Dispersive dark optical soliton with Tzitzéica type nonlinear evolution equations arising in nonlinear optics

Abstract

A improvement of the expansion methods namely the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method for solving the Tzitzéica type nonlinear evolution equations is proposed. In this work, the dispersive optical solitons that are governed by the Tzitzéica type nonlinear evolution equations. As a result, many new and more general exact travelling wave solutions are obtained including periodic function solutions, soliton-like solutions and trigonometric function solutions. The exact particular solutions containing four types hyperbolic function solution, trigonometric function solution, exponential solution and rational solution. We obtained the further solutions comparing with other methods. Recently this method is developed for searching exact travelling wave solutions of nonlinear partial differential equations. Abundant exact travelling wave solutions including solitons, kink, periodic and rational solutions have been found. These solutions might play important role in engineering fields. It is shown that this method, with the help of symbolic computation, provides a straightforward and powerful mathematical tool for solving the nonlinear problems.

1 Introduction

Nonlinear evolution equations (NLEEs) are very important equations in mathematical physics, engineering and applied mathematics for describing diverse types of physical mechanisms of natural phenomena in the field of applied sciences, biochemical and engineering. Nonlinear wave equations play a substantial role in various fields such as plasma physics, fluid mechanics, optical fibers, solid state physics, chemical kinetics, geochemistry, nonlinear optics and so on (Gray and Scott 1990; Kabir and Khajeh 2009). Much work has been down over the years on the subject of obtaining the analytical solutions to the nonlinear partial differential equations (NPDEs). One of the most exciting advances of nonlinear science and theoretical physics has been a development of methods to look for exact solutions for NPDEs. With the rapid development of nonlinear sciences based on computer algebraic system, many effective methods have been presented, such as, the homotopy analysis method (Dehghan et al. 2010), the variational iteration method (He 1999; Mohyud-Din et al. 2009; Dehghan et al. 2010; Jafari et al. 2014), the homotopy perturbation method (Dehghan and Manafian 2009; Mohyud-Din et al. 2011; Zhang et al. 2014), the sine-cosine method (Wazwaz 2006), the tanh-coth method (Manafian Heris and Lakestani 2013), the modified extended tanh-function method (Abdou and Soliman 2006; El-Wakil and Abdou 2007), the Exp-function method (Dehghan et al. 2011; Ebrahimi Ghogdia et al. 2015; Yildirim and Pinar 2010; Abdou et al. 2007; Noor et al. 2008, 2010; Mohyud-Din et al. 2010, 2012; Manafian and Lakestani 2015b), Taylor expansion method (Ma and You 2004), the exp(-$\varPhi (\xi )$)-expansion method (Roshid and Rahman 2014; Hafez et al. 2015), the $(\frac{G'}{G})$-expansion method (Naher et al. 2013; Saba et al. 2015; Manafian and Lakestani 2015a), the modified simple equation method (Jawad et al. 2010), the novel $(\frac{G'}{G})$-expansion method (Alam et al. 2014), the new approach of the generalized $(\frac{G'}{G})$-expansion method (Naher and Abdullah 2013), the Jacobi elliptic function method (Chen and Wang 2005), the homogeneous balance method (Zhao et al. 2006) and so on. Here, we use of an effective method namely the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method (Manafian et al. 2015; Manafian and Lakestani 2016) for constructing a range of exact solutions for the following ordinary partial differential equations that in this paper we developed solutions as well. In this paper, we put forth the new approach of improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method to construct exact travelling wave solutions including solitons, kink, periodic and rational solutions to the Tzitzéica type nonlinear evolution equations. Wazwaz (Wazwaz 2005) applied the tanh method to the Dodd–Bullough–Mikhailov and the Tzitzéica–Dodd–Bullough equations. Huber (2008) applied a class of solitary-like solutions of the Tzitzéica equation where generated by a similarity reduction. Borhanifar and Moghanlu (2011) obtained the exact traveling wave solutions of the DBM equation and TDB equation are using the (G’/G)-expansion method. The Tzitzéica equation (Tzitzéica 1910)

$$u_{tt}-u_{xx}-e^{u}+e^{-2u}=0,$$

(1)

has been investigated by Tzitzéica in the study of integrable surfaces. The Dodd–Bullough–Mikhailov (DBM) equation is given as

$$u_{xt}+e^{u}+e^{-2u}=0,$$

(2)

and consider the Tzitzéica–Dodd–Bullough (TDB) equation as follows

$$u_{xt}-e^{-u}-e^{-2u}=0,$$

(3)

where these equations play a significant role in many scientific applications such as solid state physics, nonlinear optics and the quantum field theory (Abazari 2010). Also, Mikhailov (1910) deducted the N-soliton solutions for Tzitzéica equation by the inverse scattering method. The purpose of this paper is to obtain exact solutions of the Tzitzéica type nonlinear evolution equations and to determine the accuracy of the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method in solving these kind of problems. The paper is organized as follows: In Sect. 2, we describe the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method. In Sect. 3, we examine the Tzitzéica equation, DBM equation and TDB equation respectively with method introduced in Sect. 2. In Sect. 4, we examine the physical interpretations of the solutions of the Tzitzéica type nonlinear evolution equations. Also conclusion is given in Sect. 5. Finally some references are given at the end of this paper.

2 Description of improved $\tan \left( \varPhi (\xi )/2\right)$-expansion technique

The $\tan \left( \varPhi (\xi )/2\right)$-expansion method is well-known analytical method which first presented and developed in Manafian et al. (2015). Also Manafian and Lakestani (2015c) applied the stated method to solving the generalized Fitzhugh–Nagumo equation with time-dependent coefficients.

Step 1. We suppose that the given nonlinear partial differential equation for u(x, t) to be in the form

$${\mathcal {N}}(u, u_x, u_t, u_{xx}, u_{tt},\ldots )=0,$$

(4)

which can be converted to an ODE

$${\mathcal {Q}}(u, u', -\mu u', u'', \mu ^2u'',\ldots )=0,$$

(5)

by the transformation $\xi =x-\mu t$, is wave variable. Also, $\mu$ is constant to be determined later.

Step 2. Suppose the traveling wave solution of Eq. (5) can be expressed as follows

$$u(\xi )=S(\varPhi )=\sum _{k=0}^{m}A_k\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^k+ \sum _{k=1}^{m}B_k\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-k},$$

(6)

where $A_k (0\le k \le m)$ and $B_k (1\le k \le m)$ are constants to be determined, such that $A_m\ne 0, B_m\ne 0$ and $\varPhi =\varPhi (\xi )$ satisfies the following ordinary differential equation:

$$\varPhi '(\xi )=a\sin (\varPhi (\xi ))+b\cos (\varPhi (\xi ))+c.$$

(7)

We will consider the following special solutions of equation (7):

• Family 1: When $a^2+b^2-c^2<0$ and $b-c\ne 0$, then $\varPhi (\xi )=2\arctan \left[ \frac{a}{b-c}-\frac{\sqrt{c^2-b^2-a^2}}{b-c}\tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}(\xi +C)\right) \right]$.

• Family 2: When $a^2+b^2-c^2>0$ and $b-c\ne 0$, then $\varPhi (\xi )=2\arctan \left[ \frac{a}{b-c}+\frac{\sqrt{b^2+a^2-c^2}}{b-c}\tanh \left( \frac{\sqrt{b^2+a^2-c^2}}{2}(\xi +C)\right) \right] .$

• Family 3: When $a^2+b^2-c^2>0$, $b\ne 0$ and $c=0,$ then $\varPhi (\xi )=2\arctan \left[ \frac{a}{b}+\frac{\sqrt{b^2+a^2}}{b}\tanh \left( \frac{\sqrt{b^2+a^2}}{2}(\xi +C)\right) \right] .$

• Family 4: When $a^2+b^2-c^2<0$, $c\ne 0$ and $b=0,$ then $\varPhi (\xi )=2\arctan \left[ -\frac{a}{c}+\frac{\sqrt{c^2-a^2}}{c}\tan \left( \frac{\sqrt{c^2-a^2}}{2}(\xi +C)\right) \right] .$

• Family 5: When $a^2+b^2-c^2>0$, $b-c\ne 0$ and $a=0$, then $\varPhi (\xi )=2\arctan \left[ \sqrt{\frac{b+c}{b-c}}\tanh \left( \frac{\sqrt{b^2-c^2}}{2}(\xi +C)\right) \right] .$

• Family 6: When $a=0$ and $c=0,$ then $\varPhi (\xi )=\arctan \left[ \frac{e^{2b(\xi +C)}-1}{e^{2b(\xi +C)}+1}, \ \frac{2e^{b(\xi +C)}}{e^{2b(\xi +C)}+1}\right]$.

• Family 7: When $b=0$ and $c=0,$ then $\varPhi (\xi )=\arctan \left[ \frac{2e^{a(\xi +C)}}{e^{2a(\xi +C)}+1}, \ \frac{e^{2a(\xi +C)}-1}{e^{2a(\xi +C)}+1}\right]$.

• Family 8: When $a^2+b^2=c^2$, then $\varPhi (\xi )=-2\arctan \left[ \frac{(b+c)(a(\xi +C)+2)}{a^2(\xi +C)}\right] .$

• Family 9: When $a=b=c=ka$, then $\varPhi (\xi )=2\arctan \left[ e^{ka(\xi +C)}-1\right]$.

• Family 10: When $a=c=ka$ and $b=-ka$, then $\varPhi (\xi )=2\arctan \left[ \frac{e^{ka(\xi +C)}}{1-e^{ka(\xi +C)}}\right]$.

• Family 11: When $c=a$, then $\varPhi (\xi )=-2\arctan \left[ \frac{(a+b)e^{b(\xi +C)}-1}{(a-b)e^{b(\xi +C)}-1}\right]$.

• Family 12: When $a=c$, then $\varPhi (\xi )=2\arctan \left[ \frac{(b+c)e^{b(\xi +C)}+1}{(b-c)e^{b(\xi +C)}-1}\right]$.

• Family 13: When $c=-a$, then $\varPhi (\xi )=2\arctan \left[ \frac{e^{b(\xi +C)}+b-a}{e^{b(\xi +C)}-b-a}\right]$.

• Family 14: When $b=-c$, then $\varPhi (\xi )=-2\arctan \left[ \frac{ae^{a(\xi +C)}}{ce^{a(\xi +C)}-1}\right]$.

