Full Paper Maejo International Journal of Science and Technology ISSN 905-7873 Available online at www.mijst.mju.ac.th New eact travelling wave solutions of generalised sinh- ordon and ( + )-dimensional ZK-BBM equations Rajesh Kumar upta, Sachin Kumar, * and Bhajan Lal School of Mathematics and Computer Applications, Thapar University, Patiala-47004, Punjab, India Department of Applied Sciences, Bahra Faculty of Engineering, Technical University, Patalia- 4700, Punjab, India * Corresponding authors, e-mail: sachinjan@yahoo.com (S. Kumar) rajeshgupta@thapar.edu (R. K. upta) Received: August 0 / Accepted: 5 September 0 / Published: October 0 Abstract: Eact travelling wave solutions have been established for generalised sinh-ordon and generalised (+) dimensional ZK-BBM equations by using epansion method where ( ) satisfies a second-order linear ordinary differential equation. The travelling wave solutions are epressed by hyperbolic, trigonometric and rational functions. Keywords: epansion method, travelling wave solutions, generalised sinh-ordon equation, (+) dimensional ZK-BBM equation INTRODUCTION Non-linear partial differential equations (PDEs) are widely used as models to describe comple physical phenomena in various fields of sciences, especially fluid mechanics, solid state physics, plasma physics, plasma wave and chemical physics. Particularly, various methods have been utilised to eplore different kinds of solutions of physical models described by non-linear PDEs. One of the basic physical problems for these models is to obtain their eact solutions. In recent years various methods for obtaining eact travelling wave solutions to non-linear equations have been presented, such as the homogeneous balance method [], the tanh function method [, 3], the Jacobi elliptic function method [4, 5] and the F-epansion method [6, 7].
345 In this paper, we use epansion method [8, 9] to establish eact travelling wave solutions for generalised sinh-ordon and generalised (+) dimensional ZK-BBM equations. The main idea of this method is that the travelling wave solutions of non-linear equations can be epressed by a polynomial in, where ( ) satisfies the second-order linear ordinary differential equation (LODE): ( ) ( ) ( ) 0, where ct and, and c are arbitrary constants. The degree of this polynomial can be determined by considering the homogeneous balance between higher-order derivatives and the non-linear term appearing in the given non-linear equations. The coefficients of this polynomial can be obtained by solving a set of algebraic equations resulting from the process of using the proposed method. The sinh-ordon equation, viz. ut sinh( u), () appears in many branches of non-linear science and plays an important role in physics [0], being an important model equation studied by several authors [ 4]. Wazwaz [4] studied the generalised sinh-ordon equation given by u au bsinh( nu) 0, () tt where a and b are arbitrary constants and n is a positive integer. By using the tanh method, Wazwaz derived eact travelling wave solutions of Eq.(), which provides a more powerful model than the Eq.(). The generalised form of the ( + ) dimensional ZK-BBM equation is given as: u u a u bu ku (3) t ( ) ( t yt ) 0, where a, b and k are arbitrary constants. It arises as a description of gravity water waves in the long-wave regime [5, 6]. A variety of eact solutions for the ( + ) dimensional ZK-BBM equation [7, 8] are developed by means of the tanh and the sine-cosine methods. In this paper we construct new travelling wave solutions of Eqs.() and (3). ( /)-EXPANSION METHOD In this section we describe the epansion method [8, 9, 9] to find the travelling wave solutions of non-linear PDEs. Suppose that a non-linear equation with two independent variables and t is given by P( u, u, ut, u, ut, utt,...) 0, (4) where u u(, t) is an unknown function, P is a polynomial in u u(, t) and its various partial derivatives, in which the highest order derivatives and non-linear terms are involved. In the following the main steps of the epansion method are given. Step. Combining the independent variables and t into one variable Vt,
346 we suppose that u (, t ) u ( ), Vt. (5) The travelling wave variables (5) permits us to reduce Eq.(4) to an ordinary differential equation (ODE): P( u, Vu, u, V u, Vu, u,...) 0. (6) Step. Suppose that the solution of the ODE (6) can be epressed by a polynomial in follows: as m u( ) m ( )..., m 0, (7) where ( ) satisfies the second-order LODE in the following form: 0. (8) The constants m, m... and are to be determined later. The unwritten part in Eq.(7) is also a polynomial in, the degree of which is, however, generally equal to or less than m-. The positive integer m can be determined by considering the homogenous balance between the highest order derivatives and the non-linear terms appearing in ODE (6). Step 3. By substituting (7) into Eq.(6) and using second-order LODE (8), the left-hand side of Eq.(6) is converted into another polynomial in. Equating each coefficient of this polynomial to zero yields a set of algebraic equations for,..., V,,. m Step 4. The constants,..., V,, can be obtained by solving the system of algebraic equations m obtained in Step 3. Since the general solutions of the second-order LODE (8) is well known depending on the sign of the discriminant 4, the eact solutions of the given Eq.(4) can be obtained. ENERALISED SINH-ORDON EQUATION To find the eplicit eact solutions of the sinh-ordon Eq.(), we proceed with the methodology eplained in the above section. First we make the transformation: u(, t) u( ) u( ct), (9) where c is the wave speed. Substituting the above travelling wave transformation into (), we get the following ODE: du bsinh( nu) 0, (0) where d c a and b is arbitrary constant. Appling the transformation using relations v e nu to Eq.(0) and
