# Symmetries, Traveling Wave Solutions, and Conservation Laws of a(3+1)-Dimensional Boussinesq Equation

Authors: Letlhogonolo Daddy Moleleki; Chaudry Masood Khalique
Journal: Advances in Mathematical Physics
 (2014)

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## Abstract

We analyze the(3+1)-dimensional Boussinesq equation, which has applications in fluid mechanics. We find exact solutions of the (3+1)-dimensional Boussinesq equation by utilizing the Lie symmetry method along with the simplest equation method. The solutions obtained are traveling wave solutions. Moreover, we construct the conservation laws of the (3+1)-dimensional Boussinesq equation using the new conservation theorem, which is due to Ibragimov.


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## Body

## 1. Introduction

It is well known that the(1+1)-dimensional Boussinesq equation [1],
(1)utt-uxx-(u2)xx-uxxxx=0,
describes the propagation of long waves on the surface of water with a small amplitude and plays a vital part in fluid mechanics [2]. It is completely integrable and admits multiple soliton solutions.The(2+1)-dimensional Boussinesq equation
(2)utt-uxx-uyy-α(u2)xx-uxxxx=0,
which describes the propagation of gravity waves on the surface of water, has been extensively studied by several authors (see, e.g., [3–7]).The(3+1)-dimensional Boussinesq equation is given by
(3)utt-uxx-uyy-uzz-α(u2)xx-uxxxx=0.In [8], the author obtained one-periodic wave solution, two-periodic wave solution, and soliton solution for (3) by means of Hirota’s bilinear method and the Riemann theta function. Wazwaz [5] employed a combination of Hirota’s method and Hereman’s method to formally study (3) and derived two soliton solutions of (3). Some other work concerning symmetries and exact solutions of some Boussinesq equations can be seen in [9–12].In the last few decades several methods have appeared in the literature, which can be used to find exact solutions of nonlinear evolution equations (NLEEs). Some of these methods are the inverse scattering transform method [13], the Darboux transformation method [14], the sine-cosine method [15], Hirota’s bilinear method [16], Jacobi elliptic function expansion method [17], Lie group analysis [18–20], and the exp-function expansion method [21].In this paper we use Lie group method along with the simplest equation method [22, 23] to construct some exact solutions of (3). Furthermore, we employ the new conservation theorem due to Ibragimov [24] to derive conservation laws for (3).Lie group method, which was developed by Sophus Lie (1842–1899) in the nineteenth century, is a systematic method that can be used to find solutions of nonlinear partial differential equations (PDEs). It is based upon the study of the invariance under one-parameter Lie group of point transformations [18, 19].Conservation laws play a very important role in the solution process and the reduction of PDEs [25–27]. They have been used in investigating the existence, uniqueness, and stability of solutions of certain nonlinear PDEs [28–30] and also in the development of numerical methods [31, 32].


## 2. Traveling Wave Solutions of (3)

We obtain exact solutions of (3) using Lie group method along with the simplest equation method.


