Interactions and Soliton Dynamics for a (2+1)-Dimensional Nonlinear Integrable Model Arising in Shallow Water

Abstract

In this study, we consider a (2+1)-dimensional integrable Boussinesq equation, where the Hirota method of positive logarithmic transformation is used to convert it into a bilinear form. We proceeded by employing different test functions, through which we obtained breather solutions, two-wave solutions, lump-periodic solutions, and new interaction solutions. The resulting soliton dynamics for the governing model are also derived using the enhanced modified extended tanh function method, where varieties of solutions, such as trigonometric, hyperbolic, and rational forms, were obtained. The derived solutions may hold significant potential for explaining real-world physical phenomena in fields like mathematical physics, plasma physics, and nonlinear optics. The accuracy and reliability of the solutions were tested by substituting them back into the original equation using Python, highlighting the method’s robustness, precision, and reliability. By choosing appropriate physical parameters, we showcased the rich diversity and dynamic behavior of the obtained soliton structures. In other words, the graphical representations in 3D, contour, and 2D were provided for some of the obtained results. The modulation instability analysis and gain spectrum of the model are also provided. The importance of the obtained results in the area of (2+1)-dimensional integrable equation application was also highlighted.

1. Introduction

The fundamental generalized wave equation [1], which describes the flow within a shallow, inviscid fluid layer, is given by

$$u_{tt}-u_{xx}-\beta {\left(u^2\right)}_{xx}-\gamma u_{xxxx}=0.$$

(1)

In Equation (1), the coefficients $\beta$ and $\gamma$ are non-zero. Specifically, $\beta$ is associated with the vertical extent of the fluid, while $\gamma$ corresponds to the characteristic speed of elongated wave profiles in shallow water. The standard Boussinesq Equation (1) is notable for preserving both quadratic nonlinearity and weak dispersion. Equation (1) has been extended to numerous Boussinesq-type models [2], leading to a wide variety of system structures developed to suit different research objectives.

The Boussinesq equation [3,4,5,6] has garnered considerable interest in various fields, including coastal and ocean engineering, tsunami modeling, and solid mechanics, particularly where nonlinear and dispersive effects play a critical role. It also emerges in the study of thin inviscid fluid layers with free surfaces, nonlinear lattice waves, ion-acoustic waves in plasma, wave propagation in elastic rods, and in the continuum limits of lattice dynamics or electrically coupled circuits. Additionally, it is relevant in the analysis of Rayleigh–Bénard convection. Remarkably, the equation also admits algebraically decaying rational solutions that resemble localized waves that appear abruptly and disappear without a trace [7,8,9].

The study of nonlinear partial differential equations has seen rapid growth in the scientific and engineering communities [10,11,12,13], due to their ability to capture key features of various physical processes [14,15,16,17,18,19]. Numerous established analytical methods have been employed to derive multiple soliton solutions [20,21,22,23], providing unified approaches to solving various newly proposed integrable equations [24,25,26,27,28,29].

The Boussinesq Equation (1) was first introduced by Boussinesq in 1871 [30], to model the propagation of long surface waves in shallow water. Over time, it has been recognized as a soliton-bearing equation that can be solved using various analytical techniques, including the test function method, Hirota’s bilinear approach, lump solutions, Bäcklund transformations, and the inverse scattering method [31,32,33,34,35,36]. Beyond shallow water theory, this equation finds applications in numerous physical settings, such as one-dimensional nonlinear lattice waves, nonlinear string vibrations, and ion-acoustic waves in plasma [28,37,38]. Nowadays, bifurcation theory and chaos analysis have become essential tools in the study of nonlinear partial differential equations (NLPDEs), prompting extensive research in this area [39]. Additionally, it has been shown in [40] that the Boussinesq equation can be reduced, via symmetry transformations, to several Painlevé equations, including the first, second, and fourth types.

In recent years, considerable research [31,32,33] has focused on the Boussinesq equation, with particular emphasis on rogue waves [34,35,36]. However, in [40], the primary focus was on rational solutions, that is, solutions that decay algebraically and closely resemble the rogue-wave solutions of the nonlinear Schrödinger equation. Over the past few decades, rational solutions of soliton equations have attracted significant interest. For certain soliton equations, such as the Benjamin–Ono equation, solitons can be explicitly described by rational solutions [40].

In [30], an extension of Equation (1) is presented, known as a (2+1)-dimensional integrable Boussinesq equation.

$$u_{tt}-u_{xx}-\beta {\left(u^2\right)}_{xx}-\gamma u_{xxxx}+{\displaystyle \frac{{\alpha }^2}{4}}{u}_{yy}+\alpha u_{yt}=0.$$

(2)

In Equation (2), the parameters $\beta$ and $\gamma$ are not zero. Dipankar et al. [41] delved into the (2+1)-dimensional Boussinesq equation and unveiled extraordinary localized waves, interaction solutions, and N-soliton solutions by examining surface water waves in shallow seas and harbors, tsunami wave propagation, wave overtopping, inundation, and near-shore wave phenomena. Md et al. [42] explored the lump, lump-stripe, and breather wave solutions of a groundbreaking integrable (2+1)-dimensional Boussinesq equation, which were derived through the ingenious Hirota bilinear method. Abdul et al. [30] examined the proposed model using a variety of functions, including Painlevé analysis, as well as real and complex multiple soliton solutions. Juniye Zhu [6] investigated the (2+1)-dimensional Boussinesq equation through the lens of the Dbar-problem, focusing on line solitons and rational solutions. In this exploration, we aim to uncover breather, two-wave, lump-periodic, and soliton dynamics for Equation (2) by applying the Hirota method of positive logarithmic transformation and the enhanced modified extended tanh function method (EMETFM), respectively.

The study will be arranged as follows: In Section 2, we present the interaction solutions. In Section 3, we present the soliton solutions. In Section 4, we present the graphical representations for some of the obtained results and discussion. In Section 5, we provide the modulation instability analysis of the governing model. In Section 6, we give the concluding remarks of the study.