• Family 15: When ${\mathrm {b}}=0$ and $a=c$, then $\varPhi (\xi )=-2\arctan \left[ \frac{c(\xi +C)+2}{c(\xi +C)}\right] .$

• Family 16: When $a=0$ and $b=c$, then $\varPhi (\xi )=2\arctan \left[ c(\xi +C)\right] .$

• Family 17: When $a=0$ and $b=-c$, then $\varPhi (\xi )=-2\arctan \left[ \frac{1}{c(\xi +C)}\right] .$

• Family 18: When $a=b=0$ then $\varPhi (\xi )=c\xi +C.$

• Family 19: When $b=c$ then $\varPhi (\xi )=2\arctan \left[ \frac{e^{a(\xi +C)}-b}{a}\right] ,$

where $A_k, B_k (k=1,2,\ldots ,m), a, b$ and c are constants to be determined later. But, the positive integer ${\mathrm {m}}$ can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in Eq. (7). If m is not an integer, then a transformation formula should be used to overcome this difficulty.

Step 3. Substituting (6) into Eq. (5) with the value of m obtained in Step 2. Collecting the coefficients of $\tan ^k\left( \frac{\varPhi (\xi )}{2}\right)$, $\cot ^k\left( \frac{\varPhi (\xi )}{2}\right) (k=0,1,2,\ldots )$, then setting each coefficient to zero, we can get a set of over-determined equations for $A_0, A_k, B_k (k = 1, 2,\ldots , m)$ a, b, c and p with the aid of symbolic computation Maple.

Step 4. Solving the algebraic equations in Step 3, then substituting $A_0, A_1, B_1, \ldots , A_m, B_m, \mu , p$ in (6).

3 The Tzitzéica type nonlinear evolution equations

In this section, we will exert the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion technique to obtain new and more general exact solutions of the Tzitzéica type nonlinear evolution  equations.

3.1 The Tzitzéica equation

We consider the Tzitzéica equation as follows

$$u_{tt}-u_{xx}-e^{u}+e^{-2u}=0,$$

(8)

by using the transformation $v(x,t)=e^{-u(x,t)}$, Eq. (8) transforms into the following partial differential equation,

$$vv_{tt}-v_t^2-vv_{xx}+v_x^2-v^3+1=0.$$

(9)

The travelling wave transformation $\xi =kx+wt$ reduces Eq. (9) to the following ODE

$$(w^2-k^2)(vv''-(v')^2)-v^3+1=0,$$

(10)

where $v'=\frac{d v}{d \xi }$ and $v''=\frac{d^2 v}{d \xi ^2}$. Balancing the terms $vv''$ and $v^3$ by using homogenous principle the following result could be obtained

$$2M+2=3M,\qquad M=2.$$

(11)

Here, for simplicity, we set $p=0$. Then (6) reduces to

$$v(\xi )=A_{0}+A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) + B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(12)

Substituting (12) and (7) into Eq. (9) and by using the well-known Maple software, we obtain the following sets of non-trivial solutions

Set I: We have the following:

$$a=\sqrt{c^2-b^2-\frac{3}{\varDelta }}, \quad \varDelta =k^2-w^2, \quad b=b, \quad c=c, \quad A_{0}=\frac{1}{2}\varDelta (b^2-c^2)+1, \quad A_{1} =0, \quad A_2=0, \quad k=k, \quad w=w,$$

(13)

$$B_1=(b+c)\sqrt{-\varDelta ^2(b^2-c^2)-3\varDelta }, \quad B_2=-\frac{1}{2}\varDelta (b+c)^2, \quad v(\xi )=A_{0}+ B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) ,$$

(14)

where a, b and c are arbitrary constants. Using the transformations $u=\ln v$ and according to (12), we obtain several types of travelling wave solutions of Eq. (9) as follows:

By using of the Families 1, 2, 3, 4, 11 and 19 can be written, respectively, as

$$\begin{aligned} u_{1}(\xi )= & {} \ln \left\{ 1+\frac{1}{2}\varDelta (b^2-c^2)+(b^2-c^2)\varDelta \left[ 1-\sqrt{-\frac{3}{3+\varDelta (b^2-c^2)}} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\varDelta ^2(b^2-c^2)^2\left[ \sqrt{-\varDelta (b^2-c^2)-3}-\sqrt{3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-2}\right\} , \end{aligned}$$

(15)

Fig. 1

Graphs of $u_1$ real values of (15) are demonstrated at a $k=5, w=3, b=2, c=3,$ b $k=3, w=2, b=2, c=4,$ when $-10\,<\,x\,<\,10,\,0\,<\,t\,<\,10$

$$\begin{aligned} u_{2}(\xi )= & {} \ln \left\{ 1+\frac{1}{2}\varDelta (b^2-c^2)+(b^2-c^2)\varDelta \left[ 1+\sqrt{\frac{3}{3+\varDelta (b^2-c^2)}} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\varDelta ^2(b^2-c^2)^2\left[ \sqrt{-\varDelta (b^2-c^2)-3}+\sqrt{-3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-2}\right\} , \end{aligned}$$

(16)

$$\begin{aligned} u_{3}(\xi )= & {} \ln \left\{ 1+\frac{1}{2}\varDelta b^2+b^2\varDelta \left[ 1+\sqrt{\frac{3}{3+\varDelta b^2}} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\varDelta ^2 b^4\left[ \sqrt{-\varDelta b^2-3}+\sqrt{-3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-2}\right\} , \end{aligned}$$

(17)

$$\begin{aligned} u_{4}(\xi )= & {} \ln \left\{ 1-\frac{1}{2}\varDelta c^2-c^2\varDelta \left[ 1-\sqrt{\frac{3}{\varDelta c^2-3}} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\varDelta ^2c^4\left[ \sqrt{\varDelta c^2-3}-\sqrt{3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^{-2}\right\} , \end{aligned}$$

(18)

$$\begin{aligned} u_{5}(\xi )= & {} \ln \left\{ 1+\frac{1}{2}(3-a^2\varDelta )-\left( a+\sqrt{\frac{3}{\varDelta }}\right) \sqrt{-\varDelta (6-a^2\varDelta )} \left[ \frac{\left( a+\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}{\left( a-\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}\right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\varDelta \left( a+\sqrt{\frac{3}{\varDelta }}\right) ^2\left[ \frac{\left( a+\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}{\left( a-\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}\right] ^{-2}\right\} , \end{aligned}$$

(19)

$$u_{6}(\xi )=\ln \left\{ 1+6b \left[ e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-b\right] ^{-1}-6b^2\left[ e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-b\right] ^{-2}\right\} ,$$

(20)

where $\xi =kx+wt$.

Fig. 2

Graphs of $u_2$ real values of (16) are demonstrated at c $k=1, w=3, b=3, c=1,$ d $k=2, w=5, b=5, c=3,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

Set II: We have the following:

$$a=\sqrt{c^2-b^2-\frac{3}{\varDelta }}, \quad \varDelta =k^2-w^2, \quad b=b, \quad c=c, \quad A_{0}=1+\frac{1}{2}\varDelta (b^2-c^2), \quad B_{1} =0, \quad B_2=0, \quad k=k, \quad w=w,$$

(21)

$$A_1= (b-c)\sqrt{-\varDelta ^2(b^2-c^2)-3\varDelta }, \quad A_2=-\frac{1}{2}\varDelta (b-c)^2, \quad v(\xi )=A_{0}+ A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(22)

By using of the Families 1, 2, 3, 4, 11 and 14 can be written, respectively, as

$$\begin{aligned} u_{7}(\xi )= & {} \ln \left\{ -2-\frac{1}{2}\varDelta (b^2-c^2)-\sqrt{-3\varDelta (b^2-c^2)-9} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right. \\&\left. -\,\frac{1}{2}\left[ \sqrt{-\varDelta (b^2-c^2)-3}-\sqrt{3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} , \end{aligned}$$

(23)

Fig. 3

Graphs of $u_6$ real values of (20) are demonstrated at a1 $k=1, w=2, b=2,$ b1 $k=1, w=2, b=5,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$\begin{aligned} u_{8}(\xi )= & {} \ln \left\{ -2-\frac{1}{2}\varDelta (b^2-c^2)+\sqrt{3\varDelta (b^2-c^2)+9} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right. \\&\left. -\,\frac{1}{2}\left[ \sqrt{-\varDelta (b^2-c^2)-3}-\sqrt{-3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} , \end{aligned}$$

(24)

$$\begin{aligned} u_{9}(\xi )= & {} \ln \left\{ -2-\frac{1}{2}\varDelta b^2-\sqrt{-3\varDelta b^2-9} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right. \\&\left. -\,\frac{1}{2}\left[ \sqrt{-\varDelta b^2-3}-\sqrt{3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} , \end{aligned}$$

(25)

$$\begin{aligned} u_{10}(\xi )= & {} \ln \left\{ -2+\frac{1}{2}\varDelta c^2+\sqrt{-3\varDelta c^2+9} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right. \\&\left. -\frac{1}{2}\left[ \sqrt{\varDelta c^2-3}-\sqrt{-3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} , \\ \end{aligned}$$

(26)

Fig. 4

Graphs of $u_{28}$ real values of (54) are demonstrated at c1 $k=1, w=2$ and d1 $k=2, w=1,$ when $-10\,<\,x\,<\,10, \,0\,<\,t<\,10$

$$\begin{aligned} u_{11}(\xi )= & {} \ln \left\{ 1+\frac{1}{2}(3-a^2\varDelta )-\left( \sqrt{\frac{3}{\varDelta }}-a\right) \sqrt{-\varDelta (6-a^2\varDelta )} \left[ \frac{\left( a+\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}{\left( a-\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}\right] \right. \\&\left. -\,\frac{1}{2}\varDelta \left( \sqrt{\frac{3}{\varDelta }}-a\right) ^2\left[ \frac{\left( a+\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}{\left( a-\sqrt{\frac{3}{\varDelta }}\right) e^{\sqrt{\frac{3}{\varDelta }}(\xi +C)}-1}\right] ^{2}\right\} , \end{aligned}$$

(27)

$$\begin{aligned} u_{12}(\xi )= & {} \ln \left\{ 1-2b\sqrt{-3\varDelta } \left[ \frac{\sqrt{-\frac{3}{\varDelta }}e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}}{be^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-1}\right] - 2b^2\varDelta \left[ \frac{\sqrt{-\frac{3}{\varDelta }}e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}}{be^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-1}\right] ^{2}\right\} , \end{aligned}$$

(28)

where $\xi =kx+wt$.

