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Introduction
Most natural phenomena of the real world are modelled by nonlinear partial differential equations (NLPDEs). Such equations can seldom be solved by an analytic method. In contrast the linear differential equations have a particularly good algebraic structure to their solutions, which makes them solvable. Unfortunately, for NLPDEs there is no general theory which can be applied to obtain exact closed-form solutions. However, scientists have developed geometric methods and dynamical systems theory which play prominent roles in the study of differential equations. Such theories deal with the long-term qualitative behaviour of dynamical systems and do not focus on finding precise solutions to the equations. Nevertheless, various methods have also been established by the researchers which provide exact solutions to NLPDEs. Some of these methods are Hirota’s bilinear transformation method [1], the inverse scattering method [2], Kudryashov’s method [3, 4], the sine-cosine method [5], the tanh-coth method [6], the simplest equation method [7, 8], the tanh-function method [9], the Darboux transformation [10], the (G′/G)–expansion method [11, 12], the Bäcklund transformation [13], and Lie symmetry methods [14, 15, 16, 17, 18, 19].
One of the NLPDEs is the (3 + 1)-dimensional Jimbo-Miwa equation
which is the second member of a Kadomtsev-Petviashvili hierarchy. This equation has been studied extensively by researchers because of the fact that it can be used to describe some fascinating (3+1)-dimensional waves in physics. See for example [20, 21, 22, 23] and references therein.
Recently equation (1) has been extended to the equation [24]
where the term uyt was extended to uxt + uyt + uzt and because of this reason it is called the extended (3+1)-dimensional Jimbo-Miwa equation. Applying the simplified Hirota’s method multiple soliton solutions of (2) were derived and it was shown that the dispersion relations and the phase shifts of (2) were distinct compared to the dispersion and shifts of (1). By using bilinear forms Sun and Chen [25] obtained the lump solutions and their dynamics of (1) and (2). Furthermore, the lump-kink solution which contains interaction between a lump and a kink wave were also obtained in [25].
In this paper we consider a generalized version of the second extended (3+1)-dimensional Jimbo-Miwa equation, namely
$$\begin{array}{}
\displaystyle u_{xxxy}+ k \left( u_{y} u_{x}\right) _x + h (u_{xt} +u_{yt}+u_{zt})-k u_{xz} = 0 ,
\end{array} $$
where h and k are constants. We obtain exact solutions of (3) using symmetry reductions along with simplest equation method. Furthermore, we derive conservation laws for (3) using the conservation theorem due to Ibragimov.
Lie symmetry theory, originally developed by Marius Sophus Lie (1842-1899), a Norwegian mathematician, around the middle of the nineteenth century, is based upon the study of the invariance under one parameter Lie group of point transformations [14, 15, 16, 17, 18, 19] The theory is highly algorithmic and is one of the most powerful methods to find exact solutions of differential equations be it linear or nonlinear. It has been applied to many scientific fields such as classical mechanics, relativity, control theory, quantum mechanics, numerical analysis, to name but a few.
Conservation laws can be described as fundamental laws of nature, which have extensive applications in various fields of scientific study such as physics, chemistry, biology, engineering, and so on. They have many uses in the study of differential equations. Conservation laws have been used to prove global existence theorems and shock wave solutions to hyperbolic systems. They have been applied to problems of stability and have been used in scattering theory and elasticity [17, 26, 27, 28, 29].
The paper is organized as follows. In Section 2 we first perform symmetry reductions of the generalized second extended (3+1)-dimensional Jimbo-Miwa equation (3) and reduce it to a nonlinear fourth-order ordinary differential equation. Thereafter we find the general solution of the ordinary differential equation in terms of the Weierstrass zeta function. We also find travelling wave solutions of (3) using the simplest equation method. Conservation laws of (3) are obtained by employing the conservation theorem due to Ibragimov in Section 3. Finally we present concluding remarks in Section 4.
In this section we present exact solutions to the generalized second extended (3+1)-dimensional Jimbo-Miwa equation (3).
Lie point symmetries and symmetry reductions of (3)
We apply the algorithm for computing Lie point symmetries of (3) and then use them to perform symmetry reductions several times until we arrive at an ordinary differential equation (ODE).
where ξi, i = 1, 2, 3, 4 and η depend on x, y, z, t and u, will generate a symmetry group of (3) provided
$$\begin{array}{}
\displaystyle \mbox{pr}^{(4)}X(u_{xxxy}+ k \left( u_{y} u_{x}\right) _x + h (u_{xt} +u_{yt}+u_{zt})-k u_{xz} )|_{(3)} =0,
\end{array} $$
where pr(4)X is the fourth prolongation of X [17]. Expanding the determining equation (4) and splitting on derivatives of u, we obtain an overdetermined system of linear homogeneous partial differential equations. Solving this resultant system one obtains the values of ξi, i = 1, 2, 3, 4 and η. Consequently, we have the following nine Lie point symmetries of (3):
We now make use of the four translation symmetries and perform symmetry reductions. Solving the associated Lagrange system for X = X1 + αX2+ X3+X4, where α is a constant, we obtain four invariants
Utilizing the symmetry Γ = Γ1 + Γ2 + βΓ3, where β is a constant, we reduce equation (7) to a PDE in two independent variables. From the associated Lagrange system for Γ, we obtain three invariants
$$\begin{array}{}
\displaystyle (\beta-\alpha) \phi _{rrrr}+ \left( \alpha \beta h -\beta ^2 h -\beta h \right) \phi _{rr} +\alpha h \phi _{rs}-2 \beta h \phi _{rs}-h \phi _{rs}-h \phi _{ss}-2 \alpha k \phi _{r} \phi _{rr} \\\displaystyle +2 \beta k \phi _{r} \phi _{rr}-k \phi _{rs}=0,
\end{array} $$
which is a nonlinear PDE in two independent variables. We perform further symmetry reduction on equation (9). This equation has five symmetries including the two translation symmetries Σ1 = ∂/∂r and Σ2 = ∂/∂s. The combination Σ = ν Σ1 + Σ2, yields the two invariants
which give rise to a group-invariant solution ϕ = F(q) and consequently, equation (9) is transformed into the fourth-order nonlinear ODE
$$\begin{array}{}
\displaystyle A F''''(q)+ B F'(q)F''(q)+CF''(q)=0,
\end{array} $$
where A = α −β, B = 2k(α −β), C = h (β −ν) (−α +β −ν +1)−kν and q = x+(β - α)y−νz +(ν − β)t. Integration of the above equation twice with respect to q gives
Thus integrating equation (11) and reverting to our original variables we obtain the solution of (3), which is given by
$$\begin{array}{}
\displaystyle u(x,y,z,t)=\frac{12 A }{B} \, \zeta \left(q;g_1,g_2\right)-\frac{C }{B} \, q,
\end{array} $$
where ζ (q;g1,g2) is the Weierstrass zeta function defined as ζ′ (q;g1,g2) = −℘ (q;g1,g2 ) [30] and A = α −β, B = 2k(α −β), C = h (β −ν) (−α +β −ν +1)−kν and q = x+(β − α)y−νz +(ν − β)t.
Exact solutions of (3) using simplest equation method
In this subsecrion we use the simplest equation method [7, 8] to solve the ODE (10) and henceforth one obtains the exact solutions of the generalized second extended (3+1)-dimensional Jimbo-Miwa equation (3). We use the Bernoulli and Riccati equations as the simplest equations. The Bernoulli equation
where H(z) solves the Bernoulli or Riccati equation, M is a positive integer which is determined by the balancing procedure and Ai, (i = 0, 1, ⋯, M) are parameters to be determined.
