1. Introduction
The study of the exact solutions of nonlinear partial differential equations (NLPDEs) plays a vital role in the investigation of nonlinear wave phenomena. The nonlinear wave phenomena acting a significant role in our real life and in various research area such as fluid mechanics [1], nonlinear optics [2], plasma physics [3], optical fibers [4], propagation of shallow water wave [5], fluid dynamics [6,7], quantum mechanics [8], electromagnetism [9] and numerous other surfaces. John Scott Russel was firstly investigated the soliton structure in 1834 and in Ref. [10], Zabusky and Kruskal used the name soliton for first time in 1965. Then, solitary waves are being used on a lot of different subjects in the following years. In recent years, many researchers have studied the soliton solutions of NLPDEs using a variety of methods such as the modified Kudryashov scheme [11], the sine-Gordon expansion scheme [12,13], the extended sinh-Gordon expansion scheme [14], the (${G}'/G,1/G$)- expansion technique [15], the Hirota bilinear approach [16], the first integral method [17], the enhanced (G′/G)-expansion method [18], [19], [20], [21], the He's variational principle [22], the extended Fan sub-equation technique [23], the extended rational sin-cos and sinh-cosh methods [24], the $\text{exp}\left( -\varphi \left( \xi \right) \right)$-expansion approach [25], the Bernoulli sub-ODE method [26], the improved (G′/G)-expansion method [27], the Ansatz approach [28], the improved Bernoulli sub-equation method [29], the complex method [30], the modified simple equation method [31], the Ricatti-Bernoulli sub-ODE method [32], the new $\Phi^{6}$-model expansion method [33], the generalized Kudryashov approach [34], the Jacobi-elliptic function method [35], [36], [37], [38], [39], [40], the Backlund transformation [41] and so on.
The study of shallow water wave propagation has become a powerful field of research in nonlinear sciences. The wave propagation of surface is influenced by the lashing procedure of the wave solutions which types of waves are corporate in lakes, rivers, beaches and oceans, and the processes that cause them can be effective in ocean engineering. Recently, some basic research in ocean engineering and science have been done such as the Kudryashov method applied to the fifth-order nonlinear water wave equation [42], symmetrical regularized long wave equation and Ostrovsky equation via the modified extended tanh-function method [43], the fourth-order nonlinear partial differential equation by using the Riccati-Bernoulli sub-ODE method [44], the tanh-coth and sine-cosine function schemes applied to the (2+1)-dimensional Chaffee-Infante equation [45], the weakly coupled B-type Kadomtsev-Petviashvili equation using the Lie classical method [46], the Mikhailov-Novikov-Wang equation via the three different techniques [47], the generalized Hirota- Satsuma-Ito equation and the newly proposed extended (3 + 1)-dimensional Jimbo-Miwa equation via the Hirota's bilinear form [48], the family of Boussinesq-like equations using the sine-Gordon expansion and the hyperbolic function approaches [49]. In 1993, the Camassa-Holm (CH) equation was derived by Camassa and Holm in [50], which is for shallow water waves due to the integrable bi-Hamiltonian structure. Also, various forms of the CH equation have already been discussed in references [51], [52], [53]. Irshad et al. [54] in 2012, have firstly studied the SMCH equation
${{u}_{t}}+2\alpha {{u}_{x}}-{{u}_{xxt}}+\beta {{u}^{2}}{{u}_{x}}=0,$
where $\ \beta >0$, $\alpha \in R$ and $u\left( x,t \right)$ represents the fluid velocity in the x-direction. Irshad et al. [54] explored the exact traveling wave solutions of the SMCH equation using exp-function method. Najafi et al. [55] investigated the exact solutions of the SMCH equation by using He's semi-inverse method. Recently, the soliton solutions of the SMCH equation have explored with the help of using the generalized (G′/G)-expansion method by Alam and Akbar [56]. In Ref. [57], Naher and Begum have investigated the exact traveling wave solution from the SMCH equation by using the improved (G′/G)-expansion method. Ali et al. [58] have discovered the soliton solutions of the SMCH equation using the exp(–φ(η))-expansion method. More recently, Lu et al. [59] have constructed solitary wave solutions of the SMCH equation by using the modified extended auxiliary equation method. Using the modified simple equation method, the exact solutions to the SMCH equation have been explored by Islam et al. [60]. According to the above discussion on the previous literature, the investigators have stated a number of novel exact solutions such as kink, singular kink, periodic and singular soliton solutions. However, no studies have yet been found to examine the exact solutions of the SMCH equation via NAE method. On the other hand, many researchers have proposed the various methods to apply the NLPDEs. The NAE method is one of the most important analytical methods for NLPDEs. Khater et al. [61] have considered the NAE method, applied to the higher order nonlinear Sasa-Satsuma equation in mono-mode fibers and obtained a lot of various types of optical soliton solutions. Al-Raeei and El-Daher [62] have used the NAE method and attained optical soliton solutions from the nonlinear complex fractional Schrödinger equation. Recently, Kumar et al. [63,64] have explored the optical soliton solution from the Kundu-Mukherjee-Naskar and Sawada-Kotera equation by using NAE method. As a result, we have considered the NAE method and applied it to the SMCH equation. We have found the various types of wave solutions such as single spike, multiple spikes, kink, anti-peakon, singular spike soliton, anti-bell shape soliton solutions that can provide new directions for understanding nonlinear wave phenomena that show an important role in ocean engineering owing the mutual relationship between integrity and water waves phenomenon.
