Table of Contents
ISRN Mathematical Physics
Volume 2012 (2012), Article ID 837241, 9 pages
Research Article

Symmetry Reduction, Exact Solutions, and Conservation Laws of the ZK-BBM Equation

1Taizhou Institute of Science and Technology, (NUST), Taizhou, Jiangsu 225300, China
2Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu 212013, China
3Department of Mathematics, National Institute of Technology, Jamshedpur, Jharkhand 831014, India

Received 22 April 2012; Accepted 4 July 2012

Academic Editors: A. Aghamohammadi and S. del Campo

Copyright © 2012 Wenbin Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Employing the classical Lie method, we obtain the symmetries of the ZK-BBM equation. Applying the given Lie symmetry, we obtain the similarity reduction, group invariant solution, and new exact solutions. We also obtain the conservation laws of ZK-BBM equation of the corresponding Lie symmetry.

1. Introduction

Symmetry is one of the most important concepts in the area of partial differential equations, especially in integrable systems, which exist infinitely in many symmetries. These symmetry group techniques provide one method for obtaining exact and special solutions of a given partial differential equation. To find the Lie point symmetry of a nonlinear equation, some effective methods have been introduced, such as the classical Lie group approach, the nonclassical Lie group approach, the Clarkson-Kruskal (CK) direct symmetry method, and the compatibility method [1, 2].

The purpose of this paper is to apply the classical Lie method to the following Zakharov-Kuznetsov-Benjamin-Bona-Mahony (ZK-BBM) equation: 𝑢𝑡+𝑢𝑥+𝑎𝑢𝑢𝑥𝑢+𝑏𝑥𝑥𝑡+𝑢𝑦𝑦𝑥=0,(1.1) which was presented by Wazwaz [3, 4]. Wazwaz obtained some solitons solutions, periodic solutions, and complex solutions by using the sine-cosine method and the extended tanh method. Abdou [5] used the extended F-expansion method to construct exact solutions. Song and Yang [6] employed the bifurcation method of dynamical systems to obtain traveling wave solutions.

This paper is arranged as follows. In Section 2, we get the symmetries and group invariant solutions of (1.1). In Section 3, the reductions of (1.1) are obtained, and in Section 4, we derive some new exact solutions of (1.1). Finally, the conservation laws of (1.1) will be presented in Section 5.

2. Symmetry of the ZK-BBM Equation

The main task of the classical Lie method is to seek some symmetries and look for exact solutions of a given partial differential equation. The vector field of (1.1) can be expressed as 𝜕𝑉=𝜉(𝑥,𝑦,𝑡,𝑢)𝜕𝜕𝑥+𝜁(𝑥,𝑦,𝑡,𝑢)𝜕𝜕𝑦+𝜏(𝑥,𝑦,𝑡,𝑢)𝜕𝜕𝑡+𝜂(𝑥,𝑦,𝑡,𝑢),𝜕𝑢(2.1) and its third prolongation can be given as Pr(3)𝑉=𝑉+𝜙𝑡𝜕𝜕𝑢𝑡+𝜙𝑥𝜕𝜕𝑢𝑥+𝜙𝑥𝑥𝑡𝜕𝜕𝑢𝑥𝑥𝑡+𝜙𝑦𝑦𝑥𝜕𝜕𝑢𝑦𝑦𝑥,(2.2) where 𝜙𝑡,𝜙𝑥,𝜙𝑥𝑥𝑡,𝜙𝑦𝑦𝑥 are given explicitly in terms of 𝜉,𝜁,𝜏,𝜙, and the derivatives of 𝜙.

From Pr(3)𝑉𝐹|𝐹=0=0, it follows 𝜙𝑡+𝜙𝑥+𝑎𝑢𝑥𝜙+𝑎𝑢𝜙𝑥+𝑏𝜙𝑥𝑥𝑡+𝑏𝜙𝑦𝑦𝑥=0.(2.3) Letting the coefficients of the polynomial be zero yields a set of differential equations of the functions 𝜉,𝜁,𝜏, and 𝜙. Solving these equations, one can arrive at 𝜉=𝑐31,𝜁=2𝑐1𝑦+𝑐4,𝜏=𝑐1𝑡+𝑐2,𝜙=𝑐1𝑎𝑢+1𝑎,(2.4) where 𝑐1,𝑐2,𝑐3, and 𝑐4 are arbitrary constants.

