Abstract

By using the first integral method, the traveling wave solutions for the generalized Pochhammer-Chree (PC) equations are constructed. The obtained results include complex exponential function solutions, complex traveling solitary wave solutions, complex periodic wave solutions, and complex rational function solutions. The power of this manageable method is confirmed.

1. Introduction

In this paper, we study the generalized Pochhammer-Chree (PC) equations:𝑢𝑡𝑡−𝑢𝑡𝑡𝑥𝑥−𝛼𝑢+𝛽𝑢𝑛+1+𝜈𝑢2𝑛+1𝑥𝑥=0,𝑛≥1,(1.1) where 𝛼,𝛽, and 𝜈 are constants. Equation (1.1) represents a nonlinear model of longitudinal wave propagation of elastic rods [1–14].The model for 𝛼=1,𝛽=1/(𝑛+1), and 𝜈=0 was studied in [4, 7, 8] where solitary wave solutions for this model were obtained for 𝑛=1,2, and 4. A second model for 𝛼=0,𝛽=−1/2, and 𝜈=0 was studied by [9], and solitary wave solutions were obtained as well.

However, a third model was investigated in [10–13] for 𝑛=1,2 where explicit solitary wave solutions and kinks solutions were derived.

It is the objective of this work to further complement studies on a generalized PC equations in [1–14].

The first integral method, which is based on the ring theory of commutative algebra, was first proposed by Feng [15]. This method was further developed by the same author in [16–21] and some other mathematicians [22–26]. Our first interest in the present work is to implement the first integral method to stress its power in handling nonlinear equations, so that one can apply it for solving various types of nonlinearity. The next interest is in the determination of exact traveling wave solutions for the generalized PC equations. The remaining structure of this paper is organized as follows: Section 2 is a brief introduction to the first integral method. In Section 3, by implementing the first integral method, new exact traveling wave solutions to the generalized PC equations are reported with the aid of mathematical software Mathematica 8.0. This describes the ability and reliability of the method. A conclusion is given in Section 4.

2. The First Integral Method

Consider a general nonlinear partial differential equation in the form𝑃𝑢,𝑢𝑡,𝑢𝑥,𝑢𝑥𝑥,𝑢𝑡𝑡,𝑢𝑥𝑡,𝑢𝑥𝑥𝑥,…=0.(2.1) Using the wave variable 𝜉=𝑥−𝑐𝑡 carries (2.1) into the following ordinary differential equation (ODE):𝑄𝑈,𝑈,𝑈,𝑈,…=0,(2.2) where prime denotes the derivative with respect to the same variable 𝜉.

Next, we introduce new independent variables 𝑥=𝑢,𝑦=𝑢𝜉 which change (2.2) to a system of ODEs:𝑥𝑦=𝑦,=𝑓(𝑥,𝑦).(2.3) According to the qualitative theory of differential equations [27], if one can find the first integrals to System (2.3) under the same conditions, the analytic solutions to (2.3) can be solved directly. However, in general, it is difficult to realize this even for a single first integral, because for a given plane autonomous system, there is no general theory telling us how to find its first integrals in a systematic way. A key idea of this approach here to find the first integral is to utilize the Division Theorem. For convenience, first let us recall the Division Theorem for two variables in the complex domain 𝐶 [15].

Division Theorem
Suppose that 𝑃(𝑥,𝑦) and 𝑄(𝑥,𝑦) are polynomials of two variables 𝑥 and 𝑦 in 𝐶[𝑥,𝑦] and 𝑃(𝑥,𝑦) is irreducible in 𝐶[𝑥,𝑦]. If 𝑄(𝑥,𝑦) vanishes at all zero points of 𝑃(𝑥,𝑦), then there exists a polynomial 𝐺(𝑥,𝑦) in 𝐶[𝑥,𝑦] such that 𝑄(𝑥,𝑦)=𝑃(𝑥,𝑦)𝐺(𝑥,𝑦).(2.4)

3. The Generalized PC Equations

We conduct our analysis by examining all possible cases of 𝜈 for the generalized PC equations (1.1).

