Traveling Wave Solution of the Modified Benjamin-Bona-Mahony Equation

Abstract

In this paper, the ansatze method is implemented to study the exact solutions for the modified Benjamin-Bona-Mahony equation (mBBM). The singular-shaped traveling wave solution, the Bell-shape is traveling wave solution, the kink-shaped traveling wave solution and the periodic traveling wave solution is obtained. With the assist of computational software MATLAB, the graphical exemplifications of solutions are illustrated of the two-dimension (2D) and three-dimension (3D) plots.

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Zhu, Y. , Liu, X. , Huang, X. and Ye, F. (2022) Traveling Wave Solution of the Modified Benjamin-Bona-Mahony Equation. Journal of Applied Mathematics and Physics, 10, 3143-3155. doi: 10.4236/jamp.2022.1010209.

1. Introduction

Nonlinear evolution equations are extensively used as models to describe intricate physical phenomena in diverse fields of the sciences, peculiarly in fluid mechanics, solid state physics, plasma physics, and chemical physics [1]. Furthermore, the study of exact solutions of nonlinear evolution equations plays a significant part in soliton theory. In the past, many available methods have been established to obtain exact solutions of nonlinear evolution equations, for instance the Hirota’s bilinear transformation method [2] [3], the tanh-function method [4] [5], the Exp-function method [6] - [13], the $\left({G}^{\prime }/G\right)$ -expansion method [14] - [21], discrete Galerkin approximations method [22], the Jacobi elliptic function method [23], the homogeneous balance method [24], the modified simple equation method [25] [26] and so on.

The Benjamin-Bona-Mahony (BBM) equation

${u}_{t}+{u}_{x}+{u}^{n}{u}_{x}+{u}_{xxt}=0$ (1)

is the most famous model in physical applications. This equation model long waves in a nonlinear dispersive system. The solution of the BBM equation exhibits definite soliton-like behavior that is not explainable by any known theory [27]. The BBM equation is used in the analysis of the surface waves of long wavelength in liquids, hydromagnetic waves in cold plasma, acoustic-gravity waves in compressible fluids and acoustic waves in anharmonic crystals. Where n = 2 in (1), the BBM equation is called the modified Benjamin-Bona-Mahony equation (mBBM) [28]. In the article, we apply the ansatze method to study the exact traveling wave solutions of the following mBBM equation [29].

${u}_{t}+{u}_{x}+a{u}^{2}{u}_{x}+b{u}_{xxt}=0,$ (2)

Here a and b are nonzero constants, $u\left(x,t\right)$ is an unknown function, with respect to the spatial variable x and temporal variable t.

Kamruzzaman Khan and M.Ali Akbar [30] apply the $\left(-\Phi \left(\xi \right)\right)$ -expansion method to find the exact solitary wave solutions of mBBM equation, including hyperbolic function solutions, trigonometric function solutions and rational solutions. Kamruzzaman Khan, M. Ali Akbar and Md. Nur Alam [29] applies the MSE method to find the traveling wave solutions of mBBM equation. A K Gupta, J Hazarika [31] investigate the exact solutions by using Kudryashov method, Furthermore, we use four-term approximate solution of OHAM and compare it with the solution of the Kudryashov method.

The article is prepared as follows: In Section 2, we discuss the exact traveling wave solutions of modified Benjamin-Bona-Mahoney equation, In Section 3, we provide the dynamic behaviors of the traveling wave solutions of mBBM equation, and the short conclusions are given in Section 4.

2. Solutions of the Modified Benjamin-Bona-Mahoney Equation

Using the Following Traveling Wave Transformation

$u\left(x,t\right)=u\left(\xi \right)=u\left(x+\omega t\right),$ (3)

Equation (2) can be reduced into the following ordinary differential equation (ODE).

$\left(\omega +1\right){u}^{\prime }+a{u}^{2}{u}^{\prime }+b\omega {u}^{‴}=0.$ (4)

Integrating (4) once with respect to $\xi$, and choosing constant of integration as zero, we can obtain the following ODE.

$\left(\omega +1\right)u+\frac{1}{3}a{u}^{3}+b{\omega }^{″}=0.$ (5)

2.1. The Hyperbolic Function Solution

Suppose that the solutions of Equation (5) have the following form.

