Research article Recurring Topics

The Role of Short-term Consolidation in Memory Persistence

  • Short-term memory, often described as working memory, is one of the most fundamental information processing systems of the human brain. Short-term memory function is necessary for language, spatial navigation, problem solving, and many other daily activities. Given its importance to cognitive function, understanding the architecture of short-term memory is of crucial importance to understanding human behavior. Recent work from several laboratories investigating the entry of information into short-term memory has uncovered a dissociation between encoding processes, those that register information into short-term memory, and consolidation processes, those that solidify the representation within short-term memory. Here I describe the key differences between short-term encoding and consolidation and briefly review what is known about the short-term consolidation process itself. Cognitive function, plausible neural instantiation, and open questions are addressed.

    Citation: Timothy J. Ricker. The Role of Short-term Consolidation in Memory Persistence[J]. AIMS Neuroscience, 2015, 2(4): 259-279. doi: 10.3934/Neuroscience.2015.4.259

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  • Short-term memory, often described as working memory, is one of the most fundamental information processing systems of the human brain. Short-term memory function is necessary for language, spatial navigation, problem solving, and many other daily activities. Given its importance to cognitive function, understanding the architecture of short-term memory is of crucial importance to understanding human behavior. Recent work from several laboratories investigating the entry of information into short-term memory has uncovered a dissociation between encoding processes, those that register information into short-term memory, and consolidation processes, those that solidify the representation within short-term memory. Here I describe the key differences between short-term encoding and consolidation and briefly review what is known about the short-term consolidation process itself. Cognitive function, plausible neural instantiation, and open questions are addressed.


    Nonlinear evolution equations (NLEEs) play an important part in the study of nonlinear science, particular in plasma physics, quantum field theory, nonlinear wave propagation and nonlinear optical fibers so that it attracted the attention of a large number of scholars. The extended auxiliary equation technique [1], the Bernoulli's equation approach [2], the Exp-function technique [3], the homotopy analysis technique [4], the homotopy perturbation technique [5], the improved $ tan(\phi/2) $-expansion technique ([6,7]), the Hirota's bilinear technique [8,9,10,11,12,13,14,15], the He's variational principle [16,17], the binary Darboux transformation [18], the Lie group analysis [19,20], the Bäcklund transformation method [21], optimal galerkin-homotopy asymptotic method applied [22], and the multiple rogue waves method ([23,24]) have been proposed to solve NLEEs. By using these approaches, various exact solutions including soliton solution, lump solution, rogue wave solution, periodic solution, interaction solution, rational solution and high-order rational solution were obtained ([25,26]).

    In this paper, we mainly consider the following dynamical model, which can be used to describe some interesting (3+1)-dimensional waves of physics, namely, the generalized Kadomtsev-Petviashvili (gKP) equation [27]. That is

    $ (Ψt+6ΨΨx+Ψxxx)x+aΦyy=0,
    $
    (1.1)

    and also above equation is integrable. Author of [28] introduced the modification of KP (mKP) equation [29] given

    $ 4Ψt6Ψ2Ψx+Ψxxx+6Ψx1xΨy+Ψ1xΦyy=0.
    $
    (1.2)

    The generalized KP (gKP) equation has been researched by some scholars [30,31,32] in which is given as

    $ (Ψt+αΨx+βΨΨx+γΨΨxxt)x+Ψyy=0.
    $
    (1.3)

    The another type of gKP equation is given in [33] as below

    $ Ψxxxy+3(ΨxΨy)x+Ψtx+Ψty+ΨtzΨzz=0.
    $
    (1.4)

    We first present the bilinear form for Eq (1.4), by taking the following first-order logarithmic transformation

    $ Ψ=2(lnf)x,
    $
    (1.5)

    then, Eq (1.4) is turned into the bilinear form

    $ (D3xDy+DxDt+DyDt+DzDtD2z)f.f=0,
    $
    (1.6)

    in which $ D_t, D_x, D_y $ and $ D_z $ are Hirota's bilinear frames. Cao [34] investigated the generalized B-type KP equation as follows

    $ Ψxxxy+3(ΨxΨy)x3ΨxzΨty=0.
    $
    (1.7)

    Guan et al. [35] derived a (3+1)-dimensional gKP equation in below form

    $ Ψxxxy+3(ΨxΨy)x+αΨxxxz+3α(ΨxΨz)x+λ1Ψxt+λ2Φyt+λ3Φzt+ω1Φxz+ω2Φyz+ω3Φzz=0,
    $
    (1.8)

    and some lump soliton solutions have been constructed using the Hirota bilinear method in [36]. Via transformation $ \Psi = 2(\ln f)_{x}, $ the bilinear form of equation (1.8) reads:

    $ (D3xDy+αD3xDz+λ1DxDt+λ2DyDt+λ3DzDt+ω1DxDz+ω2DyDz+ω3D2z)f.f=0.
    $
    (1.9)

    Most classical test functions for solving NLPDEs by using the several particular functions can be constructed via Hirota bilinear technique. In other words, Hirota operator covers most of the classical hypothesis function method. For example, the fractional generalized CBS-BK equation [37], the generalized Bogoyavlensky-Konopelchenko equation [38], the (2+1)-dimensional asymmetrical Nizhnik-Novikov-Veselov equation [39], and the (2+1)-dimensional generalized variable-coefficient KP-Burgers-type equation [40]. Diverse kinds of studies on solve NLPDEs were perused via mighty authors in which some of them can be stated, for instance, multivariate rogue wave to some PDEs [41], interaction lump solutions the gKP equation [42], the (1+1)-dimensional coupled integrable dispersionless equations [43]. Therefore, we embark on the new research topic of constructing the analytic solutions of nonlinear PDEs by exploring the bilinear method. According to recent studies, we can obtain some of the new exact analytic solutions of nonlinear PDEs by way of constructing their corresponding bilinear differential equations in [44,45,46,47]. Here, we will study the multiple Exp-function method (MEFM) for determining the multiple soliton solutions (MSSs). The MEFM employed by some of powerful authors for various nonlinear equations have been surveyed in more studies in [31,48,49,50]. Authors of [51] utilized the reduced differential transform method for solving partial differential equations. Also, Yu and Sun [52] studied the dimensionally reduced generalized KP equations by help of Hirita bilinear method and obtained of lump solutions.

    The outline of our paper is as follows: the multiple Exp-function scheme has been summarized in section 2. In sections 3, the KP equation, will investigate to finding 1-wave, 2-wave, and three-wave solutions. The Sections 4-6 devote to determined the periodic, cross-kink, and solitary wave solutions. Moreover, in Section 7, the modulation instability analysis is investigated. Finally in Section 8, the SIVP technique is considered with four cases for finding the solitary, bright, dark and singular wave solutions. A few of conclusions and outlook will be given in the final section.

    This method was summarized and improved for achieving the analytic solutions of NLPDEs:

    Step 1. Assume a nonlinear PDE is given in general frame as follows

    $ N(x,y,t,Ψ,Ψx,Ψy,Ψz,Ψt,Ψxx,Ψtt,...)=0.
    $
    (2.1)

    Take the novel variables $ \xi _{i} = \xi _{i}(x, y, z, t), 1\leq i\leq n $, by differentiable frames:

    $ ξi,x=αiξi,  ξi,y=βiξi,  ξi,z=γiξi,  ξi,t=δiξi,  1in,
    $
    (2.2)

    where $ \alpha _{i}, \beta _{i}, \gamma _{i}, 1\leq i\leq n $, are unfound amounts. It noted that one can get as the following function

    $ ξi=ϖieθi,  θi=αix+βiy+γizδit,  1in,
    $
    (2.3)

    where $ \varpi _{i}, 1\leq i\leq n $, unspecified amounts.

