# Lie Symmetry Analysis, Exact Solutions, and Conservation Laws of Variable-Coefficients Boiti-Leon-Pempinelli Equation

Authors: Feng Zhang; Yuru Hu; Xiangpeng Xin
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this article, we study the generalized (2+1)-dimensional variable-coefficients Boiti-Leon-Pempinelli (vcBLP) equation. Using Lie’s invariance infinitesimal criterion, equivalence transformations and differential invariants are derived. Applying differential invariants to construct an explicit transformation that makes vcBLP transform to the constant coefficient form, then transform to the well-known Burgers equation. The infinitesimal generators of vcBLP are obtained using the Lie group method; then, the optimal system of one-dimensional subalgebras is determined. According to the optimal system, the (1+1)-dimensional reduced partial differential equations (PDEs) are obtained by similarity reductions. Through G′/G-expansion method leads to exact solutions of vcBLP and plots the corresponding 3-dimensional figures. Subsequently, the conservation laws of vcBLP are determined using the multiplier method.


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## Body

## 1. Introduction

Nonlinear issues are widespread in some natural disciplines, and many difficult problems in some disciplines can be reduced to solving a certain PDE or investigating some properties of a PDE [1, 2]. With the rapid development of research fields like hydrodynamics and quantum physics, it has become increasingly important to investigate the exact solutions and certain properties of nonlinear evolution equations [3–5]. Compared with the PDEs with constant coefficients, the PDEs with variable coefficients can describe richer natural phenomena and construct more detailed and complex physical models [6–8].In this paper, we focus on the generalized (2+1)-dimensional Boiti-Leon-Pempinelli equation with time-part variable coefficients as
(1)F1=uyt+atuxuy+atuuxy+btuxxy+ctvxxx=0,F2=vt+dtvxx+etuvx=0,where at, bt, ct, dt, and et are any functions with respect to time t. It represents the development of the components in the horizontal velocity in the x and y directions when the water wave propagates in a channel of unchanging depth and infinitely small width [9]. The vcBLP is conditionally integrable, and the necessary conditions for it to be Painlevé integrable are d′tet=dte′t and a′tbt=atb′t. Some exact solutions of vcBLP were obtained in [10] by extended tanh-function method. We can find some periodic solutions and soliton solutions of vcBLP obtained with the homogeneous balance method in [11], and the conservation laws for the constant coefficients BLP were discussed in [9]. By reviewing the relevant literatures, no one has studied vcBLP using the Lie group method.The outline of this article is as follows. In Section2, we construct the equivalence transformations and differential invariants of vcBLP. Based on these, we give an explicit transformation to its constant coefficient form. In Section 3, the infinitesimal generators of vcBLP are obtained using the Lie group method, and then, the optimal system for the one-dimensional subalgebras is constructed. In Section 4, we obtain six sets of (1+1)-dimensional reduced PDEs by similarity reductions to vcBLP. In Section 5, some exact solutions of vcBLP are shown using the G′/G-expansion method on the basis of the reduced PDEs in above section. In Section 6, we use the multiplier method to calculate the conservation laws of vcBLP. We can find the conclusions of this paper in Section 7.


