# Invariant Solutions and Nonlinear Self-Adjointness of the Two-Component Chaplygin Gas Equation

Authors: Ben Gao; Yanxia Wang
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

In this paper, the Lie group method is performed on a special dark fluid, the Chaplygin gas, which describes both dark matter and dark energy in the present universe. Based on an optimal system of one-dimensional subalgebras, similarity reductions and group invariant solutions are given. Finally, by means of Ibragimov’s method, conservation laws are obtained.


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## Body

## 1. Introduction

Nonlinear partial differential equations (PDEs) play an indispensable role in the nature. A host of natural phenomena is able to be well simulated by them. The multitude of methods have been explored to find exact solutions for the PDEs. Some of the most representative methods are Bäcklund method [1], Hirota bilinear method [2], Darboux transformation [3], Painleve expansion method [4], Lie symmetry analysis [5–8], the inverse scattering method [9], etc. Lie symmetry analysis, a quite effective method among them, can get crucially explicit solutions of PDEs. On the other hand, an important issue regarding the PDEs is to obtain their conservation laws. Conservation laws can be obtained by variational principle or Hamilton’s principle [10, 11]. For the PDEs which do not have a Lagrangian, based on admitted symmetries [12], Ibragimov proposed the concept of the adjoint equation for the study of conservation laws by using conservation law theorem in [13]. Moreover, we can apply conservation laws to obtain exact solutions of PDEs [14, 15].The aim of this paper is to study symmetries, explicit solutions, nonlinear self-adjointness, and conservation laws of the two-component Chaplygin gas equation [16]:(1)ρt=-ρvxx-ρxvx,vt=2ρ2-12vx2.This system consists of the equation of continuity for the density ρ(x,t) and Euler’s force equation for an ideal fluid of zero vorticity with the velocity potential v(x,t), where the subscripts stand for the partial derivatives. Equation (1) was introduced by Chaplygin, Tsien, and von Karman as a mathematical approximation for calculating the lifting force on a wing of an airplane in aerodynamics [16]. The two-component Chaplygin gas equation can describe both dark matter and dark energy in the present universe [17, 18]. This system has received extensive attention. For example, the Riemann problem of (1) and the interaction of waves were exhibited in [19]; the Riemann solutions with concentration were derived in [20]; the perturbative behaviour of (1) was analyzed in [21]; qualitative and quantitative aspects of (1) were investigated in [22].In the end, this paper is arranged as follows. In Section2, we focus on obtaining Lie point symmetry of (1). Section 3 mainly constructs similarity reductions and group invariant solutions. Conservation laws, using lbragimov’s method, have been set up in Section 4. Some conclusions are presented in the last section.