$u\left(\xi \right)=\underset{i=1}{\overset{m}{\sum }}{\mathrm{sinh}}^{i-1}\alpha \left({B}_{i}\mathrm{sinh}\alpha +{A}_{i}\mathrm{cosh}\alpha \right)+{A}_{0}.$ (6)

Balancing the highest nonlinear terms ${u}^{3}$ and the highest derivative term ${u}^{″}$ in Equation (5), we obtain 3m = m+ 2, which gives m = 1.

By (6) we can write as

$u\left(\xi \right)={B}_{1}\mathrm{sinh}\alpha +{A}_{1}\mathrm{cosh}\alpha +{A}_{0}.$ (7)

where ${A}_{0}$, ${A}_{1}$ and ${B}_{1}$ are undetermined constants. Let be the ansatze

$\frac{\text{d}\alpha }{\text{d}\xi }=\mathrm{sinh}\alpha .$ (8)

By Substituting (7), (8) into Equation (5), this yields the polynomial of ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$, setting the coefficients of each power ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$ to zero, we can obtain the following algebraic equations with respect to ${A}_{0}$, ${A}_{1}$, ${B}_{1}$ and $\omega$.

$\left\{\begin{array}{l}\left(\omega +1\right){A}_{0}+\frac{1}{3}a{A}_{0}^{3}+a{A}_{0}^{2}{A}_{1}=0,\\ \left(\omega +1\right){B}_{1}+a{A}_{1}^{2}{B}_{1}+a{A}_{0}^{2}{B}_{1}+b\omega {B}_{1}=0,\\ a{A}_{0}{B}_{1}^{2}+a{A}_{0}^{2}{A}_{1}=0,\\ \frac{1}{3}a{B}_{1}^{3}+a{A}_{1}^{2}{B}_{1}+2b\omega {B}_{1}=0,\\ \left(\omega +1\right){A}_{1}-a{A}_{1}{B}_{1}^{2}+a{A}_{0}^{2}{A}_{1}-2b\omega {A}_{1}=0,\\ \frac{1}{3}a{A}_{1}^{3}+a{A}_{1}{B}_{1}^{2}+2b\omega {A}_{1}=0,\\ 2a{A}_{0}{A}_{1}{B}_{1}=0.\end{array}$ (9)

Solving the above algebraic Equations (9), we can obtain the following three sets of solutions,

Case 1. ${A}_{0}=0,{A}_{1}=0,{B}_{1}=±\sqrt{\frac{-6b\omega }{a}},\omega =\frac{-1}{1+b};$ (10)

Case 2. ${A}_{0}=0,{B}_{1}=0,{A}_{1}=±\sqrt{\frac{-6b\omega }{a}},\omega =\frac{-1}{1-2b};$ (11)

Case 3. ${A}_{0}=0,{A}_{1}=±\sqrt{\frac{-\left(\omega +1\right)-b\omega }{a}},{B}_{1}=±\sqrt{\frac{\left(\omega +1\right)-2b\omega }{a}},\omega =\frac{-2}{2-b}.$ (12)

By applying the method of separating into variables to solve Equation (8), with choosing constant of integration to zero, we obtain:

$\mathrm{sinh}\alpha =-\text{csch}\xi ,\mathrm{cosh}\alpha =-\mathrm{coth}\xi .$ (13)

Using (10), (13), (3) and (7), we have:

${u}_{1,2}\left(\xi \right)=\mp \sqrt{\frac{-6b\omega }{a}}\text{csch}\xi ,$ (14)

where $\xi =x-\frac{1}{1+b}t$, ${u}_{1}$ takes “+”, ${u}_{2}$ takes “−”.

Using (11), (13), (3) and (7), we have:

${u}_{3,4}\left(\xi \right)=\mp \sqrt{\frac{-6b\omega }{a}}\mathrm{coth}\xi ,$ (15)

where $\xi =x-\frac{1}{1-2b}t$, ${u}_{3}$ takes “+”, ${u}_{4}$ takes “−”.

