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Dynamics of new truncated M-fractional derivative wave structures to the nonlinear Zhiber-Shabat equation arising in variety of fields Cover

Dynamics of new truncated M-fractional derivative wave structures to the nonlinear Zhiber-Shabat equation arising in variety of fields

By: Jan Muhammad and  Usman Younas  
Open Access
|Feb 2026

Full Article

1
Introduction

Nonlinearity in nature is fascinating, and many researchers see nonlinear science as the most promising field for enhancing our understanding of the universe [1, 2]. The mathematical characterization of complex systems with time-varying parameters necessitates the examination of a variety of nonlinear partial differential equations (NLPDEs). Since the mid-18th century, scholars have endeavored to simplify intricate physical phenomena by employing NLPDEs. Researchers frequently use NLPDEs to reconstruct various dynamic phenomena [3]. As a result, the study of NLPDEs has continued to attract a substantial quantity of attention in recent years. To explore the exact solutions to nonlinear equations is a critical component of the investigation of nonlinear physical phenomena [4]. The significance of obtaining exact solutions for nonlinear equations is paramount, as these solutions are essential for numerical solution verification and stability analysis. A mathematical structure must be used to express the physical characteristics of any nonlinear physical system in order to obtain an exact solution. This mathematical structure is frequently represented by NLPDEs. Therefore, the investigation of NLPDEs is essential; these equations provide a fundamental framework that facilitates the highly precise understanding of a diverse array of nonlinear physical phenomena. NLPDEs are basic mathematical models that help us understand many different types of phenomena in physics, photonics, optics, fluid dynamics and nonlinear fiber optics. They are used in fields like plasma physics and nonlinear optics [5].

The soliton theory has garnered significant attention in trial designs due to its function as a research domain in the disciplines of media transmission, numerical physical science, designing, and various nonlinear sciences [6]. Specifically, the current period has seen a concentration of optical solitons. Optical solitons are waves that possess the ability to generate waves without dispersing over a significant distance, i.e., they maintain their shape over a significant distance [7, 8]. The basic role of solitons in the media communications society underscores their significance in nonlinear optics. In the system of single wave-based interchanges, optical pulse generators, fiber-optic amplifiers, and many others, solitons models are extensively useful. Examining the motion of solitons through nonlinear optical fibers, extreme laser radiation into plasmas, and the hypothesis of optical solitons are among the most interesting topics [9]. The scholars have adopted a variety of nonlinear models to predict and comprehend the nature of soliton waves.

Researchers have developed a number of approaches and methods to deal with these problems and get solutions in a wide range of systems. Among the most well-known and widely used techniques developed recently are: Darboux transformation [10], the enhanced modified extended tanh expansion method [11], the advanced F-expansion function approach [12], the truncated Painlevé approach [13], the modified F-expansion method [14], the modified simple equation method [15], the iterative transform method [16], Lie classical approach [17], the new sub equation approach [18], Adomian decomposition technique [19], the simplest equation technique [20], the tan(ϕ2)\tan ({\phi \over 2}) technique [21], the multiple exp-function approach [22], Bernoulli GG{{{G^\prime }} \over G}-expansion method [23], (GG)\left( {{{{G^\prime }} \over G}} \right)-expansion method [24], the tanh-coth method [25], the the inverse scattering approach [26], the Bäcklund transformation [27], Riccati equation mapping technique [28], bifurcation analysis [29, 30] and so on [3136].

Moreover, the model under consideration has not been investigated using these novel sophisticated methodologies in the existing literature. So, this paper has the aim to study various kinds of soliton solutions of the M-fractional nonlinear ZSE by applying the new integration methods known as modified generalized Riccati equation mapping method (MGREMM) [37], KMM [38] and multivariate generalized exponential rational integral function method (MGERIFM) [39]. These methods can be employed to obtain novel and exact solutions, including exponential, trigonometric, and rational functions. The proposed solutions will markedly improve the comprehension of diverse physical processes in associated domains.

The rest of the article is organized as follows: The definition about the new truncated M-fractional derivative of \hbar is presented in Section 2. In Section 3, the applications of methods together with graphs are presented in Section 3. The conclusion obtained in this paper is reported in Section 4.

2
Preliminaries

For explaining the complex nonlinear systems and providing a more accurate representation of system dynamics, fractional derivatives are a more sophisticated approach. Compared to their integer-order counterparts, the fractional models demonstrate superior precision and accuracy, closely aligning with the experimental data. Additionally, the system’s overall efficacy is improved by the implementation of sophisticated optimization and control strategies, which are made possible by the fractional form. The truncated M-fractional derivatives, a novel idea of fractional order derivatives, is presented contemporaneously with the development of this process. Scholars have given these fractional-order derivatives a lot of attention.

Definition 1.

Let :[0,)\hbar :[0,\infty ) \to , then, the new truncated M-fractional derivative of \hbar is defined [40] as: 1𝒟Mα,ς{()(t)}=limε0(tEς(εt1α))(t)ε,ς>0,α(0,1),t>0,{\cal D}_M^{\alpha,\;\varsigma }\{ (\hbar )(t)\} = \mathop {\lim }\limits_{\varepsilon \to 0} {{\hbar (t{_\varsigma }(\varepsilon {t^{1 - \alpha }})) - \hbar (t)} \over \varepsilon },\;\;\;\varsigma > 0,\;\alpha \in (0,1),\;\forall t > 0, with 𝔼ς(.) being the one parameter truncated Mittag-Leffler function [40].

Let ς < 0, 0 > α ≥ 1, p, s ∈ ℝ, and \hbar \;, β differentiable at a point t > 0. Then,

  • 𝒟Mα,ς{(pβ+s)(t)}=p𝒟Mα,ς{β(t)}+s𝒟Mα,ς{(t)}.{\cal D}_M^{\alpha,\;\varsigma }\{ (p\beta + s\hbar )(t)\} = p{\cal D}_M^{\alpha,\;\varsigma }\{ \beta (t)\} + s{\cal D}_M^{\alpha,\;\varsigma }\{ \hbar (t)\}.

  • 𝒟Mα,ς{(β.)(t)}=β(t)𝒟Mα,ς{(t)}+(t)𝒟Mα,ς{β(t)}.{\cal D}_M^{\alpha,\;\varsigma }\{ (\beta \;.\;\hbar )(t)\} = \beta (t){\cal D}_M^{\alpha,\;\varsigma }\{ \hbar (t)\} + \hbar (t){\cal D}_M^{\alpha,\;\varsigma }\{ \beta (t)\}.

  • 𝒟Mα,ς{β(t)}=(t)𝒟Mα,ς{β(t)}β(t)𝒟Mα,ς{(t)}[(t)]2.{\cal D}_M^{\alpha,\;\varsigma }\{ {\beta \over \hbar }(t)\} = {{\hbar (t){\cal D}_M^{\alpha,\;\varsigma }\{ \beta (t)\} - \beta (t){\cal D}_M^{\alpha,\;\varsigma }\{ \hbar (t)\} } \over {{{[\hbar (t)]}^2}}}.