Solutions of (3) using Bernoulli as the simplest equation
From equation (10) the balancing procedure yields M = 1, so the solutions of (10) can be written as
Substituting (15) into (10) and invoking the Bernoulli equation (12) we obtain the algebraic equation
$$\begin{array}{}
\displaystyle \alpha A_1 c^4 H(q)-A_1 \beta c^4 H(q)+15 \alpha A_1 c^3 d H(q)^2-15 A_1 \beta c^3 d H(q)^2+2 \alpha A_1^2 c^3 k H(q)^2\\\displaystyle \,-\,2 A_1^2 \beta c^3 k H(q)^2+50 \alpha A_1 c^2 d^2H(q)^3-50 A_1 \beta c^2 d^2 H(q)^3+8 \alpha A_1^2 c^2 d k H(q)^3\\\displaystyle \,-\,8 A_1^2 \beta c^2 d k H(q)^3-\alpha A_1 \beta c^2 h H(q)+\alpha A_1 c^2 h \nu H(q)+A_1 \beta ^2 c^2 h H(q)-2 A_1 \beta c^2 h \nu H(q)\\\displaystyle \,+\,A_1 \beta c^2 h H(q)+A_1 c^2 h \nu ^2 H(q)-A_1 c^2 h \nu H(q)-A_1 c^2 k \nu H(q)+60 \alpha A_1 c d^3 H(q)^4\\\displaystyle \,-\,60 A_1 \beta c d^3 H(q)^4+10 \alpha A_1^2 c d^2 k H(q)^4-10 A_1^2 \beta c d^2 k H(q)^4-3 \alpha A_1 \beta c d h H(q)^2\\\displaystyle \,+\,3 \alpha A_1 c d h \nu H(q)^2+3 A_1 \beta ^2 c d h H(q)^2-6 A_1 \beta c d h \nu H(q)^2+3 A_1 \beta c d h H(q)^2\\\displaystyle \,+\,3 A_1 c d h \nu ^2 H(q)^2-3 A_1 c d h \nu H(q)^2-3 A_1 c d k \nu H(q)^2+24 \alpha A_1 d^4 H(q)^5\\\displaystyle \,-\,24 A_1 \beta d^4 H(q)^5+4 \alpha A_1^2 d^3 k H(q)^5-4 A_1^2 \beta d^3 k H(q)^5-2 \alpha A_1 \beta d^2 h H(q)^3\\\displaystyle \,+\,2 \alpha A_1 d^2 h \nu H(q)^3+2 A_1 \beta ^2 d^2 h H(q)^3-4 A_1 \beta d^2 h \nu H(q)^3+2 A_1 \beta d^2 h H(q)^3\\\displaystyle \,+\,2 A_1 d^2 h \nu ^2 H(q)^3-2 A_1 d^2 h \nu H(q)^3-2 A_1 d^2 k \nu H(q)^3=0.
\end{array} $$
Equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations in terms of A0 and A1:
$$\begin{array}{}
\displaystyle \alpha A_1 c^4-A_1 \beta c^4-\alpha A_1 \beta c^2 h+\alpha A_1 c^2 h \nu +A_1 \beta ^2 c^2 h-2 A_1 \beta c^2 h \nu +A_1 \beta c^2 h+A_1 c^2 h \nu ^2\\\displaystyle \,-\,A_1 c^2 h \nu -A_1 c^2 k \nu =0,\\\displaystyle 15 \alpha A_1 c^3 d-15 A_1 \beta c^3 d+2 \alpha A_1^2 c^3 k-2 A_1^2 \beta c^3 k-3 \alpha A_1 \beta c d h+3 \alpha A_1 c d h \nu +3 A_1 \beta ^2 c d h\\\displaystyle \,-\,6 A_1 \beta c d h \nu +3 A_1 \beta c d h+3 A_1 c d h \nu ^2-3 A_1 c d h \nu -3 A_1 c d k \nu =0,\\\displaystyle 50 \alpha A_1 c^2 d^2-50 A_1 \beta c^2 d^2+8 \alpha A_1^2 c^2 d k-8 A_1^2 \beta c^2 d k-2 \alpha A_1 \beta d^2 h+2 \alpha A_1 d^2 h \nu \\\displaystyle \,+\,2 A_1 \beta ^2 d^2 h-4 A_1 \beta d^2 h \nu +2 A_1 \beta d^2 h+2 A_1 d^2 h \nu ^2-2 A_1 d^2 h \nu -2 A_1 d^2 k \nu =0\\\displaystyle 60 \alpha A_1 c d^3-60 A_1 \beta c d^3+10 \alpha A_1^2 c d^2 k-10 A_1^2 \beta c d^2 k=0\\\displaystyle 24 \alpha A_1 d^4-24 A_1 \beta d^4+4 \alpha A_1^2 d^3 k-4 A_1^2 \beta d^3 k=0.
\end{array} $$
Solving the above system with the aid of Mathematica, we obtain
where q = x+(β − α)y−νz +(ν − β)t and C is an arbitrary constant.
Solutions of (3) using Riccati as the simplest equation
Substituting (15) into (10) and using the Riccati equation (13) we obtain
$$\begin{array}{}
\displaystyle 4 d^3 k \alpha A_1^2 H(q)^5-4 d^3 k \beta A_1^2 H(q)^5+24 d^4 \alpha A_1 H(q)^5-24 d^4 \beta A_1 H(q)^5 \\\displaystyle \,+\,10 c d^2 k \alpha A_1^2 H(q)^4-10 c d^2 k \beta A_1^2 H(q)^4+60 c d^3 \alpha A_1 H(q)^4-60 c d^3 \beta A_1 H(q)^4 \\\displaystyle \,+\,8 c^2 d k \alpha A_1^2 H(q)^3+8 d^2 e k \alpha A_1^2 H(q)^3-8 c^2 d k \beta A_1^2 H(q)^3-8 d^2 e k \beta A_1^2 H(q)^3 \\\displaystyle \,+\,2 d^2 h \beta ^2 A_1 H(q)^3+2 d^2 h \nu ^2 A_1 H(q)^3+50 c^2 d^2 \alpha A_1 H(q)^3+40 d^3 e \alpha A_1 H(q)^3 \\\displaystyle \,-\,50 c^2 d^2 \beta A_1 H(q)^3-40 d^3 e \beta A_1 H(q)^3+2 d^2 h \beta A_1 H(q)^3-2 d^2 h \alpha \beta A_1 H(q)^3 \\\displaystyle \,-\,2 d^2 h \nu A_1 H(q)^3-2 d^2 k \nu A_1 H(q)^3+2 d^2 h \alpha \nu A_1 H(q)^3-4 d^2 h \beta \nu A_1 H(q)^3 \\\displaystyle \,+\,2 c^3 k \alpha A_1^2 H(q)^2+12 c d e k \alpha A_1^2 H(q)^2-2 c^3 k \beta A_1^2 H(q)^2-12 c d e k \beta A_1^2 H(q)^2 \\\displaystyle \,+\,3 c d h \beta ^2 A_1 H(q)^2+3 c d h \nu ^2 A_1 H(q)^2+15 c^3 d \alpha A_1 H(q)^2+60 c d^2 e \alpha A_1 H(q)^2 \\\displaystyle \,-\,15 c^3 d \beta A_1 H(q)^2-60 c d^2 e \beta A_1 H(q)^2+3 c d h \beta A_1 H(q)^2-3 c d h \alpha \beta A_1 H(q)^2 \\\displaystyle \,-\,3 c d h \nu A_1 H(q)^2-3 c d k \nu A_1 H(q)^2+3 c d h \alpha \nu A_1 H(q)^2-6 c d h \beta \nu A_1 H(q)^2 \\\displaystyle \,+\,4 d e^2 k \alpha A_1^2 H(q)+4 c^2 e k \alpha A_1^2 H(q)-4 d e^2 k \beta A_1^2 H(q)-4 c^2 e k \beta A_1^2 H(q) \\\displaystyle \,+\,c^2 h \beta ^2 A_1 H(q)+2 d e h \beta ^2 A_1 H(q)+c^2 h \nu ^2 A_1 H(q)+2 d e h \nu ^2 A_1 H(q)+c^4 \alpha A_1 H(q) \\\displaystyle \,+\,16 d^2 e^2 \alpha A_1 H(q)+22 c^2 d e \alpha A_1 H(q)-c^4 \beta A_1 H(q)-16 d^2 e^2 \beta A_1 H(q) \\\displaystyle \,-\,22 c^2 d e \beta A_1 H(q)+c^2 h \beta A_1 H(q)+2 d e h \beta A_1 H(q)-c^2 h \alpha \beta A_1 H(q) \\\displaystyle \,-\,2 d e h \alpha \beta A_1 H(q)-c^2 h \nu A_1 H(q)-2 d e h \nu A_1 H(q)-c^2 k \nu A_1 H(q)-2 d e k \nu A_1 H(q) \\\displaystyle \,+\,c^2 h \alpha \nu A_1 H(q)+2 d e h \alpha \nu A_1 H(q)-2 c^2 h \beta \nu A_1 H(q)-4 d e h \beta \nu A_1 H(q)+2 c e^2 k \alpha A_1^2 \\\displaystyle \,-\,2 c e^2 k \beta A_1^2+c e h \beta ^2 A_1+c e h \nu ^2 A_1+8 c d e^2 \alpha A_1+c^3 e \alpha A_1-8 c d e^2 \beta A_1-c^3 e \beta A_1 \\\displaystyle \,+\,c e h \beta A_1-c e h \alpha \beta A_1-c e h \nu A_1-c e k \nu A_1+c e h \alpha \nu A_1-2 c e h \beta \nu A_1=0.