    Step 2. Assuming the solution of the Eq (2.1) is function of variables $ \xi _{i}, 1\leq i\leq n $:

    $ Ψ=Δ(ξ1,ξ2,...,ξn)Ω(ξ1,ξ2,...,ξn),  Δ=nr,s=1Mi,j=0Δrs,ijξirξjs,  Ω=nr,s=1Ni,j=0Ωrs,ijξirξjs,  
    $
    (2.4)

    in which $ \Delta _{rs, ij} $ and $ \Omega _{rs, ij} $ are amounts to be remained. Replacing Eq (2.4) into Eq (2.1) can be achieved the below form as:

    $ Ψ=Δ(ϖ1eα1x+β1y+γ1zδ1t,...,ϖneαnx+βny+γnzδnt)Ω(ϖ1eα1x+β1y+γ1zδ1t,...,ϖneαnx+βny+γnzδnt),
    $
    (2.5)

    and also we have

    $ Δt=ni=1Δξiξi,t,  Ωt=ni=1Ωξiξi,t,  Δx=ni=1Δξiξi,x,  Ωx=ni=1Ωξiξi,x,  Δy=ni=1Δξiξi,y,  Δz=ni=1Δξiξi,z, Ωy=ni=1Ωξiξi,y,  Ωz=ni=1Ωξiξi,z, 
    $
    (2.6)
    $ \ \ \ \Psi _{t} = \frac{\Omega \sum\limits_{i = 1}^{n}\Delta _{\xi _{i}}\xi _{i, t}-\Delta \sum\limits_{i = 1}^{n}\Omega _{\xi _{i}}\xi _{i, t}}{\Omega ^{2}}, \ \ \Psi _{x} = \frac{\Omega \sum\limits_{i = 1}^{n}\Delta _{\xi _{i}}\xi _{i, x}-\Delta \sum\limits_{i = 1}^{n}\Omega _{\xi _{i}}\xi _{i, x}}{\Omega ^{2}}, $
    $ \Psi _{y} = \frac{\Omega \sum\limits_{i = 1}^{n}\Delta _{\xi _{i}}\xi _{i, y}-\Delta \sum\limits_{i = 1}^{n}\Omega _{\xi _{i}}\xi _{i, y}}{\Omega ^{2}}, \ \ \Psi _{z} = \frac{ \Omega \sum\limits_{i = 1}^{n}\Delta _{\xi _{i}}\xi _{i, z}-\Delta \sum\limits_{i = 1}^{n}\Omega _{\xi _{i}}\xi _{i, z}}{\Omega ^{2}}. $

    The one-wave function of the solution will be reduced as below form

    $ Ψ=2Δ1Ω1,  Ω1=1+ρ1+ρ2 eα1x+β1y+γ1zδ1t,  Δ1=σ1+σ2 eα1x+β1y+γ1zδ1t,
    $
    (3.1)

    in which $ \rho _{1}, \rho _{2}, \sigma _{1} $ and $ \sigma _{2} $ are unspecified amounts. Substituting (3.1) into Eq (1.8), the below cases will be concluded as:

    Case I:

    $ α1=α1,  β1=β1,  ρ1=ρ1,  ρ2=  σ2(ρ1+1)σ1  ,  σ1=σ1,  σ2=σ2,  δ1=δ1,  γ1=γ1,  γ2=γ2.
    $
    (3.2)

    Case II:

    $ α1=α1,  β1=  αα1  3γ1α1δ1λ1+α1γ1ω1δ1γ1λ3+γ12ω3  α13δ1λ2+γ1ω2,  ρ1=1,  ρ2=ρ2,  
    $
    (3.3)
    $ \sigma _{1} = \sigma _{1}, \ \ \sigma _{{2}} = \sigma _{2}, \ \ \delta _{{1} } = \delta _{1}, \ \ \gamma _{1} = \gamma _{1}, \ \ \gamma _{2} = \gamma _{2}. $

    Case III:

    $ α1=  γ1(αδ1λ2αγ1ω2δ1λ3+γ1ω3)δ1λ1γ1ω1,  β1=αγ1,  ρ1=ρ1,  ρ2=ρ2,  
    $
    (3.4)
    $ \sigma _{1} = \sigma _{1}, \ \ \sigma _{{2}} = \sigma _{2}, \ \ \delta _{{1} } = \delta _{1}, \ \ \ \ \gamma _{1} = \gamma _{1}, \ \ \gamma _{2} = \gamma _{2}. $

    Case IV:

    $ α1=ϑ,  β1=β1, or  β1=αγ1,   ρ1=1,  ρ2=ρ2,  σ1=σ1,  σ2=σ2, δ1=ϑ3+γ1ω2λ2,  γ1=γ1,  γ2=γ2, 
    $
    (3.5)

    in which $ \vartheta $, solves the equation $ \lambda_1\vartheta ^4+(\gamma _1\lambda _3-\alpha \gamma _1\lambda_2) {\vartheta }^{3}+(\gamma _1\lambda _1\omega _2-\gamma_1\lambda _2\omega _1) \vartheta -\gamma _1 ^2\omega _3\lambda _2+\gamma_1^2\lambda _3\omega_2 = 0. $

    For example, the 1-wave solution for Case III will be considered as

    $ Ψ=2σ1+σ2e  γ1(αδ1λ2αγ1ω2δ1λ3+γ1ω3)δ1λ1γ1ω1  xαγ1y+γ1zδ1t1+ρ1+ρ2e  γ1(αδ1λ2αγ1ω2δ1λ3+γ1ω3)δ1λ1γ1ω1    xαγ1y+γ1zδ1t.
    $
    (3.6)

    The two-wave function of the solution will be reduced as below form

    $ Ψ=2Δ2Ω2,
    $
    (3.7)
    $ Ω2=1+σ1eα1x+β1y+γ1zδ1t+σ2eα2x+β2y+γ2zδ2t+σ1σ2σ12e(α1+α2)x+(β1+β2)y+(γ1+γ2)z(δ1+δ2)t,  
    $
    (3.8)
    $ \Delta _{2} = \rho _{1}e^{\alpha _{1}x+\beta _{1}y+\gamma _{1}z-\delta _{1}t}+\rho _{2}e^{\alpha _{2}x+\beta _{2}y+\gamma _{2}z-\delta _{2}t}+\rho _{1}\rho _{2}\rho _{12}e^{(\alpha _{1}+\alpha _{2})x+(\beta _{1}+\beta _{2})y+(\gamma _{1}+\gamma _{2})z-(\delta _{1}+\delta _{2})t}. $

    Substituting (3.7) in terms of (3.8) into Eq (1.8), the below cases will be resulted as:

    Case I:

    $ α1=0,  α2=α2,  β1=β1,  β2=αγ2,  δ1=  γ1(β1ω2+ω3γ1)β1λ2+γ1λ3,  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=0,  ρ2=ρ2,
    $
    (3.9)
    $ \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 1. $

    Case II:

    $ α1=α1,  α2=0,  β1=αγ1,  β2=β2,  δ1=  γ1(αγ1ω 2α1ω1ω3γ1)αγ1λ2α1λ1γ1λ3,  δ2=  γ2(β2ω2+γ2ω3)β2λ2+γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=0, 
    $
    (3.10)
    $ \ \rho _{2} = \rho _{2}, \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 1. $

    Case III:

    $ α1=α1,  α2=α2,  β1=αγ1,  β2=αγ2,  δ1=  γ1(αγ1ω 2α1ω1ω3γ1)αγ1λ2α1λ1γ1λ3,  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=0,
    $
    (3.11)
    $ \ \ \rho _{2} = \rho _{2}, \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 1. $

    Case IV:

    $ α1=0,  α2=α2,  β1=β1,  β2=αγ2,  δ1=  γ1(β1ω2+ω3γ1)β1λ2+γ1λ3,  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=ρ1,  ρ2=0,
    $
    (3.12)
    $ \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 1. $

    Case V:

    $ α1=α1,  α2=α2,  β1=αγ1,  β2=αγ2,  δ1=  γ1(αγ1ω 2α1ω1ω3γ1)αγ1λ2α1λ1γ1λ3,  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=ρ1,  ρ2=0,
    $
    (3.13)
    $ \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 1. $

    Case VI:

    $ α1=α1,  α2=α2,  β1=αα1γ2  α2,  β2=αγ2,  δ1=  α1γ2(αγ 2ω2α2ω1γ2ω3)α2(αγ2λ2α2λ1γ2λ3),  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=γ1,  γ2=γ2,  ρ1=ρ1,  ρ2=0,
    $
    (3.14)
    $ \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = \sigma _{12}. $

    Case VII:

    $ α1=α1,  α2=α2,  β1=αα1γ2  α2,  β2=αγ2,  δ1=  α1γ2(αγ 2ω2α2ω1γ2ω3)α2(αγ2λ2α2λ1γ2λ3),  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=  α1γ2  α2  ,  γ2=γ2,  ρ1=ρ1,  ρ2=ρ2,
    $
    (3.15)
    $ \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = 0, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = \sigma _{12}. $

    Case VIII:

    $ α1=12α2,  α2=α2,  β1=12αγ2,  β2=αγ2,  δ1=12  γ2(αγ2ω2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,  δ2=  γ2(αγ2ω 2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3,γ1=12γ2,  γ2=γ2,  ρ1=ρ1, 
    $
    (3.16)
    $ \ \rho _{2} = \rho _{2}, \ \ \rho _{12} = \rho _{12}, \ \ \sigma _{1} = \sigma _{1}, \ \ \sigma _{2} = \sigma _{2}, \ \ \sigma _{12} = 0. $