  • 𝒟Mα,ς{c}=0{\cal D}_M^{\alpha,\;\varsigma }\{ c\} = 0, where β(t) = c is a constant.

  • If β is differentiable, then, 𝒟Mα,ς{β(t)}=t1αΓ(ς+1)dβ(t)dt.{\cal D}_M^{\alpha,\;\varsigma }\{ \beta (t)\} = {{{t^{1 - \alpha }}} \over {\Gamma (\varsigma + 1)}}{{d\beta (t)} \over {dt}}.

3
Applications

In this section, we apply the powerful methods such as MGREMM, KMM, and MGERIFM to obtain desired solutions for the nonlinear ZSE [4143] in truncated M-fractional derivative given as: 2𝒟M,xt2ε,βu+peu+qeu+re2u=0,D_{M,xt}^{2\varepsilon,\;\beta }u + p{e^u} + q{e^{ - u}} + r{e^{ - 2u}} = 0, where u = u(x,t) is an unknown function, while the parameters p, q, r are real constants. Eq.(2) transforms to sinh-Gordon equation by taking r = 0, while for q = r = 0 it transforms to the Liouville equation. Similarly, for q = 0, Eq.(2) transforms to the Dodd-Bullough-Mikhailov equation and for p = 0, q = –1, r = 1, it converts to the form of Tzitzeica-Dodd-Bullough equation. Moreover, the subscripts x and t are the partial derivatives and 𝒟Mε,ς{\cal D}_M^{,\;\varsigma } denote the M-fractional derivative with β ∈ (0,1) and ς > 0. Furthermore, the suggested model has been examined in the literature from different perspectives. In [41], various solitary wave solutions have been extracted by using tanh method, while a variety of new exact solutions have recovered in [42] by applying (1G)\left( {{1 \over {{G^\prime }}}} \right) expansion method, (GG)\left( {{{{G^\prime }} \over G}} \right) and (1G)\left( {{1 \over G}} \right) methods. In [43], the existence of soliton solutions are derived by employing ansatz approach.

In this work, we study the exact solutions of the proposed model using the suggested techniques with truncated M-fractional derivatives. First of all, let’s consider the following transformation as: 3u(x,t)=Φ(ξ),   ξ=Γ(β+1)ε(xεctε),u(x,t) = \Phi (\xi ),\xi = {{\Gamma (\beta + 1)} \over \varepsilon }\left( {{x^\varepsilon } - c{t^\varepsilon }} \right), where c is the wave speed. On manipulating the Eq.(3) into Eq.(2) provides 4peu+qeu+re2ucu=0.p{e^u} + q{e^{ - u}} + r{e^{ - 2u}} - c{u^{\prime \prime }} = 0.

Furthermore, for solving the Eq.(4), the following assumption is used as: 5v=eu   or u=lnv.v = {e^u}{\rm{or }}u = \ln v.

Manipulating Eq.(5) into Eq.(4), we get 6c((v)2vv)+pv3+qv+r=0.c\left( {{{\left( {{v^\prime }} \right)}^2} - v{v^{\prime \prime }}} \right) + p{v^3} + qv + r = 0.

Now, applying the homogeneous balance principle between the terms v3 and vv in Eq.(6) gives n = 2.

3.1
Application of the MGREMM

The solution for MGREMM [37] is expressed as: 7v(ξ)=ϕ0+r=1nϕrΩr(ξ)+r=1nψr(ξ)(Ω(ξ)Ω(ξ))r.v(\xi ) = {\phi _0} + \sum\limits_{r = 1}^n {{\phi _r}} {\Omega ^r}(\xi ) + \sum\limits_{r = 1}^n {{\psi _r}} (\xi ){\left( {{{{\Omega ^\prime }(\xi )} \over {\Omega (\xi )}}} \right)^r}.

For n = 2, we get 8v(ξ)=ϕ0+ϕ1Ω(ξ)+ϕ2(Ω(ξ))2+ψ1(ξ)Ω(ξ)Ω(ξ)+ψ2(ξ)(Ω(ξ)Ω(ξ))2,v(\xi ) = {\phi _0} + {\phi _1}\Omega (\xi ) + {\phi _2}{\left( {\Omega (\xi )} \right)^2} + {\psi _1}(\xi ){{\Omega '(\xi )} \over {\Omega (\xi )}} + {\psi _2}(\xi ){\left( {{{\Omega '(\xi )} \over {\Omega (\xi )}}} \right)^2}, and Ω(ξ)=δ0+δ1Ω(ξ)+δ2Ω(ξ)2v(\xi ) = {\phi _0} + {\phi _1}\Omega (\xi ) + {\phi _2}{\left( {\Omega (\xi )} \right)^2} + {\psi _1}(\xi ){{\Omega '(\xi )} \over {\Omega (\xi )}} + {\psi _2}(\xi ){\left( {{{\Omega '(\xi )} \over {\Omega (\xi )}}} \right)^2},. On putting Eq.(8) into Eq.(6), the general solutions are given as: (I): When Δ=δ124δ0δ2>0\Delta = \delta _1^2 - 4{\delta _0}{\delta _2} > 0, δ1 δ2 ≠ 0, δ0 δ2 ≠ 0 and ϕ1=δ1δ2ψ2, p=2cψ2{\phi _1} = - {\delta _1}{\delta _2}{\psi _2},p = {{2c} \over {{\psi _2}}},  q=c(δ0δ2ψ2+ϕ0)((δ1210δ0δ2)ψ26ϕ0)ψ2q = {{c\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\left( {\left( {\delta _1^2 - 10{\delta _0}{\delta _2}} \right){\psi _2} - 6{\phi _0}} \right)} \over {{\psi _2}}}, r=c(δ0δ2ψ2+ϕ0)2(4ϕ0(δ128δ0δ2)ψ2)ψ2r = {{c{{\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)}^2}\left( {4{\phi _0} - \left( {\delta _1^2 - 8{\delta _0}{\delta _2}} \right){\psi _2}} \right)} \over {{\psi _2}}},  ϕ2=δ22ψ2, ψ1=δ1ψ2{\phi _2} = - \delta _2^2{\psi _2},{\psi _1} = - {\delta _1}{\psi _2}, the soliton solutions are 9u1(x,t)=ln(δ0δ2ψ2sech2(Δξ2)(δ1Δsinh(Δξ)+(δ122δ0δ2)cosh(Δξ)δ12+6δ0δ2)(δ1+Δtanh(Δξ2))2+ϕ0),{u_1}(x,t) = \ln \left( {{{{\delta _0}{\delta _2}{\psi _2}{\rm{sec}}{{\rm{h}}^2}\left( {{{\sqrt \Delta \xi } \over 2}} \right)\left( {{\delta _1}\sqrt \Delta \sinh \left( {\sqrt \Delta \xi } \right) + \left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\cosh \left( {\sqrt \Delta \xi } \right) - \delta _1^2 + 6{\delta _0}{\delta _2}} \right)} \over {{{\left( {{\delta _1} + \sqrt \Delta \tanh \left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right), 10u2(x,t)=ln(δ0δ2ψ2csch2(Δξ2)(δ1Δsinh(Δξ)+(δ122δ0δ2)cosh(Δξ)+δ126δ0δ2)(δ1+Δcoth(Δξ2))2+ϕ0),{u_2}(x,t) = \ln \left( {{{{\delta _0}{\delta _2}{\psi _2}{\rm{csc}}{{\rm{h}}^2}\left( {{{\sqrt \Delta \xi } \over 2}} \right)\left( {{\delta _1}\sqrt \Delta \sinh \left( {\sqrt \Delta \xi } \right) + \left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\cosh \left( {\sqrt \Delta \xi } \right) + \delta _1^2 - 6{\delta _0}{\delta _2}} \right)} \over {{{\left( {{\delta _1} + \Delta \coth \left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right), 11u3(x,t)=ln(ϕ0+δ0δ2ψ2(1+i(δ124δ0δ2)δ12sinh(Δξ)2δ0δ2(sinh(Δξ)+i)+δ1Δcosh(Δξ))),{u_3}(x,t) = \ln \left( {{\phi _0} + {\delta _0}{\delta _2}{\psi _2}\left( {1 + {{i\left( {\delta _1^2 - 4{\delta _0}{\delta _2}} \right)} \over {\delta _1^2\sinh \left( {\sqrt \Delta \xi } \right) - 2{\delta _0}{\delta _2}\left( {\sinh \left( {\sqrt \Delta \xi } \right) + i} \right) + {\delta _1}\sqrt \Delta \cosh \left( {\sqrt \Delta \xi } \right)}}} \right)} \right), 12u4(x,t)=ln(δ1Δ(δ0δ2ψ2+ϕ0)sinh(Δξ)+(δ122δ0δ2)(δ0δ2ψ2+ϕ0)cosh(Δξ)+δ0δ2((δ126δ0δ2)ψ22ϕ0)(δ1sinh(Δξ2)+Δcosh(Δξ2))2).{u_4}(x,t) = \ln \left( {{{{\delta _1}\sqrt \Delta \left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\sinh \left( {\sqrt \Delta \xi } \right) + \left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\cosh \left( {\sqrt \Delta \xi } \right) + {\delta _0}{\delta _2}\left( {\left( {\delta _1^2 - 6{\delta _0}{\delta _2}} \right){\psi _2} - 2{\phi _0}} \right)} \over {{{\left( {{\delta _1}\sinh \left( {{{\sqrt \Delta \xi } \over 2}} \right) + \sqrt \Delta \cosh \left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right)}^2}}}} \right).

