Research article

A new hybrid form of the skew-t distribution: estimation methods comparison via Monte Carlo simulation and GARCH model application

  • In this work, estimating the exponentiated half logistic skew-t model parameters using some classical estimation procedures is considered. The finite sample performance of the EHLST parameter estimates is examined through extensive Monte Carlo simulations. The ordering performance of the six criterions was based on the partial and overall ranks of the estimation procedures for all parameter combinations. The criterions performance ordering from finest to poorest, using the overall ranks is maximum likelihood, maximum product of spacing, Anderson-Darling, Cramer-von Mises, least squares and weighted least squares estimators for all the parameter combinations. The simulation results confirm the dominance of the maximum likelihood estimation method over other methods with the least overall rank but shows that the maximum product of spacing is most advantageous when the sample size is 200. More so, the EHLST model efficacy is demonstrated through its application on Nigeria inflation rates dataset using the maximum likelihood and maximum product of spacing estimation procedures. Furthermore, the volatility modeling of the Nigeria inflation log-returns using the GARCH-type models with the EHLST innovation density relative to ten commonly used innovation densities validates the superiority of the GARCH (1, 1) and GJRGARCH (1, 1) models with EHLST innovation density in both in-sample and out-samples performance over other models.

    Citation: Obinna D. Adubisi, Ahmed Abdulkadir, Chidi. E. Adubisi. A new hybrid form of the skew-t distribution: estimation methods comparison via Monte Carlo simulation and GARCH model application[J]. Data Science in Finance and Economics, 2022, 2(2): 54-79. doi: 10.3934/DSFE.2022003

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  • In this work, estimating the exponentiated half logistic skew-t model parameters using some classical estimation procedures is considered. The finite sample performance of the EHLST parameter estimates is examined through extensive Monte Carlo simulations. The ordering performance of the six criterions was based on the partial and overall ranks of the estimation procedures for all parameter combinations. The criterions performance ordering from finest to poorest, using the overall ranks is maximum likelihood, maximum product of spacing, Anderson-Darling, Cramer-von Mises, least squares and weighted least squares estimators for all the parameter combinations. The simulation results confirm the dominance of the maximum likelihood estimation method over other methods with the least overall rank but shows that the maximum product of spacing is most advantageous when the sample size is 200. More so, the EHLST model efficacy is demonstrated through its application on Nigeria inflation rates dataset using the maximum likelihood and maximum product of spacing estimation procedures. Furthermore, the volatility modeling of the Nigeria inflation log-returns using the GARCH-type models with the EHLST innovation density relative to ten commonly used innovation densities validates the superiority of the GARCH (1, 1) and GJRGARCH (1, 1) models with EHLST innovation density in both in-sample and out-samples performance over other models.



    Many problems in the real world, such as in the area of physics, engineering, natural sciences, economics and so on, can be represented by many nonlinear partial differential equations. Therefore, it is of great significance to determine the exact solutions of such equations to better understand their physical properties. Generally speaking, no direct expression formulas are available for these solutions, even for simple nonlinear partial differential equations. For the past few decades, many effective techniques have been developed by researchers for searching explicit solutions including the Lie symmetry analysis method [1,2,3,4], the Darboux transformation method [5], the inverse scattering transform method [6], the Jacobi elliptic function expansion method [7], the Kudryashov method [8], the (G/G)expansion method [9,10], the (G/G2)expansion method [11], the Painlevé analysis method [12] and so on.

    In the middle of the 19th century, Norwegian mathematician Sophus Lie (1842–1899) pioneered the Lie symmetry analysis. Lie inaugurated the theory of Lie symmetries and applied it to differential equations. These research results have formed an important branch of mathematics in the 20th century, which is known as a Lie group and Lie algebra theory. By utilizing the invariance of solutions under a symmetry group for a partial differential equation, one can obtain solutions of a PDE via solving a reduced differential equation with lesser independent variables. Unfortunately, Lie symmetry analysis involves in a large number of complex calculations. With the rapid development of computing software, the Lie symmetry analysis method has become the most effective and powerful technique for obtaining closed form solutions to nonlinear partial differential equations. Many exact solutions with an important physical significance, such as the similarity solutions and travelling wave solutions, can be obtained by using the Lie symmetry analysis method [13,14,15,16].

    During the investigation of differential equations, conservation laws play an important role and have been receiving increased attention [17]. On one hand, they can be used to judge whether a partial differential equation is complete integrable. On the other hand, conservation laws can also be used to verify the validity of numerical solution methods. Moreover, exact solutions of partial differential equations can be constructed by using conservation laws [18,19,20]. Therefore, the most important thing is how to derive conservation laws for a given differential equation. With the development of research, some methods for constructing conservation laws of equations have emerged. For example, Noether's theorem provides an efficient approach to construct conservation laws for systems with the Lagrangian formulation [21,22]. Furthermore, the Ibragimov theorem [23,24,25] and the multiplier method [26] have more wider applications in deriving conservation laws, as they can be used for arbitrary differential equations without the limitation of the Lagrangian formulation.