### 2.1. Non-Topological Soliton Solutions Using Lie Point Symmetries

The vector field(4)X=ξ1∂∂t+ξ2∂∂x+ξ3∂∂y+ξ4∂∂z+η∂∂u,
where ξi, i=1,2,3,4 and η depend on t, x, y, z, and u, is a generator of Lie point symmetries of the (3+1)-dimensional Boussinesq equation (3) if and only if
(5)pr(4)X(utt-uxx-uyy-uzz-α(u2)xx-uxxxx)|(3)=0.
Here pr(4)X is the fourth prolongation of the vector field X. The invariance condition (5) yields the determining equations, which are a system of linear partial differential equations. Solving this system we obtain the following eight Lie point symmetries:
(6)X1=∂∂x,X2=∂∂t,X3=∂∂y,X4=∂∂z,X5=y∂∂z-z∂∂y,X6=z∂∂t+t∂∂z,X7=y∂∂t+t∂∂y,X8=-2αt∂∂t-αx∂∂x-2αy∂∂y-2αz∂∂z+(1+2αu)∂∂u.
To obtain the nontopological soliton solution of (3), we use the combination of the four translation symmetries, namely, X=X1+X2+X3+μX4, where μ is a constant. Solving the associated Lagrange system for X, we obtain the four invariants
(7)g=t-x,f=t-y,h=μt-z,θ=u.
Now considering θ as the new dependent variable and g, f, and h as new independent variables, (3) transforms to a nonlinear PDE in three independent variables, namely,
(8)2μθfh+2μθgh+2θfg+(μ2-1)θhh-2αθg2-2αθθgg-θgggg=0.
The Lie point symmetries of (8) are
(9)Γ1=∂∂g,Γ2=∂∂f,Γ3=∂∂h,Γ4=(2αfμ3-4αhμ2-2αfμ)∂∂h+(2μ2αf-αμ2g-αf-3hαμ)∂∂g+(2αμ2θ+μ2+1)∂∂θ,Γ5=(αfμ3-αfμ-3αhμ2)∂∂h+(μ2αf-αμ2g-αf-2hαμ)∂∂g-(αfμ2)∂∂f+(2αμ2θ+μ2+1)∂∂θ.
The use of the combination Γ=Γ1+Γ2+βΓ3, (β is a constant) of the three translation symmetries, gives us the three invariants
(10)r=f-g,w=βf-h,θ=ϕ.
Treating ϕ as the new dependent variable and r and w as new independent variables, (8) transforms to
(11)(μ2-2μβ-1)ϕww-2βϕrw-2ϕrr-2αϕr2-2αϕϕrr-ϕrrrr=0,
which is a nonlinear PDE in two independent variables. Equation (11) has three Lie point symmetries, namely,
(12)Σ1=∂∂w,Σ2=∂∂r,Σ3=(4wμαβ+2wα-2wμ2α)∂∂w+(wαβ+2μrαβ+rα-μ2rα)∂∂r+(β2-4μβ-2+2μ2-4αϕμβ-2αϕ+2αμ2ϕ)∂∂ϕ,
and the symmetry Σ=Σ1+δΣ2 (δ is a constant) provides the two invariants
(13)ξ=δw-r,ϕ=ψ,
which gives rise to a group invariant solution ψ=ψ(ξ). Using these invariants, the PDE (11) transforms to
(14)(μ2δ2-2μβδ2-δ2+2βδ-2)ψ′′-2αψ′2-2αψψ′′-ψ′′′′=0,
which is a fourth-order nonlinear ODE. This ODE can be integrated easily. Integrating it four times while choosing the constants of integration to be zero (because we are looking for soliton solutions) and then reverting back to our original variables t,x,y,z,u, we obtain the following group-invariant (nontopological soliton) solutions of the Boussinesq equation (3):
(15)u(x,y,t,z)=A1A2sech2[A12(B±ξ)],
where B is a constant of integration and
(16)A1=μ2δ2-2μβδ2-δ2+2βδ-2,A2=2α3,ξ=δz+(1-βδ)y-x+(δβ-δμ)t.


### 2.2. Exact Solutions of (3) Using Simplest Equation Method

We now use the simplest equation method to obtain more solutions of the nonlinear ODE (14), which will then give us more exact solutions for our Boussinesq equation (3). Bernoulli and Riccati equations will be used as the simplest equations [22, 23].


#### 2.2.1. Solutions of (3) Using the Bernoulli Equation as the Simplest Equation

In this case the balancing procedure yieldsM=2 so the solutions of (14) have the form
(17)F(k)=A0+A1G+A2G2.
Inserting (17) into (14) and using the Bernoulli equation [23] and then equating the coefficients of powers of Gi to zero gives us the following algebraic system of six equations:
(18)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-36αA22ab-24αA1b2A2=0,-2A1a2-2A1a2δ2βμ-A1a4+2A1a2δβ+A1a2δ2μ2-δ2A1a2-2αA0A1a2=0,-6αA12b2+6A2b2δ2μ2-12αA0A2b2-6δ2A2b2+12A2b2δβ-12A2b2-16αA22a2-12A2b2δ2βμ-330A2a2b2-42αA1aA2b-60A1ab3=0,-6A1abδ2βμ-16A2a4-6αA0A1ab+3A1abδ2μ2-8αA0A2a2-8A2a2δ2βμ-4αA12a2-3δ2A1ab+6A1abδβ-15A1a3b-8A2a2-6A1ab+4A2a2δ2μ2-4δ2A2a2+8A2a2δβ=0,10A2abδ2μ2-130A2a3b-20A2ab-4A1b2δ2βμ+4A1b2δβ-4A1b2-4αA0A1b2+20A2abδβ-50A1a2b2-10δ2A2ab-10αA12ab-18αA1a2A2-2δ2A1b2+2A1b2δ2μ2-20αA0A2ab-20A2abδ2βμ=0.
These equations can be solved with the aid of Mathematica and one possible solution for A0, A1, and A2 is
(19)A0=-2-2δ2βμ-a2+2δβ+δ2μ2-δ22α,A1=-6abα,A2=-6b2α.
Consequently, returning back to the original variables, a solution of (3) is [23]
(20)u(t,x,y,z)=A0+A1a{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}+A2a2{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