When ϕ1=2cδ1δ2p{\phi _1} = {{2c{\delta _1}{\delta _2}} \over p},  q=(cδ12+2cδ0δ23pϕ0)(2cδ0δ2pϕ0)pq = - {{\left( {c\delta _1^2 + 2c{\delta _0}{\delta _2} - 3p{\phi _0}} \right)\left( {2c{\delta _0}{\delta _2} - p{\phi _0}} \right)} \over p},  r=(pϕ02cδ0δ2)2(2pϕ0cδ12)p2r = {{{{\left( {p{\phi _0} - 2c{\delta _0}{\delta _2}} \right)}^2}\left( {2p{\phi _0} - c\delta _1^2} \right)} \over {{p^2}}}, ψ2 = 0,  ϕ2=2cδ22p{\phi _2} = {{2c\delta _2^2} \over p}, ψ1 = 0, we have the solitary wave solutions: 13u5(x,t)=ln(4cδ0δ2+cΔcsch2(Δξ2)+2pϕ02p),{u_5}(x,t) = \ln \left( {{{ - 4c{\delta _0}{\delta _2} + c\Delta {\rm{csc}}{{\rm{h}}^2}\left( {{{\sqrt \Delta \xi } \over 2}} \right) + 2p{\phi _0}} \over {2p}}} \right), 14u6(x,t)=ln( 1p(d+e s i n h (Δξ))2( cδ12e(edsinh(Δξ))cΔeΔ(d2+e2)cosh(Δξ) pϕ0(d+esinh(Δξ))2cδ0δ2(2d2+e2cosh(2Δξ)+3e2) ) ),{u_6}(x,t) = \ln \left( {{1 \over {p{{(d + e s i n h (\sqrt \Delta \xi ))}^2}}}\left( {c\delta _1^2e(e - d\sinh (\sqrt \Delta \xi )) - c\sqrt \Delta e\sqrt {\Delta \left( {{d^2} + {e^2}} \right)} \cosh (\sqrt \Delta \xi )} \right.} \right.\left. {\left. {p{\phi _0}{{(d + e\sinh (\sqrt \Delta \xi ))}^2} - c{\delta _0}{\delta _2}\left( {2{d^2} + {e^2}\cosh (2\sqrt \Delta \xi ) + 3{e^2}} \right)} \right)} \right), 15u7(x,t)=ln(4cδ0δ1δ2δ1pΔptanh(Δξ2)+4cδ02δ22(cosh(Δξ)+1)p(δ1cosh(Δξ2)Δsinh(Δξ2))2+ϕ0).{u_7}(x,t) = \ln \left( { - {{4c{\delta _0}{\delta _1}{\delta _2}} \over {{\delta _1}p - \sqrt \Delta p\tanh \left( {{{\sqrt \Delta \xi } \over 2}} \right)}} + {{4c\delta _0^2\delta _2^2\left( {\cosh \left( {\sqrt \Delta \xi } \right) + 1} \right)} \over {p{{\left( {{\delta _1}\cosh \left( {{{\sqrt \Delta \xi } \over 2}} \right) - \sqrt \Delta \sinh \left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right).