    For instance, the 2-wave solution for Case I will be taken as

    $ Ψ1=2ρ2e  tγ2(αγ 2ω2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3  +xα2yαγ2+zγ2  /(1+σ1etγ1(β1ω2+ω3γ1)β1λ2+γ1λ3  +yβ1+zγ1  +
    $
    (3.17)
    $ \left. \sigma _{{2}}{\mathrm{e}^{- \ \ {\frac{t\gamma _{{2}}\left( \alpha \, \gamma _{{2}}\omega _{{2}}-\alpha _{{2}}\omega _{{1}}-\gamma _{{2}}\omega _{{3}}\right) }{\alpha \, \gamma _{{2}}\lambda _{{2}}-\alpha _{{2}}\lambda _{{ 1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+x\alpha _{{2}}-y\alpha \, \gamma _{{2} }+z\gamma _{{2}}} \ \ }+\sigma _{{1}}\sigma _{{2}}{\mathrm{e}^{- \ \ {\frac{t\gamma _{{ 1}}\left( \beta _{{1}}\omega _{{2}}+\omega _{{3}}\gamma _{{1}}\right) }{ \beta _{{1}}\lambda _{{2}}+\gamma _{{1}}\lambda _{{3}}} \ \ }- \ \ {\frac{t\gamma _{{2} }\left( \alpha \, \gamma _{{2}}\omega _{{2}}-\alpha _{{2}}\omega _{{1} }-\gamma _{{2}}\omega _{{3}}\right) }{\alpha \, \gamma _{{2}}\lambda _{{2} }-\alpha _{{2}}\lambda _{{1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+x\alpha _{{2} }+y\beta _{{1}}-y\alpha \, \gamma _{{2}}+z\gamma _{{1}}+z\gamma _{{2}}} } \right) . $

    Also, the 2-wave solution for Case III will be considered as

    $ Ψ2=2ρ2e  tγ2(αγ 2ω2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3  +xα2yαγ2+zγ2  /(1+σ1etγ1(αγ1ω2α1ω1ω3γ1)αγ1λ2α1λ1γ1λ3  +xα1yαγ1+zγ1  +
    $
    (3.18)
    $ \left. \sigma _{{2}}{\mathrm{e}^{- \ \ {\frac{t\gamma _{{2}}\left( \alpha \gamma _{{2}}\omega _{{2}}-\alpha _{{2}}\omega _{{1}}-\gamma _{{2}}\omega _{{3} }\right) }{\alpha \gamma _{{2}}\lambda _{{2}}-\alpha _{{2}}\lambda _{{1} }-\gamma _{{2}}\lambda _{{3}}} \ \ }+x\alpha _{{2}}-y\alpha \gamma _{{2}}+z\gamma _{{2}}} \ \ }+\sigma _{{1}}\sigma _{{2}}{\mathrm{e}^{- \ \ {\frac{t\gamma _{{1}}\left( \alpha \gamma _{{1}}\omega _{{2}}-\alpha _{{1}}\omega _{{1}}-\omega _{{3} }\gamma _{{1}}\right) }{\alpha \gamma _{{1}}\lambda _{{2}}-\alpha _{{1} }\lambda _{{1}}-\gamma _{{1}}\lambda _{{3}}} \ \ }- \ \ {\frac{t\gamma _{{2}}\left( \alpha \gamma _{{2}}\omega _{{2}}-\alpha _{{2}}\omega _{{1}}-\gamma _{{2} }\omega _{{3}}\right) }{\alpha \gamma _{{2}}\lambda _{{2}}-\alpha _{{2} }\lambda _{{1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+x\alpha _{{1}}+x\alpha _{{2} }-y\alpha \gamma _{{1}}-y\alpha \gamma _{{2}}+z\gamma _{{1}}+z\gamma _{{2}}} } \right) . $

    And finally, the resulting two-wave solution for Case VIII will be read as

    $ Ψ3(x,y,z,t)=2(ρ1e12  tγ2(αγ2ω2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3  +12xα212yαγ2+12zγ2  +ρ2e  tγ2(αγ2ω2α2ω1γ2ω3)αγ2λ2α2λ1γ2λ3  +xα2yαγ2+zγ2  +
    $
    (3.19)
    $ \left. \left. \rho _{{1}}\rho _{{2}}\rho _{{12}}{\ \mathrm{e}^{-\frac{3}{2}{ \frac{t\gamma _{{2}}\left( \alpha \, \gamma _{{2}}\omega _{{2}}-\alpha _{{2} }\omega _{{1}}-\gamma _{{2}}\omega _{{3}}\right) }{\alpha \, \gamma _{{2} }\lambda _{{2}}-\alpha _{{2}}\lambda _{{1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+ \frac{3}{2}x\alpha _{{2}}-\frac{3}{2}y\alpha \, \gamma _{{2}}+\frac{3}{2} z\gamma _{{2}}} \ \ }\right) \right/ \left( 1+\sigma _{{1}}{\mathrm{e}^{-\frac{1}{ 2} \ \ {\frac{t\gamma _{{2}}\left( \alpha \, \gamma _{{2}}\omega _{{2}}-\alpha _{{2 }}\omega _{{1}}-\gamma _{{2}}\omega _{{3}}\right) }{\alpha \, \gamma _{{2} }\lambda _{{2}}-\alpha _{{2}}\lambda _{{1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+ \frac{1}{2}x\alpha _{{2}}-\frac{1}{2}y\alpha \gamma _{{2}}+\frac{1}{2} z\gamma _{{2}}} \ \ }\right. $
    $ \left. +\sigma _{{2}}{\mathrm{e}^{- \ \ {\frac{t\gamma _{{2}}\left( \alpha \, \gamma _{{2}}\omega _{{2}}-\alpha _{{2}}\omega _{{1}}-\gamma _{{2}}\omega _{{3}}\right) }{\alpha \, \gamma _{{2}}\lambda _{{2}}-\alpha _{{2}}\lambda _{{ 1}}-\gamma _{{2}}\lambda _{{3}}} \ \ }+x\alpha _{{2}}-y\alpha \, \gamma _{{2} }+z\gamma _{{2}}} \ \ }\right) . $

    The triple-wave function of the solution will be reduced as below form

    $ Ψ=Δ3Ω3,
    $
    (3.20)
    $ Ω3=1+ρ1eΛ1+ρ2eΛ2+ρ3eΛ3+ρ1ρ2ρ12eΛ1+Λ2+ρ1ρ3ρ13eΛ1+Λ3+ρ2ρ3ρ23eΛ2+Λ3+ρ1ρ2ρ3ρ12ρ13ρ23eΛ1+Λ2+Λ3,  
    $
    (3.21)
    $ \Delta _{3} = 2\sigma _{1}e^{\Lambda _{1}}+2\sigma _{2}e^{\Lambda _{2}}+2\sigma _{1}\sigma _{2}\sigma _{12}e^{\Lambda _{1}+\Lambda _{2}}+2\sigma _{1}\sigma _{3}\sigma _{13}e^{\Lambda _{1}+\Lambda _{3}}+2\sigma _{2}\sigma _{3}\sigma _{23}e^{\Lambda _{2}+\Lambda _{3}}+ $
    $ 2\sigma _{1}\sigma _{2}\sigma _{3}\sigma _{12}\sigma _{13}\sigma _{23}e^{\Lambda _{1}+\Lambda _{2}+\Lambda _{3}}, $

    in which $ \Lambda _{i} = \alpha _{i}x+\beta _{i}y+\gamma _{i}x-\delta _{i}t, i = 1, 2, 3 $. Inserting (3.20) in terms of (3.21) into Eq (1.8), the below case will be reached:

    $ αi=αi,  γi=γi,  σi=σi,  ρ1=0,  ρ2=ρ2,  ρ3=ρ3,  βi=αγi,  δi=  γi(αγiω 2αiω1ω3γi)αγiλ2αiλ1γiλ3,  i=1,2,3,  ρij=ρij,  σij=1,  i,j=1,2,3, ij.
    $
    (3.22)

    Then, the solution is

    $ Ψ1=2σ1eΛ1+2σ2eΛ2+2σ1σ2eΛ1+Λ2+2σ1σ3eΛ1+Λ3+2σ2σ3eΛ2+Λ3+2σ1σ2σ3eΛ1+Λ2+Λ3  1+ρ2eΛ2+ρ3eΛ3+ρ2ρ3ρ23eΛ2+Λ3  ,
    $
    (3.23)

    in which $ \Lambda _{i} = \alpha _{i}x-\alpha \, \gamma _{{i}}y+\gamma _{i}x-{ \frac{\gamma _{{i}}\left(\alpha \, \gamma _{{i}}\omega _{{\ 2}}-\alpha _{{i} }\omega _{{1}}-\omega _{{3}}\gamma _{{i}}\right) }{\alpha \, \gamma _{{i} }\lambda _{{2}}-\alpha _{{i}}\lambda _{{1}}-\gamma _{{i}}\lambda _{{3}}} \ \ } t, i = 1, 2, 3 $ and $ \Psi = \Psi (x, y, z, t) $.