    The investigations of shallow water waves and solitary have gained a great significance to describe characteristics of nonlinear wave phenomena. Stable dynamics and excitations of single- and double-hump solitons in the Kerr nonlinear media have been achieved in Ref. [27]. Bidirectional solitons and interaction solutions for a new integrable fifth-order nonlinear equation have been investigated in Ref. [28]. Nonlinear dynamic behaviors of the generalized (3+1)-dimensional KP equation have been studied in Ref. [29]. Ten years ago, Wazwaz [30] derived a (2+1)-dimensional Korteweg–de Vries 4 (KdV4) equation by using the recursion operator of the KdV equation, which reads

    vxy+vxxxt+vxxxx+3(v2x)x+4vxvxt+2vxxvt=0. (1.1)

    As a KdV type equation, Eq (1.1) can be used to model the shallow-water waves, surface, and internal waves. In Ref. [30], Wazwaz derived multiple soliton solutions by invoking Hirota's bilinear method and other travelling wave solutions by using hyperbolic functions and trigonometric function methods. The Bäcklund transformations and soliton solutions have been investigated in Ref. [31]. In Ref. [32], the authors derived breather wave solutions by using the extend homoclinic test method and some travelling wave solutions by using the (G/G2)expansion method.

    In this work, we aim to investigate the (2+1)-dimensional KdV4 equation (1.1) by invoking the Lie symmetry analysis method and consider its conservation laws. In Section 2, we perform the Lie symmetry analysis method to Eq (1.1) and present two symmetry reductions. In Section 3, we present solutions of symmetry reductions obtained in Section 2 by different methods. In Section 4, we present the conservation laws of the equation by invoking the multiplier method and Noether's theorem, respectively. A few concluding remarks will be given in the final section.

    First, we employ the Lie classical method on Eq (1.1) to find its symmetries. In this end, we consider a parameter (ϵ) Lie symmetry group of infinitesimal transformations as:

    x=x+ϵφ(x,y,t,v)+O(ϵ2),y=y+ϵω(x,y,t,v)+O(ϵ2),t=t+ϵτ(x,y,t,v)+O(ϵ2),v=v+ϵη(x,y,t,v)+O(ϵ2), (2.1)

    in which φ,ω,τ and η are infinitesimal generators. The corresponding vector field on transformation (2.1) is given by

    R=φ(x,y,t,v)x+ω(x,y,t,v)y+τ(x,y,t,v)t+η(x,y,t,v)v. (2.2)

    Noticeably, symmetries of Eq (1.1) can be derived according to the symmetry conditions

    pr(4)R(vxy+vxxxt+vxxxx+3(v2x)x+4vxvxt+2vxxvt)|(1.1)=0, (2.3)

    where pr(4)R is the fourth prolongation formula of Eq (2.2), which is given by

    pr(4)R=R+ηxvx+ηxxvxx+ηxxxxvxxxx+ηtvt+ηxtvxt+ηxyvxy+ηxxxtvxxxt.

    By utilizing the invariance condition (2.3), we have

    ηx(6vxx+4vxt)+ηxx(6vx+2vt)+ηxxxx+2ηtvxx+4ηxtvx+ηxy+ηxxxt=0. (2.4)

    As the original equation is vxy+vxxxt+vxxxx+3(v2x)x+4vxvxt+2vxxvt=0, when substituting differential prolongations of η into (2.4) and separating on different derivatives of v, we set the coefficients of vxy, vxxxt, vxxxx, 3(v2x)x, 4vxvxt, 2vxxvt to be equal and other derivatives of v to be zero. Thereafter, we obtain the following homogeneous PDEs:

    φv=0,ωv=0,τv=0,ηvv=0,τx=0,ωx=0,ωt=0,φxx=0,φxt=0,ηxx=0,ηxy=0,ηxv=0,ηtv=0,6ηx+2ηtφy=0,4ηxτy=0,4ηxt+ηyvφxy=0,ηv3φxτt=ηv4φxφt=2ηvφt3φx=2ηv2φxτt=ηvφxωy.

    By calculating the above PDEs, the symmetries of the KdV4 equation (1.1) can be given as

    σ1=f(y)x+y+f(y)t2v,σ2=g(y)x+t+g(y)t2v,σ3=h(y)x+4yt+(h(y)t23t+x)v,σ4=(tx)x2yy+vv,σ5=(3tx)x+2tt+vv,σ6=2(t+x)yx+4y2y+4tyt+(xtt22yv)v.