## 2. Lie Point Symmetries

In order to use the Lie group method to analyze the model, we assumed that the gas is incompressible. Let the pressureu=-4/ρ, and we can transform (1) to(2)ut=uvxx-uxvx,vt=18u2-12vx2.In this section, we use Lie point symmetry method for (2) and acquire its infinitesimal generators, commutation table of Lie algebra.First of all, let us consider an infinitesimal generator admitted by (2):(3)X=ξx,t,u,v∂x+τx,t,u,v∂t+ϕx,t,u,v∂u+φx,t,u,v∂v.According to the invariance conditions [5], we have(4)prX2ut-uvxx+uxvx=0,prX1vt-18u2+12vx2=0.The prolongations for (2) of the infinitesimal generator (3) are [5](5)prX2=X+ϕx1∂∂ux+ϕt1∂∂ut+φx1∂∂vx+φxx2∂∂vxx,prX1=X+φx1∂∂vx+φt1∂∂vt,where(6)ϕx1=Dxϕ-uxDxξ-utDxτ,ϕt1=Dtϕ-uxDtξ-utDtτ,φx1=Dxφ-vxDxξ-vtDxτ,φt1=Dtφ-vxDtξ-vtDtτ,φxx2=Dxφx1-vxxDxξ-vxtDxτ,and Dx,Dt denote the total derivatives, e.g.,(7)Dt=∂∂t+ut∂∂u+vt∂∂v+utx∂∂ux+vtx∂∂vx+utt∂∂ut+vtt∂∂vt+⋯.Substituting prX(2) and prX(1) into (2), we acquire the following system of overdetermined equations:(8)ξu=ξv=0,ξxx=0,ξtx=0,ξtt=0,τx=τu=τv=0,τtt=0,ϕ=ξx-τtu,φt=φu=0,φx=ξt,φv=-τt+2ξx.Solving (8), we get(9)ξ=c5t+c3x+c4,τ=c1t+c2,ϕ=c3-c1u,φ=c5x+2c3-c1v+c6,where c1, c2, c3, c4, c5, and c6 are arbitrary constants. The Lie algebra of infinitesimal symmetry of (2) is spanned by the following generators:(10)X1=t∂t-u∂u-v∂v,X2=∂t,X3=x∂x+u∂u+2v∂v,X4=∂x,X5=t∂x+x∂v,X6=∂v.To classify all the group-invariant solutions, we construct an optimal system of one-dimensional subalgebras of (2) by using the method in [23], which only depends on the commutator table. The commutator relations about X1,X2,X3,X4,X5,X6 are listed in Table 1.Table 1
Table of Lie brackets.
[ X i , X j ] X 1 X 2 X 3 X 4 X 5 X 6 X 1 0 - X 2 0 0 X 5 X 6 X 2 X 2 0 0 0 X 4 0 X 3 0 0 0 - X 4 - X 5 - 2 X 6 X 4 0 0 X 4 0 X 6 0 X 5 - X 5 - X 4 X 5 - X 6 0 0 X 6 - X 6 0 2 X 6 0 0 0An arbitrary operatorX∈L6 is of the form(11)X=l1X1+l2X2+l3X3+l4X4+l5X5+l6X6,In order to discuss the linear transformations of the vectorl=(l1,l2,l3,l4,l5,l6), the following generator is given as(12)Ei=cijklj∂lk,i=1,2,3,4,5,6 where cijk is determined by [Xi,Xj]=cijkXk. Based on (12) and Table 1, E1,E2,E3,E4,E5,E6 can be represented as(13)E1=-l2∂l2+l5∂l5+l6∂l6,E2=l1∂l2+l5∂l4,E3=-l4∂l4-l5∂l5-2l6∂l6,E4=l3∂l4+l5∂l6,E5=-l1+l3∂l5-l2∂l4-l4∂l6,E6=-l1+2l3∂l6.For E1,E2,E3,E4,E5,E6, the Lie equations which have parameters a1,a2,a3,a4,a5,a6 with the initial condition l~ai=0=l,i=1,2,3,4,5,6 are expressed as(14)dl1~da1=0,dl2~da1=-l2~,dl3~da1=0,dl4~da1=0,dl5~da1=l5~,dl6~da1=l6~,dl1~da2=0,dl2~da2=l1~,dl3~da2=0,dl4~da2=l5~,dl5~da2=0,dl6~da2=0,dl1~da3=0,dl2~da3=0,dl3~da3=0,dl4~da3=-l4~,dl5~da3=-l5~,dl6~da3=-2l6~,dl1~da4=0,dl2~da4=0,dl3~da4=0,dl4~da4=l3~,dl5~da4=0,dl6~da4=l5~,dl1~da5=0,dl2~da5=0,dl3~da5=0,dl4~da5=-l2~,dl5~da5=-l1~+l3~,dl6~da5=-l4~,dl1~da6=0,dl2~da6=0,dl3~da6=0,dl4~da6=0,dl5~da6=0,dl6~da6=-l1~+2l3~.The transformations of the solutions