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01 Jan 2016
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# Symmetry Analysis and Exact Solutions of Extended Kadomtsev-Petviashvili Equation in Fluids

###### Accettato: 18 Oct 2022
Dettagli della rivista
Formato
Rivista
eISSN
2444-8656
Prima pubblicazione
01 Jan 2016
Frequenza di pubblicazione
2 volte all'anno
Lingue
Inglese
Introduction

Computing solutions of nonlinear partial differential equations (NLPDEs) is of importance to solve and understand the physical meaning of a given problem. To obtain the exact solutions of NLPDEs, researchers have come up with several methods, such as Kudryshov method [1], double reduction method [2], simplest equation method [35], homotopy analysis transform method [6], direct method [7], etc.

The Lie’s theory [8, 9], introduced by Sophus Lie (1842–1899), is used to construct a symmetry reduction, and it can also play a major role in finding the exact solutions of partial differential equations (PDEs). The application of the Lie’s method can also be extended to fields such as relativity, classical mechanics, numerical analysis and so on [1012].

Conservation laws [1318] play a significant role in finding the solutions of PDEs. Noether’s theorem helps us with an innovation formula for deriving conservation laws by appealing to the symmetries of the action.

The Kadomtsev-Petviashvili (KP) equation emerged pursuant to the efforts made when Kadomtsev and Petviashvili were learning the growth of lengthy ion-acoustic surfs with minor amplitude broadcasting in plasmas falling short of the result of lengthy diagonal worries [19]. The KP equation defines the development of non-linear, lengthy surfs of minor amplitude with sluggish requirement on the transversal coordinate. It is the simplification of the Kortewegde Vries (KdV) equation.

In this paper, we investigate the extended Kadomtsev-Petviashvili (eKP) equation [20], which reads as: $(ut+6uux+uxxx)x−uyy−α24utt+αuty+βuxt=0,$ $$\matrix{ {{{({u_t} + 6u{u_x} + {u_{xxx}})}_x} - {u_{yy}} - {{{\alpha ^2}} \over 4}{u_{tt}} + \alpha {u_{ty}} + \beta {u_{xt}} = 0,} \cr }$$

where α and β are system parameters. Eq. (1) describes the wave propaganda in the more general nonlinear fluids.

The Lie symmetry technique is used to achieve symmetry reduction of the eKP equation and solve the resulting ordinary differential equation we obtain by applying straightforward integration in conjunction with simplest equation technique to derive results for the eKP equation. Furthermore, we derive the conservation laws of the eKP equation using multiplier methods, and then give concluding remarks.

Exact solutions of Eq. (1) by using symmetry reductions

The direction ground can be expressed as follows: $P=S1∂∂t+S2∂∂x+S3∂∂y+G∂∂u$ $$\matrix{ {P = {S^1}{\partial \over {\partial t}} + {S^2}{\partial \over {\partial x}} + {S^3}{\partial \over {\partial y}} + G{\partial \over {\partial u}}} \cr }$$

where S1, S2, S3 and G depend on t, x and y, and u is a Lie point symmetry of Eq. (1) if the following expression stands fulfilled: $P[4]{(ut+6uux+uxxx)x−uyy−α24utt+αuty+βuxt}|(1)=0,$ $$\matrix{ {{P^{[4]}}{{\left. {\left\{ {{{({u_t} + 6u{u_x} + {u_{xxx}})}_x} - {u_{yy}} - {{{\alpha ^2}} \over 4}{u_{tt}} + \alpha {u_{ty}} + \beta {u_{xt}}} \right\}} \right|}_{(1)}} = 0,} \cr }$$

where P[4] is defined as the fourth prolongation of Eq. (2). Writing Eq. (3) in detail and splitting the resulting determining equations on derivatives of u leads to linear PDEs, and with the aid of Maple, these can be used to solve the infinitesimal, which gives four symmetries, as follows: $P1=F1(t,y)∂∂t,P2=F2(t,y)∂∂y,P3=[ ∂F3(t,y)∂t+β∂F3(t,y)∂t ]∂∂u+6∂F3(t,y)∂x,P4=[ 12uF4(t,y)−βx∂F4(t,y)∂t−x∂F4(t,y)∂t ]∂∂u−6xF4(t,y)∂∂x,$ \eqalign{ & {P_1} = {F_1}(t,y){\partial \over {\partial t}}, \cr & {P_2} = {F_2}(t,y){\partial \over {\partial y}}, \cr & {P_3} = \left[ {{{\partial {F_3}(t,y)} \over {\partial t}} + \beta {{\partial {F_3}(t,y)} \over {\partial t}}} \right]{\partial \over {\partial u}} + 6{{\partial {F_3}(t,y)} \over {\partial x}}, \cr & {P_4} = \left[ {12u{F_4}(t,y) - \beta x{{\partial {F_4}(t,y)} \over {\partial t}} - x{{\partial {F_4}(t,y)} \over {\partial t}}} \right]{\partial \over {\partial u}} - 6x{F_4}(t,y){\partial \over {\partial x}}, \cr}

where F1 to F4 are arbitrary functions of t and y. Considering the symmetry (P = P1 +aP2), where a is a constant, with F1 = 1 and F2 = 1, we perform symmetry reduction. By solving the individual equations of PDE, we contract the following invariants: $g=x, h=at−y, u=Θ(g,h),$ $$\matrix{ {g = x,\quad h = at - y,\quad u = {\rm{\Theta }}(g,h),} \cr }$$