## 2. Equivalence Transformations and Differential Invariants of vcBLP

In this part, we construct the equivalence transformations [12] of Equation (1). The equivalence transformation of Equation (1) is a nondegenerate point transformation which from x,y,t,u,v to x~,y~,t~,u~,v~ [13]. It has the same differential structure but different coefficient functions a~t~,b~t~,c~t~,d~t~,e~t~ than the original equation. First, we assume that the auxiliary conditions are
(2)au=av=0,bu=bv=0,cu=cv=0,du=dv=0,eu=ev=0,with the one-parameter group of equivalence transformations is determined on the basis of
(3)x~=x+ε·ξx,y,t,u,v,a,b,c,d,e+oε2,y~=y+ε·ηx,y,t,u,v,a,b,c,d,e+oε2,t~=t+ε·τx,y,t,u,v,a,b,c,d,e+oε2,u~=u+ε·ϕ1x,y,t,u,v,a,b,c,d,e+oε2,v~=v+ε·ϕ2x,y,t,u,v,a,b,c,d,e+oε2,a~=a+ε·Δ1x,y,t,u,v,a,b,c,d,e+oε2,b~=b+ε·Δ2x,y,t,u,v,a,b,c,d,e+oε2,c~=c+ε·Δ3x,y,t,u,v,a,b,c,d,e+oε2,d~=d+ε·Δ4x,y,t,u,v,a,b,c,d,e+oε2,e~=e+ε·Δ5x,y,t,u,v,a,b,c,d,e+oε2,and ε is the group parameter. The vector field or generators of Equation (1) which corresponds to transformations (3) as
(4)Y=ξ∂∂x+η∂∂y+τ∂∂t+ϕ1∂∂u+ϕ2∂∂v+Δ1∂∂a+Δ2∂∂b+Δ3∂∂c+Δ4∂∂d+Δ5∂∂e.Since Equation (1) is invariant under the above transformations (3) and there exists the 3rd derivatives, we have to use the 3rd prolongation Y3. We define that
(5)f1,f2,f3,f4,f5≡at,bt,ct,dt,et,x1,x2,x3≡x,y,t,y1,y2≡u,v,ψ1,ψ2,ψ3≡ξ,η,τ,yji=∂yi∂xj,yjki=∂2yi∂xj∂xk,j,k=1,2,3,i=1,2.On the basis of the above Equation (5) and Y, the 3rd prolongation Y3 can be written as
(6)Y3=Y+ζji∂∂yji+ζjσi∂∂yjσi+ζjjσi∂∂yjjσi+ω~jr∂∂fxjr+ω¯yjr∂∂fyjr,r=1,2,⋯,5,where
(7)ζji=Djϕj−ykiDjψk,ζjσi=Dσζji−yjkiDσψk,ζjjσi=Dσζjji−yjjkiDσψk,ω~jr=D~jΔr−fxkrD~jψk−fyirD~jϕi,with
(8)Dj=∂∂xj+yji∂∂yi+yjki∂∂yki,D~j=∂∂xj+fxjr∂∂fr,D¯j=∂∂yj+fyjr∂∂fr.Under transformations (3), the invariance of Equation (1) requires that to satisfy the conditions
(9)Y3F1=0,Y3F2=0,Y3au=Y3av=0,Y3bu=Y3bv=0,Y3cu=Y3cv=0,Y3du=Y3dv=0,Y3eu=Y3ev=0.We can obtain a determining equation by bringing Equation (6), Equation (7), and Equation (8) into Equation (9). Subsequently, solving this determining equation, we get
(10)ξ=C1x+C2,η=ρy,τ=βt,ϕ1=C3u,ϕ2=gy,Δ1=aC1−C3−βt,Δ2=b2C1−βt,Δ3=−cρy+g+βt−3C1−C3,Δ4=d2C1−βt,Δ5=eC1−C3−βt,where β=βt,ρ=ρy, and g=gy are arbitrary smooth functions. The corresponding infinite dimensional equivalence group is generated by the following operators:
(11)Y1=x∂∂x+a∂∂a+2b∂∂b+3c∂∂c+2d∂∂d+e∂∂e,Y2=∂∂x,Y3=u∂∂u−a∂∂a+c∂∂c−e∂∂e,Yβ=β∂∂t−aβt∂∂a−bβt∂∂b−cβt∂∂c−dβt∂∂d−eβt∂∂e,Yρ=ρ∂∂y−cρy∂∂c,Yg=gv∂∂v−cg∂∂c.In the following, our task is to derive the zero-order differential invariants and the first-order differential invariants. We assume that the form of the zero-order differential invariant is(12)J=Jx,y,t,a,b,c,d,e.Applying the invariant testYJ=0 to the operators (11), we can obtain
(13)YkJ=0,k=1,2,3,β,ρ,g,and the above equations can be reduced to