Using (12), (13), (3) and (7), we have:

${u}_{5,6,7,8}\left(\xi \right)=\mp \sqrt{\frac{\left(\omega +1\right)-2b\omega }{a}}\text{csch}\xi \mp \sqrt{\frac{-\left(\omega +1\right)-b\omega }{a}}\mathrm{coth}\xi ,$ (16)

where $\xi =x-\frac{2}{2-b}t$, ${u}_{5}$ takes “+, +”, ${u}_{6}$ takes “−, −”, ${u}_{7}$ takes “+, −”, ${u}_{8}$ takes “−, +”.

By the above analysis, we can obtain the solutions of Equation (2), as follows:

Proposition 1.

1) If $ab\left(1+b\right)>0$, Equation (2) have the following hyperbolic function solutions

${u}_{1,2}\left(x,t\right)=\mp \sqrt{\frac{6b}{a\left(1+b\right)}}\text{csch}\left(x-\frac{t}{1+b}\right),$

2) If $ab\left(2b-1\right)<0$, Equation (2) have the following hyperbolic function solutions

${u}_{3,4}\left(x,t\right)=\mp \sqrt{\frac{6b}{a\left(1-2b\right)}}\mathrm{coth}\left(x-\frac{t}{1-2b}\right),$

3) If $ab\left(b-2\right)<0$, Equation (2) have the following hyperbolic function solutions

${u}_{5,6,7,8}\left(x,t\right)=\mp \sqrt{\frac{3b}{a\left(2-b\right)}}\left(\text{csch}\left(x+\frac{2}{b-2}t\right)\mp \mathrm{coth}\left(x+\frac{2}{b-2}t\right)\right).$

Suppose that the solutions of Equation (5) have the following form.

$u\left(\xi \right)=\underset{i=1}{\overset{m}{\sum }}{\mathrm{sin}}^{i-1}\alpha \left({B}_{i}\mathrm{sin}\alpha +{A}_{i}\mathrm{cos}\alpha \right)+{A}_{0}.$ (17)

Balancing the highest nonlinear terms ${u}^{3}$ and the highest derivative terms ${u}^{″}$ in Equation (5), we obtain m = 1.

By (17) we can write as

$u\left(\xi \right)={B}_{1}\mathrm{sin}\alpha +{A}_{1}\mathrm{cos}\alpha +{A}_{0},$ (18)

where ${A}_{0}$, ${A}_{1}$ and ${B}_{1}$ are undetermined constants. Let be the ansatze

$\frac{\text{d}\alpha }{\text{d}\xi }=\mathrm{sin}\alpha .$ (19)

By Substituting (18), (19) into Equation (5), this yields the polynomial of ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$, setting the coefficients of each power ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$ to zero, we can obtain the following algebraic equations with respect to ${A}_{0}$, ${A}_{1}$, ${B}_{1}$ and $\omega$.

$\left\{\begin{array}{l}\left(\omega +1\right){A}_{0}+\frac{1}{3}a{A}_{0}^{3}+a{A}_{0}{A}_{1}^{2}=0,\\ \left(\omega +1\right){B}_{1}+a{A}_{1}^{2}{B}_{1}+a{A}_{0}^{2}{B}_{1}+b\omega {B}_{1}=0,\\ a{A}_{0}{B}_{1}^{2}-a{A}_{0}{A}_{1}^{2}=0,\\ \frac{1}{3}a{B}_{1}^{3}-a{A}_{1}^{2}{B}_{1}-2b\omega {B}_{1}=0,\\ \left(\omega +1\right){A}_{1}+a{A}_{1}{B}_{1}^{2}+a{A}_{0}^{2}{A}_{1}-2b\omega {A}_{1}=0,\\ \frac{1}{3}a{A}_{1}^{3}-a{A}_{1}{B}_{1}^{2}+2b\omega {A}_{1}=0,\\ 2a{A}_{0}{A}_{1}{B}_{1}=0.\end{array}$ (20)

Solving the above algebraic Equations (20), we can obtain the following two sets of solutions,

Case 1. ${A}_{0}=0,{A}_{1}=0,{B}_{1}=±\sqrt{\frac{6b\omega }{a}},\omega =\frac{-1}{1+b}$ (21)

Case 2. ${A}_{0}=0,{B}_{1}=0,{A}_{1}=±\sqrt{\frac{-6b\omega }{a}},\omega =\frac{-1}{1-2b}.$ (22)

By applying the method of separating into variables to Equation (19), with choosing constant of integration to zero, we obtain:

$\mathrm{sin}\alpha =\mathrm{sec}h\xi ,\mathrm{cos}\alpha =±\mathrm{tanh}\xi .$ (23)

Using (21), (23), (3) and (18), we have:

${u}_{1,2}\left(\xi \right)=±\sqrt{\frac{6b\omega }{a}}\text{sech}\xi ,$ (24)

where $\xi =x-\frac{1}{1+b}t$, ${u}_{1}$ takes “+”, ${u}_{2}$ takes “−”.