    According to the results of the infinitesimal transformation, the corresponding single parameter transformation groups are:

    G1:(x,y,t,v)(x+y+εf(a)da,y+ε,t,v+f(y+ε)t2),G2:(x,y,t,v)(x+y+εg(a)da,y,t+ε,v+g(y+ε)t2),G3:(x,y,t,v)(x+y+εh(a)da,y,t+4εy,v+h(y+ε)t23εt+εx),G4:(x,y,t,v)(εt+eεx,e2εy,t,eεv),G5:(x,y,t,v)(3εt+eεx,y,e2εt,eεv),G6:(x,y,t,v)(2εty+x12εy,y14εy,t14εy,ε(xtt2)14εy+v14εy).

    As a result, if v=ρ(x,y,t) is a solution of the KdV4 equation (1.1), so are

    v1=f(y+ε)t2+ρ(xy+εf(a)da,yε,t),v2=g(y+ε)t2+ρ(xy+εg(a)da,y,tε),v3=h(y+ε)t23εt+εx+ρ(xy+εh(a)da,y,t4εy),v4=eερ(eε(xεt),e2εy,t),v5=eερ(eε(x3εt),y,e2εt),v6=ε(xtt2)14εy+114εyρ((12εy)(x2εty),y1+4εy,(14εy)t).

    In what follows, we implement symmetry reductions of the KdV4 equation (1.1) by employing its translation symmetries. We choose two combinations kσ1+σ2 and 2σ3+σ4σ5 as examples to perform symmetry reductions.

    (ⅰ) For symmetry kσ1+σ2, by taking f(y)=1,g(y)=0, then it gives three invariants

    X=xkt,Y=ykt,v=H(X,Y). (2.5)

    By substituting (2.5) into Eq (1.1), we have

    HXYkHXXXY+(1k)HXXXX+3(HX)2X4kHX(HXX+HXY)2kHXX(HX+HY)=0. (2.6)

    By calculation, Γ1=(k1)X+kY and Γ2=X are Lie point symmetries of Eq (2.6). Considering the symmetry Γ=Γ1+lΓ2, where l being an arbitrary constant, it provides us with two invariants

    χ=Xl+k1kY,H=F(χ). (2.7)

    By substituting (2.7) into Eq (2.6), we have

    klFχχχχ+6klFχFχχ(l+k1)Fχχ=0. (2.8)

    (ⅱ) For symmetry 2σ4+σ5, it admits three invariants

    α=xty14,β=ty,v=P(α,β)y14. (2.9)

    By substituting (2.9) into Eq (1.1), we have

    14Pαα12βPαβ+12Pα+Pαααβ+4PαPαβ+2PααPβ=0. (2.10)

    By calculation, Γ=12αα+ββ+(116β+12P)P is a Lie point symmetrie of Eq (2.10), and here Γ provides us with two invariants

    γ=α2β,P=Q(γ)α+β8. (2.11)

    By substituting (2.11) into Eq (2.10), we have

    8γ3Qγγγγ+24γ2QγγQγγ2Qγγ+24γ2Qγγγ8γQQγγ+4γQ2γ+6γQγγ32γQγ+12Q=0. (2.12)

    Upon integrating Eq (2.8) twice, we get

    12klF2χχ+klF3χ(l+k1)2F2χ+mFχ+n=0. (3.1)

    where m and n are constants. Denote Fχ as Ω, then Eq (3.1) can be written as

    Ω2χ=2Ω3+(k+l1)klΩ22mklΩ2nkl. (3.2)

    If the polynomial equation

    2Ω3+(k+l1)klΩ22mklΩ2nkl=0.

    has three roots p1,p2,p3 and p1>p2>p3, then Eq (3.2) can be rewritten as

    Ω2χ=2(Ωp1)(Ωp2)(Ωp3), (3.3)

    whose solution is

    Ω=p2+(p1p2)cn2(12(p1p3)χ,p1p2p1p3), (3.4)

    in which cn denotes cosine elliptic function. As Ω=Fχ, integrating (3.4) once and going back to variables x,y and t, we acquire a periodic solution of the KdV4 equation

    v(x,y,t)=2(p1p3)[EllipticE(sn(12(p1p3)χ,p1p2p1p3),p1p2p1p3)]+p3χ+C, (3.5)

    where χ=xl+k1ky+(l1)t, and C is a constant of integration. Comparing Eqs (3.2) and (3.3), we can easily find that the following relationship between p1, p2, p3 and k, l needs to satisfy

    2p1+2p2+2p3=k+l1kl. (3.6)

    Furthermore

    EllipticE(z,μ)=z01μ2m21m2dm

    denotes the incomplete elliptic integral. By selecting k=2, l=3, p1=5, p2=3, p3=1, C=0 in (3.7), we present the profiles of Eq (3.5) in Figure 1 with t=0.

