Bifurcation Analysis and Soliton Behavior of New Combined Kairat-II-X Differential Equation Using Analytical Methods

Abstract

The exact analytical solutions of a new combined Kairat-II-X differential equation are presented. The related model is investigated by combining the enhanced modified extended tanh function method and the modified $tan(\varphi /2)$-expansion method. Then, a wide range of solitary wave solutions with unknown coefficients are extracted in a variety of shapes, including dark, bright, bell-shaped, kink-type, combine, and complex solitons, exponential, hyperbolic, and trigonometric function solutions. To offer physical insight, some of the identified solutions are presented in figures. Also, 3D, 2D, and 2D density profiles of the obtained outcomes are illustrated in order to examine their dynamics with the choices of parameters involved. Based on the obtained findings, we can assert that the suggested computational approaches are efficient, dynamic, well-structured, and valuable for tackling complex nonlinear problems in several fields, including symbolic computations. The bifurcation analysis and sensitivity analysis are employed to comprehend the dynamical system. We assume that our findings will be very beneficial in improving our understanding of the waves that manifest in solids.

1. Introduction

The majority of physical processes seen in the real world exhibit nonlinearity, and as a result, nonlinear partial differential equations (NLPDEs) were used to accurately represent complicated occurrences such as the semi-global stabilization problem of parabolic PDE-ODE systems [1], the monotone iterative and quasilinearization method [2], the dynamic complex matrix equation [3], the bifurcation diagrams and Lyapunov exponents [4], and an algorithm-based verification method [5]. It is consequently of the highest significance to researchers to find their accurate, analytical, or closed-form solutions. It is difficult to acquire these solutions since there are no conventional techniques that can be used to solve nonlinear partial differential equations. There has been a significant increase in the number of powerful methods that have been implemented in the area of developing accurate solutions for NLPDEs. Several analytical methods have been proposed for solving nonlinear evolution equations, including the $exp(-\Omega (\eta \left)\right)$-expansion method [6], the homotopy analysis method [7], the homotopy perturbation method [8], the Exp-function method [9], the Hirota bilinear method [10], the first integral method [11], the inverse scattering method [12], the Lie group method [13], the Jacobi elliptic function expansion method [14], the variational iteration method [15], the Laplace-residual power series method [16], the Legendre approximation method [17], the generalized Bogoyavlensky–Konopelchenko equation [18], the generalized Hietarinta equation [19], the dispersive wave systems [20], the variable-coefficient generalized nonlinear wave equation [21], the symmetric nonlinear dispersive wave model [22], the (3 + 1)-dimensional KPB-like equation [23], and many other methods that have yet to be identified. Some studies are full of solutions that were found after a lot of work. The research findings encompass a diverse array of solutions, including breather wave solutions, solitary wave solutions, rogue wave solutions, periodic solutions, hyperbolic solutions, and many more.

Integrable systems arise in various fields such as physics, applied mathematics, and geometry, often describing nonlinear wave phenomena, fluid dynamics, quantum [24], and other physical systems [25]. In [26], the authors proposed some new integrable systems such as Kairat equations and the Zhanbhota equation. Kairat equations can be used in wave analysis in fluid mechanics and optics. The Kairat-II equation and Kairat-X equation are represented in their standard integrable forms as follows [26]:

$$h_{xt}+h_{xxxt}+2h_t{h}_{xx}+4h_x{h}_{xt}=0,$$

(1)

and

$$h_{tt}+h_{xxxt}-3{\left(h_x{h}_t\right)}_x=0,$$

(2)

where $h=h(x,t)$ symbolizes the wave profile. There are several studies on Equations (1) and (2) in the literature [27,28,29,30]. Very recently, Wazwaz et al. [31] proposed a new equation, Kairat-II-X, resulting from combining Equations (1) and (2). The Kairat-II-X equation is expressed as

$$u_0{h}_{tt}+u_1{h}_{xt}+u_2{h}_{xxxt}+u_3(h_t{h}_{xx}+h_x{h}_{xt})+u_4{h}_{xx}=0,$$

(3)

where $h=h(x,t)$ gives the wave field and $u_0$ scales the pure temporal acceleration term $h_{tt}$; $u_1$ represents the advection coefficient, which physically controls the coupling between space and time evolution $h_{xxxt}$; $u_2$ gives the higher-order dispersion coefficient, which controls third-order dispersion $h_{xxxt}$; $u_3$ provides nonlinear interaction strength, scaling the nonlinear term $h_t{h}_{xx}+h_x{h}_{xt}$; and $u_4$ represents linear dispersion, controlling classical dispersion or wave stiffness. It has been demonstrated that when $u_0=0,u_4=0$ and the remaining parameters are redefined, Equation (3) simplifies to the Kairat-II Equation (1). Additionally, by setting $u_1=0$, the Kairat-X Equation (2) is obtained. Thus, Equation (3) is the generalization of Equations (1) and (2). It is important to emphasize that the coefficients $u_0,u_1,u_2,u_3$, and $u_4$ are real numbers whose magnitudes vary according to the specific model being studied. Kairat-II-X illustrates how the principles of differential geometry of curves are connected to the idea of equivalence. The Kairat-II-X equation can be used to study wave dynamics in hydrodynamics, nonlinear optics, or plasma physics.

The variational principle, Hamiltonian, solitary, and periodic wave solutions of the fractional Benjamin Ono equation were investigated in [32]. Also, a variational principle for a thin film equation was studied in [33]. Among the mentioned analytical techniques that generate soliton solutions to NLPDEs, the enhanced modified extended tanh function method (eMETFM) and the modified $tan(\varphi /2)$-expansion method (MTEM) played an important role in solving nonlinear equations. The transformation of NLPDEs to NLODEs through wave transportation enables eMETFM and MTEM, creating a set of nonlinear algebraic equations where by solving that nonlinear system, we will reach analytical solutions [34,35].

The generalized (G′/G) method, generalized projective Riccati equation method, and Riccati modified extended simple equation method were used for extracting a diverse array of soliton solutions of the Combined Kairat–II–X differential equation [36]. Wazwaz investigated two newly developed (3 + 1)-dimensional Kairat–II and Kairat–X equations that illustrate relations with the differential geometry of curves and equivalence aspects [37]. Rizvi et al. [38] studied stochastic variants of the Kairat-II and Kairat-X equations in (3 + 1)-dimensions by using the new projective Riccati equation approach of soliton solutions. The Kairat–II–X equation provided the connections between the differential geometry of curves, the concept of equivalence was applied using the Lie symmetry analysis method to investigate the vector field, and symmetry reductions of the equation were applied to find dark solitons and various combo soliton solutions in [39]. The Kairat–II equation and the Kairat–X equation were studied using $(G^{\prime }/G)$, the Bernoulli Sub-ODE, and the modified auxiliary equation methods, where the constructed solutions were trigonometric, hyperbolic, exponential, and rational functions [40].