    The triangular periodic waves for Eq (1.8) can be assumed as below:

    $ f=exp(τ1)+a16exp(τ1)+cosh(τ2)+cos(τ3)+a17,   τ1=4i=1aixi+a5,   τ2=9i=6aixi5+a10,   
    $
    (4.1)
    $ τ3=14i=11aixi10+a15,  (x1,x2,x3,x4)=(x,t)=(x,y,z,t),   Ψ(x,t)=v0+2ln(f)x,
    $
    (4.2)

    in which $ a_{i}, i = 1, ..., 17 $ are unfound values. Substituting (4.1) and (4.2) into Eq (1.8) the below consequences will be gained:

    Case I:

    $ f=ea4t+a1x  y(a3a14ω3a4a12ω2a4a13ω3)a14ω2  +a3z+a5  +cosh(a9ty(a8a14ω3a9a12ω2a 9a13ω3)a14ω2  +a8z+a 10)+cos(ta14+ya12+za13+a15).
    $
    (4.3)

    Appending (4.3) into (4.1) and (4.2), the soliton-periodic wave solution of Eq (1.8) as below will be achieved:

    $ Ψ1=v0+2a1ea4t+a1x  y(a3a14ω3a4a12ω2a4a13ω3)a14ω2  +a3z+a5  f.
    $
    (4.4)

    By selecting the suitable values of parameters including

    $ a1=1,a3=1.5,a4=2,a5=1.5,a8=2,a9=1.5,a10=1,a12=2,a13=2.5,a14=1,a15=3.2,ω2=1.5,ω3=1.2,
    $

    the graphical display of soliton-periodic wave solution is offered in Figure 1 such as 3D plot and density plot.

    Figure 1.  The soliton-periodic solution (4.4) at (f1, f2) $ x = -2, y = -2 $, (f3, f4) $ x = 0, y = 0 $, and (f5, f6) $ x = 2, y = 2 $.

    Case II:

    $ f=ea4t+a1x+ya2  (a1  3a 14a4a13ω2)za14ω2  +a5  +cosh(a9t+ya7+  a9a13za14  +a10)+cos(ta14+ya12+za13+a 15).
    $
    (4.5)

    Plugging (4.5) into (4.1) and (4.2), obtain a soliton-periodic wave solution of Eq (1.8) as below case:

    $ Ψ2=v0+2a1ea4t+a1x+ya2  (a13a14a4a13ω2)za14ω2  +a5  f.
    $
    (4.6)

    Case III:

    $ f=ea4t+a1x  y(a3ω3+a4λ3)ω2  +a3z+a5  +cosh(a9t  y(a8ω3+a9λ3)ω2  +a8z+a10)+cos(  a13ω3yω2  +za13+a15).
    $
    (4.7)

    Incorporating (4.7) into (4.1) and (4.2), the soliton-periodic wave solution of Eq (1.8) will be gained as below:

    $ Ψ3=v0+2a1ea4t+a1x  y(a3ω3+a4λ3)ω2  +a3z+a5  f.
    $
    (4.8)

    Case IV:

    $ f=ea4t  y(a3a9ω3a4a7ω2a4a8ω3)a9ω2  +a3z+a5  +cosh(a9t+ya7+a8z+a10)+cos(xa11+a15).
    $
    (4.9)

    Plugging (4.9) into (4.1) and (4.2), the soliton-periodic wave solution of Eq (1.8) will be achieved as below:

    $ Ψ4=v02sin(xa11+a15)a11f.
    $
    (4.10)

    Case V:

    $ f=ea4t+ya2+  a4a8za9  +a5+cosh(a9t+ya7+a8z+a10)+cos(xa11  (a112ω3+ω 1ω2)a11yω22  +  a 11  3zω2  +a15).
    $
    (4.11)

    Incorporating (4.11) into (4.1) and (4.2), we capture a soliton-periodic wave solution of Eq (1.8) as below:

    $ Ψ5=v02sin(xa11  (a11  2ω3+ω1ω2)a11yω2  2  + a113zω2  +a15)a11f.
    $
    (4.12)

    Case VI:

    $ f=ea4t+  ω2(a3a9a4a8)xa9a112  13  y(2αa9a116ω2+2a9a11  6ω3+3αa32a9ω2  33αa3a4a8ω23+2a9a114ω1ω2)ω23(a3a9a4a8)  +a3z+a5+cosh(a9t13  yΩa4ω2  3(a3a9a4a8)(a92a116+ω22(a3a9a4a8)2)+a8z+a10)+cos(xa11  αa113yω2  +a113zω2  +a15),
    $
    (4.13)
    $ Ω=3αa4a8ω25(a3a9a4a8)3+αa92a116ω23(a3a9+2a4a8)(a3a9a4a8)a92a114ω22(a3a9a4a8)2(a112ω3+ω1ω2)a92a1112(3a42a9  2)(αω2ω3).
    $

    Appending (4.13) into (4.1) and (4.2), the soliton-periodic wave solution of Eq (1.8) will be obtained as below:

    $ Ψ5=v0+2f[  ω2(a3a9a4a8)a9a112ea4t+  ω2(a3a9a4a8)xa9a112  13  y(2αa9a116ω2+2a9a116ω3+3αa32a9ω233αa3a4a8ω23+2a9a114ω1ω2)ω23(a3a9a4a8)+a3z+a5  +sin(xa11+  αa113yω2    a113zω2  a15)a11].
    $
    (4.14)

    By selecting suitable values of parameters including

    $ α=0.5,a3=1,a4=1.5,a5=2,a8=2,a9=1.5,a10=1,a11=2,a13=2.5,a14=1,a15=3.2,ω1=1.5,ω2=1.2,ω3=1.5,
    $

    the graphical display of soliton-periodic wave solution is offered in Figure 2 such as 3D chart and density chart.

    Figure 2.  The soliton-periodic solution (4.14) at (f1, f2) $ x = -2, y = -2 $, (f3, f4) $ x = 0, y = 0 $, and (f5, f6) $ x = 2, y = 2 $.

    Case VII:

    $ f=e  Ωa9ta113  +a1x16  y(6a14a115ω33a14a113ω1ω2+2a117ω1ω2+6Ωa1a8a112ω2ω3+3Ωa1a8ω1ω2  2)a115ω22a1  +  (a13a113+Ωa8ω2)za11  3ω2  +a5  +
    $
    (4.15)
    $ \cosh \left( a_{{9}}t+ya_{{7}}+a_{{8}}z+a_{{10}}\right) +\cos \left( xa_{{11} }-\frac{1}{2}\, \ \ {\frac{\left( 2\, {a_{{11}}} ^{2}\omega _{{3}}+\omega _{{1} }\omega _{{2}}\right) a_{{11}}y}{{\omega _{{2}}} ^{2}}} \ \ + \ \ {\frac{{a_{{11}}} ^{3}z }{\omega _{{2}}} \ \ }+a_{{15}}\right) , $
    $ \Omega = \sqrt{-{a_{{1}}} ^{6}-2\, {a_{{1}}} ^{4}{a_{{11}}} ^{2}+{a_{{11}}} ^{6}}. $

    Incorporating (4.15) into (4.1) and (4.2), the soliton-periodic wave solution of Eq (1.8) will be received as below:

    $ Ψ5=v0+2f[a1e  Ωa9ta113  +a1x16  y(6a14a115ω33a14a11  3ω1ω2+2a117ω1ω2+6Ωa1a8a112ω2ω3+3Ωa1a8ω1ω22)a115ω22a1  +  (a13a11  3+Ωa8ω2)za113ω2  +a5  sin(xa1112  (2a11  2ω3+ω1ω2)a11yω22  +  a113zω2  +a15)a11].
    $
    (4.16)

    By selecting the specific amounts of parameters including

    $ α=0.5,a1=1,a4=a9=ω1=ω3=1.5,,a5=2,a7=1.5,a8=2,a10=1,a11=2,a13=2.5,a14=1,a15=3.2,ω2=1.2,
    $

    the graphical display of soliton-periodic wave solution is offered in Figure 3 such as 3D chart, density chart, and 2D chart and below cases:

    $ (f3)\ y = 1, 2, 3, \ \ (f6)\ y = 1, 2, 3, \ and\ \ (f9)\ y = -1, -2, -3. $
    Figure 3.  The soliton-periodic solution (4.16) at (f1, f2) $ z = -2, t = -2 $, (f4, f5) $ z = 0, t = 0 $, and (f7, f8) $ z = 2, t = 2 $.