(14)∂J∂x=0,∂J∂y=0,∂J∂c=0,∂J∂t=0,b∂J∂b+d∂J∂d=0,a∂J∂a+e∂J∂e=0,solving this system, we get that the zero-order differential invariants are
(15)J10=db,J20=ea,with invariant equations are
(16)a=0,b=0,d=0,e=0.Next, we derive the interesting first-order differential invariants. Similar to the above case, we suppose that the first-order differential invariant is the form(17)J=Jx,y,t,a,b,c,d,e,at,bt,ct,dt,et.In order to get the first-order differential invariants, we need to make the first prolongation of the operatorsYk,k=1,2,3,β,ρ,g using Y1=Y+ω~jr∂/∂fxjr, and Equation (11) can be rewritten as
(18)Y11=x∂∂x+a∂∂a+2b∂∂b+3c∂∂c+2d∂∂d+e∂∂e+at∂∂at+2bt∂∂bt+3ct∂∂ct+2dt∂∂dt+et∂∂et,Y21=∂∂x,Y31=u∂∂u−a∂∂a+c∂∂c−e∂∂e−at∂∂at+ct∂∂ct−et∂∂et,Yβ1=β∂∂t−aβt∂∂a−bβt∂∂b−cβt∂∂c−dβt∂∂d−eβt∂∂e−aβtt+2atβt∂∂at−bβtt+2btβt∂∂bt−cβtt+2ctβt∂∂ct−dβtt+2dtβt∂∂dt−eβtt+2etβt∂∂et,Yρ1=ρ∂∂y−cρy∂∂c−ctρy∂∂ct,Yg1=vg∂∂v−cg∂∂c−ctg∂∂ct.The invariance test associated with Equation (18) is
(19)Yk1J=0,k=1,2,3,β,ρ,g,solving this system yields
(20)J11=bact−atccabt−atb,J21=adt−atdabt−atb,J31=baet−ateaabt−atb,with invariant equations are
(21)abt−atb=0,act−atc=0,adt−atd=0,aet−ate=0.Based on the above facts, we use differential invariants to give the transformation of vcBLP to its constant coefficient form. We take the constant coefficient form of Equation (1) as
(22)uyt+m1uxuy+m1uuxy+m2uxxy+m3vxxx=0,vt+m4vxx+m5uvx=0,where mi,i=1,⋯,5 are arbitrary nonzero constants. To obtain the transformation from Equation (1) to Equation (22), we need a necessary condition that the coefficient functions of Equation (1) must satisfy the following equations
(23)db=m4m2,ea=m5m1,abt−atb=0,act−atc=0,adt−atd=0,aet−ate=0.Under this condition, the more general form of the coefficient functions are(24)at=m1γ3m5ct,bt=m2γ4m4ct,dt=γ4ct,et=γ3ct,therefore, Equation (1) becomes
(25)uyt+m1γ3m5ctuxuy+m1γ3m5ctuuxy+m2γ4m4ctuxxy+ctvxxx=0,vt+γ4ctvxx+γ3ctuvx=0,and there exists transformation
(26)u=m5m4·γ1γ4γ3u~x~,y~,t~,v=δ′yv~x~,y~,t~,x~=γ1x+γ2,y~=m42m3m5·γ3γ42δy,t~=γ12γ4m4∫ctdt+γ5,where γi,i=1,⋯,5 are arbitrary constants and δy is an arbitrary function. It is easy to verify that the above transformation maps Equation (1) into
(27)u~y~t~+m1u~x~u~y~+m1u~u~x~y~+m2u~x~x~y~+m3v~x~x~x~=0,(28)v~t~+m4v~x~x~+m5u~v~x~=0,so the constant coefficient form in other literature [9] is a special case of Equation (1), and it is easy to find that Equation (24) satisfies the conditions for integrability.We takev~=u~,y~=x~, then the above Equation (27) and Equation (28) become
(29)u~x~t~+m1u~x~2+m1u~u~x~x~+m2+m3u~x~x~x~=0,(30)u~t~+2m5u~u~x~+m4u~x~x~=0,subsequently, we integrate Equation (29) once with respect to x yields
(31)u~t~+m1u~u~x~+m2+m3u~x~x~=0,therefore, when m1=2m5,m4=m2+m3, the above Equation (27) and Equation (28) can be converted into the well-known constant coefficient Burgers equation.