which, when used, reduce the eKP equation to the following form: $(a+aβ)∂2∂h∂gΘ+6(∂∂gΘ)2+6Θ∂2∂g2Θ+∂4∂g4Θ−∂2∂h2Θ−a2α24∂2∂h2Θ−aα∂2∂h2θ=0.$ $$\matrix{ {(a + a\beta ){{{\partial ^2}} \over {\partial h\partial g}}\Theta + 6{{\left( {{\partial \over {\partial g}}\Theta } \right)}^2} + 6\Theta {{{\partial ^2}} \over {\partial {g^2}}}\Theta + {{{\partial ^4}} \over {\partial {g^4}}}\Theta - {{{\partial ^2}} \over {\partial {h^2}}}\Theta - {{{a^2}{\alpha ^2}} \over 4}{{{\partial ^2}} \over {\partial {h^2}}}\Theta - a\alpha {{{\partial ^2}} \over {\partial {h^2}}}\theta = 0.} \cr }$$

We further perform symmetry reductions by utilising the Lie symmetries of Eq. (4). Eq. (4) has three Lie symmetries, namely: $ϒ1=∂∂g,ϒ2=∂∂h,ϒ3=(6a2α2Θ+24aαΘ+a2+2aαβ+aβ2+24Θ)∂∂Θ+(−6a2α2h−24aαh−24h)∂∂g+(−3a2α2h−12aαg+6ah+6aβh−12g)∂∂h.$ \eqalign{ & {\Upsilon _1} = {\partial \over {\partial g}}, \cr & {\Upsilon _2} = {\partial \over {\partial h}}, \cr & {\Upsilon _3} = (6{a^2}{\alpha ^2}{\rm{\Theta }} + 24a\alpha {\rm{\Theta }} + {a^2} + 2a\alpha \beta + a{\beta ^2} + 24{\rm{\Theta }}){\partial \over {\partial {\rm{\Theta }}}} + ( - 6{a^2}{\alpha ^2}h - 24a\alpha h - 24h){\partial \over {\partial g}} \cr & + ( - 3{a^2}{\alpha ^2}h - 12a\alpha g + 6ah + 6a\beta h - 12g){\partial \over {\partial h}}. \cr}

By considering the symmetry (Υ = Υ1 +µΥ), where µ is a constant, two invariants are resultantly obtained, i.e. $ξ=h−μg, Θ=Φ(ξ)$ $$\matrix{ {\xi = h - \mu g,\quad {\rm{\Theta }} = {\rm{\Phi }}(\xi )} \cr }$$

which reduces the eKP equation to the following ODE: $−aμΦ ″+6μ2(Φ′)2+6μ2ΦΦ″+μ4Φ″″−Φ″−a2α24Φ″−aαΦ″−aβμΦ″=0$ $$\matrix{ { - a\mu {\rm{\Phi ''}} + 6{\mu ^2}{{({\rm{\Phi '}})}^2} + 6{\mu ^2}{\rm{\Phi \Phi ''}} + {\mu ^4}{{\rm{\Phi }}^{\prime \prime \prime \prime }} - {\rm{\Phi ''}} - {{{a^2}{\alpha ^2}} \over 4}{\rm{\Phi ''}} - a\alpha {\rm{\Phi ''}} - a\beta \mu {\rm{\Phi ''}} = 0.} \cr }$$

Eq. (5) can be written as $AΦ″+6μ2(Φ′2+ΦΦ″)+μ4Φ″″=0$ $$\matrix{ {A{\rm{\Phi ''}} + 6{\mu ^2}({{\rm{\Phi }}^{\prime 2}} + {\rm{\Phi \Phi ''}}) + {\mu ^4}{{\rm{\Phi }}^{\prime \prime \prime \prime }} = 0} \cr }$$

where $A=−aμ−1−aα−aβμ−14a2α2$ $A = - a\mu - 1 - a\alpha - a\beta \mu - {1 \over 4}{a^2}{\alpha ^2}$ .

Solutions of Eq. (1) by straightforward integration

Integrating Eq. (6) yields the following expression: $AΦ′+6μΦΦ′+μ4Φ″′+C1=0$ $$\matrix{ {A{\rm{\Phi '}} + 6\mu {\rm{\Phi \Phi '}} + {\mu ^4}{\rm{\Phi '''}} + {C_1} = 0} \cr }$$

where C1 is a constant of integration.

Furthermore, the integral of Eq. (7) is expressed as: $AΦ+3μ2Φ2+μ4Φ″+C1ξ+C2$ $$\matrix{ {A{\rm{\Phi }} + 3{\mu ^2}{{\rm{\Phi }}^2} + {\mu ^4}{\rm{\Phi ''}} + {C_1}{\rm{\xi }} + {C_2}} \cr }$$

where C2 is an integration constant.