This manuscript is organized as follows: The NAE method is discussed in Section 2. In the next section, the method is applied to the SMCH equation and found the soliton solutions. Graphical and Physical behaviors are discussed in Section 4. In Section 5, stability analysis and comparison are discussed. Finally, conclusion is given in Section 6.
2. Procedure of the NAE mechanism
Suppose NLPDEs is in the following structure:
$R\left( u,{{u}_{t}},{{u}_{x}},{{u}_{y}},{{u}_{xx}},{{u}_{tx}},{{u}_{yy}},{{u}_{ty}},{{u}_{tt}},{{u}_{xy}},\ldots \ldots \right)=0,$
where R represents nonlinear polynomial including wave function $u\left( x,y,t \right)$ and it's partial derivatives with respect to x,y and t.
We suppose that
$u\left( x,y,t \right)=u\left( \xi \right)and\ \xi =x+y\pm \omega t$
The travelling wave variable in Eq. (2.2) reducing the NLPDEs (2.1) to following ordinary differential equation (ODE) as follows:
$Q\left( u,{u}',{u}'',\cdots \cdots \cdots \right)=0,$
Wherein symbol (’) indicates the derivative with respect to $\ \xi $. we consider the exact soliton solutions of the Eq. (2.3) is to be
$u\left( \xi \right)=\underset{k=0}{\overset{N}{\mathop \sum }}\,{{C}_{k}}{{a}^{kg\left( \xi \right)}},$
Where ${{C}_{k}}\left( k=0,1,2,\cdots \cdots \cdots,N \right)$ are constants to be calculated, such that ${{C}_{N}}\ne 0$ and $g\left( \xi \right)$ is called the solutions of first order auxiliary equation as
${g}'\left( \xi \right)=\frac{1}{\ln \left( a \right)}\left\{ {{l}_{1}}{{a}^{-g\left( \xi \right)}}+{{l}_{2}}+{{l}_{3}}{{a}^{g\left( \xi \right)}} \right\}.$
The balancing rule needs to apply to discover the value of N in Eq. (2.4).
In this step, we substituting the Eq. (2.4) and Eq. (2.5) into Eq. (2.3) and we get an algebraic equation which are equated left and right side based on powers of ${{a}^{kg\left( \xi \right)}},\left( k=0,1,2,3,\cdots \cdots \right)$. As a result, we attain an algebraic equation, solving this equation and also find out the values of and so on. The accomplishments of N, a, b, c, $\omega $ the Eq. (2.5) are derived as follows:
Case-1: When ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}<0$ and ${{l}_{3}}\ne 0$
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{3}}}+\frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2{{l}_{3}}}\text{tan}\left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{3}}}-\frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2{{l}_{3}}}\text{cot}\left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2}\ \xi \right),$
Case-2: When ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}>0$ and ${{l}_{3}}\ne 0$
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{3}}}-\frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2{{l}_{3}}}tanh\left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{3}}}-\frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2{{l}_{3}}}\text{coth}\left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-{{l}_{2}}^{2}}}{2}\ \xi \right),$
Case-3: When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}<0$,${{l}_{3}}\ne 0$ and ${{l}_{3}}=-{{l}_{1}}$
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{2}}}{2{{l}_{1}}}-\frac{\sqrt{-{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{tan}\left( \frac{\sqrt{-{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{2}}}{2{{l}_{1}}}+\frac{\sqrt{-{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{cot}\left( \frac{\sqrt{-{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Case-4: When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}>0$, ${{l}_{3}}\ne 0$ and ${{l}_{3}}=-{{l}_{1}}$