#### 2.2.2. Solutions of (3) Using the Riccati Equation as the Simplest Equation

Here the balancing procedure givesM=2 so the solutions of (14) are of the form
(21)F(z)=A0+A1G+A2G2.
Substituting (21) into (14) and using the Riccati equation [23], as before, we obtain the following algebraic system of equations in terms of A0, A1, and A2:
(22)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-24αA1b2A2-36αA22ab=0,-42αA1aA2b-6δ2A2b2-12A2b2-330A2a2b2-6αA12b2-16αA22a2+12A2b2δβ-32αA22db+6A2b2δ2μ2-12A2b2δ2βμ-240A2b3d-60A1ab3-12αA0A2b2=0,-2A1adδ2βμ-2δ2A2d2+2A1adδβ-4A2d2-A1a3d-4αA0A2d2-2αA12d2-2αA0A1ad-δ2A1ad-16A2bd3+2A2d2δ2μ2+4A2d2δβ-8A1abd2-4A2d2δ2βμ-2A1ad+A1adδ2μ2-14A2a2d2=0,-20A2abδ2βμ-18αA1a2A2-36αA1dA2b-10δ2A2ab-20A2ab-10αA12ab-2δ2A1b2+4A1b2δβ-4A1b2δ2βμ-4A1b2-130A2a3b-440A1adb2+10A2abδ2μ2-28αA22da-4αA0A1b2-50A1a2b2+2A1b2δ2μ2-20αA0A2ab+20A2abδβ-40A1db3=0,-12A1daδ2βμ-4A1bdδ2βμ+2A1bdδ2μ2+6A2daδ2μ2+4A1bdδβ+12A2daδβ-4αA0A1bd-12αA0A2da-2A1a2δ2βμ+A1a2δ2μ2+2A1a2δβ-2δ2A1bd-6δ2A1da-6αA12da-12αA1d2A2-2αA0A1a2-120A2ad2b-22A1a2bd-A1a4-2A1a2-4A1bd-δ2A1a2-12A2da-16A1b2d2-30A2a3d=0,-16A2dbδ2βμ-6A1abδ2βμ+8A2dbδ2μ2-8A2a2δ2βμ+6A1abδβ+16A2dbδβ-30αA1dA2a-6αA0A1ab-16αA0A2db+3A1abδ2μ2+4A2a2δ2μ2+8A2a2δβ-3δ2A1ab-8δ2A2db-8αA12db-8αA0A2a2-60A1adb2-232A2a2db-8A2a2-16A2a4-4αA22a2-15A1a3b-16A2db-136A2b2d2-12αA22d2-4δ2A2a2-6A2ab=0.
Solving the above equations yields
(23)A0=-a2-8bd-δ2-2+2δβ+δ2μ2-2δ2βμ2α,A1=-6abα,A2=-6b2α,
and, consequently, the solutions of (3) are
(24)u(t,x,y,z)=A0+A1{-a2b-θ2btanh[12θ(ξ+C)]}+A2{-a2b-θ2btanh[12θ(ξ+C)]}2,u(t,x,y,z)=A0+A1{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}+A2{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