    Figure 1.  (Color online) Profiles of solution (3.5). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=1.

    By analysis, we can see the physical characteristics in Figure 1. On one hand, the features of the elliptic function are shown in Figure 1(a), (c); on the other hand, it exhibits certain periodicities in Figure 1(b).

    If the roots satisfy p2=p3, then the solution (3.5) can be reduced to the following form

    v(x,y,t)=2p12p2tanh(2(p1p2)χ2)+p2χ+C.

    When p1=2h2(h is a constant) and p2=p3=0, then the solution (3.5) can be reduced by

    v(x,y,t)=2htanh(hχ)+C, (3.7)

    which is similar to the solution obtained in [30]. By selecting h=2, k=3, l=4, C=2 in (3.5), we present the profiles of Eq (3.7) in Figure 2 with t=1. By observing Figure 2, we can see that Eq (3.7) is a kink-type hyperbolic function solution.

    Figure 2.  (Color online) Profiles of solution (3.7). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=2.

    In what follows, we investigate the solutions of Eq (2.8) by invoking the (G/G) expansion method [9]. By balancing the highest order derivative Fχχχχ and the nonlinear term of the highest order FχFχχ, we assume the solution of Eq (2.8) in this form

    F=b1(G(χ)G(χ))1+a0+a1(G(χ)G(χ)), (3.8)

    where G(χ) satisfies the following Riccati equation

    G(χ)+λG(χ)+μG(χ)=0. (3.9)

    Substituting Eq (3.8) into Eq (2.8), using Eq (3.9) for reduction, collecting all the powers of (G(χ)G(χ))1 and (G(χ)G(χ)), then equating all the obtained coefficients to zero, eleven algebraic equations can be obtained:

    12a21kl+24a1kl=0,30a21klλ+60a1klλ=0,12b21klμ3+24b1klμ4=0,30b21klλμ2+60b1klλμ3=0,24a21klλ224a21klμ+50a1klλ2+12a1b1kl+40a1klμ+2a1k2a1l2a1=0,12a1b1klμ3+24b21klλ2μ+50b1klλ2μ2+24b21klμ2+40b1klμ3+2b1kμ22b1lμ22b1μ2=0,24a1b1klλμ2+6b21klλ3+15b1klλ3μ+36b21klλμ+60b1klλμ2+3b1kλμ3b1lλμ3b1λμ=0,6a21klλ336a21klλμ+15a1klλ3+24a1b1klλ+60a1klλμ+3a1kλ3a1lλ3a1λ=0,6a21klλμ2+a1klλ3μ+8a1klλμ2+b1klλ3+6b21klλ+8b1klλμ+a1kλμa1lλμa1λμ+b1kλb1lλb1λ=0,12a1b1klλ2μ+b1klλ412a1b1klμ2+12b21klλ2+22b1klλ2μ+12b21klμ+16b1klμ2+b1kλ2b1lλ2+2b1kμ2b1lμb1λ22b1μ=0,12a21klλ2μ+a1klλ412a21klμ2+12a1b1klλ2+22a1klλ2μ+12a1b1klμ+16a1klμ2+a1kλ2a1lλ2+2a1kμ2a1lμa1λ22a1μ=0.

    Solving above equations with Maple, two kinds of solutions are obtained:

    Case 1:

    b1=2μ,a0=a0,a1=0,k=l+1lλ24lμ+1,l=l.

    Case 2:

    a0=a0,a1=2,b1=0,k=l+1lλ24lμ+1,l=l.

    As a result, six kinds of solutions to the KdV4 equation (1.1) can be obtained accordingly.

    For the first case, when λ24μ>0, we get one hyperbolic function solution

    v(x,y,t)=λμ4μλ24μ(C1cosh(Λ1χ)+C2sinh(Λ1χ)C1sinh(Λ1χ)+C2cosh(Λ1χ))+a0, (3.10)

    where Λ1=λ24μ2, χ=xl+k1ky+(l1)t with k=l+1lλ24lμ+1 and C1,C2 are arbitrary constants. By selecting λ=3, μ=2, l=3, c1=13, c2=14, a0=1, we present the profiles of Eq (3.10) in Figure 3 with t=1.

    Figure 3.  (Color online) Profiles of solution (3.10). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=0.