2. Interactions Solutions

For the interaction solutions, we use the transformation [30]:

$$u(x,y,t)={\displaystyle \frac{6\gamma }{\beta }}{\displaystyle \frac{{\partial }^{2}log\left(f(x,y,t)\right)}{\partial x^2}},$$

(3)

to convert Equation (2) into bilinear form as follows:

$$-4\alpha f_t{f}_y-4f_t^2+f\left(4f_{\mathrm{tt}}-4f_{\mathrm{xx}}-4\gamma f_{\mathrm{xxxx}}+4\alpha f_{\mathrm{yt}}+{\alpha }^2{f}_{\mathrm{yy}}\right)+16\gamma f_x{f}_{\mathrm{xxx}}+4f_x^2-12\gamma f_{\mathrm{xx}}^2-{\alpha }^2{f}_y^2=0.$$

(4)

2.1. The Breather Solution

For the breather solution, we used the test function [43,44,45]:

$$f=q_{2}exp\left(p_1{\chi }_1\right)+exp\left(-p_1{\chi }_1\right)+q_{1}cos\left(p_0{\chi }_2\right),$$

(5)

where ${\chi }_1={\rho }_{1}t+x+y,{\chi }_2={\rho }_{2}t+x+y.$

Furthermore, ${\rho }_1$, and ${\rho }_2$ are the speed of soliton waves. By intertwining Equation (5) with Equation (4) and executing the essential transformations, we obtained a polynomial involving exponential, trigonometric, and hyperbolic functions. By gathering like powers of these enchanting functions and setting them to zero, we unveil a tapestry of algebraic equations. Through the artful resolution of this intricate system, we reveal the following intriguing cases of solutions.

Result 1.

$$p_1=-\sqrt{-p_0^2},{\rho }_1={\displaystyle \frac{-{\alpha }^2-16\gamma p_0^2+\frac{2\alpha \sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}}{q_1^2-4q_2}+\frac{{\alpha }^2{q}_1^2}{q_1^2-4q_2}-\frac{4{\alpha }^2{q}_2}{q_1^2-4q_2}+4}{2\alpha -\frac{4\sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}}{q_1^2-4q_2}-\frac{2\alpha q_1^2}{q_1^2-4q_2}+\frac{8\alpha q_2}{q_1^2-4q_2}}},$$

$${\rho }_2={\displaystyle \frac{-2\sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}-\alpha q_1^2+4\alpha q_2}{2\left(q_1^2-4q_2\right)}}.$$

(6)

Substituting Equation (6) into Equation (5), we obtained

$$f=q_2{e}^{k\left(-\sqrt{-p_0^2}\right)}+e^{k\sqrt{-p_0^2}}+q_{1}cos\left(\mathrm{k}1p_0\right).$$

(7)

Substituting Equation (7) into Equation (3), we obtained the exact breather solution to Equation (2) as follows

$$u_1(x,y,t)={\displaystyle \frac{6\gamma \left(\frac{-p_0^2{q}_2{e}^{k\left(-\sqrt{-p_0^2}\right)}+p_0^2\left(-e^{k\sqrt{-p_0^2}}\right)-p_0^2{q}_{1}cos\left(\mathrm{k}1p_0\right)}{q_2{e}^{k\left(-\sqrt{-p_0^2}\right)}+e^{k\sqrt{-p_0^2}}+q_{1}cos\left(\mathrm{k}1p_0\right)}-\frac{{\left(-\sqrt{-p_0^2}{q}_2{e}^{k\left(-\sqrt{-p_0^2}\right)}+\sqrt{-p_0^2}{e}^{k\sqrt{-p_0^2}}-p_0{q}_{1}sin\left(\mathrm{k}1p_0\right)\right)}^2}{{\left(q_2{e}^{k\left(-\sqrt{-p_0^2}\right)}+e^{k\sqrt{-p_0^2}}+q_{1}cos\left(\mathrm{k}1p_0\right)\right)}^2}\right)}{\beta }},$$

(8)

where $$\begin{gathered} k={\displaystyle \frac{t\left(-{\alpha }^2-16\gamma p_0^2+\frac{2\alpha \sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}}{q_1^2-4q_2}+\frac{{\alpha }^2{q}_1^2}{q_1^2-4q_2}-\frac{4{\alpha }^2{q}_2}{q_1^2-4q_2}+4\right)}{2\alpha -\frac{4\sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}}{q_1^2-4q_2}-\frac{2\alpha q_1^2}{q_1^2-4q_2}+\frac{8\alpha q_2}{q_1^2-4q_2}}}+x+y, \\ \mathrm{k}1={\displaystyle \frac{t\left(-2\sqrt{-4\gamma p_0^2{q}_1^4+32\gamma p_0^2{q}_2{q}_1^2-64\gamma p_0^2{q}_2^2+q_1^4-8q_2{q}_1^2+16q_2^2}-\alpha q_1^2+4\alpha q_2\right)}{2\left(q_1^2-4q_2\right)}}+x+y. \end{gathered}$$

Result 2.

$$\begin{array}{cc}{p}_0={\displaystyle \frac{\sqrt{3}\sqrt{q_2}}{\sqrt{\gamma q_1^2+8\gamma q_2}}},p_1=-{\displaystyle \frac{\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}},{\rho }_1={\displaystyle \frac{-\alpha q_1^2-2\alpha q_2+\frac{\sqrt{2}\sqrt{q_1^6-6q_2{q}_1^4+32q_2^3}}{\sqrt{q_1^2+8q_2}}}{2\left(q_1^2+2q_2\right)}},\hfill & {\rho }_2=-{\displaystyle \frac{\alpha }{2}}.\hfill \end{array}$$

(9)

Substituting Equation (9), into Equation (5), we obtained

$$f=exp\left({\displaystyle \frac{k_1\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}}\right)+q_{2}exp\left(-{\displaystyle \frac{k_1\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}}\right)+q_{1}cos\left(k_2\right).$$

(10)

Substituting Equation (10), into Equation (3), we obtained the exact breather solution to Equation (2), as

$$u_2(x,y,t)={\displaystyle \frac{6\gamma \left(k_4-k_5\right)}{\beta }},$$

(11)