## 2.1. Non-Topological Soliton Solutions Using Lie Point Symmetries

The vector field(4)X=ξ1∂∂t+ξ2∂∂x+ξ3∂∂y+ξ4∂∂z+η∂∂u,
where ξi, i=1,2,3,4 and η depend on t, x, y, z, and u, is a generator of Lie point symmetries of the (3+1)-dimensional Boussinesq equation (3) if and only if
(5)pr(4)X(utt-uxx-uyy-uzz-α(u2)xx-uxxxx)|(3)=0.
Here pr(4)X is the fourth prolongation of the vector field X. The invariance condition (5) yields the determining equations, which are a system of linear partial differential equations. Solving this system we obtain the following eight Lie point symmetries:
(6)X1=∂∂x,X2=∂∂t,X3=∂∂y,X4=∂∂z,X5=y∂∂z-z∂∂y,X6=z∂∂t+t∂∂z,X7=y∂∂t+t∂∂y,X8=-2αt∂∂t-αx∂∂x-2αy∂∂y-2αz∂∂z+(1+2αu)∂∂u.
To obtain the nontopological soliton solution of (3), we use the combination of the four translation symmetries, namely, X=X1+X2+X3+μX4, where μ is a constant. Solving the associated Lagrange system for X, we obtain the four invariants
(7)g=t-x,f=t-y,h=μt-z,θ=u.
Now considering θ as the new dependent variable and g, f, and h as new independent variables, (3) transforms to a nonlinear PDE in three independent variables, namely,
(8)2μθfh+2μθgh+2θfg+(μ2-1)θhh-2αθg2-2αθθgg-θgggg=0.
The Lie point symmetries of (8) are
(9)Γ1=∂∂g,Γ2=∂∂f,Γ3=∂∂h,Γ4=(2αfμ3-4αhμ2-2αfμ)∂∂h+(2μ2αf-αμ2g-αf-3hαμ)∂∂g+(2αμ2θ+μ2+1)∂∂θ,Γ5=(αfμ3-αfμ-3αhμ2)∂∂h+(μ2αf-αμ2g-αf-2hαμ)∂∂g-(αfμ2)∂∂f+(2αμ2θ+μ2+1)∂∂θ.
The use of the combination Γ=Γ1+Γ2+βΓ3, (β is a constant) of the three translation symmetries, gives us the three invariants
(10)r=f-g,w=βf-h,θ=ϕ.
Treating ϕ as the new dependent variable and r and w as new independent variables, (8) transforms to
(11)(μ2-2μβ-1)ϕww-2βϕrw-2ϕrr-2αϕr2-2αϕϕrr-ϕrrrr=0,
which is a nonlinear PDE in two independent variables. Equation (11) has three Lie point symmetries, namely,
(12)Σ1=∂∂w,Σ2=∂∂r,Σ3=(4wμαβ+2wα-2wμ2α)∂∂w+(wαβ+2μrαβ+rα-μ2rα)∂∂r+(β2-4μβ-2+2μ2-4αϕμβ-2αϕ+2αμ2ϕ)∂∂ϕ,
and the symmetry Σ=Σ1+δΣ2 (δ is a constant) provides the two invariants
(13)ξ=δw-r,ϕ=ψ,
which gives rise to a group invariant solution ψ=ψ(ξ). Using these invariants, the PDE (11) transforms to
(14)(μ2δ2-2μβδ2-δ2+2βδ-2)ψ′′-2αψ′2-2αψψ′′-ψ′′′′=0,
which is a fourth-order nonlinear ODE. This ODE can be integrated easily. Integrating it four times while choosing the constants of integration to be zero (because we are looking for soliton solutions) and then reverting back to our original variables t,x,y,z,u, we obtain the following group-invariant (nontopological soliton) solutions of the Boussinesq equation (3):
(15)u(x,y,t,z)=A1A2sech2[A12(B±ξ)],
where B is a constant of integration and
(16)A1=μ2δ2-2μβδ2-δ2+2βδ-2,A2=2α3,ξ=δz+(1-βδ)y-x+(δβ-δμ)t.


## 2.2. Exact Solutions of (3) Using Simplest Equation Method

We now use the simplest equation method to obtain more solutions of the nonlinear ODE (14), which will then give us more exact solutions for our Boussinesq equation (3). Bernoulli and Riccati equations will be used as the simplest equations [22, 23].


### 2.2.1. Solutions of (3) Using the Bernoulli Equation as the Simplest Equation

In this case the balancing procedure yieldsM=2 so the solutions of (14) have the form
(17)F(k)=A0+A1G+A2G2.
Inserting (17) into (14) and using the Bernoulli equation [23] and then equating the coefficients of powers of Gi to zero gives us the following algebraic system of six equations:
(18)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-36αA22ab-24αA1b2A2=0,-2A1a2-2A1a2δ2βμ-A1a4+2A1a2δβ+A1a2δ2μ2-δ2A1a2-2αA0A1a2=0,-6αA12b2+6A2b2δ2μ2-12αA0A2b2-6δ2A2b2+12A2b2δβ-12A2b2-16αA22a2-12A2b2δ2βμ-330A2a2b2-42αA1aA2b-60A1ab3=0,-6A1abδ2βμ-16A2a4-6αA0A1ab+3A1abδ2μ2-8αA0A2a2-8A2a2δ2βμ-4αA12a2-3δ2A1ab+6A1abδβ-15A1a3b-8A2a2-6A1ab+4A2a2δ2μ2-4δ2A2a2+8A2a2δβ=0,10A2abδ2μ2-130A2a3b-20A2ab-4A1b2δ2βμ+4A1b2δβ-4A1b2-4αA0A1b2+20A2abδβ-50A1a2b2-10δ2A2ab-10αA12ab-18αA1a2A2-2δ2A1b2+2A1b2δ2μ2-20αA0A2ab-20A2abδ2βμ=0.
These equations can be solved with the aid of Mathematica and one possible solution for A0, A1, and A2 is
(19)A0=-2-2δ2βμ-a2+2δβ+δ2μ2-δ22α,A1=-6abα,A2=-6b2α.
Consequently, returning back to the original variables, a solution of (3) is [23]
(20)u(t,x,y,z)=A0+A1a{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}+A2a2{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