## 3. Lie Classical Symmetry Analysis of vcBLP

Lie group method is an effective way to find invariant solutions and to explore certain properties by reducing the dimensionality of the equations [14]. It has been described in sufficient detail in many literatures [15–18]. To begin with, we suppose that the one-parameter ε Lie group in Equation (1) is
(32)x∗=x+ε·ξx,y,t,u,v+oε2,y∗=y+ε·ηx,y,t,u,v+oε2,t∗=t+ε·τx,y,t,u,v+oε2,u∗=u+ε·ϕ1x,y,t,u,v+oε2,v∗=v+ε·ϕ2x,y,t,u,v+oε2,and Equation (1) remains invariant under transformations (32).The vector field or infinitesimal generator of Equation (1) which corresponds to transformations (32) is
(33)V=ξx,y,t,u,v∂∂x+ηx,y,t,u,v∂∂y+τx,y,t,u,v∂∂t+ϕ1x,y,t,u,v∂∂u+ϕ2x,y,t,u,v∂∂v,its 3rd prolongation is written as Pr3V. Equation (1) remains invariant under transformations (32), which requires that
(34)Pr3VF1F1=0=0,Pr3VF2F2=0=0.Expanding Equation (34), the invariant conditions are redefined as
(35)φyt1+τa′tuuyx+b′tuxxy+c′tvxxx+atφx1uy+φy1ux+φxy1u+ϕ1uxy+btφxxy1+ctφxxx2=0,φt2+τd′tvxx+e′tuvx+etϕ1vx+ϕ2ux+φx1v+φx2u+dtφxx2=0,where
(36)φx1=Dxϕ1−Dxξux−Dxηuy−Dxτut,φx2=Dxϕ2−Dxξvx−Dxηvy−Dxτvt,φy1=Dyϕ1−Dyξux−Dyηuy−Dyτut,φt2=Dtϕ2−Dtξvx−Dtηvy−Dtτvt,φxx1=Dxφx1−Dxξuxx−Dxηuxy−Dxτuxt,φxx2=Dxφx2−Dxξvxx−Dxηvxy−Dxτvxt,φxy1=Dyφx1−Dxξuxy−Dxηuyy−Dxτuyt,φyt1=Dtφy1−Dyξuxt−Dyηuyt−Dyτutt,φxxy1=Dyφxx1−Dxξuxxy−Dxηuxyy−Dxτuxyt,φxxx2=Dxφxx2−Dxξvxxx−Dxηvxxy−Dxτvxxt,with Dx, Dy, and Dt are the total differentiation of x, y, and t, respectively.Solving Equation (35) yields the following results as
(37)ξ=C1x+C2,η=C1y+C3,ϕ1=−C1u,τ=C4bt+2C1bt·∫btdt,ϕ2=−C1v,where Cii=1,2,3,4 are any constants. Furthermore, the coefficient functions at, bt, ct, dt, and et in Equation (1), which depend on time t and have to satisfy the conditions
(38)a′tτ+atτt−2C1at=0,b′tτ+btτt−2C1bt=0,c′tτ+ctτt−2C1ct=0,d′tτ+dtτt−2C1dt=0,e′tτ+etτt−2C1et=0.Thus, the infinitesimal generators of Equation (1) are expanded by the below vector field
(39)V1=∂∂x,V2=∂∂y,V3=1bt∂∂t,V4=x∂∂x+y∂∂y+2bt∫btdt∂∂t−u∂∂u−v∂∂v.Depending on theVii=1,2,3,4, we have the following four groups
(40)G1:x,y,t,u,v⟶x+ε,y,t,u,v,G2:x,y,t,u,v⟶x,y+ε,t,u,v,G3:x,y,t,u,v⟶x,y,t+εbt,u,v,G4:x,y,t,u,v⟶xeε,yeε,t+2εbt∫btdt,ue−ε,ve−ε,where Gii=1,2,3,4 are one-parameter Lie point symmetry groups. It is not difficult to find that G1 and G2 are space translations and G4 is a dependent variable translation. For G3, it is a time translation when bt is an arbitrary constant. For the combination of G1 and G2 can be understood as a translation along a certain direction, the group invariant solutions are the traveling wave solutions. The most important application of the traveling wave solutions is to construct soliton solutions of the PDE. The soliton reflects a rather common nonlinear phenomenon in nature, which is mainly characterized by its superstability, i.e., the wave shape remains stable after the collision of two solitary waves with different velocities. We can also understand solitons as local traveling wave solutions of nonlinear development equations. Also, symmetry has a great connection with conservation laws in physics; for example, space translation corresponds to momentum conservation, and time translation corresponds to energy conservation.Generally speaking, it is possible to construct it group-invariant solutions for arbitrary subgroup or subalgebra. However, the Lie group has infinitely many subgroups with the same dimension, and it is impossible to compute the group-invariant solutions of all subgroups [19]. We have to sort them into some mutual equivalence, which requires the optimal system of one-dimensional subalgebras to be computed.To get the optimal system, we start by constructing the commutator table as Table1 with the help of Lie algebra Vi,Vi=ViVj−VjVi [15, 20].Table 1
Commutator table.
Vi,VjV1V2V3V4V1000V1V2000V2V30002V3V4−V1−V2−2V30With reference to Table1, the adjoint relationship table can be acquired as Table 2, where
(41)AdexpεViVj=Vj−εVi,Vj+ε22Vi,Vi,Vj−⋯,with ε is an infinitesimal real number.Table 2
Adjoint representation table.
AdV1V2V3V4V1V1V2V3V4−εV1V2V1V2V3V4−εV2V3V1V2V3V4−2εV3V4V1eεV2eεV3e2εV4Through Tables1 and 2, it is quite simple to obtain the optimal system for the one-dimensional subalgebras of Equation (1), which is given by the following forms
(42)1V1+α1V2+α2V3,2V2+α3V3,3V3,4V4.