where $k_1={\displaystyle \frac{t\left(-\alpha q_1^2-2\alpha q_2+\frac{\sqrt{2}\sqrt{q_1^6-6q_2{q}_1^4+32q_2^3}}{\sqrt{q_1^2+8q_2}}\right)}{2\left(q_1^2+2q_2\right)}}+x+y,$ $k_4={\displaystyle \frac{\frac{e^{-k_3}\left(-q_1^2-2q_2\right)q_2}{2\left(\gamma q_1^2+8\gamma q_2\right)}+\frac{e^{k_3}\left(-q_1^2-2q_2\right)}{2\left(\gamma q_1^2+8\gamma q_2\right)}-\frac{3q_1{q}_{2}cos\left(k_2\right)}{\gamma q_1^2+8\gamma q_2}}{e^{-k_3}{q}_2+q_{1}cos\left(k_2\right)+e^{k_3}}},$ $k_2={\displaystyle \frac{\sqrt{3}\sqrt{q_2}\left(-\frac{\alpha t}{2}+x+y\right)}{\sqrt{\gamma q_1^2+8\gamma q_2}}},k_3={\displaystyle \frac{k_1\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}},$ $k_5={\displaystyle \frac{{\left(\frac{e^{k_3}\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}-\frac{e^{-k_3}{q}_2\sqrt{-q_1^2-2q_2}}{\sqrt{2}\sqrt{\gamma q_1^2+8\gamma q_2}}-\frac{\sqrt{3}{q}_1\sqrt{q_2}sin\left(k_2\right)}{\sqrt{\gamma q_1^2+8\gamma q_2}}\right)}^2}{{\left(e^{-k_3}{q}_2+q_{1}cos\left(k_2\right)+e^{k_3}\right)}^2}}.$

2.2. The Two-Waves Solution

For the two-waves solution, we used the test function [43,44,45]:

$$f={\delta }_{3}sin\left({\chi }_2\right)+{\delta }_{4}sinh\left({\chi }_3\right)+{\delta }_{1}exp\left({\chi }_1\right)+{\delta }_{2}exp\left(-{\chi }_1\right),$$

(12)

where ${\chi }_1={\rho }_{1}t+x+y,{\chi }_2={\rho }_{2}t+x+y,{\chi }_3={\rho }_{3}t+x+y.$

By substituting Equation (12) into Equation (4) and carrying out the required algebraic manipulations, we derive a polynomial involving exponential, trigonometric, and hyperbolic functions. Grouping like terms according to their functional powers and setting their coefficients to zero yields a system of algebraic equations. Solving this system leads to the following sets of solutions.

Result 1.

$${\delta }_3=0,{\rho }_1={\displaystyle \frac{1}{2}}\left(2\sqrt{4\gamma +1}-\alpha \right),{\rho }_3=\sqrt{4\gamma +1}-{\displaystyle \frac{\alpha }{2}}.$$

(13)

Substituting Equation (13) into Equation (12), we obtained

$$f={\delta }_1{e}^{{\displaystyle \frac{1}{2}}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y}+{\delta }_2{e}^{-{\displaystyle \frac{1}{2}}t\left(2\sqrt{4\gamma +1}-\alpha \right)-x-y}+{\delta }_{4}sinh\left(t\left(\sqrt{4\gamma +1}-{\displaystyle \frac{\alpha }{2}}\right)+x+y\right).$$

(14)

Substituting Equation (14) into Equation (3), we obtained the exact two-waves solution to Equation (2) as

$$u_3(x,y,t)={\displaystyle \frac{6\gamma \left(1-\frac{{\left({\delta }_1{e}^{\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y}-{\delta }_2{e}^{-\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)-x-y}+{\delta }_{4}cosh\left(t\left(\sqrt{4\gamma +1}-\frac{\alpha }{2}\right)+x+y\right)\right)}^2}{{\left({\delta }_1{e}^{\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y}+{\delta }_2{e}^{-\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)-x-y}+{\delta }_{4}sinh\left(t\left(\sqrt{4\gamma +1}-\frac{\alpha }{2}\right)+x+y\right)\right)}^2}\right)}{\beta }}.$$

(15)

Result 2.

$${\delta }_1={\displaystyle \frac{3\sqrt{4{\gamma }^2+1}{\delta }_3^2-10\gamma {\delta }_3^2}{(32\gamma +12){\delta }_2}},{\delta }_4=0,{\rho }_1={\displaystyle \frac{1}{2}}\left(2\sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }-\alpha \right),$$

$${\rho }_2={\displaystyle \frac{1}{2}}\left(-\alpha +2{\left(-\sqrt{4{\gamma }^2+1}-2\gamma \right)}^{3/2}+8\gamma \sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }\right).$$

(16)

Substituting Equation (16) into Equation (12), we obtained

$$\begin{array}{cc}\hfill & f={\delta }_{2}exp\left(-{\displaystyle \frac{1}{2}}t\left(2\sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }-\alpha \right)-x-y\right)\hfill \\ \hfill & +{\displaystyle \frac{(3\sqrt{4{\gamma }^2+1}{\delta }_3^2-10\gamma {\delta }_3^2)exp\left(\frac{1}{2}t\left(2\sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }-\alpha \right)+x+y\right)}{(32\gamma +12){\delta }_2}}\hfill \\ \hfill & +{\delta }_{3}sin\left({\displaystyle \frac{1}{2}}t\left(-\alpha +2{\left(-\sqrt{4{\gamma }^2+1}-2\gamma \right)}^{3/2}+8\gamma \sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }\right)+x+y\right)\hfill \end{array}$$

(17)

Substituting Equation (17) into Equation (3), we obtained the exact two-waves solution to Equation (2) as

$$u_4(x,y,t)={\displaystyle \frac{6\gamma \left(\frac{{\delta }_2{k}_1-{\delta }_3{k}_3+k_5}{{\delta }_2{k}_1+{\delta }_3{k}_3+k_5}-\frac{{\left(-{\delta }_2{k}_1+{\delta }_3{k}_4+k_5\right)}^2}{{\left({\delta }_2{k}_1+{\delta }_3{k}_3+k_5\right)}^2}\right)}{\beta }},$$

(18)

where $k_1=exp\left(-{\displaystyle \frac{1}{2}}t\left(2\sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }-\alpha \right)-x-y\right)$,

$k_2=exp\left({\displaystyle \frac{1}{2}}t\left(2\sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }-\alpha \right)+x+y\right)$,

$k_3=sin\left({\displaystyle \frac{1}{2}}t\left(-\alpha +2{\left(-\sqrt{4{\gamma }^2+1}-2\gamma \right)}^{3/2}+8\gamma \sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }\right)+x+y\right),$

$k_4=cos\left({\displaystyle \frac{1}{2}}t\left(-\alpha +2{\left(-\sqrt{4{\gamma }^2+1}-2\gamma \right)}^{3/2}+8\gamma \sqrt{-\sqrt{4{\gamma }^2+1}-2\gamma }\right)+x+y\right),$

$k_5={\displaystyle \frac{k_2\left(3\sqrt{4{\gamma }^2+1}{\delta }_3^2-10\gamma {\delta }_3^2\right)}{(32\gamma +12){\delta }_2}}.$

2.3. The Lump-Periodic Solution

For the lump-periodic solution, we used the test function [46]:

$$f={\delta }_{2}cos\left({\chi }_2\right)+{\delta }_{1}cosh\left({\chi }_1\right)+{\delta }_{3}cosh\left({\chi }_3\right),$$

(19)

where ${\chi }_1={\rho }_{1}t+x+y,{\chi }_2={\rho }_{2}t+x+y,{\chi }_3={\rho }_{3}t+x+y.$

Substituting Equation (19) into Equation (4) and then performing the necessary manipulations, we obtained a polynomial in exponential, trigonometric, and hyperbolic functions. Collecting the same powers of these functions and equating them to zero, we obtained a system of algebraic equations. By solving this system, we obtained the following cases of solutions.