Case 1. 𝛽≠0,𝜈≠0.(3.1) Using the wave variable 𝜉=𝑥−𝑐𝑡 and integrating twice, we obtain 𝑐2−𝛼𝑢−𝑐2𝑢−𝛽𝑢𝑛+1−𝜈𝑢2𝑛+1=0,(3.2) where prime denotes the derivative with respect to the same variable 𝜉. Making the following transformation: 𝑣=𝑢𝑛,(3.3) then (3.2) becomes 𝑐2𝑛−𝛼2𝑣2−𝑛𝑐2𝑣𝑣−𝑐2𝑣(1−𝑛)2−𝑛2𝛽𝑣3−𝑛2𝜈𝑣4=0,(3.4) where 𝑣and 𝑣 denote 𝑑𝑣/𝑑𝜉 and 𝑑2𝑣/𝑑𝜉2, respectively. Equation (3.4) is a nonlinear ODE, and we can rewrite it as 𝑣𝑣−𝑎𝑣+𝑏2𝑣+𝑑𝑣2+𝑓𝑣3=0,(3.5) where 𝛼𝑎=1−𝑐2𝑛,𝑏=1−𝑛𝑛,𝑑=𝑛𝛽𝑐2,𝑓=𝑛𝜈𝑐2.(3.6) Let 𝑥=𝑣,let𝑦=𝑑𝑣/𝑑𝜉, and let (3.5) be equivalent to the following two-dimensional autonomous system 𝑑𝑥𝑑𝜉=𝑦,𝑑𝑦𝑦𝑑𝜉=𝑎𝑥−𝑏2𝑥−𝑑𝑥2−𝑓𝑥3.(3.7) Assume that 𝑑𝜏=𝑑𝜉𝑥,(3.8) thus system (3.7) becomes 𝑑𝑥𝑑𝜏=𝑥𝑦,𝑑𝑦𝑑𝜏=𝑎𝑥2−𝑏𝑦2−𝑑𝑥3−𝑓𝑥4.(3.9) Now, we are applying the Division Theorem to seek the first integral to system (3.9). Suppose that 𝑥=𝑥(𝜏),𝑦=𝑦(𝜏) are the nontrivial solutions to (3.9), and ∑𝑝(𝑥,𝑦)=𝑚𝑖=0𝑎𝑖(𝑥)𝑦𝑖 is an irreducible polynomial in 𝐶[𝑥,𝑦], such that 𝑝[]=𝑥(𝜏),𝑦(𝜏)𝑚𝑖=0𝑎𝑖(𝑥(𝜏))𝑦(𝜏)𝑖=0,(3.10) where 𝑎𝑖(𝑥) (𝑖=0,1,…,𝑚) are polynomials of 𝑥 and 𝑎𝑚(𝑥)≠0. We call (3.10) the first integral of polynomial form to system (3.9). We start our study by assuming 𝑚=1 in (3.10). Note that 𝑑𝑝/𝑑𝜏 is a polynomial in 𝑥 and 𝑦, and 𝑝[𝑥(𝜏),𝑦(𝜏)]=0 implies 𝑑𝑝/𝑑𝜏|(3.9)=0. According to the Division Theorem, there exists a polynomial 𝐻(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑥)𝑦 in 𝐶[𝑥,𝑦] such that 𝑑𝑝|||𝑑𝜏(3.9)=𝜕𝑝𝜕𝑥𝜕𝑥+𝜕𝜏𝜕𝑝𝜕𝑦𝜕𝑦||||𝜕𝜏(3.9)=1𝑖=0𝑎𝑖(𝑥)𝑦𝑖+⋅𝑥𝑦1𝑖=0𝑖𝑎𝑖(𝑥)𝑦𝑖−1⋅𝑎𝑥2−𝑏𝑦2−𝑑𝑥3−𝑓𝑥4=(ℎ(𝑥)+𝑔(𝑥)𝑦)1𝑖=0𝑎𝑖(𝑥)𝑦𝑖,(3.11) where prime denotes differentiation with respect to the variable 𝑥. On equating the coefficients of 𝑦𝑖(𝑖=2,1,0) on both sides of (3.11), we have 𝑥𝑎1(𝑥)−𝑏𝑎1(𝑥)=𝑔(𝑥)𝑎1(𝑥),(3.12)𝑥𝑎0(𝑥)=ℎ(𝑥)𝑎1(𝑥)+𝑔(𝑥)𝑎0(𝑥),(3.13)𝑎1(𝑥)𝑎𝑥2−𝑑𝑥3−𝑓𝑥4=ℎ(𝑥)𝑎0(𝑥).