Set III: We have the following:

$$a= \frac{1}{\sqrt{-\varDelta }}, \quad \varDelta =k^2-w^2, \quad b=\sqrt{c^2-\frac{2}{\varDelta }}, \quad c=c, \quad A_{0}=0, \quad B_{1} =0, \quad B_2=0, \quad k=k, \quad w=w,$$

(29)

$$A_1= (c-b)\sqrt{-\varDelta }, \quad A_2=1-c^2\varDelta +c\sqrt{2\varDelta -c^2\varDelta ^2}, \quad v(\xi )= A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(30)

By using of the Families 1, 2, 11 and 13 can be written, respectively, as

$$u_{13}(\xi )=\ln \left\{ -1+\sqrt{-3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) -\frac{1}{2}\left[ 1-\sqrt{-3} \tan \left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} ,$$

(31)

Fig. 5

Graphs of $u_{1}$ real values of (73) are demonstrated at a $k=1, w=-2, b=1, c=2$ and b $k=3, w=-6, b=2, c=4,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$u_{14}(\xi )= \ln \left\{ -1-\sqrt{3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) -\frac{1}{2}\left[ 1+\sqrt{3} \tanh \left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(\xi +C)\right) \right] ^2\right\} ,$$

(32)

$$u_{15}(\xi )= \ln \left\{ (1-\sqrt{3}) \left[ \frac{\left( 1+\sqrt{3}\right) e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\sqrt{-\varDelta }}{\left( 1-\sqrt{3}\right) e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\sqrt{-\varDelta }}\right] + (2+\sqrt{3})\left[ \frac{\left( 1+\sqrt{3}\right) e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\sqrt{-\varDelta }}{\left( 1-\sqrt{3}\right) e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\sqrt{-\varDelta }}\right] ^{2}\right\} ,$$

(33)

$$u_{16}(\xi )= \ln \left\{ -(1+\sqrt{3})\left[ \frac{\sqrt{-\varDelta } e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\left( \sqrt{3}-1\right) }{\sqrt{-\varDelta } e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\left( \sqrt{3}+1\right) }\right] + (2+\sqrt{3})\left[ \frac{\sqrt{-\varDelta } e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\left( \sqrt{3}-1\right) }{\sqrt{-\varDelta } e^{\sqrt{-\frac{3}{\varDelta }}(\xi +C)}-\left( \sqrt{3}+1\right) }\right] ^{2}\right\} ,$$

(34)

where $\xi =kx+wt$.

Set IV: We have the following:

$$a= \sqrt{\frac{1-\sqrt{3}i}{2\varDelta }}, \quad \varDelta =k^2-w^2, \quad b=\sqrt{c^2+\frac{1-\sqrt{3}i}{\varDelta }}, \quad c=c, \quad A_{0}=0, \quad B_{1} =0, \quad B_2=0, \quad k=k, \quad w=w,$$

(35)

$$A_1= (b-c)\sqrt{\frac{1-\sqrt{3}i}{2}\varDelta }, \quad A_2=-\frac{1}{2}\varDelta (b-c)^2, \quad v(\xi )= A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(36)

By using of the Families 1, 2 and 11 can be written, respectively, as

$$u_{17}(x,t)=\ln \left\{ \frac{1-\sqrt{3}i}{4}\left( 2 \left[ 1-\sqrt{-3} \tan \left( \sqrt{\frac{3\sqrt{3}i-3}{8\varDelta }}(kx+wt)\right) \right] - \left[ 1-\sqrt{-3} \tan \left( \sqrt{\frac{3\sqrt{3}i-3}{8\varDelta }}(kx+wt)\right) \right] ^2\right) \right\} ,$$

(37)

Fig. 6

Graphs of $u_{2}$ real values of (74) are demonstrated at c $k=1, w=2, b=3, c=2$ and d $k=2, w=1, b=4, c=2,$ when $-10\,<\,x\,<\,10, \,0\,<\,t<\,10$

$$u_{18}(x,t)= \ln \left\{ \frac{1-\sqrt{3}i}{4}\left( 2 \left[ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3-3\sqrt{3}i}{8\varDelta }}(kx+wt)\right) \right] - \left[ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3-3\sqrt{3}i}{8\varDelta }}(kx+wt)\right) \right] ^2\right) \right\} ,$$

(38)

$$\begin{aligned} u_{19}(x,t)= & {} \ln \left\{ \frac{1}{2}(\sqrt{3}-1)(1-\sqrt{3}i) \left[ \frac{\left( 1+\sqrt{3}\right) \sqrt{1-\sqrt{3}i} e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-\sqrt{2\varDelta }}{\left( 1-\sqrt{3}\right) \sqrt{1-\sqrt{3}i}e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-\sqrt{2\varDelta }}\right] \right. \\&\left. -\,\frac{1}{4}(\sqrt{3}-1)^2(1-\sqrt{3}i)\left[ \frac{\left( 1+\sqrt{3}\right) \sqrt{1-\sqrt{3}i} e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-\sqrt{2\varDelta }}{\left( 1-\sqrt{3}\right) \sqrt{1-\sqrt{3}i}e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-\sqrt{2\varDelta }}\right] ^{2}\right\} . \end{aligned}$$

(39)

Set V: We have the following:

$$a= \sqrt{(b^2-c^2)i-\frac{3\sqrt{3}-3i}{2\varDelta }}, \quad \varDelta =k^2-w^2, \quad b=b, \quad c=c, \quad A_{0}=\frac{1-2i}{6}\varDelta (b^2-c^2)-\frac{\sqrt{3}-i}{2}, \quad A_{1} =0,$$

(40)

$$\begin{aligned} A_2= & {} 0, \quad B_1=-(b+c)\varDelta \sqrt{(b^2-c^2)i-\frac{3\sqrt{3}-3i}{2\varDelta }}, B_2=-\frac{1}{2}\varDelta (b+c)^2, k=k, \quad w=w, \\&v(\xi )=A_{0}+ B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) . \end{aligned}$$

(41)

By using of the Families 1, 2, 14 and 18 can be written, respectively, as

Fig. 7

Graphs of $u_{4}$ real values of (76) are demonstrated at a1 $k=5, w\,=\,6, c=4$ and b1 $k=-10, w=6, c=10,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$\begin{aligned} u_{20}(x,t)= & {} \ln \left\{ \frac{1-2i}{6}\varDelta (b^2-c^2)-\frac{\sqrt{3}-i}{2}+(b^2-c^2) \left[ 1-\sqrt{\frac{2\varDelta (c^2-b^2)(1+i)+3\sqrt{3}-3i}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}} \right. \right. \\&\left. \left. \times \, \tan \left( \frac{\sqrt{\frac{3\sqrt{3}-3i}{2\varDelta }-(b^2-c^2)(1+i)}}{2}(kx+wt)\right) \right] ^{-1} -\frac{\varDelta ^2(b^2-c^2)^2}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}\right. \\&\left. \times \,\left[ 1-\sqrt{\frac{2\varDelta (c^2-b^2)(1+i)+3\sqrt{3}-3i}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}} \tan \left( \frac{\sqrt{\frac{3\sqrt{3}-3i}{2\varDelta }-(b^2-c^2)(1+i)}}{2}(kx+wt)\right) \right] ^{-2}\right\} , \end{aligned}$$

(42)

$$\begin{aligned} u_{21}(x,t)= & {} \ln \left\{ \frac{1-2i}{6}\varDelta (b^2-c^2)-\frac{\sqrt{3}-i}{2}+(b^2-c^2) \left[ 1+\sqrt{\frac{2\varDelta (b^2-c^2)(1+i)-3\sqrt{3}+3i}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}} \right. \right. \\&\left. \left. \times \, \tanh \left( \frac{\sqrt{\frac{-3\sqrt{3}+3i}{2\varDelta }+(b^2-c^2)(1+i)}}{2}(kx+wt)\right) \right] ^{-1} -\frac{\varDelta ^2(b^2-c^2)^2}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}\right. \\&\left. \times \,\left[ 1+\sqrt{\frac{2\varDelta (b^2-c^2)(1+i)-3\sqrt{3}+3i}{2\varDelta (b^2-c^2)i-3\sqrt{3}+3i}} \tanh \left( \frac{\sqrt{\frac{-3\sqrt{3}+3i}{2\varDelta }+(b^2-c^2)(1+i)}}{2}(kx+wt)\right) \right] ^{-2}\right\} , \end{aligned}$$

(43)

Fig. 8

Graphs of $u_{24}$ real values of (108) are demonstrated at c1 $w=2, b=2$ and d1 $w=10, b=-5,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$\begin{aligned} u_{22}(x,t)= & {} \ln \left\{ -\frac{\sqrt{3}-i}{2}-2c\sqrt{\varDelta \left( \frac{3\sqrt{3}-3i}{2}\right) } \left[ \frac{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }}e^{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }}(\xi +C)}}{ce^{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }}(\xi +C)}-1}\right] ^{-1} \right. \\&\left. -\,2\varDelta c^2\left[ \frac{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }} e^{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }}(\xi +C)}}{ce^{\sqrt{-\frac{3\sqrt{3}-3i}{2\varDelta }}(\xi +C)}-1}\right] ^{-2}\right\} , \end{aligned}$$

(44)

$$\begin{aligned} u_{23}(x,t)= & {} \ln \left\{ \frac{(4-\sqrt{3})i-1-4\sqrt{3}}{4}+ (\sqrt{3}+i) \cot \left( \sqrt{\frac{3+3\sqrt{3}i}{8\varDelta }}(kx+wt)\right) \right. \\&\left. +\,\frac{3+3\sqrt{3}i}{4}\cot ^2\left( \sqrt{\frac{3+3\sqrt{3}i}{8\varDelta }}(kx+wt)\right) \right\} . \end{aligned}$$

(45)

Set VI: We have the following:

$$a= 0, \quad b=\sqrt{c^2-\frac{3}{\varDelta }}, \quad c=c, \quad A_{1} =0, \quad A_2=\frac{3}{2}-c^2\varDelta +c\varDelta \sqrt{c^2-\frac{3}{\varDelta }}, \quad A_{0}=-\frac{1}{2},$$

(46)

$$B_1= 0, \quad B_2=0, \quad k=k, \quad w=w, \quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(47)

By using of the Families 5, 6 and 18 can be written, respectively, as

$$u_{24}(x,t)= \ln \left\{ -\frac{1}{2}+\left( \frac{3}{2}+c\varDelta \sqrt{c^2-\frac{3}{\varDelta }}-c^2\varDelta \right) \frac{\sqrt{c^2-\frac{3}{\varDelta }}+c}{\sqrt{c^2-\frac{3}{\varDelta }}-c}\tanh ^2\left( \frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(kx+wt)\right) \right\} ,$$