When ϕ1 = –δ1 δ2 ψ2,  q=12p(δ0δ2ψ2+ϕ0)(6ϕ0(δ1210δ0δ2)ψ2)q = - {1 \over 2}p\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\left( {6{\phi _0} - \left( {\delta _1^2 - 10{\delta _0}{\delta _2}} \right){\psi _2}} \right),  r=12p(δ0δ2ψ2+ϕ0)2(4ϕ0(δ128δ0δ2)ψ2)r = {1 \over 2}p{\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)^2}\left( {4{\phi _0} - \left( {\delta _1^2 - 8{\delta _0}{\delta _2}} \right){\psi _2}} \right), c=pψ22c = {{p{\psi _2}} \over 2},  ϕ2=δ22ψ2{\phi _2} = - \delta _2^2{\psi _2}, ψ1 = –δ1 ψ2, we have the following solutions: 16u8(x,t)=ln(14ψ2(4δ0δ2+Δcsch2(Δξ2))+ϕ0),{u_8}(x,t) = \ln \left( {{1 \over 4}{\psi _2}\left( {4{\delta _0}{\delta _2} + \Delta {\rm{csc}}{{\rm{h}}^2}\left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right) + {\phi _0}} \right), 17u9(x,t)=ln(δ0δ2ψ2Δψ22+2sinh(Δξ)+ϕ0),{u_9}(x,t) = \ln \left( {{\delta _0}{\delta _2}{\psi _2} - {{\Delta {\psi _2}} \over {2 + 2\sinh \left( {\sqrt \Delta \xi } \right)}} + {\phi _0}} \right), 18u10(x,t)=ln(14ψ2(4δ0δ2+Δcsch2(Δξ2))+ϕ0).{u_{10}}(x,t) = \ln \left( {{1 \over 4}{\psi _2}\left( {4{\delta _0}{\delta _2} + \Delta {\rm{csc}}{{\rm{h}}^2}\left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right) + {\phi _0}} \right). (II): When Δ=δ124δ0δ2<0,δ1δ20,δ0δ20\Delta = \delta _1^2 - 4{\delta _0}{\delta _2} < 0,\;{\delta _1}{\delta _2} \ne 0,\;{\delta _0}{\delta _2} \ne 0 and ϕ1=δ1δ2ψ2,q=12p(δ0δ2ψ2+ϕ0)(6ϕ0(δ1210δ0δ2)ψ2){\phi _1} = - {\delta _1}{\delta _2}{\psi _2},\;q = - {1 \over 2}p\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\left( {6{\phi _0} - \left( {\delta _1^2 - 10{\delta _0}{\delta _2}} \right){\psi _2}} \right), r=12p(δ0δ2ψ2+ϕ0)2(4ϕ0(δ128δ0δ2)ψ2)r = {1 \over 2}p{\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)^2}\left( {4{\phi _0} - \left( {\delta _1^2 - 8{\delta _0}{\delta _2}} \right){\psi _2}} \right), c=pψ22c = {{p{\psi _2}} \over 2},  ϕ2=δ22ψ2{\phi _2} = - \delta _2^2{\psi _2}, ψ1 = –δ1 ψ2, we have the following periodic solutions: 19u11(x,t)=ln(ϕ0(Δtan(Δξ2)δ1)24δ0δ2+δ1ϕ0(Δtan(Δξ2)δ1)2δ0δ2+ϕ0),{u_{11}}(x,t) = \ln \left( {{{{\phi _0}{{\left( {\sqrt { - \Delta } \tan \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right) - {\delta _1}} \right)}^2}} \over {4{\delta _0}{\delta _2}}} + {{{\delta _1}{\phi _0}\left( {\sqrt { - \Delta } \tan \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right) - {\delta _1}} \right)} \over {2{\delta _0}{\delta _2}}} + {\phi _0}} \right), 20u12(x,t)=ln(δ0δ2ψ2csc2(Δξ2)(δ1Δsin(Δξ)δ12+6δ0δ2Δcos(Δξ))(δ1+Δcot(Δξ2))2+ϕ0),{u_{12}}(x,t) = \ln \left( {{{{\delta _0}{\delta _2}{\psi _2}{{\csc }^2}\left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)\left( {{\delta _1}\sqrt { - \Delta } \sin \left( {\sqrt { - \Delta } \xi } \right) - \delta _1^2 + 6{\delta _0}{\delta _2} - \Delta \cos \left( {\sqrt { - \Delta } \xi } \right)} \right)} \over {{{\left( {{\delta _1} + \sqrt { - \Delta } \cot \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right), 21u13(x,t)=ln( δ0δ1δ2Δψ2cosh(Δξ)+δ1Δϕ0cos(Δξ)+sin(Δξ)(δ122δ0δ2)sin(Δξ)+δ1Δcos(Δξ)2δ0δ2 (δ122δ0δ2)(δ0δ2ψ2+ϕ0)+δ0δ2((δ126δ0δ2)ψ22ϕ0) ),{u_{13}}(x,t) = \ln \left( {{{{\delta _0}{\delta _1}{\delta _2}\sqrt { - \Delta } {\psi _2}\cosh (\sqrt \Delta \xi ) + {\delta _1}\sqrt { - \Delta } {\phi _0}\cos (\sqrt { - \Delta } \xi ) + \sin (\sqrt { - \Delta } \xi )} \over {\left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\sin (\sqrt { - \Delta } \xi ) + {\delta _1}\sqrt { - \Delta } \cos (\sqrt { - \Delta } \xi ) - 2{\delta _0}{\delta _2}}}} \right.\left. {\left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right) + {\delta _0}{\delta _2}\left( {\left( {\delta _1^2 - 6{\delta _0}{\delta _2}} \right){\psi _2} - 2{\phi _0}} \right)} \right), 22u14(x,t)=ln(δ1Δ(δ0δ2ψ2+ϕ0)sin(Δξ)+(δ122δ0δ2)(δ0δ2ψ2+ϕ0)cosh(Δξ)+δ0δ2((δ126δ0δ2)ψ22ϕ0)(δ1sin(Δξ2)+Δcosh(Δξ2))2).{u_{14}}(x,t) = \ln \left( {{{{\delta _1}\sqrt \Delta \left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\sin \left( {\sqrt { - \Delta } \xi } \right) + \left( {\delta _1^2 - 2{\delta _0}{\delta _2}} \right)\left( {{\delta _0}{\delta _2}{\psi _2} + {\phi _0}} \right)\cosh \left( {\sqrt \Delta \xi } \right) + {\delta _0}{\delta _2}\left( {\left( {\delta _1^2 - 6{\delta _0}{\delta _2}} \right){\psi _2} - 2{\phi _0}} \right)} \over {{{\left( {{\delta _1}\sin \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right) + \sqrt \Delta \cosh \left( {{{\sqrt \Delta \xi } \over 2}} \right)} \right)}^2}}}} \right).