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{2}}}{2{{l}_{1}}}+\frac{\sqrt{{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{tanh}\left( \frac{\sqrt{{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{2}}}{2{{l}_{1}}}+\frac{\sqrt{{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{coth}\left( \frac{\sqrt{{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Case-5: When ${{l}_{2}}^{2}-4{{l}_{1}}^{2}<0$ and ${{l}_{3}}={{l}_{1}}$
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{1}}}-\frac{\sqrt{-{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{tan}\left( \frac{\sqrt{-{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{1}}}-\frac{\sqrt{-{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{cot}\left( \frac{\sqrt{-{{l}_{2}}^{2}+4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Case-6: When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}>0$ and ${{l}_{3}}={{l}_{1}}$
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{1}}}-\frac{\sqrt{{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{tanh}\left( \frac{\sqrt{{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=\frac{-{{l}_{2}}}{2{{l}_{1}}}-\frac{\sqrt{{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2{{l}_{1}}}\text{coth}\left( \frac{\sqrt{{{l}_{2}}^{2}-4{{l}_{1}}^{2}}}{2}\ \xi \right),$
Case-7: When ${{l}_{2}}^{2}=4{{l}_{1}}{{l}_{3}}$
${{a}^{g\left( \xi \right)}}=-\frac{2+{{l}_{2}}\xi }{2n\xi }$
Case-8: When ${{l}_{1}}{{l}_{3}}<0$,${{l}_{2}}=0$ and ${{l}_{3}}\ne 0$
${{a}^{g\left( \xi \right)}}=-\sqrt{\frac{-{{l}_{1}}}{{{l}_{3}}}}\text{tanh}\left( \sqrt{-{{l}_{1}}{{l}_{3}}}\ \xi \right),$
Or
${{a}^{g\left( \xi \right)}}=-\sqrt{\frac{-{{l}_{1}}}{{{l}_{3}}}}\text{coth}\left( \sqrt{-{{l}_{1}}{{l}_{3}}}\ \xi \right),$
Case-9: When ${{l}_{2}}=0$ and ${{l}_{1}}=-{{l}_{3}}$
${{a}^{g\left( \xi \right)}}=\frac{1+{{e}^{\left( -2{{l}_{3}}\xi \right)}}}{-1+{{e}^{\left( -2{{l}_{3}}\xi \right)}}},$
Case-10: When ${{l}_{1}}={{l}_{3}}=0$
${{a}^{g\left( \xi \right)}}=\cosh \left( {{l}_{2}}\xi \right)+\text{sinh}\left( {{l}_{2}}\xi \right),$
Case-11: When ${{l}_{1}}={{l}_{2}}=K$ and ${{l}_{3}}=0$
${{a}^{g\left( \xi \right)}}={{e}^{\text{K}\xi }}-1,$
Case-12: When ${{l}_{2}}={{l}_{3}}=\varphi $ and ${{l}_{1}}=0$
${{a}^{g\left( \xi \right)}}=\frac{{{e}^{\varphi \xi }}}{1-{{e}^{\varphi \xi }}},$
Case-13: When ${{l}_{2}}=\left( {{l}_{1}}+{{l}_{3}} \right)$
${{a}^{g\left( \xi \right)}}=-\frac{1-{{l}_{1}}{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}}{1-{{l}_{3}}{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}},$
Case-14: When ${{l}_{2}}=-\left( {{l}_{1}}+{{l}_{3}} \right)$
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{1}}-{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}}{{{l}_{3}}-{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}},$
Case-15: When ${{l}_{1}}=0$
${{a}^{g\left( \xi \right)}}=\frac{{{l}_{2}}{{e}^{{{l}_{2}}\xi }}}{1-{{l}_{3}}{{e}^{{{l}_{2}}\xi }}},$
Case-16: When ${{l}_{3}}={{l}_{2}}={{l}_{1}}\ne 0$
${{a}^{g\left( \xi \right)}}=\frac{1}{2}\left\{ \sqrt{3}\ tan\left( \frac{\sqrt{3}}{2}{{l}_{1}}\xi \right)-1 \right\},$
Case-17: When ${{l}_{3}}={{l}_{2}}=0$
${{a}^{g\left( \xi \right)}}={{l}_{1}}\xi,$
Case-18: When ${{l}_{1}}={{l}_{2}}=0$
${{a}^{g\left( \xi \right)}}=\frac{-1}{{{l}_{3}}\xi },$
Case-19: When ${{l}_{3}}={{l}_{1}}$ and ${{l}_{2}}=0$
${{a}^{g\left( \xi \right)}}=\tan \left( {{l}_{1}}\xi \right),$
Case-20: When ${{l}_{3}}=0$
${{a}^{g\left( \xi \right)}}={{e}^{{{l}_{2}}\xi }}-\frac{p}{q},$
Finally, the constants ${{C}_{k}}$, ${{l}_{1}}$, ${{l}_{2}}$ and ${{l}_{3}}$ gained, that constant and $g\left( \xi \right)$ setting inside (2.4), as a result we obtained abundant soliton solutions from the NLPDEs (2.1).