\end{array} $$
As before, equating coefficients of Hi to zero, we obtain
$$\begin{array}{}
\displaystyle e \alpha A_1 c^3-e \beta A_1 c^3+2 e^2 k \alpha A_1^2 c-2 e^2 k \beta A_1^2 c+e h \beta ^2 A_1 c+e h \nu ^2 A_1 c+8 d e^2 \alpha A_1 c \\\displaystyle \,-\,8 d e^2 \beta A_1 c+e h \beta A_1 c-e h \alpha \beta A_1 c-e h \nu A_1 c-e k \nu A_1 c+e h \alpha \nu A_1 c-2 e h \beta \nu A_1 c=0,\\\displaystyle \alpha A_1 c^4-\beta A_1 c^4+4 e k \alpha A_1^2 c^2-4 e k \beta A_1^2 c^2+h \beta ^2 A_1 c^2+h \nu ^2 A_1 c^2+22 d e \alpha A_1 c^2 \\\displaystyle \,-\,22 d e \beta A_1 c^2+h \beta A_1 c^2-h \alpha \beta A_1 c^2-h \nu A_1 c^2-k \nu A_1 c^2+h \alpha \nu A_1 c^2-2 h \beta \nu A_1 c^2 \\\displaystyle \,+\,4 d e^2 k \alpha A_1^2-4 d e^2 k \beta A_1^2+2 d e h \beta ^2 A_1+2 d e h \nu ^2 A_1+16 d^2 e^2 \alpha A_1-16 d^2 e^2 \beta A_1 \\\displaystyle \,+\,2 d e h \beta A_1-2 d e h \alpha \beta A_1-2 d e h \nu A_1-2 d e k \nu A_1+2 d e h \alpha \nu A_1-4 d e h \beta \nu A_1=0,\\\displaystyle 2 k \alpha A_1^2 c^3-2 k \beta A_1^2 c^3+15 d \alpha A_1 c^3-15 d \beta A_1 c^3+12 d e k \alpha A_1^2 c-12 d e k \beta A_1^2 c \\\displaystyle \,+\,3 d h \beta ^2 A_1 c+3 d h \nu ^2 A_1 c+60 d^2 e \alpha A_1 c-60 d^2 e \beta A_1 c+3 d h \beta A_1 c-3 d h \alpha \beta A_1 c \\\displaystyle \,-\,3 d h \nu A_1 c-3 d k \nu A_1 c+3 d h \alpha \nu A_1 c-6 d h \beta \nu A_1 c=0,\\\displaystyle 40 e \alpha A_1 d^3-40 e \beta A_1 d^3+8 e k \alpha A_1^2 d^2-8 e k \beta A_1^2 d^2+2 h \beta ^2 A_1 d^2+2 h \nu ^2 A_1 d^2 \\\displaystyle \,+\,50 c^2 \alpha A_1 d^2-50 c^2 \beta A_1 d^2+2 h \beta A_1 d^2-2 h \alpha \beta A_1 d^2-2 h \nu A_1 d^2-2 k \nu A_1 d^2 \\\displaystyle \,+\,2 h \alpha \nu A_1 d^2-4 h \beta \nu A_1 d^2+8 c^2 k \alpha A_1^2 d-8 c^2 k \beta A_1^2 d=0,\\\displaystyle 60 c \alpha A_1 d^3-60 c \beta A_1 d^3+10 c k \alpha A_1^2 d^2-10 c k \beta A_1^2 d^2=0,\\\displaystyle 24 \alpha A_1 d^4-24 \beta A_1 d^4+4 k \alpha A_1^2 d^3-4 k \beta A_1^2 d^3=0.
\end{array} $$
Solving the above system of algebraic equations we obtain
where Wα is the Lie characteristic function given by $\begin{array}{}
W^{\alpha}=\eta^{\alpha}-\xi^ju^{\alpha}_j,
\end{array} $α = 1, 2 and j runs from 1, ⋯, 4 in this particular case. Thus the conserved vectors corresponding to the nine Lie point symmetries are given by, respectively
$$\begin{array}{}
\displaystyle T _x^1 =\, \frac{1}{2} k u_{zt} v-\frac{1}{2} h u_{tt} v-\frac{1}{2} k u_{x} u_{yt} v-k u_{y} u_{xt} v+\frac{1}{2} k u_{t} u_{xy} v-\frac{3}{4} u_{xxyt} v+\frac{1}{2} h u_{t} v_{t} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k u_{t} u_{x} v_{y}+k u_{t} u_{y} v_{x}-\frac{1}{2} k u_{t} v_{z}-\frac{1}{2} u_{xt} v_{xy}+\frac{1}{2} v_{x} u_{xyt}-\frac{1}{4} v_{xx} u_{yt} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} v_{y} u_{xxt}+\frac{3}{4} u_{t} v_{xxy} , \\\displaystyle T _y^1 =\, -\frac{1}{2} h u_{tt} v-\frac{1}{2} k u_{x} u_{xt} v-\frac{1}{2} k u_{t} u_{xx} v-\frac{1}{4} u_{xxxt} v+\frac{1}{2} h u_{t} v_{t}+\frac{1}{2} k u_{t} u_{x} v_{x} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} v_{xx} u_{xt}+\frac{1}{4} v_{x} u_{xxt}+\frac{1}{4} u_{t} v_{xxx} , \\\displaystyle T _z^1 =\, -\frac{1}{2} h u_{tt} v+\frac{1}{2} k u_{xt} v+\frac{1}{2} h u_{t} v_{t}-\frac{1}{2} k u_{t} v_{x} , \\\displaystyle T _t^1 =\, \frac{1}{2} h u_{zt} v+\frac{1}{2} h u_{yt} v+\frac{1}{2} h u_{xt} v-k u_{xz} v+k u_{x} u_{xy} v+k u_{xx} u_{y} v+u_{xxxy} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h u_{t} v_{x}+\frac{1}{2} h u_{t} v_{y}+\frac{1}{2} h u_{t} v_{z} ; \\\\\\\\\\\displaystyle T _x^2 =\, h u_{zt} v+h u_{yt} v+\frac{1}{2} h u_{xt} v+k u_{x} u_{xy} v-\frac{1}{2} k u_{xz} v+\frac{1}{4} u_{xxxy} v+\frac{1}{2} h v_{t} u_{x} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k u_{x}^2 v_{y}+k u_{x} u_{y} v_{x}-\frac{1}{2} k u_{x} v_{z}+\frac{3}{4} u_{x} v_{xxy}-\frac{1}{2} u_{xx} v_{xy}-\frac{1}{4} v_{xx} u_{xy} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} v_{x} u_{xxy}+\frac{1}{4} u_{xxx} v_{y} , \\\displaystyle T _y^2 =\, \frac{1}{2} h v_{t} u_{x}-\frac{1}{2} h u_{xt} v-k u_{xx} u_{x} v-\frac{1}{4} u_{xxxx} v+\frac{1}{2} k u_{x}^2 v_{x}+\frac{1}{4} u_{x} v_{xxx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} u_{xx} v_{xx}+\frac{1}{4} u_{xxx} v_{x} , \\\displaystyle T _z^2 =\, -\frac{1}{2} h u_{xt} v+\frac{1}{2} k u_{xx} v+\frac{1}{2} h v_{t} u_{x}-\frac{1}{2} k u_{x} v_{x} , \\\displaystyle T _t^2 =\, -\frac{1}{2} h u_{xz} v-\frac{1}{2} h u_{xy} v-\frac{1}{2} h u_{xx} v+\frac{1}{2} h u_{x} v_{y}+\frac{1}{2} h u_{x} v_{z}+\frac{1}{2} h u_{x} v_{x} ; \\\\\\\\\\\displaystyle T _x^3 =\, -\frac{1}{2} h u_{yt} v-\frac{1}{2} k u_{y} u_{xy} v+\frac{1}{2} k u_{yz} v-\frac{1}{2} k u_{x} u_{yy} v-\frac{3}{4} u_{xxyy} v+\frac{1}{2} h v_{t} u_{y} \\\displaystyle \qquad\,\,\,\,+\,k u_{y}^2 v_{x}+\frac{1}{2} k u_{x} u_{y} v_{y}-\frac{1}{2} k u_{y} v_{z}+\frac{3}{4} u_{y} v_{xxy}-\frac{1}{2} u_{xy} v_{xy}+\frac{1}{2} v_{x} u_{xyy} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} u_{yy} v_{xx}+\frac{1}{4} v_{y} u_{xxy} , \\\displaystyle T _y^3 =\, h u_{zt} v+\frac{1}{2} h u_{yt} v+h u_{xt} v-k u_{xz} v+\frac{1}{2} k u_{x} u_{xy} v+\frac{1}{2} k