Introduction

Nonlinear wave equations represent a wide range of physical phenomena found in physics, engineering and in applied mathematics. As a result, it’s critical to look into the specific solutions. In the literature, several approaches have been proposed, including the inverse scattering transform method, Hirota’s bilinear method, tanh and sine-cosine methods, and so on [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58]. There is no one approach that can be used to solve nonlinear evolution equations, despite the fact that numerous attempts have been made in this direction [40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58]. Over the last several decades, significant advancements have been achieved in the theory of nonlinear wave equations [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58]. Numerous research articles that have been published in the literature show that these equations’ integrability characteristics have been thoroughly explored. Since numerical approaches are often inappropriate, precise exact solutions are frequently needed. Exact solutions to nonlinear wave equations emerging in fluid dynamics, continuum mechanics, and general relativity are very important because they provide insight into extreme situations that cannot be handled numerically. Finding exact solutions is a challenging goal. Despite this, several fresh approaches to nonlinear wave equations integration have lately been created such as the inverse scattering transform, Hirota’s bilinear approach, homogeneous balancing method, auxiliary ordinary differential equation method, He’s variational iteration method, sine-cosine method, extended tanh method, Lie symmetry method, etc. [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58].

We focus on a generalised (3+1)-dimensional nonlinear wave [59] in this study

$$\begin{aligned} u_{tx}+2 h_{1}u_{xx}u+h_{1}\left( u_{x}^2\right) + h_{2} u_{xxxx}+h_{3}u_{xx}+h_{4}u_{yy}+h_{5}u_{zz}=0. \end{aligned}$$
(1.1)

Here \(u = u(x, y, z, t)\) is a real-valued function and \( h_i (i = 1\cdots 5)\) are nonzero constants. Equation (1.1) leads to many nonlinear wave equations [59] that can be retrieved after making appropriate cosmetic adjustments. The (3+1)-dimensional Kadomtsev-Petviashvili equation

$$\begin{aligned} (u_t - 6 u u_x + u_{xxx})_x + 3u_{yy}+3u_{zz} =0 \end{aligned}$$
(1.2)

was elaborated in [31,32,33]. The (3+1)-dimensional nonlinear wave equation [34]

$$\begin{aligned} (u_t + u u_x + u_{xxx})_x + \frac{1}{2}\left( u_{yy}+u_{zz}\right) =0 \end{aligned}$$
(1.3)

captures the physics of pressure waves in mixture liquid and gas bubbles by taking into consideration the viscosity of liquid and heat transfer. The distinguished Korteweg-de Vries (KdV) equation [47]

$$\begin{aligned} u_t + 6 u u_x + u_{xxx} =0 \end{aligned}$$
(1.4)

is an illustration of a nonlinear wave equation. Originally it was developed to portray shallow water waves of long wavelength and small amplitude. It is a considerable equation in the theory of integrable systems since it has an infinite number of conservation laws, multiple-soliton solutions, and numerous other material assets.

The main purpose of this work is to study a generalized (3+1)-dimensional nonlinear wave equation (1.1). The work is organized as follows. In Sects. 2-3, Lie Symmetry generators and corresponding symmetry reductions of equation (1.1) will be constructed. In Sect. 4, travelling wave solutions of equation (1.1) will be computed. Finally conserved densities and fluxes are shown and concluding remarks are given.

Symmetry Analysis of (1.1)

The Lie symmetry approach has emerged during the last several decades as a flexible method for resolving nonlinear issues posed by differential equations in physics, mathematics, and many other scientific disciplines. For further information on the theory and use of the Lie symmetry approach, see, for instance [60,61,62].

The Lie point symmetries of (1.1) is generated using the vector field

$$\begin{aligned} {\varvec{\Gamma }}= & {} \xi ^1 (t,x,y,z,u) \displaystyle \frac{\partial }{\partial t} + \xi ^2 (t, x,y,z, u) \displaystyle \frac{\partial }{\partial x} + \xi ^3 (t, x,y,z, u) \displaystyle \frac{\partial }{\partial y} +\xi ^4 (t, x,y,z, u) \displaystyle \frac{\partial }{\partial z}\nonumber \\ {}{} & {} + \eta (t, x,y,z,u) \displaystyle \frac{\partial }{\partial u}. \end{aligned}$$
(2.1)

By applying the fourth extension \(\hbox {pr}^{(4)} {\varvec{\Gamma }}\) to (1.1), an overdetermined system of linear partial differential equations is obtained as follows.