of these equations are given as follows:(15)T1:l1~=l1,l2~=e-a1l2,l3~=l3,l4~=l4,l5~=ea1l5,l6~=ea1l6,T2:l1~=l1,l2~=a2l1+l2,l3~=l3,l4~=a2l5+l4,l5~=l5,l6~=l6,T3:l1~=l1,l2~=l2,l3~=l3,l4~=e-a3l4,l5~=e-a3l5,l6~=e-2a3l6,T4:l1~=l1,l2~=l2,l3~=l3,l4~=a4l3+l4,l5~=l5,l6~=a4l5+l6,T5:l1~=l1,l2~=l2,l3~=l3,l4~=-a5l2+l4,l5~=a5-l1+l3+l5,l6~=a512a5l2-l4+l6,T6:l1~=l1,l2~=l2,l3~=l3,l4~=l4,l5~=l5,l6~=a6-l1+2l3+l6.The structure of the optimal system needs a simplification of the vector:(16)l=l1,l2,l3,l4,l5,l6, by using the transformations T1-T6. We mainly concentrate on finding a simplest representative of vector (16). The structure is classified into two cases.Case 1 (l5 = 0).
Consider vector (16) of the form(17)l=l1,l2,l3,l4,0,l6.
(1) l 1 ≠ 0
By takinga2=-l2/l1 in the transformation T2, we have l2~=0. Vector (17) is hence reduced as(18)l=l1,0,l3,l4,0,l6.
(1.1) l 3 ≠ 0
By takinga4=-l4/l3 in the transformation T4, we have l4~=0. Vector (18) is simplified as(19)l=l1,0,l3,0,0,l6.
(1.1.1) l 1 ≠ 2 l 3
By takinga6=l6/l1-2l3 in the transformation T6, we have l6~=0. Vector (19) is simplified as(20)l=l1,0,l3,0,0,0.
We obtain the following representatives:(21)X1±X3.
(1.1.2) l 1 = 2 l 3
We get the following representatives:(22)2X1+X3,2X1+X3±X6.
(1.2) l 3 = 0
Vector (18) is simplified as(23)l=l1,0,0,l4,0,l6.
By takinga6=l6/l1 in the transformation T6, we have l6~=0. Vector (23) is simplified as(24)l=l1,0,0,l4,0,0.
As a result, we get the following representatives:(25)X1,X1±X4.
(2) l 1 = 0
We obtain vector (17) of the form(26)l=0,l2,l3,l4,0,l6.
(2.1) l 3 ≠ 0
By takinga4=-l4/l3 and a6=-l6/2l3 in the transformations T4 and T6, we have l4~=l6~=0. Vector (26) is simplified as(27)l=0,l2,l3,0,0,0.
We get the following representatives:(28)X3,X2±X3.
(2.2) l 3 = 0
We obtain vector (26) of the form(29)l=0,l2,0,l4,0,l6.
(2.2.1) l 2 ≠ 0
By takinga5=l4/l2 in the transformation T5, we have l4~=0. Vector (29) is simplified as(30)l=0,l2,0,0,0,l6.
We get the following representatives:(31)X2,X2±X6.
(2.2.2) l 2 = 0
We obtain vector (29) of the form(32)l=0,0,0,l4,0,l6.
(2.2.2.1) l 4 ≠ 0
By takinga5=l6/l4 in the transformation T5, we have l6~=0. Vector (32) is simplified as(33)l=0,0,0,l4,0,0.
We get the representative(34)X4.
(2.2.2.1) l 4 = 0
We obtain vector (32) of the form(35)l=0,0,0,0,0,l6.
We get the representative(36)X6.Case 2 (l5≠0).
By takinga2=-l4/l5 and a4=-l6/l5 in the transformations T2 and T4, we have l4~=l6~=0. Vector (16) is simplified as(37)l=l1,l2,l3,0,l5,0.
Then, we obtain the following representatives:(38)X5,X5±X1,X5±X2,X5±X3,X5±X1±X2,X5±X1±X3,X5±X2±X3,X5±X1±X2±X3.
Hence, by gathering the operators (21), (22), (25), (28), (31), (34), (36), and (38), we get to the following theorem.Theorem 1.
The operators in{X1,X2,X3,X4,X5,X6} create an optimal system:(39)X5,X5±X1,X5±X2,X5±X3,X5±X1±X2,X5±X1±X3,X5±X2±X3,X5±X1±X2±X3,X1±X3,2X1+X3,2X1+X3±X6,X1,X1±X4,X3,X2±X3,X2,X2±X6,X4,X6.