Result 1.

$${\delta }_2=0,{\rho }_1={\displaystyle \frac{1}{2}}\left(2\sqrt{4\gamma +1}-\alpha \right),{\rho }_3=\sqrt{4\gamma +1}-{\displaystyle \frac{\alpha }{2}}.$$

(20)

Substituting Equation (20) into Equation (19), we obtained:

$$f={\delta }_{1}cosh\left({\displaystyle \frac{1}{2}}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y\right)+{\delta }_{3}cosh\left(t\left(\sqrt{4\gamma +1}-{\displaystyle \frac{\alpha }{2}}\right)+x+y\right).$$

(21)

Substituting Equation (21) into Equation (3), we obtained the exact lump-periodic solution to Equation (2) as follows:

$$u_5(x,y,t)={\displaystyle \frac{6\gamma \left(1-\frac{{\left({\delta }_{1}sinh\left(\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y\right)+{\delta }_{3}sinh\left(t\left(\sqrt{4\gamma +1}-\frac{\alpha }{2}\right)+x+y\right)\right)}^2}{{\left({\delta }_{1}cosh\left(\frac{1}{2}t\left(2\sqrt{4\gamma +1}-\alpha \right)+x+y\right)+{\delta }_{3}cosh\left(t\left(\sqrt{4\gamma +1}-\frac{\alpha }{2}\right)+x+y\right)\right)}^2}\right)}{\beta }}.$$

(22)

2.4. The New Interaction Solution

For the new interaction solution, we used the test function [43,46]:

$$f=q_{2}exp\left(p_1{\chi }_1\right)+q_{3}exp\left(-p_1{\chi }_1\right)+q_{1}sin\left(p_0{\chi }_2\right)+q_{4}sinh\left(p_2{\chi }_3\right),$$

(23)

where ${\chi }_1={\rho }_{1}t+x+y,{\chi }_2={\rho }_{2}t+x+y,{\chi }_3={\rho }_{3}t+x+y.$

Substituting Equation (23) into Equation (4) and then performing the necessary manipulations, we obtained a polynomial in exponential, trigonometric, and hyperbolic functions. Collecting the same powers of these functions and equating them to zero, we obtained a system of algebraic equations. By solving this system, we obtained the following cases of solutions.

Result 1.

$$p_0=-\sqrt{-p_2^2},p_1=-p_2,{\rho }_1={\displaystyle \frac{1}{2}}\left(-\alpha -2\sqrt{4\gamma p_2^2+1}\right),$$

$$\begin{array}{c}\hfill {\rho }_2={\displaystyle \frac{-\alpha -4\alpha \gamma p_2^2-8\gamma p_2^2\sqrt{4\gamma p_2^2+1}-2\sqrt{4\gamma p_2^2+1}}{2\left(4\gamma p_2^2+1\right)}},\\ \hfill {\rho }_3={\displaystyle \frac{-\alpha -4\alpha \gamma p_2^2-8\gamma p_2^2\sqrt{4\gamma p_2^2+1}-2\sqrt{4\gamma p_2^2+1}}{2\left(4\gamma p_2^2+1\right)}}.\end{array}$$

(24)

Substituting Equation (24) into Equation (23), we obtained

$$f=q_1\left(-sin\left(k\sqrt{-p_2^2}\right)\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1\left(-p_2\right)}+q_3{e}^{\mathrm{k}1p_2}.$$

(25)

Substituting Equation (25) into Equation (3), we obtained the exact new interaction solution to Equation (2) as follows:

$u_6(x,y,t)=-{\displaystyle \frac{6\gamma {\left(\sqrt{-p_2^2}{q}_1\left(-cos\left(k\sqrt{-p_2^2}\right)\right)+p_2{q}_{4}cosh\left(k{p}_2\right)-p_2{q}_2{e}^{\mathrm{k}1\left(-p_2\right)}+\mathrm{k}1p_2{q}_3\right)}^2}{\beta {\left(q_1\left(-sin\left(k\sqrt{-p_2^2}\right)\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1\left(-p_2\right)}+q_3{e}^{\mathrm{k}1p_2}\right)}^2}}$

$${\displaystyle \frac{6+\frac{1}{\beta }\gamma \left(p_2^2{q}_1\left(-sin\left(k\sqrt{-p_2^2}\right)\right)+p_2^2{q}_{4}sinh\left(k{p}_2\right)+p_2^2{q}_2{e}^{\mathrm{k}1\left(-p_2\right)}+p_2^2{q}_3{e}^{\mathrm{k}1p_2}\right)}{q_1\left(-sin\left(k\sqrt{-p_2^2}\right)\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1\left(-p_2\right)}+q_3{e}^{\mathrm{k}1p_2}}},$$

(26)

where

$k={\displaystyle \frac{t\left(-\alpha -4\alpha \gamma p_2^2-8\gamma p_2^2\sqrt{4\gamma p_2^2+1}-2\sqrt{4\gamma p_2^2+1}\right)}{2\left(4\gamma p_2^2+1\right)}}+x+y,\mathrm{k}1={\displaystyle \frac{1}{2}}t\left(-\alpha -2\sqrt{4\gamma p_2^2+1}\right)+x+y.$

Result 2.

$$p_0=i{p}_2,p_1=p_2,{\rho }_1={\displaystyle \frac{1}{2}}\left(2\sqrt{4\gamma p_2^2+1}-\alpha \right),{\rho }_2={\displaystyle \frac{1}{2}}\left(-\alpha +{\displaystyle \frac{8\gamma p_2^2}{\sqrt{4\gamma p_2^2+1}}}+{\displaystyle \frac{2}{\sqrt{4\gamma p_2^2+1}}}\right),$$

$${\rho }_3=-{\displaystyle \frac{\alpha }{2}}+{\displaystyle \frac{4\gamma p_2^2}{\sqrt{4\gamma p_2^2+1}}}+{\displaystyle \frac{1}{\sqrt{4\gamma p_2^2+1}}}.$$