## 4. Similarity Reductions of vcBLP

In the first step, we reduce Equation (1) to the (1+1)-dimensional PDEs based on the above optimal system using the similarly reductions. The similarity variables and the (1+1)-dimensional reduced PDEs can be found in Table 3, and the expressions for the coefficient functions depending on time t can be found in Table 4. We only show the process of calculation with the example of case V1+α1V2+α2V3; the results of other cases are in Tables 3 and 4.Table 3
Similarity reductions.
CaseSimilarity variablesReduced PDEsIV1+α1V2α1≠0X=y−α1x,Y=t,u=PX,Y,v=QX,Y.PXY−α1atPPXX+α12btPXXX−α13ctQXXX=0,QY−α1etPQX+α12dtQXX=0.IIV1+α2V3α2≠0X=x−1/α2∫1/τtdt,Y=y,u=PX,Y,v=QX,Y.α2k1PPYX+α2k2PXXY+α2k3QXXX−PXY=0,α2k5PQX+α2k4QXX−QX=0.IIIV1+α1V2+α2V3α1≠0,α2≠0X=x−1/α2∫1/τtdt,Y=y−α1/α2∫1/τtdt,u=PX,Y,v=QX,Y.α2k1PPYX+α2k2PXXY+α2k3QXXX−α1PYY−PXY=0,α2k5PQX+α2k4QXX−α1QY−QX=0.IVV2+α3V3α3≠0X=x,Y=y−1/α3∫1/τtdt,u=PX,Y,v=QX,Y.α3k1PPYX+α3k2PXXY+α3k3QXXX−PYY=0,α3k5PQX+α3k4QXX−QY=0.VV3X=x,Y=y,u=PX,Y,v=QX,Y.atPPYX+btPXXY+ctQXXX=0,dtQXX+etPQX+etQPX=0.VIV4X=xe−∫1/τtdt,Y=ye−∫1/τtdt,u=PX,Ye∫1/τtdt,v=QX,Ye∫1/τtdt.−XPXY+k2PXXY+k3QXXX−YPYY−2PY=0,k4QXX−XQX−YQY−Q=0.Table 4
The expressions of the coefficient functions.
CaseThe expressions of the coefficient functionsIV1+α1V2α1≠0at, bt, ct, dt, and et are any functions that depend on t.IIV1+α2V3α2≠0at=k1/τt, bt=k2/τt, ct=k3/τt, dt=k4/τt, et=k5/τt.IIIV1+α1V2+α2V3α1≠0,α2≠0at=k1/τt, bt=k2/τt, ct=k3/τt, dt=k4/τt, et=k5/τt.IVV2+α3V3α3≠0at=k1/τt, bt=k2/τt, ct=k3/τt, dt=k4/τt, et=k5/τt.VV3at, bt, ct, dt, and et are any functions that depend on t.VIV4at=et=0,bt=k2τte∫2/τtdt,ct=k3τte∫2/τtdt,dt=k4τte∫2/τtdt.For this Lie vector, its corresponding characteristic equation is(43)dx1=dyα1=dtα2·τt=du0=dv0,solving this equation to obtain the relative similarity variables are
(44)X=x−1α2∫1τtdt,Y=y−α1α2∫1τtdt,u=PX,Y,v=QX,Y.Through Equation (44), Equation (1) is reduced to the following forms:
(45)atPPYX+btPXXY+ctQXXX−1α2·τtPXY−α1α2·τtPYY=0,etPQX+dtQXX−α1α2·τtQY−1α2·τtQX=0,here to ensure that there are only two independent variables X and Y in Equation (45), and the coefficient functions satisfy conditions (38), so the expressions for the coefficient functions are
(46)at=k1τt,bt=k2τt,ct=k3τt,dt=k4τt,et=k5τt,where kii=1,⋯,5 are arbitrary constants. Substituting Equation (46) to Equation (45) yields the reduced PDEs as
(47)α2k1PPYX+α2k2PXXY+α2k3QXXX−α1PYY−PXY=0,α2k5QPX+α2k4QXX−α1QY−QX=0.