When ϕ1=2cδ1δ2p{\phi _1} = {{2c{\delta _1}{\delta _2}} \over p},  q=(cδ12+2cδ0δ23pϕ0)(2cδ0δ2pϕ0)pq = - {{\left( {c\delta _1^2 + 2c{\delta _0}{\delta _2} - 3p{\phi _0}} \right)\left( {2c{\delta _0}{\delta _2} - p{\phi _0}} \right)} \over p},  r=(pϕ02cδ0δ2)2(2pϕ0cδ12)p2r = {{{{\left( {p{\phi _0} - 2c{\delta _0}{\delta _2}} \right)}^2}\left( {2p{\phi _0} - c\delta _1^2} \right)} \over {{p^2}}}, ψ2 = 0,  ϕ2=2cδ22p{\phi _2} = {{2c\delta _2^2} \over p}, ψ1 = 0, we have: 23u15(x,t)=ln(2cδ0δ2p+cΔp(cos(Δξ)1)+ϕ0),{u_{15}}(x,t) = \ln \left( { - {{2c{\delta _0}{\delta _2}} \over p} + {{c\Delta } \over {p\left( {\cos \left( {\sqrt { - \Delta } \xi } \right) - 1} \right)}} + {\phi _0}} \right), 24u16(x,t)=ln( 1p(d+e s i n (Δξ))2( cδ12e(dsin(Δξ)+e)cΔeΔ(de)(d+e)cos(Δξ) +cδ0δ2(2d2+e2cosh(2Δξ)+3e2)+12pϕ0(2d2+4desin(Δξ)e2cos(2Δξ)+e2) ) ),{u_{16}}(x,t) = \ln \left( {{1 \over {p{{(d + e s i n (\sqrt { - \Delta } \xi ))}^2}}}\left( { - c\delta _1^2e(d\sin (\sqrt { - \Delta } \xi ) + e) - c\sqrt { - \Delta } e\sqrt {\Delta (d - e)(d + e)} \cos (\sqrt { - \Delta } \xi )} \right.} \right.\left. {\left. { + c{\delta _0}{\delta _2}\left( { - 2{d^2} + {e^2}\cosh (2\sqrt \Delta \xi ) + 3{e^2}} \right) + {1 \over 2}p{\phi _0}\left( {2{d^2} + 4de\sin (\sqrt { - \Delta \xi } ) - {e^2}\cos (2\sqrt { - \Delta \xi } ) + {e^2}} \right)} \right)} \right), 25u17(x,t)=ln(4cδ0δ1δ2δ1p+Δptan(Δξ2)+4cδ02δ22(cos(Δξ)+1)p(δ1cos(Δξ2)+Δsin(Δξ2))2+ϕ0),{u_{17}}(x,t) = \ln \left( { - {{4c{\delta _0}{\delta _1}{\delta _2}} \over {{\delta _1}p + \sqrt { - \Delta } p\tan \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)}} + {{4c\delta _0^2\delta _2^2\left( {\cos \left( {\sqrt { - \Delta } \xi } \right) + 1} \right)} \over {p{{\left( {{\delta _1}\cos \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right) + \sqrt { - \Delta } \sin \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right), 26u18(x,t)=ln(4cδ0δ1δ2δ1pΔpcot(Δξ2)4cδ02δ22(cos(Δξ)1)p(δ1sin(Δξ2)Δcos(Δξ2))2+ϕ0),{u_{18}}(x,t) = \ln \left( { - {{4c{\delta _0}{\delta _1}{\delta _2}} \over {{\delta _1}p - \sqrt { - \Delta } p\cot \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)}} - {{4c\delta _0^2\delta _2^2\left( {\cos \left( {\sqrt { - \Delta } \xi } \right) - 1} \right)} \over {p{{\left( {{\delta _1}\sin \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right) - \sqrt { - \Delta } \cos \left( {{{\sqrt { - \Delta } \xi } \over 2}} \right)} \right)}^2}}} + {\phi _0}} \right), we have:where ξ is expressed in Eq.(3). All the obtained solutions are written in the form of Eq.(5), such that u = In v. We simulate the wave behaviours of the solutions of the governing model.

3.2
Application of the KMM

For n = 2, the general solution forms in KMM [38] can be described as 27v(ξ)=ρ0+ρ1Ω(ξ)+ρ2(Ω(ξ))2,v(\xi ) = {\rho _0} + {\rho _1}\Omega (\xi ) + {\rho _2}{\left( {\Omega (\xi )} \right)^2}, with 28Ω(ξ)=γ1Ω(ξ)4+γ2Ω(ξ)3+γ3Ω(ξ)2+γ4Ω(ξ)+γ5.\Omega '(\xi ) = \sqrt {{\gamma _1}\Omega {{(\xi )}^4} + {\gamma _2}\Omega {{(\xi )}^3} + {\gamma _3}\Omega {{(\xi )}^2} + {\gamma _4}\Omega (\xi ) + {\gamma _5}}.

The following solutions are obtained by employing (27) along with (28) in (6):

• Case-1 For γ5=(4γ1γ3γ22)264γ13, γ4=γ2(4γ1γ3γ22)8γ12\Omega '(\xi ) = \sqrt {{\gamma _1}\Omega {{(\xi )}^4} + {\gamma _2}\Omega {{(\xi )}^3} + {\gamma _3}\Omega {{(\xi )}^2} + {\gamma _4}\Omega (\xi ) + {\gamma _5}}. offers ρ2=2γ1ρ1γ2,p=cγ2ρ1,r=c(8γ1ρ0γ2ρ1)(γ22ρ1+4γ1γ2ρ04γ1γ3ρ1)264γ13γ2ρ1{\gamma _5} = {{{{\left( {4{\gamma _1}{\gamma _3} - \gamma _2^2} \right)}^2}} \over {64\gamma _1^3}},\;{\gamma _4} = {{{\gamma _2}\left( {4{\gamma _1}{\gamma _3} - \gamma _2^2} \right)} \over {8\gamma _1^2}}, q=c(3γ2ρ0γ3ρ1)(γ22ρ1+4γ1γ2ρ04γ1γ3ρ1)4γ1γ2ρ1.{\rho _2} = {{2{\gamma _1}{\rho _1}} \over {{\gamma _2}}},\;p = {{c{\gamma _2}} \over {{\rho _1}}},\;r = {{c\left( {8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \right){{\left( {\gamma _2^2{\rho _1} + 4{\gamma _1}{\gamma _2}{\rho _0} - 4{\gamma _1}{\gamma _3}{\rho _1}} \right)}^2}} \over {64\gamma _1^3{\gamma _2}{\rho _1}}}. As a result we have the following solutions:

When γ1 > 0 and 8γ1γ33γ22<08{\gamma _1}{\gamma _3} - 3\gamma _2^2 < 0, we get the solutions as follows:

The dark type soliton solution 29u1(x,t)=ln((3γ228γ1γ3)ρ1tanh2(3γ228γ1γ3Γ(β+1)(xεctε)4γ1ε)γ2+8γ1ρ0γ2ρ18γ1).{u_1}(x,t) = \ln \left( {{{{{\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right){\rho _1}{{\tanh }^2}\left( {{{\sqrt {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \Gamma (\beta + 1)\left( {{x^} - c{t^}} \right)} \over {4\sqrt {{\gamma _1}} }}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right).

The singular soliton solution 30u2=ln(γ3ρ1coth2(γ1(3γ228γ1γ3)Γ(β+1)(xεctε)4γ1ε)γ2+γ2ρ1(3coth2(γ1(3γ228γ1γ3)Γ(β+1)(xεctε)4γ1ε)1)8γ1+ρ0).{u_2} = \ln \left( { - {{{\gamma _3}{\rho _1}{{\coth }^2}\left( {{{\sqrt {{\gamma _1}\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right)} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {4{\gamma _1}\varepsilon }}} \right)} \over {{\gamma _2}}} + {{{\gamma _2}{\rho _1}\left( {3{{\coth }^2}\left( {{{\sqrt {{\gamma _1}\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right)} \Gamma (\beta + 1)\left( {{x^{\rm{\varepsilon }}} - c{t^{\rm{\varepsilon }}}} \right)} \over {4{\gamma _1}\varepsilon }}} \right) - 1} \right)} \over {8{\gamma _1}}} + {\rho _0}} \right).