    Noticeably, Eq (3.10) is the combination of hyperbolic sine function and hyperbolic cosine function. When choosing special parameters, its expression may have discontinuities, and multiple peaks and valleys appeared in Figure 3.

    When λ24μ<0, we acquire one trigonometric function solution

    v(x,y,t)=λμ4μ4μλ2(C1cos(Λ2χ)+C2sin(Λ2χ)C1sin(Λ2χ)+C2cos(Λ2χ))+a0, (3.11)

    where Λ2=4μλ22, χ=xl+k1ky+(l1)t with k=l+1lλ24lμ+1 and C1,C2 are arbitrary constants. By selecting λ=2, μ=2, l=2, c1=1, c2=3, a0=0, we present the profiles of Eq (3.11) in Figure 4 with t=1.

    Figure 4.  (Color online) Profiles of solution (3.11). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=3.

    Noticeably, Eq (3.11) is the combination of sine function and cosine function. When choosing special parameters, its denominator is similar to its numerator, so features of cotangent function appeared in Figure 4.

    When λ24μ=0, we achieve one rational function solution

    v(x,y,t)=λμ2μ(C1+C2(x2ll+1y+(l1)t)C2)+a0, (3.12)

    where C1,C2 are arbitrary constants. Noticeably, Eq (3.12) is a plane wave solution.

    For the second case, when λ24μ>0, we get the other hyperbolic function solution

    v(x,y,t)=λ+λ24μ(C1sinh(Λ1χ)+C2cosh(Λ1χ)C1cosh(Λ1χ)+C2sinh(Λ1χ))+a0, (3.13)

    where Λ1=λ24μ2, χ=xl+k1ky+(l1)t with k=l+1lλ24lμ+1 and C1,C2 are arbitrary constants. By selecting λ=3, μ=2, l=3, c1=14, c2=13, a0=2, we present the profiles of Eq (3.13) in Figure 5 with t=0.

    Figure 5.  (Color online) Profiles of solution (3.13). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=0.

    Noticeably, Eq (3.13) is the combination of hyperbolic sine function and hyperbolic cosine function and is seeming like the inverse of Eq (3.10). When choosing special parameters, its expression may also have discontinuities, and multiple peaks and valleys appeared in Figure 5.

    When λ24μ<0, we acquire the other trigonometric function solution

    v(x,y,t)=λ+4μλ2(C1sin(Λ2χ)+C2cos(Λ2χ)C1cos(Λ2χ)+C2sin(Λ2χ))+a0, (3.14)

    where Λ2=4μλ22, χ=xl+k1ky+(l1)t with k=l+1lλ24lμ+1 and C1,C2 are arbitrary constants. By selecting λ=2, μ=2, l=2, c1=1, c2=3, a0=0, we present the profiles of Eq (3.14) in Figure 6 with t=1.

    Figure 6.  (Color online) Profiles of solution (3.14). (a) Perspective view of 3D profile. (b) Overhead view of 2D profile. (c) The wave propagation pattern along the x axis with y=3.

    Noticeably, Eq (3.14) is the combination of sine function and cosine function and is seeming like the inverse of Eq (3.11). When choosing special parameters, its denominator is similar to its numerator, so features of cotangent function appeared in Figure 6.

    When λ24μ=0, we achieve the other rational function solution

    v(x,y,t)=λ+2μC2C1+C2(x2ll+1y+(l1)t)+a0, (3.15)

    where C1,C2 are arbitrary constants. Evidently the denominator of Eq (3.15) is a plane, so Eq (3.15) is a curved surface.

    First, we suppose the solution of (2.12) is given by

    Q=k=0Qkγk,

    where Q0,Q1,,Qk, are constants to be determined. By substituting the expressions of Qγ,Qγγ, Qγγγ, Qγγγγ into Eq (2.12), we have

    12Q0+(12Q2+4Q21Q116Q0Q2)γ+(48a2a148a3a0+180a392a2)γ2+(96a0a4+120a1a3+96a2210a3+840a4)γ3+(224a4a1+416a2a3160a5a0+2520a5352a4)χ4++((8k4+8k32k22k)Qk+1+(24k+1l=1(k+1l)(kl)lQlQk+1l8k+1l=1(k+1l)(kl)QlQk+1l+4k+1l=1(k+1l)lQlQk+1l)(k2+k12)Qk)γk+=0, (3.16)

    By equating the coefficients of the powers of γ in Eq (3.16) to be zero, we get

    Q0=0,Q2=112Q113Q21,Q3=140Q2415Q1Q2,Qk+1=1(8k4+8k32k22k)(24k+1l=1(k+1l)(kl)lQlQk+1l+8k+1l=1(k+1l)(kl)QlQk+1l4k+1l=1(k+1l)lQlQk+1l)+(k2+k12)Qk).

