On the study of solutions of Bogoyavlenskii equation via improved $ G'/G^2 $ method and simplified $ \tan(\phi(\xi)/2) $ method

Guowei Zhang; Jianming Qi; Qinghao Zhu; Guowei Zhang; Jianming Qi; Qinghao Zhu

doi:10.3934/math.20221078

AIMS Mathematics

2022, Volume 7, Issue 11: 19649-19663. doi: 10.3934/math.20221078

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Research article

On the study of solutions of Bogoyavlenskii equation via improved $G'/G^2$ method and simplified $\tan(\phi(\xi)/2)$ method

1.
School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan, 455000, China
2.
School of Business, Shanghai Dianji University, Shanghai, 201306, China

Received: 21 June 2022 Revised: 27 August 2022 Accepted: 28 August 2022 Published: 06 September 2022
MSC : 34A05, 35C07, 35G20

The Bogoyavlenskii equation is used to describe some kinds of waves on the sea surface and discussed by many researchers. Recently, the $G'/G^2$ method and simplified $\tan(\frac{\phi(\xi)}{2})$ method are introduced to find novel solutions to differential equations. To the best of our knowledge, the Bogoyavlenskii equation has not been investigated by these two methods. In this article, we applied these two methods to the Bogoyavlenskii equation in order to obtain the novel exact traveling wave solutions. Consequently, we found that some new rational functions, trigonometric functions, and hyperbolic functions can be the traveling wave solutions of this equation. Some of these solutions we obtained have not been reported in the former literature. Through comparison, we see that the two methods are more effective than the previous methods for this equation. In order to make these solutions more obvious, we draw some 3D and 2D plots of them.

Keywords:

Citation: Guowei Zhang, Jianming Qi, Qinghao Zhu. On the study of solutions of Bogoyavlenskii equation via improved $G'/G^2$ method and simplified $\tan(\phi(\xi)/2)$ method[J]. AIMS Mathematics, 2022, 7(11): 19649-19663. doi: 10.3934/math.20221078

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Abstract

1. Introduction

Many phenomena in physics, biology, and chemistry etc. can be studied with the aid of nonlinear differential equations. When the researchers investigate a system, they firstly seek the underlying feature, and then convert the problem into mathematical equations by mathematical modeling. Most of time these equations are nonlinear differential equation. Therefore, we need to use some suitable methods to get the exact solutions of these equations. In the past few years, many systematic methods have formed, for example ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25]}.

We know that the mathematical equations describing waves in the ocean are nonlinear and their solutions are solitonic. In 1990, the following nonlinear partial differential equation is given ^[26], later called Bogoyavlenskii equation, to describe the fluctuation of sea waves.

$\begin{eqnarray} \left\{ \begin{aligned} &4u_t+u_{xxy}-4u^2u_y-4u_xv = 0;\\ &uu_y = v_x. \end{aligned} \right. \end{eqnarray}$

(1.1)

This classical equation has been studied for a long time. For the spectral parameter of (1.1), the authors of ^[26] studied nonisospectral condition. Kudryashov and Pickering ^[27] obtained the Schwarzian breaking soliton hierarchy of (1.1). It also has relation with non-isospectral scattering problems ^[28] and possesses Painlevé property ^[29]. As a modified breaking soliton equation, it describes the interaction of waves along the x-axis and y-axis. These results describe the phenomenon of waves on the sea surface.

In many years, seeking the exact solution of this nonlinear PDE by different methods is one of the main research aspects for researchers. Here we list some of the major approaches and results, but not all of them. By the singular manifold method, Peng and Shen ^[30] studied the analytical solutions of (1.1). In aid of modified extended tanh-function method the authors ^[31] got some exact traveling wave solutions. Using( $G'/G$ )-expansion method Malik et al ^[32] investigated Eq (1.1), and Yu, Sun ^[33] also treated this equation by modified technique of simplest equation. In 2020, Yokus et al. ^[34] constructed some exact solutions of (1.1) by $(1/G')$ -expansion and $(G'/G, 1/G)$ -expansion method.