Using (22), (23), (3) and (18), we have:

${u}_{3,4}\left(\xi \right)=±\sqrt{\frac{-6b\omega }{a}}\mathrm{tanh}\xi ,$ (25)

where $\xi =x-\frac{1}{1-2b}t$, ${u}_{3}$ takes “+”, ${u}_{4}$ takes “−”.

By the above analysis, we can obtain the solutions of Equation (2), as follows:

Proposition 2.

1) If $ab\left(1+b\right)<0$, Equation (2) have the following hyperbolic function solutions

${u}_{1,2}\left(x,t\right)=\mp \sqrt{\frac{-6b}{a\left(1+b\right)}}\text{sech}\left(x-\frac{t}{1+b}\right),$

2) If $ab\left(1-2b\right)>0$, Equation (2) have the following hyperbolic function solutions

${u}_{3,4}\left(x,t\right)=\mp \sqrt{\frac{6b}{a\left(1-2b\right)}}\mathrm{tanh}\left(x-\frac{t}{1-2b}\right).$

2.2. The Trigonometric Function Solutions

For (6) and (7), Let be the ansatze

$\frac{\text{d}\alpha }{\text{d}\xi }=\mathrm{cosh}\alpha .$ (26)

By Substituting (7), (26) into Equation (5), this yields the polynomial of ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$, setting the coefficients of each power ${\mathrm{sinh}}^{i}\alpha {\mathrm{cosh}}^{i}\alpha \text{\hspace{0.17em}}\left(i=0,1,2,3\right)$ to zero, we can obtain the following algebraic equations with respect to ${A}_{0}$, ${A}_{1}$, ${B}_{1}$ and $\omega$.

$\left\{\begin{array}{l}\left(\omega +1\right){A}_{0}+\frac{1}{3}a{A}_{0}^{3}+a{A}_{0}^{2}{A}_{1}=0,\\ \left(\omega +1\right){B}_{1}+a{A}_{1}^{2}{B}_{1}+a{A}_{0}^{2}{B}_{1}+2b\omega {B}_{1}=0,\\ a{A}_{0}{B}_{1}^{2}+a{A}_{0}^{2}{A}_{1}=0,\\ \frac{1}{3}a{B}_{1}^{3}+a{A}_{1}^{2}{B}_{1}+2b\omega {B}_{1}=0,\\ \left(\omega +1\right){A}_{1}-a{A}_{1}{B}_{1}^{2}+a{A}_{0}^{2}{A}_{1}-b\omega {A}_{1}=0,\\ \frac{1}{3}a{A}_{1}^{3}+a{A}_{1}{B}_{1}^{2}+2b\omega {A}_{1}=0,\\ 2a{A}_{0}{A}_{1}{B}_{1}=0.\end{array}$ (27)

Solving the above algebraic Equations (27), we can obtain the following three sets of solutions,

Case 1. ${A}_{0}=0,{A}_{1}=0,{B}_{1}=±\sqrt{\frac{-6b\omega }{a}},\omega =\frac{-1}{1+2b};$ (28)

Case 2. ${A}_{0}=0,{B}_{1}=0,{A}_{1}=±\sqrt{\frac{-6b\omega }{a}},\omega =\frac{-1}{1-b};$ (29)

Case 3. ${A}_{0}=0,{A}_{1}=±\sqrt{\frac{-\left(\omega +1\right)-2b\omega }{a}},{B}_{1}=±\sqrt{\frac{\left(\omega +1\right)-b\omega }{a}},\omega =\frac{-2}{2+b}.$ (30)

By applying the method of separating into variables to Equation (26), with choosing constant of integration to zero, we obtain:

$\mathrm{sinh}\alpha =-\mathrm{cot}\xi ,\mathrm{cosh}\alpha =-\mathrm{csc}\xi .$ (31)

Using (28), (31), (3) and (7), we have:

${u}_{1,2}\left(\xi \right)=\mp \sqrt{\frac{-6b\omega }{a}}\mathrm{cot}\xi ,$ (32)

where $\xi =x-\frac{1}{1+2b}t$, ${u}_{1}$ takes “+”, ${u}_{2}$ takes “−”.

