Exact Solutions of 𝜙^𝟒 Equation Using Lie Symmetry Approach along with the Simplest Equation and Exp-Function Methods

Abstract

This paper obtains the exact solutions of the 𝜙⁴ equation. The Lie symmetry approach along with the simplest equation method and the Exp-function method are used to obtain these solutions. As a simplest equation we have used the equation of Riccati in the simplest equation method. Exact solutions obtained are travelling wave solutions.

1. Introduction

The research area of nonlinear equations has been very active for the past few decades. There are several kinds of nonlinear equations that appear in various areas of physics and mathematical sciences. Much effort has been made on the construction of exact solutions of nonlinear equations as they play an important role in many scientific areas, such as, in the study of nonlinear physical phenomena [1, 2]. Nonlinear wave phenomena appear in various scientific and engineering fields, such as fluid mechanics, plasma physics, optical fiber, biology, oceanology [3], solid state physics, chemical physics, and geometry. In recent years, many powerful and efficient methods to find analytic solutions of nonlinear equation have drawn a lot of interest by a diverse group of scientists. These methods include, the tanh-function method, the extended tanh-function method [2, 4, 5], the sine-cosine method [6], and the (𝐺^′/𝐺)-expansion method [7, 8].

In this paper, we study the 𝜑⁴ equation, namely, 𝜙_𝑡𝑡−𝜙_𝑥𝑥−𝜙+𝜙³=0. (1.1) The purpose of this paper is to use the Lie symmetry method along with the simplest equation method (SEM) and the Exp-function method to obtain exact solutions of the 𝜑⁴ equation. The simplest equation method was developed by Kudryashov [9–12] on the basis of a procedure analogous to the first step of the test for the Painlevéproperty. The Exp-function method is a very powerful method for solving nonlinear equations. This method was introduced by He and Wu [13] and since its appearance in the literature it has been applied by many researchers for solving nonlinear partial differential equations. See for example, [14, 15].

The outline of this paper is as follows. In Section 2 we discuss the methodology of Lie symmetry analysis and obtain the Lie point symmetries of the 𝜑⁴ equation. We then use the translation symmetries to reduce this equation to an ordinary differential equation (ODE). In Section 3 we describe the SEM and then we obtain the exact solutions of the reduced ODE using SEM. In Section 4 we explain the basic idea of the Exp-function method and obtain exact solutions of the reduced ODE using the Exp-function method. Concluding remarks are summarized in Section 5.

2. Lie Symmetry Analysis

We recall that a Lie point symmetry of a partial differential equation (PDE) is an invertible transformation of the independent and dependent variables that keep the equation invariant. In general determining all the symmetries of a partial differential equation is a daunting task. However, Sophus Lie (1842–1899) noticed that if we confine ourselves to symmetries that depend continuously on a small parameter and that form a group (continuous one-parameter group of transformations), one can linearize the symmetry condition and end up with an algorithm for calculating continuous symmetries [16–19].