For the past few years, many researchers ^[35,36,37] have introduced the modified ( $G'/G^2$ ) method, which is powerful and effective, to seek novel soliton solutions for different kinds of nonlinear PDE. Moreover, Manafian et al. ^[38,39] proposed a powerful technique called the improved $\tan(\phi(\xi)/2)$ approach to get exact soliton solutions of various PDE. Later, in ^[40] this method has been simplified. This approach also can be used to treat the Kundu-Eckhaus equation ^[40], Konopelchenko-Dubrovsky equation and Boussinesq equation ^[41]. In this work we will apply these two methods to Bogoyavlenskii equation for finding new solutions. Some of these solutions we obtained have not been reported in former literature. Comparing these two methods to the previous results, we find that our approaches are more effective.

In the nonlinear waves theory, the investigation of traveling wave solutions with a fixed velocity is one of the most important aspect. We aim to look for the new traveling wave solution of (1.1) such as

$\begin{eqnarray} u(x, y, t) = u(\xi), \, \, v(x, y, t) = v(\xi), \, \, \xi = x+y-kt. \end{eqnarray}$

(1.2)

We substitute (1.2) into (1.1) and integrate once the second equation of (1.1). For simplicity we choose the integral constant as zero, thus we obtain

$\begin{eqnarray} \left\{ \begin{aligned} &-4ku'+u'''-4u^2u'-4u'v = 0;\\ & \frac{u^2}{2} = v. \end{aligned} \right. \end{eqnarray}$

(1.3)

Combining the two equations of (1.3) together, and then integrating it once, we obtain the following ordinary differential equation

$\begin{eqnarray} u''-2u^3-4ku = 0. \end{eqnarray}$

(1.4)

Then we just need to find the solutions $u(\xi)$ of (1.4), then $v(\xi)$ is easy to obtain by the second equation of (1.3).

2. Via $G'/G^2$ method

At first, we introduce the steps of the modified $(G'/G^2)$ -expansion approach, and then employ this method to find new traveling wave solutions of (1.1).

Step 1. We think about the following PDE

$\begin{eqnarray} P(u, u_x, u_y, u_t, u_{xx}, u_{yy}, u_{tt}, u_{xy}, u_{xt}, u_{yx}, u_{yt}, \cdots) = 0, \end{eqnarray}$

(2.1)

as to $u(x, y, t)$ . In order to get the traveling wave solution of (1.1), we use the wave variables $\xi = x+y-kt$ to convert (2.1) into

$\begin{eqnarray} P(u, u_{\xi}, u_{\xi\xi}, u_{\xi\xi\xi}, \cdots) = 0. \end{eqnarray}$

(2.2)

Step 2. We express the exact solution of (1.4) to be a polynomial in $(G'/G^2)$ , that is,

$\begin{align} u({\xi}) = A_0+\sum\limits_{i = 1}^n\left[A_i\left(\frac{G'}{G^2}\right)^i+A_{-i}\left(\frac{G'}{G^2}\right)^{-i}\right], \end{align}$

(2.3)

here ${G} = {G}({\xi})$ is a solution of

$\begin{align} \left(\frac{G'}{G^2}\right)' = a +b \left(\frac{G'}{G^2}\right)+c \left(\frac{G'}{G^2}\right)^2, \end{align}$

(2.4)

the coefficients $a, b$ and $c$ are any of the constants. Since the highest order derivatives terms and the non-linear terms should be homogenous balance in (2.2), we can obtain the value of integer number $n$ .

Step 3. By taking (2.3) into (2.2) and utilizing (2.4), Eq (2.2) yields to a polynomial as to ${G'}/{G^2}$ . Then we collect all the terms with the same power of ${G'}/{G^2}$ and let all the coefficients of this collected polynomial to be zero. Hence, we deduce a system algebraic equations for $a, b, c$ , $A_{0}, A_{i}, (i = \pm 1, \pm2, \cdots)$ .

Step 4. By Solving this system of algebraic equations, we can get some families of the values of $A_0, A_{i}, (i = \pm 1, \pm2, \cdots)$ and $a, b, c$ . Since ordinary differential Eq (2.4) have five kinds solutions as follows:

Solution 1: If $ac > 0, b = 0$ , then

$\left(\frac{G'}{G^2}\right)(\xi) = \sqrt{\frac{a}{c}} \left(\frac{C_1\cos\sqrt{ac}\xi+C_2\sin\sqrt{ac}\xi} {C_2\cos\sqrt{ac}\xi-C_1\sin\sqrt{ac}\xi}\right);$