(27)

Substituting Equation (27) into Equation (23), we obtained

$$f=i{q}_{1}sinh\left(k{p}_2\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1p_2}+q_3{e}^{\mathrm{k}1\left(-p_2\right)}.$$

(28)

Substituting Equation (28) into Equation (3), we obtained the exact new interaction solution to Equation (2) as follows:

$$u_7(x,y,t)=-{\displaystyle \frac{6\gamma {\left(i{p}_2{q}_{1}cosh\left(k{p}_2\right)+p_2{q}_{4}cosh\left(k{p}_2\right)+p_2{q}_2{e}^{\mathrm{k}1p_2}-p_2{q}_3{e}^{\mathrm{k}1\left(-p_2\right)}\right)}^2}{\beta {\left(i{q}_{1}sinh\left(k{p}_2\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1p_2}+q_3{e}^{\mathrm{k}1\left(-p_2\right)}\right)}^2}}$$

$$+{\displaystyle \frac{6\gamma \left(i{p}_2^2{q}_{1}sinh\left(k{p}_2\right)+p_2^2{q}_{4}sinh\left(k{p}_2\right)+p_2^2{q}_2{e}^{\mathrm{k}1p_2}+p_2^2{q}_3{e}^{\mathrm{k}1\left(-p_2\right)}\right)}{\beta \left(i{q}_{1}sinh\left(k{p}_2\right)+q_{4}sinh\left(k{p}_2\right)+q_2{e}^{\mathrm{k}1p_2}+e^{-\mathrm{k}1}{q}_3\right)}},$$

(29)

where

$k={\displaystyle \frac{1}{2}}t\left(-\alpha +{\displaystyle \frac{8\gamma p_2^2}{\sqrt{4\gamma p_2^2+1}}}+{\displaystyle \frac{2}{\sqrt{4\gamma p_2^2+1}}}\right)+x+y,\mathrm{k}1={\displaystyle \frac{1}{2}}t\left(2\sqrt{4\gamma p_2^2+1}-\alpha \right)+x+y.$

3. Solitary, Kink/Antikink and Periodic Wave Solution

In this section, we will explore the solitary, Kink/anti-kink, and periodic wave solution Equation (2) via the approach of EMETFM with the help of Python 3.10 software in Visual Studio Code version 1.105.

Consider the traveling wave transformation

$$U(x,y,t)=u\left(\xi \right);\xi =-tv+x\omega +\theta y.$$

(30)

Substituting Equation (30) into Equation (2), we obtained the following ODE:

$$-\gamma {\omega }^4{u}^{\left(4\right)}\left(\xi \right)+{\displaystyle \frac{1}{4}}{\alpha }^2{\theta }^2{u}^{″}\left(\xi \right)-{\omega }^2{u}^{″}\left(\xi \right)+v^2{u}^{″}\left(\xi \right)-\alpha \theta v{u}^{″}\left(\xi \right)-\beta \left(2{\omega }^{2}u\left(\xi \right)u^{″}\left(\xi \right)+2{\omega }^2{u}^{\prime }{\left(\xi \right)}^2\right).$$

(31)

Integrating Equation (31) w.r.t $\xi$ twice and considering the constants of integration equal to zero, we have

$$-\gamma {\omega }^4{u}^{″}\left(\xi \right)-\beta {\omega }^{2}u{\left(\xi \right)}^2+u\left(\xi \right)\left({\displaystyle \frac{{\alpha }^2{\theta }^2}{4}}+v^2-\alpha \theta v-{\omega }^2\right).$$

(32)

Balancing between $u^{″}\left(\xi \right)$ and $u{\left(\xi \right)}^2$, that is $2n=n+2,⟹n=2.$

The general solution of Equation (32) is given as follows [47]:

$$u\left(\xi \right)=a_0+\sum _{i=1}^n{a}_i{\left(\psi \left(\xi \right)\right)}^i+\sum _{i=1}^n{b}_i{\left(\psi \left(\xi \right)\right)}^{-i}$$

(33)

By substituting the value of $n=2$ into Equation (33), we obtained the following results:

$$u\left(\xi \right)=a_2\psi {\left(\xi \right)}^2+a_1\psi \left(\xi \right)+a_0+{\displaystyle \frac{b_2}{\psi {\left(\xi \right)}^2}}+{\displaystyle \frac{b_1}{\psi \left(\xi \right)}}.$$

(34)

Substituting Equation (34) together with the differential equation,

$${\psi }^{\prime }\left(\xi \right)={\delta }_0+\psi {\left(\xi \right)}^2.$$

(35)

Equation (35) into Equation (32), we obtained a polynomial in $\psi \left(\xi \right)$. Collecting the same powers of $\psi \left(\xi \right)$ and equating them to zero, we have a system of over-determined equations.

$-\beta b_2^2{\omega }^2-6b_2\gamma {\delta }_0^2{\omega }^4=0,$

$-2\beta b_1{b}_2{\omega }^2-2b_1\gamma {\delta }_0^2{\omega }^4=0,$

$-2a_0\beta b_2{\omega }^2+{\displaystyle \frac{1}{4}}{\alpha }^2{b}_2{\theta }^2-\beta b_1^2{\omega }^2-8b_2\gamma {\delta }_0{\omega }^4+b_2{v}^2-\alpha b_2\theta v-b_2{\omega }^2=0,$

$-2a_0\beta b_1{\omega }^2-2a_1\beta b_2{\omega }^2+{\displaystyle \frac{1}{4}}{\alpha }^2{b}_1{\theta }^2-2b_1\gamma {\delta }_0{\omega }^4+b_1{v}^2-\alpha b_1\theta v-b_1{\omega }^2=0,$

${\displaystyle \frac{1}{4}}{\alpha }^2{a}_0{\theta }^2-2a_1\beta b_1{\omega }^2-2a_2\beta b_2{\omega }^2-a_0^2\beta {\omega }^2-2a_2\gamma {\delta }_0^2{\omega }^4+a_0{v}^2-\alpha a_0\theta v-a_0{\omega }^2-2b_2\gamma {\omega }^4=0,$

${\displaystyle \frac{1}{4}}{\alpha }^2{a}_1{\theta }^2-2a_2\beta b_1{\omega }^2-2a_0{a}_1\beta {\omega }^2-2a_1\gamma {\delta }_0{\omega }^4+a_1{v}^2-\alpha a_1\theta v-a_1{\omega }^2=0,$

${\displaystyle \frac{1}{4}}{\alpha }^2{a}_2{\theta }^2-a_1^2\beta {\omega }^2-2a_0{a}_2\beta {\omega }^2-8a_2\gamma {\delta }_0{\omega }^4+a_2{v}^2-\alpha a_2\theta v-a_2{\omega }^2=0,$

$-2a_1{a}_2\beta {\omega }^2-2a_1\gamma {\omega }^4=0,$

$-a_2^2\beta {\omega }^2-6a_2\gamma {\omega }^4=0.$

By solving this system, we obtained the following cases of solutions.