## 5. Exact Solutions of vcBLP

In this section, the exact solutions of corresponding reduced PDEs are found for some cases in Table3 using the G′/G-expansion method [21, 22]. For computational simplicity, we let ki=1i=1,⋯,5 for Tables 3 and 4.Case 1.
IIV1+α2V3
First, we assume that the traveling wave variables are [23].
(48)PX,Y=Pς,QX,Y=Qς,ς=X−VY,where V is the traveling wave speed to be determined.
Next, using Equation (48), the (1+1)-dimensional reduced PDEs are transformed into ordinary differential equations (ODEs)
(49)−α2VPP′′−α2VP′′+VP′′−α2VP′′′+α2Q′′′=0,α2PQ′−Q′+α2Q′′+α2QP′=0,where ′ represents the derivative of ς. Through homogeneous balance, we assume that the solutions of the ODEs (49) can be expressed as
(50)Pς=a1G′G+a0,a1≠0,Qς=b1G′G+b0,b1≠0,where a0,a1,b0,b1 are coefficients to be determined and G=Gς satisfies
(51)G′′+λG′+μG=0,with λ and μ are arbitrary constants.
Substituting Equation (50) and Equation (51) into Equation (49), then collecting the coefficients of the same order of G′/G and making them equal to zero yields
(52)μb1+α2λμb1−μa0b1−μa1b0=0,2α2b1−a1b1=0,b1+α23λb1−2λa1b1−a0b1−a1b0=0,λb1+α2λ2b1+2μb1−λa0b1−λa1b0−2μa1b1=0.
Solving Equation (52), we get
(53)a0=12λ+1α2,a1=1,b0=14Vλ,b1=12V,with V,λ,μ,α2 are any constants.
By applying Equation (53), Equation (50) can be rewritten as
(54)Pς=G′G+12λ+1α2,Qς=12VG′G+14Vλ,where ς=X−VY. Substituting the solutions of the Equation (51) into Equation (54), we obtain that the three types of exact solutions of reduced PDEs are as follows [24]:
Whenλ2−4μ>0,(55)P1ς=12λ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς+1α2,Q1ς=14Vλ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς,where ς=X−VY, l1, and l2 are any constants.
Whenλ2−4μ<0,
(56)P2ς=124μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς+1α2,Q2ς=14V4μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς,where ς=X−VY, l1, and l2 are any constants.
Whenλ2−4μ=0,
(57)P3ς=l2l1+l2ς+1α2,Q3ς=12l2Vl1+l2ς,where ς=X−VY, l1, and l2 are any constants.
By substituting the corresponding similarity variables in Table3 into the above solutions, the exact solutions of vcBLP are obtained as follows:
Whenλ2−4μ>0,
(58)u1ς=12λ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς+1α2,v1ς=14Vλ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς,where ς=x−1/α2·∫1/τtdt−Vy, l1, and l2 are any constants.
Whenλ2−4μ<0,
(59)u2ς=124μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς+1α2,v2ς=14V4μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς,where ς=x−1/α2·∫1/τtdt−Vy, l1, and l2 are any constants.
Whenλ2−4μ=0,