Solution 2: If $ac < 0, b = 0$ , then

$\left(\frac{G'}{G^2}\right)(\xi) = -\sqrt{\left|\frac{a}{c}\right|} \left(\frac{C_1\sinh2\sqrt{|ac|}\xi+C_1\cosh2\sqrt{|ac|}\xi+C_2} {C_1\cosh2\sqrt{|ac|}\xi+C_1\sinh2\sqrt{|ac|}\xi-C_2}\right);$

Solution 3: If $a = 0, c\neq 0, b = 0,$ then

$\left(\frac{G'}{G^2}\right)(\xi) = -\frac{C_1}{c(C_1\xi+C_2)};$

Solution 4: If $b\neq 0, \Delta\geq 0,$ then

$\left(\frac{G'}{G^2}\right)(\xi) = -\frac{b}{2c}-\frac{\sqrt{\Delta}(C_1\cosh(\frac{\sqrt{\Delta}}{2})\xi +C_2\sinh(\frac{\sqrt{\Delta}}{2})\xi)} {2c(C_2\cosh(\frac{\sqrt{\Delta}}{2})\xi+C_1\sinh(\frac{\sqrt{\Delta}}{2})\xi)};$

Solution 5: If $b\neq 0, \Delta < 0,$ then

$\left(\frac{G'}{G^2}\right)(\xi) = -\frac{b}{2c}-\frac{\sqrt{-\Delta} (C_1\cos(\frac{\sqrt{-\Delta}}{2})\xi-C_2\sin(\frac{\sqrt{-\Delta}}{2})\xi)} {2c(C_2\cos(\frac{\sqrt{-\Delta}}{2})\xi+C_1\sin(\frac{\sqrt{-\Delta}}{2})\xi)};$

where $C_1, C_2$ are free constants and $\Delta = b^2-4ac$ , then the exact solutions of (2.2) can be obtained.

Step 5. Applying the inverse transformation to $u(\xi)\, (\xi = x+y-kt)$ , then all exact solutions $u(x, y, t)$ of (2.1) can be gotten.

Since $u''$ and $u^3$ in (1.4) should be homogeneous balance, we get $n = 1$ . Then $u(\xi)$ in (2.3) has the form

$\begin{eqnarray} u(\xi) = A_1\left(\frac{G'}{G^2}\right)+A_0+A_{-1}\left(\frac{G'}{G^2}\right)^{-1}, \end{eqnarray}$

(2.5)

where ${G'}/{G^2}$ satisfies Eq (2.4). For simplicity, set $g: = {G'}/{G^2}$ , then (2.4) can be rewritten as

$\begin{eqnarray} g' = a+b g+c g^2. \end{eqnarray}$

(2.6)

Taking derivatives of (2.5), we obtain

$\begin{eqnarray} u'(\xi) = A_1c g^2+A_1b g+(A_1a-A_{-1}c)-A_{-1}b g^{-1}-A_{-1}a g^{-2} \end{eqnarray}$

(2.7)

and

$\begin{eqnarray} u''(\xi)& = &2A_1c^2 g^3+3A_1bc g^2+(2A_1ac+A_1b^2)g+(A_1ab+A_{-1}bc)\\ &&+(A_{-1}b^2+2A_{-1}ac)g^{-1}+(3A_{-1}ab)g^{-2}+2A_{-1}a^2g^{-3}. \end{eqnarray}$

(2.8)

By (2.5) and calculation we have

$\begin{eqnarray} 2u^3+4ku& = &2A_1^3g^3+6A_0A_1^2 g^2+(6A_{-1}A_1^2+6A_0^2A_1+4kA_1)g+(12A_{-1}A_0A_1+2A_0^3+4kA_0)\\ &&(6A_{-1}^2A_1+6A_{-1}A_0^2+4kA_{-1})g^{-1}+6A_{-1}^2A_0 g^{-2}+2A_{-1}^3 g^{-3}. \end{eqnarray}$

(2.9)

Now, we put (2.8) and (2.9) into (1.4), and collect all the same power terms together. Then we extract its undetermined coefficients of the power of $g$ , and set them to be zero. Therefore, we have the following equations

$\begin{eqnarray} g^3:& 2A_1c^2 = 2A_1^3, \\ g^2:&3A_1bc = 6A_0A_1^2, \\ g:& 2A_1ac +A_1b^2 = 6A_{-1}A_1^2+6A_0^2A_1+4kA_1, \\ const:& A_1ab+A_{-1}bc = 12A_{-1}A_0A_1+2A_0^3+4kA_0, \\ g^{-1}:&A_{-1}b^2+2A_{-1}ac = 6A_{-1}^2A_1+6A_{-1}A_0^2+4kA_{-1}, \\ g^{-2}:&3A_{-1}ab = 6A_{-1}^2A_0, \\ g^{-3}:&2A_{-1}a^2 = 2A_{-1}^3. \end{eqnarray}$

By solving the above equations, we have some cases as follows.

