa  + 6δa μ − Va = 0, 0 1 1 1 1 1 2 1 2αa a + 10δa  + 2δa + −2βVa − 10βVa  = 0, 2 1 2 1 1 2 2αa a + 10a  + 2a + −2βVa − 10βVa  = 0, 2 1 2 1 1 2 (15) αa + 6δa − 6βVa = 0, 2 2 By solving the algebraic equations, the required solution is given below. 1 αa + 6δ 1 1 V = ,  = 0, a = a , a = 0, μ = 2 βα 6Cβ + 6αβa − αa a − 6a δ , 0 0 1 0 2 0 6 β 2 βαa where  and μ are any constants. Now putting the values in Eq. (14), we obtain −2ϕ(η) u = a + a e , (16) 0 2 where η = x − Vt . By putting (6–10) in (16), we obtain the solutions which are given below. Case 1 When  − 4μ> 0 and μ �= 0, we have, 4a μ u (η) = a + , 1 0 −4μ −  − 4μtanh (η + c ) − 2 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 4 of 10 1 αa +6δ where and where c is any constant. η = x − t 6 β Case 2 When  − 4μ< 0 and μ �= 0, we have, 4a μ u (η) = a + , 2 0 − +4μ − + 4μtan (η + c ) − αa +6δ 1 2 where η = x − t and where c is any constant. 6 β Case 3 When μ = 0 and  �= 0, we have, u (η) = a + , 3 0 exp (η + c ) − 1 1 αa +6δ where η = x − t and where c is any constant. 6 β Case 4 When  − 4μ = 0,  �= 0, and μ �= 0, we obtain, 2 4 a (η + c ) 2 1 u (η) = a + , 4 0 2(η + c ) + 2 1 αa +6δ where η = x − t and where c is any constant. 6 β a αa +6δ 2 1 2 Case 5 When  = 0, and μ = 0, we have, u η = a + , where η = x − t ( ) 5 0 6 β (η+c ) and where c is any constant. Graphical demonstration The graphs are given in Figs. 1, 2, 3, 4 and 5. (2 + 1)‑dimensional Burger’s equation Consider the equation, u − uu − u − u = 0, t x xx yy (17) Fig. 1 Kink wave solution of u when a = 1, a = 2, y = 0,  = 3, μ = 2, c = 1 1 2 0 1 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 5 of 10 Fig. 2 Singular kink wave solution u when a = 10, a = 8, y = 0,  = 7, μ = 5, c = −10 2 2 0 1 Fig. 3 Singular kink wave solution u when a = 1, a = 2, y = 0,  = 1, c =−1 3 2 0 1 Fig. 4 Singular kink wave solution u when a = 3, a = 2, y = 0,  = 5,μ = 4, c = −2 4 2 0 1 where α, β and δ are some nonzero parameters. We have, u = u(η), η = x + y − Vt , we can convert Eq. (17) into an ODE. ′ ′′ ′ −Vu − 2u − uu = 0, (18) where dash denotes the derivative w. r. t.η. Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 6 of 10 Fig. 5 Singular kink wave solution u when a = 0.5, a = 0.2, y = 0,  = 0.1, c = −0.1 5 2 0 1 Integrating Eq. (18), we have, ′ 2 −Vu − 2u − u + C = 0, (19) ′ 2 u u , M = 1 Using homogenous principle, balancing the and we have, . The trial solution of Eq. (18) can be stated as, u(η) = a (exp(−ϕ(η))) + a , 1 0 (20) ′ ′′ 2 where a �= 0, a is a constant, while ,μ are any constants. By putting u, u , u , u in 1 0 Eq. (19) and comparing, we get − a + 2a μ + C − Va = 0, 1 0 − a a + 2a  − Va = 0, 0 1 1 1 (21) − a + 2a = 0, By solving the algebraic equations, the required solution is given below. 2 2 = V + 2C + 16μ, a =−V + V + 2C + 16μ, a = 4, 0 1 where  and μ are any constants. Now putting the values in Eq. (20), we obtain, −ϕ(η) u =−V + V + 2C + 16μ + 4e , (22) where η = x − Vt . Now putting (6–10) in (22), we obtain the solutions as. Case 1 When  − 4μ> 0 and μ �= 0, we have, 8μ u (η) =−1 + 1 + 2C + 16μ + √ , −4μ −  − 4μtanh (η + c ) − 2 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 7 of 10 where η = x − Vt and where c is any constant. Case 2 When  − 4μ< 0 and μ �= 0, we obtain, 8μ u (η) == −1+ 1 + 2C + 16μ+ , − +4μ + − + 4μtanh (η + c ) − where η = x − Vt and where c is any constant. Case 3 When μ = 0 and  �= 0, we have, u (η) === −1 + 1 + 2C + 16μ +   , η + c − 1 ( ) where η = x − Vt and where c is any constant. Case 4 When  − 4μ = 0,  �= 0, and μ �= 0, we obtain, 4(η + c ) u (η) =−1 + 1 + 2C + 16μ + , 2(η + c ) + 2 where η = x − Vt and where c is any constant. Case 5 When  = 0, and μ = 0, we have, u (η) =−1 + 1 + 2C + 16μ + , (η + c ) where η = x − Vt and where c is any constant. Graphical illustration The graphs are given in Figs. 6, 7, 8, 9 and 10. Fig. 6 Kink wave solution u when C = 1, a = 1, y = 0,  = 3,μ = 1, c = 1 6 0 1 Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 8 of 10 Fig. 7 Periodic solution u (η) when a = 2, C = 1, y = 0,  = 1,μ = 2, c = −1 7 2 1 Fig. 8 Singular kink wave solution u when μ = 1, C = 1, y = 0,  = 3, c =−1 8 1 Fig. 9 Singular kink wave solution u when a = 1, C = 1, y = 0,  = 13,μ = 1, c = −1 9 2 1 Conclusions The exp (−ϕ(η))-expansion method has been successfully applied to find the exact solu - tions of (ZK-MEW) equation and the Burger’s equation. The attained results show that Mohyud‑Din et al. Asia Pac. J. Comput. Engin. (2015) 2:2 Page 9 of 10 Fig. 10 Singular kink wave solution u when C = 15, y = 0, μ = 12, c = −1 the proposed technique is effective and capable for solving nonlinear partial differential equations. In this study, some exact solitary wave solutions, mostly solitons and kink solutions, are obtained through the hyperbolic and rational functions. This study shows that the proposed method is quite proficient and practically well organized in finding exact solutions of other physical problems. Authors’ contributions The work was carried out in cooperation among all the authors (STM‑D, AA and MAI). All authors have a good involve ‑ ment to plan the paper, and to execute the analysis of this research work together. All authors read and approved the final manuscript. Compliance with ethical guidelines Competing interests The authors declare that they have no competing interests. Received: 1 June 2015 Accepted: 20 September 2015 References 1. Ablowitz MJ, Clarkson PA (1991) Solitons, nonlinear evolution equations and inverse scattering. Cambridge Univer‑ sity Press, New York 2. Wazwaz AM (2004) The tanh‑method for traveling wave solutions of nonlinear equations. Appl Math Comput 154:713–723 3. Malfliet W, Hereman W (1996) The tanh method: exact solutions of nonlinear evolution and wave equations. Phys Scr 54:563–568 4. Wazwaz AM (2007) The tanh‑method for traveling wave solutions of nonlinear wave equations. Appl Math Comput 187:1131–1142 5. 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Asia Pacific Journal on Computational Engineering – Springer Journals

Published: Oct 7, 2015

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