$$\begin{aligned}{} & {} \xi ^{1}_{z}=0,\end{aligned}$$
(2.2)
$$\begin{aligned}{} & {} \xi ^{4}_{x}=0,\end{aligned}$$
(2.3)
$$\begin{aligned}{} & {} \xi ^{4}_{u}=0,\end{aligned}$$
(2.4)
$$\begin{aligned}{} & {} \xi ^{1}_{y}=0,\end{aligned}$$
(2.5)
$$\begin{aligned}{} & {} \xi ^{3}_{x}=0,\end{aligned}$$
(2.6)
$$\begin{aligned}{} & {} \xi ^{3}_{u}=0,\end{aligned}$$
(2.7)
$$\begin{aligned}{} & {} \xi ^{1}_{x}=0,\end{aligned}$$
(2.8)
$$\begin{aligned}{} & {} \xi ^{2}_{u}=0,\end{aligned}$$
(2.9)
$$\begin{aligned}{} & {} \xi ^{1}_{u}=0,\end{aligned}$$
(2.10)
$$\begin{aligned}{} & {} \xi ^{2}_{xx}=0,\end{aligned}$$
(2.11)
$$\begin{aligned}{} & {} \eta _{xu}=0,\end{aligned}$$
(2.12)
$$\begin{aligned}{} & {} \eta _{uu}=0,\end{aligned}$$
(2.13)
$$\begin{aligned}{} & {} 2\xi ^{2}_{x}+\eta _{u}=0,\end{aligned}$$
(2.14)
$$\begin{aligned}{} & {} -2 h_{4}\xi ^{2}_{y}-\xi ^{3}_{t}=0,\end{aligned}$$
(2.15)
$$\begin{aligned}{} & {} -2h_{5}\xi ^{2}_{z}-\xi ^{4}_{t}=0,\end{aligned}$$
(2.16)
$$\begin{aligned}{} & {} -h_{5}\xi ^{3}_{z}-h_{4}\xi ^{4}_{y}=0,\end{aligned}$$
(2.17)
$$\begin{aligned}{} & {} -\xi ^{1}_{t}+3\xi ^{2}_{x}=0,\end{aligned}$$
(2.18)
$$\begin{aligned}{} & {} 2\xi ^{2}_{x}-\xi ^{4}_{z}=0,\end{aligned}$$
(2.19)
$$\begin{aligned}{} & {} -h_{4}\xi ^{3}_{yy}+2 h_{4}\eta _{yu}-h_{5}\xi ^{3}_{zz}=0,\end{aligned}$$
(2.20)
$$\begin{aligned}{} & {} 2h_{5}\eta _{zu}-h_{4}\xi ^{4}_{yy}-h_{5}\xi ^{4}_{zz}=0,\end{aligned}$$
(2.21)
$$\begin{aligned}{} & {} 2h_{1}\eta +4h_{1}u\xi ^{2}_{x}+2h_{3}\xi ^{2}_{x}-\xi ^{2}_{t}=0,\end{aligned}$$
(2.22)
$$\begin{aligned}{} & {} 2h_{1}\eta _{x}-\xi ^{2}_{tx}-h_{5}\xi ^{2}_{zz}-h_{4}\xi ^{2}_{yy}+\eta _{tu}=0,\end{aligned}$$
(2.23)
$$\begin{aligned}{} & {} 2h_{1}u\eta _{xx}+h_{3}\eta _{xx}+h_{2}\eta _{xxxx}+h_{5}\eta _{zz}+h_{4}\eta _{yy}+\eta _{tx}=0. \end{aligned}$$
(2.24)

By constraining the arbitrary functions to quadratic polynomials, the aforementioned system has a special solution of the form

$$\begin{aligned} \xi ^{1}( t,x,y,z,u)= & {} -3{ C_4}h_{1}h_{4}h_{5}t^{2}-3{C_2}h_{1}t+{C_3},\\ \xi ^{2}( t,x,y,z,u)= & {} -{C_2}h_{1}x-2{C_4}h_{1}h_{4}h_{5}tx+{C_4}h_{1}h_{4}z^{2}+{ C_4}h_{1}h_{5}y^{2}+\nonumber \\ {}{} & {} 4{C_5}h_{1}ty+2{C_6}h_{1}y+4{C_8}h_{1}tz+2{C_9}h_{1}z+2C_{11}h_{1}t^{2}+2 C_{12}h_{1}t\nonumber \\ {}{} & {} +2C_{13}h_{1},\\ \xi ^{3}( t,x,y,z,u)= & {} -4{ C_4}h_{1}h_{4}h_{5}ty-2{ C_2}h_{1}y-4{ C_5}h_{1}h_{4}t_{2}-4{ C_6}h_{1}h_{4}t-4{C_7}h_{1}h_{4}\nonumber \\ {}{} & {} +{C_1}h_{4}z,\\ \xi ^{4}( t,x,y,z,u)= & {} -4{ C_4}h_{1}h_{4}h_{5}tz-2{ C_2}h_{1}z-4{ C_8}h_{1}h_{5}t^{2}-4{ C_9}h_{1}h_{5}t-4 C_{10}h_{1}h_{5}\nonumber \\ {}{} & {} -{C_1}h_{5}y,\\ \eta (t,x,y,z,u)= & {} 2{ C_2}h_{1}u+{C_2}h_{3}+4{C_4}h_{1}h_{4}h_{5}tu+2{C_4}h_{3}h_{4}h_{5}t-{ C_4}h_{4}h_{5}x+\nonumber \\ {}{} & {} 2{ C_5}y+2{C_8}z +2C_{11} t +C_{12}. \end{aligned}$$