Case 1.

When $a_0=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }},a_1=0,a_2=-{\displaystyle \frac{6\gamma {\omega }^2}{\beta }};b_1=0,b_2=-{\displaystyle \frac{6\gamma {\delta }_0^2{\omega }^2}{\beta }},v={\displaystyle \frac{1}{2}}\left(\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4}\right).$

For ${\delta }_0<0$, the following solution cases were obtained:

$$\begin{array}{c}\hfill u_8(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{tanh}^2\left(\sqrt{-{\delta }_0}\left(-\frac{1}{2}t\left(\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4}\right)+x\omega +\theta y\right)\right)}{\beta }}\hfill \\ \hfill +{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{coth}^2\left(\sqrt{-{\delta }_0}\left(-\frac{1}{2}t\left(\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4}\right)+x\omega +\theta y\right)\right)}{\beta }}.\hfill \end{array}$$

(36)

$$u_9(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(tanh\left(k\right)+i\sec \mathrm{h}\left(k\right))}^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2}{\beta {(tanh\left(k\right)+i\sec \mathrm{h}\left(k\right))}^2}}.$$

(37)

$$u_{10}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\omega }^2{\left({\delta }_0-\sqrt{-{\delta }_0}tanh\left(k\right)\right)}^2}{\beta {\left(\sqrt{-{\delta }_0}tanh\left(k\right)+1\right)}^2}}-{\displaystyle \frac{6\gamma {\delta }_0^2{\omega }^2{\left(\sqrt{-{\delta }_0}tanh\left(k\right)+1\right)}^2}{\beta {\left({\delta }_0-\sqrt{-{\delta }_0}tanh\left(k\right)\right)}^2}}.$$

(38)

$$u_{11}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(4sinh\left(k\right)+3)}^2}{\beta {(5-4cosh\left(k\right))}^2}}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(5-4cosh\left(k\right))}^2}{\beta {(4sinh\left(k\right)+3)}^2}}.$$

(39)

$$u_{12}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\omega }^2{\left(\sqrt{13}\sqrt{-{\delta }_0}-2\sqrt{-{\delta }_0}cosh\left(k\right)\right)}^2}{\beta {(2sinh\left(k\right)+3)}^2}}$$

$$-{\displaystyle \frac{6\gamma {\delta }_0^2{\omega }^2{(2sinh\left(k\right)+3)}^2}{\beta {\left(\sqrt{13}\sqrt{-{\delta }_0}-2\sqrt{-{\delta }_0}cosh\left(k\right)\right)}^2}}.$$

(40)

$$u_{13}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{\left(1-\frac{4}{-sinh\left(k\right)+cosh\left(k\right)+2}\right)}^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2}{\beta {\left(1-\frac{4}{-sinh\left(k\right)+cosh\left(k\right)+2}\right)}^2}}.$$

(41)

where $k=2\sqrt{-{\delta }_0}\left(-{\displaystyle \frac{1}{2}}t\left(\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4}\right)+x\omega +\theta y\right)$ in $u_8(x,y,t),u_9(x,y,t),u_{10}(x,y,t),$$u_{11}(x,y,t),u_{12}(x,y,t),u_{13}(x,y,t)$.

For ${\delta }_0>0$, we have the following exact periodic solutions

$$u_{14}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{tan}^2\left(k\right)}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{cot}^2\left(k\right)}{\beta }}.$$

(42)

$$u_{15}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(tan\left(k\right)+sec\left(k\right))}^2}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2}{\beta {(tan\left(k\right)+sec\left(k\right))}^2}}.$$

(43)

$$u_{16}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(tan\left(k\right)+1)}^2}{\beta {(1-tan\left(k\right))}^2}}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(1-tan\left(k\right))}^2}{\beta {(tan\left(k\right)+1)}^2}}.$$

(44)

$$u_{17}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(5sin\left(k\right)+3)}^2}{\beta {(4-5cos\left(k\right))}^2}}-{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{(4-5cos\left(k\right))}^2}{\beta {(5sin\left(k\right)+3)}^2}},$$

(45)

$$u_{18}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}-{\displaystyle \frac{6\gamma {\omega }^2{\left(\sqrt{5}\sqrt{-{\delta }_0}-2\sqrt{{\delta }_0}cos\left(k\right)\right)}^2}{\beta {(2sin\left(k\right)+3)}^2}}-{\displaystyle \frac{6\gamma {\delta }_0^2{\omega }^2{(2sin\left(k\right)+3)}^2}{\beta {\left(\sqrt{5}\sqrt{-{\delta }_0}-2\sqrt{{\delta }_0}cos\left(k\right)\right)}^2}}.$$

(46)

$$u_{19}(x,y,t)=-{\displaystyle \frac{12\gamma {\delta }_0{\omega }^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2{\left(1-\frac{4}{-isin\left(k\right)+cos\left(k\right)+2}\right)}^2}{\beta }}+{\displaystyle \frac{6\gamma {\delta }_0{\omega }^2}{\beta {\left(1-\frac{4}{-isin\left(k\right)+cos\left(k\right)+2}\right)}^2}}.$$

(47)

where $k=2\sqrt{{\delta }_0}\left(-{\displaystyle \frac{1}{2}}t\left(\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4}\right)+x\omega +\theta y\right)$ in $u_{14}(x,y,t)u_{15}(x,y,t),u_{16}(x,y,t),$

$u_{17}(x,y,t),u_{18}(x,y,t),u_{19}(x,y,t)$.

Case 2.