(3.14) Since, 𝑎1(𝑥) is a polynomial of 𝑥, from (3.12) we conclude that 𝑎1(𝑥) is a constant and 𝑔(𝑥)=−𝑏. For simplicity, we take 𝑎1(𝑥)=1, and balancing the degrees of ℎ(𝑥) and 𝑎0(𝑥) we conclude that deg(ℎ(𝑥))=2 and deg(𝑎0(𝑥))=2 only. Now suppose that ℎ(𝑥)=𝐴2𝑥2+𝐴1𝑥+𝐴0,𝑎0(𝑥)=𝐵2𝑥2+𝐵1𝑥+𝐵0𝐴2≠0,𝐵2≠0,(3.15) where 𝐴𝑖,𝐵𝑖,(𝑖=0,1,2) are all constants to be determined. Substituting (3.15) into (3.13), we obtain ℎ(𝑥)=(𝑏+2)𝐵2𝑥2+(𝑏+1)𝐵1𝑥+𝑏𝐵0.(3.16) Substituting 𝑎0(𝑥),𝑎1(𝑥),and ℎ(𝑥) in (3.14) and setting all the coefficients of powers 𝑥 to be zero, we obtain a system of nonlinear algebraic equations, and by, solving it, we obtain the following solutions: √𝑑=−√𝑎(3+2𝑏)𝑓√√−2−𝑏1+𝑏,𝐵0=0,𝐵1√=−𝑎√1+𝑏,𝐵2√=−𝑓√−2−𝑏,(3.17)√𝑑=−√𝑎(3+2𝑏)𝑓√√−2−𝑏1+𝑏,𝐵0=0,𝐵1=√𝑎√1+𝑏,𝐵2=√𝑓√−2−𝑏,(3.18)√𝑑=√𝑎(3+2𝑏)𝑓√√−2−𝑏1+𝑏,𝐵0=0,𝐵1√=−𝑎√1+𝑏,𝐵2=√𝑓√−2−𝑏,(3.19)√𝑑=√𝑎(3+2𝑏)𝑓√√−2−𝑏1+𝑏,𝐵0=0,𝐵1=√𝑎√1+𝑏,𝐵2√=−𝑓√−2−𝑏.(3.20) Setting (3.17) and (3.18) in (3.10), we obtain that System (3.9) has one first integral √𝑦∓𝑓√𝑥−2−𝑏2+√𝑎√𝑥1+𝑏=0,(3.21) respectively. Combining this first integral with (3.9), the second-order differential equation (3.5) can be reduced to 𝑑𝑣√𝑑𝜉=±𝑓√𝑣−2−𝑏2+√𝑎√𝑣1+𝑏.(3.22) Solving (3.22) directly and changing to the original variables, we obtain the following complex exponential function solutions to (1.1): 𝑢1⎛⎜⎜⎝(𝑥,𝑡)=𝑖𝑅√exp−𝑛1−𝛼/𝑐2(𝑥−𝑐𝑡)−𝑖𝑅𝑐1−√⎞⎟⎟⎠𝜈/𝑐1/𝑛,(3.23)𝑢2⎛⎜⎜⎝(𝑥,𝑡)=𝑖𝑅exp𝑖𝑅𝑐1𝑛√exp1−𝛼/𝑐2−√(𝑥−𝑐𝑡)𝜈/𝑐exp𝑖𝑅𝑐1⎞⎟⎟⎠1/𝑛.(3.24) Similarly, for the cases of (3.19) and (3.20), we have another complex exponential function solutions: 𝑢3⎛⎜⎜⎜⎝(𝑥,𝑡)=𝑖𝑅−exp(−𝑛1−𝛼/𝑐2(𝑥−𝑐𝑡)−𝑖𝑅𝑐1√)+⎞⎟⎟⎟⎠𝜈/𝑐1/𝑛,(3.25)𝑢4⎛⎜⎜⎝(𝑥,𝑡)=𝑖𝑅exp𝑖𝑅𝑐1𝑛√−exp1−𝛼/𝑐2+√(𝑥−𝑐𝑡)𝜈/𝑐exp𝑖𝑅𝑐1⎞⎟⎟⎠1/𝑛,(3.26) where, √𝑅=1−𝛼/𝑐2√1+𝑛, 𝑐1 is an arbitrary constant. These solutions are all new exact solutions. Now we assume that 𝑚=2 in (3.10). By the Division