Taking C1 = 0 and multiplying Eq. (8) by Φ1 yields the following result: $AΦΦ′+3μ2Φ2Φ′+μ4Φ′Φ″+C2Φ′=0.$ $$\matrix{ {A{\rm{\Phi \Phi '}} + 3{\mu ^2}{{\rm{\Phi }}^2}{\rm{\Phi '}} + {\mu ^4}{\rm{\Phi '\Phi ''}} + {C_2}{\rm{\Phi '}} = 0.} \cr }$$

The integral of Eq. (9) yields: $12AΦ2+μ2Φ3+12μ4Φ′2+C2Φ+C3=0$ $$\matrix{ {{1 \over 2}A{{\rm{\Phi }}^2} + {\mu ^2}{{\rm{\Phi }}^3} + {1 \over 2}{\mu ^4}{{\rm{\Phi }}^{\prime 2}} + {C_2}{\rm{\Phi }} + {C_3} = 0} \cr }$$

which in its simplest form can be expressed as: $Φ′2+1μΦ3+Aμ2Φ2+2C2μ4Φ+2C3μ4=0.$ $$\matrix{ {{{\rm{\Phi }}^{\prime 2}} + {1 \over \mu }{{\rm{\Phi }}^3} + {A \over {{\mu ^2}}}{{\rm{\Phi }}^2} + {{2{C_2}} \over {{\mu ^4}}}{\rm{\Phi }} + {{2{C_3}} \over {{\mu ^4}}} = 0.} \cr }$$

Assuming that the cubic equation $1μΦ3+Aμ2Φ2+2C2μ4Φ+2C3μ4=0$ $$\matrix{ {{1 \over \mu }{{\rm{\Phi }}^3} + {A \over {{\mu ^2}}}{{\rm{\Phi }}^2} + {{2{C_2}} \over {{\mu ^4}}}{\rm{\Phi }} + {{2{C_3}} \over {{\mu ^4}}} = 0} \cr }$$

has real roots a1, a2, a3 such that a1 > a2 > a3, the solution for Eq. (10) would then be: $Φ(ξ)=a2+(a1−a2)cn2{(a1−a2)μ2ξ,M2},M2=a1−a2a1−a3$ $$\matrix{ {{\rm{\Phi }}(\xi ) = {a_2} + ({a_1} - {a_2}){\rm{c}}{{\rm{n}}^2}\left\{ {\sqrt {{{({a_1} - {a_2})} \over {{\mu ^2}}}\xi } ,{M^2}} \right\},{M^2} = {{{a_1} - {a_2}} \over {{a_1} - {a_3}}}} \cr }$$

where cn is the Jacobi cosine function. Subsequently, the solution for Eq. (1) is obtained as: $u(t,x,y)=a2+(a1−a2)cn2{(a1−a2)μ2(at−y−μx),M2},$ $$\matrix{ {u(t,x,y) = {a_2} + ({a_1} - {a_2}){\rm{c}}{{\rm{n}}^2}\left\{ {\sqrt {{{({a_1} - {a_2})} \over {{\mu ^2}}}(at - y - \mu x)} ,{M^2}} \right\},} \cr }$$

where $M2=a1−a2a1−a3$ $${{M^2} = {{{a_1} - {a_2}} \over {{a_1} - {a_3}}}}$$

Exact solutions using the simplest equation method

We now solve Eq. (5) by employing the simplest equation method [21]. By doing so, we are facilitated to ascertain the exact solutions for Eq. (1). We will use the Bernoulli and Riccati equations as our simplest equations.

Recalling the simplest equation method briefly, let $Φ(q)=∑r=0NAr(K(q))r,$ $$\Phi (q) = \mathop \sum \limits_{r = 0}^N {A_r}{(K(q))^r},$$

be the solutions of Eq. (1). In this case, K(q) fulfils the Bernoulli or Riccati equation. The positive integer N can be calculated by using the matching procedure. A0, ···, AN are the parameters needing to be calculated.

By considering the Riccati equation $K′(q)=eK2(q)+dK(q)+p,$ $$\matrix{ {K'(q) = e{K^2}(q) + dK(q) + p,} \cr }$$

where e, d and p are just constants, this gives the solutions $K(q)=−d2e−Θ2etanh[ Θ2(q+B) ]$ $$\matrix{ {K(q) = - {d \over {2e}} - {{\rm{\Theta }} \over {2e}}\tanh \left[ {{{\rm{\Theta }} \over 2}(q + B)} \right]} \cr }$$

and $K(q)=−d2e−Θ2etanh(Θq2)+sech(Θq2)cosh(Θq2)−2eΘsinh(Θq2)$ $$\matrix{ {K(q) = - {d \over {2e}} - {{\rm{\Theta }} \over {2e}}\tanh \left( {{{{\rm{\Theta }}q} \over 2}} \right) + {{sech\left( {{{{\rm{\Theta }}q} \over 2}} \right)} \over {\cosh \left( {{{{\rm{\Theta }}q} \over 2}} \right) - {{2e} \over {\rm{\Theta }}}\sinh \left( {{{{\rm{\Theta }}q} \over 2}} \right)}}} \cr }$$

where Θ2 = d2 –4ep and B is an integration constant.