### 2.2.2. Solutions of (3) Using the Riccati Equation as the Simplest Equation

Here the balancing procedure givesM=2 so the solutions of (14) are of the form
(21)F(z)=A0+A1G+A2G2.
Substituting (21) into (14) and using the Riccati equation [23], as before, we obtain the following algebraic system of equations in terms of A0, A1, and A2:
(22)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-24αA1b2A2-36αA22ab=0,-42αA1aA2b-6δ2A2b2-12A2b2-330A2a2b2-6αA12b2-16αA22a2+12A2b2δβ-32αA22db+6A2b2δ2μ2-12A2b2δ2βμ-240A2b3d-60A1ab3-12αA0A2b2=0,-2A1adδ2βμ-2δ2A2d2+2A1adδβ-4A2d2-A1a3d-4αA0A2d2-2αA12d2-2αA0A1ad-δ2A1ad-16A2bd3+2A2d2δ2μ2+4A2d2δβ-8A1abd2-4A2d2δ2βμ-2A1ad+A1adδ2μ2-14A2a2d2=0,-20A2abδ2βμ-18αA1a2A2-36αA1dA2b-10δ2A2ab-20A2ab-10αA12ab-2δ2A1b2+4A1b2δβ-4A1b2δ2βμ-4A1b2-130A2a3b-440A1adb2+10A2abδ2μ2-28αA22da-4αA0A1b2-50A1a2b2+2A1b2δ2μ2-20αA0A2ab+20A2abδβ-40A1db3=0,-12A1daδ2βμ-4A1bdδ2βμ+2A1bdδ2μ2+6A2daδ2μ2+4A1bdδβ+12A2daδβ-4αA0A1bd-12αA0A2da-2A1a2δ2βμ+A1a2δ2μ2+2A1a2δβ-2δ2A1bd-6δ2A1da-6αA12da-12αA1d2A2-2αA0A1a2-120A2ad2b-22A1a2bd-A1a4-2A1a2-4A1bd-δ2A1a2-12A2da-16A1b2d2-30A2a3d=0,-16A2dbδ2βμ-6A1abδ2βμ+8A2dbδ2μ2-8A2a2δ2βμ+6A1abδβ+16A2dbδβ-30αA1dA2a-6αA0A1ab-16αA0A2db+3A1abδ2μ2+4A2a2δ2μ2+8A2a2δβ-3δ2A1ab-8δ2A2db-8αA12db-8αA0A2a2-60A1adb2-232A2a2db-8A2a2-16A2a4-4αA22a2-15A1a3b-16A2db-136A2b2d2-12αA22d2-4δ2A2a2-6A2ab=0.
Solving the above equations yields
(23)A0=-a2-8bd-δ2-2+2δβ+δ2μ2-2δ2βμ2α,A1=-6abα,A2=-6b2α,
and, consequently, the solutions of (3) are
(24)u(t,x,y,z)=A0+A1{-a2b-θ2btanh[12θ(ξ+C)]}+A2{-a2b-θ2btanh[12θ(ξ+C)]}2,u(t,x,y,z)=A0+A1{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}+A2{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


## 2.2.1. Solutions of (3) Using the Bernoulli Equation as the Simplest Equation

In this case the balancing procedure yieldsM=2 so the solutions of (14) have the form
(17)F(k)=A0+A1G+A2G2.
Inserting (17) into (14) and using the Bernoulli equation [23] and then equating the coefficients of powers of Gi to zero gives us the following algebraic system of six equations:
(18)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-36αA22ab-24αA1b2A2=0,-2A1a2-2A1a2δ2βμ-A1a4+2A1a2δβ+A1a2δ2μ2-δ2A1a2-2αA0A1a2=0,-6αA12b2+6A2b2δ2μ2-12αA0A2b2-6δ2A2b2+12A2b2δβ-12A2b2-16αA22a2-12A2b2δ2βμ-330A2a2b2-42αA1aA2b-60A1ab3=0,-6A1abδ2βμ-16A2a4-6αA0A1ab+3A1abδ2μ2-8αA0A2a2-8A2a2δ2βμ-4αA12a2-3δ2A1ab+6A1abδβ-15A1a3b-8A2a2-6A1ab+4A2a2δ2μ2-4δ2A2a2+8A2a2δβ=0,10A2abδ2μ2-130A2a3b-20A2ab-4A1b2δ2βμ+4A1b2δβ-4A1b2-4αA0A1b2+20A2abδβ-50A1a2b2-10δ2A2ab-10αA12ab-18αA1a2A2-2δ2A1b2+2A1b2δ2μ2-20αA0A2ab-20A2abδ2βμ=0.
These equations can be solved with the aid of Mathematica and one possible solution for A0, A1, and A2 is
(19)A0=-2-2δ2βμ-a2+2δβ+δ2μ2-δ22α,A1=-6abα,A2=-6b2α.
Consequently, returning back to the original variables, a solution of (3) is [23]
(20)u(t,x,y,z)=A0+A1a{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}+A2a2{cosh⁡⁡[a(ξ+C)]+sinh⁡⁡[a(ξ+C)]1-bcosh⁡[a(ξ+C)]-bsinh⁡[a(ξ+C)]}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