With the rapid development of nonlinear science, soliton theory has become increasingly important and lots of effective methods to develop soliton solutions have emerged [18,19,20]. Motivated by these advances, the present work employs the modified extended direct algebraic method to derive a comprehensive family of analytical solutions for a combined Kairat-II-X differential equation with constant coefficients, including bright, dark, mixed, and singular structures. Bifurcation analysis is an essential mathematical approach used to examine how the qualitative behavior of solutions in nonlinear dynamical and chaotic systems [41] evolves with changes in system parameters. In nonlinear optical wave propagation, especially in the realm of the higher-order nonlinear fractional two-mode Nizhnik–Novikov–Veselov equation and the higher-dimensional Date–Jimbo–Kashiwara–Miwa model, it offers critical understanding of the stability and dynamics of soliton solutions [42,43]. Bifurcation analysis, chaotic phenomena, the variational principle, the Hamiltonian, and solitary and periodic wave solutions of the fractional Benjamin Ono equation were investigated in [32]. Ahmed et al. [44] investigated the chaotic features of a novel variable-order fractional Rössler system with Liouville–Caputo derivatives of variable order. Dynamic, bifurcation, and Lyapunov analysis of fractional Rössler chaos were probed using the modified numerical approximation method and Laplace decomposition method [45]. The Bifurcation and sensitivity analysis, chaotic behaviors, variational principle, Hamiltonian and diverse wave solutions to investigate the extended integrable Kadomtsev-Petviashvili equation were studied in [46]. The vibration of the capillary in a nanoscale deformable tube was studied based on the variational principle and Hamiltonian [47].

In this study, we aimed to use the enhanced modified extended tanh function method, the modified $tan(\varphi /2)$-expansion method, and bifurcation analysis to investigate new combined Kairat-II-X differential equations. As demonstrated by applying the eMETFM and MTEM, generating exact, implicit solutions for this nonlinear partial differential equation turned out to be highly effective, providing clear evidence for the utility of this method in solving NLPDEs. The obtained soliton and periodic solutions can therefore provide insights into the design of stable optical transmission systems and study wave dynamics in hydrodynamics and nonlinear optics. The novelty of this work lies in extending the Kairat–X equation to include connections between the differential geometry of curves and the concept of equivalence aspects, and in employing the enhanced modified extended tanh function method and the modified $tan(\varphi /2)$-expansion method to obtain new families of analytical solutions including mixed bright–dark solitons and singular periodic structures that have not been reported in previous studies.

This paper is organized into six sections. Section 1 introduces the combined Kairat-II-X differential equation and some related nonlinear equations. The eMETFM and its application are offered in Section 2. Section 3 discusses the modified $tan(\varphi /2)$-expansion method and some exact solutions. The bifurcation analysis is analyzed in Section 4. The physical interpretation of solutions is investigated in Section 5. Section 6 provides the conclusion.

2. The eMETFM for Combined Kairat-II-X Differential Equation

This section explains the enhanced modified extended tanh function method [48]. Let the nonlinear partial differential equation be expressed as follows:

$$\mathcal{P}(h,h_x,h_t,h_{xx},h_{xt},h_{tt},h_{xxxt},\dots )=0,$$

(4)

where $\mathcal{P}$ is a function of the partial differential equations. Using the transformation $\theta =mx+nt$, Equation (4) is transformed as

$$\mathcal{O}(h,m{h}^{\prime },n{h}^{\prime },m^2{h}^{″},mn{h}^{″},n^2{h}^{″},m^{3}n{h}^{⁗},\dots )=0,$$

(5)

where $\mathcal{O}$ is a function of the ordinary differential equations. According to [49], the solution of Equation (5) is obtained as

$$h\left(\theta \right)=k_0+\sum _{i=1}^N{k}_{i}X{\left(\theta \right)}^i+l_{i}X{\left(\theta \right)}^{-i},$$

(6)

where $k_0,k_1,\dots ,k_N$ and $l_1,l_2,\dots ,l_N$ are constants to be determined. N is a positive integer that can be found by balancing linear and nonlinear terms in Equation (5). Function $X\left(\theta \right)$ satisfies the following Riccati equation:

$${\displaystyle \frac{dX\left(\theta \right)}{d\theta }}=\delta +X^2\left(\theta \right),$$

(7)

where $\delta$ is a parameter. Inserting (6) along with (7) in Equation (5) obtains the nonlinear algebra system. Solving the algebraic system of this equation, we easily obtain the required parameters.

• First Case:

Theorem 1.

Suppose $\delta <0$; hence, Equation (7) has hyperbolic solutions [48,49].

Proof. 

When $\delta <0$, then Riccati equation creates hyperbolic solutions with $\rho =\pm 1$ in the frame below:

$$\begin{array}{c}{X}_1\left(\theta \right)=-\sqrt{-\delta }tanh\left(\sqrt{-\delta }(\theta +{\theta }_0)\right),\hfill \\ X_2\left(\theta \right)=-\sqrt{-\delta }coth\left(\sqrt{-\delta }(\theta +{\theta }_0)\right),\hfill \\ X_3\left(\theta \right)=-\sqrt{-\delta }tanh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right)+i\rho sech\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right),\hfill \\ X_4\left(\theta \right)={\displaystyle \frac{\delta -\sqrt{-\delta }tanh\left(\sqrt{-\delta }(\theta +{\theta }_0)\right)}{1+\sqrt{-\delta }tanh\left(\sqrt{-\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_5\left(\theta \right)={\displaystyle \frac{\sqrt{-\delta }(5-4cosh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right))}{3+4sinh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_6\left(\theta \right)={\displaystyle \frac{\rho \sqrt{-\delta (a_1^2+a_2^2)}-a_1\sqrt{-\delta }cosh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right))}{a_2+a_{1}sinh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_7\left(\theta \right)=\rho \sqrt{-\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cosh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right)-\rho sinh\left(2\sqrt{-\delta }(\theta +{\theta }_0)\right)}}\right].\hfill \end{array}$$

(8)

□

• Second Case:

Theorem 2.

Let $\delta >0$, then Equation (7) has trigonometric-type solutions [48,49].

Proof. 

When $\varphi >0$, the Riccati equation creates trigonometric solutions with $\rho =\pm 1$ in the frame below:

$$\begin{array}{c}{X}_8\left(\theta \right)=\sqrt{\delta }tan\left(\sqrt{\delta }(\theta +{\theta }_0)\right),\hfill \\ X_9\left(\theta \right)=\sqrt{\delta }cot\left(\sqrt{\delta }(\theta +{\theta }_0)\right),\hfill \\ X_{10}\left(\theta \right)=\sqrt{\delta }tan\left(2\sqrt{\delta }(\theta +{\theta }_0)\right)+\rho sec\left(2\sqrt{\delta }(\theta +{\theta }_0)\right),\hfill \\ X_{11}\left(\theta \right)=-{\displaystyle \frac{\sqrt{\delta }(1-tan\left(\sqrt{\delta }(\theta +{\theta }_0)\right))}{1+tan\left(\sqrt{\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_{12}\left(\theta \right)={\displaystyle \frac{\sqrt{\delta }(4-5cos\left(2\sqrt{\delta }(\theta +{\theta }_0)\right))}{3+5sin\left(2\sqrt{\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_{13}\left(\theta \right)={\displaystyle \frac{\rho \sqrt{\delta (a_1^2-a_2^2)}-a_1\sqrt{\delta }cos\left(2\sqrt{\delta }(\theta +{\theta }_0)\right))}{a_2+a_{1}sin\left(2\sqrt{\delta }(\theta +{\theta }_0)\right)}},\hfill \\ X_{14}\left(\theta \right)=i\rho \sqrt{\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cos\left(2\sqrt{\delta }(\theta +{\theta }_0)\right)-i\rho sin\left(2\sqrt{\delta }(\theta +{\theta }_0)\right)}}\right].\hfill \end{array}$$

(9)

□

• Third Case:

Theorem 3.