When γ1 > 0 and 8γ1γ33γ22>08{\gamma _1}{\gamma _3} - 3\gamma _2^2 > 0, we get the periodic solutions as follows: 31u3(x,t)=ln((8γ1γ33γ22)ρ1tan2(8γ1γ33γ22Γ(β+1)(xεctε)4γ1ε)γ2+8γ1ρ0γ2ρ18γ1),{u_3}(x,t) = \ln \left( {{{{{\left( {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \right){\rho _1}{{\tan }^2}\left( {{{\sqrt {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {4\sqrt {{\gamma _1}} \varepsilon }}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right), 32u4(x,t)=ln(γ3ρ1cot2(γ1(8γ1γ33γ22)Γ(β+1)(xεctε)4γ1ε)γ2γ2ρ1(3cot2(γ1(8γ1γ33γ22)Γ(β+1)(xεctε)4γ1ε)+1)8γ1+ρ0).{u_4}(x,t) = \ln \left( {{{{\gamma _3}{\rho _1}{{\cot }^2}\left( {{{\sqrt {{\gamma _1}\left( {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \right)} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {4{\gamma _1}\varepsilon }}} \right)} \over {{\gamma _2}}} - {{{\gamma _2}{\rho _1}\left( {3{{\cot }^2}\left( {{{\sqrt {{\gamma _1}\left( {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \right)} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {4{\gamma _1}\varepsilon }}} \right) + 1} \right)} \over {8{\gamma _1}}} + {\rho _0}} \right).

• Case-2 For γ5=γ22(16γ1γ35γ22)256γ13, γ4=γ2(4γ1γ3γ22)8γ12{\gamma _5} = {{\gamma _2^2\left( {16{\gamma _1}{\gamma _3} - 5\gamma _2^2} \right)} \over {256\gamma _1^3}},\;{\gamma _4} = {{{\gamma _2}\left( {4{\gamma _1}{\gamma _3} - \gamma _2^2} \right)} \over {8\gamma _1^2}} offers ρ2=2γ1ρ1γ2, p=cγ2ρ1{\rho _2} = {{2{\gamma _1}{\rho _1}} \over {{\gamma _2}}},\;p = {{c{\gamma _2}} \over {{\rho _1}}},  r=c(8γ1ρ0γ2ρ1)2(5γ22ρ18γ1γ2ρ0+16γ1γ3ρ1)256γ13ρ1r = - {{c{{\left( {8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \right)}^2}\left( { - 5\gamma _2^2{\rho _1} - 8{\gamma _1}{\gamma _2}{\rho _0} + 16{\gamma _1}{\gamma _3}{\rho _1}} \right)} \over {256\gamma _1^3{\rho _1}}},  q=c(8γ1ρ0γ2ρ1)(9γ22ρ124γ1γ2ρ0+32γ1γ3ρ1)64γ12ρ1q = {{c\left( {8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \right)\left( { - 9\gamma _2^2{\rho _1} - 24{\gamma _1}{\gamma _2}{\rho _0} + 32{\gamma _1}{\gamma _3}{\rho _1}} \right)} \over {64\gamma _1^2{\rho _1}}}.. As a result we have the following solutions:

When γ1 < 0 and 8γ1γ33γ22<08{\gamma _1}{\gamma _3} - 3\gamma _2^2 < 0, we get: The bright soliton solution 33u5(x,t)=ln(2(3γ228γ1γ3)ρ1sech2(8γ1γ33γ22Γ(β+1)(xεctε)22γ1ε)γ2+8γ1ρ0γ2ρ18γ1).{u_5}(x,t) = \ln \left( {{{{{2\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right){\rho _1}{\rm{sec}}{{\rm{h}}^2}\left( {{{\sqrt {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {2\sqrt 2 \sqrt {{\gamma _1}} \varepsilon }}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right).

When γ > 0 and 8γ1γ33γ22>08{\gamma _1}{\gamma _3} - 3\gamma _2^2 > 0, we get the hyperbolic solution as follows: 34u6(x,t)=ln(2(8γ1γ33γ22)ρ1csch2(8γ1γ33γ22Γ(β+1)(xεctε)22γ1ε)γ2+8γ1ρ0γ2ρ18γ1).{u_6}(x,t) = \ln \left( {{{{{2\left( {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \right){\rho _1}{\rm{csc}}{{\rm{h}}^2}\left( {{{\sqrt {8{\gamma _1}{\gamma _3} - 3\gamma _2^2} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {2\sqrt 2 \sqrt {{\gamma _1}} }}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right).

When γ > 0 and 8γ1γ33γ22<08{\gamma _1}{\gamma _3} - 3\gamma _2^2 < 0, we get the periodic solutions as follows: 35u7(x,t)=ln(2(3γ228γ1γ3)ρ1sec2(3γ228γ1γ3Γ(β+1)(xεctε)22γ1ε)γ2+8γ1ρ0γ2ρ18γ1),{u_7}(x,t) = \ln \left( {{{{{2\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right){\rho _1}{{\sec }^2}\left( {{{\sqrt {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {2\sqrt 2 {{\sqrt {{\gamma _1}} }^\varepsilon }}}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right), 36u8(x,t)=ln(2(3γ228γ1γ3)ρ1csc2(3γ228γ1γ3Γ(β+1)(xεctε)22γ1ε)γ2+8γ1ρ0γ2ρ18γ1).{u_8}(x,t) = \ln \left( {{{{{2\left( {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \right){\rho _1}{{\csc }^2}\left( {{{\sqrt {3\gamma _2^2 - 8{\gamma _1}{\gamma _3}} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {2\sqrt 2 \sqrt {{\gamma _1}} \varepsilon }}} \right)} \over {{\gamma _2}}} + 8{\gamma _1}{\rho _0} - {\gamma _2}{\rho _1}} \over {8{\gamma _1}}}} \right).

• Case-3 For γ2 = 0, γ4 = 0, γ5 = 0 and γ3 > 0, offers ρ1 = 0, c=pρ22γ1,r=2(γ1pρ03γ3pρ02ρ2)γ1,q=2γ3pρ0ρ23γ1pρ02γ1c = {{p{\rho _2}} \over {2{\gamma _1}}},\;r = {{2\left( {{\gamma _1}p\rho _0^3 - {\gamma _3}p\rho _0^2{\rho _2}} \right)} \over {{\gamma _1}}},\;q = {{2{\gamma _3}p{\rho _0}{\rho _2} - 3{\gamma _1}p\rho _0^2} \over {{\gamma _1}}}. As a result we have the following exponential solution: 37u9(x,t)=ln(16γ32ρ2ρ2e2γ3Γ(β+1)(xεpρ2tε2γ1)ε(γ1γ34ρ2e2γ3Γ(β+1)(xεpρ2tε2γ1)ε)2+ρ0).{u_9}(x,t) = \ln \left( {{{16\gamma _3^2{\rho ^2}{\rho _2}{e^{{{2\sqrt {{\gamma _3}} \Gamma (\beta + 1)\left( {{x^\varepsilon } - {{p{\rho _2}{t^\varepsilon }} \over {2{\gamma _1}}}} \right)} \over \varepsilon }}}} \over {{{\left( {{\gamma _1}{\gamma _3} - 4{\rho ^2}{e^{{{2\sqrt {{\gamma _3}} \Gamma (\beta + 1)\left( {{x^\varepsilon } - {{p{\rho _2}{t^\varepsilon }} \over {2{\gamma _1}}}} \right)} \over \varepsilon }}}} \right)}^2}}} + {\rho _0}} \right).