Case 1:

$\begin{equation} \left\{ \begin{aligned} &A_1 = 0;\\ &A_{-1} = 0;\\ &A_0 = 0, \sqrt{-2k}, \rm{or}, -\sqrt{-2k}, \end{aligned} \right. \end{equation}$

(2.10)

then $u(\xi)$ is a constant.

Case 2:

$\begin{equation} \left\{ \begin{aligned} &A_1 = 0;\\ &A_{-1} = \pm a;\\ &A_0 = \pm \frac{b}{2}, \end{aligned} \right. \end{equation}$

(2.11)

where $k = -\frac{b^2}{8}+\frac{ac}{2}$ , that is $\Delta = b^2-4ac = -8k$ .

Subcase 2.1: If $ac > 0, b = 0$ , then

$\begin{eqnarray} u(\xi) = \pm\sqrt{{a}{c}}\left(\frac {C_2\cos\sqrt{ac}\xi-C_1\sin\sqrt{ac}\xi}{C_1\cos\sqrt{ac}\xi+C_2\sin\sqrt{ac}\xi}\right). \end{eqnarray}$

(2.12)

Subcase 2.2: If $ac < 0, b = 0$ , then

$\begin{eqnarray} u(\xi) = \mp\sqrt{\left|{a}{c}\right|} \left(\frac{C_1\cosh2\sqrt{|ac|}\xi+C_1\sinh2\sqrt{|ac|}\xi-C_2}{C_1\sinh2\sqrt{|ac|}\xi+C_1\cosh2\sqrt{|ac|}\xi+C_2}\right). \end{eqnarray}$

(2.13)

Subcase 2.3: If $a = 0, c\neq 0, b = 0$ , then

$\begin{eqnarray} u(\xi) = 0; \end{eqnarray}$

(2.14)

Subcase 2.4: If $b\neq 0, \Delta = -8k\geq 0$ , then

$\begin{eqnarray} u(\xi) = \pm \frac{b}{2}\mp\left(\frac{b}{2c}+\frac{\sqrt{\Delta}(C_1\cosh(\frac{\sqrt{\Delta}}{2})\xi +C_2\sinh(\frac{\sqrt{\Delta}}{2})\xi)} {2c(C_2\cosh(\frac{\sqrt{\Delta}}{2})\xi+C_1\sinh(\frac{\sqrt{\Delta}}{2})\xi)}\right)^{-1}. \end{eqnarray}$

(2.15)

Subcase 2.5: If $b\neq 0, \Delta = -8k < 0$ , then

$\begin{eqnarray} u(\xi) = \pm \frac{b}{2}\mp\left(\frac{b}{2c}+\frac{\sqrt{-\Delta} (C_1\cos(\frac{\sqrt{-\Delta}}{2})\xi-C_2\sin(\frac{\sqrt{-\Delta}}{2})\xi)} {2c(C_2\cos(\frac{\sqrt{-\Delta}}{2})\xi+C_1\sin(\frac{\sqrt{-\Delta}}{2})\xi)}\right)^{-1}. \end{eqnarray}$

(2.16)

Case 3:

$\begin{equation} \left\{ \begin{aligned} &A_1 = \pm c;\\ &A_{-1} = 0;\\ &A_0 = \pm \frac{b}{2}; \end{aligned} \right. \end{equation}$

(2.17)

where $k = -\frac{b^2}{8}+\frac{ac}{2}$ , that is $\Delta = b^2-4ac = -8k$ .