u_{xx} u_{y} v+\frac{3}{4} u_{xxxy} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h v_{t} u_{y}+\frac{1}{2} k u_{x} u_{y} v_{x}-\frac{1}{4} v_{xx} u_{xy}+\frac{1}{4} v_{x} u_{xxy}+\frac{1}{4} u_{y} v_{xxx} , \\\displaystyle T _z^3 =\, -\frac{1}{2} h u_{yt} v+\frac{1}{2} k u_{xy} v+\frac{1}{2} h v_{t} u_{y}-\frac{1}{2} k u_{y} v_{x} , \\\displaystyle T _t^3 =\, -\frac{1}{2} h u_{yz} v-\frac{1}{2} h u_{yy} v-\frac{1}{2} h u_{xy} v+\frac{1}{2} h u_{y} v_{x}+\frac{1}{2} h u_{y} v_{z}+\frac{1}{2} h u_{y} v_{y} ; \\\\\\\\\\\displaystyle T _x^4 =\, k v_{x} u_{y}^2+\frac{1}{2} h v_{t} u_{y}-\frac{1}{2} k v_{z} u_{y}+\frac{1}{2} k v_{y} u_{x} u_{y}+k u_{z} v_{x} u_{y}+k u_{x} v_{x} u_{y}-k v u_{xz} u_{y} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k v u_{xy} u_{y}+\frac{3}{4} v_{xxy} u_{y}+\frac{1}{2} k v_{y} u_{x}^2+\frac{1}{2} h v_{t} u_{z}-\frac{1}{2} k u_{z} v_{z}+\frac{1}{2} h v u_{zt} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k v u_{zz}+\frac{1}{2} h v u_{yt}+\frac{1}{2} k v u_{yz}+\frac{1}{2} h v_{t} u_{x}-\frac{1}{2} k v_{z} u_{x}+\frac{1}{2} k u_{z} v_{y} u_{x} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k v u_{yz} u_{x}-\frac{1}{2} k v u_{yy} u_{x}+\frac{1}{2} h v u_{xt}-\frac{1}{2} k v u_{xz}+\frac{1}{2} k v u_{z} u_{xy}+k v u_{x} u_{xy} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} u_{xz} v_{xy}-\frac{1}{2} u_{xy} v_{xy}+\frac{1}{2} v_{x} u_{xyz}+\frac{1}{2} v_{x} u_{xyy}-\frac{1}{2} v_{xy} u_{xx}-\frac{1}{4} u_{yz} v_{xx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} u_{yy} v_{xx}-\frac{1}{4} u_{xy} v_{xx}+\frac{1}{4} v_{y} u_{xxz}+\frac{1}{4} v_{y} u_{xxy}+\frac{1}{2} v_{x} u_{xxy}+\frac{3}{4} u_{z} v_{xxy} \\\displaystyle \qquad\,\,\,\,+\,\frac{3}{4} u_{x} v_{xxy}-\frac{3}{4} v u_{xxyz}-\frac{3}{4} v u_{xxyy}+\frac{1}{4} v_{y} u_{xxx}+\frac{1}{4} v u_{xxxy} , \\\displaystyle T _y^4 =\, \frac{1}{2} h u_{zt} v+\frac{1}{2} h u_{yt} v+\frac{1}{2} h u_{xt} v-\frac{1}{2} k u_{x} u_{xz} v+\frac{1}{2} k u_{x} u_{xy} v-k u_{xx} u_{x} v \\\displaystyle \qquad\,\,\,\,-\,k u_{xz} v-\frac{1}{2} k u_{xx} u_{z} v+\frac{1}{2} k u_{xx} u_{y} v-\frac{1}{4} u_{xxxz} v+\frac{3}{4} u_{xxxy} v-\frac{1}{4} u_{xxxx} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h v_{t} u_{x}+\frac{1}{2} h v_{t} u_{y}+\frac{1}{2} h v_{t} u_{z}+\frac{1}{2} k u_{x} u_{y} v_{x}+\frac{1}{2} k u_{x} u_{z} v_{x}+\frac{1}{2} k u_{x}^2 v_{x} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} v_{xx} u_{xy}+\frac{1}{4} v_{x} u_{xxy}+\frac{1}{4} u_{y} v_{xxx}-\frac{1}{4} v_{xx} u_{xz}+\frac{1}{4} v_{x} u_{xxz}+\frac{1}{4} u_{z} v_{xxx} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} u_{x} v_{xxx}-\frac{1}{4} u_{xx} v_{xx}+\frac{1}{4} u_{xxx} v_{x} , \\\displaystyle T _z^4 =\, \frac{1}{2} h u_{zt} v+\frac{1}{2} h u_{yt} v+\frac{1}{2} h u_{xt} v-\frac{1}{2} k u_{xz} v+\frac{1}{2} k u_{xy} v+k u_{x} u_{xy} v+\frac{1}{2} k u_{xx} v \\\displaystyle \qquad\,\,\,\,+\,k u_{xx} u_{y} v+u_{xxxy} v+\frac{1}{2} h v_{t} u_{x}+\frac{1}{2} h v_{t} u_{y}+\frac{1}{2} h v_{t} u_{z}-\frac{1}{2} k u_{y} v_{x} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k u_{z} v_{x}-\frac{1}{2} k u_{x} v_{x} , \\\displaystyle T _t^4 =\, \frac{1}{2} h u_{y} v_{x}-\frac{1}{2} h u_{zz} v-h u_{yz} v-\frac{1}{2} h u_{yy} v-h u_{xz} v-h u_{xy} v-\frac{1}{2} h u_{xx} v+\frac{1}{2} h u_{x} v_{y} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h u_{x} v_{z}+\frac{1}{2} h u_{z} v_{x}+\frac{1}{2} h u_{x} v_{x}+\frac{1}{2} h u_{y} v_{z}+\frac{1}{2} h u_{z} v_{y}+\frac{1}{2} h u_{y} v_{y}+\frac{1}{2} h u_{z} v_{z} ;\\\\\\\\\\\displaystyle T _x^5 =\, \frac{1}{2} h f_1'(t) v+ f_1(t) \left( \frac{1}{2} k v_{{}z}-\frac{1}{2} h v_{t} -\frac{1}{2} k u_{x} v_{y}-k u_{y} v_{x} -\frac{3}{4} v_{xxy}-\frac{1}{2} k u_{xy} v \right) , \\\displaystyle T _y^5 =\, \frac{1}{2} k f_1(t) u_{xx} v+\frac{1}{2} h f_1'(t) v-\frac{1}{2} h f(t) v_{t}-\frac{1}{2} k f(t) u_{x} v_{x}-\frac{1}{4} f(t) v_{xxx} , \\\displaystyle T _z^5 =\, \frac{1}{2} h f_1'(t) v-\frac{1}{2} h f_1(t) v_{t}+\frac{1}{2} k f_1(t) v_{x} , \\\displaystyle T _t^5 =\, -\frac{1}{2} h f_1(t) v_{x}-\frac{1}{2} h f_1(t) v_{y}-\frac{1}{2} h f_1(t) v_{z} ; \\\\\\\\\\\displaystyle T _x^6 =\, -\frac{1}{2} k f_2'(z) v+ f_2(z)\left( \frac{1}{2} k v_{{}z}-\frac{1}{2} h v_{t}-\frac{1}{2} k u_{x} v_{y}-k u_{y} v_{x} -\frac{3}{4} v_{xxy}-\frac{1}{2} k u_{xy} v\right) , \\\displaystyle T _y^6=\, \frac{1}{2} k f_2(z) u_{xx} v-\frac{1}{2} h f_2(z) v_{t}-\frac{1}{2} k f_2(z) u_{x} v_{x}-\frac{1}{4} f_2(z) v_{xxx} , \\\displaystyle T _z^6 =\, \frac{1}{2} k f_2(z) v_{x}-\frac{1}{2} h f_2(z) v_{t} , \\\displaystyle T _t^6 =\, \frac{1}{2} h f_2'(z) v-\frac{1}{2} h f_2(z) v_{x}-\frac{1}{2} h f_2(z) v_{y}-\frac{1}{2} h f_2(z) v_{z} ; \\\\\\\\\\\displaystyle T _x^7 =\, 2 v u_{t} h^2-x v_{t} h^2-\frac{1}{2} y v_{t} h^2+2 z v_{t} h^2-\frac{1}{2} u v_{t} h^2-\frac{3}{2} t u_{t} v_{t} h^2+\frac{3}{2} t v u_{tt} h^2 \\\displaystyle \qquad\,\,\,\,-\,x v u_{zt} h^2+z v u_{zt} h^2-x v u_{yt} h^2+z v u_{yt} h^2-\frac{1}{2} x v_{t} u_{x} h^2+\frac{1}{2} z v_{t} u_{x} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} x v u_{xt} h^2+\frac{1}{2} z v u_{xt} h^2-\frac{1}{2} k