The special infinitesimal symmetries of (1.1) in operator form are

$$\begin{aligned}{} & {} {{\varvec{\Gamma }}_1} = zh_{4}\frac{\partial }{\partial y}-yh_{5}\frac{\partial }{\partial z},\\{} & {} {{\varvec{\Gamma }}_2} = -h_{1}h_{5} \frac{\partial }{\partial z},\\{} & {} {{\varvec{\Gamma }}_3}= t\frac{\partial }{\partial u}+h_{1}t^{2}\frac{\partial }{\partial x},\\{} & {} {{\varvec{\Gamma }}_4}= \frac{\partial }{\partial u}+2h_{1}t\frac{\partial }{\partial x},\\{} & {} {{\varvec{\Gamma }}_5}= h_{1} \frac{\partial }{\partial x},\\{} & {} {{\varvec{\Gamma }}_6}= \left( 2h_{1}u+h_{3}\right) \frac{\partial }{\partial u}-h_{1}x\frac{\partial }{\partial x}-2h_{1}y\frac{\partial }{\partial y}-2h_{1}z\frac{\partial }{\partial z}-3h_{1}t\frac{\partial }{\partial t},\\{} & {} {{\varvec{\Gamma }}_7}= \frac{\partial }{\partial t},\\{} & {} {{\varvec{\Gamma }}_8}= \left( 4h_{1}h_{4}h_{5}tu+2h_{3}h_{4}h_{5}t-h_{4}h_{5}x\right) \frac{\partial }{\partial u}-3h_{1}h_{4}h_{5} t^{2}\frac{\partial }{\partial t}\nonumber \\{} & {} +\left( -2h_{1}h_{4}h_{5}tx+h_{1}h_{4}z^{2}+h_{1}h_{5}y^{2}\right) \frac{\partial }{\partial x}-4h_{1}h_{4}h_{5}ty\frac{\partial }{\partial y}-4h_{1}h_{4}h_{5}tz\frac{\partial }{\partial z},\\{} & {} {{\varvec{\Gamma }}_9}= 2 h_{1}ty\frac{\partial }{\partial x}-2h_{1}h_{4}t^{2}\frac{\partial }{\partial y}+y\frac{\partial }{\partial u},\\{} & {} {{\varvec{\Gamma }}_{10}}= -2 h_{1}h_{4}t\frac{\partial }{\partial y}+ h_{1}y\frac{\partial }{\partial x},\\{} & {} {{\varvec{\Gamma }}_{11}}= -h_{1}h_{4}\frac{\partial }{\partial y},\\{} & {} {{\varvec{\Gamma }}_{12}}= h_{1}tz \frac{\partial }{\partial x}-2h_{1}h_{5} t^{2}\frac{\partial }{\partial z}+z\frac{\partial }{\partial u},\\{} & {} {{\varvec{\Gamma }}_{13}}= -2h_{1}h_{5}t\frac{\partial }{\partial z}+h_{1}z\frac{\partial }{\partial x}. \end{aligned}$$

Symmetry Reductions of (1.1)

In this subsection, we’ll create symmetry reductions and solutions to equation (1.1).

Case 1.

Considering combination of translational symmetry \({{\varvec{\Upsilon }}_1}\), where \( {\varvec{\Upsilon }}_1 = {\varvec{\Gamma }}_2+ {\varvec{\Gamma }} _5+{\varvec{\Gamma }}_7+{\varvec{\Gamma }}_{11} \), we turn equation (1.1) into a partial differential equation with three independent variables. This symmetry \({{\varvec{\Upsilon }}_1}\) generates the following invariants.

$$\begin{aligned} p=h_{4} x+y,\quad q= h_{5} x+z, \quad r= \frac{h_{1} t - x}{h_{1}}, \quad \theta =u. \end{aligned}$$

Equation (1.1) is then transformed into the following nonlinear partial differential equation using the above invariants.