When $a_0={\displaystyle \frac{-{\alpha }^2{\theta }^2-4v^2+4\alpha \theta v+4{\omega }^2}{8\beta {\omega }^2}},a_1=0,a_2=-{\displaystyle \frac{6\gamma {\omega }^2}{\beta }},b_1=0,b_2=0,{\delta }_0={\displaystyle \frac{{\alpha }^2{\theta }^2+4v^2-4\alpha \theta v-4{\omega }^2}{16\gamma {\omega }^4}}.$

For ${\delta }_0<0$, we have the following exact solutions

$$u_{20}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right){tanh}^2\left(\frac{1}{4}\sqrt{-\frac{{\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2}{\gamma {\omega }^4}}\left(-\frac{kt}{2}+x\omega +\theta y\right)\right)}{8\beta {\omega }^2}}.$$

(48)

$$u_{21}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right){coth}^2\left(\frac{1}{4}\sqrt{-\frac{{\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2}{\gamma {\omega }^4}}\left(-\frac{kt}{2}+x\omega +\theta y\right)\right)}{8\beta {\omega }^2}}.$$

(49)

$$u_{22}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3{(tanh\left(\mathrm{k}1\right)+i\sec \mathrm{h}\left(\mathrm{k}1\right))}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2}}.$$

(50)

$$u_{23}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$-{\displaystyle \frac{6\gamma {\omega }^2{\left(\frac{{\alpha }^2{\theta }^2+\alpha {\mathrm{k}}^2-2\alpha \theta k-4{\omega }^2}{16\gamma {\omega }^4}-\mathrm{k}2tanh\left(\mathrm{k}1\right)\right)}^2}{\beta {(\mathrm{k}2tanh\left(\mathrm{k}1\right)+1)}^2}}.$$

(51)

$$u_{24}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3{(5-4cosh\left(\mathrm{k}1\right))}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2{(4sinh\left(\mathrm{k}1\right)+3)}^2}}.$$

(52)

$$u_{25}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$-{\displaystyle \frac{6\gamma {\omega }^2{\left(\mathrm{k}2-\frac{1}{2}cosh\left(\mathrm{k}1\right)\sqrt{-\frac{{\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2}{\gamma {\omega }^4}}\right)}^2}{\beta {(2sinh\left(\mathrm{k}1\right)+3)}^2}}.$$

(53)

$$u_{26}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3{\left(1-\frac{4}{-sinh\left(\mathrm{k}1\right)+cosh\left(\mathrm{k}1\right)+2}\right)}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2}}.$$

(54)

For ${\delta }_0>0$, we have the following exact solutions.

$$u_{27}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2+2\alpha \alpha \mathrm{k}\theta -k^2+4{\omega }^2}{8\beta {\omega }^2}}$$

$$-{\displaystyle \frac{3{tan}^2\left(\mathrm{k}1\right)\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2}}.$$

(55)

$$u_{27}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$-{\displaystyle \frac{3{cot}^2\left(\mathrm{k}1\right)\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2}}$$

(56)

$$-{\displaystyle \frac{3{(tan\left(\mathrm{k}1\right)+sec\left(\mathrm{k}1\right))}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2}}.$$

(57)

$$u_{28}(x,y,t)={\displaystyle \frac{3{(1-tan\left(\mathrm{k}1\right))}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2{(tan\left(\mathrm{k}1\right)+1)}^2}}.$$

(58)

$$-{\displaystyle \frac{3{(4-5cos\left(\mathrm{k}1\right))}^2\left({\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2\right)}{8\beta {\omega }^2{(5sin\left(\mathrm{k}1\right)+3)}^2}}.$$

(59)

$$u_{30}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}-{\displaystyle \frac{6\gamma {\omega }^2{\left(\frac{\sqrt{5}\mathrm{k}2}{4}-\frac{1}{2}\mathrm{k}2cos\left(\mathrm{k}1\right)\right)}^2}{\beta {(2sin\left(\mathrm{k}1\right)+3)}^2}}.$$

(60)

$$u_{31}(x,y,t)={\displaystyle \frac{-{\alpha }^2{\theta }^2-k^2+2\alpha \theta k+4{\omega }^2}{8\beta {\omega }^2}}$$

$$+{\displaystyle \frac{3{\left(1-\frac{4}{-isin\left(\mathrm{k}1\right)+cos\left(\mathrm{k}1\right)+2}\right)}^2\left({\alpha }^2{\theta }^2-2\alpha \alpha \mathrm{k}\theta +k^2-4{\omega }^2\right)}{8\beta {\omega }^2}}.$$

(61)

where $k=\alpha \theta -2\sqrt{{\omega }^2-16\gamma {\delta }_0{\omega }^4},$ $\mathrm{k}1={\displaystyle \frac{1}{4}}\sqrt{-{\displaystyle \frac{{\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2}{\gamma {\omega }^4}}}\left(-{\displaystyle \frac{kt}{2}}+x\omega +\theta y\right),$ $\mathrm{k}2={\displaystyle \frac{1}{4}}\sqrt{-{\displaystyle \frac{{\alpha }^2{\theta }^2+k^2-2\alpha \theta k-4{\omega }^2}{\gamma {\omega }^4}}}$ for all solutions under case two.

4. Graphical Representation

In this section, we present the graphs for some of the obtained solutions by choosing appropriate values for the parameters that are involved, which serves as a bridge between the theoretical model and real shallow-water phenomena, offering visual and quantitative insight into how nonlinear, dispersive, and stability features manifest physically.

In Figure 1, the surface profile of Equation (11) is shown, in 3D, contour and 2D plot, that is $u_2(x,y,t)$ by taking the following parameter values: $\alpha =1.1,\beta =3.26,\gamma =1,q_1=2,q_2=0.5,y=1.$ Here is a wave packet obtained for the chosen parameter values.

In Figure 2, we show the surface profile of Equation (15), in 3D, contour and 2D plot, that is $u_3(x,y,t)$ by taking the following parameter values: $\alpha =0.03,\beta =0.34,\gamma =0.05,{\delta }_4=0.05,{\delta }_1=0.5,{\delta }_2=2.12,y=1.$ Here, we have a perfect wave packet that represents this particular solution under the chosen values of the parameters.

In Figure 3, we illustrate the surface profile of Equation (18), in 3D, contour and 2D plot, that is $u_4(x,y,t)$ by taking the following parameter values: $\alpha =1.12,\beta =3.3,\gamma =0.09,{\delta }_3=0.74,{\delta }_2=3.1,y=1.$ Here, we have a perfect wave packet that represents this particular solution under the chosen values of the parameters.