The symmetry group of (1.1) will be generated by the vector field of the form 𝜕𝑋=𝜏𝑡,𝑥,𝜙𝜕𝜕𝑡+𝜉𝑡,𝑥,𝜙𝜕𝜕𝑥+𝜂𝑡,𝑥,𝜙.𝜕𝜙 (2.1) Applying the second prolongation 𝑋^[2] to (1.1) we obtain 𝑋^[2]𝜙_𝑡𝑡−𝜙_𝑥𝑥−𝜙+𝜙³_(1.1)=0, (2.2) where 𝑋^[2]=𝑋+𝜁₁𝜕𝜕𝜙_𝑡+𝜁₂𝜕𝜕𝜙_𝑥+𝜁₁₁𝜕𝜕𝜙_𝑡𝑡+𝜁₁₂𝜕𝜕𝜙_𝑡𝑥+𝜁₂₂𝜕𝜕𝜙_𝑥𝑥,𝜁₁=𝐷_𝑡𝜂−𝜙_𝑡𝐷_𝑡(𝜏)−𝜙_𝑥𝐷_𝑡(𝜉),𝜁₂=𝐷_𝑥𝜂−𝜙_𝑡𝐷_𝑥(𝜏)−𝜙_𝑥𝐷_𝑥(𝜉),𝜁₁₁=𝐷_𝑡𝜁₁−𝜙_𝑡𝑡𝐷_𝑡(𝜏)−𝜙_𝑡𝑥𝐷_𝑡(𝜉),𝜁₁₂=𝐷_𝑥𝜁₁−𝜙_𝑡𝑡𝐷_𝑥(𝜏)−𝜙_𝑡𝑥𝐷_𝑥(𝜉),𝜁₂₂=𝐷_𝑥𝜁₂−𝜙_𝑡𝑥𝐷_𝑡(𝜏)−𝜙_𝑥𝑥𝐷_𝑡(𝜉),𝐷_𝑡=𝜕𝜕𝑡+𝜙_𝑡𝜕𝜕𝜙+𝜙_𝑡𝑥𝜕𝜕𝜙_𝑥+𝜙_𝑡𝑡𝜕𝜕𝜙_𝑡𝐷+…,_𝑥=𝜕𝜕𝑥+𝜙_𝑥𝜕𝜕𝜙+𝜙_𝑥𝑥𝜕𝜕𝜙_𝑥+𝜙_𝑡𝑥𝜕𝜕𝜙_𝑡+…. (2.3) Expanding the (2.2) we obtain the following overdetermined system of linear partial differential equations: 𝜂−𝜂_𝑡𝑡+𝜂_𝑥𝑥=0,𝜂_𝑢−2𝜏_𝑡=0,2𝜂_𝑡𝑢−𝜏_𝑡𝑡+𝜏_𝑥𝑥=0,𝜏_𝑡−𝜉_𝑥𝜂=0,_𝑢𝑢−2𝜏_𝑡𝑢=0,𝜏_𝑢=0,𝜉_𝑡𝑡+2𝜂_𝑥𝑢−𝜉_𝑥𝑥=0,𝜏_𝑢𝑢=0,𝜉_𝑡−𝜏_𝑥𝜉=0,_𝑡𝑢−𝜏_𝑥𝑢𝜉=0,_𝑢𝑢𝜉=0,_𝑢𝜂=0,_𝑢𝑢−2𝜉_𝑥𝑢=0. (2.4)

Solving the above system we obtain the following infinitesimal generators: 𝑋₁=𝜕𝜕𝑡,𝑋₂=𝜕𝜕𝑥,𝑋₃𝜕=𝑥𝜕𝜕𝑡+𝑡.𝜕𝑥 (2.5) We now use a linear combination of the translation symmetries 𝑋₁ and 𝑋₂, namely, 𝑋=𝑋₁+𝑐𝑋₂ and reduce (1.1) to an ordinary differential equation. The symmetry 𝑋 yields the following two invariants: 𝜒=𝑥−𝑐𝑡,𝑢=𝜙, (2.6) which gives a group invariant solution 𝑢=𝑢(𝜒) and consequently using these invariants (1.1) is transformed into the second-order nonlinear ODE 𝑐²−1𝑢^′′−𝑢+𝑢³=0. (2.7)

3. Solution of (2.7) Using the Simplest Equation Method

We now use the simplest equation method to solve (2.7). The simplest equation that will be used is the Ricatti equation 𝐺^′𝜒=𝑏𝐺𝜒+𝑑𝐺𝜒², (3.1) where 𝑏 and 𝑑 are arbitrary constants. This equation is a well-known nonlinear ordinary differential equation which possess exact solutions given by elementary functions. The solutions can be expressed as 𝐺𝜒=𝑏exp𝑏𝜒+𝐶𝑏,1−𝑑exp𝜒+𝐶 (3.2) for the case when 𝑑<0, 𝑏>0, and 𝐺𝜒𝑏=−𝑏exp𝜒+𝐶𝑏,1+𝑑exp𝜒+𝐶 (3.3) for 𝑑>0, 𝑏<0. Here 𝐶 is a constant of integration.

Let us consider the solution of (2.7) of the form 𝑢𝜒=^𝑀_𝑖=0𝐴_𝑖𝐺𝜒^𝑖, (3.4) where 𝐺(𝜒) satisfies the Riccati equation (3.1), 𝑀 is a positive integer that can be determined by balancing procedure, and 𝐴₀,𝐴₁,𝐴₂,…,𝐴_𝑀 are parameters to be determined.

The balancing procedure yields 𝑀=1, so the solution of (2.7) is of form 𝑢𝜒=𝐴₀+𝐴₁𝐺𝜒. (3.5)

3.1. Solution of (2.7) When 𝑑<0 and 𝑏>0

Substituting (3.5) into (2.7) and making use of the Ricatti equation (3.1) and then equating all coefficients of the functions 𝐺^𝑖 to zero, we obtain an algebraic system of equations in terms of 𝐴₀ and 𝐴₁. Solving these algebraic equations, with the aid of Mathematica, we obtain the following values of 𝐴₀ and 𝐴₁.