## 2.2.2. Solutions of (3) Using the Riccati Equation as the Simplest Equation

Here the balancing procedure givesM=2 so the solutions of (14) are of the form
(21)F(z)=A0+A1G+A2G2.
Substituting (21) into (14) and using the Riccati equation [23], as before, we obtain the following algebraic system of equations in terms of A0, A1, and A2:
(22)-20αA22b2-120A2b4=0,-336A2ab3-24A1b4-24αA1b2A2-36αA22ab=0,-42αA1aA2b-6δ2A2b2-12A2b2-330A2a2b2-6αA12b2-16αA22a2+12A2b2δβ-32αA22db+6A2b2δ2μ2-12A2b2δ2βμ-240A2b3d-60A1ab3-12αA0A2b2=0,-2A1adδ2βμ-2δ2A2d2+2A1adδβ-4A2d2-A1a3d-4αA0A2d2-2αA12d2-2αA0A1ad-δ2A1ad-16A2bd3+2A2d2δ2μ2+4A2d2δβ-8A1abd2-4A2d2δ2βμ-2A1ad+A1adδ2μ2-14A2a2d2=0,-20A2abδ2βμ-18αA1a2A2-36αA1dA2b-10δ2A2ab-20A2ab-10αA12ab-2δ2A1b2+4A1b2δβ-4A1b2δ2βμ-4A1b2-130A2a3b-440A1adb2+10A2abδ2μ2-28αA22da-4αA0A1b2-50A1a2b2+2A1b2δ2μ2-20αA0A2ab+20A2abδβ-40A1db3=0,-12A1daδ2βμ-4A1bdδ2βμ+2A1bdδ2μ2+6A2daδ2μ2+4A1bdδβ+12A2daδβ-4αA0A1bd-12αA0A2da-2A1a2δ2βμ+A1a2δ2μ2+2A1a2δβ-2δ2A1bd-6δ2A1da-6αA12da-12αA1d2A2-2αA0A1a2-120A2ad2b-22A1a2bd-A1a4-2A1a2-4A1bd-δ2A1a2-12A2da-16A1b2d2-30A2a3d=0,-16A2dbδ2βμ-6A1abδ2βμ+8A2dbδ2μ2-8A2a2δ2βμ+6A1abδβ+16A2dbδβ-30αA1dA2a-6αA0A1ab-16αA0A2db+3A1abδ2μ2+4A2a2δ2μ2+8A2a2δβ-3δ2A1ab-8δ2A2db-8αA12db-8αA0A2a2-60A1adb2-232A2a2db-8A2a2-16A2a4-4αA22a2-15A1a3b-16A2db-136A2b2d2-12αA22d2-4δ2A2a2-6A2ab=0.
Solving the above equations yields
(23)A0=-a2-8bd-δ2-2+2δβ+δ2μ2-2δ2βμ2α,A1=-6abα,A2=-6b2α,
and, consequently, the solutions of (3) are
(24)u(t,x,y,z)=A0+A1{-a2b-θ2btanh[12θ(ξ+C)]}+A2{-a2b-θ2btanh[12θ(ξ+C)]}2,u(t,x,y,z)=A0+A1{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}+A2{sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)-a2b-θ2btanh⁡(12θξ)+sech(θξ/2)Ccosh⁡(θξ/2)-(2b/θ)sinh⁡(θξ/2)}2,
where ξ=δz+(1-αδ)y-x+(δβ-δμ)t and C is an arbitrary constant of integration.