The Bernoulli equation is expressed as: $K′(q)=eK(q)2+dK(q),$ $$\matrix{ {K'(q) = eK{{(q)}^2} + dK(q),} \cr }$$

where e and d are constants. This gives the solution of the equation as: $K(q)=e{cosh[e(q+B)]+sinh[e(q+B)]1−dcosh[e(q+B)]−dsinh[e(q+B)]}.$ $$\matrix{ {K(q) = e\left\{ {{{\cosh [e(q + B)] + \sinh [e(q + B)]} \over {1 - d\cosh [e(q + B)] - d\sinh [e(q + B)]}}} \right\}.} \cr }$$

Solution by Bernoulli equation

The matching technique produces N = 2, and this yields the solutions of Eq. (5) as: $Φ(q)=A0+A1K+A2K2.$ $$\matrix{ {{\rm{\Phi }}(q) = {A_0} + {A_1}K + {A_2}{K^2}.} \cr }$$

We now proceed to replace Φ(q) into Eq. (5), concomitant with the usage of Eq. (15), and then compare the factors of Kr to zero; then, with the aid of Maple, we obtain systems of equations in relation to A0, A1 and A2: $μ4A1d4−A1d2−14α2a2A1d2−αaA1d2−μaA1d2+6μ2A0A1d2−βμaA1d2=0,60μ4A1e3d+36μ2A0A2e2−6μaA2e2+330μ4A2e2d2−32α2a2A2e2−6αaA2e2+18μ2A12e2+48μ2A22d2−6A2e2−6βμaA2e2+126μ2A1eA2d=0,16μ4A2d4−4μaA2d2+24μ2A0A2d2+15μ4A1ed3−α2a2A2d2−4αaA2d2−3A1ed+12μ2A12d2−3βμaA1ed−4A2d2−34α2a2A1ed−3αaA1ed−4βμaA2d2−3μaA1ed+18μ2A0A1ed=0,−2μaA1e2+30μ2A12ed+54μ2A1d2A2+12μ2A0A1e2+50μ4A1e2d2+130μ4A2ed3−12α2a2A1e2−2αaA1e2−10A2ed−10βμaA2ed−2A1e2−52α2a2A2ed−10αaA2ed−2βμaA1e2−10μaA2ed+60μ2A0A2ed=0,120e4μ4A2+60e2μ2A22=0,336μ4A2e3d+24μ4A1e4+108μ2A22ed+72μ2A1e2A2=0.$ \eqalign{ & {\mu ^4}{A_1}{d^4} - {A_1}{d^2} - {1 \over 4}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}{d^2} - \alpha {\mkern 1mu} a{A_1}{d^2} - \mu {\mkern 1mu} a{A_1}{d^2} + 6{\mkern 1mu} {\mu ^2}{A_0}{A_1}{d^2} - \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}{d^2} = 0, \cr & 60{\mkern 1mu} {\mu ^4}{A_1}{e^3}d + 36{\mkern 1mu} {\mu ^2}{A_0}{A_2}{e^2} - 6{\mkern 1mu} \mu {\mkern 1mu} a{A_2}{e^2} + 330{\mkern 1mu} {\mu ^4}{A_2}{e^2}{d^2} - {3 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}{e^2} - 6{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}{e^2} + 18{\mkern 1mu} {\mu ^2}{A_1}^2{e^2} + 48{\mkern 1mu} {\mu ^2}{A_2}^2{d^2} \cr & - 6{\mkern 1mu} {A_2}{e^2} - 6{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}{e^2} + 126{\mkern 1mu} {\mu ^2}{A_1}e{A_2}d = 0, \cr & 16{\mkern 1mu} {\mu ^4}{A_2}{d^4} - 4{\mkern 1mu} \mu {\mkern 1mu} a{A_2}{d^2} + 24{\mkern 1mu} {\mu ^2}{A_0}{A_2}{d^2} + 15{\mkern 1mu} {\mu ^4}{A_1}e{d^3} - {\alpha ^2}{a^2}{A_2}{d^2} - 4{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}{d^2} - 3{\mkern 1mu} {A_1}ed + 12{\mkern 1mu} {\mu ^2}{A_1}^2{d^2} \cr & - 3{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}ed - 4{\mkern 1mu} {A_2}{d^2} - {3 \over 4}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}ed - 3{\mkern 1mu} \alpha {\mkern 1mu} a{A_1}ed - 4{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}{d^2} - 3{\mkern 1mu} \mu {\mkern 1mu} a{A_1}ed + 18{\mkern 1mu} {\mu ^2}{A_0}{A_1}ed = 0, \cr & - 2{\mkern 1mu} \mu {\mkern 1mu} a{A_1}{e^2} + 30{\mkern 1mu} {\mu ^2}{A_1}^2ed + 54{\mkern 1mu} {\mu ^2}{A_1}{d^2}{A_2} + 12{\mkern 1mu} {\mu ^2}{A_0}{A_1}{e^2} + 50{\mkern 1mu} {\mu ^4}{A_1}{e^2}{d^2} + 130{\mkern 1mu} {\mu ^4}{A_2}e{d^3} - {1 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}{e^2} \cr & - 2{\mkern 1mu} \alpha {\mkern 1mu} a{A_1}{e^2} - 10{\mkern 1mu} {A_2}ed - 10{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}ed - 2{\mkern 1mu} {A_1}{e^2} - {5 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}ed - 10{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}ed - 2{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}{e^2} - 10{\mkern 1mu} \mu {\mkern 1mu} a{A_2}ed \cr & + 60{\mkern 1mu} {\mu ^2}{A_0}{A_2}ed = 0, \cr & 120{\mkern 1mu} {e^4}{\mu ^4}{A_2} + 60{\mkern 1mu} {e^2}{\mu ^2}{A_2}^2 = 0, \cr & 336{\mkern 1mu} {\mu ^4}{A_2}{e^3}d + 24{\mkern 1mu} {\mu ^4}{A_1}{e^4} + 108{\mkern 1mu} {\mu ^2}{A_2}^2ed + 72{\mkern 1mu} {\mu ^2}{A_1}{e^2}{A_2} = 0. \cr}