    Three function containing exponential, hyperbolic, and triangular periodic waves for Eq (1.8) can be assumed as the following:

    $ f=exp(τ1)+a16exp(τ1)+sinh(τ2)+sin(τ3)+a17,   τ1=4i=1aixi+a5,   τ2=9i=6aixi5+a10,
    $
    (5.1)
    $ τ3=14i=11aixi10+a15,  (x1,x2,x3,x4)=(x,t)=(x,y,z,t),   Ψ(x,t)=v0+2ln(f)x,
    $
    (5.2)

    in which $ a_{i}, i = 1, ..., 17 $ are unfound values. Substituting (5.2) into Eq (1.8) the below consequences will be gained:

    Case I:

    $ f=ea4t+a1x  (a3a14ω3a4a12ω2a4a13ω3)ya14ω2  +a3z+a5  +sinh(a9t  (a8a14ω3a9a12ω2a9a13ω3)ya14ω2  +a8z+a10)
    $
    (5.3)
    $ +\sin \left( ta_{{14}}+ya_{{12}}+za_{{13}}+a_{{15}}\right) . $

    Substituting (5.3) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be gained as the following:

    $ Ψ1=v0+2a1ea4t+a1x  (a3a14ω3a4a12ω2a4a13ω3)ya14ω2  +a3z+a5  f.
    $
    (5.4)

    By selecting the suitable values of parameters including

    $ a1=1,a3=3,a4=2,a5=1.5,a8=1.7,a9=1.5,a10=1.5,a12=2.5,a13=1.1,a14=2.1,a15=3.2,ω2=1,ω3=1.5,
    $

    the graphical representation of cross-kink wave solution is offered in Figure 4 such 3D plot and density plot.

    Figure 4.  The cross-kink wave solution (5.4) at (f1, f2) $ x = 3, y = 2 $, (f3, f4) $ x = 0, y = 2 $, and (f5, f6) $ x = -3, y = 2 $.

    Case II:

    $ f=ea4t+a1x+a2y  (a1  3a14a4a13ω2)za14ω2  +a5  +sinh(a9t+a7y+  a9a13za14  +a10)+sin(ta14+ya12+za13+a15).
    $
    (5.5)

    Putting (5.5) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be received as the following:

    $ Ψ2=v0+2a1ea4t+a1x+a2y(a13a14a4a13ω2)za14ω2  +a5  f.
    $
    (5.6)

    By selecting the suitable values of parameters including

    $ a1=1,a3=3,a4=2,a5=1.5,a8=1.7,a9=1.5,a10=1.5,a12=2.5,a13=1.1,a14=2.1,a15=3.2,ω2=1,ω3=1.5,
    $

    the graphical exhibition of cross-kink solution is offered in Figure 5 such as 3D chart and density chart.

    Figure 5.  The cross-kink wave solution (5.6) at (f1, f2) $ x = 3, y = 2 $, (f3, f4) $ x = 0, y = 2 $, and (f5, f6) $ x = -3, y = 2 $.

    Case III:

    $ f=ea4t+a1x  (a3ω3+a4λ3)yω2  +a3z+a5  +sinh(a9t  (a8ω3+a9λ 3)yω2  +a8z+a10)+sin(  a13ω3yω2  +a13z+a15).
    $
    (5.7)

    Plugging (5.7) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be obtained as the following:

    $ Ψ3=v0+2a1ea4t+a1x  (a3ω3+a4λ3)yω2  +a3z+a5  f.
    $
    (5.8)

    Case IV:

    $ f=e  a9(a112(ω3a12ω1ω2)(a1  3+a3ω2)a15ω1ω2)ta12a7a112ω22  +a1x  a3(ω3a12ω1ω2)ya12ω2  +a3z+a5  +sin(xa11+a15)+sinh(a9t+a7y  (a13+a3ω2)a12a7a112ω2za11  2(ω3a12ω1ω2)(a13+a3ω2)a15ω1ω2  +a10).
    $
    (5.9)

    Inserting (5.9) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be gained as the following:

    $ Ψ4=v0+2a1e  a9(a11  2(ω3a12ω1ω2)(a13+a3ω2)a15ω1ω2)ta12a7a112ω2  2  +a1x  a3(ω3a12ω1ω2)ya12ω2  +a3z+a5  +cos(xa11+a15)a11  f.
    $
    (5.10)

    Case V:

    $ f=ea4t+a2y+  a4a8za9  +a5  +sinh(a9t+a7y+a8z+a10)+sin(xa11  a11(a112ω3+ω1ω2)yω22  +  a113zω2  +a15).
    $
    (5.11)

    Substituting (5.11) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be received as the following:

    $ Ψ5=v0+2cos(xa11  a11(a112ω3+ω1ω2)yω22+  a113zω2  +a15)a11  f.
    $
    (5.12)

    By selecting the suitable values of parameters including

    $ a1=1,a3=3,a4=2,a5=1.5,a8=1.7,a9=1.5,a10=1.5,a12=2.5,a13=1.1,a14=2.1,a15=3.2,ω2=1,ω3=1.5,
    $

    the graphical exhibition of cross-kink solution is offered in Figure 6 such as 3D chart and density chart.

    Figure 6.  The cross-kink wave solution (5.6) at (f1, f2) $ x = 3, y = 2 $, (f3, f4) $ x = 0, y = 2 $, and (f5, f6) $ x = -3, y = 2 $.

    Case VI:

    $ f=e  Ωa9ta113  +a1x112  (3Ωa1a8ω2(4a112ω3+3ω1ω2)12a14a115ω3a113ω1ω2(9a142a112a122a11  4))ya115ω22a1  +  (a13a113+Ωa8ω2)zω2a113  +a5  sinh(a9t+a7y+a8z+a10)+sin(xa11112  a11(12a112ω3+7ω1ω2)yω22  +  a311zω2+a15),
    $
    (5.13)
    $ \Omega = \sqrt{-3\, {a_{{1}}} ^{6}-4\, {a_{{1}}} ^{4}{a_{{11}}} ^{2}+{a_{{11}}} ^{6} }. $

    Putting (5.13) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be concluded as the following:

    $ Ψ5=v0+2f[a1e  Ωa9ta113  +a1x112  (3Ωa1a8ω2(4a112ω3+3ω1ω2)12a14a11  5ω3a113ω1ω2(9a1  42a112a122a114))ya115ω22a1  +  (a13a11  3+Ωa8ω2)zω2a113  +a5  +cos(xa11112  a11(12a112ω3+7ω1ω2)yω2  2  +  a113zω2  +a15)a11].
    $
    (5.14)

    By selecting suitable values of parameters including

    $ α=0.5,a3=1,a4=a9=ω1=ω3=1.5,a5=2,a8=2,a10=1,a11=2,a13=2.5,a14=1,a15=3.2,ω2=1.2,
    $

    the graphical representation of cross-kink solution is offered in Figure 7 such as 3D chart and density chart.

    Figure 7.  The cross-kink wave solution (5.14) at (f1, f2) $ x = -2, y = -2 $, (f3, f4) $ x = 0, y = 0 $, and (f5, f6) $ x = 2, y = 2 $.

    Case VII:

    $ f=e  Ωa9ta113  +a1x16 y(6a14a115ω33a14a113ω1ω2+2a11  7ω1ω2+6Ωa1a8a11  2ω2ω3+3Ωa1a8ω1 ω22)a115ω22a1  + (a13a113+Ωa8ω2)za113ω2  +a5  +cosh(a9t+ya7+a8z+a10)+cos(xa 1112  (2a112ω3+ω1ω2)a11yω22  +  a113zω2  +a15),
    $
    (5.15)
    $ \Omega = \sqrt{-{a_{{1}}} ^{6}-2\, {a_{{1}}} ^{4}{a_{{11}}} ^{2}+{a_{{11}}} ^{6}}. $

    Inserting (5.15) into (5.1) and (5.2), the cross-kink solution of Eq (1.8) will be gained as the following:

    $ Ψ5=v0+2f[a1e  Ωa9ta113  +a1x16  y(6a14a115ω33a14a11  3ω1ω2+2a117ω1ω2+6Ωa1a8a112ω2ω3+3Ωa1a8ω1ω22)a115ω22a1  +  (a13a11  3+Ωa8ω2)za113ω2  +a5  sin(xa1112  (2a11  2ω3+ω1ω2)a11yω22  +  a113zω2  +a15)a11].
    $
    (5.16)

    By selecting suitable values of parameters including

    $ \alpha = 0.5, a_1 = 1, a_5 = 1.5, a_8 = 2, a_9 = 0.5, a_{10} = 1.5, a_{15} = 3.2, \omega_1 = 1.5, \omega_2 = 1.2, \omega_3 = 1.5, $

    the graphical representation of cross-kink solution is offered in Figure 8 such as 3D chart and density chart.

    Figure 8.  The lump-periodic solution (5.16) at (f1, f2) $ z = -2 $, (f4, f5) $ z = 0 $, and (f7, f8) $ z = 2 $.