Theorem, there exists a polynomial 𝐻(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑥)𝑦 in 𝐶[𝑥,𝑦] such that 𝑑𝑝|||𝑑𝜏(3.9)=𝜕𝑝𝜕𝑥𝜕𝑥+𝜕𝜏𝜕𝑝𝜕𝑦𝜕𝑦||||𝜕𝜏(3.9)=2𝑚=1𝑎𝑖(𝑥)𝑦𝑖+⋅𝑥𝑦2𝑚=1𝑖𝑎𝑖(𝑥)𝑦𝑖−1⋅𝑎𝑥2−𝑏𝑦2−𝑑𝑥3−𝑓𝑥4=(ℎ(𝑥)+𝑔(𝑥)𝑦)2𝑚=1𝑎𝑖(𝑥)𝑦𝑖,(3.27)On equating the coefficients of 𝑦𝑖(𝑖=3,2,1,0) on both sides of (3.11), we have 𝑥𝑎2(𝑥)−2𝑏𝑎2(𝑥)=𝑔(𝑥)𝑎2(𝑥),(3.28)𝑥𝑎1(𝑥)−𝑏𝑎1(𝑥)=ℎ(𝑥)𝑎2(𝑥)+𝑔(𝑥)𝑎1(𝑥),(3.29)𝑥𝑎0(𝑥)+2𝑎2(𝑥)𝑎𝑥2−𝑑𝑥3−𝑓𝑥4=ℎ(𝑥)𝑎1(𝑥)+𝑔(𝑥)𝑎0(𝑥),(3.30)𝑎1(𝑥)𝑎𝑥2−𝑑𝑥3−𝑓𝑥4=ℎ(𝑥)𝑎0(𝑥).(3.31) Since 𝑎2(𝑥) is a polynomial of 𝑥, from (3.28) we conclude that 𝑎2(𝑥) is a constant and 𝑔(𝑥)=−2𝑏. For simplicity, we take 𝑎2(𝑥)=1, and balancing the degrees of ℎ(𝑥),𝑎0(𝑥), and 𝑎1(𝑥) we conclude that deg(ℎ(𝑥))=2 and deg(𝑎1(𝑥))=2. In this case, we assume that ℎ(𝑥)=𝐴2𝑥2+𝐴1𝑥+𝐴0,𝑎1(𝑥)=𝐵2𝑥2+𝐵1𝑥+𝐵0𝐴2≠0,𝐵2≠0,(3.32) where 𝐴𝑖,𝐵𝑖(𝑖=0,1,2) are constants to be determined. Substituting (3.32) into (3.29) and (3.30), we have ℎ(𝑥)=(2+𝑏)𝐵2𝑥2+(1+𝑏)𝐵1𝑥+𝑏𝐵0,𝑎0(𝑥)=2𝑓+(2+𝑏)𝐵22𝑥2(2+𝑏)4+2𝑑+(3+2𝑏)𝐵1𝐵2𝑥3+2𝑏3+−2𝑎+(1+𝑏)𝐵21+2(1+𝑏)𝐵0𝐵2𝑥2(1+𝑏)2+𝐵0𝐵1𝐵𝑥+202+𝐹𝑥−2𝑏,(3.33) where 𝐹 is an arbitrary integration constant. Substituting 𝑎0(𝑥), 𝑎1(𝑥), and ℎ(𝑥) in (3.31) and setting all the coefficients of powers 𝑥 to be zero, we obtain a system of nonlinear algebraic equations, and by solving it we obtain 4𝐹=0,𝑎=(1+𝑏)𝑑2(3+2𝑏)2𝐵221,𝑓=−4(2+𝑏)𝐵22,𝐵0=0,𝐵1=−4𝑑(3+2𝑏)𝐵2.(3.34) Setting (3.34) in (3.10), we obtain 𝑦=4𝑑𝑥−(3+2𝑏)𝐵22𝑥22(3+2𝑏)𝐵2.(3.35) Using this first integral, the second-order ODE (3.5) reduces to 𝑑𝑣=𝑑𝜉4𝑑𝑣−(3+2𝑏)𝐵22𝑣22(3+2𝑏)𝐵2.(3.36) Similarly, solving (3.36) and changing to the original variables, we obtain the exponential function solutions: 𝑢5(𝑥,𝑡)=2𝛽(2+𝑛)𝐵2𝑆𝛽𝑛exp2𝐵2(1+2/𝑛)𝑐1𝑆−𝑥+𝑐𝑡+(2+𝑛)𝐵221/𝑛,(3.37) where 𝑆=2𝑛2/(2+𝑛)𝑐2𝐵2, 𝑐1 is an arbitrary constant. These solutions are all new exact solutions.