(48)

$$u_{25}(x,t)= \ln \left\{ -\frac{1}{2}+\frac{3}{2} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{-\frac{12}{\varDelta }}(kx+wt)}-1}{e^{\sqrt{-\frac{12}{\varDelta }}(kx+wt)}+1}, \ \frac{2e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}}{e^{\sqrt{-\frac{12}{\varDelta }}(kx+wt)}+1}\right] \right) \right\} ,$$

(49)

$$u_{26}(x,t)= \ln \left\{ -\frac{1}{2}-\frac{3}{2} \tan ^2\left( \sqrt{\frac{3}{4\varDelta }}(kx+wt)\right) \right\} .$$

(50)

Set VII: We have the following:

$$a= 0, \quad b=\sqrt{ic^2-\frac{3i+3\sqrt{3}}{2\varDelta }}, \quad c=c, \quad A_{1} =0, \quad A_2=-\frac{1}{2}\varDelta (b-c)^2,$$

(51)

$$A_{0}= \frac{c^2(i-1)\varDelta }{6}+\frac{i+\sqrt{3}}{4}, \quad B_1=0, \quad B_2=0, \quad k=k, \quad w=w, \quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(52)

By using of the Families 5, 6 and 18 can be written, respectively, as

$$u_{27}(x,t)= \ln \left\{ \frac{c^2(i-1)\varDelta }{6}+\frac{i+\sqrt{3}}{4}-\frac{1}{2}\varDelta \left( (i-1)c^2-\frac{3i+3\sqrt{3}}{2\varDelta }\right) \tanh ^2\left( \frac{1}{2}\sqrt{(i-1)c^2-\frac{3i+3\sqrt{3}}{2\varDelta }}(kx+wt)\right) \right\} ,$$

(53)

$$u_{28}(x,t)= \ln \left\{ \frac{i+\sqrt{3}}{4}+\frac{3i+3\sqrt{3}}{4} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{-\frac{6i+6\sqrt{3}}{\varDelta }}(kx+wt)}-1}{e^{\sqrt{-\frac{6i+6\sqrt{3}}{\varDelta }}(kx+wt)}+1}, \ \frac{2e^{\sqrt{-\frac{3i+3\sqrt{3}}{2\varDelta }}(2kx+2wt)}}{e^{\sqrt{-\frac{6i+6\sqrt{3}}{\varDelta }}(kx+wt)}+1}\right] \right) \right\} ,$$

(54)

$$u_{29}(x,t)= \ln \left\{ \frac{(2\sqrt{3}-1)+(2+\sqrt{3})i}{4}-\frac{3-3\sqrt{3}i}{4} \tan ^2\left( \sqrt{\frac{3-3\sqrt{3}i}{8\varDelta }}(kx+wt)\right) \right\} .$$

(55)

Set VIII: We have the following:

$$a= 0, \quad b=\sqrt{c^2-\frac{3}{4\varDelta }}, \quad c=c, \quad A_{0}=\frac{1}{4}, \quad A_{1} =0, \quad A_2=\frac{3}{8}-\varDelta c^2+c\varDelta \sqrt{c^2-\frac{3}{4\varDelta }}, \quad B_1=0,$$

(56)

$$B_2= \frac{3}{8}-\varDelta c^2-c\varDelta \sqrt{c^2-\frac{3}{4\varDelta }}, \quad k=k, \quad w=w, \quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) ,$$

(57)

where $\varDelta =k^2-w^2$. By using of the Families 5, 6 and 18 can be written, respectively, as

$$\begin{aligned} u_{30}(x,t)= & {} \ln \left\{ \frac{1}{4}+\left( \frac{3}{8}-\varDelta c^2+c\varDelta \sqrt{c^2-\frac{3}{4\varDelta }}\right) \frac{\sqrt{c^2-\frac{3}{4\varDelta }}+c}{\sqrt{c^2-\frac{3}{4\varDelta }}-c} \tanh ^2\left( \frac{1}{4}\sqrt{-\frac{3}{\varDelta }}(kx+wt)\right) \right. \\&\left. +\,\left( \frac{3}{8}-\varDelta c^2-c\varDelta \sqrt{c^2-\frac{3}{4\varDelta }}\right) \frac{\sqrt{c^2-\frac{3}{4\varDelta }}-c}{\sqrt{c^2-\frac{3}{4\varDelta }}+c} \coth ^2\left( \frac{1}{4}\sqrt{-\frac{3}{\varDelta }}(kx+wt)\right) \right\} , \end{aligned}$$

(58)

$$\begin{aligned} u_{31}(x,t)= & {} \ln \left\{ \frac{1}{4}+\frac{3}{8} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-1}{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}+1}, \frac{2e^{\frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(kx+wt)}}{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}+1}\right] \right) \right. \\&\left. +\,\frac{3}{8} \cot ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}-1}{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}+1}, \frac{2e^{\frac{1}{2}\sqrt{-\frac{3}{\varDelta }}(kx+wt)}}{e^{\sqrt{-\frac{3}{\varDelta }}(kx+wt)}+1}\right] \right) \right\} , \end{aligned}$$

(59)

$$u_{32}(x,t)= \ln \left\{ \frac{1}{4}-\frac{3}{8}\tan ^2\left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(kx+wt)\right) - \frac{3}{8}\cot ^2\left( \frac{1}{2}\sqrt{\frac{3}{\varDelta }}(kx+wt)\right) \right\} .$$

(60)

Set IX: We have the following:

$$a=0,b=\sqrt{c^2+\frac{3-3\sqrt{3}i}{8\varDelta }}, \quad c=c, \quad A_{1} =0,\quad A_2=-\frac{1}{2}\varDelta (b-c)^2,\quad A_{0}=\frac{1}{3}(c^2-b^2)\varDelta , \quad B_1=0,$$

(61)

$$B_2=-\frac{1}{2}\varDelta (b+c)^2,\quad k=k, \quad w=w, \quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) ,$$

(62)

where $\varDelta =k^2-w^2$. By using of the Families 5, 6 and 18 can be written, respectively, as

$$u_{33}(x,t)= \ln \left\{ \frac{3\sqrt{3}i-3}{24}+\frac{3\sqrt{3}i-3}{16}\left[ \tanh ^2\left( \frac{1}{4}\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)\right) +\coth ^2\left( \frac{1}{4}\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)\right) \right] \right\} ,$$

(63)

$$\begin{aligned} u_{34}(x,t)= & {} \ln \left\{ \frac{3\sqrt{3}i-3}{24}+\frac{3\sqrt{3}i-3}{16} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}-1}{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}+1}, \frac{2e^{\frac{1}{2}\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}}{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}+1}\right] \right) \right. \\&\left. +\,\frac{3\sqrt{3}i-3}{16} \cot ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}-1}{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}+1}, \frac{2e^{\frac{1}{2}\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}}{e^{\sqrt{\frac{3-3\sqrt{3}i}{2\varDelta }}(kx+wt)}+1}\right] \right) \right\} , \end{aligned}$$

(64)

$$u_{35}(x,t)= \ln \left\{ \frac{3\sqrt{3}i-3}{24}-\frac{3\sqrt{3}i-3}{16}\tan ^2\left( \frac{1}{4}\sqrt{\frac{3\sqrt{3}i-3}{2\varDelta }}(kx+wt)\right) - \frac{3\sqrt{3}i-3}{16}\cot ^2\left( \frac{1}{4}\sqrt{\frac{3\sqrt{3}i-3}{2\varDelta }}\xi +C\right) \right\} .$$

(65)

3.2 Dodd–Bullough–Mikhailov (DBM) equation

Now we consider the Dodd–Bullough–Mikhailov (DBM) equation as follows

$$u_{xt}+e^{u}+e^{-2u}=0,$$

(66)

by using the transformation $v(x,t)=e^{-u(x,t)}$, Eq. (66) transforms into the following partial differential equation,

$$vv_{xt}-v_xv_{t}+v^3+1=0.$$

(67)

The travelling wave transformation ${\xi =kx+wt}$ reduces Eq. (67) to the following ODE

$$kw(vv''-(v')^2)+v^3+1=0,$$

(68)

where $v'=\frac{d v}{d \xi }$ and $v''=\frac{d^2 v}{d \xi ^2}$. Balancing the terms $vv''$ and $v^3$ by using homogenous principle one can be found

$$2M+2=3M,\quad M=2.$$

(69)

Here, for simplicity, we set $p=0$. Then (6) reduces to

$$v(\xi )=A_{0}+A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) + B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(70)

Substituting (70) and (7) into Eq. (68), the following solutions are obtained.

Set I: We have the following:

$$a=\sqrt{c^2-b^2+\frac{3}{kw}}, \quad b=b, \quad c=c, A_{0}=-1+\frac{1}{2}kw(b^2-c^2), \quad A_{1} =0, \quad A_2=0, \quad k=k, \quad w=w,$$

(71)

$$B_1=-(b+c)kw\sqrt{c^2-b^2+\frac{3}{kw}}, \quad B_2=-\frac{1}{2}kw(b+c)^2, \quad v(\xi )=A_{0}+ B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) ,$$

(72)

By using of the Families 1, 2, 12 and 19 can be written, respectively, as

$$\begin{aligned} u_{1}(x,t)= & {} \ln \left\{ -1+\frac{1}{2}kw(b^2-c^2)-(b^2-c^2)kw \left[ 1-\sqrt{-\frac{3}{3-kw(b^2-c^2)}} \tan \left( \frac{1}{2}\sqrt{-\frac{3}{kw}}(kx+wt)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\frac{k^2w^2(b^2-c^2)^2}{3-(b^2-c^2)kw}\left[ 1-\sqrt{-\frac{3}{3-kw(b^2-c^2)}} \tan \left( \frac{1}{2}\sqrt{-\frac{3}{kw}}(kx+wt)\right) \right] ^{-2}\right\} ,\end{aligned}$$

(73)

$$\begin{aligned} u_{2}(x,t)= & {} \ln \left\{ -1+\frac{1}{2}kw(b^2-c^2)-(b^2-c^2)kw \left[ 1+\sqrt{\frac{3}{3-kw(b^2-c^2)}} \tanh \left( \frac{1}{2}\sqrt{\frac{3}{kw}}(kx+wt)\right) \right] ^{-1}\right. \\&\left. -\,\frac{1}{2}\frac{k^2w^2(b^2-c^2)^2}{3-(b^2-c^2)kw}\left[ 1+\sqrt{\frac{3}{3-kw(b^2-c^2)}} \tanh \left( \frac{1}{2}\sqrt{\frac{3}{kw}}(kx+wt)\right) \right] ^{-2}\right\} , \end{aligned}$$