## 3. Exact Solutions

In this section, we cope with the reductions and derive group invariant solutions of (2).Case 1 (reduction byX5).
For generatorX5, we have similarity variables(40)z=t,p=u,q=-x2+2vt2t,and the group-invariant solution is p=f(z), q=g(z), i.e.,(41)u=ft,v=x2+2tgt2t.
Using (41) in (2), we have(42)f-zf′=0,f2-8g′=0,where f′=df/dz, g′=dg/dz. Therefore, (2) has a solution u=c1t, v=(12x2+24c2t+c12t4)/24t, where c1,c2 are arbitrary constants.Case 2 (reduction byX5+X1).
Similar to Case1, we have z=x-t, u=f(z)/t, v=1/2x2+g(z)/t. And the reduced equation is(43)f-zf′+fg′′-f′g′=0,g+18f2-12g′2-zg′=0, where f′=df/dz, g′=dg/dz. Therefore, (2) has a solution u=-22c1/t, v=2xt-t2+2c12/2t, where c1 is an arbitrary constant.Case 3.
For generatorX5+X2, we have u=f(z), v=g(z)+tx-1/3t3, where z=-2x+t2. The reduced equation is(44)fg′′-f′g′=0,-z-14f2+4g′2=0, where f′=df/dz, g′=dg/dz. Therefore, (2) has a solution u=2(9c12-1)(-2x+t2), v=tx-(1/3)t3 + c1(-2x+t2)3+c2, where c1,c2 are arbitrary constants.Some of the reductions for the optimal system of one-dimensional subalgebras and exact solutions are represented in Tables2 and 3, respectively.Table 2
The reductions of the two-component Chaplygin gas equation.
Generators Similarity variables Reduced equations X 5 + X 3 z = t, u=(x+t)f(z), v=2x+t2g(z)-t-2x2. f ′ = 0, 18f2-2g2-g′=0. X 1 + X 3 z = t x, u=f(z), v=xgz. f ′ - z 2 f g ′ ′ - z f ′ g + z 2 f ′ g ′ = 0, g′-18f2+12g2-zgg′+12z2g′2=0. 2 X 1 + X 3 z = t x 2, u=f(z)x, v=g(z). f ′ - 4 z f g ′ - 4 z 2 f g ′ ′ + 4 z 2 f ′ g ′ = 0, g′-18f2+2z2g′2=0. X 1 z = x, u=f(z)t, v=g(z)t. f + f g ′ ′ - f ′ g ′ = 0, g+18f2-12g′2=0. X 3 z = t, u=f(z)x, v=g(z)x2. f ′ = 0, g′-18f2+2g2=0. X 2 + X 6 z = x, u=f(z), v=g(z)+t. f g ′ ′ - f ′ g ′ = 0, 1-18f2+12g′2=0.Table 3
Exact solutions of the two-component Chaplygin gas equation.
Generators Exact solutions X 5 + X 3 u = c 1 ( x + t ), v=c1(x+t)2(ec1(t+c2)+1)4(ec1(t+c2)-1)-x-t2. X 1 + X 3 u = 2 2 c 1 + c 2 2, v=c1t+c2x. 2 X 1 + X 3 u = - 2 2 c 1 x 3 x 4 - 2 c 1 2 t 2, v=22c1arctanh⁡2c1tx2+c2. X 1 u = 0, v=(c1-x)22t. X 3 u = c 1 x, v=c1x24ec1(c2+t)+1ec1(c2+t)-1. X 2 + X 6 u = 2 2 + c 1 2, v=c1x+t+c2.


## 4. Conservation Laws

In this section, we prove that (2) is nonlinear self-adjoint. Moreover, using Ibragimov’s method, we construct conservation laws.


### 4.1. Preliminaries

Consider anrth-order system of m PDEs(45)Eαx,u,u1,…,ur=0,α=1,2,…,m,where x=(x1,…,xn), u=(u1,…,um), uiα=∂uα/∂xi, uijα=∂2uα/∂xi∂xj, and u(i) denotes the collection of all ith-order partial derivatives of u with respect to x.The adjoint equations of (45) are defined by(46)Eα∗x,u,v,u1,v1,…,ur,vr=δLδuσ=0,where v=(v1,…,vm), L=vβEβ(x,u,u(1),…,u(r)) is the formal Lagrangian, and δ/δuσ is the Euler-Lagrange operator written as(47)δδuσ=∂∂uσ+∑s=1∞-1sDi1…Dis∂∂ui1…isσ,where Di denotes the total derivative with respect to xi.Definition 2 (see [12]).
System (45) is said to be nonlinearly self-adjoint if the adjoint system (46) is satisfied for all solutions u of (45) upon a substitution:(48)vα=φαx,u,α=1,…,m,such that φ(x,u)=(φ1,…,φm)≠0.Definition2 means that(49)Eα∗x,u,v,u1,v1,…,ur,vr=λαβEβx,u,u1,…,ur,α,β=1,…,m,where λαβ is a particular function.The following theorem brings about conservation laws [13].Theorem 3.
Any infinitesimal symmetry(50)X=ξix,u,u1,…∂∂xi+ηαx,u,u1,…∂∂uαadmitted by (45) gives rise to a conservation law Di(Ci)=0, where Ci is(51)Ci=Wα∂L∂uiα-Dj∂L∂uijα+DjDk∂L∂uijkα-⋯+DjWα∂L∂uijα-Dk∂L∂uijkα+⋯+DjDkWα∂L∂uijkα-⋯+⋯,where Wα=ηα-ξjujα. The Lagrangian L should be represented with respect to all mixed derivatives uijα,uijkα,….