Case 1. 𝐴₀=−1, 𝐴₁=−𝑏𝑑+𝑏𝑐²𝑑, 𝑏=±√2/√1−𝑐², 1−𝑐²≠0.

Case 2. 𝐴₀=1, 𝐴₁=𝑏𝑑−𝑏𝑐²𝑑, 𝑏=±√2/√1−𝑐², 1−𝑐²≠0.
Therefore, when 𝑑<0, 𝑏>0 the solution of (2.7) and hence the solution of (1.1) for Case 1 is given by 𝜙₁()𝑏𝑥,𝑡=−1+²𝑑𝑐²−1exp[𝑏(𝑥−𝑐𝑡+𝐶)][𝑏(,1−𝑑exp𝑥−𝑐𝑡+𝐶)] (3.6) and the solution of (1.1) for Case 2 is given by 𝜙₂()𝑏𝑥,𝑡=1−²𝑑𝑐²−1exp[𝑏(𝑥−𝑐𝑡+𝐶)][𝑏(.1−𝑑exp𝑥−𝑐𝑡+𝐶)] (3.7)

3.2. Solution of (2.7) When 𝑑>0 and 𝑏<0

If 𝑑>0, 𝑏<0, substituting (3.5) into (2.7) and making use of (3.1) and then proceeding as above, we obtain the following values of 𝐴₀ and 𝐴₁.

Case 3. 𝐴₀=−1, 𝐴₁=−𝑏𝑑+𝑏𝑐²𝑑, 𝑏=±√2/√1−𝑐², 1−𝑐²≠0.

Case 4. 𝐴₀=1, 𝐴₁=𝑏𝑑−𝑏𝑐²𝑑, 𝑏=±√2/√1−𝑐², 1−𝑐²≠0.
Therefore, when 𝑑>0, 𝑐<0 the solution of (2.7) and hence the solution of (1.1) for Case 3 is given by 𝜙₃()𝑏𝑥,𝑡=−1−²𝑑𝑐²−1exp[𝑏(𝑥−𝑐𝑡+𝐶)][𝑏(,1+𝑑exp𝑥−𝑐𝑡+𝐶)] (3.8) and the solution of (1.1) for Case 4 is given by 𝜙₄()𝑏𝑥,𝑡=1+²𝑑𝑐²−1exp[𝑏(𝑥−𝑐𝑡+𝐶)][𝑏(.1+𝑑exp𝑥−𝑐𝑡+𝐶)] (3.9)

4. Solution of (2.7) Using the Exp-Function Method

In this section we use the Exp-function method for solving (2.7). According to the Exp-function method [13–15], we consider solutions of (2.7) in the form 𝑢𝜒=∑^𝑑_𝑛=−𝑏𝑎_𝑛exp𝑛𝜒∑^𝑞_𝑚=−𝑝𝑏_𝑚,exp𝑚𝜒 (4.1) where 𝑏, 𝑑, 𝑝, and 𝑞 are positive integers which are unknown to be further determined, 𝑎_𝑛 and 𝑏_𝑚 are unknown constants. By the balancing procedure of the Exp-function method, we obtain 𝑝=𝑏 and 𝑞=𝑑. Furthermore, for simplicity, we set 𝑝=𝑏=1 and 𝑞=𝑑=1, so (4.1) reduces to 𝑢𝜒=𝑎₋₁exp−𝜒+𝑎₀+𝑎₁exp𝜒𝑏₋₁exp−𝜒+𝑏₀+𝑏₁𝜒.exp (4.2) Substituting (4.2) into (2.7) and by the help of Mathematica, we obtain 𝑐=±√2,𝑎₋₁=0,𝑎₁𝑏=0,₋₁=𝑎²₀8,𝑏₀=0,𝑏₁=1, (4.3) where 𝑎₀ is a free parameter. Substituting these results into (4.2), we obtain the exact solution 𝑢𝜒=𝑎₀exp𝜒𝑎²₀,/8+exp2𝜒 (4.4) of (2.7). Consequently, if we choose that 𝑎₀=√8 then this solution, in terms of the variables 𝑥 and 𝑡 becomes 𝜙(𝑥,𝑡)=√2sech1𝑐²(),−1𝑥−𝑐𝑡 (4.5) which is a soliton solution of our 𝜑⁴ equation (1.1).

5. Conclusion

In this paper, Lie symmetry analysis in conjunction with the simplest equation method and the Exp-function method have been successfully used to obtain exact solutions of the 𝜑⁴ equation. As a simplest equation, we have used the Riccati equation. The solutions obtained were travelling wave solutions. In particular, a soliton solution was also obtained.