$$\begin{aligned}{} & {} -4h_{1}^{3} h_{2} h_{5}^{3} \theta _{qqqr} +6 h_{1}^{2} h_{2} h_{4}^{2}\theta _{pprr}+6 h_{1}^{2} h_{2} h_{5}^{2} \theta _{qqrr}+h_{1}^{4} h_{3} h_{4}^{2}\theta _{pp}\\{} & {} +2 h_{1}^{5} h_{5}^{2} \theta \theta _{qq}+2 h_{1}^{5} h_{4}^{2} \theta \theta _{p p} -4 h_{1}^{4} h_{5}\theta \theta _{q r} -4h_{1}^{4} h_{4} \theta \theta _{p r}\\{} & {} -4h_{1} h_{2} h_{4} \theta _{p r r r} -4 h_{1} h_{2} h_{5} \theta _{q r r r}+h_{1}^{4} h_{3} h_{5}^{2}\theta _{q q} +h_{1}^{4} h_{2} h_{4}^{4}\theta _{p p p p} -2h_{1}^{4} h_{5} \theta _{q} \theta _{r} \\{} & {} -2h_{1}^{4} h_{4} \theta _{p} \theta _{r} -2h_{1}^{3} h_{3} h_{4} \theta _{p r} -2h_{1}^{3} h_{3} h_{5} \theta _{q r} +h_{1}^{4} h_{2} h_{5}^{4}\theta _{q q q q} -4h_{1}^{3} h_{2} h_{4}^{3} \theta _{p p p r} \\{} & {} +6 h_{1}^{4} h_{2} h_{4}^{2} h_{5}^{2}\theta _{p p q q}+4h_{1}^{4} h_{2} h_{4}^{3} h_{5} \theta _{p p p q} +12 h_{1}^{2} h_{2} h_{4} h_{5} \theta _{p q r r} \\{} & {} -12 h_{1}^{3} h_{2} h_{4} h_{5}^{2}\theta _{p q q r} +4 h_{1}^{4} h_{2} h_{4} h_{5}^{3}\theta _{p q q q} -12h_{1}^{3} h_{2} h_{4}^{2} h_{5} \theta _{p p q r} +2 h_{1}^{4} h_{3} h_{4} h_{5} \theta _{p q}\\{} & {} +2h_{1}^{5} h_{4} h_{5} \theta _{p} \theta _{q} +4 h_{1}^{5} h_{4} h_{5}\theta \theta _{p q} +h_{1}^{3}\theta _{r}^{2} -h_{1}^{3}\theta _{r r} +h_{2}\theta _{r r r r} \\{} & {} +h_{1}^{5} h_{5}^{2}\theta _{q}^{2} +h_{1}^{5} h_{4}^{2}\theta _{p}^{2} +h_{1}^{4} h_{4}\theta _{p r} +h_{1}^{4} h_{5}\theta _{q r} +h_{1}^{4}h_{5} \theta _{q q} + h_{1}^{4}h_{4} \theta _{p p}+h_{1}^{2} h_{3}\theta _{r r}\\{} & {} +2h_{1}^{3} \theta \theta _{r r} = 0. \end{aligned}$$

The solution to the above is

$$\begin{aligned} \theta (p,q,r)= & {} \tanh \biggl (C_{2} p -\frac{C_{2} q \sqrt{-h_{4} h_{5}}}{h_{5}}+r\bigg (C_{2}h_{1}h_{4}-C_{2} h_{1}\sqrt{-h_{4}h_{5}}\bigg )+C_{1}\biggr )C_6 \nonumber \\ {}{} & {} +C_5, \end{aligned}$$
(2.25)

where \(C_1\), \(C_2\), \(C_5\) and \(C_6\) are parameters. By reverting back to the original variables x, y, z, t the invariant solution of (1.1) is

$$\begin{aligned}{} & {} u(t,x,y,z)=\tanh \biggl (C_{2} (h_{4} x+y) -\frac{C_{2} (h_{5} x+z) \sqrt{-h_{4} h_{5}}}{h_{5}}\nonumber \\{} & {} +\frac{(h_{1} t-x)(C_{2} h_{1}h_{4}-C_{2}h_{1}\sqrt{-h_{4}h_{5}})}{h_{1}}+C_{1}\biggr )C_6+C_5. \end{aligned}$$
(2.26)
Fig. 1
figure 1

Kink waves profile (2.26)

Full size image

Kink waves [24] are asymptotic waves that rise or fall from one state to the next. At infinity, the kink solution approaches a constant.

Case 2.

In the case of \( {{\varvec{\Gamma }} _1}\), the invariants are \(p=t, \quad q=x, \quad r=\frac{h_{4} z^{2}+h_{5} y^{2}}{h_{5}}\) and the group-invariant solution consolidates to

$$\begin{aligned} u=\theta (p,q,r) \end{aligned}$$
(2.27)

where the function \(\theta (p,q,r)\) satisfies the nonlinear partial differential equation

$$\begin{aligned} 4 h_{4} r\theta _{{rr}}+ h_{1}{\theta _{{q}}}^{2}+2 h_{1}\theta \theta _{{qq}}+h_{3} \theta _{{qq}}+h_{2} \theta _{{qqqq}}+4 h_{4} \theta _{{r}} + \theta _{{pq}}=0. \end{aligned}$$

Case 3.

When we take \(a {{\varvec{\Gamma }}_3} + b {{\varvec{\Gamma }}_4}\) into account, we get the invariants \({\displaystyle p=t, {\displaystyle q=y}, {\displaystyle r=z}}\) and the group-invariant solution of the type

$$\begin{aligned} u(t,x,y,z)={\frac{a h_{1} t^{2} \theta +2b h_{1}t \theta -atx-bx}{h_{1} t(at+2b)}}, \end{aligned}$$
(2.28)

where the function \(\theta \) satisfies the nonlinear partial differential equation

$$\begin{aligned}{} & {} a^2h_{1}h_{4}p^4\theta _{qq}+a^2h_{1}h_{5}p^4\theta _{rr}+4abh_{1}h_{4}p^3\theta _{qq}\\{} & {} +4 abh_{1}h_{5}p^3\theta _{rr}+4b^2h_{1}h_{4}p^{2}\theta _{qq}+4b^2h_{1}h_{5}p^2\theta _{rr}-b^2=0. \end{aligned}$$

Case 4.