In Figure 4, we show the surface profile of Equation (22), in 3D, contour and 2D plot, that is $u_5(x,y,t)$ by taking the following parameter values: $\alpha =2.3,\beta =2.74,\gamma =1.02,{\delta }_3=3.3,{\delta }_1=0.4,y=1.$ This represents a bright soliton wave obtained for the selected parameters.

In Figure 5, we show the surface profile of Equation (36), in 3D, contour and 2D plot, that is $u_8(x,y,t)$ by taking the following parameter values: $\alpha =1.3,\beta =-0.74,\gamma =2,{\delta }_0=0.01,\theta =0.3,y=1,\omega =1.0.$ Here, we have a perfect wave packet that represents this particular solution under the chosen values of the parameters.

In Figure 6, we illustrate the surface profile of Equation (43), in 3D, contour and 2D plot, that is $u_{14}(x,y,t)$ by taking the following parameter values: $\alpha =1.3,\beta =0.74,\gamma =2,{\delta }_0=0.05,\theta =0.3,y=1,\omega =0.34.$ Here, we have a perfect wave packet that represents this particular solution under the chosen values of the parameters.

5. Modulation Instability Analysis

Modulation Instability (MI) is a fundamental nonlinear phenomenon observed in various physical systems, most commonly in optics and fluid dynamics. It occurs when a continuous wave (CW) or plane wave becomes unstable due to small perturbations, leading to the exponential growth of these perturbations and the formation of localized structures, such as solitons or pulse trains.

Assume the steady state solution of Equation (2) as

$$u(x,y,t)=e^{itm}\left(U(x,y,t)+\sqrt{m}\right),$$

(62)

where m is the normalized optical power. Upon putting Equation (62) into Equation (2) and linearizing, we obtained

$$-m^{2}U+2im{U}_t+i\alpha m{U}_y+U_{\mathrm{tt}}-U_{\mathrm{xx}}-\gamma U_{\mathrm{xxxx}}+\alpha U_{\mathrm{yt}}+{\displaystyle \frac{{\alpha }^2{U}_{\mathrm{yy}}}{4}}.$$

(63)

Assume the solution of Equation (63) to be of the form

$$U(x,y,t)=c_{1}exp\left(i(ky-t\omega +\theta x)\right)+c_{2}exp(-i(ky-t\omega +\theta x)).$$

(64)

where $\theta$ and k are the normalized wave numbers, while $\omega$ is the frequency of perturbation.

Substituting Equation (64) into Equation (63), splitting the coefficients of $c_1{e}^{i(ky-t\omega +\theta x)}$ and $e^{-i(ky-t\omega +\theta x)}$ and then solving the coefficients of the resulting determinant matrix, we obtained the following dispersion relation.

$$-{\gamma }^2{\theta }^8+2\gamma {\theta }^6-2\gamma {\theta }^4{\omega }^2-{\theta }^4+2{\theta }^2{\omega }^2-{\displaystyle \frac{{\alpha }^4{k}^4}{16}}+{\displaystyle \frac{1}{2}}{\alpha }^3{k}^3\omega -{\displaystyle \frac{1}{2}}{\alpha }^2\gamma {\theta }^4{k}^2+{\displaystyle \frac{1}{2}}{\alpha }^2{\theta }^2{k}^2-{\displaystyle \frac{3}{2}}{\alpha }^2{k}^2{\omega }^2$$

$$+{\displaystyle \frac{1}{2}}{\alpha }^2{k}^2{m}^2+2\alpha \gamma {\theta }^{4}k\omega -2\alpha {\theta }^{2}k\omega +2\alpha k{\omega }^3-2\alpha k{m}^2\omega -m^4-2\gamma {\theta }^4{m}^2+2{\theta }^2{m}^2+2m^2{\omega }^2-{\omega }^4.$$

(65)

Solving Equation (65), which is the dispersion relation for $\omega$, we obtained

$$\omega ={\displaystyle \frac{1}{2}}\left(\alpha k\pm 2\sqrt{-\gamma {\theta }^4+{\theta }^2-m}-2m\right).$$

(66)

In a scenario where $-\gamma {\theta }^4+{\theta }^2-m\ge 0$, $\omega$ is real for any real value of m, and the steady state is stable against small perturbations. In contrast, the steady state is unstable, when $-\gamma {\theta }^4+{\theta }^2-m<0$, that is, $\omega$ is imaginary and hence the perturbation grows exponentially leading to a MI.

Under this condition, the gain spectrum is expressed as

$$G\left(m\right)=2Im\left(\omega \right)=2Im\left({\displaystyle \frac{1}{2}}\left(\alpha k\pm 2\sqrt{-\gamma {\theta }^4+{\theta }^2-m}-2m\right)\right).$$

(67)

In Figure 7, we present the gain spectrum of modulation instability for the (2+1)-dimensional Boussinesq equation, plotted as a function of the normalized modulation parameter m. The surface and contour plots display the variation of the gain magnitude $G\left(m\right)$ with respect to m. Warm color regions correspond to high gain, indicating strong modulation instability, while cool color regions denote low gain, representing stable zones. The pattern clearly demonstrates the presence of modulation instability under the selected parameter values.

6. Conclusions

In this study, a comprehensive analysis of a (2+1)-dimensional integrable equation was conducted by employing the Hirota bilinear method and EMETFM. The interaction solutions, which include breather, two-wave, lump-periodic, and new interaction solutions, were obtained by the Hirota method under different tests functions, while soliton solutions were obtained by applying the enhanced modified extended tanh function method. These exact solutions enrich our understanding of the nonlinear dynamics governed by the equation and reveal intricate phenomena such as localized wave interactions, periodic wave structures, and stable traveling wave packets.

The novelty of this study lies in the comprehensive analysis of multidimensional soliton dynamics and interaction patterns within a unified bilinear framework. The findings reveal new mechanisms of multidirectional energy transport and nonlinear pattern formation, offering valuable theoretical insights applicable to shallow water dynamics, optical systems, plasma waves, and other nonlinear dispersive media.

Furthermore, the obtained solutions are of significant importance, as they not only offer analytical insight into the underlying physical processes but also provide benchmark results for numerical simulations. Such solutions find applications in various branches of physics and engineering, including fluid dynamics, plasma physics, optical fiber communications, and nonlinear lattice models, where understanding wave propagation, interaction, and stability is crucial. The methods and solutions presented herein contribute valuable tools and references for further theoretical development and practical modeling of complex nonlinear wave phenomena in higher-dimensional settings.