(74)

$$\begin{aligned} u_{3}(x,t)= & {} \ln \left\{ -1+\frac{1}{2}(3-kwc^2)-\left( \sqrt{\frac{3}{kw}}+c\right) kwc \left[ \frac{\left( \sqrt{\frac{3}{kw}}+c\right) e^{\sqrt{\frac{3}{kw}}(kx+wt)}+1}{\left( \sqrt{\frac{3}{kw}}-c\right) e^{\sqrt{\frac{3}{kw}}(kx+wt)}-1}\right] ^{-1}\right. \\&\left. -\,\frac{1}{2}kw\left( \sqrt{\frac{3}{kw}}+c\right) ^2 \left[ \frac{\left( \sqrt{\frac{3}{kw}}+c\right) e^{\sqrt{\frac{3}{kw}}(kx+wt)}+1}{\left( \sqrt{\frac{3}{kw}}-c\right) e^{\sqrt{\frac{3}{kw}}(kx+wt)}-1}\right] ^{-2}\right\} , \end{aligned}$$

(75)

$$u_{4}(x,t)=\ln \left\{ -1-6c \left[ \frac{1}{e^{\sqrt{\frac{3}{kw}}(kx+wt)}-c}\right] -6c^2 \left[ \frac{1}{e^{\sqrt{\frac{3}{kw}}(kx+wt)}-c}\right] ^{2}\right\} .$$

(76)

Set II: We have the following:

$$a=\sqrt{c^2-b^2-\frac{3+3\sqrt{3}i}{2kw}}, \quad b=b, \quad c=c, A_{0}=\frac{1+\sqrt{3}i}{2}+\frac{1}{2}kw(b^2-c^2),\quad A_{1} =0,\quad A_2=0, \quad k=k, \quad w=w,$$

(77)

$$B_1=-(b+c)kw\sqrt{c^2-b^2-\frac{3+3\sqrt{3}i}{2kw}},\quad B_2=-\frac{1}{2}kw(b+c)^2,\quad v(\xi )=A_{0}+ B_{1}\cot \left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(78)

By using of the Families 1, 2, 12 and 19 can be written, respectively, as

$$\begin{aligned} u_{5}(x,t)= & {} \ln \left\{ \frac{1+\sqrt{3}i+kw(b^2-c^2)}{2}-(b^2-c^2)kw\left[ 1-\sqrt{-\frac{3+3\sqrt{3}i}{3+3\sqrt{3}i+2kw(b^2-c^2)}} \tan \left( \sqrt{\frac{3+3\sqrt{3}i}{8kw}}(kx+wt)\right) \right] ^{-1}\right. \\&\left. +\,\frac{1}{2}\frac{k^2w^2(b^2-c^2)^2}{(b^2-c^2)kw+3+3\sqrt{3}i}\left[ 1-\sqrt{-\frac{3+3\sqrt{3}i}{3+3\sqrt{3}i+2kw(b^2-c^2)}} \tan \left( \sqrt{\frac{3+3\sqrt{3}i}{8kw}}(kx+wt)\right) \right] ^{-2}\right\} , \end{aligned}$$

(79)

$$\begin{aligned} u_{6}(x,t)= & {} \ln \left\{ \frac{1+\sqrt{3}i+kw(b^2-c^2)}{2}-(b^2-c^2)kw\left[ 1+\sqrt{\frac{3+3\sqrt{3}i}{3+3\sqrt{3}i+2kw(b^2-c^2)}} \tanh \left( \sqrt{-\frac{3+3\sqrt{3}i}{8kw}}(kx+wt)\right) \right] ^{-1}\right. \\&\left. +\,\frac{1}{2}\frac{k^2w^2(b^2-c^2)^2}{(b^2-c^2)kw+3+3\sqrt{3}i}\left[ 1+\sqrt{\frac{3+3\sqrt{3}i}{3+3\sqrt{3}i+2kw(b^2-c^2)}} \tanh \left( \sqrt{-\frac{3+3\sqrt{3}i}{8kw}}(kx+wt)\right) \right] ^{-2}\right\} , \end{aligned}$$

(80)

$$\begin{aligned} u_{7}(x,t)= & {} \ln \left\{ -\frac{1+\sqrt{3}i+2kwc^2}{4}-\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}+c\right) kwc \left[ \frac{\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}+c\right) e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}+1}{\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}-c\right) e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}-1}\right] ^{-1}\right. \\&\left. -\,\frac{1}{2}kw\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}+c\right) ^2 \left[ \frac{\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}+c\right) e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}+1}{\left( \sqrt{-\frac{3+3\sqrt{3}i}{2kw}}-c\right) e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}-1}\right] ^{-2}\right\} , \end{aligned}$$

(81)

$$u_{8}(x,t)=\ln \left\{ \frac{1+\sqrt{3}i}{2}+ \left[ \frac{c(3+3\sqrt{3}i)}{e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}-c}\right] +c^2(3+3\sqrt{3}i) \left[ \frac{1}{e^{\sqrt{-\frac{3+3\sqrt{3}i}{2kw}}(kx+wt)}-c}\right] ^{2} \right\} .$$

(82)

Set III: We have the following:

$$a=a,\quad b=b,\quad c=c, A_{0}=-\frac{2a^2+c^2-b^2}{2(a^2+b^2-c^2)},\quad A_{1} =\frac{3a(b-c)}{a^2+b^2-c^2}, A_2=-\frac{3(b-c)^2}{2(a^2+b^2-c^2)},$$

(83)

$$B_1=0,\quad B_2=0,\quad k=\frac{3}{w(a^2+b^2-c^2)},\quad w=w, v(\xi )=A_{0}+ A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(84)

By using of the Families 1, 2, 6, 10, 12 and 18 can be written, respectively, as

$$\begin{aligned} u_{9}(x,t)= & {} \ln \left\{ \frac{6a^2+b^2-c^2}{2(a^2+b^2-c^2)}+\frac{3a}{\sqrt{c^2-b^2-a^2}} \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( \frac{3}{w(a^2+b^2-c^2)}x+wt\right) \right) \right. \\&\left. -\,\frac{3}{2(a^2+b^2-c^2)} \left[ a-\sqrt{c^2-b^2-a^2}\tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( \frac{3}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^2\right\} , \end{aligned}$$

(85)

$$\begin{aligned} u_{10}(x,t)= & {} \ln \left\{ \frac{6a^2+b^2-c^2}{2(a^2+b^2-c^2)}-\frac{3a}{\sqrt{a^2+b^2-c^2}} \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( \frac{3}{w(a^2+b^2-c^2)}x+wt\right) \right) \right. \\&\left. -\,\frac{3}{2(a^2+b^2-c^2)} \left[ a+\sqrt{a^2+b^2-c^2}\tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( \frac{3}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^2\right\} , \end{aligned}$$

(86)

$$u_{11}(x,t)=\ln \left\{ \frac{1}{2}-\frac{3}{2}tan^2 \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{wb^2}x+wt\right) }-1}{e^{2b\left( \frac{3}{wb^2}x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{wb^2}x+wt\right) }}{e^{2b\left( \frac{3}{wb^2}x+wt\right) }+1}\right] \right) \right\} ,$$

(87)

$$u_{12}(x,t)=\ln \left\{ -1-6 \left[ \frac{e^{ka\left( \frac{3}{wk^2a^2}x+wt\right) }}{1-e^{ka\left( \frac{3}{wk^2a^2}x+wt\right) }}\right] -6 \left[ \frac{e^{ka\left( \frac{3}{wk^2a^2}x+wt\right) }}{1-e^{ka\left( \frac{3}{wk^2a^2}x+wt\right) }}\right] ^2\right\} ,$$

(88)

$$u_{13}(x,t)=\ln \left\{ \frac{b^2-3c^2}{2b^2}+\frac{3c(b-c)}{b^2} \left[ \frac{(b+c)e^{b\left( \frac{3}{wb^2}x+wt\right) }+1}{(b-c)e^{b\left( \frac{3}{wb^2}x+wt\right) }-1}\right] -\frac{3(b-c)^2}{2b^2} \left[ \frac{(b+c)e^{b\left( \frac{3}{wb^2}x+wt\right) }+1}{(b-c)e^{b\left( \frac{3}{wb^2}x+wt\right) }-1}\right] ^2\right\} ,$$

(89)

$$u_{14}(x,t)=\ln \left\{ -\frac{1}{2}-\frac{3}{2} \tan ^2\left[ \frac{1}{2}\left( -\frac{3}{cw}x+cwt+C\right) \right] \right\} .$$

(90)

Set IV: We have the following:

$$a=\sqrt{c^2-b^2-\frac{3-3\sqrt{3}i}{2kw}},\quad b=b,\quad c=c, A_{0}=\frac{1}{2}kw(b^2-c^2)+\frac{3-3\sqrt{3}i}{6},\quad B_{1} =0,\quad B_2=0,\quad k=k,$$

(91)

$$w=w,\quad A_1=(b-c)kwa,\quad A_2=-\frac{1}{2}kw(b-c)^2,\quad v(\xi )=A_{0}+ A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(92)

By using of the Families 1, 2, 14 and 18 can be written, respectively, as

$$\begin{aligned} u_{15}(x,t)= & {} \ln \left\{ \frac{1}{2}kw(c^2-b^2)-\frac{3-3\sqrt{3}i}{3}-\frac{1}{2}\sqrt{18(1+\sqrt{3}i)-6kw(1-\sqrt{3}i)(b^2-c^2)} \tan \left( \sqrt{\frac{3-3\sqrt{3}i}{8kw}}(kx+wt)\right) \right. \\&\left. -\,\frac{1}{2}kw\left[ \sqrt{c^2-b^2-\frac{3-3\sqrt{3}i}{2kw}}-\sqrt{\frac{3-3\sqrt{3}i}{2kw}} \tan \left( \sqrt{\frac{3-3\sqrt{3}i}{8kw}}(kx+wt)\right) \right] ^{2}\right\} , \end{aligned}$$