By solving the corresponding Lagrange system for the symmetry \( {{\varvec{\Gamma }}_{13}}\), one obtains invariants \({\displaystyle p=t,q=y, r=4h_{5}tx +z^2}\) and the group-invariant solution of the form

$$\begin{aligned} u=\theta \end{aligned}$$
(2.29)

is obtained by solving the associated Lagrange system for the symmetry \( {{\varvec{\Gamma }}_{13}}\). Here the function \(\theta \) satisfies

$$\begin{aligned} 256 h_{2} {h_{5}}^4 p^4\theta _{{rrrr}}+32 h_{1}{h_{5}}^2 p^2 \theta \theta _{{rr}}+16 h_{1} {h_{5}}^2 p^2 {\theta _{{r}}}^2 +16 h_{3} {h_{5}}^2 p^2\theta _{{rr}}+4 h_{5}p \theta _{{pr}}\nonumber \\+4h_{5} r\theta _{{rr}}+h_{4}\theta _{{qq}}+6h_{5}\theta _{{r}}=0. \end{aligned}$$

Case 5.

\({\varvec{\Upsilon }}_2={\varvec{\Gamma }}_{11}+ {\varvec{\Gamma }} _{13}\) prompts the following:

$$\begin{aligned}{} & {} p=t,\quad r=4h_{5}tx+z^{2}, \quad s= \frac{1}{2}\frac{2h_{5}ty+h_{4}z}{h_{5}t}\\{} & {} 256 h_{5}^{4} p^4 \theta _{{rrrr}}+\frac{1}{4} \frac{128 h_{1} h_{5}^3 p^4}{h_{2} h_{5} p^2} \theta \theta _{{rr}}-64 h_{1} {h_{5}}^3 p^4 {\theta _{{r}}}^2 -64 h_{3} {h_{5}}^3 p^4\theta _{{rr}}-64 h_{5}^3 p^3 x \theta _{{rr}}\\{} & {} -16 h_{5}^2 p^2 z^2 \theta _{{rr}}-16h_{5}^2 p^3 \theta _{{rp}}-24 h_{5}^2 p^2\theta _{{r}}-4h_{4}h_{5}p^2\theta _{{ss}}-h_{4}^2 \theta _{{ss}}=0,\\{} & {} u(t,x,y,z)=\theta (t,4h_{5}tx+z^{2},\frac{1}{2}\frac{2h_{5}ty+h_{4}z}{h_{5}t})=\theta (p,q,r,s). \end{aligned}$$

Case 6.

\({\varvec{\Upsilon }}_3={\varvec{\Gamma }}_{7}+ {\varvec{\Gamma }} _{13}\) generates the following:

$$\begin{aligned}{} & {} p=y,\quad r= h_{1}h_{5} t^{2},\quad s=\frac{1}{3} h_{1}^2 h_{5} t^{3}-(h_{1}h_{5}t^2+z)h_{1}t+x,\\{} & {} \theta _{{ssss}}+\frac{2 h_{1} \theta }{h_{2}}\theta _{{ss}}+ h_{1} r \theta _{{ss}} - h_{1} {\theta _{s}}^2- h_{3} \theta _{{ss}}\\{} & {} - h_{4} \theta _{{pp}}-h_{5} \theta _{{rr}}=0, u=\theta (p,q,r). \end{aligned}$$

Case 7.

The symmetry \({\varvec{\Upsilon }}_4={\varvec{\Gamma }}_{7}+ {\varvec{\Gamma }} _{11}+ {\varvec{\Gamma }} _{13}\) leads to the following:

$$\begin{aligned}{} & {} p=h_{1}h_{4}t+y,\quad r=h_{1}h_{5}t^2+z, \quad s=\frac{1}{3}h_{1}^2h_{5}t^3-(h_{1}h_{5}t^2+z)h_{1}t+x, \quad \theta =u,\\{} & {} \theta _{{ssss}}+\frac{2 h_{1} \theta }{h_{2}}\theta _{{ss}}+ h_{1} r \theta _{{ss}} - h_{1} {\theta _{s}}^2- h_{1} h_{4}\theta _{{sp}}\\{} & {} - h_{3} \theta _{{ss}}-h_{4} \theta _{{pp}}-h_{5}\theta _{{rr}}=0, \end{aligned}$$
$$\begin{aligned} u(t,x,y,z)=\theta (h_{1}h_{4}t+y,h_{1}h_{5}t^2+z,\frac{1}{3}h_{1}^2h_{5}t^3-(h_{1}h_{5}t^2+z)h_{1}t+x). \end{aligned}$$
(2.30)

In many applications, group invariant solutions capture the limiting behaviour of problems that far away from their initial or boundary conditions.