### 4.2. Nonlinear Self-Adjointness

Following Definition2 and Theorem 3, we prove that (2) is nonlinear self-adjoint.Theorem 4.
Equation (2) is nonlinearly self-adjoint under the substitutions(52)Λ1=c1u2,Λ2=0,where c1 is an arbitrary constant.Proof.
Let the formal Lagrangian of (2) be of the form(53)L=Λ1ut-uvxx+uxvx+Λ2vt-18u2+12vx2,where Λ1,Λ2 are two new dependent variables.
Using the equivalent formula (49), the identities(54)δLδu=λ11ut-uvxx+uxvx+λ12vt-18u2+12vx2,δLδv=λ21ut-uvxx+uxvx+λ22vt-18u2+12vx2,are established under the substitutions Λ1=Λ1(x,t,u,v), Λ2=Λ2(x,t,u,v).
Equation (54) can be split by using the coefficients of different order derivatives of u and v, and then we obtain the system in unknown variables Λ1, Λ2 whose solutions are(55)Λ1=c1u2,Λ2=0,where c1 is an arbitrary constant.


### 4.3. Construction of Conservation Laws

For the generatorX=ξ∂x+τ∂t+ϕ∂u+φ∂v, by Theorem 3, the conservation law of (2) is expressed in the form DxCx+DtCt=0, and the conserved vectors Ct,Cx are(56)Cx=Wu∂L∂ux+Wv∂L∂vx-Dx∂L∂vxx+DxWv∂L∂vxx,Ct=Wu∂L∂ut+Wv∂L∂vt,where Wu=ϕ-ξux-τut, Wv=φ-ξvx-τvt, L=Λ1(ut-uvxx+uxvx) + Λ2(vt-1/8u2+1/2vx2) = c1(ut-uvxx+uxvx)/u2.By simplifyingCx, Ct, we have(57)Cx=Wuc1vxu2-DxWvc1u,(58)Ct=Wuc1u2.For the generatorX1=t∂t-u∂u-v∂v, we obtain Wu=-u-tut, Wv=-v-tvt. On the basis of formulas (57) and (58), we obtain(59)Cx=-c1vxu+tutu2+c1vx+tvxtu,Ct=-c1u+tutu2.The remaining conservation laws of other generators for (2) are shown in Table 4.Table 4
Conservation laws of the two-component Chaplygin gas equation.
Generators C x C t X 2 - c 1 u t v x u 2 + c 1 v x t u - c 1 u t u 2 X 3 c 1 ( u - x u x ) v x u 2 - c 1 ( v x - x v x x ) u c 1 ( u - x u x ) u 2 X 4 - c 1 u x v x u 2 + c 1 v x x u - c 1 u x u 2 X 5 - c 1 t u x v x u 2 - c 1 ( 1 - t v x x ) u - c 1 t u x u 2 X 6 0 0