(93)

$$\begin{aligned} u_{16}(x,t)= & {} \ln \left\{ \frac{1}{2}kw(c^2-b^2)-\frac{3-3\sqrt{3}i}{3}+\frac{1}{2}\sqrt{6kw(1-\sqrt{3}i)(b^2-c^2)-18(1+\sqrt{3}i)} \tanh \left( \sqrt{\frac{3\sqrt{3}i-3}{8kw}}(kx+wt)\right) \right. \\&\left. -\,\frac{1}{2}kw\left[ \sqrt{c^2-b^2-\frac{3-3\sqrt{3}i}{2kw}}+\sqrt{\frac{3\sqrt{3}i-3}{2kw}} \tanh \left( \sqrt{\frac{3\sqrt{3}i-3}{8kw}}(kx+wt)\right) \right] ^{2}\right\} , \end{aligned}$$

(94)

$$\begin{aligned} u_{17}(x,t)= & {} \ln \left\{ \frac{3-3\sqrt{3}i}{6}+2ckw\sqrt{-\frac{3-3\sqrt{3}i}{2kw}} \left[ \frac{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}e^{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}(kx+wt)}}{ce^{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}(kx+wt)}-1}\right] \right. \\&\left. - \,2kw c^2\left[ \frac{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}e^{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}(kx+wt)}}{ce^{\sqrt{-\frac{3-3\sqrt{3}i}{2kw}}(kx+wt)}-1}\right] ^{2}\right\} , \end{aligned}$$

(95)

$$u_{18}(x,t)= \ln \left\{ \frac{1-\sqrt{3}i-kwc^2}{2}-\frac{1}{2}kwc^2 \tan ^2\left( \sqrt{\frac{3-3\sqrt{3}i}{8kw}}(kx+wt)+C\right) \right\} .$$

(96)

Set V: We have the following:

$$a=\sqrt{\frac{b^2-c^2}{2}},\quad b=b,\quad c=c,\quad A_{0}=0,\quad B_{1} =0,\quad B_2=0,\quad A_1=\frac{2}{b+c}\sqrt{\frac{b^2-c^2}{2}},\quad w=w,$$

(97)

$$A_2=-\frac{b-c}{b+c},\quad k=\frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad v(\xi )= A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(98)

By using of the Families 1 and 2 can be written, respectively, as

$$\begin{aligned}&u_{19}(x,t)=\ln \left\{ 1-\sqrt{-3} \tan \left( \sqrt{-\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right. \\&\left. -\,\frac{1}{2}\left[ 1-\sqrt{-3} \tan \left( \sqrt{-\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] ^2\right\} , \end{aligned}$$

(99)

$$\begin{aligned} u_{20}(x,t)= & {} \ln \left\{ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right. \\&\left. -\,\frac{1}{2}\left[ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] ^2\right\} . \end{aligned}$$

(100)

Set VI: We have the following:

$$a=\sqrt{\frac{b^2-c^2}{2}},\quad b=b,\quad c=c, A_{0}=0,\quad B_{1} =0,\quad B_2=0, A_1=\frac{2}{b+c}\sqrt{\frac{b^2-c^2}{2}}\left( \frac{-1+\sqrt{3}i}{2}\right) , w=w,$$

(101)

$$A_2=-\frac{b-c}{b+c}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad k=\frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) , \quad v(\xi )= A_{1}\tan \left( \frac{\varPhi (\xi )}{2}\right) +A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(102)

By using of the Families 1 and 2 can be written, respectively, as

$$\begin{aligned} u_{21}(x,t)= & {} \ln \left\{ \frac{-1+\sqrt{3}i}{2}\left[ 1-\sqrt{-3} \tan \left( \sqrt{-\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] \right. \\&\left. -\,\frac{-1+\sqrt{3}i}{4}\left[ 1-\sqrt{-3} \tan \left( \sqrt{-\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] ^2\right\} , \end{aligned}$$

(103)

$$\begin{aligned} u_{22}(x,t)= & {} \ln \left\{ \frac{-1+\sqrt{3}i}{2}\left[ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] \right. \\&\left. -\,\frac{-1+\sqrt{3}i}{4}\left[ 1+\sqrt{3} \tanh \left( \sqrt{\frac{3}{8}(b^2-c^2)}\left( \frac{2}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) \right) \right] ^2\right\} . \end{aligned}$$

(104)

Set VII: We have the following:

$$a=0,\quad b=b,\quad c=c,\quad A_{0}=\frac{1}{2},\quad A_{1} =0,\quad A_2=-\frac{3}{2}\frac{b-c}{b+c},\quad B_1=0,\quad B_2=0,$$

(105)

$$k=\frac{3}{w(b^2-c^2)},\quad w=w,\quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(106)

By using of the Families 5 and 6 can be written, respectively, as

$$u_{23}(x,t)=\ln \left\{ \frac{1}{2}-\frac{3}{2} \tan ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{w(b^2-c^2)}x+wt+C\right) \right) \right\} ,$$

(107)

$$u_{24}(x,t)=\ln \left\{ \frac{1}{2}-\frac{3}{2} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{wb^2}x+wt+C\right) }-1}{e^{2b\left( \frac{3}{wb^2}x+wt+C\right) }+1}, \frac{2e^{b\left( \frac{3}{wb^2}x+wt+C\right) }}{e^{2b\left( \frac{3}{wb^2}x+wt+C\right) }+1}\right] \right) \right\} .$$

(108)

Set VIII: We have the following:

$$a=0,\quad b=b,\quad c=c,\quad A_{0}=\left( \frac{-1+\sqrt{3}i}{4}\right) ,\quad A_{1} =0,\quad A_2=-\frac{3}{2}\frac{b-c}{b+c}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad B_1=0,$$

(109)

$$B_2=0,\quad k=\frac{3}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad w=w,\quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(110)

By using of the Families 5 and 6 can be written, respectively, as

$$u_{25}(x,t)=\ln \left\{ \frac{-1+\sqrt{3}i}{4}-\frac{3}{2}\left( \frac{-1+\sqrt{3}i}{4}\right) \tan ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{4}\right) x+wt+C\right) \right) \right\} ,$$

(111)

$$u_{26}(x,t)=\ln \left\{ \frac{-1+\sqrt{3}i}{4}\left[ 1-\frac{3}{2} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }-1}{e^{2b\left( \frac{3}{wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }}{e^{2b\left( \frac{3}{wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}\right] \right) \right] \right\} .$$

(112)

Set IX: We have the following:

$$a=0,\quad b=b,\quad c=c,\quad A_{0}=-\frac{1}{4},\quad A_{1} =0,\quad A_2=-\frac{3}{8}\frac{b-c}{b+c},\quad B_1=0, B_2=-\frac{3}{8}\frac{b+c}{b-c},$$

(113)

$$k=\frac{3}{4w(b^2-c^2)},\quad w=w,\quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(114)

By using of the Families 5 and 6 can be written, respectively, as

$$u_{27}(x,t)=\ln \left\{ -\frac{1}{4}-\frac{3}{8}\left[ \tan ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{4w(b^2-c^2)}x+wt\right) \right) + \cot ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{4w(b^2-c^2)}x+wt\right) \right) \right] \right\} ,$$

(115)

$$\begin{aligned} u_{28}(x,t)= & {} \ln \left\{ -\frac{1}{4}-\frac{3}{8} \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }-1}{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{4wb^2}x+wt\right) }}{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }+1}\right] \right) \right. \\&\left. -\,\frac{3}{8} \cot ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }-1}{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{4wb^2}x+wt\right) }}{e^{2b\left( \frac{3}{4wb^2}x+wt\right) }+1}\right] \right) \right\} . \end{aligned}$$

(116)

Set X: We have the following:

$$a=0,\quad b=b,\quad c=c,\quad A_{1} =0,\quad A_{0}=-\frac{1}{4}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad A_2=B_2=-\frac{3}{8}\frac{b-c}{b+c}\left( \frac{-1+\sqrt{3}i}{2}\right) ,$$

(117)

$$B_1=0, k=\frac{3}{4w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) ,\quad w=w,\quad v(\xi )=A_{0}+ A_{2}\tan ^2\left( \frac{\varPhi (\xi )}{2}\right) +B_{2}\cot ^2\left( \frac{\varPhi (\xi )}{2}\right) .$$

(118)

By using of the Families 5 and 6 can be written, respectively, as

$$\begin{aligned} u_{29}(x,t)= & {} \ln \left\{ -\frac{1}{4}\left( \frac{-1+\sqrt{3}i}{2}\right) -\frac{3}{8}\left( \frac{-1+\sqrt{3}i}{2}\right) \left[ \tan ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{4w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt+C\right) \right) +\right. \right. \\&\left. \left. \times \, \cot ^2\left( \frac{\sqrt{b^2-c^2}}{2}\left( \frac{3}{4w(b^2-c^2)}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt+C\right) \right) \right] \right\} , \end{aligned}$$

(119)

$$\begin{aligned} u_{30}(x,t)= & {} \ln \left\{ -\frac{1}{4}-\frac{3}{8}\left( \frac{-1+\sqrt{3}i}{2}\right) \tan ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }-1}{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }}{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}\right] \right) \right. \\&\left. -\,\frac{3}{8}\left( \frac{-1+\sqrt{3}i}{2}\right) \cot ^2\left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }-1}{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}, \frac{2e^{b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }}{e^{2b\left( \frac{3}{4wb^2}\left( \frac{-1+\sqrt{3}i}{2}\right) x+wt\right) }+1}\right] \right) \right\} . \end{aligned}$$

(120)

Fig. 9

Graphs of $u_{1}$ real values of (127) are demonstrated at a $k=10, a=3, b=2, c=5$ and b $k=-10, a=4, b=5, c=7,$ when $-10\,<\,x<\,10, \,0\,<\,t\,<\,10$

3.3 Tzitzéica–Dodd–Bullough (TDB) equation

As last example, we consider the Tzitzéica–Dodd–Bullough equation as follows:

$$u_{xt}-e^{-u}-e^{-2u}=0,$$

(121)

by using the transformation $v(x,t)=e^{-u(x,t)}$, Eq. (121) transforms into the following partial differential equation,

$$-vv_{xt}+v_xv_{t}-v^3-v^4=0.$$

(122)

The travelling wave transformation $\xi =kx+wt$ reduces Eq.  (122) to the following ODE

$$kw((v')^2-vv'')-v^3-v^4=0.$$

(123)

Balancing the terms $vv''$ and $v^4$ by using homogenous principle one can be found $M=1$. Then (6) reduces to

$$v(\xi )=A_{0}+A_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] + B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1}.$$

(124)

Substituting (124) and (7) into Eq. (122), the following solutions are obtained.