Let $\delta =0$, then Equation (7) has a rational solution.

Proof. 

When $\delta =0$, the Riccati equation creates a rational solution in the frame below:

$$X_{15}\left(\theta \right)=-{\displaystyle \frac{1}{\theta +{\theta }_0}}.$$

(10)

□

Three Types of Solutions Using eMETFM for the Combined Kairat-II-X Equation

By using the transformation wave $\theta =mx+Nt$, Equation (3) changes to

$$\begin{array}{c}(n^2{u}_0+mn{u}_1+m^2{u}_4)h^{″}+m^{3}n{h}^{⁗}+2m^{2}n{u}_3{h}^{\prime }{h}^{″}=0.\hfill \end{array}$$

(11)

After integrating Equation (11) once with respect to $\theta$, we get

$$\begin{array}{c}(n^2{u}_0+mn{u}_1+m^2{u}_4)h^{\prime }+m^{3}n{h}^{‴}+m^{2}n{u}_3{h}^{\prime 2}=0.\hfill \end{array}$$

(12)

Employing the principle of homogeneous balance to the terms ${\left(h^{\prime }\right)}^2$ and $h^{‴}$ in Equation (12) yields $N=1$. So, Equation (6) can be re-written as follows:

$$h\left(\theta \right)=k_0+k_{1}X\left(\theta \right)+{\displaystyle \frac{l_1}{X\left(\theta \right)}},$$

(13)

where $k_1$ and $l_1$ cannot be zero simultaneously. Appending Equation (13) along with Equation (7) onto Equation (12), we get the polynomial based on functions of $X\left(\theta \right)$ as

$$\begin{array}{c}{X}^1\left(\theta \right)=6m^{3}n{k}_1{u}_2+m^{2}n{k}_1^2{u}_3=0,\hfill \\ X^2\left(\theta \right)=8\delta m^{3}n{k}_1{u}_2+2\delta m^{2}n{k}_1^2{u}_3-2m^{2}n{k}_1{l}_1{u}_3+m^2{k}_1{u}_4+mn{k}_1{u}_1+n^2{k}_1{u}_0=0,\hfill \\ X^3\left(\theta \right)=2{\delta }^2{m}^{3}n{k}_1{u}_2+{\delta }^2{m}^{2}n{k}_1^2{u}_3-2\delta m^{3}n{l}_1{u}_2-4\delta m^{2}n{k}_1{l}_1{u}_3+m^{2}n{l}_1^2{u}_3+\delta m^2{k}_1{u}_4+\delta mn{k}_1{u}_1+\hfill \\ \delta n^2{k}_1{u}_0-m^2{l}_1{u}_4-mn{l}_1{u}_1-n^2{l}_1{u}_0=0,\hfill \\ X^4\left(\theta \right)=-8{\delta }^2{m}^{3}n{l}_1{u}_2-2{\delta }^2{m}^{2}n{k}_1{l}_1{u}_3+2\delta m^{2}n{l}_1^2{u}_3-\delta m^2{l}_1{u}_4-\delta mn{l}_1{u}_1-\delta n^2{l}_1{u}_0=0,\hfill \\ X^5\left(\theta \right)=-6{\delta }^3{m}^{3}n{l}_1{u}_2+{\delta }^2{m}^{2}n{l}_1^2{u}_3=0.\hfill \end{array}$$

(14)

Solving the above system, we get the following results:

• Solution 1: 

$$\begin{array}{c}\delta =0,m=m,n=n,k_0=k_0,k_1=0,l_1={\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{m^{2}n{u}_3}},\theta =mx+nt.\hfill \end{array}$$

(15)

According to (10), we obtain the rational solution

$$h_{\mathbf{1}}\left(\theta \right)=k_0+{\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{m^{2}n{u}_{3}X\left(\theta \right)}},\Rightarrow h_{\mathbf{1}}(x,t)=k_0-{\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{m^{2}n{u}_3}}(mx+nt+{\theta }_0).$$

(16)

• Solution 2: 

$$\begin{array}{c}\delta =0,m={\displaystyle \frac{\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n}{2u_4}},n=n,k_0=k_0,l_1=0,k_1=-{\displaystyle \frac{3\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n{u}_2}{u_4{u}_3}},\hfill \\ \theta ={\displaystyle \frac{\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n}{2u_4}}x+nt.\hfill \end{array}$$

(17)

According to (10), we obtain the rational solution

$$h_{\mathbf{2}}\left(\theta \right)=k_0-{\displaystyle \frac{3\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n{u}_{2}X\left(\theta \right)}{u_4{u}_3}},\Rightarrow h_{\mathbf{2}}(x,t)=k_0+{\displaystyle \frac{3\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n{u}_2}{u_4{u}_3\left(\frac{\left(-u_1+\sqrt{-4u_4{u}_0+u_1^2}\right)n}{2u_4}x+nt+{\theta }_0\right)}}.$$

(18)

• Solution 3: 

$$\begin{array}{c}\delta ={\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}},m=m,n=n,k_0=k_0,k_1=0,l_1={\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}},\hfill \\ \theta =mx+nt,h_{\mathbf{3}}\left(\theta \right)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_{3}X\left(\theta \right)}},\hfill \end{array}$$

(19)

According to (8), we obtain the hyperbolic solutions

$$\begin{array}{c}{h}_{\mathbf{3}}(x,t)=k_0-{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{4}}(x,t)=k_0-{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{-\delta }coth\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{5}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\left(-\sqrt{-\delta }tanh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+i\rho sech\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)\right)}},\hfill \\ h_{\mathbf{6}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{1+\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)}{\delta -\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{7}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{3+4sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{\sqrt{-\delta }(5-4cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{8}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{a_2+a_{1}sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{\rho \sqrt{-\delta (a_1^2+a_2^2)}-a_1\sqrt{-\delta }cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{9}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\rho \sqrt{-\delta }}}{\left[1-{\displaystyle \frac{2a_1}{a_1+cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))-\rho sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}\right]}^{-1},\hfill \end{array}$$

(20)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}}<0$.