It is remarkable to point out that the NAE method is a special case of the transformed rational function method by Ma and Lee [65] because the expansion in Eq. (2.4) is a polynomial and Eq. (2.5) is a rational function of a solution to an integrable ODE, which is coincide with the Riccati equation in Eq. (39) by Ma and Fuchssteiner [66]. Ma and Lee [65] used the transformed rational function method to the (3+1)-dimensional Jimbo-Miwa equation and obtained some soliton solutions. Ma and Fuchssteiner [66] have investigated the exact solutions to the Kolmogorov-Petrovskii-Piskunov equation. On the other hand, Ma [67], [68], [69], [70], [71] have analyzed the N-soliton solution and verify the Hirota N-soliton condition of the nonlinear evolution equations.
3. Determination of the solutions
In this segment, we execute a common, relevant and widespread explicit soliton solutions to the SMCH equation uses the above mechanism. We apply the following wave transformation:
$u\left( x,t \right)=u\left( \xi \right),\ \xi =x-\omega t,$
From Eqs. (1.1) and (3.1), yields:
$-\omega {u}'+2\alpha {u}'+\omega {{u}^{\prime\prime \prime }}+\beta {{u}^{2}}{u}'=0$.
Integrating and simplifying the above equation:
$\omega u-2\alpha u-\omega {u}''-\frac{\beta }{3}{{u}^{3}}=0.$
Balancing between ${u}''$ and ${{u}^{3}}$ in Eq. (3.2), gives N=1. Eq. (2.4) yields
$u\left( x,t \right)={{C}_{0}}+{{C}_{1}}{{a}^{g\left( \xi \right)}},$
where ${{C}_{0}}$ and ${{C}_{1}}$ are constant such that ${{C}_{0}}\ne 0$ and ${{C}_{1}}\ne 0$. and $g\left( \xi \right)$ is the solution of the nonlinear Eq. (2.5).
With the help of Maple, we attain from (3.2), using (3.3) and (2.5)
$\begin{align} & \left( -2\omega {{C}_{1}}l_{3}^{2}-\frac{1}{3}\beta C_{1}^{3} \right){{\left\{ {{a}^{g\left( \xi \right)}} \right\}}^{3}}+\left( -\beta {{C}_{0}}C_{1}^{2}-3\omega {{C}_{1}}{{l}_{2}}{{l}_{3}} \right){{\left\{ {{a}^{g\left( \xi \right)}} \right\}}^{2}} \\ & +\left( -\beta C_{0}^{2}{{C}_{1}}-2\omega {{C}_{1}}{{l}_{1}}{{l}_{3}}-\omega {{C}_{1}}l_{2}^{2}-2\alpha {{C}_{1}}+\omega {{C}_{1}} \right)\left\{ {{a}^{g\left( \xi \right)}} \right\} \\ & +\left( \omega {{C}_{0}}-2\alpha {{C}_{0}}-\omega {{C}_{1}}{{l}_{1}}{{l}_{2}}-\frac{\beta C_{0}^{3}}{3} \right)=0 \\ \end{align}$
Equating the coefficients of the powers ${{a}^{g\left( \xi \right)}}$, then we held a scheme of algebraic equations:
$-2\omega {{C}_{1}}l_{3}^{2}-\frac{1}{3}\beta C_{1}^{3}=0$
$-\beta {{C}_{0}}C_{1}^{2}-3\omega {{C}_{1}}{{l}_{2}}{{l}_{3}}=0$
$-\beta C_{0}^{2}{{C}_{1}}-2\omega {{C}_{1}}{{l}_{1}}{{l}_{3}}-\omega {{C}_{1}}l_{2}^{2}-2\alpha {{C}_{1}}+\omega {{C}_{1}}=0$
$\omega {{C}_{0}}-2\alpha {{C}_{0}}-\omega {{C}_{1}}{{l}_{1}}{{l}_{2}}-\frac{\beta C_{0}^{3}}{3}=0$
After solution the above algebraic equation we have following solution sets:
$\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2},{{C}_{0}}=\pm \sqrt{6\ }\sqrt{\frac{\alpha }{4\beta {{l}_{1}}{{l}_{3}}-\beta l_{2}^{2}-2\beta }}\ {{l}_{2}},$
${{C}_{1}}=\pm 2\sqrt{6\ }\sqrt{\frac{\alpha }{4\beta {{l}_{1}}{{l}_{3}}-\beta l_{2}^{2}-2\beta }}\ {{l}_{3}}$
Substituting the above values in Eq. (3.2) along with the Eqs. (2.6)-(2.32), we get the following solutions to the SMCH equation.
When ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}<0$ and ${{l}_{3}}\ne 0$, then the trigonometric solutions obtained as
${{u}_{{{1}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2 \right)}}\ \sqrt{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}}\ \tan \left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{1}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2 \right)}}\ \sqrt{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}}\ \cot \left( \frac{\sqrt{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}}}{2}\xi \right)$where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}>0$ and ${{l}_{3}}\ne 0$, then the hyperbolic solutions attained as
${{u}_{{{2}_{1}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2 \right)}}\ \sqrt{-4{{l}_{1}}{{l}_{3}}+l_{2}^{2}}\ \tanh \left( \frac{\sqrt{-4{{l}_{1}}{{l}_{3}}+l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{2}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2 \right)}}\ \sqrt{-4{{l}_{1}}{{l}_{3}}+l_{2}^{2}}\ \coth \left( \frac{\sqrt{-4{{l}_{1}}{{l}_{3}}+l_{2}^{2}}}{2}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}<0$,${{l}_{3}}\ne 0$ and ${{l}_{3}}=-l$, then we attained the following solutions
${{u}_{{{3}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( 4l_{1}^{2}+l_{2}^{2}+2 \right)}}\ \sqrt{-4l_{1}^{2}-l_{2}^{2}}\ \tan \left( \frac{\sqrt{-4l_{1}^{2}-l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{3}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( 4l_{1}^{2}+l_{2}^{2}+2 \right)}}\ \sqrt{-4l_{1}^{2}-l_{2}^{2}}\ \cot \left( \frac{\sqrt{-4l_{1}^{2}-l_{2}^{2}}}{2}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$. When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}>0$, ${{l}_{3}}\ne 0$ and ${{l}_{3}}=-{{l}_{1}}$, then the solutions are
${{u}_{{{4}_{1}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( 4l_{1}^{2}+l_{2}^{2}+2 \right)}}\ \sqrt{4l_{1}^{2}+l_{2}^{2}}\ \tanh \left( \frac{\sqrt{4l_{1}^{2}+l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{4}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( 4l_{1}^{2}+l_{2}^{2}+2 \right)}}\ \sqrt{4l_{1}^{2}+l_{2}^{2}}\ \coth \left( \frac{\sqrt{4l_{1}^{2}+l_{2}^{2}}}{2}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}^{2}-4{{l}_{1}}^{2}<0$ and ${{l}_{3}}={{l}_{1}}$, then the trigonometric solutions are given as
${{u}_{{{5}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}^{2}-l_{2}^{2}-2 \right)}}\ \sqrt{4{{l}_{1}}^{2}-l_{2}^{2}}\ \tan \left( \frac{\sqrt{4{{l}_{1}}^{2}-l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{5}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4{{l}_{1}}^{2}-l_{2}^{2}-2 \right)}}\ \sqrt{4{{l}_{1}}^{2}-l_{2}^{2}}\ \cot \left( \frac{\sqrt{4{{l}_{1}}^{2}-l_{2}^{2}}}{2}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}^{2}+4{{l}_{1}}^{2}>0$ and ${{l}_{3}}={{l}_{1}}$, then the solutions are obtained as
${{u}_{{{6}_{1}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4l_{1}^{2}-l_{2}^{2}-2 \right)}}\ \sqrt{-4l_{1}^{2}+l_{2}^{2}}\ \tanh \left( \frac{\sqrt{-4l_{1}^{2}+l_{2}^{2}}}{2}\xi \right)$
Or
${{u}_{{{6}_{2}}}}\left( \xi \right)=\mp \sqrt{6\ }\sqrt{\frac{\alpha }{\beta \left( 4l_{1}^{2}-l_{2}^{2}-2 \right)}}\ \sqrt{-4l_{1}^{2}+l_{2}^{2}}\ \coth \left( \frac{\sqrt{-4l_{1}^{2}+l_{2}^{2}}}{2}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}^{2}=4{{l}_{1}}{{l}_{3}}$, then the rational function solution as
${{u}_{{{7}_{1}}}}\left( \xi \right)=\mp \frac{2\sqrt{3\ }\sqrt{-\frac{\alpha }{\beta }}}{\xi }$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{1}}{{l}_{3}}<0$, ${{l}_{2}}=0$ and ${{l}_{3}}\ne 0$, then we attained the hyperbolic solution as
${{u}_{{{8}_{1}}}}\left( \xi \right)=\mp 2\sqrt{3\ }\sqrt{-{{l}_{1}}{{l}_{3}}}\ \sqrt{\frac{\alpha }{\beta \left( 2\ {{l}_{1}}{{l}_{3}}-1 \right)}}\ \tanh \left( \sqrt{-{{l}_{1}}{{l}_{3}}}\xi \right)$
Or
${{u}_{{{8}_{2}}}}\left( \xi \right)=\mp 2\sqrt{3\ }\sqrt{-{{l}_{1}}{{l}_{3}}}\ \sqrt{\frac{\alpha }{\beta \left( 2\ {{l}_{1}}{{l}_{3}}-1 \right)}}\ \coth \left( \sqrt{-{{l}_{1}}{{l}_{3}}}\xi \right)$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}=0$ and ${{l}_{1}}=-{{l}_{3}}$, then the exponential function solution as
${{u}_{{{9}_{1}}}}\left( \xi \right)=\pm \frac{2\sqrt{3\ }\left( 1+{{e}^{-2{{l}_{3}}\xi }} \right)\sqrt{-\frac{\alpha }{\beta \left( 2l_{3}^{2}+1 \right)}}\ {{l}_{3}}}{-1+{{e}^{-2{{l}_{3}}\xi }}}$
Where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{1}}={{l}_{3}}=0$, then we attain
${{u}_{{{10}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( l_{2}^{2}+2 \right)}}\ {{l}_{2}}$
This solution is the constant function solution. So, it has no physical meaning.