Solving the system of equations, we get $A0=−4μ4d2+α2a2+4βμa+4αa+4μa+424μ2,A1=−2deμ2,A2=−2e2μ2.$ \eqalign{ & {A_0} = {{ - 4{\mu ^4}{d^2} + {\alpha ^2}{a^2} + 4\beta \mu a + 4\alpha a + 4\mu a + 4} \over {24{\mu ^2}}}, \cr & {A_1} = - 2de{\mu ^2}, \cr & {A_2} = - 2{e^2}{\mu ^2}. \cr}

As a result, a solution for Eq. (1) is obtained, as the following: $u(x,t,y)=A0+A1e{ cosh[e(q+B)]+sinhe[(q+B)]1−dcosh[e(q+B)]−dsinh[e(q+B)] }+A2{ ecosh[e(q+B)]+sinh[e(q+B)]1−dcosh[e(q+B)]−dsinh[e(q+B)] }2,$ $$\displaylines{ u(x,t,y) = {A_0} + {A_1}e\left\{ {{{\cosh [e(q + B)] + \sinh e[(q + B)]} \over {1 - d\cosh [e(q + B)] - d\sinh [e(q + B)]}}} \right\} \cr + {A_2}{\left\{ {e{{\cosh [e(q + B)] + \sinh [e(q + B)]} \over {1 - d\cosh [e(q + B)] - d\sinh [e(q + B)]}}} \right\}^2}, \cr}$$

where q = atµxy and B represent the integration constant.

Solution by Ricatti equation

We obtain N = 2 as our matching technique, and so the solutions of Eq. (5) assume the form: $Φ(q)=A0+A1K+A2K2.$ $$\matrix{ {{\rm{\Phi }}(q) = {A_0} + {A_1}K + {A_2}{K^2}.} \cr }$$

We replace Φ(q) into Eq. (5), concomitant with the usage of Eq. (15), and then compare the factors of Kr to zero; then, with the aid of Maple, we obtain systems of equations in relation to A0, A1 and A2:

$18μ2A12e2+48μ2A22d2−6A2e2+36μ2A0A2e2+60μ4A1e3d+240μ4A2e3p+330μ4A2e2d2−32α2a2A2e2−6αaA2e2−6μaA2e2+96μ2A22ep+126μ2A1eA2d−6βμaA2e2−2μaA2p2+12μ2A0A2p2+μ4A1d3p+14μ4A2d2p2+16μ4A2ep3−12α2a2A2p2−2αaA2p2+6μ2A12p2−A1dp−2A2p2+6μ2A0A1dp+8μ4A1edp2−14α2a2A1dp−αaA1dp−2βμaA2p2−μaA1dp−βμaA1dp=0,−10A2ed−2A1e2+40μ4A1e3p+50μ4A1e2d2+130μ4A2ed3−12α2a2A1e2−2αaA1e2+54μ2A1d2A2+84μ2A22dp+12μ2A0A1e2−2μaA1e2+30μ2A12ed−52α2a2A2ed−10αaA2ed−2βμaA1e2−10μaA2ed+108μ2A1eA2p+60μ2A0A2ed+440μ4A2e2dp−10βμaA2ed=0,−14α2a2A1d2−αaA1d2+18μ2A12dp+36μ2A1p2A2+6μ2A0A1d2−μaA1d2+120μ4A2ep2d−6βμaA2dp−2βμaA1ep+16μ4A1e2p2+30μ4A2d3p+22μ4A1ed2p+12μ2A0A1ep+36μ2A0A2dp−6μaA2dp−βμaA1d2−2μaA1ep−2αaA1ep−6αaA2dp−1/2α2a2A1ep−32α2a2A2dp−2A1ep+μ4A1d4−6A2dp−A1d2=0,60μ4A1e2dp+48μ2A0A2ep+90μ2A1dA2p+18μ2A0A1ed−8μaA2ep−4βμaA2d2−3μaA1ed−8αaA2ep−2α2a2A2ep−3αaA1ed+232μ4A2ed2p−34α2a2A1ed−4A2d2+36μ2A22p2+12μ2A12d2−8A2ep−3A1ed+16μ4A2d4−3βμaA1ed−8βμaA2ep+24μ2A12ep+24μ2A0A2d2−4μaA2d2+15μ4A1ed3+136μ4A2e2p2−α2a2A2d2−4αaA2d2,120e4μ4A2+60e2μ2A22=0,336de3μ4A2+24e4μ4A1+108,deμ2A22+72e2μ2A1A2=0.$ \eqalign{ & 18{\mkern 1mu} {\mu ^2}{A_1}^2{e^2} + 48{\mkern 1mu} {\mu ^2}{A_2}^2{d^2} - 6{\mkern 1mu} {A_2}{e^2} + 36{\mkern 1mu} {\mu ^2}{A_0}{A_2}{e^2} + 60{\mkern 1mu} {\mu ^4}{A_1}{e^3}d + 240{\mkern 1mu} {\mu ^4}{A_2}{e^3}p + 330{\mkern 1mu} {\mu ^4}{A_2}{e^2}{d^2} \cr & - {3 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}{e^2} - 6{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}{e^2} - 6{\mkern 1mu} \mu {\mkern 1mu} a{A_2}{e^2} + 96{\mkern 1mu} {\mu ^2}{A_2}^2ep + 126{\mkern 1mu} {\mu ^2}{A_1}e{A_2}d - 6{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}{e^2} \cr & - 2{\mkern 1mu} \mu {\mkern 1mu} a{A_2}{p^2} + 12{\mkern 1mu} {\mu ^2}{A_0}{A_2}{p^2} + {\mu ^4}{A_1}{d^3}p + 14{\mkern 1mu} {\mu ^4}{A_2}{d^2}{p^2} + 16{\mkern 1mu} {\mu ^4}{A_2}e{p^3} - {1 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}{p^2} - 2{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}{p^2} \cr & + 6{\mkern 1mu} {\mu ^2}{A_1}^2{p^2} - {A_1}dp - 2{\mkern 1mu} {A_2}{p^2} + 6{\mkern 1mu} {\mu ^2}{A_0}{A_1}dp + 8{\mkern 1mu} {\mu ^4}{A_1}ed{p^2} - {1 \over 4}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}dp - \alpha {\mkern 1mu} a{A_1}dp \cr & - 2{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}{p^2} - \mu {\mkern 1mu} a{A_1}dp - \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}dp = 0, \cr & - 10{\mkern 1mu} {A_2}ed - 2{\mkern 1mu} {A_1}{e^2} + 40{\mkern 1mu} {\mu ^4}{A_1}{e^3}p + 50{\mkern 1mu} {\mu ^4}{A_1}{e^2}{d^2} + 130{\mkern 1mu} {\mu ^4}{A_2}e{d^3} - {1 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}{e^2} - 2{\mkern 1mu} \alpha {\mkern 1mu} a{A_1}{e^2} \cr & + 54{\mkern 1mu} {\mu ^2}{A_1}{d^2}{A_2} + 84{\mkern 1mu} {\mu ^2}{A_2}^2dp + 12{\mkern 1mu} {\mu ^2}{A_0}{A_1}{e^2} - 2{\mkern 1mu} \mu {\mkern 1mu} a{A_1}{e^2} + 30{\mkern 1mu} {\mu ^2}{A_1}^2ed - {5 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}ed \cr & - 10{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}ed - 2{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}{e^2} - 10{\mkern 1mu} \mu {\mkern 1mu} a{A_2}ed + 108{\mkern 1mu} {\mu ^2}{A_1}e{A_2}p + 60{\mkern 1mu} {\mu ^2}{A_0}{A_2}ed + 440{\mkern 1mu} {\mu ^4}{A_2}{e^2}dp \cr & - 10{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}ed = 0, \cr & - {1 \over 4}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}{d^2} - \alpha {\mkern 1mu} a{A_1}{d^2} + 18{\mkern 1mu} {\mu ^2}{A_1}^2dp + 36{\mkern 1mu} {\mu ^2}{A_1}{p^2}{A_2} + 6{\mkern 1mu} {\mu ^2}{A_0}{A_1}{d^2} - \mu {\mkern 1mu} a{A_1}{d^2} + 120{\mkern 1mu} {\mu ^4}{A_2}e{p^2}d \cr & - 6{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}dp - 2{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}ep + 16{\mkern 1mu} {\mu ^4}{A_1}{e^2}{p^2} + 30{\mkern 1mu} {\mu ^4}{A_2}{d^3}p + 22{\mkern 1mu} {\mu ^4}{A_1}e{d^2}p + 12{\mkern 1mu} {\mu ^2}{A_0}{A_1}ep \cr & + 36{\mkern 1mu} {\mu ^2}{A_0}{A_2}dp - 6{\mkern 1mu} \mu {\mkern 1mu} a{A_2}dp - \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}{d^2} - 2{\mkern 1mu} \mu {\mkern 1mu} a{A_1}ep - 2{\mkern 1mu} \alpha {\mkern 1mu} a{A_1}ep - 6{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}dp - 1/2{\mkern 1mu} {\alpha ^2}{a^2}{A_1}ep \cr & - {3 \over 2}{\mkern 1mu} {\alpha ^2}{a^2}{A_2}dp - 2{\mkern 1mu} {A_1}ep + {\mu ^4}{A_1}{d^4} - 6{\mkern 1mu} {A_2}dp - {A_1}{d^2} = 0, \cr & 60{\mkern 1mu} {\mu ^4}{A_1}{e^2}dp + 48{\mkern 1mu} {\mu ^2}{A_0}{A_2}ep + 90{\mkern 1mu} {\mu ^2}{A_1}d{A_2}p + 18{\mkern 1mu} {\mu ^2}{A_0}{A_1}ed - 8{\mkern 1mu} \mu {\mkern 1mu} a{A_2}ep - 4{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}{d^2} - 3{\mkern 1mu} \mu {\mkern 1mu} a{A_1}ed \cr & - 8{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}ep - 2{\mkern 1mu} {\alpha ^2}{a^2}{A_2}ep - 3{\mkern 1mu} \alpha {\mkern 1mu} a{A_1}ed + 232{\mkern 1mu} {\mu ^4}{A_2}e{d^2}p - {3 \over 4}{\mkern 1mu} {\alpha ^2}{a^2}{A_1}ed - 4{\mkern 1mu} {A_2}{d^2} + 36{\mkern 1mu} {\mu ^2}{A_2}^2{p^2} \cr & + 12{\mkern 1mu} {\mu ^2}{A_1}^2{d^2} - 8{\mkern 1mu} {A_2}ep - 3{\mkern 1mu} {A_1}ed + 16{\mkern 1mu} {\mu ^4}{A_2}{d^4} - 3{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_1}ed - 8{\mkern 1mu} \beta {\mkern 1mu} \mu {\mkern 1mu} a{A_2}ep + 24{\mkern 1mu} {\mu ^2}{A_1}^2ep \cr & + 24{\mkern 1mu} {\mu ^2}{A_0}{A_2}{d^2} - 4{\mkern 1mu} \mu {\mkern 1mu} a{A_2}{d^2} + 15{\mkern 1mu} {\mu ^4}{A_1}e{d^3} + 136{\mkern 1mu} {\mu ^4}{A_2}{e^2}{p^2} - {\alpha ^2}{a^2}{A_2}{d^2} - 4{\mkern 1mu} \alpha {\mkern 1mu} a{A_2}{d^2}, \cr & 120{\mkern 1mu} {e^4}{\mu ^4}{A_2} + 60{\mkern 1mu} {e^2}{\mu ^2}{A_2}^2 = 0, \cr & 336{\mkern 1mu} d{e^3}{\mu ^4}{A_2} + 24{\mkern 1mu} {e^4}{\mu ^4}{A_1} + 108,de{\mu ^2}{A_2}^2 + 72{\mkern 1mu} {e^2}{\mu ^2}{A_1}{A_2} = 0. \cr}