x v_{y} u_{x}^2 h+\frac{1}{2} k z v_{y} u_{x}^2 h+2 k v h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k v u_{z} h+k x v_{z} h+\frac{1}{2} k y v_{z} h-2 k z v_{z} h+\frac{1}{2} k u v_{z} h+\frac{3}{2} k t u_{t} v_{z} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k t v u_{zt} h+2 k v u_{y} h+2 k t v u_{yt} h+k t v_{t} u_{x} h+\frac{1}{2} k x v_{z} u_{x} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k z v_{z} u_{x} h+\frac{5}{2} k v u_{y} u_{x} h-k x v_{y} u_{x} h-\frac{1}{2} k y v_{y} u_{x} h+2 k z v_{y} u_{x} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k u v_{y} u_{x} h-\frac{3}{2} k t u_{t} v_{y} u_{x} h+\frac{3}{2} k t v u_{yt} u_{x} h-2 k x u_{y} v_{x} h-k y u_{y} v_{x} h \\\displaystyle \qquad\,\,\,\,+\,4 k z u_{y} v_{x} h-k u u_{y} v_{x} h-3 k t u_{t} u_{y} v_{x} h-k x u_{y} u_{x} v_{x} h+k z u_{y} u_{x} v_{x} h \\\displaystyle \qquad\,\,\,\,+\,k t v u_{xt} h+3 k t v u_{y} u_{xt} h+\frac{1}{2} k x v u_{xz} h-\frac{1}{2} k z v u_{xz} h-k x v u_{xy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k y v u_{xy} h+2 k z v u_{xy} h-\frac{1}{2} k u v u_{xy} h-\frac{3}{2} k t v u_{t} u_{xy} h-k x v u_{x} u_{xy} h \\\displaystyle \qquad\,\,\,\,+\,k z v u_{x} u_{xy} h-v_{x} u_{xy} h+u_{x} v_{xy} h+\frac{3}{2} t u_{xt} v_{xy} h+v_{xy} h-\frac{3}{2} t v_{x} u_{xyt} h \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{4} v_{y} u_{xx} h+\frac{1}{2} x v_{xy} u_{xx} h-\frac{1}{2} z v_{xy} u_{xx} h+\frac{1}{4} u_{y} v_{xx} h+\frac{3}{4} t u_{yt} v_{xx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} x u_{xy} v_{xx} h-\frac{1}{4} z u_{xy} v_{xx} h+\frac{1}{4} v_{xx} h-\frac{3}{4} t v_{y} u_{xxt} h+\frac{9}{4} v u_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} x v_{x} u_{xxy} h+\frac{1}{2} z v_{x} u_{xxy} h-\frac{3}{2} x v_{xxy} h-\frac{3}{4} y v_{xxy} h+3 z v_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{4} u v_{xxy} h-\frac{9}{4} t u_{t} v_{xxy} h-\frac{3}{4} x u_{x} v_{xxy} h+\frac{3}{4} z u_{x} v_{xxy} h+\frac{9}{4} t v u_{xxyt} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} x v_{y} u_{xxx} h+\frac{1}{4} z v_{y} u_{xxx} h-\frac{1}{4} x v u_{xxxy} h+\frac{1}{4} z v u_{xxxy} h+k^2 t v_{y} u_{x}^2 \\\displaystyle \qquad\,\,\,\,-\,k^2 t v_{z} u_{x}+2 k^2 t u_{y} u_{x} v_{x}-k^2 t v u_{xz}+2 k^2 t v u_{x} u_{xy}-k t v_{xy} u_{xx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t u_{xy} v_{xx}+k t v_{x} u_{xxy}+\frac{3}{2} k t u_{x} v_{xxy}+\frac{1}{2} k t v_{y} u_{xxx}+\frac{1}{2} k t v u_{xxxy} , \\\displaystyle T _y^7 =\, 2 v u_{t} h^2-x v_{t} h^2-\frac{1}{2} y v_{t} h^2+2 z v_{t} h^2-\frac{1}{2} u v_{t} h^2-\frac{3}{2} t u_{t} v_{t} h^2+\frac{3}{2} t v u_{tt} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} x v_{t} u_{x} h^2+\frac{1}{2} z v_{t} u_{x} h^2+\frac{1}{2} x v u_{xt} h^2-\frac{1}{2} z v u_{xt} h^2+k v u_{x}^2 h+k t v_{t} u_{x} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k x u_{x}^2 v_{x} h+\frac{1}{2} k z u_{x}^2 v_{x} h-k x u_{x} v_{x} h-\frac{1}{2} k y u_{x} v_{x} h+2 k z u_{x} v_{x} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k u u_{x} v_{x} h-\frac{3}{2} k t u_{t} u_{x} v_{x} h-k t v u_{xt} h+\frac{3}{2} k t v u_{x} u_{xt} h+k x v u_{xx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k y v u_{xx} h-2 k z v u_{xx} h+\frac{1}{2} k u v u_{xx} h+\frac{3}{2} k t v u_{t} u_{xx} h+k x v u_{x} u_{xx} h \\\displaystyle \qquad\,\,\,\,-\,k z v u_{x} u_{xx} h-\frac{3}{4} v_{x} u_{xx} h+\frac{1}{2} u_{x} v_{xx} h+\frac{3}{4} t u_{xt} v_{xx} h+\frac{1}{4} x u_{xx} v_{xx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} z u_{xx} v_{xx} h+\frac{1}{2} v_{xx} h -\frac{3}{4} t v_{x} u_{xxt} h+v u_{xxx} h -\frac{1}{4} x v_{x} u_{xxx} h+\frac{1}{4} z v_{x} u_{xxx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} x v_{xxx} h-\frac{1}{4} y v_{xxx} h+z v_{xxx} h-\frac{1}{4} u v_{xxx} h-\frac{3}{4} t u_{t} v_{xxx} h-\frac{1}{4} x u_{x} v_{xxx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} z u_{x} v_{xxx} h+\frac{3}{4} t v u_{xxxt} h+\frac{1}{4} x v u_{xxxx} h-\frac{1}{4} z v u_{xxxx} h+k^2 t u_{x}^2 v_{x} \\\displaystyle \qquad\,\,\,\,-\,2 k^2 t v u_{x} u_{xx}-\frac{1}{2} k t u_{xx} v_{xx}+\frac{1}{2} k t v_{x} u_{xxx}+\frac{1}{2} k t u_{x} v_{xxx}-\frac{1}{2} k t v u_{xxxx} , \\\displaystyle T _z^7 =\, 2 h^2 u_{t} v-\frac{1}{2} h^2 v_{t} u+\frac{3}{2} h^2 t u_{tt} v+\frac{1}{2} h^2 x u_{xt} v-\frac{1}{2} h^2 z u_{xt} v-2 h ku_{x} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h k v_{x} u-\frac{5}{2} h k t u_{xt} v-\frac{1}{2} h k x u_{xx} v+\frac{1}{2} h k z u_{xx} v+k^2 t u_{xx} v-h k v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h^2 z v_{t} u_{x}-\frac{1}{2} h^2 x v_{t} u_{x}-\frac{3}{2} h^2 t u_{t} v_{t}-h^2 x v_{t}-\frac{1}{2} h^2 y v_{t}+2 h^2 z v_{t}+h k t v_{t} u_{x} \\\displaystyle \qquad\,\,\,\,+\,\frac{3}{2} h k t u_{t} v_{x} -\frac{1}{2} h k z u_{x} v_{x}+\frac{1}{2} h k x u_{x} v_{x} +\frac{1}{2} h k y v_{x}-2 h k z v_{x}+h k x v_{x}-k^2 t u_{x} v_{x} , \\\displaystyle T _t^7 =\, -\frac{1}{2} v h^2+\frac{1}{2} v u_{z} h^2-x v_{z} h^2-\frac{1}{2} y v_{z} h^2+2 z v_{z} h^2-\frac{1}{2} u v_{z} h^2-\frac{3}{2} t u_{t} v_{z} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{2} t