The corresponding symmetries are 𝜎=𝑐3𝑢𝑥+12𝑐1𝑦+𝑐4𝑢𝑦+𝑐1𝑡+𝑐2𝑢𝑡𝑐1𝑎𝑢+1𝑎𝑢.(2.5) The vector field can be written as 𝜈=𝑐3𝜕+1𝜕𝑥2𝑐1𝑦+𝑐4𝜕+𝑐𝜕𝑦1𝑡+𝑐2𝜕𝜕𝑡𝑐1𝑎𝑢+1𝑎𝜕.𝜕𝑢(2.6) In order to get some exact solutions from the known ones of (1.1), we will find the corresponding Lie symmetry groups. To this end, solving the following initial problems: 𝑑𝑑𝜖𝑥,𝑦,𝑡,𝑢=(𝑥,𝑦,𝑡,𝑢),𝑥,𝑦,𝑡,𝑢|𝜖=0=(𝑥,𝑦,𝑡,𝑢),(2.7) where 𝜖 is a parameter.

From (2.7) we can obtain the Lie symmetry group 𝑔(𝑥,𝑦,𝑡,𝑢)(𝑥,𝑦,𝑡,𝑢). According to different 𝜉,𝜁,𝜏, and 𝜙 in 𝜎, we have the following group by solving (2.7): 𝑔1𝑥,𝑦,𝑡,𝑢𝑔(𝑥+𝜖,𝑦,𝑡,𝑢),2𝑥,𝑦,𝑡,𝑢𝑔(𝑥,𝑦+𝜖,𝑡,𝑢),3𝑥,𝑦,𝑡,𝑢𝑔(𝑥,𝑦,𝑡+𝜖,𝑢),4𝑥,𝑦,𝑡,𝑢𝑥,𝑦+𝑒(1/2)𝜖,𝑡+𝑒𝜖,𝑢+𝑒𝜖1𝑎.(2.8) We can obtain the corresponding new solutions by applying the above groups 𝑔1,𝑔2,𝑔3, and 𝑔4 as follows: 𝑢1𝑢=𝑓(𝑥𝜖,𝑦,𝑡),2𝑢=𝑓(𝑥,𝑦𝜖,𝑡),3𝑢=𝑓(𝑥,𝑦,𝑡𝜖),4=𝑓𝑥,𝑦𝑒(1/2)𝜖,𝑡𝑒𝜖1𝑎+𝑒𝜖.(2.9) For example, taking the following periodic wave solution [3] of (1.1) 𝑢(𝑥,𝑦,𝑡)=3(𝑐1)2𝑎sec212𝑏𝜉,(2.10) where 𝑏>0 and 𝜉=𝑥+𝑦𝑐𝑡.

One can obtain new exact solution of (1.1) by applying 𝑢4 as follows: 𝑢(𝑥,𝑦,𝑡)=3(𝑐1)2𝑎sec212𝑏𝜉1𝑎+𝑒𝜖,(2.11) where 𝑏>0 and 𝜉=𝑥+𝑦𝑒(1/2)𝜖𝑐(𝑡𝑒𝜖).

3. Reductions of ZK-BBM Equation

Now we will reduce (1.1) by means of the symmetries (2.4). Here we discuss the following cases.

Case 1. Letting 𝑐10,𝑐2=0,𝑐3=0,𝑐4=0. Then 1𝜎=2𝑐1𝑦𝑢𝑦+𝑐1𝑡𝑢𝑡𝑐1𝑎𝑢+1𝑎𝑢.(3.1) Solving the differential equation 𝜎=0, one can get 𝑦𝜉=𝑥,𝜂=2𝑡1,𝑢=𝑡1𝑓(𝜉,𝜂)𝑎.(3.2) Substituting (3.2) into (1.1), we have 𝑓𝜂𝑓𝜂+𝑓𝑓𝜉+𝑏𝑓𝜉𝜉𝜂𝑓𝜉𝜉𝜂+2𝑓𝜂𝜉+4𝜂𝑓𝜂𝜂𝜉=0.(3.3)