Travelling Wave Solutions

Travelling wave soultions are computed in this section courtesy of the simplest equation method [52, 63,64,65,66,67,68]. By employing the transformation \(u(x,y,z,t)=F(p), p=k_{1}x+k_{2}y+k_{3}z+k_{4}t+k_{5}\) one would obtain the following nonlinear ordinary differential equation

$$\begin{aligned}{} & {} k_{1}k_{4} F^{''}(p) +2 h_{1}k_{1}^2 F(p)F^{''}(p)+h_{1}k_{1}^2 (F^{'}(p))^2 \nonumber \\{} & {} +h_{2}k_{1}^{4} F^{''''}(p)+h_{3}k_{1}^{2} F^{''}(p)+h_{4}k_{2}^{2} F^{''}(p)+ h_{5}k_{3}^2 F^{''}(p)=0. \end{aligned}$$
(3.1)

Assume that the solution of (3.1) can be stated as

$$\begin{aligned} F(p)=A_0+A_1 H(p)+A_2 H(p)^2. \end{aligned}$$
(3.2)

Furthermore we assume that H(p) is solution of the ordinary differential equation below.

$$\begin{aligned} H'(p)=aH^2(p)+bH(p)+c. \end{aligned}$$
(3.3)

The solutions to the above equation are written as follows

$$\begin{aligned}{} & {} H(p)=-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \bigg [\frac{1}{2} \theta (p+C)\bigg ], \end{aligned}$$
(3.4)
$$\begin{aligned}{} & {} H(p)=-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \bigg (\frac{1}{2} \theta p\bigg )+ \frac{\text {sech}\left( \frac{\theta p}{2}\right) }{C \cosh \left( \frac{\theta p}{2}\right) -\frac{2 a}{\theta } \sinh \left( \frac{\theta p}{2}\right) }{ },\end{aligned}$$
(3.5)
$$\begin{aligned}{} & {} \theta =b^2-4ac. \end{aligned}$$
(3.6)

After some mechanical calculations one ends up with following travelling wave solutions:

$$\begin{aligned}{} & {} a=\frac{b^{2}}{4 c},\\{} & {} \mathcal {A}_0 = -\frac{-8 c^{2} A_{2} h_{1} k_{1}^{2}+b^{2} h_{3} k_{1}^{2}+b^{2} h_{4} k_{2}^{2}+b^{2} h_{5} k_{3}^{2}+b^{2} k_{1} k_{4}}{2 b^{2} h_{1} k_{1}^{2}},\\{} & {} \mathcal {A}_1=\frac{4 A_{2} c}{b},\\{} & {} {h}_2=-\frac{32 A_{2} h_{1} c^{2}}{15 b^{4} k_{1}^{2}},\\ \end{aligned}$$
$$\begin{aligned}{} & {} u(t,x,y,z)=\mathcal {A}_{0}+\mathcal {A}_{1}\bigg \{-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \bigg [\frac{1}{2} \theta (p+C)\bigg ]\bigg \}\nonumber \\ {}{} & {} \qquad \qquad +\mathcal {A}_{2}\bigg \{-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \bigg [\frac{1}{2} \theta (p+C)\bigg ]\bigg \}^2, \end{aligned}$$
(3.7)
$$\begin{aligned}{} & {} u(t,x,y,z)=\mathcal {A}_0+\mathcal {A}_1 \bigg \{-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \left( \frac{\theta p}{2}\right) + \frac{\text {sech}\left( \frac{\theta p}{2}\right) }{C \cosh \left( \frac{\theta p}{2}\right) -\frac{2 a}{\theta } \sinh \left( \frac{\theta p}{2}\right) }{ }\bigg \}\nonumber \\ {}{} & {} \qquad +\mathcal {A}_2\bigg \{-\frac{b}{2 a}-\frac{\theta }{2a}\tanh \bigg (\frac{1}{2} \theta p\bigg )+ \frac{\text {sech}\left( \frac{\theta p}{2}\right) }{C \cosh \left( \frac{\theta p}{2}\right) -\frac{2 a}{\theta } \sinh \left( \frac{\theta p}{2}\right) }{ }\bigg \}^2. \end{aligned}$$
(3.8)
Fig. 2
figure 2

Evolution of kink waves (3.7)

Full size image

Conservation Laws of (1.1)

This section examines the local conserved currents of a generalised (3+1)-dimensional nonlinear wave equation (1.1). Consider a partial differential equation \(E=0\) that has four independent variables (t, x, y, z) and one dependent field variable u. Given \(T=(T^t, T^x, T^y,T^z)\) such that the divergence \(\partial _t T^t+\partial _x T^x+\partial _y T^y+\partial _z T^z=0\), \((T^t, T^x, T^y,T^z)\) are known as conserved currents, and the divergence \(\partial _t T^t+\partial _x T^x+\partial _y T^y+\partial _z T^z=0\) is known as local conservation law. It is important to note that \((T^t, T^x, T^y, T^z)\) are functions of the field variable, derivatives of the field variable, space variables, and temporal variables.

We now recall that there exists a function \(\Lambda \) termed the multiplier such that \(\Lambda E=\partial _t T^t+\partial _x T^x+\partial _y T^y+\partial _z T^z\) if the divergence \(\partial _t T^t+\partial _x T^x+\partial _y T^y+\partial _z T^z\) holds for all solutions of \(E=0\). \(\Lambda \) is function of (t, x, y, z, u) and the derivatives of the field variable. Here \(\varepsilon _u\) is used as the Euler-Lagrange operator in \(\varepsilon _u( \Lambda E)=0\), the function \(\Lambda \) is computed. The lemma below may now be stated without losing generality.