Set I: We have the following:

$$k=k,\quad \varLambda =a^2+b^2-c^2 p=p,\quad A_{0}=\frac{p(c-b)-a}{2\sqrt{\varLambda }},\quad A_{1} =\frac{b-c}{2\sqrt{\varLambda }},\quad B_1=0,$$

(125)

$$w=-\frac{1}{k\varLambda },\quad v(\xi )=A_{0}+A_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] .$$

(126)

Using the transformation ${u=\ln v}$ and according to (126), the following solutions are obtained.

By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

Fig. 10

Graphs of $u_{2}$ real values of (128) are demonstrated at c $k=5, a=3, b=3, c=4$ and d $k=5, a=4, b=4, c=5,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$u_{1}(x,t)=\ln \left\{ -\frac{\sqrt{-1}}{2} \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( kx-\frac{1}{k(a^2+b^2-c^2)}t\right) \right) \right\} ,$$

(127)

$$u_{2}(x,t)=\ln \left\{ \frac{1}{2} \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( kx-\frac{1}{k(a^2+b^2-c^2)}t\right) \right) \right\} ,$$

(128)

$$u_{3}(x,t)=\ln \left\{ \frac{1}{2} \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( kx-\frac{1}{kb^2}t\right) }-1}{e^{2b\left( kx-\frac{1}{kb^2}t\right) }+1}, \frac{2e^{b\left( kx-\frac{1}{kb^2}t\right) }}{e^{2b\left( kx-\frac{1}{kb^2}t\right) }+1}\right] \right) \right\} ,$$

(129)

$$u_{4}(x,t)=\ln \left\{ -\frac{1}{2}+\frac{e^{a\left( kx-\frac{1}{ka^2}t\right) }}{-1+e^{a\left( kx-\frac{1}{ka^2}t\right) }}\right\} ,\quad u_{5}(\xi )=\ln \left\{ \frac{-a}{2b}-\frac{b-a}{2b} \frac{(a+b)e^{b\left( kx-\frac{1}{kb^2}t\right) }-1}{(a-b)e^{b\left( kx-\frac{1}{kb^2}t\right) }-1}\right\} ,$$

(130)

$$u_{6}(x,t)=\ln \left\{ \frac{1}{2}i\tan \left[ \frac{1}{2}\left( kx+\frac{1}{kc^2}t+C\right) \right] \right\} .$$

(131)

Set II: We have the following:

$$w=w,\quad p=-\frac{a}{b-c},\quad A_{0}=\frac{-1}{2},\quad A_{1} =0,\quad B_1=\frac{\sqrt{\varDelta }}{2(b-c)},\quad k=-\frac{1}{w\varDelta },\quad v(\xi )=A_{0}+B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1},$$

(132)

where $\varDelta =a^2+b^2-c^2$. By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

$$u_{7}(x,t)=\ln \left\{ \frac{-1}{2}-\frac{1}{2}i \cot \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right\} ,$$

(133)

Fig. 11

Graphs of $u_{4}$ real values of (130) are demonstrated at a1 $k=5, a=2$ and b1 $k=10, a=2,$ when $-10\,<\,x\,<\,10, \,0\,<\,t\,<\,10$

$$u_{8}(x,t)=\ln \left\{ \frac{-1}{2}+\frac{1}{2} \coth \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right\} ,$$

(134)

$$u_{9}(x,t)=\ln \left\{ -\frac{1}{2}+\frac{1}{2} \cot \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }-1}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}, \frac{2e^{b\left( -\frac{1}{wb^2}x+wt\right) }}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}\right] \right) \right\} ,$$

(135)

$$u_{10}(x,t)=\ln \left\{ -\frac{1}{2}-\frac{1}{4}\left[ \frac{1}{2}- \frac{e^{a\left( -\frac{1}{wa^2}x+wt\right) }}{e^{a\left( -\frac{1}{wa^2}x+wt\right) -1}}\right] ^{-1}\right\} ,$$

(136)

$$\begin{aligned} u_{11}(x,t)= & {} \ln \left\{ -\frac{1}{2}+\frac{b}{2(b-a)}\left[ -\frac{a}{b-a}- \frac{(a+b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}{(a-b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}\right] ^{-1}\right\} , \\ u_{12}(x,t)= & {} \ln \left\{ -\frac{1}{2}-\frac{1}{2}i\cot \left[ \frac{1}{2}\left( \frac{1}{wc^2}x+wt+C\right) \right] \right\} . \end{aligned}$$

(137)

Set III: We have the following:

$$p=\frac{-a+3\sqrt{a^2+b^2-c^2}}{b-c},\quad A_{0}=-2,\quad A_{1} =0,\quad B_1=\frac{4}{b-c}\sqrt{a^2+b^2-c^2},$$

(138)

$$w=w, k=-\frac{1}{w(a^2+b^2-c^2)},\quad v(\xi )=A_{0}+B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1}.$$

(139)

By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

$$u_{13}(x,t)=\ln \left\{ -2+4\left[ 3-i \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^{-1}\right\} ,$$

(140)

$$u_{14}(x,t)=\ln \left\{ -2+4\left[ 3+ \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^{-1}\right\} ,$$

(141)

$$u_{15}(x,t)=\ln \left\{ -2+4\left[ 3+ \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }-1}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}, \frac{2e^{b\left( -\frac{1}{wb^2}x+wt\right) }}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}\right] \right) \right] ^{-1}\right\} .$$

(142)

$$\begin{aligned} u_{16}(x,t)= & {} \ln \left\{ -2+2\left[ 1+\frac{e^{a\left( -\frac{1}{wa^2}x+wt\right) }}{1-e^{a\left( -\frac{1}{wa^2}x+wt\right) }}\right] ^{-1}\right\} , \\ u_{17}(x,t)= & {} \ln \left\{ -2+\frac{4b}{b-a}\left[ \frac{3b-a}{b-a}-\frac{(a+b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}{(a-b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}\right] ^{-1}\right\} , \end{aligned}$$

(143)

$$u_{18}(x,t)= \ln \left\{ -2-4i\left[ -3i+\tan \left( \frac{1}{2}\left( \frac{1}{wc}x+cwt\right) \right) \right] ^{-1}\right\} .$$

(144)

Set IV: We have the following:

$$p=-\frac{2(2A_0+1)(a^2+b^2-c^2)+2a\sqrt{a^2+b^2-c^2}}{2(b-c)\sqrt{a^2+b^2-c^2}},\quad A_{1} =0,\quad B_1=\frac{2A_0(A_0+1)}{b-c}\sqrt{a^2+b^2-c^2},$$

(145)

$$A_{0}=A_0,\quad w=w,\quad k=-\frac{1}{w(a^2+b^2-c^2)},\quad v(\xi )=A_{0}+B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1}.$$

(146)

By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

$$u_{19}(x,t)=\ln \left\{ A_0-2A_0(A_0+1)\left[ 2A_0+1+i \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^{-1}\right\} ,$$

(147)

$$u_{20}(x,t)=\ln \left\{ A_0-2A_0(A_0+1)\left[ 2A_0+1- \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( -\frac{1}{w(a^2+b^2-c^2)}x+wt\right) \right) \right] ^{-1}\right\} ,$$

(148)

$$u_{21}(x,t)=\ln \left\{ A_0-2A_0(A_0+1)\left[ 2A_0+1- \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }-1}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}, \frac{2e^{b\left( -\frac{1}{wb^2}x+wt\right) }}{e^{2b\left( -\frac{1}{wb^2}x+wt\right) }+1}\right] \right) \right] ^{-1}\right\} ,$$

(149)

$$\begin{aligned} u_{22}(x,t)= & {} \ln \left\{ A_0-A_0(A_0+1)\left[ 2A_0+2-\frac{e^{a\left( -\frac{1}{wa^2}x+wt\right) }}{e^{a\left( -\frac{1}{wa^2}x+wt\right) }-1}\right] ^{-1}\right\} , \\ u_{23}(x,t)= & {} \ln \left\{ A_0+\frac{2bA_0(A_0+1)}{b-a}\left[ \frac{(2A_0+1)b+a}{a-b}-\frac{(a+b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}{(a-b)e^{b\left( -\frac{1}{wb^2}x+wt\right) }-1}\right] ^{-1}\right\} , \end{aligned}$$

(150)

$$u_{24}(x,t)=\ln \left\{ A_0-2A_0(A_0+1)i\left[ (2A_0+1)i+\tan \left( \frac{1}{2wc}x+\frac{cw}{2}t+C\right) \right] ^{-1}\right\} .$$

(151)

Set V: We have the following:

$$p=\frac{b-c-2aA_1}{2(b-c)A_1},\quad A_{0}=-1,\quad A_{1} =A_1,\quad B_1=\frac{(b-c)^2-4A_1^2(a^2+b^2-c^2)}{4A_1(b-c)^2},$$

(152)

$$w=-\frac{4A_1^2}{k(b-c)^2},\quad k=k, \quad v(\xi )=A_{0}+A_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] + B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1}.$$

(153)

By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

$$\begin{aligned} u_{25}(x,t)= & {} \ln \left\{ -\frac{1}{2}-A_1 \frac{\sqrt{c^2-b^2-a^2}}{b-c} \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( kx-\frac{4A_1^2}{k(b-c)^2}t\right) \right) \right. \\&\left. +\,\frac{(b-c)^2-4A_1^2(a^2+b^2-c^2)}{4A_1(b-c)^2}\left[ \frac{1}{2A_1} -\frac{\sqrt{c^2-b^2-a^2}}{b-c} \tan \left( \frac{\sqrt{c^2-b^2-a^2}}{2}\left( kx-\frac{4A_1^2}{k(b-c)^2}t\right) \right) \right] ^{-1}\right\} , \end{aligned}$$

(154)

$$\begin{aligned} u_{26}(x,t)= & {} \ln \left\{ -\frac{1}{2}+A_1 \frac{\sqrt{a^2+b^2-c^2}}{b-c} \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( kx-\frac{4A_1^2}{k(b-c)^2}t\right) \right) \right. \\&\left. +\,\frac{(b-c)^2-4A_1^2(a^2+b^2-c^2)}{4A_1(b-c)^2}\left[ \frac{1}{2A_1} +\frac{\sqrt{a^2+b^2-c^2}}{b-c} \tanh \left( \frac{\sqrt{a^2+b^2-c^2}}{2}\left( kx-\frac{4A_1^2}{k(b-c)^2}t\right) \right) \right] ^{-1}\right\} , \end{aligned}$$