    Therefore, the power series solution of Eq (2.12) is

    Q(γ)=Q1γ+(112Q113Q21)γ2+(140Q2415Q1Q2)χ3++1(8k4+8k32k22k)(24k+1l=1(k+1l)(kl)lQlQk+1l+8k+1l=1(k+1l)(kl)QlQk+1l4k+1l=1(k+1l)lQlQk+1l+(k2+k12)Qk)γk+1+. (3.17)

    Hence, the particular solution of Eq (1.1) is given by

    v(x,y,t)=y14Q((xt)2ty)xt+t8y,

    where Q((xt)2ty) is determined by Eq (3.17).

    In this section, we derive the conservation laws of the KdV4 equation (1.1) by means of the multiplier method and Noether's theorem, respectively.

    In the beginning, we invoke the multiplier method to determine conserved quantities of the KdV4 equation (1.1). In order to obtain the first order multipliers R, namely

    R=R(x,y,t,v,vx,vy,vt).

    These multipliers are derived by

    δδv[R(vxy+vxxxt+vxxxx+3(v2x)x+4vxvxt+2vxxvt)]=0. (4.1)

    in which

    δδv=vDxvxDyvyDtvt+D2xvxx+DxDyvxy+DxDtvxt+D4xvxxxx+D3xDtvxxxt

    denotes the Euler operator and Dx,Dy,Dt are total derivative operators. Expanding (4.1) and collecting on derivatives of v from second order to fourth order, then equating their coefficients to be zero, we get twenty-four PDEs:

    Rxx=0,Rxv=0,Rtv=0,Rxtvx=0,Rxvy=0,Rxvt=0,Rtvy=0,Rvv=0,Rvvx=0,Rvvy=0,Rvvt=0,Rvxvy=0,Rvxvt=0,Rvyvt=0,Rvxvx=0,Rvyvy=0,Rvtvt=0,2Rxvx+2RvRyvy=0,2Rv+RtvxRtvt+3RxvxRyvy=0,4Rxtvx+Ryvvx+Rxy=0,2RvRtvt=0,(6Rv2Rtvt+2Rxvx2Ryvy)vt+(18Rv6Rtvt+4Rtvx+6Rxvx6Ryvy)vx+6Rx+2Rt+Ryvx=0,4Rxtvx+Ryvvx+Rxy=0,12Rvvx4Ryvyvx+4Rx+Ryvt=0.

    By solving above equations with Maple, we obtain

    R=g(y)t2vxg(y)+f(y)+C1(tvx+2tvt+xvx+6yvx+4yvy+vx)+C2vx+C3vy+C4vt.

    where C1, C2, C3, C4 are arbitrary constants and g,f are functions of y. The conserved quantities of Eq (1.1) can be derived by employing the following divergence expression

    DxCx+DyCy+DtCt=R(vxy+vxxxt+vxxxx+3(v2x)x+4vxvxt+2vxxvt),

    with Cx,Cy are spatial fluxes and Ct is the conserved density. Therefore, we obtain the following conserved vectors (Cx,Cy,Ct) with regard to the six multipliers:

    Case 1. For R1=tvx+2tvt+xvx+6yvx+4yvy+vx, we get the corresponding conserved vector corresponding to R1 as follows

    Cx1=tvxvxx+tvxvxxx12tv2xx+2tv3x+7tv2xvt+tvyvt+2tvxxtvttv2xt+2tvxxxvt2tvxxvxt+4tvxv2t+xvxvxxt+xvxvxxx12xv2xx2vxvxx+2xv3x+xv2xvt+6yvxvxxt+6yvxvxxx3yv2xx+12yv3x+6yv2xvt+2yv2y+2yvxxtvy+2yvxxyvt2yvxyvxt2vvxxt+4yvxxxvy4yvxxvxy+12yv2xvy83yvxyvtv+83yvxtvyv+8yvxvyvt83vxvtv+vxxtv+vxxxv+3v2xv+2vxvtvxvyxvxxt+vxtxvxxx+vxx3xv2x2xvxvt+2vxvtdx,Cy1=12tv2x+tvxvt+12xv2x+3yv2x+2yvvxxxt+2yv2xx4yv3x+83yvxxvtv+163yvxvxtv+v+vxv,Ct1=tv2xxtv3xtvxvy+tv2xx12xv2xx+xv3x3yv2xx+6yv3x2yvvxxxy163yvxvxyv83yvxxvyv13v2xvxv2x.

    Case 2. For R2=vx, we get the corresponding conserved vector corresponding to R2 as follows

    Cx2=2v3x+vxvxxx12v2xx+vxvxxt+v2xvt,Cy2=12v2x,Ct2=v3x12v2xx.