Case 2. 𝛽=0,𝜈≠0.(3.38) We now investigate the generalized PC equation (1.1) for 𝛽=0, then, we obtain 𝑐2−𝛼𝑢−𝑐2𝑢−𝜈𝑢2𝑛+1=0,(3.39) where prime denotes the derivative with respect to 𝜉. Similarly as in Case 1, making then the following transformation: 𝑣=𝑢𝑛,(3.40) then (3.39) becomes 𝑐2𝑛−𝛼2𝑣2−𝑛𝑐2𝑣𝑣−𝑐2𝑣(1−𝑛)2−𝑛2𝜈𝑣4=0,(3.41) where 𝑣′and 𝑣 denote 𝑑𝑣/𝑑𝜉 and 𝑑2𝑣/𝑑𝜉2, respectively. Let us rewrite (3.41) as 𝑣𝑣−𝑎𝑣+𝑏2𝑣+𝑓𝑣3=0,(3.42) where 𝑎,𝑏,𝑓 are as given in (3.6). Let 𝑥=𝑣,let𝑦=𝑑𝑣/𝑑𝜉, and (3.42) become the following two-dimensional autonomous system: 𝑑𝑥𝑑𝜉=𝑦,𝑑𝑦𝑦𝑑𝜉=𝑎𝑥−𝑏2𝑥−𝑓𝑥3.(3.43) Assume that 𝑑𝜏=𝑑𝜉𝑥,(3.44) thus system (3.43) becomes 𝑑𝑥𝑑𝜏=𝑥𝑦,𝑑𝑦𝑑𝜏=𝑎𝑥2−𝑏𝑦2−𝑓𝑥4.(3.45) Following the same procedures as in Case 1, so we are applying the Division Theorem to seek the first integral to system (3.45). Suppose that 𝑥=𝑥(𝜏)and𝑦=𝑦(𝜏) are the nontrivial solutions to (3.45), and ∑𝑝(𝑥,𝑦)=𝑚𝑖=0𝑎𝑖(𝑥)𝑦𝑖 is an irreducible polynomial in 𝐶[𝑥,𝑦], such that 𝑝[]=𝑥(𝜏),𝑦(𝜏)𝑚𝑖=0𝑎𝑖(𝑥(𝜏))𝑦(𝜏)𝑖=0,(3.46) where 𝑎𝑖(𝑥)(𝑖=0,1,…,𝑚) are polynomials of 𝑥 and 𝑎𝑚(𝑥)≠0. We call (3.46) the first integral of polynomial form to system (3.45). We start by assuming 𝑚=1 in (3.46). Note that 𝑑𝑝/𝑑𝜏 is a polynomial in 𝑥 and 𝑦, and 𝑝[𝑥(𝜏),𝑦(𝜏)]=0 implies 𝑑𝑝/𝑑𝜏|(3.44)=0. According to the Division Theorem, there exists a polynomial 𝐻(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑥)𝑦 in 𝐶[𝑥,𝑦] such that 𝑑𝑝|||𝑑𝜏(3.45)=𝜕𝑝𝜕𝑥𝜕𝑥+𝜕𝜏𝜕𝑝𝜕𝑦𝜕y||||𝜕𝜏(3.45)=1𝑖=0𝑎𝑖(𝑥)𝑦𝑖+⋅𝑥𝑦1𝑖=0𝑖𝑎𝑖(𝑥)𝑦𝑖−1⋅𝑎𝑥2−𝑏𝑦2−𝑓𝑥4=(ℎ(𝑥)+𝑔(𝑥)𝑦)1𝑖=0𝑎𝑖(𝑥)𝑦𝑖,(3.47) where prime denotes differentiation with respect to the variable 𝑥. On equating the coefficients of 𝑦𝑖(𝑖=2,1,0) on both sides of (3.47), we have 𝑥𝑎1(𝑥)−𝑏𝑎1(𝑥)=𝑔(𝑥)𝑎1(𝑥),(3.48)𝑥𝑎0(𝑥)=ℎ(𝑥)𝑎1(𝑥)+𝑔(𝑥)𝑎0(𝑥),(3.49)𝑎1(𝑥)𝑎𝑥2−𝑓𝑥4=ℎ(𝑥)𝑎0(𝑥).