Case 2. Letting 𝑐1=0,𝑐20,𝑐30,𝑐40. Then 𝜎=𝑐3𝑢𝑥+𝑐4𝑢𝑦+𝑐2𝑢𝑡.(3.4) Solving the differential equation 𝜎=0, one can get 𝑐𝜉=𝑥3𝑐2𝑐𝑡,𝜂=𝑦4𝑐2𝑡,𝑢=𝑓(𝜉,𝜂).(3.5) Substituting (3.5) into (1.1), one can reduce (1.1) to the following equation: 𝑐13𝑐2𝑓𝜉𝑐4𝑐2𝑓𝜂+𝑎𝑓𝑓𝜉𝑐+𝑏3𝑐2𝑓𝜉𝜉𝜉𝑐4𝑐2𝑓𝜉𝜉𝜂+𝑓𝜂𝜂𝜉=0.(3.6)

Case 3. Letting 𝑐1=0,𝑐20,𝑐3=0,𝑐40. Then 𝜎=𝑐4𝑢𝑦+𝑐2𝑢𝑡.(3.7) Solving the differential equation 𝜎=0, one can get 𝑐𝜉=𝑥,𝜂=𝑦4𝑐2𝑡,𝑢=𝑓(𝜉,𝜂).(3.8) Substituting (3.8) into (1.1), we have 𝑓𝜉𝑐4𝑐2𝑓𝜂+𝑎𝑓𝑓𝜉𝑐+𝑏4𝑐2𝑓𝜉𝜉𝜂+𝑓𝜂𝜂𝜉=0.(3.9)

4. Similarity Solutions of ZK-BBM Equation

By solving the reduced equations (3.3), (3.6), and (3.9), we can get some new explicit solutions of (1.1). Now we discuss some cases in the following.

Case 1. Assuming (3.3) has the following solution: 𝑓=𝑟(𝜂)𝜉+𝜔(𝜂),(4.1) where 𝑟(𝜂) and 𝜔(𝜂) are the functions to be determined. Substituting (4.1) into (3.3) yields 1𝑓=1+𝑐1𝜂𝜉+𝜔(𝜂),(4.2) then the new exact solution of (1.1) is expressed as 𝑢1=1𝑎𝑡+𝑑1𝑦2𝑥+2𝑏𝑑1ln𝑡+𝑑1𝑦24𝑏𝑑1ln𝑦+𝑑28𝑏𝑡𝑑1𝑎𝑡+𝑑1𝑦221𝑎,(4.3) where 𝑑1 and 𝑑2 are constants.

Case 2. To solve (3.6), we apply the 𝐺/𝐺-expansion method [7] and look for the travelling wave solution of (3.6) by setting 𝑓=𝑁𝑖=1𝛼𝑖𝐺(𝜔)𝐺(𝜔)𝑖+𝛼0,𝛼𝑚0,(4.4) where 𝜔=𝑘𝜉+𝑙𝜂,𝛼𝑖 are constants to be determined later. It can be seen 𝑚=2 by balancing 𝑓𝜉 and 𝑓𝑓𝜉 in (3.4). Suppose that the solutions of (3.4) are of the form 𝑓=𝛼2𝐺(𝜔)𝐺(𝜔)2+𝛼1𝐺(𝜔)𝐺(𝜔)+𝛼0,𝛼20(4.5) with 𝐺(𝜔) satisfying the second-order linear ordinary differential equation 𝐺(𝜔)+𝜆𝐺(𝜔)+𝜇𝐺(𝜔)=0,(4.6) where 𝜔=𝑘𝜉+𝑙𝜂, 𝛼0,𝛼1,𝛼2, 𝜆, and 𝜇 are constants to be determined later. Substituting (4.5) into (3.6) along with (4.6) and setting the coefficients of (𝐺(𝜔)/𝐺(𝜔))𝑖,(𝑖=0,,4) to zero yields a system of equations with respect to 𝛼0,𝛼1,𝛼2,, and 𝑙. Solving the corresponding algebraic equations and using (3.5), (4.5), and (4.6), we get three types of traveling wave solutions of (1.1) as follows.

When 𝜆24𝜇>0, 𝑢2=3𝑏𝑎𝑐2𝜆2𝑐4𝜇3𝑘+𝑐4𝑙𝑐2𝑙2𝑘𝑑1sin𝜔+𝑑2cos𝜔𝑑1cos𝜔+𝑑2+sin𝜔2𝑏𝑎𝑐2𝜆24𝜇𝑐3𝑘𝑐4𝑙+𝑐2𝑙2𝑘+1𝑎𝑐2𝑐4𝑙𝑘𝑐2+𝑐3.(4.7)