According to (9), we obtain the trigonometric solutions

$$\begin{array}{c}{h}_{\mathbf{10}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{\delta }tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{11}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{\delta }cot\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{12}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3\left(\left(\sqrt{\delta }tanh\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)+\rho sec\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right)\right)}},\hfill \\ h_{\mathbf{13}}(x,t)=k_0-{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{1+tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\sqrt{\delta }(1-tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)))}},\hfill \\ h_{\mathbf{14}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{3+5sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\sqrt{\delta }(4-5cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{15}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{a_2+a_{1}sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\rho \sqrt{\delta (a_1^2-a_2^2)}-a_1\sqrt{\delta }cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{16}}(x,t)=k_0+{\displaystyle \frac{\frac{3}{2}{m}^2{u}_4+\frac{3}{2}mn{u}_1+\frac{3}{2}{n}^2{u}_0}{m^{2}n{u}_{3}i\rho \sqrt{\delta }}}{\left[1-{\displaystyle \frac{2a_1}{a_1+cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)-i\rho sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}\right]}^{-1},\hfill \end{array}$$

(21)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}}>0$.

• Solution 4: 

$$\begin{array}{c}\delta ={\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}},m=m,n=n,k_0=k_0,k_1=-{\displaystyle \frac{6m{u}_2}{u_3}},l_1=0,\hfill \\ \theta =mx+nt,h_{\mathbf{4}}\left(\theta \right)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}X\left(\theta \right),\hfill \end{array}$$

(22)

According to (8), we obtain the hyperbolic solutions

$$\begin{array}{c}{h}_{\mathbf{17}}(x,t)=k_0+{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right),\hfill \\ h_{\mathbf{18}}(x,t)=k_0+{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{-\delta }coth\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right),\hfill \\ h_{\mathbf{19}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\left(\left(-\sqrt{-\delta }tanh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+i\rho sech\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)\right),\hfill \\ h_{\mathbf{20}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\delta -\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{1+\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{21}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{-\delta }(5-4cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{3+4sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{22}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\rho \sqrt{-\delta (a_1^2+a_2^2)}-a_1\sqrt{-\delta }cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{a_2+a_{1}sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{23}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\rho \sqrt{-\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))-\rho sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}\right],\hfill \end{array}$$

(23)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}}<0$.

According to (9), we obtain the trigonometric solutions

$$\begin{array}{c}{h}_{\mathbf{24}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{\delta }tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right),\hfill \\ h_{\mathbf{25}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{\delta }cot\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right),\hfill \\ h_{\mathbf{26}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\left(\sqrt{\delta }tanh\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)+\rho sec\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right),\hfill \\ h_{\mathbf{27}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{\delta }(1-tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)))}{1+tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{28}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{\delta }(4-5cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{3+5sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{29}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\rho \sqrt{\delta (a_1^2-a_2^2)}-a_1\sqrt{\delta }cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{a_2+a_{1}sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{30}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}i\rho \sqrt{\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)-i\rho sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}\right],\hfill \end{array}$$

(24)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{4m^{3}n{u}_2}}>0$.

• Solution 5: 

$$\begin{array}{c}\delta ={\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{16m^{3}n{u}_2}},m=m,n=n,k_0=k_0,k_1=-{\displaystyle \frac{6m{u}_2}{u_3}},l_1={\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}},\hfill \\ \theta =mx+nt,h_{\mathbf{5}}\left(\theta \right)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}X\left(\theta \right)+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_{3}X\left(\theta \right)}},\hfill \end{array}$$

(25)

According to (8), we obtain the hyperbolic solutions

$$\begin{array}{c}{h}_{\mathbf{31}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{32}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{-\delta }coth\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{-\delta }coth\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{33}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}(\left(-\sqrt{-\delta }tanh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+i\rho sech\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3(\left(-\sqrt{-\delta }tanh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)+i\rho sech\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{34}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\delta -\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{1+\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)}}+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{1+\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right)}{\delta -\sqrt{-\delta }tanh\left(\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{35}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{-\delta }(5-4cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{3+4sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{3+4sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{\sqrt{-\delta }(5-4cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{36}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\rho \sqrt{-\delta (a_1^2+a_2^2)}-a_1\sqrt{-\delta }cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{a_2+a_{1}sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{a_2+a_{1}sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}{\rho \sqrt{-\delta (a_1^2+a_2^2)}-a_1\sqrt{-\delta }cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}},\hfill \\ h_{\mathbf{37}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\rho \sqrt{-\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))-\rho sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}\right]+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\rho \sqrt{-\delta }}}{\left[1-{\displaystyle \frac{2a_1}{a_1+cosh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))-\rho sinh\left(2\sqrt{-\delta }(mx+nt+{\theta }_0)\right))}}\right]}^{-1},\hfill \end{array}$$

(26)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{16m^{3}n{u}_2}}<0$.

According to (9), we obtain the trigonometric solutions

$$\begin{array}{c}{h}_{\mathbf{38}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{\delta }tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\sqrt{\delta }tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{39}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\sqrt{\delta }cot\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\sqrt{\delta }cot\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{40}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}\left(\sqrt{\delta }tan\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)+\rho sec\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right)+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3\left(\sqrt{\delta }tanh\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)+\rho sec\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)\right)}},\hfill \\ h_{\mathbf{41}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{\delta }(1-tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)))}{1+tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{1+tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\sqrt{\delta }(1-tan\left(\sqrt{\delta }(mx+nt+{\theta }_0)\right)))}},\hfill \\ h_{\mathbf{42}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\sqrt{\delta }(4-5cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{3+5sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}+{\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{3+5sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\sqrt{\delta }(4-5cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{43}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}{\displaystyle \frac{\rho \sqrt{\delta (a_1^2-a_2^2)}-a_1\sqrt{\delta }cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{a_2+a_{1}sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_3}}{\displaystyle \frac{a_2+a_{1}sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}{\rho \sqrt{\delta (a_1^2-a_2^2)}-a_1\sqrt{\delta }cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}},\hfill \\ h_{\mathbf{44}}(x,t)=k_0-{\displaystyle \frac{6m{u}_2}{u_3}}i\rho \sqrt{\delta }\left[1-{\displaystyle \frac{2a_1}{a_1+cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)-i\rho sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}\right]+\hfill \\ {\displaystyle \frac{\frac{3}{8}{m}^2{u}_4+\frac{3}{8}mn{u}_1+\frac{3}{8}{n}^2{u}_0}{m^{2}n{u}_{3}i\rho \sqrt{\delta }}}{\left[1-{\displaystyle \frac{2a_1}{a_1+cos\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)-i\rho sin\left(2\sqrt{\delta }(mx+nt+{\theta }_0)\right)}}\right]}^{-1},\hfill \end{array}$$

(27)

so that ${\displaystyle \frac{m^2{u}_4+mn{u}_1+n^2{u}_0}{16m^{3}n{u}_2}}>0$.