Subcase 3.1: If $ac > 0, b = 0$ , then

$\begin{eqnarray} u(\xi) = \sqrt{{a}{c}} \left(\frac{C_1\cos\sqrt{ac}\xi+C_2\sin\sqrt{ac}\xi} {C_2\cos\sqrt{ac}\xi-C_1\sin\sqrt{ac}\xi}\right). \end{eqnarray}$

(2.18)

Subcase 3.2: If $ac < 0, b = 0$ , then

$\begin{eqnarray} u(\xi) = \mp\sqrt{\left|{a}{c}\right|} \left(\frac{C_1\sinh2\sqrt{|ac|}\xi+C_1\cosh2\sqrt{|ac|}\xi+C_2} {C_1\cosh2\sqrt{|ac|}\xi+C_1\sinh2\sqrt{|ac|}\xi-C_2}\right). \end{eqnarray}$

(2.19)

Subcase 3.3: If $a = 0, c\neq 0$ , then

$\begin{eqnarray} u(\xi) = \mp\frac{C_1}{C_1\xi+C_2}; \end{eqnarray}$

(2.20)

Subcase 3.4: If $b\neq 0, \Delta = -8k\geq 0$ , then

$\begin{eqnarray} u(\xi) = \mp\frac{\sqrt{\Delta}}{2}\cdot\frac{C_1\cosh(\frac{\sqrt{\Delta}}{2})\xi +C_2\sinh(\frac{\sqrt{\Delta}}{2})\xi} {C_2\cosh(\frac{\sqrt{\Delta}}{2})\xi+C_1\sinh(\frac{\sqrt{\Delta}}{2})\xi}. \end{eqnarray}$

(2.21)

Subcase 3.5: If $b\neq 0, \Delta = -8k < 0$ , then

$\begin{eqnarray} u(\xi) = \mp\frac{\sqrt{-\Delta}}{2}\cdot\frac{C_1\cos(\frac{\sqrt{-\Delta}}{2})\xi-C_2\sin(\frac{\sqrt{-\Delta}}{2})\xi} {C_2\cos(\frac{\sqrt{-\Delta}}{2})\xi+C_1\sin(\frac{\sqrt{-\Delta}}{2})\xi}. \end{eqnarray}$

(2.22)

3. Via $\tan(\frac{\phi(\xi)}{2})$ method

At first, we give the outline of $\tan(\frac{\phi(\xi)}{2})$ method.

Step 1. The same as $G'/G^2$ method.

Step 2. Suppose that the solutions can be written as follows

$\begin{eqnarray} u(\xi) = \sum\limits_{j = 0}^mA_j\left(p+\tan(\frac{\phi(\xi)}{2})\right)^j+\sum\limits_{j = 1}^mA_{-j}\left(p+\tan(\frac{\phi(\xi)}{2})\right)^{-j}, \end{eqnarray}$

(3.1)

$\phi(\xi)$ satisfies

$\begin{eqnarray} \phi'(\xi) = a\sin(\phi(\xi))+b\cos(\phi(\xi))+c, \end{eqnarray}$

(3.2)

here $a, b, c, A_j$ and $A_{-j}$ are unknown constants. (3.2) has five kinds of solutions.

Solution 1: If $b = c, a = 0,$ then

$\tan\left(\frac{\phi}{2}\right) = b\xi+C_1-p;$

Solution 2: If $b = c, \, a\neq0,$ then

$\tan\left(\frac{\phi}{2}\right) = C_1\exp(a\xi)-\frac{b}{a};$

Solution 3: If $b\neq c, \Delta = a^2+b^2-c^2 > 0,$ then

$\tan\left(\frac{\phi}{2}\right) = \frac{2}{b-c}\cdot\frac{C_1r_1\exp(r_1\xi)+C_2r_2\exp(r_2\xi)}{C_1\exp(r_1\xi)+C_2\exp(r_2\xi)}-p;$

Solution 4: If $b\neq c, \Delta = a^2+b^2-c^2 = 0,$ then

$\tan\left(\frac{\phi}{2}\right) = \frac{a}{b-c}+\frac{2}{b-c}\cdot\frac{C_2}{C_1+C_2\xi};$

Solution 5: If $b\neq c, \Delta = a^2+b^2-c^2 < 0,$ then

$\tan\left(\frac{\phi}{2}\right) = \frac{a}{b-c}+\frac{\sqrt{-\Delta}}{b-c}\cdot\frac{ -C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)} {C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)};$

where $C_1$ and $C_2$ are arbitrary constants, $r_1 = \frac{a+p(b-c)+\sqrt{\Delta}}{2}, r_2 = \frac{a+p(b-c)-\sqrt{\Delta}}{2}$ .