When 𝜆24𝜇<0, 𝑢3=3𝑏𝑎𝑐24𝜇𝜆2𝑐3𝑘+𝑐4𝑙𝑐2𝑙2𝑘𝑑1sin𝜔+𝑑2cos𝜔𝑑1cos𝜔+𝑑2+sin𝜔2𝑏𝑎𝑐24𝜇𝜆2𝑐3𝑘𝑐4𝑙+𝑐2𝑙2𝑘+1𝑎𝑐2𝑐4𝑙𝑘𝑐2+𝑐3.(4.8)

When 𝜆24𝜇=0, 𝑢4=12𝑏𝑎𝑐2𝑐3𝑘+𝑐4𝑙𝑐2𝑙2𝑘𝑑1sin𝜔+𝑑2cos𝜔𝑑1cos𝜔+𝑑2+1sin𝜔𝑎𝑐2𝑐4𝑙𝑘𝑐2+𝑐3,(4.9) where 𝜔=𝑘𝑥+𝑙𝑦((𝑘𝑐3+𝑙𝑐4)/𝑐2)𝑡.

Case 3. Let 𝑓=𝑓(𝜔), where 𝜔=𝑘𝜉+𝑙𝜂. Equation (3.9) becomes the following ordinary partial differential equation: 𝑐𝑘4𝑐2𝑙𝑓+𝑎𝑘𝑓𝑓𝑐+𝑏4𝑐2𝑘2𝑙+𝑘𝑙2𝑓=0.(4.10)
Integrating twice with respect to 𝜔 in (4.10), we get 𝑓2=𝐴1𝑓3+𝐴2𝑓2+𝐴3𝑓,(4.11) where 𝐴1=𝑐2𝑎/3𝑏𝑙(𝑐4𝑘𝑐2𝑙),  𝐴2=(𝑐2𝑘𝑐4𝑙)/𝑏𝑘𝑙(𝑐4𝑘𝑐2𝑙), and 𝐴3 is a constant.

Since solutions of (4.11) have been given [8], we have some similarity solutions of (1.1) as follows.

When 𝐴1=4𝑚2,𝐴2=4(𝑚2+1),𝐴3=4, 𝑢5=𝑠𝑛2𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.12)

When 𝐴1=4𝑚2,𝐴2=4(2𝑚21),𝐴3=4(1𝑚2), 𝑢6=𝑐𝑛2𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.13)

When 𝐴1=4,𝐴2=4(2𝑚2),𝐴3=4(1𝑚2), 𝑢7=𝑑𝑛2𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.14)

When 𝐴1𝑓<0,𝐴2>0,𝐴3=0, 𝑢8𝐵=2𝐵1sech2𝐵22𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.15)

When 𝐴1𝑓>0,𝐴2>0,𝐴3=0, 𝑢9=𝐵2𝐵1csch2𝐵22𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.16)

When 𝐴2<0,𝐴3=0, 𝑢10=𝐵2𝐵1sec2𝐵22𝑐𝑘𝑥+𝑙𝑦4𝑐2𝑙𝑡.(4.17)

5. Conservation Laws of ZK-BBM Equation

In this section we will study the conservation laws by using the adjoint equation and symmetries of ZK-BBM equation. For (1.1), the adjoint equation has the form 𝑣𝑡+𝑣𝑥+𝑎𝑢𝑣𝑥+𝑏𝑣𝑥𝑥𝑡+𝑏𝑣𝑦𝑦𝑥=0,(5.1) and the Lagrangian is 𝑢𝐿=𝑣𝑡+𝑢𝑥+𝑎𝑢𝑢𝑥+𝑏𝑢𝑥𝑥𝑣𝑡+𝑏𝑢𝑦𝑦𝑣𝑥.(5.2)

Since every Lie point, Lie-backlund and nonlocal symmetry of (1.1), provides a conservation law for (1.1) and the adjoint equation [9], the elements of conservation vector (𝐶1,𝐶2,𝐶3) are defined by the following expression: 𝐶𝑖=𝜉𝑖𝐿+𝑊𝛼𝜕𝐿𝜕𝑢𝛼𝑖𝐷𝑗𝜕𝐿𝜕𝑢𝛼𝑖𝑗+𝐷𝑗𝐷𝑘𝜕𝐿𝜕𝑢𝛼𝑖𝑗𝑘+𝐷𝑗(𝑊𝛼)𝜕𝐿𝜕𝑢𝛼𝑖𝑗𝐷𝑘𝜕𝐿𝜕𝑢𝛼𝑖𝑗𝑘+𝐷𝑗𝐷𝑘(𝑊𝛼)𝜕𝐿𝜕𝑢𝛼𝑖𝑗𝑘(𝑖=1,2,3),(5.3) where 𝑊𝛼=𝜂𝛼𝜉𝑗𝑢𝛼𝑗.