Using (29), (31), (3) and (7), we have:

${u}_{3,4}\left(\xi \right)=\mp \sqrt{\frac{-6b\omega }{a}}\mathrm{csc}\xi ,$ (33)

where $\xi =x-\frac{1}{1-b}t$, ${u}_{3}$ takes “+”, ${u}_{4}$ takes “−”.

Using (30), (31), (3) and (7), we have:

${u}_{5,6,7,8}\left(\xi \right)=\mp \sqrt{\frac{\left(\omega +1\right)-2b\omega }{a}}\mathrm{cot}\xi \mp \sqrt{\frac{-\left(\omega +1\right)-b\omega }{a}}\mathrm{csc}\xi ,$ (34)

where $\xi =x-\frac{2}{2+b}t$, ${u}_{5}$ takes “+, +”, ${u}_{6}$ takes “−, −”, ${u}_{7}$ takes “+, −”, ${u}_{8}$ takes “−, +”.

By the above analysis, we can obtain the solutions of Equation (2), as follows:

Proposition 3.

1) If $ab\left(1+2b\right)>0$, Equation (2) have the following trigonometric function solutions

${u}_{1,2}\left(x,t\right)=\mp \sqrt{\frac{6b}{a\left(1+2b\right)}}\mathrm{cot}\left(x-\frac{t}{1+2b}\right),$

2) If $ab\left(b-1\right)<0$, Equation (2) have the following trigonometric function solutions

${u}_{3,4}\left(x,t\right)=\mp \sqrt{\frac{6b}{a\left(1-b\right)}}\mathrm{csc}\left(x-\frac{t}{1-b}\right),$

3) If $ab\left(b+2\right)>0$, Equation (2) have the following trigonometric function solutions

${u}_{5,6,7,8}\left(x,t\right)=\mp \sqrt{\frac{5b}{a\left(2+b\right)}}\mathrm{cot}\left(x-\frac{2}{2+b}t\right)\mp \sqrt{\frac{b}{a\left(2+b\right)}}\mathrm{csc}\left(x-\frac{2}{2+b}t\right).$

3. Physical and Graphical Explanation

The solution (14) is the singular-shape traveling wave solution. Figure 1 shows the shape of the exact singular solution ${u}_{1}$ of Equation (2) (only shows the shape of solution (14) with wave speed a = 1, b = 1, $\omega =-\frac{1}{2}$ into the interval $-5\le x,t\le 5$ ). By Figure 1, we know that the solution ${u}_{1}$ have a valley and a peak; at the same time, the highest points and lowest points are nonsmooth. The solution (15) is the singular-shape traveling wave solution. Figure 2 shows the shape of the exact singular solution ${u}_{3}$ of Equation (2) (only shows the shape of solution (15) with wave speed a = −2, b = 1, $\omega =1$ into the interval $-20\le x,t\le 20$ ). By Figure 2, we know that the solution ${u}_{3}$ have a valley and a peak and these are not smooth. The solution (16) is the singular-shape traveling wave solution. Figure 3 shows the shape of the exact singular solution ${u}_{5}$ of Equation (2) (only shows the shape of solution (16) with wave speed a = 2, b = 1, $\omega =-2$ into the interval $-10\le x,t\le 10$ ). By Figure 3, we know that the solution ${u}_{5}$ have a valley and a peak, meantime, the highest points and lowest points are nonsmooth, the valleys and peaks are shorter compared to Figure 1.