When ${{l}_{1}}={{l}_{2}}=\text{K}$ and ${{l}_{3}}=0$, we obtain
${{u}_{{{11}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( {{\text{K}}^{2}}+2 \right)}}\text{K}$
This solution is the constant function solution. So, it has no physical meaning.
When ${{l}_{2}}={{l}_{3}}=\varphi $ and ${{l}_{1}}=0$, then the exponential function solution as
${{u}_{{{12}_{1}}}}\left( \xi \right)=\mp \frac{\sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( {{\varphi }^{2}}+2 \right)}}\varphi \ \left( 1+{{e}^{\varphi \xi }} \right)}{-1+{{e}^{\varphi \xi }}}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}=\left( {{l}_{1}}+{{l}_{3}} \right)$, then we got the exponential solution as
${{u}_{{{13}_{1}}}}\left( \xi \right)=\mp \frac{\sqrt{6\ }\left( {{l}_{1}}-{{l}_{3}} \right)\sqrt{-\frac{\alpha }{\beta \left( l_{1}^{2}-2{{l}_{1}}{{l}_{3}}+l_{3}^{2}+2 \right)}}\ \left( {{l}_{3}}{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}+1 \right)}{{{l}_{3}}{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}-1}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{2}}=-\left( {{l}_{1}}+{{l}_{3}} \right)$, then the solutions as
${{u}_{{{14}_{1}}}}\left( \xi \right)=\pm \frac{\sqrt{6\ }\left( {{l}_{1}}-{{l}_{3}} \right)\sqrt{-\frac{\alpha }{\beta \left( l_{1}^{2}-2{{l}_{1}}{{l}_{3}}+l_{3}^{2}+2 \right)}}\ \left( {{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}+{{l}_{3}} \right)}{{{l}_{3}}-{{e}^{\left( {{l}_{1}}-{{l}_{3}} \right)\xi }}}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{1}}=0$, we attain the following solutions as
${{u}_{{{15}_{1}}}}\left( \xi \right)=\mp \frac{\sqrt{6\ }\sqrt{-\frac{\alpha }{\beta \left( l_{2}^{2}+2 \right)}}\ {{l}_{2}}\left( {{l}_{3}}{{e}^{{{l}_{2}}\xi }}+1 \right)}{{{l}_{3}}{{e}^{{{l}_{2}}\xi }}-1}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{3}}={{l}_{2}}={{l}_{1}}\ne 0$, then we obtain the solution as
${{u}_{{{16}_{1}}}}\left( \xi \right)=\pm 3\sqrt{2\ }{{l}_{1}}\tan \left( \frac{\sqrt{3\ }{{l}_{1}}\xi }{2} \right)\sqrt{\frac{\alpha }{\beta \left( 3l_{1}^{2}-2 \right)}}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{3}}={{l}_{1}}$ and ${{l}_{2}}=0$, then the solution as
${{u}_{{{17}_{1}}}}\left( \xi \right)=\pm 2\sqrt{3\ }{{l}_{1}}\tan \left( {{l}_{1}}\xi \right)\sqrt{\frac{\alpha }{\beta \left( 2l_{1}^{2}-1 \right)}}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
When ${{l}_{3}}=0$, then
${{u}_{{{18}_{1}}}}\left( \xi \right)=\pm \sqrt{6\ }{{l}_{2}}\sqrt{-\frac{\alpha }{\beta \left( l_{2}^{2}+2 \right)}}$
where $\xi =x-\omega t$ and $\omega =-\frac{4\alpha }{4{{l}_{1}}{{l}_{3}}-l_{2}^{2}-2}$.
This solution is the constant function solution. So, it has no physical meaning.
Remark: All obtained solutions have been simplified and it's satisfied the original equation using Maple.