By solving the resultant system of equations, we acquire: $A0=−4μ4d2−32eμ2p+α2a2+4βμa+4αa+4μa+424μ2,A1=−2deμ2,A2=−2e2μ2.$ \eqalign{ & {A_0} = {{ - 4{\mu ^4}{d^2} - 32e{\mu ^2}p + {\alpha ^2}{a^2} + 4\beta \mu a + 4\alpha a + 4\mu a + 4} \over {24{\mu ^2}}}, \cr & {A_1} = - 2de{\mu ^2}, \cr & {A_2} = - 2{e^2}{\mu ^2}. \cr}

As a consequence, results for Eq. (1) are obtained in the following form: $u(x,t,y)=A0+A1{ −d2e−Θ2etanh[ Θ2(q+B) ] }+A2{ −d2e−Θ2etanh[ Θ2(q+B) ] }2,$ $$\matrix{ {u(x,t,y) = {A_0} + {A_1}\left\{ { - {d \over {2e}} - {\Theta \over {2e}}\tanh \left[ {{\Theta \over 2}(q + B)} \right]} \right\} + {A_2}{{\left\{ { - {d \over {2e}} - {\Theta \over {2e}}\tanh \left[ {{\Theta \over 2}(q + B)} \right]} \right\}}^2},} \cr }$$

and $u(x,t,y)=A0+A1e{−d2e−Θ2etanh(Θq2)+sech(Θq2)cosh(Θq2)−2eΘsinh(Θq2)}+A2{−d2e−Θ2etanh(Θq2)+sech(Θq2)cosh(Θq2)−2eΘsinh(Θq2)]}2$ $$\matrix{ {u(x,t,y)} & { = {A_0} + {A_1}e\left\{ { - {d \over {2e}} - {{\rm{\Theta }} \over {2e}}\tanh ({{{\rm{\Theta }}q} \over 2}) + {{sech({{{\rm{\Theta }}q} \over 2})} \over {\cosh ({{{\rm{\Theta }}q} \over 2}) - {{2e} \over {\rm{\Theta }}}\sinh ({{{\rm{\Theta }}q} \over 2})}}} \right\}} \cr {} & { + {A_2}{{\left\{ { - {d \over {2e}} - {{\rm{\Theta }} \over {2e}}\tanh ({{{\rm{\Theta }}q} \over 2}) + {{sech({{{\rm{\Theta }}q} \over 2})} \over {\cosh ({{{\rm{\Theta }}q} \over 2}) - {{2e} \over {\rm{\Theta }}}\sinh ({{{\rm{\Theta }}q} \over 2})}}]} \right\}}^2}} \cr }$$

where q = atµzy and B represent the integration constant.

Conservation laws of eKP Eq. (1)

The resident conservation law for eKP Eq. (1) can be given as: $G=(ut+6uux+uxxx)x−uyy−α24utt+αuty+βuxt$ $$G = {({u_t} + 6u{u_x} + {u_{xxx}})_x} - {u_{yy}} - {{{\alpha ^2}} \over 4}{u_{tt}} + \alpha {u_{ty}} + \beta {u_{xt}}$$

being a space-time deviation. $DtTt+DxTx+DyTy|(1)=0,$ $$\matrix{ {{D_t}{T^t} + {D_x}{T^x} + {D_y}{T^y}{|_{(1)}} = 0,} \cr }$$

which holds good for all official results of Eq. (1). We achieve a multiplier Λ, which is expressed in terms of the equation: $Λ(t,x,y,u)=14C1α2tx+14C1α2βtx+C4x+12C1t2+C2t+C3.$ $$\matrix{ {{\rm{\Lambda }}(t,x,y,u) = {1 \over 4}{\mkern 1mu} {C_1}{\alpha ^2}tx + {1 \over 4}{\mkern 1mu} {C_1}{\alpha ^2}\beta tx + {C_4}x + {1 \over 2}{\mkern 1mu} {C_1}{t^2} + {C_2}t + {C_3}.} \cr }$$