    Case 3. For R3=vy, we get the corresponding conserved vector corresponding to R3 as follows

    Cx3=12v2y+12vxxtvy+12vxxyvt12vxyvxt+vxxxvyvxxvxy+3v2xvy23vvxyvt+23vvxtvy+2vxvyvt,Cy3=12vvxxxt+12v2xxv3x+23vvxxvt+43vvxvxt,Ct3=12vvxxxy43vvxvxy23vvxxvy.

    Case 4. For R4=vt, we get the corresponding conserved vector corresponding to R4 as follows

    Cx4=12vyvt+vxxtvt12v2xt+vxxxvtvxxvxt+3v2xvt+2vxv2t,Cy4=12vxvt,Ct4=12vxvy+12v2xxv3x.

    Case 5. For R5=g(y)t2vxg(y), we get the corresponding conserved vector corresponding to R5 as follows

    Cx5=vyg(y)t+vxxtg(y)t+vxxxg(y)t+3v2xg(y)t+2vxvtg(y)t2vxvxxtg(y)2vxvxxxg(y)+v2xxg(y)4v3xg(y)2v2xvtg(y),Cy5=2vxvxtg(y)t,Ct5=2vxyvxg(y)t+v2xxg(y)2v3xg(y).

    Case 6. For R6=f(y), we get the corresponding conserved vector corresponding to R6 as follows

    Cx6=f(y)v+f(y)vxxx+3f(y)v2x2f(y)vvxt,Cy6=f(y)vx,Ct6=f(y)vxxx+3f(y)v2x+2f(y)vvxx.

    Next, we construct the conservation laws of Eq (1.1) by utilizing the classical Noether's theorem [20]. The second order Lagrangian of equation (1.1) is

    L=12vxvy+12vxxvxt+12v2xxv3xv2xvt, (4.2)

    Therefore, the Noether symmetries

    R=φ(x,y,t,v)x+ω(x,y,t,v)y+τ(x,y,t,v)t+η(x,y,t,v)v

    of the KdV4 equation (1.1) are established by applying the Lagrangian equation (4.2) on the following equation

    R[2]L+L(Dx(φ)+Dy(ω)+Dt(τ))+Dx(Bx)+Dy(By)+Dt(Bt)=0, (4.3)

    where R[2] is the second prolongation of R and Bx, By, Bt are gauge functions. Expanding the above equation (4.3) and seeking solutions for the resulting system of partial differential equations, the following Noether symmetries as well as their corresponding gauge functions can be obtained:

    σ1=f(y)x+y+f(y)t2v,Bx=14f(y)tv,By=0,Bt=0,σ2=g(y)x+t+g(y)t2v,Bx=14g(y)tv,By=0,Bt=0,σ3=h(y)x+4yt+(h(y)t23t+x)v,Bx=14h(y)tv,By=12v,Bt=0,σ4=(t+x)x+4yy+2ttvv,Bx=0,By=0,Bt=0,σ5=2(t+x)yx+4y2y+4tyt+(xtt22yv)v,Bx=12v2+12vx,By=12tv,Bt=0.

    By using the above Noether symmetries and invoking the following formula [21]

    Ck=Lξk+(ηuxjξj)(LuxkΣkl=1Dxl(Luxlxk))+Σnl=k(ζluxlxjξj)LuxkxlBk.