v u_{zt} h^2+\frac{1}{2} v u_{y} h^2-x v_{y} h^2-\frac{1}{2} y v_{y} h^2+2 z v_{y} h^2-\frac{1}{2} u v_{y} h^2-\frac{3}{2} t u_{t} v_{y} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{2} t v u_{yt} h^2+\frac{1}{2} v u_{x} h^2-\frac{1}{2} x v_{z} u_{x} h^2 + \frac{1}{2} z v_{z} u_{x} h^2-\frac{1}{2} x v_{y} u_{x} h^2+\frac{1}{2} z v_{y} u_{x} h^2 \\\displaystyle \qquad\,\,\,\,-\,x v_{x} h^2-\frac{1}{2} y v_{x} h^2+2 z v_{x} h^2- \frac{1}{2} uv_{x} h^2-\frac{3}{2} t u_{t} v_{x} h^2-\frac{1}{2} x u_{x} v_{x} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} z u_{x} v_{x} h^2-\frac{3}{2} t v u_{xt} h^2+\frac{1}{2} x v u_{xz} h^2-\frac{1}{2} z v u_{xz} h^2+\frac{1}{2} x v u_{xy} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v u_{xy} h^2+\frac{1}{2} x v u_{xx}h^2- \frac{1}{2} z v u_{xx} h^2+k t v_{z} u_{x} h+k t v_{y} u_{x} h+k t u_{x} v_{x} h \\\displaystyle \qquad\,\,\,\,+\,kht\left( 2 v u_{xz} - v u_{xy} -3 v u_{x} u_{xy} - v u_{xx} -3 v u_{y} u_{xx} \right) -3 ht v u_{xxxy} ; \\\\\\\\\\\displaystyle T _x^{8} =\, \frac{1}{2} t u_{t} v_{t} h^2-\frac{1}{2} v u_{t} h^2 -\frac{1}{2} t v u_{tt} h^2-\frac{1}{2} z v_{t} u_{z} h^2 - \frac{1}{2} z v u_{zt} h^2-\frac{1}{2} z v_{t} u_{y} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v u_{yt} h^2-\frac{1}{2} z v_{t} u_{x} h^2-\frac{1}{2} z v u_{xt} h^2-\frac{1}{2} k z v_{y} u_{x}^2 h-\frac{1}{2} k v h-\frac{1}{2} k t v_{t} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k v u_{z} h+k z v_{z} h-\frac{1}{2} k t u_{t} v_{z} h+\frac{1}{2} k z u_{z} v_{z} h-\frac{1}{2} k t v u_{zt} h-\frac{1}{2} k z v u_{zz} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k v u_{y} h+\frac{1}{2} k z v_{z} u_{y} h-k t v u_{yt} h-\frac{1}{2} k z v u_{yz} h-\frac{1}{2} k t v_{t} u_{x} h+\frac{1}{2} k z v_{z} u_{x} h \\\displaystyle \qquad\,\,\,\,-\,k z v_{y} u_{x} h+\frac{1}{2} k t u_{t} v_{y} u_{x} h-\frac{1}{2} k z u_{z} v_{y} u_{x} h-\frac{1}{2} k z u_{y} v_{y} u_{x} h-\frac{1}{2} k t v u_{yt} u_{x} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k z v u_{yz} u_{x} h+\frac{1}{2} k z v u_{yy} u_{x} h-k z u_{y}^2 v_{x} h-2 k z u_{y} v_{x} h+k t u_{t} u_{y} v_{x} h \\\displaystyle \qquad\,\,\,\,-\,k z u_{z} u_{y} v_{x} h-k z u_{y} u_{x} v_{x} h-\frac{1}{2} k t v u_{xt} h-k t v u_{y} u_{xt} h+\frac{1}{2} k z v u_{xz} h \\\displaystyle \qquad\,\,\,\,+\,k z v u_{y} u_{xz} h-k z v u_{xy} h+\frac{1}{2} k t v u_{t} u_{xy} h-\frac{1}{2} k z v u_{z} u_{xy} h+\frac{1}{2} k z v u_{y} u_{xy} h \\\displaystyle \qquad\,\,\,\,-\,k z v u_{x} u_{xy} h-\frac{1}{2} t u_{xt} v_{xy} h+\frac{1}{2} z u_{xz} v_{xy} h+\frac{1}{2} z u_{xy} v_{xy} h+\frac{1}{2} t v_{x} u_{xyt} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v_{x} u_{xyz} h-\frac{1}{2} z v_{x} u_{xyy} h+\frac{1}{2} z v_{xy} u_{xx} h-\frac{1}{4} t u_{yt} v_{xx} h+\frac{1}{4} z u_{yz} v_{xx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} z u_{yy} v_{xx} h+\frac{1}{4} z u_{xy} v_{xx} h+\frac{1}{4} t v_{y} u_{xxt} h-\frac{1}{4} z v_{y} u_{xxz} h-\frac{1}{4} z v_{y} u_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v_{x} u_{xxy} h-\frac{3}{2} z v_{xxy} h+\frac{3}{4} t u_{t} v_{xxy} h-\frac{3}{4} z u_{z} v_{xxy} h-\frac{3}{4} z u_{y} v_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{4} z u_{x} v_{xxy} h-\frac{3}{4} t v u_{xxyt} h+\frac{3}{4} z v u_{xxyz} h+\frac{3}{4} z v u_{xxyy} h-\frac{1}{4} z v_{y} u_{xxx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} z v u_{xxxy} h-\frac{1}{2} k^2 t v_{y} u_{x}^2+\frac{1}{2} k^2 t v_{z}+\frac{1}{2} k^2 t v_{z} u_{x}-\frac{1}{2} k^2 tv_{y} u_{x}-k^2 t u_{y} v_{x} \\\displaystyle \qquad\,\,\,\,-\,k^2 t u_{y} u_{x} v_{x}+\frac{1}{2} k^2 t v u_{xz}-\frac{1}{2} k^2 t v u_{xy}-k^2 t v u_{x} u_{xy}+\frac{1}{2} k t v_{xy} u_{xx}+\frac{1}{4} k t u_{xy} v_{xx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v_{x} u_{xxy}-\frac{3}{4} k t v_{xxy}-\frac{3}{4} k t u_{x} v_{xxy}-\frac{1}{4} k t v_{y} u_{xxx}-\frac{1}{4} k t v u_{xxxy}-z v_{t} h^2 , \\\displaystyle T _y^{8} =\, -\frac{1}{2} v u_{t} h^2-z v_{t} h^2+\frac{1}{2} t u_{t} v_{t} h^2-\frac{1}{2} t v u_{tt} h^2-\frac{1}{2} z v_{t} u_{z} h^2-\frac{1}{2} z v u_{zt} h^2-\frac{1}{2} z v_{t} u_{y} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v u_{yt} h^2-\frac{1}{2} z v_{t} u_{x} h^2-\frac{1}{2} z v u_{xt} h^2+\frac{1}{2} k v h-\frac{1}{2} k t v_{t} h+\frac{1}{2} k v u_{x} h-\frac{1}{2} k t v_{t} u_{x} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k z u_{x}^2 v_{x} h-k z u_{x} v_{x} h+\frac{1}{2} k t u_{t} u_{x} v_{x} h-\frac{1}{2} k z u_{z} u_{x} v_{x} h-\frac{1}{2} k z u_{y} u_{x} v_{x} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k t v u_{xt} h-\frac{1}{2} k t v u_{x} u_{xt} h+k z v u_{xz} h+\frac{1}{2} k z v u_{x} u_{xz} h-\frac{1}{2} k z v u_{x} u_{xy} h+k z v u_{xx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v u_{t} u_{xx} h+\frac{1}{2} k z v u_{z} u_{xx} h-\frac{1}{2} k z v u_{y} u_{xx} h+k z v u_{x} u_{xx} h-\frac{1}{4} t u_{xt} v_{xx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} z u_{xz} v_{xx} h+\frac{1}{4} z u_{xy} v_{xx} h+\frac{1}{4} z u_{xx} v_{xx} h+\frac{1}{4} t v_{x} u_{xxt} h-\frac{1}{4} z