The solution (24) is the Bell-shape traveling wave solution. Figure 4 shows the shape of the exact Bell-shape isolated solution ${u}_{1}$ of Equation (2) (only shows the shape of solution (24)) with wave speed a = −1, b = 1, $\omega =1$ into

Figure 1. The solution $u={u}_{1}\left(x,t\right)$ in (14) with a = 1, b = 1, $\omega =-\frac{1}{2}$, $-5\le x,t\le 5$ : (a) the 2D plot for t = 1; (b) the 3D plot.

Figure 2. The solution $u={u}_{3}\left(x,t\right)$ in (15) with a = −2, b = 1, $\omega =1$, $-20\le x,t\le 20$ : (a) the 2D plot for t = 1; (b) the 3D plot.

Figure 3. The solution $u={u}_{5}\left(x,t\right)$ in (16) with a = 2, b = 1, $\omega =-2$, $-10\le x,t\le 10$ : (a) the 2D plot fort = 1; (b) the 3D plot.

the interval $-5\le x,t\le 5$ ). By Figure 4, we know that the solution ${u}_{1}$ approaches the x-axis infinitely and smooth everywhere. The solution (25) is the kink-shape traveling wave solution. Figure 5 shows the shape of the exact Kink-type solution ${u}_{3}$ of Equation (2) (only shows the shape of solution (25) with wave speed a = −1, b = 1, $\omega =1$ into the interval $-20\le x,t\le 20$ ). By Figure 5, we know that the solution ${u}_{3}$ is continuous and monotonically increasing in the minddle.

The solution (32) is the periodic traveling wave solution. Figure 6 shows the shape of the exact singular solution ${u}_{1}$ of Equation (2) (only shows the shape of solution (32) with wave speed a = 3, b = 2, $\omega =-\frac{1}{5}$ into the interval $-10\le x,t\le 10$ ). By Figure 6, we know that the solution ${u}_{1}$ have multiple valley

Figure 4. The solution $u={u}_{1}\left(x,t\right)$ in (24) with a = −1, b = 1, $\omega =-\frac{1}{2}$, $-5\le x,t\le 5$ : (a) the 2D plot for t = 1; (b) the 3D plot.

Figure 5. The solution $u={u}_{3}\left(x,t\right)$ in (25) with a = −1, b = 1, $\omega =1$, $-20\le x,t\le 20$ : (a) the 2D plot for t = 1; (b) the 3D plot.

and multiple peaks and varies in height. The solution (33) is the periodic traveling wave solution. Figure 7 shows the shape of the exact singular solution ${u}_{3}$ of Equation (2) (only shows the shape of solution (33) with wave speed a = −1, b = −2, $\omega =-\frac{1}{3}$ into the interval $-10\le x,t\le 10$ ). By Figure 7, we know that the solution ${u}_{3}$ have multiple valley and multiple peaks and varies in height. The solution (34) is the periodic traveling wave solution. Figure 8 shows the shape of the exact singular solution ${u}_{5}$ of Equation (2) (only shows the shape of solution (34) with wave speed a = 1, b = 1, $\omega =-\frac{2}{3}$ into the interval $-4\le x\le 4$, $-5\le t\le 5$ ). By Figure 8, we know that the solution ${u}_{5}$ have multiple valley and multiple peaks and varies in height.

Figure 6. The solution $u={u}_{1}\left(x,t\right)$ in (32) with a = 3, b = 2, $\omega =-\frac{1}{5}$, $-10\le x,t\le 10$ : (a) the 2D plot for t = 1; (b) the 3D plot.

Figure 7. The solution $u={u}_{3}\left(x,t\right)$ in (33) with a = −1, b = −2, $\omega =-\frac{1}{3}$, $-10\le x,t\le 10$ : (a) the 2D plot for t = 1;(b) the 3D plot.

Figure 8. The solution $u={u}_{5}\left(x,t\right)$ in (34) with a = 1, b = 1, $\omega =-\frac{2}{3}$, $-4\le x\le 4$, $-5\le t\le 5$ : (a) the 2D plot for t = 1; (b) the 3D plot.

4. Conclusion

In this paper, we apply the ansatze method to find the exact solutions of the mBBM equation. Including the singular-shape traveling wave solution, the Bell-shape traveling wave solution, the kink-shape traveling wave solution and the periodic traveling wave solution, the dynamic behaviors of solutions are given.

Acknowledgements

Sincere thanks to the members of JAMP for their professional performance.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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