Step 3. We need to balance the highest order derivative and the nonlinear terms because of homogenous, then the value of positive integer $m$ we will get. Substituting (3.1) into (2.2) yields a equation with the power of $\tan(\frac{\phi}{2})$ . We firstly collect the terms with the same power of $\tan(\frac{\phi}{2})$ , and then set the coefficients of it to be zero, then a system of equations for unknown $A_j$ , $B_j$ , $a, b, c$ and $p$ we will obtain.

Step 4. Solving the above equations we just obtain, and then substituting $A_0, A_1, B_1,$ $\cdots, A_m, B_m, p$ into (3.1), we get the expression $u(\xi)$ .

Step 5. This step is similar to Step 5 of $G'/G^2$ .

In the following, we utilize $\tan(\frac{\phi(\xi)}{2})$ method to seek new solutions of (1.1). Substituting (3.1) into (1.4) and by Step 3 we get $m = 1$ . Then (3.1) can be written as

$\begin{eqnarray} u(\xi) = A_0+A\left(p+\tan\left(\frac{\phi(\xi)}{2}\right)\right)+A_{-1}\left(p+\tan\left(\frac{\phi(\xi)}{2}\right)\right)^{-1}. \end{eqnarray}$

(3.3)

Substituting (3.3) into the left side and right side of Eq (1.4) and combining with (3.2), then we have

$\begin{eqnarray} &&(p+t)^3\cdot u''(\xi)\\ & = &{(-b+c)^{2} A \, t^{6}}{2}\\ &+&\left(\frac{3 (-b+c)^{2} A p}{2}+\frac{3 a (-b+c) A}{2}\right) t^{5}\\ &+&\left(\frac{(3 A \, p^{2}-A_{-1}) (-b+c)^{2}}{2}+\frac{9 A p a (-b+c)}{2}\right. \\&+&\left. \frac{A (4 a^{2}-2 b^{2}+2 c^{2})}{4}+\frac{A_{-1} (-b+c)^{2}}{2}\right) t^{4}\\ &+&\left(\frac{(A \, p^{3}-A_{-1} p) (-b+c)^{2}}{2}+\frac{3 (3 A \, p^{2}-A_{-1}) a (-b+c)}{2}\right. \\ &+&\left.\frac{3 A p (4 a^{2}-2 b^{2}+2 c^{2})}{4}+\frac{A a (b+c)}{2}+2 A_{-1} a (-b+c)\right) t^{3}\\ &+&\left(\frac{3 (A \, p^{3}-A_{-1} p) a (-b+c)}{2}+\frac{(3 A \, p^{2}-A_{-1}) (4 a^{2}-2 b^{2}+2 c^{2})}{4}\right. \\ &+&\left.\frac{3 A p a (b+c)}{2}+\frac{A_{-1} (2 (b+c) (-b+c)+4 a^{2})}{2}\right) t^{2}\\ &+&\left(\frac{(A \, p^{3}-A_{-1} p) (4 a^{2}-2 b^{2}+2 c^{2})}{4}+\frac{(3 A \, p^{2}-A_{-1}) a (b+c)}{2}+2 A_{-1} a (b+c)\right) t\\ &+&\frac{(A \, p^{3}-A_{-1} p) a (b+c)}{2}+\frac{A_{-1} (b+c)^{2}}{2}; \end{eqnarray}$

(3.4)