(3.50) Since, 𝑎1(𝑥) is a polynomial of 𝑥, from (3.48) we conclude that 𝑎1(𝑥) is a constant and 𝑔(𝑥)=−𝑏. For simplicity, we take 𝑎1(𝑥)=1, and balancing the degrees of ℎ(𝑥) and 𝑎0(𝑥) we conclude that deg(ℎ(𝑥))=2 and deg(𝑎0(𝑥))=2 only. Now suppose that ℎ(𝑥)=𝐴2𝑥2+𝐴1𝑥+𝐴0,𝑎0(𝑥)=𝐵2𝑥2+𝐵1𝑥+𝐵0𝐴2≠0,𝐵2≠0,(3.51) where 𝐴𝑖,𝐵𝑖,(𝑖=0,1,2) are constants to be determined. Substituting (3.51) into (3.49), we have ℎ(𝑥)=(2+𝑏)𝐵2𝑥2+(1+𝑏)𝐵1𝑥+𝑏𝐵0.(3.52) Substituting 𝑎0(𝑥),𝑎1(𝑥), and ℎ(𝑥) in (3.50) and setting all the coefficients of powers 𝑥 to be zero, we obtain a system of nonlinear algebraic equations, and, by solving it, we obtain the following solutions: 2√𝑎=−𝑓𝐵0√−2−𝑏,𝐵2√=−𝑓√−2−𝑏,𝐵12√=0,𝑎=𝑓𝐵0√−2−𝑏,𝐵2=√𝑓√−2−𝑏,𝐵1=0.(3.53) Thus, by the similar procedure explained above in Case 1, the complex traveling solitary wave and the complex periodic wave solutions to the generalized PC equations in this Case 2 are given, respectively, by 𝑢1⎛⎜⎜⎝−𝑞√(𝑥,𝑡)=𝐵0√𝑝√𝑐tanh𝑥−𝑐𝑡−𝑖1+1/𝑛𝑐1√𝐵0/𝑞√𝑐𝑝⎞⎟⎟⎠1/𝑛,𝑢2⎛⎜⎜⎝−𝑞√(𝑥,𝑡)=𝐵0√𝑝√𝑐tan𝑥−𝑐𝑡−𝑖1+1/𝑛𝑐1√𝐵0/𝑞√𝑐𝑝⎞⎟⎟⎠1/𝑛,(3.54) where 𝑝=𝑛1/4𝜈1/4,𝑞=𝑖1/4(1+1/𝑛)1/4,𝑐1 is an arbitrary constant. These solutions are all new exact solutions. Now we assume that 𝑚=2 in (3.46). By the Division Theorem, there exists a polynomial 𝐻(𝑥,𝑦)=ℎ(𝑥)+𝑔(𝑥)𝑦 in 𝐶[𝑥,𝑦] such that 𝑑𝑝|||𝑑𝜏(3.45)=𝜕𝑝𝜕𝑥𝜕𝑥+𝜕𝜏𝜕𝑝𝜕𝑦𝜕𝑦||||𝜕𝜏(3.45)=2𝑖=0𝑎𝑖(𝑥)𝑦𝑖+⋅𝑥𝑦2𝑖=0𝑖𝑎𝑖(𝑥)𝑦𝑖−1⋅𝑎𝑥2−𝑏𝑦2−𝑓𝑥4=(ℎ(𝑥)+𝑔(𝑥)𝑦)2𝑖=0𝑎𝑖(𝑥)𝑦𝑖(3.55) On equating the coefficients of 𝑦𝑖(𝑖=3,2,1,0) on both sides of (3.55), we have 𝑥𝑎2(𝑥)−2𝑏𝑎2(𝑥)=𝑔(𝑥)𝑎2(𝑥),(3.56)𝑥𝑎1(𝑥)−𝑏𝑎1(𝑥)=ℎ(𝑥)𝑎2(𝑥)+𝑔(𝑥)𝑎1(𝑥),(3.57)𝑥𝑎0(𝑥)+2𝑎2(𝑥)𝑎𝑥2−𝑓𝑥4=ℎ(𝑥)𝑎1(𝑥)+𝑔(𝑥)𝑎0(𝑥),(3.58)𝑎1(𝑥)𝑎𝑥2−𝑓𝑥4=ℎ(𝑥)𝑎0(𝑥).(3.59) Since 𝑎2(𝑥) is a polynomial of 𝑥, from (3.56) we conclude that 𝑎2(𝑥) is a constant and 𝑔(𝑥)=−2𝑏. For simplicity, we take 𝑎2(𝑥)=1, and balancing the degrees of ℎ(𝑥),𝑎0(𝑥) and 𝑎1(𝑥) we conclude that deg(ℎ(𝑥))=1, deg(𝑎1(𝑥))=1anddeg(ℎ(𝑥))=2, deg(𝑎1(𝑥))=2.