v_{x} u_{xxz} h-\frac{1}{4} z v_{x} u_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} z v_{x} u_{xxx} h-\frac{1}{2} z v_{xxx} h+\frac{1}{4} t u_{t} v_{xxx} h - \frac{1}{4} z u_{z} v_{xxx} h-\frac{1}{4} z u_{y} v_{xxx} h-\frac{1}{4} z u_{x} v_{xxx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} t v u_{xxxt} h+\frac{1}{4} z v u_{xxxz} h-\frac{3}{4} z v u_{xxxy} h+\frac{1}{4} z v u_{xxxx} h-\frac{1}{2} k^2 t u_{x}^2 v_{x}-\frac{1}{2} k^2 t u_{x} v_{x} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k^2 t v u_{xx}+k^2 t v u_{x} u_{xx}+\frac{1}{4} k t \left( u_{xx} v_{xx}- v_{x} u_{xxx}- v_{xxx}- u_{x} v_{xxx}+ v u_{xxxx}\right) , \\\displaystyle T _z^{8} =\, -\frac{1}{2} h^2 u_{t} v - \frac{1}{2} h^2 t u_{tt} v-\frac{1}{2} h^2 z u_{zt} v-\frac{1}{2} h^2 z u_{yt} v-\frac{1}{2} h^2 z u_{xt} v+\frac{1}{2} h k u_{x} v+h k t u_{xt} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h k z u_{xz} v-\frac{1}{2} h k z u_{xy} v-h k z u_{x} u_{xy} v -\frac{1}{2} h k z u_{xx} v-h k z u_{xx} u_{y} v-h z u_{xxxy} v \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k^2 t u_{xx} v+\frac{1}{2} h k v-\frac{1}{2} h^2 z v_{t} u_{x}-\frac{1}{2} h^2 z v_{t} u_{y} -\frac{1}{2} h^2 z v_{t} u_{z} + \frac{1}{2} h^2 t u_{t} v_{t}-h^2 z v_{t} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} h k t v_{t} u_{x}-\frac{1}{2} h k t u_{t} v_{x}-\frac{1}{2} h k t v_{t}+\frac{1}{2} h k z u_{y} v_{x}+\frac{1}{2} h k z u_{z} v_{x} + \frac{1}{2} h k z u_{x} v_{x} \\\displaystyle \qquad\,\,\,\,+\,h k z v_{x}+\frac{1}{2} k^2 t u_{x} v_{x}+\frac{1}{2} k^2 t v_{x} , \\\displaystyle T _t^{8} =\, v h^2+\frac{1}{2} v u_{z} h^2-z v_{z} h^2+\frac{1}{2} t u_{t} v_{z} h^2-\frac{1}{2} z u_{z} v_{z} h^2+\frac{1}{2} t v u_{zt} h^2+\frac{1}{2} z v u_{zz} h^2+\frac{1}{2} v u_{y} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z v_{z} u_{y} h^2-z v_{y} h^2+\frac{1}{2} t u_{t} v_{y} h^2-\frac{1}{2} z u_{z} v_{y} h^2-\frac{1}{2} z u_{y} v_{y} h^2+\frac{1}{2} t v u_{yt} h^2+z v u_{yz} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} z v u_{yy} h^2+\frac{1}{2} v u_{x} h^2-\frac{1}{2} z v_{z} u_{x} h^2-\frac{1}{2} z v_{y} u_{x} h^2-z v_{x} h^2+\frac{1}{2} t u_{t} v_{x} h^2-\frac{1}{2} z u_{z} v_{x} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} z u_{y} v_{x} h^2-\frac{1}{2} z u_{x} v_{x} h^2+\frac{1}{2} t v u_{xt} h^2+z v u_{xz} h^2+z v u_{xy} h^2+\frac{1}{2} z v u_{xx} h^2-\frac{1}{2} k t v_{z} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v_{y} h-\frac{1}{2} k t v_{z} u_{x} h-\frac{1}{2} k t v_{y} u_{x} h - \frac{1}{2} k t v_{x} h-\frac{1}{2} k t u_{x} v_{x} h-\frac{1}{2} k t v u_{xz} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k t v u_{xy} h+k t v u_{x} u_{xy} h+\frac{1}{2} k t v u_{xx} h+k t v u_{y} u_{xx} h+t v u_{xxxy} h ; \\\\\\\\\\\displaystyle T _x^{9} =\, -\frac{1}{2} v u_{t} h^2+\frac{1}{2} y v_{t} h^2-z v_{t} h^2+\frac{1}{2} t u_{t} v_{t} h^2-\frac{1}{2} t v u_{tt} h^2+\frac{1}{2} y v_{t} u_{y} h^2-\frac{1}{2} z v_{t} u_{y} h^2 \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} y v u_{yt} h^2+\frac{1}{2} z v u_{yt} h^2-\frac{1}{2} k v h-\frac{1}{2} k t v_{t} h-\frac{1}{2} k y v_{z} h+k z v_{z} h-\frac{1}{2} k t u_{t} v_{z} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v u_{zt} h-\frac{1}{2} k v u_{y} h-\frac{1}{2} k y v_{z} u_{y} h+\frac{1}{2} k z v_{z} u_{y} h-k t v u_{yt} h+\frac{1}{2} k y v u_{yz} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k z v u_{yz} h-\frac{1}{2} k t v_{t} u_{x} h-\frac{1}{2} k v u_{y} u_{x} h+\frac{1}{2} k y v_{y} u_{x} h-k z v_{y} u_{x} h+\frac{1}{2} k t u_{t} v_{y} u_{x} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k y u_{y} v_{y} u_{x} h-\frac{1}{2} k z u_{y} v_{y} u_{x} h-\frac{1}{2} k t v u_{yt} u_{x} h-\frac{1}{2} k y v u_{yy} u_{x} h+ \frac{1}{2} k z v u_{yy} u_{x} h \\\displaystyle \qquad\,\,\,\,+\,k y u_{y}^2 v_{x} h-k z u_{y}^2 v_{x} h+k y u_{y} v_{x} h-2 k z u_{y} v_{x} h+k t u_{t} u_{y} v_{x} h-\frac{1}{2} k t v u_{xt} h \\\displaystyle \qquad\,\,\,\,-\,k t v u_{y} u_{xt} h+\frac{1}{2} k y v u_{xy} h -k z v u_{xy} h+\frac{1}{2} k t v u_{t} u_{xy} h-\frac{1}{2} k y v u_{y} u_{xy} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k z v u_{y} u_{xy} h+\frac{1}{2} v_{x} u_{xy} h-\frac{1}{2} t u_{xt} v_{xy} h-\frac{1}{2} y u_{xy} v_{xy} h+\frac{1}{2} z u_{xy} v_{xy} h+\frac{1}{2} t v_{x} u_{xyt} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} y v_{x} u_{xyy} h-\frac{1}{2} z v_{x} u_{xyy} h-\frac{1}{4} u_{y} v_{xx} h-\frac{1}{4} t u_{yt} v_{xx} h-\frac{1}{4} y u_{yy} v_{xx} h+\frac{1}{4} z u_{yy} v_{xx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} v_{xx} h+\frac{1}{4} t v_{y} u_{xxt} h-\frac{3}{4} v u_{xxy} h+\frac{1}{4} y v_{y} u_{xxy} h-\frac{1}{4} z v_{y} u_{xxy} h+\frac{3}{4} y v_{xxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{2} z v_{xxy} h+\frac{3}{4} t u_{t} v_{xxy} h+\frac{3}{4} y u_{y} v_{xxy} h-\frac{3}{4} z u_{y} v_{xxy} h-\frac{3}{4} t v u_{xxyt} h-\frac{3}{4} y v u_{xxyy} h \\\displaystyle \qquad\,\,\,\,+\,\frac{3}{4} z v u_{xxyy} h-\frac{1}{2} k^2 t v_{y} u_{x}^2+\frac{1}{2} k^2 t v_{z}+\frac{1}{2} k^2 t v_{z} u_{x}-\frac{1}{2} k^2 t v_{y} u_{x}-k^2 t