3. New Modified $\mathbf{tan}\mathbf{(}\mathit{\varphi }\mathbf{/}\mathbf{2}\mathbf{)}$-Expansion Method

The approach of the modified $tan(\varphi /2)$-expansion method (MTEM) is related to a first-order equation. We explain the auxiliary equation below before describing the main steps of the considered technique [50]. The equation is given below:

$$h\left(\theta \right)=k_0+\sum _{i=1}^N{\left({\displaystyle \frac{T}{1+T^2}}\right)}^{i-1}\left(k_i{\displaystyle \frac{T}{1+T}}+l_i{\displaystyle \frac{1-T^2}{1+T^2}}\right),$$

(28)

where $T=tan(\varphi /2)$, $k_i(0\le i\le N),l_i(1\le i\le N)$ are the parameters to be decided and $k_N\ne 0,l_N\ne 0$ and $\varphi =\varphi \left(\theta \right)$ are fulfilled within the ODE according to

$${\varphi }^{\prime }\left(\theta \right)=q_{1}sin\left(\varphi \left(\theta \right)\right)+q_{2}cos\left(\varphi \left(\theta \right)\right)+q_3.$$

(29)

The specific arrangements of condition (29) will be examined as

• Product 1: With $\mathsf{\Delta }=q_1^2+q_2^2-q_3^2<0$ and $q_2-q_3\ne 0$, we get $\varphi \left(\theta \right)=2arctan[{\displaystyle \frac{q_1}{q_2-q_3}}-{\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{q_2-q_3}}$$tan\left({\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{2}}\theta \right)].$
• Product 2: With $\mathsf{\Delta }>0$ and $q_2-q_3\ne 0$, we get $\varphi \left(\theta \right)=2$$arctan\left[{\displaystyle \frac{q_1}{q_2-q_3}}+{\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{q_2-q_3}}tanh\left({\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{2}}\theta \right)\right].$
• Product 3: With $\mathsf{\Delta }>0$, $q_2\ne 0$, and $q_3=0,$ we get $\varphi \left(\theta \right)=2$$arctan[{\displaystyle \frac{q_1}{q_2}}+{\displaystyle \frac{\sqrt{q_1^2+q_2^2}}{q_2}}$$tanh\left({\displaystyle \frac{\sqrt{q_1^2+q_2^2}}{2}}\theta \right)].$
• Product 4: With $\mathsf{\Delta }<0$, $q_3\ne 0$, and $q_2=0,$ we get $\varphi \left(\theta \right)=2arctan[-{\displaystyle \frac{q_1}{q_3}}+{\displaystyle \frac{\sqrt{q_3^2-q_1^2}}{q_3}}$$tan\left({\displaystyle \frac{\sqrt{q_3^2-q_1^2}}{2}}\theta \right)].$
• Product 5: With $\mathsf{\Delta }>0$, $q_2-q_3\ne 0$, and $q_1=0,$ we get $\varphi \left(\theta \right)=2arctan[\sqrt{{\displaystyle \frac{q_2+q_3}{q_2-q_3}}}$$tanh\left({\displaystyle \frac{\sqrt{q_2^2-q_3^2}}{2}}\theta \right)].$
• Product 6: With $q_1=0$ and $q_3=0,$ we get $\varphi \left(\theta \right)=arctan[{\displaystyle \frac{e^{2q_2\theta }-1}{e^{2q_2\theta }+1}},{\displaystyle \frac{2e^{q_2\theta }}{e^{2q_2\theta }+1}}].$
• Product 7: With $q_2=0$ and $q_3=0,$ we get $\varphi \left(\theta \right)=arctan\left[{\displaystyle \frac{2e^{q_1\theta }}{e^{2q_1\theta }+1}},{\displaystyle \frac{e^{2q_1\theta }-1}{e^{2q_1\theta }+1}}\right].$
• Product 8: With $\mathsf{\Delta }=0$, we get $\varphi \left(\theta \right)=-2arctan\left[{\displaystyle \frac{(q_2+q_3)(q_1\theta +2)}{q_1^2\theta }}\right].$
• Product 9: With $q_1=0,$ and $q_2=-q_3$, we get $\varphi \left(\theta \right)=-2arctan\left[{\displaystyle \frac{1}{q_3\theta }}\right].$
• Product 10: With $q_1=0,$ and $q_2=q_3$, we get $\varphi \left(\theta \right)=2arctan\left[C_1+q_3\theta \right].$
• Product 11: With $q_3=0,q_2=i$ and $q_1$, we get $\varphi \left(\theta \right)=-2iarctanh\left[{\displaystyle \frac{q_1\theta +2}{q_1\theta }}\right].$

To see the rest of the cases, refer to Refs. [51,52,53]. Also, $A_k,B_k(k=1,2,\dots ,N),q_1,q_2$, and $q_3$ are the values to be explored later.

Application of MTEM for the Combined Kairat-II-X Equation

According to Equation (12), which yields $N=1$, Equation (28) can be re-written as follows:

$$h\left(\theta \right)=k_0+k_1{\displaystyle \frac{T}{1+T^2}}+l_1{\displaystyle \frac{1-T^2}{1+T^2}}.$$

(30)

Inserting Equation (12) and its derivatives into Equation (30), we get the related nonlinear arithmetical condition. The sets of solutions are as follows:

• Merge I.

  $$\begin{array}{c}{k}_1=0,l_1={\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{q}_3{u}_3}},\theta =mx+nt,q_1=0,q_2=-q_3,h\left(\theta \right)=k_0+{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{q}_3{u}_3}}{T}^{-1}.\hfill \end{array}$$

  (31)

  Through Product 9, the exact cuspon solution is

  $$\begin{array}{c}{h}_1(x,t)=k_0-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{u}_3}}(mx+nt).\hfill \end{array}$$

  (32)

Merge II.

$$\begin{array}{c}{k}_1=0,l_1={\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}},\theta =mx+nt,q_1=0,q_3=q_3,q_2={\displaystyle \frac{\sqrt{G}}{n{u}_2{m}^2}},\hfill \\ h\left(\theta \right)=k_0+{\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}}{T}^{-1},G=n{u}_{2}m\left(m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right).\hfill \end{array}$$

(33)

Through Product 5, the exact singular soliton solution is

$$\begin{array}{c}{h}_2(x,t)=k_0+{\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}}\sqrt{{\displaystyle \frac{-q_{3}n{u}_2{m}^2+\sqrt{G}}{q_{3}n{u}_2{m}^2+\sqrt{G}}}}coth\left(\sqrt{-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}}(mx+nt)\right),\hfill \end{array}$$

(34)

so that $G=n{u}_{2}m\left(m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right)>0$ and ${\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}<0$.

• Merge III.

  $$\begin{array}{c}{k}_1=0,l_1={\displaystyle \frac{3m{u}_2\left(q_2+q_3\right)}{u_3}},q_1={\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2}},\hfill \\ q_2=q_2,q_3=q_3,\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}},\theta =mx+nt,h\left(\theta \right)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2+q_3\right)}{u_3}}{T}^{-1}.\hfill \end{array}$$

  (35)

  Through Product 1, the exact periodic wave solution is

  $$\begin{array}{c}{h}_3(x,t)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2+q_3\right)}{u_3}}{\left[{\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2\left(q_2-q_3\right)}}-{\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{q_2-q_3}}tan\left({\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{2}}(mx+nt)\right)\right]}^{-1},\hfill \end{array}$$

  (36)

  so that $\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}<0$.