Taking γ1=4ρ2γ3{\gamma _1} = - {{4{\rho ^2}} \over {{\gamma _3}}} in Eq. (37), we have bright soliton solution 38u10(x,t)=ln(γ32ρ2sech2(γ3Γ(β+1)(xεctε)ε)4ρ2+ρ0).{u_{10}}(x,t) = \ln \left( {{{\gamma _3^2{\rho _2}{\rm{sec}}{{\rm{h}}^2}\left( {{{\sqrt {{\gamma _3}} \Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over \varepsilon }} \right)} \over {4{\rho ^2}}} + {\rho _0}} \right).

Similarly, taking γ1=4ρ2γ3{\gamma _1} = {{4{\rho ^2}} \over {{\gamma _3}}} in Eq. (37), we have 39u11(x,t)=ln(γ32ρ2csch2(γ3Γ(β+1)(xεctε)ε)4ρ2+ρ0).{u_{11}}(x,t) = \ln \left( {{{\gamma _3^2{\rho _2}{{{\mathop{\rm csch}\nolimits} }^2}\left( {{{\sqrt {{\gamma _3}\Gamma } (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over \varepsilon }} \right)} \over {4{\rho ^2}}} + {\rho _0}} \right).

Remark:

All the obtained solutions are written in the form of Eq. (5), such that u = ln v.

3.3
Application of the MGERIFM

The solution via MGERIFM [39] is described as: 40v(ξ)=c0+r=1nci(rv(ξ)dξdξdξ)r+r=1ndr(rv(ξ)dξdξdξ)r.v(\xi ) = {c_0} + \sum\limits_{r = 1}^n {{c_i}} {\left( {\underbrace {\int\!\!\!\int \cdots \smallint }_rv(\xi )d\xi d\xi \cdots d\xi } \right)^r} + \sum\limits_{r = 1}^n {{d_r}} {\left( {\underbrace {\int\!\!\!\int \cdots \smallint }_rv(\xi )d\xi d\xi \cdots d\xi } \right)^{ - r}}.

The solution to equation (40) for n = 2 is as follows: 41v(ξ)=c0+c1ϕ(ξ)dξ+d1(ϕ(ξ)dξ)1+c2(ϕ(ξ)dξ)2+d2(ϕ(ξ)dξ)2,v(\xi ) = {c_0} + {c_1}\int \phi (\xi )d\xi + {d_1}{\left( {\int \phi (\xi )d\xi } \right)^{ - 1}} + {c_2}{\left( {\int \phi (\xi )d\xi } \right)^2} + {d_2}{\left( {\int \phi (\xi )d\xi } \right)^{ - 2}}, where v(ξ) is defined by 42ϕ(ξ)=τ1eξs1+τ2eξs2τ3eξs3+τ4eξs4.\phi (\xi ) = {{{\tau _1}{e^{\xi {s_1}}} + {\tau _2}{e^{\xi {s_2}}}} \over {{\tau _3}{e^{\xi {s_3}}} + {\tau _4}{e^{\xi {s_4}}}}}.

Moreover, we proceed as:

• Case-1: Choosing [τ1, τ2, τ3, τ4 = [–i, –i, –i, –i] and [s1, s2, s3, s4] = [1, –1,0,0], Eq. (42) takes the following form 43ϕ(ξ)=cosh(ξ).\phi (\xi ) = \cosh (\xi ).

Inserting Eq. (43) into Eq. (41), implies that 44v(ξ)=c0+c1sinh(ξ)+c2sinh2(ξ)+d1sinh(ξ)+d2sinh2(ξ).v(\xi ) = {c_0} + {c_1}\sinh (\xi ) + {c_2}{\sinh ^2}(\xi ) + {{{d_1}} \over {\sinh (\xi )}} + {{{d_2}} \over {{{\sinh }^2}(\xi )}}.

The solutions are as follows when Eq. (44) is inserted into Eq. (6):

For d1 = 0, c1 = 0, c2 = 0, c=d2p2c = {{{d_2}p} \over 2}, r=2c02p(c0d2)r = 2c_0^2p\left( {{c_0} - {d_2}} \right), q = c0p (2d2 – 3c0), we have: 45u1(x,t)=ln(2ccsch2(Γ(β+1)(xεctε)ε)p+c0).{u_1}(x,t) = \ln \left( {{{2c{{{\mathop{\rm csch}\nolimits} }^2}\left( {{{\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over \varepsilon }} \right)} \over p} + {c_0}} \right).

• Case-2: Let [τ1, τ2, τ3, τ4] = [2i, –2i, 4i, 4i] and [ s1,s2,s3,s4 ]=[ 12,12,0,0 ]\left[ {{s_1},{s_2},{s_3},{s_4}} \right] = \left[ {{1 \over 2}, - {1 \over 2},0,0} \right], Eq. (42) transforms to sine hyperbolic function 46ϕ(ξ)=12sinh(ξ2).\phi (\xi ) = {1 \over 2}\sinh \left( {{\xi \over 2}} \right).

Plugging Eq. (46) into Eq. (41), offers 47v(ξ)=c0+c1cosh(ξ2)+c2cosh2(ξ2)+d1cosh(ξ2)+d2cosh2(ξ2).v(\xi ) = {c_0} + {c_1}\cosh \left( {{\xi \over 2}} \right) + {c_2}{\cosh ^2}\left( {{\xi \over 2}} \right) + {{{d_1}} \over {\cosh \left( {{\xi \over 2}} \right)}} + {{{d_2}} \over {{{\cosh }^2}\left( {{\xi \over 2}} \right)}}.

Applying Eq. (47) to Eq. (6), we get:

For d1 = 0, c1 = 0, c2 = 0, c = –2d2p, r=2c02p(c0+d2)r = 2c_0^2p\left( {{c_0} + {d_2}} \right), q = c0(–p) (3c0 + 2d2), we get the solution as follows: 48u2(x,t)=ln(c0csech2(Γ(β+1)(xεctε)2ε)2p).{u_2}(x,t) = \ln \left( {{c_0} - {{c{{{\mathop{\rm sech}\nolimits} }^2}\left( {{{\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {2\varepsilon }}} \right)} \over {2p}}} \right).

• Case-3: Taking [τ1, τ2, τ3, τ4] = [1, –1, i, i] and [s1, s2, s3, s4] = [i, –i, 0, 0], Eq. (42) converts to the periodic function 49ϕ(ξ)=sin(ξ).\phi (\xi ) = \sin (\xi )

Putting Eq. (49) into Eq. (41), gives 50v(ξ)=c0+c2(cos(ξ))2+c1(cos(ξ))+d1cos(ξ)+d2(cos(ξ))2.v(\xi ) = {c_0} + {c_2}{( - \cos (\xi ))^2} + {c_1}( - \cos (\xi )) + {{{d_1}} \over { - \cos (\xi )}} + {{{d_2}} \over {{{( - \cos (\xi ))}^2}}}.