u_{y} v_{x} \\\displaystyle \qquad\,\,\,\,-\,k^2 t u_{y} u_{x} v_{x}+\frac{1}{2} k^2 t v u_{xz}-\frac{1}{2} k^2 t v u_{xy}-k^2 t v u_{x} u_{xy}+\frac{1}{2} k t v_{xy} u_{xx} +\frac{1}{4} k t u_{xy} v_{xx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v_{x} u_{xxy}-\frac{3}{4} k t v_{xxy}-\frac{3}{4} k t u_{x} v_{xxy}-\frac{1}{4} k t v_{y} u_{xxx}-\frac{1}{4} k t v u_{xxxy} , \\\displaystyle T _y^{9} =\, -\frac{1}{2} v u_{t} h^2+\frac{1}{2} y v_{t} h^2-z v_{t} h^2+\frac{1}{2} t u_{t} v_{t} h^2-\frac{1}{2} t v u_{tt} h^2+y v u_{zt} h^2-z v u_{zt} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} y v_{t} u_{y} h^2-\frac{1}{2} z v_{t} u_{y} h^2+\frac{1}{2} y v u_{yt} h^2-\frac{1}{2} z v u_{yt} h^2+y v u_{xt} h^2-z v u_{xt} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k v h-\frac{1}{2} k t v_{t} h+\frac{1}{2} k v u_{x} h-\frac{1}{2} k t v_{t} u_{x} h+\frac{1}{2} k y u_{x} v_{x} h-k z u_{x} v_{x} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k t u_{t} u_{x} v_{x} h+\frac{1}{2} k y u_{y} u_{x} v_{x} h-\frac{1}{2} k z u_{y} u_{x} v_{x} h+\frac{1}{2} k t v u_{xt} h-\frac{1}{2} k t v u_{x} u_{xt} h \\\displaystyle \qquad\,\,\,\,-\,k y v u_{xz} h+k z v u_{xz} h+\frac{1}{2} k y v u_{x} u_{xy} h-\frac{1}{2} k z v u_{x} u_{xy} h-\frac{1}{2} k y v u_{xx} h+k z v u_{xx} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v u_{t} u_{xx} h+\frac{1}{2} k y v u_{y} u_{xx} h-\frac{1}{2} k z v u_{y} u_{xx} h-\frac{1}{4} t u_{xt} v_{xx} h-\frac{1}{4} y u_{xy} v_{xx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} z u_{xy} v_{xx} h+\frac{1}{4} t v_{x} u_{xxt} h+\frac{1}{4} y v_{x} u_{xxy} h-\frac{1}{4} z v_{x} u_{xxy} h+\frac{1}{4} y v_{xxx} h-\frac{1}{2} z v_{xxx} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{4} t u_{t} v_{xxx} h+\frac{1}{4} y u_{y} v_{xxx} h-\frac{1}{4} z u_{y} v_{xxx} h-\frac{1}{4} t v u_{xxxt} h+\frac{3}{4} y v u_{xxxy} h \\\displaystyle \qquad\,\,\,\,-\,\frac{3}{4} z v u_{xxxy} h-\frac{1}{2} k^2 t u_{x}^2 v_{x}-\frac{1}{2} k^2 t u_{x} v_{x}+\frac{1}{2} k^2 t v u_{xx}+k^2 t v u_{x} u_{xx}+\frac{1}{4} k t u_{xx} v_{xx} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{4} k t v_{x} u_{xxx}-\frac{1}{4} k t v_{xxx}-\frac{1}{4} k t u_{x} v_{xxx}+\frac{1}{4} k t v u_{xxxx} , \\\displaystyle T _z^{9} =\, -\frac{1}{2} h^2 u_{t} v-\frac{1}{2} h^2 t u_{tt} v-\frac{1}{2} h^2 y u_{yt} v+\frac{1}{2} h^2 z u_{yt} v+\frac{1}{2} h k u_{x} v+h k t u_{xt} v \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h k y u_{xy} v-\frac{1}{2} h k z u_{xy} v-\frac{1}{2} k^2 t u_{xx} v+\frac{1}{2} h k v-\frac{1}{2} h^2 z v_{t} u_{y}+\frac{1}{2} h^2 y v_{t} u_{y} \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} h^2 t u_{t} v_{t}+\frac{1}{2} h^2 y v_{t}-h^2 z v_{t}-\frac{1}{2} h k t v_{t} u_{x}-\frac{1}{2} h k t u_{t} v_{x}-\frac{1}{2} h k t v_{t}+\frac{1}{2} h k z u_{y} v_{x} \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} h k y u_{y} v_{x}-\frac{1}{2} h k y v_{x}+h k z v_{x}+\frac{1}{2} k^2 t u_{x} v_{x}+\frac{1}{2} k^2 t v_{x} , \\\displaystyle T _t^{9} =\, \frac{1}{2} v h^2+\frac{1}{2} y v_{z} h^2-z v_{z} h^2+\frac{1}{2} t u_{t} v_{z} h^2+\frac{1}{2} t v u_{zt} h^2+\frac{1}{2} y v_{z} u_{y} h^2-\frac{1}{2} z v_{z} u_{y} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} y v_{y} h^2-z v_{y} h^2+\frac{1}{2} t u_{t} v_{y} h^2+\frac{1}{2} y u_{y} v_{y} h^2-\frac{1}{2} z u_{y} v_{y} h^2+\frac{1}{2} t v u_{yt} h^2-\frac{1}{2} y v u_{yz} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} z v u_{yz} h^2-\frac{1}{2} y v u_{yy} h^2+\frac{1}{2} z v u_{yy} h^2+\frac{1}{2} y v_{x} h^2-z v_{x} h^2+\frac{1}{2} t u_{t} v_{x} h^2 \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} y u_{y} v_{x} h^2-\frac{1}{2} z u_{y} v_{x} h^2+\frac{1}{2} t v u_{xt} h^2-\frac{1}{2} y v u_{xy} h^2+\frac{1}{2} z v u_{xy} h^2-\frac{1}{2} k t v_{z} h \\\displaystyle \qquad\,\,\,\,-\,\frac{1}{2} k t v_{y} h-\frac{1}{2} k t v_{z} u_{x} h-\frac{1}{2} k t v_{y} u_{x} h-\frac{1}{2} k t v_{x} h-\frac{1}{2} k t u_{x} v_{x} h-\frac{1}{2} k t v u_{xz} h \\\displaystyle \qquad\,\,\,\,+\,\frac{1}{2} k t v u_{xy} h+k t v u_{x} u_{xy} h+\frac{1}{2} k t v u_{xx} h+k t v u_{y} u_{xx} h+t v u_{xxxy} h .
\end{array} $$
Remark. It should be noted that the above conservation laws include the energy conservation law, which corresponds to the time translation and three momentum conservation laws, which correspond to the three space translations.
Conclusions
In this paper we studied the generalized second extended (3+1)-dimensional Jimbo-Miwa equation (3). Symmetry reductions of this equation were performed several times until it was reduced to a nonlinear fourth-order ordinary differential equation. The general solution of this ordinary differential equation was obtained in terms of the Weierstrass zeta function. Travelling wave solutions of (3) were also derived using the simplest equation method. Finally, the conservation laws of (3) were computed by invoking the conservation theorem due to Ibragimov. These conservation laws included an energy conservation law, which corresponded to the time translation and three momentum conservation laws that corresponded to the three space translations.