## 3. Conservation Laws for (3)

We utilize the new conservation theorem due to Ibragimov [24] to obtain conservation laws for the (3+1)-dimensional Boussinesq equation (3) written as
(25)utt-uxx-uyy-uzz-2αux2-2αuuxx-uxxxx=0.
For details of notations, definitions, and theorems the reader is referred to [24].In Section2.1 we derived the following eight Lie point symmetries of equation (25):
(26)X1=∂∂x,X2=∂∂t,X3=∂∂y,X4=∂∂z,X5=y∂∂z-z∂∂y,X6=z∂∂t+t∂∂z,X7=y∂∂t+t∂∂y,X8=-2αt∂∂t-αx∂∂x-2αy∂∂y-2αz∂∂z+(1+2αu)∂∂u.
Corresponding to each of these eight Lie point symmetries we shall construct eight conserved vectors. By definition [24] the adjoint equation of (25) is given by
(27)E*(t,x,u,v,…,uxxxx,vxxxx)=δδu[v(utt-uxx-uyy-uzz-2αux2-2αuuxx-uxxxxv(utt-uxx-uyy-uzz-2αux2)]=0,
which gives
(28)vtt-vxx-vyy-vzz-2αuvxx-vxxxx=0.
Here v=v(t,x,y,z) is a new dependent variable. Clearly, (25) is not self-adjoint. The Lagrangian for the system of (25) and (28) is given by
(29)L=v(utt-uxx-uyy-uzz-2αux2-2αuuxx-uxxxxv(utt-uxx-uyy-uzz-2αux2).
(i) Consider first the translation symmetry X1=∂/∂x. In this case the operator Y1 [24] is the same as X1 and the Lie characteristic function W=-ux. Thus the components [24] Ti,i=1,2,3,4, of the conserved vector T=(T1,T2,T3,T4) are given by
(30)T1=uxvt-vutx,T2=vutt-vuyy-vuzz-uxvx-2αuuxvx-uxvxxx+vxxuxx-vxuxxx,T3=-uxvy+vuxy,T4=-uxvz+vuxz.
(ii) The second translation symmetry X2=∂/∂t gives W=-ut. Hence the symmetry generator X2 gives rise to the following components of the conserved vector:
(31)T1=-vuxx-vuyy-vuzz-2αvux2-2αuvuxx-vuxxxx+utvt,T2=2αvutux-utvx-2αuutvx-utvxxx+vutx+2αuvutx+vxxutx-vxutxx+vutxxx,T3=-vyut+vuty,T4=-vzut+vutz.
(iii) For the third symmetry X3=∂/∂y, we have W=-uy and the corresponding components of the conserved vector are
(32)T1=vtuy-vuty,T2=-uyvx+2αvuyux-2αuuyvx-uyvxxx+vuxy+2αuvuxy+vxxuxy-vxuxxy+vuxxxy,T3=vutt-vuxx-vuzz-2αvux2-2αuvuxx-vuxxxx-uyvy,T4=-vzuy+vuyz.
(iv) The fourth symmetry X4=∂/∂z gives W=-uz and the corresponding components of the conserved vector are
(33)T1=vtuz-vutz,T2=-uzvx+2αvuzux-2αuuzvx-uzvxxx+vuxz+2αuvuxz+vxxuxz-vxuxxz+vuxxxz,T3=-vyuz+vuyz,T4=vutt-vuxx-vuyy-2αvux2-2αuvuxx-vuxxxx-uzvz.
(v) For the symmetry X5=y∂/∂z-z∂/∂y, we have W=-yuz+zuy and the corresponding components of the conserved vector, as before, are given by
(34)T1=yuzvt-zuyvt-yvutz+zvuty,T2=2αyvuxuz-yuzvx-2αyuuzvx-yuzvxxx-4αzvuxuy+zuyvx+2αzvuyux+2αzuuyvx+zuyvxxxx+yvuxz+2αyuvuxz+yuxzvxx-zvuxy-2αzuvuxy-zuxyvxx-yvxuxxz-zvxuxxy-yvuxxxz-zvuxxxy,T3=-zvutt+zvuxx+zvuzz+2αzvux2+2αzuvuxx+zvuxxxx-yvyuz+zvyuy+vut+yvuyz+vuz,T4=yvutt-zvuxx-zvuyy-2αyvux2-2αyuvuxx-yvuxxxx-yvzuz+zvzuy-vuy-zvuyz.