(60)u3ς=l2l1+l2ς+1α2,v3ς=12l2Vl1+l2ς,where ς=x−1/α2·∫1/τtdt−Vy, l1, and l2 are any constants.For the other cases, we also obtained three types of exact solutions of the reduced PDEs by the method ofG′/G-expansion, and we will not repeat the calculation process, just list the results of the calculation.Case 2.
IIIV1+α1V2+α2V3.
Whenλ2−4μ>0,
(61)u1ς=12λ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς+1−α1Vα2,(62)v1ς=14Vλ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς,where ς=x−1/α2∫1/τtdt−y−α1/α2∫1/τtdt·V, l1, l2, and V are any constants. We choose parameters λ=3, μ=2, α1=1, α2=1, l1=2, l2=1, V=2, and τt=t; the images of (61) and (62) are, respectively, in Figures 1 and 2.
Whenλ2−4μ<0,
(63)u2ς=124μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς+1−α1Vα2,(64)v2ς=14V4μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς,where ς=x−1/α2∫1/τtdt−y−α1/α2∫1/τtdt·V, l1, l2, and V are any constants. We choose parameters λ=4, μ=5, α1=2, α2=1, l1=3.5, l2=3, V=1, and τt=t2; the images of (63) and (64) are, respectively, in Figures 3 and 4.
Whenλ2−4μ=0,(65)u3ς=l2l1+l2ς+1α2−α1α2V,(66)v3ς=12l2Vl1+l2ς,where ς=x−1/α2∫1/τtdt−y−α1/α2∫1/τtdt·V, l1, l2, and V are any constants. We choose parameters α1=2, α2=1, l1=3, l2=2, V=1, and τt=t; the images of (65) and (66) are, respectively, in Figures 5 and 6.Case 3.
IVV2+α3V3.
Whenλ2−4μ>0,
(67)u1ς=12λ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς+a0−12λ,v1ς=18α3λ−14α3a0λ2−4μl1sinh1/2λ2−4μς+l2cosh1/2λ2−4μςl1cosh1/2λ2−4μς+l2sinh1/2λ2−4μς,where ς=x−1/2α3λ−2a0·y−1/α3∫1/τtdt, l1, l2, and a0 are any constants.
Whenλ2−4μ<0,
(68)u2ς=124μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς+a0−12λ,v2ς=18α3λ−14α3a04μ−λ2−l1sin1/24μ−λ2ς+l2cos1/24μ−λ2ςl1cos1/24μ−λ2ς+l2sin1/24μ−λ2ς,where ς=x−1/2α3λ−2a0·y−1/α3∫1/τtdt, l1, l2, and a0 are any constants.
Whenλ2−4μ=0,
(69)u3ς=a0−λ2+l2l1+l2ς,v3ς=14l2α3λ−2a0l1+l2ς,where ς=x−1/2α3λ−2a0·y−1/α3∫1/τtdt, l1, l2, and a0 are any constants.Figure 1
Evolution of Equation (61) at (a) t=10−1, (b) t=102, and (c) t=105. (d) Contour plot.
(a)(b)(c)(d)Figure 2
Evolution of Equation (62) at (a) t=10−1, (b) t=102, and (c) t=105. (d) Contour plot.
(a)(b)(c)(d)Figure 3
Evolution of Equation (63) at (a) t=0.293, (b) t=0.2938, and (c) t=0.2942.
(a)(b)(c)Figure 4
Evolution of Equation (64) at (a) t=0.479, (b) t=0.48, and (c) t=0.4815.
(a)(b)(c)Figure 5
Evolution of Equation (65) at (a) t=0.4, (b) t=0.45, and (c) t=0.5.
(a)(b)(c)Figure 6
Evolution of Equation (66) at (a) t=16, (b) t=18, and (c) t=20.
(a)(b)(c)We can easily find that all the above solutions are traveling wave solutions whenτt takes any constant, and all the above solutions are group-invariant solutions when τt is an arbitrary function of t.