    Three function containing exponential, hyperbolic, and triangular periodic waves for Eq (1.8) can be assumed as the following:

    $ f=exp(τ1)+a16exp(τ1)+tanh(τ2)+tan(τ3)+a17,   τ1=4i=1aixi+a5,   τ2=9i=6aixi5+a10,   
    $
    (6.1)
    $ τ3=14i=11aixi10+a15,  (x1,x2,x3,x4)=(x,t)=(x,y,z,t),   Ψ(x,t)=v0+2ln(f)x,
    $
    (6.2)

    in which $ a_{i}, i = 1, ..., 17 $ are unfound values. Substituting (6.2) into Eq (1.8) the below consequences will be gained:

    Case I:

    $ f=ea4t+a1x+a2y  (a1  3+a4λ2)zω2  +a5  +tanh(ya7+a10)+tan(ya12+a15)+a17.
    $
    (6.3)

    Substituting (6.3) into (6.1) and (6.2), we can capture a solitary wave solution of Eq (1.8) as the following:

    $ Ψ1=v0+2a1ea4t+a1x+a2y(a13+a4λ2)zω2  +a5  f.
    $
    (6.4)

    Case II:

    $ f=ea1x+a2y  a13zω2  +a5  +tanh(ya7+a10)+tan(ya12+a15)+a17.
    $
    (6.5)

    Inserting (6.5) into (6.1) and (6.2), we can capture a solitary wave solution of Eq (1.8) as below:

    $ Ψ2=v0+2a1ea1x+a2y  a13zω2  +a5  f.
    $
    (6.6)

    Case III:

    $ f=ea4t+a1x+a2y  (a1  3ω3+a4λ3ω2)zω2ω3  +a5  +tanh(ta9+ya7  za9λ3  ω3  +a10)+tan(ya12+a15)+a17.
    $
    (6.7)

    Putting (6.7) into (6.1) and (6.2), the solitary wave solution of Eq (1.8) can be indicated as below:

    $ Ψ3=v0+2a1ea4t+a1x+a2y(a13ω3+a4λ3ω2)zω2ω3  +a5  f.
    $
    (6.8)

    Case IV:

    $ f=ea4t+a1x+a2y+za3+a5  tanh(  ya8ω3  ω2  za8a10)tan(  ya13ω3  ω2  a13za15)+a17.
    $
    (6.9)

    Putting (6.9) into (6.1) and (6.2), the solitary wave of Eq (1.8) can be stated as the following:

    $ Ψ4=v0+2a1ea4t+a1x+a2y+za3+a5  f.
    $
    (6.10)

    By selecting the suitable values of parameters including

    $ a1=1,a2=1.5,a3=2,a4=2,a5=1.5,a8=2,a9=1.5,a10=1.5,a13=2,a13=1.1,a15=3.2,a17=2,ω2=1.5,
    $
    $ \omega _{3} = 1.2, \lambda _{2} = 1, \lambda _{3} = 1.5, x = -2, t = 2, $

    with the following components

    $ α=  a13ω3a1ω1ω2a2ω22a3ω2ω3+a4λ2ω3a4λ3ω2  ω2a13  ,  λ1=  a13a2ω2+a13a3ω3+a2a4λ2ω2+a3a4λ2ω3  a1a4ω2,
    $

    the graphical representation of rational solitary solution is offered in Figure 9 such as 3D chart and density chart.

    Figure 9.  The rational solitary wave solution (6.10) at (f1) $ z = 0 $, (f2) $ z = 1 $, and (f3) $ z = 3 $.

    Case V:

    $  f=ea1x  a3ω3yω2  +za3+a5  tanh(  ya8ω3  ω2za8a10)tan(  ya13ω3  ω2  a13za15)+a17.
    $
    (6.11)

    Incorporating (6.11) into (6.1) and (6.2), the solitary solution of Eq (1.8) will be gained as below:

    $ Ψ5=v0+2a1ea1x  a3ω3yω2  +za3+a5  f,  α=  a12ω3ω1ω2  a12ω2  .
    $
    (6.12)

    Case VI:

    $ f=ea4t+a1x  (a3ω3+a4λ3)yω2  +za3+a5  +tanh(ta9  y(a8ω3+a9λ3)ω2  +za8+a10)tan( ya13ω3  ω2  a13za15)+a17.
    $
    (6.13)

    Appending (6.13) into (6.1) and (6.2), the solitary solution of Eq (1.8) can be written as the following:

    $ Ψ6=v0+2a1ea4t+a1x  (a3ω 3+a4λ3)yω2  +za3+a5  f,  α=  a12ω3ω1ω2  a12ω2  .
    $
    (6.14)

    Case VII:

    $ f=ea4t+a1x  a3ω3yω2  +za3+a5  +tanh(ta9  ya8ω3  ω2  +za8+a10)tan(  ya13ω3  ω2  a13za15)+a17.
    $
    (6.15)

    Appending (6.15) into (6.1) and (6.2), the solitary solution of Eq (1.8) will be obtained as below:

    $ Ψ7=v0+2a1ea4t+a1x  a3ω3yω2  +za3+a5  f,  α=  a12ω3ω1ω2  a1  2ω2,  λ2=  λ3ω2  ω3  .
    $
    (6.16)

    By choosing the specific amounts of parameters including

    $ a1=1,a2=1.5,a3=2,a4=2,a5=1.5,a8=2,a9=1.3,a10=1.5,a13=2,a13=2,a15=3.2,a17=2,ω2=1.5,ω3=1.2,λ2=1,λ3=1.5,x=2,t=2,
    $

    the graphical representation of rational solitary solution is offered in Figure 10 such as 3D chart and density chart.

    Figure 10.  The rational solitary wave solution (6.16) at (f1) $ z = 0 $, (f2) $ z = 1 $, and (f3) $ z = 3 $.

    Case VIII:

    $ f=ea4t+a1x+a2y  (a1  3a 14a4a13ω2)za14ω2  +a5  +tanh(ta9+ya7+  za9a13  a14  +a10)+tan(ta14+ya12+a13z+a 15)+a17.
    $
    (6.17)

    Incorporating (6.17) into (6.1) and (6.2), the solitary solution of Eq (1.8) will be received as below:

    $ Ψ8=v0+2a1ea4t+a1x+a2y(a13a14a4a13ω2)za14ω2  +a5  f,  α=  a1  2ω3ω1ω2  a12ω2,  λ2=  a13ω2  a14,  λ3=a13ω3  a14  .
    $
    (6.18)

    Case IX:

    $ f=ea4t+a1x  (a3a14ω3a4a12ω2a4a13ω3)ya14ω2  +a3z+a5  
    $
    (6.19)
    $ +tanh(ta9  y(a8a14ω3a9a12ω2a 9a13ω3)a14ω2  +za8+a 10)
    $
    (6.20)
    $ +tan(ta14+ya12+za13+a15)+a17.
    $
    (6.21)

    Appending (6.21) into (6.1) and (6.2), the solitary solution of Eq (1.8) can be reached as below:

    $ Ψ9=v0+2a1ea4t+a1x  (a3a14ω3a4a12ω2a4a13ω3)ya14ω2  +a3z+a5  f, α=  a12ω3ω1ω2  a12ω2, 
    $
    (6.22)
    $ \lambda _{{1}} = - \ \ {\frac{{a_{{1}}} ^{2}\left( a_{{12}}\omega _{{2}}+a_{{\ 13} }\omega _{{3}}\right) }{a_{{14}}\omega _{{2}}} }, \ \lambda _{{3}} = - \ \ {\frac{a_{{ 12}}\omega _{{2}}+a_{{13}}\omega _{{3}}} \ \ {a_{{\ 14}}} \ \ }. $

    Case X:

    $ f=ea4t+a1x  a3ω3yω2+a3z+a5  +tanh(  ya8ω3ω2  +za8+a10)+tan(ta14 ya13ω3  ω2  +za13+a15)+a17.
    $
    (6.23)

    Inserting (6.21) into (6.1) and (6.2), the solitary solution of Eq (1.8) will be gained as below form:

    $ Ψ10=v0+2a1ea4t+a1x  a3ω3yω2  +a3z+a5  f, α=a12ω3ω1ω2  a1  2ω2  , λ2=  λ3ω2  ω3  .
    $
    (6.24)

    By choosing the specific amounts of parameters including

    $ a1=1,a3=1.5,a4=2,a5=1.5,a7=2,a8=2,a9=2.1,a10=1.5,a13=2,a13=2,a14=2.5,a15=3.2,a17=2,ω2=1.5,ω3=1.2,λ3=1.5,x=2,t=2,
    $

    the graphical representation of rational solitary solution is offered in Figure 11 such as 3D chart and density chart.

    Figure 11.  The rational solitary wave solution (6.24) at (f1) $ z = -2 $, (f2) $ z = 0 $, and (f3) $ z = 2 $.