(vi) Likewise, the symmetry X6=z∂/∂t+t∂/∂z gives W=-zut-tuz and the corresponding components of the conserved vector are given by
(35)T1=-zvuxx-zvuyy-zvuzz-2αzvux2-2αzuvuxx-zvuxxxx+zvtut+tvtuz-vuz-tvutz,T2=2αzvuxut-zutvx-2αzuutvx-zutvxxx+2αtvuxuz-tuzvx-2αtuuzvx-tuzvxxx+2vutx+2αzuvutx+zutxvxx+tvuxz+2αtuvuxz+tuxzvxx-zvxutxx-tvxuxxz+zvutxxx+tvuxxxz,T3=-zutvy-tuzvy+zvuty+tvuyz,T4=tvutt-tvuxx-tvuyy-2αtvux2-2αtuvuxx-tvuxxxx-zvtuz-tvzuz+vut+zvutz.
(vii) As before, the symmetry X7=y∂/∂t+t∂/∂y yields W=-yut-tuy and the corresponding components of the conserved vector are given by
(36)T1=-yvuxx-yvuyy-yvuzz-2αyvux2T1=-2αyuvuxx-yvuxxxx+yvtut+tvtuyT1=-vuy-tvuty,T2=2αyvuxut-yutvx-2αyuutvx-yutvxxxT1=+4αtvuxuy-tuyvx-2αtuuyvxT1=-2αtvuxuy-tuyvxxx+yvutx+2αyuvutxT1=+yutxvxx+tvuxy+2αtuvuxyT1=+tuxyvxx-yvxutxx-tvxuxxy+yvutxxxT1=+tvuxxxy,T3=tvutt-tvuxx-tvuyy-2αtvux2T1=-2αtuvuxx-tvuxxxx-yvyut-tvyuyT1=+vut+yvuty,T4=-yutvz-tuyvz+yvutz+tvuyz.
(viii) Finally, for the symmetry
(37)X8=-2αt∂∂t-xα∂∂x-2αy∂∂y-2αz∂∂z+(1+2αu)∂∂u,
the value of Y8 is not the same as X8 and in fact is given by
(38)Y8=-2αt∂∂t-xα∂∂x-2αy∂∂y-2αz∂∂z+(1+2αu)∂∂u+αv∂∂v.
The Lie characteristic function W=1+2αu+2αtut+xαux+2αyuy+2αzuz and, consequently, the conserved vector T has components given by
(39)T1=2αtvuxx+2αtvuyy+2αtvuzz+4α2tvux2+4α2tvuuxx+2αtvuxxxx-vt-2αuvt-2αtutvt-αxuxvt-2αyuyvt-2αzuzvt+4αvut+αxvutx+2αyvuty+2αzvutz,T2=-αxvutt+αxvuyy+αxvuzz-3αvux+vx+4αuvx+vxxx-8α2uvux+4α2u2vx+2αuvxxx+2αtutvx-4α2tvutux+4α2tuutvx+2αtutvxxx+αxuxvx+2α2xuuxvx+αxuxvxxx+2αyuyvx-4α2yvuxuy+4α2yuuyvx+2αyuyvxxx+2αzuzvx-4α2zvuxuz+4α2zuuzvx+2αzuzvxxx-2αtvutx-4α2tuvutx-2αtutxvxx-2α2uvux-3αuxvxx-αxuxxvxx-2αyvuxy-4α2yuvuxy-2αyuxyvxx-2αzvuxz-4α2zuvuxz-2αzuxzvxx+4αvxuxx+2αtvxutxx+αxvxuxxx+2αyvxuxxy+2αzvxuxxz-2αtvutxxx-5αvuxxx-2αyvuxxxy-2αzvuxxxz,T3=-2αyvutt+2αyvuxx+2αyvuzz+4α2yvux2+4α2yvuuxx+2αyvuxxxx+vy+2αuvy+2αtutvy+αxuxvy+2αyuyvy+2αzuzvy-4αvuy-2αtvuty-αxvuxy-2αzvuz,T4=-2αzvutt+2αzvuxx+2αzvuyy+4α2zvux2+4α2zvuuxx+2αzvuxxxx+vz+2αuvz+2αtutvz+αxuxvz+2αyuyvz+2αzuzvz-4αvuz-2αtvutz-αxvuxz-2αyvuyz.Remark. Each conserved vector T obtained above contains the arbitrary solution v of the adjoint equation (28) and hence gives an infinite number of conservation laws.


## 4. Conclusions

Exact solutions of the(3+1)-dimensional Boussinesq equation (3) were obtained with the aid of Lie point symmetries of (3) as well as the simplest equation method. The solutions obtained were solitary waves and nontopological soliton. Furthermore, the conservation laws for the (3+1)-dimensional Boussinesq equation were also constructed by utilizing the new conservation theorem due to Ibragimov [24].



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*Source: 672679-2014-07-02.xml*