$\begin{eqnarray} &&(p+t)^3\cdot (2u^{3}(\xi)+4ku(\xi))\\ & = & 2A^{3} t^{6}\\ &+&(12 A^{3} p+6 A^{2} A_{0}) t^{5}\\ &+&(30 A^{3} p^{2}+30 A^{2} A_{0} p+6 A^{2} A_{-1}+6 A A_{0}^{2}+4 k A) t^{4}\\ &+&(40 A^{3} p^{3}+60 A^{2} A_{0} p^{2}+24 A^{2} A_{-1} p+24 A A_{0}^{2} p+12 A A_{0} A_{-1}+16 A p k+2 A_{0}^{3}+4 A_{0} k) t^{3}\\ &+&(30 A^{3} p^{4}+60 A^{2} A_{0} p^{3}+36 A^{2} A_{-1} \, p^{2}+36 A A_{0}^{2} p^{2}+36 A A_{0} A_{-1} p+24 A k \, p^{2}+6 A_{0}^{3} p\\ &+&6 A \, A_{-1}^{2}+6 A_{0}^{2} A_{-1}+12 A_{0} k p+4 A_{-1} k) t^{2}\\ &+&(12 A^{3} p^{5}+30 A^{2} A_{0} p^{4}+24 A^{2} A_{-1} \, p^{3}+24 A A_{0}^{2} p^{3}+36 A A_{0} A_{-1} \, p^{2}+16 A k \, p^{3}\\ &+&6 A_{0}^{3} p^{2}+12 A \, A_{-1}^{2} p+12 A_{0}^{2} A_{-1} p+12 A_{0} k \, p^{2}+6 A_{0} A_{-1}^{2}+8 A_{-1} k p) t\\ &+&2 A^{3} p^{6}+6 A^{2} A_{0} p^{5}+6 A^{2} A_{-1} \, p^{4}+6 A A_{0}^{2} p^{4}+12 A A_{0} A_{-1} \, p^{3}+4 A k \, p^{4}+2 A_{0}^{3} p^{3}\\ &+&6 A \, A_{-1}^{2} p^{2}+6 A_{0}^{2} A_{-1} \, p^{2}+4 A_{0} k \, p^{3}+6 A_{0} A_{-1}^{2} p+4 A_{-1} k \, p^{2}+2 A_{-1}^{3}. \end{eqnarray}$

(3.5)

Substituting (3.4) and (3.5) into (1.4), a system of algebraic equations for the unknowns $A_0, A, A_{-1}$ will be obtained. Solving the algebraic equation by the aid of Maple software, we can have some cases as follows.

Case 1: If $b = c$ , then we have two subcases.

Subcase 1.1

$\begin{equation} \left\{ \begin{aligned} &A = 0; \\ &A_0 = 0, \rm{or} \pm \sqrt{-2k};\\ &A_{-1} = 0, \end{aligned} \right. \end{equation}$

(3.6)

then $u(\xi)$ is a constant.

Subcase 1.2:

$\begin{equation} \left\{ \begin{aligned} &A = 0; \\ &A_0 = \pm\sqrt{-2k} = \frac{a}{2};\\ &A_{-1} = \pm \frac{3 a c+16 k p-pa^{2}}{6 \sqrt{-2 k}} = c-ap, \end{aligned} \right. \end{equation}$

(3.7)

with $a^2 = -8k$ , $c, p$ are arbitrary constant. If $a = 0$ , then

$\begin{eqnarray} u(\xi) = \frac{c}{c\xi+C_1}. \end{eqnarray}$

(3.8)

If $a\neq 0$ ,

$\begin{eqnarray} u(\xi) = \frac{a}{2}+(c-ap)\left(\frac{pa-c}{a}+C_1\exp(a\xi)\right)^{-1}. \end{eqnarray}$

(3.9)

Case 2: If $b\neq c$ , then

$\begin{equation} \left\{ \begin{aligned} &A = \pm\frac{c-b}{2}; \\ &A_0 = \pm\frac{a-p(c-b)}{2} = 0;\\ &A_{-1} = \pm\frac{a^{2}+b^{2}-c^{2}+8k}{6 (b-c)}, \end{aligned} \right. \end{equation}$

(3.10)

with $p = -a/(b - c)$ and $c^2 = a^2+b^2-4k$ , $c^2 = a^2+b^2+2k$ , or $c^2 = a^2+b^2+8k$ . then $\Delta = 4k, -2k$ , or $-8k$ . $A_{-1} = \pm\frac{2k}{b-c}, \pm\frac{k}{b-c}$ or 0.

Subcase 2.1: If $\Delta > 0$ , then $r_1 = \frac{\sqrt{\Delta}}{2}, r_2 = -\frac{\sqrt{\Delta}}{2}$ .