Through Product 2, the exact soliton solution is

$$\begin{array}{c}{h}_4(x,t)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2+q_3\right)}{u_3}}{\left[{\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2\left(q_2-q_3\right)}}+{\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{q_2-q_3}}tanh\left({\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{2}}(mx+nt)\right)\right]}^{-1},\hfill \end{array}$$

(37)

so that $\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}>0$.

• Merge IV.

  $$\begin{array}{c}{l}_1=0,k_1=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{q}_3{u}_3}},\theta =mx+nt,q_1=0,q_2=q_3,\hfill \\ q_3=q_3,h\left(\theta \right)=k_0-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{q}_3{u}_3}}T.\hfill \end{array}$$

  (38)

  Through Product 9, the exact cuspon solution is

  $$\begin{array}{c}{h}_5(x,t)=k_0-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{m^{2}n{u}_3}}(mx+nt+{\theta }_0).\hfill \end{array}$$

  (39)

• Merge V.

  $$\begin{array}{c}{l}_1=0,k_1={\displaystyle \frac{3u_2\left(-q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}},\theta =mx+nt,q_1=0,q_2={\displaystyle \frac{\sqrt{G}}{n{u}_2{m}^2}},\hfill \\ h\left(\theta \right)=k_0+{\displaystyle \frac{3u_2\left(-q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}}T,G=n{u}_{2}m\left(m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right).\hfill \end{array}$$

  (40)

  Through Product 5, the exact dark soliton solution is

  $$\begin{array}{c}{h}_6(x,t)=k_0+{\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}}\sqrt{{\displaystyle \frac{-q_{3}n{u}_2{m}^2+\sqrt{G}}{q_{3}n{u}_2{m}^2+\sqrt{G}}}}tanh\left(\sqrt{-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}}(mx+nt)\right),\hfill \end{array}$$

  (41)

  so that $G=n{u}_{2}m\left(m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right)>0$ and ${\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}<0$.

• Merge VI.

  $$\begin{array}{c}{l}_1=0,k_1={\displaystyle \frac{3m{u}_2\left(q_2-q_3\right)}{u_3}},q_1={\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2}},\hfill \\ q_2=q_2,q_3=q_3,\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}},\theta =mx+nt,h\left(\theta \right)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2-q_3\right)}{u_3}}T.\hfill \end{array}$$

  (42)

  Through Product 1, the exact periodic wave solution is

  $$\begin{array}{c}{h}_7(x,t)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2-q_3\right)}{u_3}}\left[{\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2\left(q_2-q_3\right)}}-{\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{q_2-q_3}}tan\left({\displaystyle \frac{\sqrt{-\mathsf{\Delta }}}{2}}(mx+nt)\right)\right],\hfill \end{array}$$

  (43)

  so that $\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}<0$.

Through Product 2, the exact dark soliton solution is

$$\begin{array}{c}{h}_8(x,t)=k_0+{\displaystyle \frac{3m{u}_2\left(q_2-q_3\right)}{u_3}}\left[{\displaystyle \frac{\sqrt{-n{u}_{2}m\left(m^{3}n{q}_2^2{u}_2-m^{3}n{q}_3^2{u}_2+u_4{m}^2+u_{1}mn+u_0{n}^2\right)}}{n{u}_2{m}^2\left(q_2-q_3\right)}}+{\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{q_2-q_3}}tanh\left({\displaystyle \frac{\sqrt{\mathsf{\Delta }}}{2}}(mx+nt)\right)\right],\hfill \end{array}$$

(44)

so that $\mathsf{\Delta }=-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}>0$.

• Merge VII.

  $$\begin{array}{c}{l}_1={\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{2n{u}_{2}m}\right)}{u_3}},\theta =mx+nt,q_1=0,k_1={\displaystyle \frac{3u_2\left(-q_{3}m+\frac{\sqrt{G}}{2n{u}_{2}m}\right)}{u_3}},q_2={\displaystyle \frac{\sqrt{G}}{2n{u}_2{m}^2}},\hfill \\ h\left(\theta \right)=k_0+{\displaystyle \frac{3u_2\left(-q_{3}m+\frac{\sqrt{G}}{n{u}_{2}m}\right)}{u_3}}T+{\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{2n{u}_{2}m}\right)}{u_3}}{T}^{-1},G=n{u}_{2}m\left(4m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right).\hfill \end{array}$$

  (45)

  Through Product 5, the exact combined dark soliton and singular solution is

  $$\begin{array}{c}{h}_6(x,t)=k_0+{\displaystyle \frac{3u_2\left(-q_{3}m+\frac{\sqrt{G}}{2n{u}_{2}m}\right)}{u_3}}\sqrt{{\displaystyle \frac{2q_{3}n{u}_2{m}^2+\sqrt{G}}{-2q_{3}n{u}_2{m}^2+\sqrt{G}}}}tanh\left(\sqrt{-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{4n{u}_2{m}^3}}}(mx+nt)\right)+\hfill \\ {\displaystyle \frac{3u_2\left(q_{3}m+\frac{\sqrt{G}}{2n{u}_{2}m}\right)}{u_3}}\sqrt{{\displaystyle \frac{-2q_{3}n{u}_2{m}^2+\sqrt{G}}{2q_{3}n{u}_2{m}^2+\sqrt{G}}}}coth\left(\sqrt{-{\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{4n{u}_2{m}^3}}}(mx+nt)\right)\hfill \end{array}$$

  (46)

  so that $G=n{u}_{2}m\left(4m^3{q}_3^{2}n{u}_2-u_4{m}^2-u_{1}mn-u_0{n}^2\right)>0$ and ${\displaystyle \frac{u_4{m}^2+u_{1}mn+u_0{n}^2}{n{u}_2{m}^3}}<0$.

4. Bifurcation Analysis

Bifurcation analysis is important in the investigation of dynamical systems, as it explains how the qualitative behavior of a system changes in response to variations in parameters. A bifurcation occurs when a small change in the value of a parameter causes a sudden topological change in the system’s behavior. On the other hand, stability analysis is used to determine the robustness of equilibrium points in a dynamical system. Consider Equation (12) again as follows:

$$\begin{array}{c}{\sigma }_1{h}^{\prime }+{\sigma }_2{h}^{‴}+{\sigma }_3{h}^{\prime 2}=0,\hfill \end{array}$$

(47)

where ${\sigma }_1=n^2{u}_0+mn{u}_1+m^2{u}_4,{\sigma }_2=m^{3}n$ and ${\sigma }_3=m^{2}n{u}_3$. Letting ${\displaystyle \frac{dh}{d\theta }}=A$, the planner dynamics of the governing problem for Equation (47) can be expressed as

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{dA}{d\theta }}=B=a(A,B),\hfill \\ {\displaystyle \frac{dB}{d\theta }}=-{\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}A-{\displaystyle \frac{{\sigma }_3}{{\sigma }_2}}{A}^2=b(A,B),\hfill \end{array}\right.\end{array}$$