4. Graphical representations and physical significance
Many of the travelling waves are physically raised naturally, usually described by partial differential equations. Solitary waves are also known as solitons, which are a particular class of travelling waves. They have additional properties that they can interact with other solitons such as they come out after a collision without changing shape, for a small stage change. We will discuss the nature of the travelling wave solutions obtained from the SMCH equation in this portion. The different types of solitons, their physical behaviors will also be discussed. We have drawn 3D and 2D figures by Mathematica.
In this paper, we attain trigonometric, hyperbolic and rational solutions and each solution has pair of solitons (positive and negative). Also, we find out single spike, multiple spikes, kink, anti-peakon, singular spike soliton, and anti-bell shape soliton solutions. Asserting the Fig. 1(a) of the solution ${{u}_{{{1}_{1}}}}\left( \xi \right)$ represent multiple spikes type wave for $\alpha =7.9$ these characteristics remains unchanged after increasing the linear part constant term $\alpha =9$ shown in Fig. 1(b) but Fig. 1(c) represent single spike for $\alpha =16$. Also, illustrated in 2D combined line plots are shown in Fig. 1(d). Applying the corresponding value of $\alpha $ in 3D we noticed that different wave velocity $\omega $ gives different wave-lengths and wave numbers. Also, we observed that for the negative value of ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}$ when wave velocity $\omega $ decrease, the wave length also small as a result wave number increasing. For the value of $\alpha =-1$ and the positive condition of ${{l}_{2}}^{2}-4{{l}_{1}}{{l}_{3}}$ the solution ${{u}_{{{2}_{1}}}}\left( \xi \right)$ represents the kink shape soliton sketched in Fig. 2(a). Also, decreasing the value of $\alpha $ the soliton profile has no changed shown in Fig. 2(b,c). Using the corresponding 3D value of $\alpha $, we see that the effect of the asymptotic line descends from the x-direction (positive) to the x-direction (negative) with different travelling wave $\omega $, which is sketched in Fig. 2(d). That means asymptotic line comes down from right to left. We also observed that decreasing the value of wave velocity $\omega $ the amplitude of the wave increasing in y-direction (positive). The profile of the result ${{u}_{{{6}_{1}}}}$ is anti-peakon soliton for the limitation of ${{l}_{2}}^{2}+4{{l}_{1}}^{2}>0$ and ${{l}_{3}}={{l}_{1}}$, which wave profile sketched in Fig. 3(a). We also see that the soliton shape has unchanged for the different values of $\alpha $ in Fig. 3(b,c). In 2D figure, we have shown the deviation of the solution for the corresponding value of 3D and we also observed that for increasing the wave velocity the deviation of the soliton is seen to move upwards and gradually to the positive side sketched in Fig. 3(d). Fig. 4(a–d) representation of the profile ${{u}_{{{8}_{1}}}}$ for the several values of $\alpha $ whose are proposed shapes looks like as anti-bell types. We have also shown in 2D figure for the same value of 3D by fulfilling the condition of ${{l}_{1}}{{l}_{3}}<0$, ${{l}_{2}}=0$ and ${{l}_{3}}\ne 0$ and showing their dispersion changes. The solution ${{u}_{{{8}_{2}}}}$ represents the singular spike soliton wave profile, which is shown in Fig. 5(a–c) for different values of $\alpha $. In Fig. 5(d), we have also observed that the same value of 3D by fulfilling the condition and showing their dispersion changes.
Fig. 1. 3D modulus plots of ${{u}_{{{1}_{1}}}}\left( x,t \right)$ for ${{l}_{1}}=0.1$, ${{l}_{2}}=2$, ${{l}_{3}}=2$, $\beta =7$ with (a) $\alpha =7.9$, (b) $\alpha =9$ and (c) $\alpha =16$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
Fig. 2. 3D modulus plots of ${{u}_{{{2}_{1}}}}\left( x,t \right)$ for ${{l}_{1}}=-0.1$, ${{l}_{2}}=1$, ${{l}_{3}}=0.5$, $\beta =0.1$ with (a) $\alpha =-1$, (b) $\alpha =-2$ and (c) $\alpha =-4$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
Fig. 3. 3D modulus plots of ${{u}_{{{6}_{1}}}}\left( x,t \right)$ for ${{l}_{1}}=-0.01$, ${{l}_{2}}=0.07$, ${{l}_{3}}=0.01$, $\beta =1$ with (a) $\alpha =0.1$, (b) $\alpha =0.3$ and (c) $\alpha =0.5$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
Fig. 4. 3D modulus plots of ${{u}_{{{8}_{1}}}}\left( x,t \right)$ for ${{l}_{1}}=-0.1$, ${{l}_{2}}=0$, ${{l}_{3}}=0.3$, $\beta =1$with (a) $\alpha =0.1$, (b) $\alpha =0.25$ and (c) $\alpha =0.5$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
Fig. 5. 3D modulus plots of ${{u}_{{{8}_{2}}}}\left( x,t \right)$ for ${{l}_{1}}=-0.1$, ${{l}_{2}}=0$, ${{l}_{3}}=25$, $\beta =0.5$ with (a) $\alpha =7.3$, (b) $\alpha =7.4$ and (c) $\alpha =7.5$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
In Fig. 6 represents the 3D wave profile of the solution ${{u}_{{{1}_{1}}}}\left( x,t \right)$ for the different values of $\beta $. It can be seen that the wave profile of Fig. 6 is almost similar to the wave profile of Fig. 1. So, we can say that the there is no effect of $\beta $. It is evocative that the gained solutions existing the manuscript would be tremendously obliging for examining the nature of the plane wave phenomena in science and ocean engineering.