Subcase 2.1. deg(ℎ(𝑥))=1 and deg(𝑎1(𝑥))=1. In this case, we assume that ℎ(𝑥)=𝐴1𝑥+𝐴0,𝑎1(𝑥)=𝐵1𝑥+𝐵0𝐴1≠0,𝐵1≠0,(3.60) where 𝐴𝑖,𝐵𝑖(𝑖=0,1) are constants to be determined. Inserting (3.60) into (3.57) and (3.58), we deduce that ℎ(𝑥)=(1+𝑏)𝐵1𝑥+𝑏𝐵0𝑎0𝑓(𝑥)=𝑥2+𝑏4+−2𝑎+(1+𝑏)𝐵21𝑥2(1+𝑏)2+𝐵0𝐵1𝐵𝑥+202+𝐹𝑥−2𝑏,(3.61) where 𝐹 is an arbitrary integration constant. Substituting 𝑎0(𝑥), 𝑎1(𝑥), and ℎ(𝑥) in (3.59) and setting all the coefficients of powers 𝑥 to be zero, we obtain a system of nonlinear algebraic equations, and by solving it we obtain 1𝑎=4𝐵21(1+𝑏),𝐹=0,𝐵0=0.(3.62) Then, by the similar procedure explained above, we get the complex exponential function solutions which can be expressed as 𝑢3(𝑥,𝑡)=𝑖𝐾exp𝐾𝐵1𝑐1𝐵−exp1(√/2𝑥−𝑐𝑡)+(2𝐾/𝑐)𝜈√𝑛exp𝐾𝐵1𝑐11/𝑛,𝑢4−(𝑥,𝑡)=𝑖𝐾exp𝐾𝐵1𝑐1𝐵−exp1(√/2𝑥−𝑐𝑡)+(2𝐾/𝑐)𝜈√𝑛exp𝐾𝐵1𝑐11/𝑛,(3.63) where √𝐾=1+1/𝑛. These solutions are all new exact solutions.
Subcase 2.2. deg(ℎ(𝑥))=2 and deg(𝑎1(𝑥))=2. Now suppose that ℎ(𝑥)=𝐴2𝑥2+𝐴1𝑥+𝐴0,𝑎1(𝑥)=𝐵2𝑥2+𝐵1𝑥+𝐵0𝐴2≠0,𝐵2≠0,(3.64) where, 𝐴𝑖,𝐵𝑖,(𝑖=0,1,2) are constants to be determined. Substituting (3.64) into (3.57) and (3.58), we have ℎ(𝑥)=(2+𝑏)𝐵2𝑥2+(1+𝑏)𝐵1𝑥+𝑏𝐵0𝑎(3.65)0(𝑥)=2𝑓+(2+𝑏)𝐵22𝑥2(2+𝑏)4+𝐵1𝐵2𝑥3+−2𝑎+(1+𝑏)𝐵21+2(1+𝑏)𝐵0𝐵2𝑥2(1+𝑏)2+𝐵0𝐵1𝐵𝑥+202+𝐹𝑥−2𝑏,(3.66) where 𝐹 is an arbitrary integration constant. Substituting 𝑎0(𝑥),𝑎1(𝑥), and ℎ(𝑥) in (3.59) and setting all the coefficients of powers 𝑥 to be zero, we obtain a system of nonlinear algebraic equations, and, by solving it, we obtain the following solutions: 𝐹=0,𝑎=0,𝐵0=0,𝐵1=0,𝐵22√=−𝑓√,−2−𝑏𝐹=0,𝑎=0,𝐵0=0,𝐵1=0,𝐵2=2√𝑓√.