(48)

with the assessed Hamiltonian function identified as

$$\begin{array}{c}\hfill H(A,B)={\displaystyle \frac{1}{2}}{B}^2+{\displaystyle \frac{{\sigma }_1}{2{\sigma }_2}}{A}^2+{\displaystyle \frac{{\sigma }_3}{3{\sigma }_2}}{A}^3=h,\end{array}$$

(49)

and h characterizes a continuous variable, also called the Hamiltonian constants or energy levels. Moreover, ${\displaystyle \frac{1}{2}}{B}^2$ denotes the kinetic energy, and ${\displaystyle \frac{1}{2}}{B}^2+{\displaystyle \frac{{\sigma }_1}{2{\sigma }_2}}{A}^2+{\displaystyle \frac{{\sigma }_3}{3{\sigma }_2}}{A}^3$ represents the potential energy for the Hamiltonian equation (49). Critical points in the system, where no variations occur, are identified by solving the condition $\nabla H(A,B)=0$. For the phase portrait trajectories of the system (48), fixed points or equilibria of Equation (48) are obtained by utilizing the following results:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{dA}{d\theta }}=0,\hfill \\ {\displaystyle \frac{dB}{d\theta }}=0,\hfill \end{array}\right.\end{array}$$

(50)

which becomes

$$\begin{array}{c}\hfill B=0,and-{\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}A-{\displaystyle \frac{{\sigma }_3}{{\sigma }_2}}{A}^2=0.\end{array}$$

(51)

Then, we have

$$\begin{array}{c}\hfill B=0,andA=0,-{\displaystyle \frac{{\sigma }_1}{{\sigma }_3}}=-{\displaystyle \frac{n^2{u}_0+mn{u}_1+m^2{u}_4}{m^{2}n{u}_3}}.\end{array}$$

(52)

Hence, we have two equilibrium points $O(A_0,0)$ and $E(A_1,0)$ with the values

$$\begin{array}{c}\hfill A_0=0,A_1=-{\displaystyle \frac{n^2{u}_0+mn{u}_1+m^2{u}_4}{m^{2}n{u}_3}}.\end{array}$$

(53)

So, the equilibrium points for the system (48) are $E_1=(0,0)$ and $E_2=(-{\displaystyle \frac{{\sigma }_1}{{\sigma }_3}},0)$. Take the linearized form of the system (48) at equilibrium point $(A_0,B_0)$ and let

$$\begin{array}{c}\hfill M(A_0,B_0)=\left(\begin{array}{cc}0& 1\\ -{\sigma }_1& 0\end{array}\right).\end{array}$$

(54)

So, we can analyze the stability of these points by computing the Jacobian matrix $J(A,B)={\displaystyle \frac{\partial (a,b)}{\partial (A,B)}}={\left(\begin{array}{cc}{\displaystyle \frac{\partial a}{\partial A}}& {\displaystyle \frac{\partial a}{\partial B}}\\ {\displaystyle \frac{\partial b}{\partial A}}& {\displaystyle \frac{\partial b}{\partial B}}\end{array}\right)}_{A=A_0,B=B_0}$ of the system (48) as

$$\begin{array}{c}\hfill J(A_0,B_0)=\left|\begin{array}{cc}0& 1\\ -{\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}-2{\displaystyle \frac{{\sigma }_3}{{\sigma }_2}}A& 0\end{array}\right|={\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}+2{\displaystyle \frac{{\sigma }_3}{{\sigma }_2}}A,{\sigma }_{1,2}={\displaystyle \frac{{\sigma }_1}{{\sigma }_2}},{\sigma }_{3,2}={\displaystyle \frac{{\sigma }_3}{{\sigma }_2}},\end{array}$$

(55)

The eigenvalues of J are given by $|J-\lambda I|=0$, which implies that

$$\begin{array}{c}\hfill {\lambda }^2-tr\left(J\right)\lambda +\left|J\right|=0.\end{array}$$

(56)

where $tr\left(J\right)=0$ and $\left|J\right|={\sigma }_{1,2}+2{\sigma }_{3,2}A$. The following conditions are utilized:

• If the Jacobian is positive, the equilibrium point offers a center point.
• If the Jacobian is negative, the equilibrium point offers a saddle point.
• If the Jacobian is zero, the equilibrium point offers a cuspidal point. We have the following cases:

(I): ${\sigma }_{1,2}>0$ and ${\sigma }_{3,2}>0.$

If we obtain the amounts $u_0=1,u_1=1,u_3=5,u_4=2,m=2$, and $n=1$, we obtain the equilibrium points of $E_1=(0,0)$ and $E_2=\left({\displaystyle \frac{11}{20}},0\right)$, where $E_1$ is a center point, since $J(0,0)={\sigma }_{1,2}={\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}={\displaystyle \frac{11}{8}}>0$, and $E_2$ is saddle point, since $J(1,0)={\sigma }_{1,2}+2{\sigma }_{3,2}={\displaystyle \frac{51}{8}}>0$, as presented in Figure 1.

(II): ${\sigma }_{1,2}>0$ and ${\sigma }_{3,2}<0.$

If we obtain the amounts $u_0=1,u_1=1,u_3=5,u_4=2,m=-2$, and $n=-1$, we achieve the equilibrium points of $E_1=(0,0)$ and $E_2=\left(-{\displaystyle \frac{11}{20}},0\right)$, where $E_1$ is a center point, since $J(0,0)={\sigma }_{1,2}={\displaystyle \frac{11}{8}}>0$, and $E_2$ is saddle point, since $J(-1/2,0)={\sigma }_{1,2}-{\sigma }_{3,2}={\displaystyle \frac{31}{8}}>0$, as shown in Figure 2.

(III): ${\sigma }_{1,2}<0$ and ${\sigma }_{3,2}<0.$

If we obtain the amounts $u_0=1,u_1=1,u_3=-5,u_4=2,m=2$, and $n=-1$, we obtain the equilibrium points of $E_1=(0,0)$ and $E_2=\left({\displaystyle \frac{7}{20}},0\right)$, where $E_1$ is a saddle point, since $J(0,0)={\sigma }_{1,2}=-{\displaystyle \frac{7}{8}}<0$, and $E_2$ is saddle point, since $J(1,0)={\sigma }_{1,2}+2{\sigma }_{3,2}=-{\displaystyle \frac{47}{8}}<0$, as shown in Figure 3.

(IV): ${\sigma }_{1,2}<0$ and ${\sigma }_{3,2}>0.$

If we obtain the amounts $u_0=1,u_1=1,u_3=5,u_4=2,m=2$, and $n=-1$, we achieve the equilibrium points of $E_1=(0,0)$ and $E_2=\left(-{\displaystyle \frac{7}{20}},0\right)$, where $E_1$ is a saddle point, since $J(0,0)={\sigma }_{1,2}=-{\displaystyle \frac{7}{8}}<0$, and $E_2$ is center point, since $J(1,0)={\sigma }_{1,2}+2{\sigma }_{3,2}={\displaystyle \frac{33}{8}}>0$, as shown in Figure 4.