(155)

$$\begin{aligned} u_{27}(x,t)= & {} \ln \left\{ -\frac{1}{2}+A_1 \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }-1}{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }+1}, \frac{2e^{b\left( kx-\frac{4A_1^2}{kb^2}t\right) }}{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }+1}\right] \right) \right. \\&\left. +\,\frac{1-4A_1^2}{4A_1}\left[ \frac{1}{2A_1}+ \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }-1}{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }+1}, \frac{2e^{b\left( kx-\frac{4A_1^2}{kb^2}t\right) }}{e^{2b\left( kx-\frac{4A_1^2}{kb^2}t\right) }+1}\right] \right) \right] ^{-1}\right\} , \end{aligned}$$

(156)

$$u_{28}(x,t)=\ln \left\{ -\frac{1}{2}+\frac{A_1}{2}+\frac{A_1e^{a\left( kx-\frac{A_1^2}{ka^2}t\right) }}{1-e^{a\left( kx-\frac{A_1^2}{ka^2}t\right) }}+ \frac{1-A_1^2}{A_1} \left[ \frac{1+A_1}{2A_1}+\frac{e^{a\left( kx-\frac{A_1^2}{ka^2}t\right) }}{1-e^{a\left( kx-\frac{A_1^2}{ka^2}t\right) }}\right] ^{-1}\right\} ,$$

(157)

$$\begin{aligned} u_{29}(x,t)= & {} \ln \left\{ -\frac{1}{2}-\frac{aA_1}{b-a}-\frac{(a+b)e^{b\left( kx-\frac{4A_1^2}{k(b-a)^2}t\right) }-1}{(a-b)e^{b\left( kx-\frac{4A_1^2}{k(b-a)^2}t\right) }-1}\right. \\&\left. +\,\frac{(b-a)^2-4A_1^2b^2}{4A_1(b-a)^2}\left[ \frac{b-a-2aA_1}{2(b-a)A_1}-\frac{(a+b)e^{b\left( kx-\frac{4A_1^2}{k(b-a)^2}t\right) }-1}{(a-b)e^{b\left( kx-\frac{4A_1^2}{k(b-a)^2}t\right) }-1}\right] ^{-1}\right\} , \end{aligned}$$

(158)

$$u_{30}(x,t)= \ln \left\{ -\frac{1}{2}+A_1\tan \left( \frac{1}{2}kcx-\frac{2A_1^2}{kc}t\right) + \frac{1+4A_1^2}{4A_1}\left[ \frac{1}{2A_1}+\tan \left( \frac{1}{2}kcx-\frac{2A_1^2}{kc}t\right) \right] ^{-1}\right\} .$$

(159)

Set VI: We have the following:

$$p=\frac{-a}{b-c},\quad A_{0}=-\frac{1}{2},\quad A_{1} =\frac{b-c}{4\sqrt{\varDelta }},\quad B_1=\frac{\sqrt{\varDelta }}{4(b-c)},$$

(160)

$$w=-\frac{1}{4k\varDelta },\quad k=k,\quad v(\xi )=A_{0}+A_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] + B_{1}\left[ p+\tan \left( \frac{\varPhi (\xi )}{2}\right) \right] ^{-1},$$

(161)

where $\varDelta =a^2+b^2-c^2$. By using of the Families 1, 2, 6, 10, 11 and 18 can be written, respectively, as

$$u_{31}(x,t)=\ln \left\{ -\frac{1}{2}-\frac{1}{4}i\left[ \tan \left( \frac{\sqrt{-\varDelta }}{2}\left( kx-\frac{1}{4k\varDelta }t\right) \right) + \cot \left( \frac{\sqrt{-\varDelta }}{2}\left( kx-\frac{1}{4k\varDelta }t\right) \right) \right] \right\} ,$$

(162)

$$u_{32}(x,t)\,=\,\ln \left\{ -\frac{1}{2}+\frac{1}{4}\left[ \tanh \left( \frac{\sqrt{\varDelta }}{2}\left( kx-\frac{1}{4k\varDelta }t\right) \right) + \coth \left( \frac{\sqrt{\varDelta }}{2}\left( kx-\frac{1}{4k\varDelta }t\right) \right) \right] \right\} ,$$

(163)

$$\begin{aligned} u_{33}(x,t)\,=\, & {} \ln \left\{ -\frac{1}{2}+\frac{1}{4} \tan \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }-1}{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }+1}, \frac{2e^{b\left( kx-\frac{1}{4kb^2}t\right) }}{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }+1}\right] \right) \right. \\&\left. +\,\frac{1}{4} \cot \left( \frac{1}{2}\arctan \left[ \frac{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }-1}{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }+1}, \frac{2e^{b\left( kx-\frac{1}{4kb^2}t\right) }}{e^{2b\left( kx-\frac{1}{4kb^2}t\right) }+1}\right] \right) \right\} , \end{aligned}$$

(164)

$$u_{34}(x,t)\,=\, \ln \left\{ -\frac{3}{4}-\frac{1}{2}\frac{e^{a\left( kx-\frac{1}{4ka^2}t\right) }}{e^{a\left( kx-\frac{1}{4ka^2}t\right) }-1}- \frac{1}{8}\left[ \frac{1}{2}-\frac{e^{a\left( kx-\frac{1}{4ka^2}t\right) }}{e^{a\left( kx-\frac{1}{4ka^2}t\right) }-1}\right] ^{-1}\right\} ,$$

(165)

$$\begin{aligned} u_{35}(x,t)\,=\, & {} \ln \left\{ -\frac{1}{2}-\frac{a}{4b}+\frac{a-b}{4b}\frac{(a+b)e^{b\left( kx-\frac{1}{4kb^2}t\right) }-1}{(a-b)e^{b\left( kx-\frac{1}{4kb^2}t\right) }-1}\right. \\&\left. +\,\frac{b}{4(b-a)}\left[ \frac{a}{a-b}-\frac{(a+b)e^{b\left( kx-\frac{1}{4kb^2}t\right) }-1}{(a-b)e^{b\left( kx-\frac{1}{4kb^2}t\right) }-1}\right] ^{-1}\right\} , \end{aligned}$$

(166)

$$u_{36}(x,t)=\ln \left\{ -\frac{1}{2}+\frac{1}{4}i\tan \left( \frac{kc}{2}x+\frac{1}{8kc}t\right) - \frac{1}{4}i\cot \left( \frac{kc}{2}x+\frac{1}{8kc}t\right) \right\} .$$

(167)

$\bullet$ Note that: All the obtained results have been checked with Maple 13 by putting them back into the original equation and found correct.

4 Physical interpretations of the solutions

In this section, we depict the graph and signify the obtained solutions to the Tzitzéica type nonlinear evolution equations. Now, we will discuss all possible physical significance for parameter.

Remark 1

In Figures 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11, we plot three dimensional graphics of real values of (15), (16), (20), (54), (73), (74), (76), (108), (127), (128), (130) respectively, which denote the dynamics of solutions with appropriate parametric selections. We plot three dimensional graphics of Figs. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11, when $-10<x<10, 0<t<10$. To the best of our knowledge, these optical soliton solutions have not been submitted to literature in advance. The analytical solutions and figures obtained in this paper give us a different physical interpretation for the Tzitzéica type nonlinear evolution equations.

Remark 2

Solutions $u_1, u_4, u_7, u_9, u_{13}, u_{17}, u_{20}, u_{23}, u_{26}, u_{29}, u_{32}, u_{35}$ of the Tzitzéica equation and $u_1, u_5, u_9, u_{14}, u_{15},$ $u_{18}, u_{19}, u_{21}, u_{23}, u_{25}, u_{29}$ of the DBM equation and $u_1, u_6, u_7, u_{12}, u_{13}, u_{18}, u_{19}, u_{24}, u_{25}, u_{30}, u_{31}, u_{36}$ of TDB equation represent the exact periodic traveling wave solutions. Periodic solutions are traveling wave solutions that are periodic, such as $sin(x+t)$. Figures 1, 5 and 9 show the shape of the periodic solutions of $u_1$ of (15), (73) and (127) respectively. The other figures are ignored for simplicity.

Remark 3

Solutions $u_2, u_3, u_8, u_{10}, u_{14}, u_{18}, u_{21}, u_{24}, u_{27}, u_{30}, u_{33}$ of the Tzitzéica equation and $u_2, u_6, u_{10}, u_{16},$ $u_{20}, u_{22}$ of the DBM equation and $u_2, u_8, u_{14}, u_{20}, u_{26}, u_{32}$ of TDB equation are the soliton solutions. Solitons are special kinds of solitary waves. Solitons have a remarkable property that keeps its identity upon interacting with other solitons. Soliton solutions have particle-like structures, for example, magnetic monopoles, and extended structures, like, domain walls and cosmic strings, that have implications in cosmology of the early universe. Figures 2, 6 and 10 show the shape of the exact soliton solutions of $u_2$ of (16), (74) and (128) respectively. The other figures are ignored for simplicity.

Remark 4

Solutions $u_5, u_6, u_{11}, u_{12}, u_{15}, u_{16}, u_{19}, u_{22}, u_{25}, u_{28}, u_{31}, u_{34}$ of the Tzitzéica equation and $u_3, u_4, u_{7}, u_{8},$ $u_{11}, u_{12}, u_{13}, u_{17}, u_{24}, u_{26}, u_{28}, u_{30}$ of the DBM equation and $u_3, u_4, u_5, u_9, u_{10}, u_{11}$, $u_{15}, u_{16}, u_{17}, u_{21}$, $u_{22}, u_{23}, u_{27}$, $u_{28}$, $u_{29}$, $u_{33}, u_{34}, u_{35}$ of TDB equation represent the singular kink-type traveling wave solutions. Figures 3, 4, 7, 8 and 11 show the shape of the exact singular kink-type solution of $u_6, u_{28}$ of (20), (54), $u_4, u_{24}$ of (76), (108) and $u_4$ of (130) respectively. We note that kink solution acquires non-vanishing values as $x\longrightarrow \infty$. For convenience the other figures are omitted.

5 Conclusion

In this paper, the new approach of expansion method namely the improved $\tan \left( \varPhi (\xi )/2\right)$-expansion method has successfully been implemented to investigate the Tzitzéica type nonlinear evolution equations. Abundant exact travelling wave solutions including solitons, kink, periodic and rational solutions are obtained. It is worth mentioning that some of newly obtained solutions are identical to already published results. It has been shown that the applied method is effective and more wide-ranging than the generalized and improved (G’/G)-expansion method because it gives many new solutions. Also figures analysis for Tzitzéica type nonlinear evolution equations have surveyed as well. Therefore, this method can be applied to study many other nonlinear partial differential equations which frequently arise in engineering, mathematical physics and nonlinear optics and the quantum field theory.