    Case XI:

    $ f=ea4t+a1x+a2y  (a1  3a 14a4a13ω2)za14ω2  +a5  +tanh(  ta14a8  a13  +ya7+za8+a10)+tan(ta14+za13+a15)+a17.
    $
    (6.25)

    Inserting (6.25) into (6.1) and (6.2), the solitary solution of Eq (1.8) can be stated as below case:

    $ Ψ11=v0+2a1ea4t+a1x+a2y(a13a14a4a13ω2)za14ω2  +a5  f, α=  a1  3a2+a1a3ω1+a2a3ω2+a3  2ω3  a13a3,
    $
    (6.26)
    $ λ1=  a13a2a12a13a1  3a3a122+a2a3a12a13ω2+a2a3a132ω3+a2a3a13a14λ3  a3a12a1a14  +a32a122ω2a32a12a13ω3a32a12a14λ3  a3a12a1a14  ,
    $
    $ \lambda _{{2}} = - \ \ {\frac{a_{{13}}\left( a_{{12}}\omega _{{2}}+a_{{13}}\omega _{ {3}}+a_{{14}}\lambda _{{3}}\right) }{a_{{12}}a_{{14}}} \ \ }. $

    Case XII:

    $ f=ea4t+a1x  a3ω3yω2  +a3z+a5  +tanh(ta9+ya7  za7ω2  ω3  +a10)+tan(ta14 ya13ω3  ω2  +za13+a15)+a17.
    $
    (6.27)

    Inserting (6.27) into (6.1) and (6.2), the solitary solution of Eq (1.8) will be gained as below form:

    $ Ψ12=v0+2a1ea4t+a1x  a3ω3yω2  +a3z+a5  f, α=a13a2+a1a3ω1+a2a3ω2+a32ω3  a13a3,
    $
    (6.28)
    $ λ1=  a13a2a12a13a1  3a3a122+a2a3a12a13ω2+a2a3a132ω3+a2a3a13a14λ3  a3a12a1a14  +a32a122ω2a32a12a13ω3a32a12a14λ3  a3a12a1a14  ,
    $
    $ \lambda _{{2}} = - \ \ {\frac{a_{{13}}\left( a_{{12}}\omega _{{2}}+a_{{13}}\omega _{ {3}}+a_{{14}}\lambda _{{3}}\right) }{a_{{12}}a_{{14}}} \ \ }. $

    In the current section, we will analyze the continuous modulational instability of the nonlinear generalized KP equation. In addition, the feasibility of the localized waves in the present system is certified by linear stability analysis. First, we search the perturbed solution for the giving Eq (1.8) of the form

    $ Ψ(x,y,z,t)=ζ+δ Θ,
    $
    (7.1)

    where $ \Theta = \Theta(x, y, z, t) $ and $ \zeta $ is a steady state solution. Inserting (7.1) into Eq (1.8), become

    $ δ  4x3yΘ+αδ  4x3zΘ+3δ2(  2x2  Θ) yΘ+3δ2(xΘ)  2xyΘ+3αδ2(  2x2  Θ)  zΘ+
    $
    (7.2)
    $ 3\, \alpha\, {\delta}^{2} \left( \ \ {\frac {\partial }{\partial x}}\Theta \right) {\ \frac {\partial ^{2}}{\partial x\partial z}}\Theta + \lambda_{{1}}\delta\, \ \ {\frac {\partial ^{2}}{\partial t\partial x}}\Theta +\lambda_{{2}}\delta\, { \frac {\partial ^{2}}{ \partial t\partial y}}\Theta + \lambda_{{3}}\delta\, { \ \frac {\partial ^{2}}{\partial t\partial z}}\Theta + \omega_{{1}}\delta\, { \frac {\partial ^{2}}{\partial x\partial z}}\Theta +\omega_{{2}}\delta\, { \frac {\partial ^{2}}{ \partial y\partial z}}\Theta +\omega_{{3}}\delta\, {\ \frac {\partial ^{2}}{\partial {z}^{2}}} \ \ \Theta = 0, $

    by linerization Eq (7.2), one gets

    $ δ  4x3yΘ+αδ  4x3zΘ+λ1δ  2txΘ+λ2δ  2tyΘ+λ3δ  2tzΘ+ω1δ  2xzΘ+ω2δ  2yzΘ+ω3δ  2z2  Θ=0.
    $
    (7.3)

    Theorem 7.1. Assume that the solution of Eq (7.3) has the following case as

    $ Θ(x,y,z,t)=ρ1 ei(Mx+Ny+Pz+Bt),
    $
    (7.4)

    in which $ M, N, P $ are the normalized wave numbers, by plugging (7.4) into Eq (7.3), separation the coefficients of $ e^{i(Mx+Ny+Pz+Wt)} $ one gets

    $ B(M,N,P)=  M3Pα+M3NMPω1NPω2P2ω3  Mλ1+Nλ2+Pλ3  .
    $
    (7.5)

    Proof. By putting (7.4) into (7.3), becomes

    $ δ  4x3y¯Θ+αδ  4x3z¯Θ+λ1δ  2tx¯Θ+λ2δ  2ty¯Φ+λ3δ  2tz¯Θ+ω1δ  2xz¯Θ+ω2δ  2yz¯Θ+ω3δ  2z2  ¯Θ
    $
    (7.6)
    $ = {\mathrm{e}^{i \left( Mx+Ny+Pz+Bt \right) }}\delta\, \rho_{{1}} \left( {M} ^{3}P\alpha+{M}^{3}N-MP\omega_{{1}}-MB\lambda_{{1}}-NP\omega_{{2}}-NB \lambda_{{2}}-{P}^{2}\omega_{{3}}-PB\lambda_{{3}} \right), $

    in which $ \overline{\Theta} = \Theta(x, y, z, t) $. By solving and simplifying we can determine the function of $ B(M, N, P) $ as the following

    $ B(M,N,P)=  M3Pα+M3NMPω1NPω2P2ω3  Mλ1+Nλ2+Pλ3  .
    $
    (7.7)

    Accordingly, the considered solution was obtained. Thereupon the proof is perfect.

    It is easy to notice that modulation stability occurs when $ M\lambda_{{1} }+N\lambda_{{2}}+P\lambda_{{3}}\neq0 $. So, the modulation stability achieve spectrum $ \Upsilon(B) $ will be as below form:

    $ Υ(B)=  M3Pα+M3NMPω1NPω2P2ω3  Mλ1+Nλ2+Pλ3  .
    $
    (7.8)

    In Figures 12-14 can be discovered that while the sign of $ B(M, N, P) $ is positive for all quantity of $ M $. Furthermore, in Figure 12 can be observed if the $ B(M, N, P) $ is positive or negative for some quantities of $ M $. Finally, in Figure 13 and 14 can be perceived that while the sign of $ B(M, N, P) $ is positive for all quantity of $ M $.

    Figure 12.  The graphic representation $ \Upsilon(B) $ for the wave number $ M $ via considering the diverse quantities $ \lambda_1 = 1, \lambda_2 = 1.5, \lambda_3 = 2, \omega_1 = 2.2, \omega_2 = -2, \omega_3 = 3, \alpha = 0.5 $.
    Figure 13.  The graphic representation $ \Upsilon(B) $ for the wave number $ M $ via considering the diverse quantities $ \lambda_1 = 1, \lambda_2 = 1.5, \lambda_3 = 2, \omega_1 = 2.2, \omega_2 = -2, \omega_3 = 3, \alpha = 0.5 $.
    Figure 14.  The graphic representation $ \Upsilon(B) $ for the wave number $ M $ via considering the diverse quantities $ \lambda_1 = 1, \lambda_2 = 1.5, \lambda_3 = 2, \omega_1 = 2.2, \omega_2 = -2, \omega_3 = 3, \alpha = 0.5 $.

    Via employing the wave alteration $ \xi = k(x+ay+bz-ct) $ in Eq (1.8) once can gain to the below ODE as

    $ k2(αb+a)Ψ+6k(αb+a)ΨΨ+(abω2acλ2+b2ω3bcλ3+bω1cλ1)Ψ=0,
    $
    (8.1)

    in which $ \Psi = \Psi(\xi) $ and $ \Psi^{\prime } = \frac{d\Psi}{d \xi} $. According to the SIVP [16,17] and by multiplying Eq (8.1) with $ \Psi^{\prime } $ and integrating once respect to $ \xi $, the following stationary integral will be arises

    $ J=0[A1(ΨΨ12(Ψ2)+13A2(Ψ3+12A3(Ψ2]dξ,
    $
    (8.2)

    in which

    $ A_1 = k^2(\alpha b+a), \ \ \ A_2 = 6k(\alpha b+a), \ \ \ A_3 = ab\omega_{{2} }-ac\lambda_{{2}}+{b}^{2}\omega_{{3}}-bc\lambda_{{3}}+b \omega_{{1} }-c\lambda_{{1}}. $

    We utilize the solitary wave function as the following

    $ u(ξ)=δ sech(μξ).
    $
    (8.3)

    Hence, the stationary integral transforms to

    $ J=130k2δ2μ21αbk2μ221ak2μ28αbδkμ8aδkμ+5abω25acλ2+5b2ω35bcλ 3+5bω15cλ1).
    $
    (8.4)

    Based on the SIVP and using derivative $ J $ respect to $ A $ and $ B $, one get

    $ JA=115k2δμ21αbk2μ221ak2μ28αbδkμ8aδkμ+5abω25acλ2+5b2ω35bcλ3+5bω15cλ1)+130k2δ2μ(8αbkμ8akμ)=0,
    $
    (8.5)

    and

    $ JB=130k2δ221αbk2μ221a k2μ28αbδkμ8aδkμ+5abω25acλ2+5b2ω35bcλ3+5bω15cλ1)+130k2δ2μ(42αbk2μ42ak2μ8αbδk8aδk)=0.
    $
    (8.6)