    We obtain the corresponding conserved vectors, which are given by

    Ct1=12f(y)tv2x+f(y)v3x+v2xvy12f(y)v2xx12vxxvxy,Cx1=12f(y)v2xx+2f(y)v3x+f(y)v2xvt+3v2xvt14f(y)tvy+12vyvt32f(y)tv2xf(y)tvxvt+2vxv2t12f(y)tvxxt+f(y)vxvxxt+vxxtvt12f(y)tvxxx+f(y)vxvxxx+vxxxvt12v2xtvxxvxt+14f(y)tv,Cy1=12vxvy+12vxxvxt+12v2xxv3xv2xvt14f(y)tvx+12f(y)v2x+12vxvt.
    Ct2=12vxvy+12v2xxv3x12g(y)tv2x+g(y)v3x12g(y)v2xx,Cx2=12g(y)v2xx+2g(y)v3x+g(y)v2xvt14g(y)tvy+12vyvt32g(y)tv2x+3v2xvtg(y)tvxvt+2vxv2t12g(y)tvxxx+g(y)vxvxxx+vxxxvt12g(y)tvxxt+g(y)vxvxxt+vxxtvt12v2xtvxxvxt+14g(y)tv,Cy2=14g(y)tvx+12g(y)v2x+12vxvt.
    Ct3=2yvxvy+2yv2xx4yv3x12h(y)tv2x+3tv2xxv2x+h(y)v3x+12vxx12h(y)v2xx,Cx3=12h(y)v2xx+2h(y)v3x+h(y)v2xvt14h(y)tvy+32tvy12xvy+2yvyvt32h(y)tv2x+9tv2x3xv2x+12yv2xvth(y)tvxvt+6tvxvt2xvxvt+8yvxv2t12h(y)tvxxt+3tvxxtxvxxt+h(y)vxvxxt+4yvxxtvt12h(y)tvxxx+3tvxxxxvxxx+h(y)vxvxxx+4yvxxxvt+12vxt+vxx2yv2xt4yvxxvxt+14h(y)tv,Cy3=14h(y)tvx+32tvx12xvx+12h(y)v2x+2yvxvt+12v.
    Ct4=tvxvy+12tv2xxtv3x+v2xv+xv3x+4yv2xvy12vxvxx12xv2xx2yvxxvxy,Cx4=12tv2xx+2tv3x+5tv2xvt+xvxvy12xv2xxxv3x+xv2xvt+12vvy+2yv2y+tvyvt+3vv2x+3xv3x+12yv2xvy+2vvxvt+2tv2xvt+8yvxvyvt+4tvxv2t+vvxxt+tvxvxxt+xvxvxxt+4yvxxtvy+2tvxxtvt+vvxxx+tvxvxxx+xvxvxxx+4yvxxxvy+2tvxxxvt12vxvxt2yvxyvxttv2xtvxvxx4yvxxvxy2tvxxvxt,Cy4=2yvxxvxt+2yv2xx4yv3x4yv2xvt+12vvx+12tv2x+12xv2x+tvxvt.
    Ct5=2tyvxvy+tyv2xx2tyv3xxtv2x+t2v2x+2yvv2x+2xyv3x+4y2v2xvy+12tvxx2yvxvxxxyv2xx2y2vxxvxy,Cx5=xyv2xx+4xyv3x+2xyv2xvtytv2xx+4ytv3x+14ytv2xvt12xtvy+12t2vy+yvvy+2y2v2y+2ytvyvt3xtv2x+3t2v2x+6yvv2x+12y2v2xvy2xtvxvt+2t2vxvt+4yvvxvt+8y2vxvyvt+8ytvxv2txtvxxt+t2vxxt+2yvvxxt+2xyvxvxxt+2ytvxvxxt+4y2vxxtvy+4ytvxxtvtxtvxxx+t2vxxx+2yvvxxx+2xyvxvxxx+2ytvxvxxx+4y2vxxxvy+4ytvxxxvt+12tvxt2yvxvxt2y2vxyvxt2ytv2xt+tvxx4yvxvxx4y2vxxvxy4ytvxxvxt12v212vx,Cy5=2y2vxxvxt+2y2v2xx4y2v3x4y2v2xvt12vxvt+12t2vx+yvvx+xyv2x+tyv2x+2tyvxvt+12tv.

    In this paper, we performed the Lie symmetry analysis method to a (2+1)-dimensional KdV4 equation and derived Lie symmetries of this equation, then used these symmetries to perform symmetry reductions. By using translation symmetries, two fourth-order ordinary differential equations were obtained. Different methods were adopted in order to derive the solutions of the obtained differential equations. For one fourth-order ordinary differential equation, we presented its solutions by using the direct integration method and the (G/G)-expansion method, respectively. Furthermore, their corresponding solutions were shown with the appropriate graphical representations. For the other fourth-order ordinary differential equation, we derived its solutions by using the power series technique. Finally, two kinds of conserved vectors of this equation are presented by invoking the multiplier method and Noether's theorem, respectively. Six multipliers were obtained from the multiplier method, and thus six local conservation laws for the KdV4 equation (1.1) were given, while five local conservation laws were obtained by Noether's theorem. It is necessary to point out that the multiplier method has more wider applications in deriving conservation laws than Noether's theorem. Although we successfully presented the power series solution for the KdV4 equation (1.1), we had difficulties in discussing the corresponding convergence analysis. Its convergence analysis is worth studying in our future work. For the Lie symmetry analysis method, many attentions have been concentrated on (1+1)-dimensional and (2+1)-dimensional differential equations. For (3+1)-dimensional differential equations, there are not many relevant documents. We will investigate the (3+1)-dimensional differential equations by using the Lie point symmetry analysis method in the future.

    This work was supported by Henan Youth Backbone Teacher Project (No. 2019GGJS204) and the Natural Science Foundation of Henan Province (No.232300420361).

    The author declares there is no conflict of interest.



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