Incorporating Eq. (50) in Eq. (6), we have:

For d1 = 0, c1 = 0, c2 = 0, c=d2p2c = {{{d_2}p} \over 2}, r=2c02p(c0+d2)r = 2c_0^2p\left( {{c_0} + {d_2}} \right), q = c0(–p) (3c0 + 2d2), we obtain: 51u3(x,t)=ln(2csec2(Γ(β+1)(xεctε)ε)p+c0).{u_3}(x,t) = \ln \left( {{{2c{{\sec }^2}\left( {{{\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over \varepsilon }} \right)} \over p} + {c_0}} \right).

• Case-4: Choosing the parameters [τ1, τ2, τ3, τ4] = [1,1,1,1] and [s1, s2, s3, s4] = [i, –i, 0, 0], Eq. (42) offers 52ϕ(ξ)=cos(ξ).\phi (\xi ) = \cos (\xi ).

Manipulating Eq. (52) and Eq. (41), we have 53v(ξ)=c0+c1sin(ξ)+c2sin2(ξ)+d1sin(ξ)+d2sin2(ξ).v(\xi ) = {c_0} + {c_1}\sin (\xi ) + {c_2}{\sin ^2}(\xi ) + {{{d_1}} \over {\sin (\xi )}} + {{{d_2}} \over {{{\sin }^2}(\xi )}}.

Plugging Eq. (53) into Eq. (6), give the following solutions:

When d1 = 0, c1 = 0, c2 = 0, c=d2p2c = {{{d_2}p} \over 2}, q = c0(–p) (3c0 + 2d2), r=2c02p(c0+d2)r = 2c_0^2p\left( {{c_0} + {d_2}} \right), we get: 54u4(x,t)=ln(2ccsc2(Γ(β+1)(xεctε)ε)p+c0).{u_4}(x,t) = \ln \left( {{{2c{{\csc }^2}\left( {{{\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over \varepsilon }} \right)} \over p} + {c_0}} \right).

• Case-5: Taking the parameters [τ1, τ2, τ3, τ4] = [2,2,2,2] and [s1, s2, [ s1,s2,s3,s4 ]=[ 25,25,0,0 ]\left[ {{s_1},{s_2},{s_3},{s_4}} \right] = \left[ {{2 \over 5},{2 \over 5},0,0} \right], Eq. (42) gives 55ϕ(ξ)=e2ξ5.\phi (\xi ) = {e^{{{2\xi } \over 5}}}.

Manipulating Eq. (55) and Eq. (41), we have 56v(ξ)=c0+12c1(5e2ξ5)+c2(52e2ξ5)2+d152e2ξ5+d2(52e2ξ5)2.v(\xi ) = {c_0} + {1 \over 2}{c_1}\left( {5{e^{{{2\xi } \over 5}}}} \right) + {c_2}{\left( {{5 \over 2}{e^{{{2\xi } \over 5}}}} \right)^2} + {{{d_1}} \over {{5 \over 2}{e^{{{2\xi } \over 5}}}}} + {{{d_2}} \over {{{\left( {{5 \over 2}{e^{{{2\xi } \over 5}}}} \right)}^2}}}.

Inserting Eq. (56) into Eq. (6), give the following solution:

When d1 = 0, c1 = 0, p = 0, r=125(16)c(c024c2d2),q=16cc025r = {1 \over {25}}( - 16)c\left( {c_0^2 - 4{c_2}{d_2}} \right),\;q = {{16c{c_0}} \over {25}}, we get: 57u5(x,t)=ln(425d2e4Γ(β+1)(xεαε)5ε+254c2e 4Γ(β+1)(xεαε)5ε )+c0).{u_5}(x,t) = \ln \left( {{4 \over {25}}{d_2}{e^{ - {{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - {\alpha ^\varepsilon }} \right)} \over {5\varepsilon }}}} + {{25} \over 4}{c_2}{e^{\left. {{{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - {\alpha ^\varepsilon }} \right)} \over {5\varepsilon }}} \right)}} + {c_0}} \right).

Next, the hyperbolic solution is written as 58u6(x,t)=ln( 254c2(cosh(4Γ(β+1)(xεctε)5ε)+sinh(4Γ(β+1)(xεctε)5ε))+c0 +425d2(cosh(4Γ(β+1)(xεctε)5ε)sinh(4Γ(β+1)(xεctε)5ε)) ).{u_6}(x,t) = \ln \left( {{{25} \over 4}{c_2}\left( {\cosh \left( {{{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {5\varepsilon }}} \right) + \sinh \left( {{{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {5\varepsilon }}} \right)} \right) + {c_0}} \right.\left. { + {4 \over {25}}{d_2}\left( {\cosh \left( {{{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {5\varepsilon }}} \right) - \sinh \left( {{{4\Gamma (\beta + 1)\left( {{x^\varepsilon } - c{t^\varepsilon }} \right)} \over {5\varepsilon }}} \right)} \right)} \right).

Remark:

All the obtained solutions are written in the form of Eq. (5), such that u = lnv.

4
Conclusions

In this paper, we have discussed the nonlinear truncated M-fractional ZSE and investigated the fractional parametric effect to the dynamics of wave movement. A variety of solutions have been secured through the implementation of the recently advanced mathematical methods. For the proposed model, our methods have effectively identified many types of propagating wave solutions, demonstrating their versatility and applicability in various scenarios. The physical characteristics of the developed solutions have been illustrated by a variety of Figures (1-7) and it has been observed that a slight change in the fractional parameter offers the different nature of the wave solutions. The obtained findings may be of interest for further investigation of frameworks that are designed to address nonlinear issues in the field of applied sciences. The results are fascinating from both a theoretical and practical perspective, especially in terms of the behavior of optical components. The presented techniques are the most practical, smoothly, and straightforward approaches to addressing different models in applied sciences, offering a wide range of exact solutions compared to previous methodologies. In future, the techniques can be extended for the solution of higher non-linear systems.

Fig. 1

Sketches of the Eq. (9) with fractional parameter.

Fig. 2

Sketches of the Eq.(11) with fractional parameter.

Fig. 3

Sketches of the Eq. (36) with fractional parameter.

Fig. 4

Sketches of the Eq. (37) with fractional parameter.

Fig. 5

Sketches of the Eq. (38) with fractional parameter.

Fig. 6

Sketches of the Eq. (51) with fractional parameter.

Fig. 7

Sketches of the Eq. (57) with fractional parameter.

Language: English
Submitted on: Oct 7, 2024
|
Accepted on: Sep 29, 2025
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Published on: Feb 2, 2026
Published by: Harran University
In partnership with: Paradigm Publishing Services
Publication frequency: 2 issues per year

© 2026 Jan Muhammad, Usman Younas, published by Harran University
This work is licensed under the Creative Commons Attribution 4.0 License.

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