When $\Delta = 4k, k > 0$

$\begin{eqnarray} u(\xi) = \mp\frac{C_1\frac{\sqrt{\Delta}}{2}\exp(\frac{\sqrt{\Delta}}{2}\xi)-C_2\frac{\sqrt{\Delta}}{2}\exp(-\frac{\sqrt{\Delta}}{2}\xi)}{C_1\exp(\frac{\sqrt{\Delta}}{2}\xi)+C_2\exp(-\frac{\sqrt{\Delta}}{2}\xi)} \pm k\frac{C_1\exp(\frac{\sqrt{\Delta}}{2}\xi)+C_2\exp(-\frac{\sqrt{\Delta}}{2}\xi)}{C_1\frac{\sqrt{\Delta}}{2}\exp(\frac{\sqrt{\Delta}}{2}\xi)-C_2\frac{\sqrt{\Delta}}{2}\exp(-\frac{\sqrt{\Delta}}{2}\xi)}. \end{eqnarray}$

(3.11)

When $\Delta = -2k, k < 0$

$\begin{eqnarray} u(\xi) = \mp\frac{C_1\frac{\sqrt{\Delta}}{2}\exp(\frac{\sqrt{\Delta}}{2}\xi)-C_2\frac{\sqrt{\Delta}}{2}\exp(-\frac{\sqrt{\Delta}}{2}\xi)}{C_1\exp(\frac{\sqrt{\Delta}}{2}\xi)+C_2\exp(-\frac{\sqrt{\Delta}}{2}\xi)} \pm \frac{k}{2}\frac{C_1\exp(\frac{\sqrt{\Delta}}{2}\xi)+C_2\exp(-\frac{\sqrt{\Delta}}{2}\xi)}{C_1\frac{\sqrt{\Delta}}{2}\exp(\frac{\sqrt{\Delta}}{2}\xi)-C_2\frac{\sqrt{\Delta}}{2}\exp(-\frac{\sqrt{\Delta}}{2}\xi)}. \end{eqnarray}$

(3.12)

When $\Delta = -8k, k < 0$

(3.13)

Subcase 2.2: If $\Delta = 0$ , then $k = 0, A_{-1} = 0$ .

$\begin{eqnarray} u(\xi) = \mp \frac{C_2}{C_1+C_2\xi}. \end{eqnarray}$

(3.14)

Subcase 2.3: When $\Delta = 4k, k < 0$ , we have

$u(\xi) = \mp\frac{\sqrt{-\Delta}}{2}\cdot\frac{-C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)}{C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)} \pm\frac{2k}{\sqrt{-\Delta}}\cdot\\ \frac{C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)}{-C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)}.$

(3.15)

When $\Delta = -2k, k > 0$ , we have

$u(\xi) = \mp\frac{\sqrt{-\Delta}}{2}\cdot\frac{-C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)}{C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)} \pm\frac{k}{\sqrt{-\Delta}}\cdot\\ \frac{C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)}{-C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)}.$

(3.16)

When $\Delta = -8k, k > 0$ , we have

$\begin{eqnarray} u(\xi) = \mp\frac{\sqrt{-\Delta}}{2}\cdot\frac{-C_1\sin(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\cos(\frac{\sqrt{-\Delta}}{2}\xi)}{C_1\cos(\frac{\sqrt{-\Delta}}{2}\xi)+C_2\sin(\frac{\sqrt{-\Delta}}{2}\xi)}. \end{eqnarray}$

(3.17)

$C_1, C_2$ are arbitrary constants when they appear.

4. Conclusions and discussion

Traveling waves play an important role in solitary wave theory. To our knowledge, this work is the first time to use these two methods to study the Bogoyavlenskii equation. We have obtained some new traveling wave soliton solutions of the Bogoyavlenskii equation, these solutions are constructed by trigonometric, exponential and rational functions with arbitrary coefficients and parameters. We draw the figures of some solutions with special values of the coefficients and parameters. These coefficients and parameters have practical physical meanings, the coefficients and parameters depend on the initial value. It is easy for us to observe the soliton behaviors of these solutions, such as to . Our results have verified that the new solutions of this equation do indeed have soliton phenomena, these will help us to study waves in the ocean. In the process of the use of the modified $G'/G^2$ and simplicity improved $\tan(\phi(\xi)/2)$ methods, we find that both methods are very useful, and can help us to get rational, trigonometric, exponential, hyperbolic solutions. By comparing these two methods, we found that the second method are more complicated, but we can get some additional solutions which we can not obtain by the modified $G'/G^2$ method, such as (3.11), (3.12), (3.15) and (3.16).