Lemma 1

Let \(\Lambda \) be a multiplier, \(u_{tx}+2 h_{1}u_{xx}u+h_{1}\left( u_{x}^2\right) + h_{2} u_{xxxx}+h_{3}u_{xx}+h_{4}u_{yy}+h_{5}u_{zz}=0\), has a multiplier of the structure

$$\begin{aligned} \Lambda= & {} C_1u_{yz}-\frac{C_{1}h_{5}yu_{z}}{h_{4}}+C_2u_{y}+C_{4}u_{t}+C_3u_{z}, \end{aligned}$$

where \(C_i\), \(i=1\cdots 4\) are arbitrary constants of integration.

Proof

A straightforward but lengthy computation from \(\varepsilon _u( \Lambda E)=0\). \(\Box \)

Thus, corresponding to the above multiplier we have the following conservation laws of (1.1):

$$\begin{aligned} D_{t}T_{1}^t+D_{x}T_{1}^x+D_{y}T_{1}^y+D_{z}T_{1}^z=0, \end{aligned}$$

where

$$\begin{aligned}{} & {} T_{1}^t=\frac{1}{2} \frac{u_{x}\left( h_{4}zu_{y}-h_{5}yu_{z}\right) }{h_{4}},\nonumber \\{} & {} \quad T_{1}^x=-\frac{1}{6} \frac{1}{h_{4}}\bigg (4h_{1}h_{4}zu^{2}u_{xy}-4h_{1}h_{5}yu^{2}u_{xz}-4h_{1}h_{4}zuu_{x}u_{y}+4h_{1}h_{5}yuu_{x}u_{z}\nonumber \\{} & {} \quad +3h_{2}h_{4}zuu_{xxxy}-3h_{2}h_{5}yuu_{xxxz}+3h_{3}h_{4}zuu_{xy}-3h_{3}h_{5}yuu_{xz}-3h_{2}h_{4}zu_{x}u_{xxy}\nonumber \\{} & {} \quad +h_{2}h_{5}yu_{x}u_{xxz}-3h_{3}h_{4}zu_{x}u_{y}+3h_{3}h_{5}yu_{x}u_{z}\nonumber \\{} & {} \quad +3h_{2}h_{4}zu_{xx}u_{xy}-3h_{2}h_{5}yu_{xx}u_{xz}-3h_{2}h_{4}zu_{y}u_{xxx}+3h_{2}h_{5}yu_{xxx}u_{z}+3h_{4}zuu_{ty}-3h_{5}yuu_{tz}\bigg ),\nonumber \\{} & {} \quad T_{1}^y=\frac{2}{3}h_{1}zu^{2}u_{xx}+\frac{1}{3}h_{1}zu{u_x}^{2}\nonumber \\{} & {} \quad +\frac{1}{2}h_{5}yuu_{yz}+\frac{1}{2} h_{5}uu_{z}+\frac{1}{2} h_{4}z{u_{y}}^{2}\nonumber \\{} & {} \quad -\frac{1}{2}h_{5}yu_{y}u_{z}+\frac{1}{2}h_{3}zuu_{xx}+\frac{1}{2}h_{2}zuu_{xxxx}+\frac{1}{2}h_{5}zuu_{zz}+\frac{1}{2}zuu_{tx},\nonumber \\{} & {} \quad T_{1}^z=-\frac{1}{6}\frac{1}{h_{4}}\biggl (h_{5}\bigg (4h_{1}yu^{2}u_{xx}+2h_{1}yu{u_{x}}^{2} \nonumber \\{} & {} \quad +3h_{3}yuu_{xx}+3h_{2}yuu_{xxxx}+3h_{4}yu u_{yy}+3h_{4}zuu_{xy}\nonumber \\{} & {} \quad -3h_{4}z u_{y}u_{z}+3h_{5}y{u_{z}}^{2} +3yuu_{tx} +3 h_{4}u u_{y}\biggr )\biggr ); \end{aligned}$$
(4.1)
$$\begin{aligned} D_{t}T_{2}^t+D_{x}T_{2}^x+D_{y}T_{2}^y+D_{z}T_{2}^z=0, \end{aligned}$$

where

$$\begin{aligned}{} & {} T_{2}^t=\frac{1}{2} u_{x} u_{y},\nonumber \\{} & {} \quad T_{2}^x=\frac{2}{3} h_{1}u u_{x}u_{y} -\frac{2}{3}h_{1} u^{2} u_{xy} -\frac{1}{2} u u_{ty}-\frac{1}{2}h_{3} u u_{xy}\nonumber \\{} & {} \quad -\frac{1}{2} h_{2} u u_{xxxy}+\frac{1}{2}h_{3} u_{x} u_{y}+ \frac{1}{2} h_{2} u_{x} u_{xxy} -\frac{1}{2} h_{2} u_{xx} u_{xy}+\frac{1}{2} h_{2}u_{y} u_{xxx},\nonumber \\{} & {} \quad T_{2}^y=\frac{2}{3} h_{1} u^{2} u_{xx} + \frac{1}{3} h_{1} u {u_{x}}^{2} +\frac{1}{2} h_{4} {u_{y}}^{2}\nonumber \\{} & {} \quad +\frac{1}{2} h_{3} u u_{xx}+\frac{1}{2} h_{2} u u_{xxxx}+\frac{1}{2} h_{5} u u_{zz}+\frac{1}{2} u u_{tx},\nonumber \\{} & {} \quad T_{2}^z=-\frac{1}{2} h_{5} u u_{yz} +\frac{1}{2} h_{5} u_{y} u_{z}; \end{aligned}$$
(4.2)
$$\begin{aligned} D_{t}T_{3}^t+D_{x}T_{3}^x+D_{y}T_{3}^y+D_{z}T_{3}^z=0, \end{aligned}$$