Sensitivity Analysis

This section provides a thorough examination of the sensitivity characteristics of the transformed dynamical system. A numerical approach is adopted to systematically assess how the system reacts to small perturbations in both initial conditions and system parameters. The objective is to quantify the extent to which such perturbations affect the overall evolution of the system. A sensitivity framework is developed to characterize the system’s stability. In this context, low sensitivity implies that minor variations in initial conditions result in proportionally small deviations in the system’s trajectory. To perform this sensitivity study, we solve the parametric form of the transformed dynamical model. The governing equations are represented by the following second-order system:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{dA}{d\theta }}=B=a(A,B),\hfill \\ {\displaystyle \frac{dB}{d\theta }}=-{\displaystyle \frac{{\sigma }_1}{{\sigma }_2}}A-{\displaystyle \frac{{\sigma }_3}{{\sigma }_2}}{A}^2=b(A,B),\hfill \end{array}\right.\end{array}$$

(57)

where the coefficients are defined in terms of system parameters as follows:

$${\sigma }_1=n^2{u}_0+mn{u}_1+m^2{u}_4,{\sigma }_2=m^{3}n,{\sigma }_3=m^{2}n{u}_3.$$

(58)

The analysis focuses on the sensitivity of the solution to controlled changes in the initial conditions and parameter values. The original system is transformed into the above planar dynamical form using a Galilean transformation, enabling a clearer examination of the solution’s behavior. To systematically evaluate the sensitivity, we consider two sets of initial conditions:

Set 1: $A\left(0\right)=0.1,B\left(0\right)=0.2$; $A\left(0\right)=0.01,B\left(0\right)=0.02$;

Set 2: $A\left(0\right)=0.001,B\left(0\right)=0.002$; $A\left(0\right)=0.001,B\left(0\right)=0.0011$.

Numerical simulations reveal that even slight changes in the initial state can lead to significantly different trajectories. Figure 5a illustrates the system’s response under Set 1, showing how an increase in the initial momentum $B\left(0\right)$ leads to a shift in amplitude and phase of the resulting oscillations. Figure 5b demonstrates similar findings for Set 1, but the divergence between trajectories is more pronounced due to simultaneous variations in both $A\left(0\right)$ and $B\left(0\right)$ in Figure 5c and finally a variation in $B\left(0\right)$ with constant $A\left(0\right)$ in Figure 5d. Physically, these figures underscore the nonlinearity of the system—where even minor perturbations result in increasingly divergent trajectories over time. This reflects the inherent instability and possible chaotic behavior under specific parameter regimes. Understanding these sensitivities is vital for predicting long-term behavior and for designing control strategies in applications where precision and robustness are critical.

5. The Physical Interpretation of Solutions

This section aims to generate graphical illustrations of the obtained solutions of the combined Kairat-II-X differential equation, which are obtained by the eMTEFM, MTEM, and bifurcation analysis. The results showcase a variety of visual depictions, including 3D, 2D, and density plots. Here, we elaborate and investigate the dynamics of some of the obtained solutions. These results generally show dark, periodic, kink, bright, M-shaped, and W-shaped soliton solutions. The dark soliton solution (20) $h_3(x,t)$ with suitable values $u_0=2,u_1=-1,u_2=2,u_3=1,u_4=1/2,m=4,n=-3,k_0=2$ is plotted in Figure 6. The behavior of soliton solution (20) $h_5(x,t)$ with suitable values $u_0=2$, $u_1=-1$, $u_2=2$, $u_3=1$, $u_4=1/2$, $m=4$, $n=-3$, $k_0=2$ is investigated in Figure 7. And the periodic wave solution (21) $h_{10}(x,t)$ is designed in Figure 8 with parameters $u_0=2$, $u_1=-1$, $u_2=2$, $u_3=1$, $u_4=1/2$, $m=4$, $n=3$, $k_0=2$. The periodic wave solution (21) $h_{12}(x,t)$ with special values is plotted in Figure 9 with parameters $u_0=2$, $u_1=-1$, $u_2=2$, $u_3=1$, $u_4=1/2$, $m=4$, $n=3$, $k_0=2$. The dark-singular soliton solution (26) $h_{31}(x,t)$ with suitable values $u_0=2,u_1=-1,u_2=2,u_3=1,u_4=1/2,m=1/4,n=-1/5,k_0=2$ is plotted in Figure 10. The behavior of periodic soliton solution (26) $h_{33}(x,t)$ with suitable values $u_0=2,u_1=-1,u_2=2,u_3=1,u_4=1/2,m=1/4,n=1/5,k_0=2$ is investigated in Figure 11. And the periodic-singular wave solution (21) $h_{38}(x,t)$ is designed in Figure 12 with parameters $u_0=2,u_1=-1,u_2=2,u_3=1,u_4=1/2,m=4,n=3,k_0=2$. The periodic wave solution (27) $h_{40}(x,t)$ with special values is plotted in Figure 13 with parameters $u_0=2,u_1=-1,u_2=2,u_3=1,u_4=1/2,m=4,n=3,k_0=2$. We integrate the solution from Equations (15)–(46) using Maple. The results show equality between the left- and right-hand sides, confirming an exact solution. Comparison with existing literature confirms the novelty of these solutions. Bright solitons are non-dispersive wave solutions, arising in a diverse range of nonlinear, one-dimensional systems, including atomic Bose–Einstein condensates with attractive interactions. Dark solitons are defined as solutions to the nonlinear Schrödinger equation that exhibit a dip in intensity within a uniform background, characterized by an abrupt phase change at the dip center. The singular solution, in mathematics, is a solution of a differential equation that cannot be obtained from the general solution obtained by the usual method of solving the differential equation. When a differential equation is solved, a general solution consisting of a family of curves is obtained.

6. Conclusions

We conclude that this study delivered a rigorous analytical examination of the combined Kairat-II-X differential equation. We first assessed the system with a combination of two equations. Upon validating the balance test, we introduced two novel solution techniques, the enhanced modified extended tanh function method and the modified $tan(\varphi /2)$-expansion method, to derive new soliton solutions characterized by distinct dynamic behaviors. Our analysis encompasses a comprehensive characterization of stable and functional soliton structures, including bright, dark, and periodic waves, illustrated via three-dimensional, two-dimensional, and density plots. In addition, the qualitative behavior of the dynamical system at the equilibrium points was investigated via bifurcation analysis. The bifurcation and sensitivity analysis provides crucial insights into the stability and dynamic behavior of these wave solutions under varying physical conditions. The effectiveness of these two methods in addressing the proposed model was demonstrated through both analytical methods and graphical representations, which reveal inherent periodic structures. The solutions were expressed in terms of exponential, trigonometric, and hyperbolic unctions, as well as their combinations, under specific parameter regimes. These soliton solutions offer valuable insights for both physics and mathematics. The results confirm that the employed methodologies were robust and reliable for generating a diverse set of stable, effective solutions applicable to a wide range of nonlinear partial differential equations, and they may be extended to other NLPDE models.