## 4.1. Preliminaries

Consider anrth-order system of m PDEs(45)Eαx,u,u1,…,ur=0,α=1,2,…,m,where x=(x1,…,xn), u=(u1,…,um), uiα=∂uα/∂xi, uijα=∂2uα/∂xi∂xj, and u(i) denotes the collection of all ith-order partial derivatives of u with respect to x.The adjoint equations of (45) are defined by(46)Eα∗x,u,v,u1,v1,…,ur,vr=δLδuσ=0,where v=(v1,…,vm), L=vβEβ(x,u,u(1),…,u(r)) is the formal Lagrangian, and δ/δuσ is the Euler-Lagrange operator written as(47)δδuσ=∂∂uσ+∑s=1∞-1sDi1…Dis∂∂ui1…isσ,where Di denotes the total derivative with respect to xi.Definition 2 (see [12]).
System (45) is said to be nonlinearly self-adjoint if the adjoint system (46) is satisfied for all solutions u of (45) upon a substitution:(48)vα=φαx,u,α=1,…,m,such that φ(x,u)=(φ1,…,φm)≠0.Definition2 means that(49)Eα∗x,u,v,u1,v1,…,ur,vr=λαβEβx,u,u1,…,ur,α,β=1,…,m,where λαβ is a particular function.The following theorem brings about conservation laws [13].Theorem 3.
Any infinitesimal symmetry(50)X=ξix,u,u1,…∂∂xi+ηαx,u,u1,…∂∂uαadmitted by (45) gives rise to a conservation law Di(Ci)=0, where Ci is(51)Ci=Wα∂L∂uiα-Dj∂L∂uijα+DjDk∂L∂uijkα-⋯+DjWα∂L∂uijα-Dk∂L∂uijkα+⋯+DjDkWα∂L∂uijkα-⋯+⋯,where Wα=ηα-ξjujα. The Lagrangian L should be represented with respect to all mixed derivatives uijα,uijkα,….


## 4.2. Nonlinear Self-Adjointness

Following Definition2 and Theorem 3, we prove that (2) is nonlinear self-adjoint.Theorem 4.
Equation (2) is nonlinearly self-adjoint under the substitutions(52)Λ1=c1u2,Λ2=0,where c1 is an arbitrary constant.Proof.
Let the formal Lagrangian of (2) be of the form(53)L=Λ1ut-uvxx+uxvx+Λ2vt-18u2+12vx2,where Λ1,Λ2 are two new dependent variables.
Using the equivalent formula (49), the identities(54)δLδu=λ11ut-uvxx+uxvx+λ12vt-18u2+12vx2,δLδv=λ21ut-uvxx+uxvx+λ22vt-18u2+12vx2,are established under the substitutions Λ1=Λ1(x,t,u,v), Λ2=Λ2(x,t,u,v).
Equation (54) can be split by using the coefficients of different order derivatives of u and v, and then we obtain the system in unknown variables Λ1, Λ2 whose solutions are(55)Λ1=c1u2,Λ2=0,where c1 is an arbitrary constant.


## 4.3. Construction of Conservation Laws

For the generatorX=ξ∂x+τ∂t+ϕ∂u+φ∂v, by Theorem 3, the conservation law of (2) is expressed in the form DxCx+DtCt=0, and the conserved vectors Ct,Cx are(56)Cx=Wu∂L∂ux+Wv∂L∂vx-Dx∂L∂vxx+DxWv∂L∂vxx,Ct=Wu∂L∂ut+Wv∂L∂vt,where Wu=ϕ-ξux-τut, Wv=φ-ξvx-τvt, L=Λ1(ut-uvxx+uxvx) + Λ2(vt-1/8u2+1/2vx2) = c1(ut-uvxx+uxvx)/u2.By simplifyingCx, Ct, we have(57)Cx=Wuc1vxu2-DxWvc1u,(58)Ct=Wuc1u2.For the generatorX1=t∂t-u∂u-v∂v, we obtain Wu=-u-tut, Wv=-v-tvt. On the basis of formulas (57) and (58), we obtain(59)Cx=-c1vxu+tutu2+c1vx+tvxtu,Ct=-c1u+tutu2.The remaining conservation laws of other generators for (2) are shown in Table 4.Table 4
Conservation laws of the two-component Chaplygin gas equation.
Generators C x C t X 2 - c 1 u t v x u 2 + c 1 v x t u - c 1 u t u 2 X 3 c 1 ( u - x u x ) v x u 2 - c 1 ( v x - x v x x ) u c 1 ( u - x u x ) u 2 X 4 - c 1 u x v x u 2 + c 1 v x x u - c 1 u x u 2 X 5 - c 1 t u x v x u 2 - c 1 ( 1 - t v x x ) u - c 1 t u x u 2 X 6 0 0


## 5. Conclusions

In this paper, we apply Lie group method to the two-component Chaplygin gas equation. Based on this method, the optimal system of one-dimensional subalgebras and similarity reductions are derived. Furthermore, exact solutions of the reduced equations are constructed. Eventually, it is shown that (2) is nonlinearly self-adjoint, and, at the same time, we obtain its conservation laws. Our results can be used to characterize the problems of both dark matter and dark energy in the present universe.



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