where

$$\begin{aligned}{} & {} T_{3}^t=\frac{1}{2} u_{x}u_{z},\nonumber \\{} & {} \quad T_{3}^x = \frac{2}{3} h_{1} u u_{x} u_{z}-\frac{2}{3} h_{1} u^{2} u_{xz} -\frac{1}{2} u u_{tz}-\frac{1}{2} h_{3} u u_{xz}\nonumber \\{} & {} \quad -\frac{1}{2} h_{2} u u_{xxxz}+\frac{1}{2} h_{3} u_{x}u_{z}+\frac{1}{2} h_{2} u_{x} u_{xxz}-\frac{1}{2} h_{2} u_{xx} u_{xz}+\frac{1}{2} h_{2} u_{z} u_{xxx},\nonumber \\{} & {} \quad T_{3}^y=-\frac{1}{2} h_{4} u u_{yz}+\frac{1}{2} h_{4} u_{y} u_{z},\nonumber \\{} & {} \quad T_{3}^z=\frac{2}{3} h_{1} u^{2} u_{xx}+\frac{1}{3} h_{1} u {u_{x}}^{2}+\frac{1}{2} h_{5} {u_{z}}^{2}\nonumber \\{} & {} \quad +\frac{1}{2} h_{3} u u_{xx}+\frac{1}{2} h_{2} u u_{xxxx}+\frac{1}{2} h_{4} u u_{yy}+\frac{1}{2} u u_{tx}; \end{aligned}$$
(4.3)
$$\begin{aligned} D_{t}T_{4}^t+D_{x}T_{4}^x+D_{y}T_{4}^y+D_{z}T_{4}^z=0, \end{aligned}$$

where

$$\begin{aligned}{} & {} T_{4}^t=\frac{2}{3} h_{1} u^{2} u_{xx}+\frac{1}{3} h_{1} u {u_{x}}^2+\frac{1}{2} u_{t} u_{x}\nonumber \\{} & {} \quad +\frac{1}{2} h_{3} u u_{xx}+\frac{1}{2} h_{2} u u_{xxxx}+\frac{1}{2} h_{4} u u_{yy} + \frac{1}{2} h_{5} u u_{zz}+\frac{1}{2} u u_{tx},\nonumber \\{} & {} \quad T_{4}^x=\frac{2}{3} h_{1} u u_{t} u_{x} -\frac{2}{3} h_{1} u^{2} u_{tx}-\frac{1}{2} u u_{tt}\nonumber \\{} & {} \quad -\frac{1}{2} h_{3} u u_{tx} -\frac{1}{2} h_{2} u u_{txxx}+\frac{1}{2} h_{3} u_{t} u_{x}+\frac{1}{2} h_{2} u_{x} u_{txx}-\frac{1}{2} h_{2} u_{xx} u_{tx}+\frac{1}{2} h_{2} u_{t} u_{xxx},\nonumber \\{} & {} \quad T_{4}^y= -\frac{1}{2} h_{4} u u_{ty}+\frac{1}{2} h_{4} u_{t} u_{y},\nonumber \\{} & {} \quad T_{4}^z= -\frac{1}{2} h_{5} u u_{tz}+\frac{1}{2} h_{5} u_{t} u_{z}; \end{aligned}$$
(4.4)

Mathematical representations of physical laws including the conservation of energy, mass, and momentum are known as conservation laws. The reduction and solution of partial differential equations both heavily rely on conservation laws. Additionally, conservation laws have been used extensively in the construction of numerical integrators for partial differential equations as well as in the research of the existence, uniqueness, and stability of solutions to nonlinear partial differential equations.

Concluding Remarks

A generalised (3+1)-dimensional nonlinear wave equation that simulates a number of nonlinear processes that occur in liquids with gas bubbles was studied. After some cosmetic adjustments to the underlying equation, it was discovered that this generalised (3+1)-dimensional nonlinear wave equation naturally degenerated to the (3+1)-dimensional Kadomtsev-Petviashvili equation, the (3+1)-dimensional nonlinear wave equation, and the Korteweg-de Vries equation. To completely investigate this fundamental equation, a clear and methodical technique was used. The Lie symmetry method, multiplier method and simplest equation method led to novel symmetry reductions, group invariant solutions, conservation laws. Our equation is a modified general equation and its solutions are new and they have never been reported in the literature. The exact solutions achieved here can be used as benchmarks for numerical simulations, and we plan to do numerical simulations of the underlying equation in the future, and these results will be presented elsewhere.