347 v v v v v v v v v sinh( nu), cosh( nu), u arccos h, u n nv in Eq.(0), we have 3 d( v v v ) bnv bnv 0. () Suppose that the solution of ODE () can be epressed by a polynomial in as follows: m v( ) m ( )..., () where ( ) satisfies the second-order LODE in the form: 0. (3) From Eqs.() and (3), we have v v v ( )... ( )... ( ) ( )... 3 3 3m m m m m m m m m Considering the homogeneous balance between m. So we can write () as: v v and (4) 3 v in Eq.(), it is required that v( ) ( ) ( ) 0; 0. (5) By using Eq.(5) and (3) in Eq.() and collecting all terms with the same power of together, the left-hand side of Eq.() is converted into another polynomial in. Equating each coefficient of this polynomial to zero yields a set of simultaneous algebraic equations for,, 0 and d as follows: 4d bn 0 3 8 d( ) 3bn d( 0 ) d 0 bn ( ) d(8 3 4 ) d( 0 ) d 0 0 0 d( 4( ) ( ) ) 0 bn(4 ( )) d (6 ) d (8 3 4 ) 0 0 d ( 0 ) d (4 ( )( )) 0 0 bn( ( ) ) bn d( ) d(6 ) 0 0 0 0 d(8 3 4 ) d( ( ) ( ) ) 0 0
348 3bn bn d( ) d(6 ) 4 d ( ) 0 0 0 d ( ) bn d bn 0. 3 0 0 0 Solving the algebraic equations above yields where 4 4 6n 0,,, d, (7) 4 4 4 4, and n are arbitrary constants. By using (7), epression (5) can be written as: (6) 4 4 v( ) ( ) ( ), (8) 4 4 4 6n where t a 4. Substituting the general solution of Eq.(3) into Eq.(8), we have two types of travelling wave solution of Eq.() as follows: Case : When 4 0, 4 4 u(, t) log ( ) ( ), (9) n 4 4 4 where C 4 4 C cosh C sinh 4 4 sinh C cosh 4 and 6n t a 4. C and C are arbitrary constants. Profile of solution (9) for n a C, C 0 and is shown in Figure.
349 u t Figure. Profile of solution (9) when n a C, C 0 and Case : When 4 0, 4 4 u(, t) log ( ) ( ), (0) n 4 4 4 where C 4 4 C cos C sin 4 4 sin C cos 4 and 6n t a 4. C and C are arbitrary constants. In Figure we have periodic solution (0) when n a C C and.
350 u t Figure. 3D plot of periodic solution (0) when n a C C and (+) DIMENSIONAL ZK-BBM EQUATION In finding eact solutions of ZK-BBM Eq.(3), first we make the transformation: u( ) u( ), y ct, () which permits us to convert Eq.(3) into an ODE for u u( ) as follows: u ( c) auu cu ( b k) 0, () where a, b and k are arbitrary parameters and c is the wave speed. Integrating it with respect to once yields: V u( c) au cu ( b k) 0, where V is an integration constant that is to be determined later. Proceeding in a similar manner as in the above section and considering the homogeneous balance between u and u in Eq.(3), we have m. So we can assume the solution of Eq.(3) as follows: (3) u( ) ( ) ( ) 0; 0, (4) where 0, and are to be determined later. By using (4) and (3) in Eq.(3) and collecting all terms with the same power of together, the left-hand side of Eq.(3) is converted into another polynomial in. Equating each
35 coefficient of this polynomial to zero yields a set of simultaneous algebraic equations for c and V as follows: a bc ( b k) 0 c( 0 )( b k) a 0 ( c) c(8 3 4 )( b k) a( ) 0 0 ( c) c( b 4 )( b k) a 0 0 V ( c) a c( )( b k) 0. 0 0 (5) Solving the algebraic equations above yields bc( b k) bc( b k) b c 8bc k c c 8kc,, 0, a a a (c 8b c 8k c 6b c k 6c k 6c k 3bc k V 4 4 4 bkc b c k c c ) 4a where,, a, b and k are arbitrary constants. By using (6), epression (4) can be written as:, (6) where bc( b k) bc( b k) b c 8bc k c c 8kc u( ) ( ) ( ) (7) a a a,,b, c and k are arbitrary constants. Substituting the general solution of Eq.(3) into Eq.(7), we have the following travelling wave solutions of Eq. (3): Case. When 4 0, bc( b k) bc( b k) b c 8bc k c c 8kc u( ) ( ) ( ), (8) a a a where C 4 4 sinh C cosh 4 4 4 C cosh sinh C and y ct. C and C are arbitrary constants. The solution (8) furnishes a bell-shaped soliton solution (8) when a k c C, b, t 0., C 0 and, as shown in Figure 3.
35 u y Figure 3. Bell-shaped soliton solution (8) when a k c C, b, t 0., C 0 and Case. When 4 0, bc( b k) bc( b k) b c 8bc k c c 8kc u( ) ( ) ( ), (9) a a a where ( ) = ( ) C sin( 4 ) C cos( 4 ) 4 C cos( 4 ) sin( 4 ) C are arbitrary constants. For a =.5, k = c = C =, b =, t = 0., = and y ct. C and C = 0 and, solution (9) gives the periodic solution as shown in Figure 4.
353 u y Figure 4. Periodic solution (9) when a.5, k c C, b, t 0., C 0 and Case 3. When 4 0, bc( b k) C C C bc( b k) C C C u( ) a ( C C ) a ( C C ) b c 8bc k c c 8kc, a where y ct. C and C are arbitrary constants. In Figure 5, we have a soliton solution (30) for a.5, k c C C, b, t 0., and. 4 (30) u y Figure 5. Soliton solution (30) for a.5, k c C C, b, t 0., and 4
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