Corresponding to Eq. (27), we then obtain the resulting conservation laws: $T1t=18+8β [ (2αt2+2αβt2+αtx)uy−(α2t2+α2βt2−12α4tx)ut+(2t2+4βt2+α2tx+2β2t2+α2βtx)ux +(α2t+α2βt+12α4x)u ],T1x18+8β [ (24t2+24βt2+12α2x)uux+(4βt2+2β2t2+2t2+α2βx+α2tx)ut +(4t2+4βt2+2α2tx)uxxx−2α2tuxx−(4β2+8β+4+6α2u+α2x+α2βx)u ],T1y=18+8β[ (2αt2+2αβt2+α3xt)ut−(4t2+4βt2−2α2tx)uy−(4αt+4αβt+α2x)u ];$ $$\matrix{ {T_1^t = {1 \over {8 + 8\beta }}\left[ {(2\alpha {t^2} + 2\alpha \beta {t^2} + \alpha tx){u_y} - ({\alpha ^2}{t^2} + {\alpha ^2}\beta {t^2} - {1 \over 2}{\alpha ^4}tx){u_t} + (2{t^2} + 4\beta {t^2} + {\alpha ^2}tx + 2{\beta ^2}{t^2} + {\alpha ^2}\beta tx){u_x}} \right.} \hfill \cr {\left. { + ({\alpha ^2}t + {\alpha ^2}\beta t + {1 \over 2}{\alpha ^4}x)u} \right],} \hfill \cr {T_1^x{1 \over {8 + 8\beta }}\left[ {(24{t^2} + 24\beta {t^2} + 12{\alpha ^2}x)u{u_x} + (4\beta {t^2} + 2{\beta ^2}{t^2} + 2{t^2} + {\alpha ^2}\beta x + {\alpha ^2}tx){u_t}} \right.} \hfill \cr {\left. { + (4{t^2} + 4\beta {t^2} + 2{\alpha ^2}tx){u_{xxx}} - 2{\alpha ^2}t{u_{xx}} - (4{\beta ^2} + 8\beta + 4 + 6{\alpha ^2}u + {\alpha ^2}x + {\alpha ^2}\beta x)u} \right],} \hfill \cr {T_1^y = {1 \over {8 + 8\beta }}\left[ {(2\alpha {t^2} + 2\alpha \beta {t^2} + {\alpha ^3}xt){u_t} - (4{t^2} + 4\beta {t^2} - 2{\alpha ^2}tx){u_y} - (4\alpha t + 4\alpha \beta t + {\alpha ^2}x)u} \right];} \hfill \cr }$$ $T2t=14[ α2tut+2αtuy+2t(1+β)ux+α2u ],T2x=12[ 12tuux+βtut+tut+2tuxxx−(1+β)u ],T2y=12α(tut+u)−tuy;$ \eqalign{ & T_2^t = {1 \over 4}\left[ {{\alpha ^2}t{u_t} + 2\alpha t{u_y} + 2t(1 + \beta ){u_x} + {\alpha ^2}u} \right], \cr & T_2^x = {1 \over 2}\left[ {12tu{u_x} + \beta t{u_t} + t{u_t} + 2t{u_{xxx}} - (1 + \beta )u} \right], \cr & T_2^y = {1 \over 2}\alpha (t{u_t} + u) - t{u_y}; \cr} $T3t=12[ 12α2ut+αuy+(1+β)ux ],T3x=12(1+β)ut+6uux+uxxx,T3y=12αut−uy;$ \eqalign{ & T_3^t = {1 \over 2}\left[ {{1 \over 2}{\alpha ^2}{u_t} + \alpha {u_y} + (1 + \beta ){u_x}} \right], \cr & T_3^x = {1 \over 2}(1 + \beta ){u_t} + 6u{u_x} + {u_{xxx}}, \cr & T_3^y = {1 \over 2}\alpha {u_t} - {u_y}; \cr} $T4t=12αxut+αxuy+(1+β)xux−12(1+β)u,T4x=12(1+β)xut+6xuux+xuxxx−3u2−uxx,T4y=xut−xuy.$ \eqalign{ & T_4^t = {1 \over 2}\alpha x{u_t} + \alpha x{u_y} + (1 + \beta )x{u_x} - {1 \over 2}(1 + \beta )u, \cr & T_4^x = {1 \over 2}(1 + \beta )x{u_t} + 6xu{u_x} + x{u_{xxx}} - 3{u^2} - {u_{xx}}, \cr & T_4^y = x{u_t} - x{u_y}. \cr}

Conclusion

In this research paper, we studied the eKP equation by using the Lie symmetry analysis, which reduced the eKP equation to an ODE; furthermore, we used straightforward integration and the simplest equation method to obtain exact solutions for the eKP equation. Straightforward integration gave us the Jacobi cosine function solution, whereas using the simplest equation method, we obtained the trigonometric hyperbolic solutions. Finally, using the multiplier approach and with the aid of Maple, conservation laws were derived for the eKP equation.

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