Fig. 6. 3D modulus plots of ${{u}_{{{1}_{1}}}}\left( x,t \right)$ for ${{l}_{1}}=-0.1$, ${{l}_{2}}=0$, ${{l}_{3}}=25$, $\alpha =7.9$ with (a) $\beta =4$, (b) $\beta =7$ and (c) $\beta =10$ within the interval $-30\le x$, $t\le 30$. (d) 2D combined plots $t=1$. |
5. Comparison and stability analysis
With ${{e}^{-\varphi \left( \eta \right)}}$-expansion method [58]
We have observed that the both equations are very much closer, if we take the general form of exponential function of the ${{e}^{-\varphi \left( \eta \right)}}$-expansion method. If we take the values ${{a}^{g\left( \xi \right)}}={{e}^{-\varphi \left( \eta \right)}}$, ${{l}_{1}}=-\mu $, ${{l}_{2}}=-\lambda $, ${{l}_{3}}=-1$, then the both methods are similar. But the -expansion method gives us only five types of solutions while the NAE method gives us thirty different solutions. We can also see that the reference [58] the ${{e}^{-\varphi \left( \eta \right)}}$-expansion method applied to the SMCH equation and explored the five types of soliton solutions. On the other hand, the NAE method applied to the SMCH equation and obtained thirty types of exact soliton solutions.
With generalized (${G}'/G$)-expansion method [56]
The generalized (${G}'/G$)-expansion method have applied to the SMCH equation in Ref. [56]. Based on this method, they have reported nine types of soliton solutions. All solutions have involved in hyperbolic and trigonometric function solutions. On the other hand, we have applied the NAE method to the SMCH equation and found the thirty types of exact traveling soliton solutions in this manuscript. More precisely, all obtained solutions have involved hyperbolic, trigonometric, exponential and rational functions type's solutions. We can say that our method is more successful than other methods in providing appropriate solutions for newly development solutions and also see the details discussion in Ref. [61].
5.2. Stability analysis
$u\left( x,t \right)=\delta \eta \left( x,t \right)+{{\eta }_{0}},$
Where ${{\eta }_{0}}$ is a steady state solution of (1.1). Inserting Eq. (5.2.1) in Eq. (1.1) and linearizing in $\delta $, yields
$\delta {{\eta }_{t}}+2\alpha \delta {{\eta }_{x}}-\delta {{\eta }_{xxt}}+\beta \eta _{0}^{2}\delta {{\eta }_{x}}=0.$
Let us considered that the Eq. (5.2.2) has a solution as
$\eta \left( x,t \right)={{e}^{i\left( Kx+\omega t \right)}},$
Where K is the normalized wave number, putting Eq. (5.2.3) into Eq. (5.2.2) and solving for $\omega $, we obtain
$\omega =-\frac{2\alpha K+\beta \eta _{0}^{2}K}{1+{{K}^{2}}}$
We observed that the denominator of right hand sides of Eq. (5.2.4) is always positive values, then $\omega $ is negative. Thus, the dispersion is stable. Otherwise, the dispersion is unstable.
6. Conclusion
In this study, we used the NAE method to significantly determine the soliton solutions of the SMCH equation. From the SMCH equation, we have found the single spike, multiple spike, kink, anti-picon, single spike soliton, anti-bell shape soliton solutions. The NAE method has also extracted some new wave solitons to the under study. It has been found that the wave profile changes with the change of dispersion coefficient $\alpha $. The 3D and 2D graphs are presented to understand the effect of changing the quality of parameters in solutions obtained for different values of different parameters. We have compared the solutions with the previous literature and found many soliton solutions of various types. Therefore, it can be said that the use of the mentioned method will be easier, more efficient, more powerful, productive, and esthetic to solve NLPDEs. Finally, this method will serve as a more powerful mathematical tool for NLPDEs to find soliton solutions and make wave phenomena more applicable in science and engineering.
Declaration of Competing Interest
None.