The conserved vector corresponding to an operator 𝑣=𝜉1𝜕(𝑥,𝑦,𝑡,𝑢)𝜕𝑥+𝜉2𝜕(𝑥,𝑦,𝑡,𝑢)𝜕𝑦+𝜉3𝜕(𝑥,𝑦,𝑡,𝑢)𝜕𝜕𝑡+𝜂(𝑥,𝑦,𝑡,𝑢).𝜕𝑢(5.4) The operator 𝑣 yields the conservation law 𝐷𝑡𝐶1+𝐷𝑥𝐶2+𝐷𝑦𝐶3=0,(5.5) where the conserved vector 𝐶=(𝐶1,𝐶2,𝐶3) is given by (5.3) and has the components 𝐶1=𝑤1𝑣(1+𝑎𝑢)𝐷𝑥𝑏𝑣𝑣𝑡+𝑘𝑢𝑦𝑦𝑣𝑤2+𝑏𝑣𝑣𝑡𝐷𝑥𝑤1,𝐶2=𝜉2𝐿𝐷𝑦𝑘𝑣𝑣𝑥𝑤1+𝑏𝑣𝑣𝑥𝐷𝑦𝑤1,𝐶3=𝜉3𝐿+𝑤1+𝑏𝑢𝑥𝑥𝑣𝑤2.(5.6)

Thus, (5.6) define components of a nonlocal conservation law for the system of (1.1), (5.1) corresponding to any operator 𝑣 admitted by (1.1).

Let us make more detailed calculations for the operator 𝑣=(1/2)𝑦𝜕𝑦+𝑡𝜕𝑡((𝑎𝑢+1)/𝑎)𝜕𝑢. For this operator, we have 𝑤1=(𝑎𝑢+1)/𝑎((1/2)𝑦)𝑢𝑦𝑡𝑢𝑡, 𝑤2=(1/2)𝑣((1/2)𝑦)𝑣𝑦𝑡𝑣𝑡 and (5.3) written for the sixth-order Lagrangian equation (5.2) yields the conserved vector 𝐶11=𝑢+𝑎+12𝑦𝑢𝑦+𝑡𝑢𝑡𝑣+𝑎𝑢𝑣𝑏𝑣𝑥𝑣𝑡𝑏𝑣𝑣𝑡𝑥+𝑏𝑢𝑦𝑦𝑣121𝑣2𝑦𝑣𝑦𝑡𝑣𝑡𝑏𝑣𝑣𝑡𝑢𝑥+12𝑦𝑢𝑦𝑥+𝑡𝑢𝑡𝑥,𝐶2=12𝑢𝑦𝑣𝑡+𝑢𝑥+𝑎𝑢𝑢𝑥+𝑏𝑢𝑥𝑥𝑣𝑡+𝑏𝑢𝑦𝑦𝑣𝑥𝑣𝑏𝑦𝑣𝑥+𝑣𝑣𝑥𝑦1𝑢+𝑎+12𝑦𝑢𝑦+𝑡𝑢𝑡𝑏𝑣𝑣𝑥𝑢𝑦+12𝑢𝑦+12𝑦𝑢𝑦𝑦+𝑡𝑢𝑡𝑦,𝐶3𝑢=𝑡𝑣𝑡+𝑢𝑥+𝑎𝑢𝑢𝑥+𝑏𝑢𝑥𝑥𝑣𝑡+𝑏𝑢𝑦𝑦𝑣𝑥1𝑢+𝑎+12𝑦𝑢𝑦+𝑡𝑢𝑡+𝑏𝑢𝑥𝑥𝑣121𝑣2𝑦𝑣𝑦𝑡𝑣𝑡.(5.7)


This work was supported by the Startup Fund for Advanced Talents of Jiangsu University (no. 09JDG013), the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (no. 09KJB110003), the National Nature Science Foundation of China (no. 71073072), the Nature Science Foundation of Jiangsu (no. BK 2010329), the Project of Excellent Discipline Construction of Jiangsu Province of China, the Taizhou Social Development project (no. 2011213), and the Priority Academic Program Development of Jiangsu Higher Education Institutions.


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