    Solve the Eqs (8.5) and (8.6), become

    $ δ=±212  abω2acλ2+b2ω3bcλ3+bω1cλ1  (21αb+21a)(abω2acλ2+b2ω3bcλ3+bω1cλ1),   
    $
    (8.7)
    $ \mu = \pm \frac{1}{21} \ \ {\frac{\sqrt{-\left( 21\, \alpha \, b+21\, a\right) \left( ab\omega _{{2}}-ac\lambda _{{2}}+{b}^{2}\omega _{{3}}-bc\lambda _{{3} }+b\omega _{{1}}-c\lambda _{{1}}\right) }}{\left( \alpha \, b+a\right) k}}. $

    The condition can be obtained as below

    $ (αb+a)(abω2acλ2+b2ω3bcλ3+bω1cλ1)<0.
    $
    (8.8)

    Finally, the solitary solution received by utilizing of SIVP can be reached as

    $ Ψ(x,y,z,t)=δ sech[kμ(x+ay+bzct)].
    $
    (8.9)

    We utilize the bright wave function as the following

    $ u(ξ)=δ sech2(μξ).
    $
    (8.10)

    Hence, the stationary integral transforms to

    $ J=  2δ2μ(120αbk2μ2+120ak2μ2+35αbδkμ+35aδkμ14abω2  105+14acλ214b2ω3+14bcλ314bω1+14cλ1)105.
    $
    (8.11)

    Based on the SIVP and using derivative $ J $ respect to $ A $ and $ B $, one get

    $ JA=  4δμ(120αbk2μ2+120ak2μ2+35αbδkμ+35aδkμ14abω2+14acλ2  105+14b2ω3+14bcλ314bω1+14cλ1)2δ2μ(35αbkμ+35akμ)105=0.
    $
    (8.12)

    and

    $ JB=  2δ2(120αbk2μ2+120a k2μ2+35αbδkμ+35aδkμ14abω2+14acλ214b2ω3  105++14bcλ314bω1+14cλ1)2δ2μ(240αbk2μ+240ak2μ+35αbδk+35aδk)105=0.
    $
    (8.13)

    Solve the Eqs (8.12) and (8.13), become

    $ δ=±485  abω2acλ2+b2ω3bcλ3+bω1cλ1  (21αb+21a)(abω2acλ2+b2ω3bcλ3+bω1cλ1),   
    $
    (8.14)
    $ \mu = \pm \frac{1}{30} \ \ {\frac{\sqrt{-\left( 21\, \alpha \, b+21\, a\right) \left( ab\omega _{{2}}-ac\lambda _{{2}}+{b}^{2}\omega _{{3}}-bc\lambda _{{3} }+b\omega _{{1}}-c\lambda _{{1}}\right) }}{\left( \alpha \, b+a\right) k}}. $

    The condition of definition of the above relations can be expressed as

    $ (αb+a)(abω2acλ2+b2ω3bcλ3+bω1cλ1)<0.
    $
    (8.15)

    Finally, the solitary solution gained by utilizing of SIVP will be as

    $ Ψ(x,y,z,t)=δ sech2[kμ(x+ay+bzct)].
    $
    (8.16)

    Assume the dark soliton wave solution be as the below case as

    $ u(ξ)=δ tanh2(μξ).
    $
    (8.17)

    Hence, the stationary integral transforms to

    $ J=  2δ2μ(120αbk2μ2120ak2μ2+35αbδkμ+35aδkμ+14abω2  105+14acλ2+14b2ω314bcλ3+14bω114cλ1)105.
    $
    (8.18)

    Based on the SIVP and using derivative $ J $ respect to $ A $ and $ B $, one get

    $ Jδ=  4δμ(120αbk2μ2120ak2μ2+35αbδkμ+35aδkμ+14abω214acλ2  105++14b2ω314bcλ3+14bω114cλ1)+2δ2μ(35αbkμ+35akμ)105=0,
    $
    (8.19)

    and

    $ Jμ=  2δ2(120αbk2μ2120a k2μ2+35αbδkμ+35aδkμ+14abω214acλ2+105+14b2ω314bcλ3+14bω114cλ1)+2δ2μ(240αbk2μ240ak2μ+35αbδk+35aδk)105=0.
    $
    (8.20)

    Solve the Eqs (8.19) and (8.20), one get

    $ δ=485  abω2acλ2+b2ω3bcλ3+bω1cλ1  (21αb+21a)(abω2acλ2+b2ω3bcλ3+bω1cλ1),   
    $
    (8.21)
    $ \mu = \pm \frac{1}{30} \ \ {\frac{\sqrt{-\left( 21\, \alpha \, b+21\, a\right) \left( ab\omega _{{2}}-ac\lambda _{{2}}+{b}^{2}\omega _{{3}}-bc\lambda _{{3} }+b\omega _{{1}}-c\lambda _{{1}}\right) }}{\left( \alpha \, b+a\right) k}}. $

    The condition of definition of the above relations can be presented the following form

    $ (αb+a)(abω2acλ2+b2ω3bcλ3+bω1cλ1)<0.
    $
    (8.22)

    Finally, the dark solution acquired by utilizing of SIVP will be as

    $ Ψ(x,y,z,t)=δ tanh2[kμ(x+ay+bzct)].
    $
    (8.23)

    Let the singular soliton wave solution be as the below case as

    $ u(ξ)=δ csch2(μξ).
    $
    (8.24)

    Then, the stationary integral transforms to

    $ J=  2δ2μ(120αbk2μ2120ak2μ270αbδkμ70aδkμ+14abω2  105+14acλ2+14b2ω314bcλ3+14bω114cλ1)105.
    $
    (8.25)

    Based on the SIVP and using derivative $ J $ respect to $ A $ and $ B $, become

    $ Jδ=  4δμ(120αbk2μ2120ak2μ270αbδkμ70aδkμ+14abω214acλ2105++14b2ω314bcλ3+14bω114cλ1)2δ2μ(70αbkμ70akμ)105=0
    $
    (8.26)

    and

    $ Jμ=  2δ2(120αbk2μ2120ak2μ270αbδkμ70aδkμ+14abω214acλ2+14b2ω3  105+14bcλ3+14bω114cλ1)2δ2μ(240αbk2μ240ak2μ70αbδk70aδk)105=0.
    $
    (8.27)

    Solve the Eqs (8.26) and (8.27), one get

    $ δ=±245  abω2acλ2+b2ω3bcλ3+bω1cλ1  (21αb+21a)(abω2acλ2+b2ω3bcλ3+bω1cλ1),   
    $
    (8.28)
    $ \mu = \pm \frac{1}{30} \ \ {\frac{\sqrt{-\left( 21\, \alpha \, b+21\, a\right) \left( ab\omega _{{2}}-ac\lambda _{{2}}+{b}^{2}\omega _{{3}}-bc\lambda _{{3} }+b\omega _{{1}}-c\lambda _{{1}}\right) }}{\left( \alpha \, b+a\right) k}}. $

    The condition of definition of the above relations can be presented as the below form as

    $ (αb+a)(abω2acλ2+b2ω3bcλ3+bω1cλ1)<0.
    $
    (8.29)

    Finally, the singular solution acquired by utilizing of SIVP will be as

    $ Ψ(x,y,z,t)=δ csch2[kμ(x+ay+bzct)].
    $
    (8.30)

    In this article, the MEFM employed for searching the MSSs for the gKP equation, which contains 1-wave, 2-wave, and 3-wave solutions. The periodic wave, cross-kink, and solitary wave solutions have been obtained. In continuing, the modulation instability applied to discuss the stability of earned solutions. It is quite visible that these novel schemes have plenty of family solutions containing rational exponential, hyperbolic, and periodic functions with selecting particular parameters. Also, the semi-inverse variational principle will be used for the gKP equation. Four major cases containing the solitary, bright, dark and singular wave solutions were studied from four different ansatzes.

    By means of symbolic computation, these analytical solutions and corresponding rogue waves are obtained. Via various curve plots, density plot and three-dimensional plots, dynamical characteristics of these rouge waves are exhibited. Because of the strong nonlinear characteristic of Hirota bilinear method, the test function constructed by the Hirota operator, which can be regarded as the test function constructed by considered model. The results are beneficial to the study of the plasma, optics, acoustics, fluid dynamics and fluid mechanics. All computations in this paper have been employed quickly with the help of the Maple 18. Moreover, the method applied in this paper provides an effective tool to obtain exact solutions of nonlinear system and can in common use for other NLEEs.

    This work is supported by the Education and scientific research project for young and middle-aged teachers of Fujian Province (No. JAT.190666-No.JAT200469)

    The authors declare that there is no conflict of interests regarding the publication of this paper.

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