## 6. Conservation Laws of vcBLP

The conservation law is extremely valuable for studying the integrability and exploring the exact solutions of PDE [25–27]. We can use it to explain many physical phenomena described by PDE [28–30]. In this section, we use the multiplier method [31–33] to calculate the conservation laws of vcBLP. The first order multipliers Λ1=Λ1x,y,t,u,v,ux,uy,ut,vx,vy,vt and Λ2=Λ2x,y,t,u,v,ux,uy,ut,vx,vy,vt of vcBLP can be obtained by the following equations with
(70)δδuΛ1·uyt+atuxuy+atuuxy+btuxxy+ctvxxx+Λ2·vt+dtvxx+etuvx=0,δδvΛ1·uyt+atuxuy+atuuxy+btuxxy+ctvxxx+Λ2·vt+dtvxx+etuvx=0,where δ/δu and δ/δv are Euler operators.By expanding Equation (70) and decomposing them according to the derivatives of u,v, we can obtain a system and solve this system can obtain
(71)Λ1=C1x2+C2x2Ht+Gy,Λ2=Ry,where C1 and C2 are arbitrary constants, Ry and Gy are arbitrary functions about y, and Ht is an arbitrary function about t.Therefore, we can obtain the following low-order conservation laws and the corresponding multipliers. The details are discussed as below.Case 1.
Λ1=Gy,Λ2=Ry.
(72)Ct=Gyuy+Ryv,Cx=Gyatuuy+Gybtuxy+Gyctvxx+Rydtvx+Ryetuv,Cy=0.Case 2.
Λ1=x2+Gy,Λ2=Ry.
(73)Ct=Gyuy+x2uy+Ryv,Cx=Ryetuv+Gyatuuy+Gyctvxx+ctx2vxx−2ctxvx+2ctv+Gybtuxy+Rydtvx,Cy=atx2uux+btx2uxx,Case 3.
Λ1=x2Ht+Gy,Λ2=Ry.
(74)Ct=Gyuy+Ryv,Cx=Gyatuuy+Gybtuxy+Gyctvxx+Htctx2vxx−2Htctxvx+2Htctv+Rydtvx+Ryetuv,Cy=Htx2ut+Htatx2uux+Htbtx2uxx.Case 4.
Λ1=x2+Gy+x2Ht,Λ2=Ry.
(75)Ct=x2uy+Gyuy+Ryv,Cx=ctx2vxx−2ctxvx+2ctv+Gyatuuy+Gybtuxy+Gyctvxx+Htctx2vxx−2Htctxvx+2Htctv+Rydtvx+Ryetuv,Cy=atx2uux+btx2uxx+Htx2ut+Htatx2uux+Htbtx2uxx.The results obtained above have been verified using Maple software to ensure that∂/∂tCt+∂/∂xCx+∂/∂yCy=0 holds.


## 7. Conclusions of This Article

In this paper, the zero-order and first-order differential invariants were determined by using the equivalence group of vcBLP. With the help of these, the explicit transformation to its constant coefficient form was given. Subsequently, we have successfully performed the Lie symmetry analysis of vcBLP, obtained some exact solutions, and plotted the corresponding 3-dimensional figures to describe the evolution of the solutions. Moreover, the four conservation laws of vcBLP were obtained using the multiplier method.



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