−2−𝑏(3.67) Thus, as above, we obtain the complex rational function solutions which can be written as 𝑢5⎛⎜⎜⎝(𝑥,𝑡)=𝑖𝐾√𝑛√𝜈√∓𝑥/𝛼+𝑡−𝑖𝐾𝑐1⎞⎟⎟⎠1/𝑛,𝑢6⎛⎜⎜⎝−(𝑥,𝑡)=𝑖𝐾√𝑛√𝜈√∓𝑥/𝛼+𝑡+𝑖𝐾𝑐1⎞⎟⎟⎠1/𝑛,(3.68) where 𝐾 as defined above. These solutions are all new exact solutions.

Notice that the results in this paper are based on the assumption of 𝑚=1,2 for the generalized PC equations. For the cases of 𝑚=3,4 for these equations, the discussions become more complicated and involves the irregular singular point theory and the elliptic integrals of the second kind and the hyperelliptic integrals. Some solutions in the functional form cannot be expressed explicitly. One does not need to consider the cases 𝑚≥5 because it is well known that an algebraic equation with the degree greater than or equal to 5 is generally not solvable.

4. Conclusion

In this work, we are concerned with the generalized PC equations for seeking their traveling wave solutions. We first transform each equation into an equivalent two-dimensional planar autonomous system then use the first integral method to find one first integral which enables us to reduce the generalized PC equations to a first-order integrable ordinary differential equations. Finally, a class of traveling wave solutions for the considered equations are obtained. These solutions include complex exponential function solutions, complex traveling solitary wave solutions, complex periodic wave solutions, and complex rational function solutions. We believe that this method can be applied widely to many other nonlinear evolution equations, and this will be done in a future work.