Noether and partial Noether approach for the nonlinear (3+1)-dimensional elastic wave equations

Abstract

The Lie group method is a powerful technique for obtaining analytical solutions for various nonlinear differential equations. This study aimed to explore the behavior of nonlinear elastic wave equations and their underlying physical properties using Lie group invariants. We derived eight-dimensional symmetry algebra for the (3+1)-dimensional nonlinear elastic wave equation, which was used to obtain the optimal system. Group-invariant solutions were obtained using this optimal system. The same analysis was conducted for the damped version of this equation. For the conservation laws, we applied Noether’s theorem to the nonlinear elastic wave equations owing to the availability of a classical Lagrangian. However, for the damped version, we cannot obtain a classical Lagrangian, which makes Noether’s theorem inapplicable. Instead, we used an extended approach based on the concept of a partial Lagrangian to uncover conservation laws. Both techniques account for the conservation laws of linear momentum and energy within the model. These novel approaches add an application of variational calculus to the existing literature. This offers valuable insights and potential avenues for further exploration of the elastic wave equations.

1 Introduction

Elasticity [1] is a fundamental concept in many areas of science and engineering such as solid mechanics, material science, and structural engineering. Common examples of elastic media include rubber, metals, springs, and many other materials that can undergo temporary deformation and restore their original shapes when subjected to mechanical loads. In an elastic medium, when stress (force per unit area) is applied, the material undergoes reversible and recoverable deformation, which means that it can return to its initial state after the force is removed. In the case of small deformations, the relationship between the stress and strain (rate of deformation) is linear. However, when the deformations are not small, for example, in the case of hyper-or hypoelasticity, this relationship does not hold.

Elastic wave equations have attracted the interest of numerous researchers owing to their significance in various fields. The Lie group methods have been successfully applied to the study of a variety of elastic wave equations. For instance, Bokhari et al. [2] discussed the Lie group’s approach to the elastic wave equation. Mustafa and Masood [3] adopted the same tool over the third-order elastic wave equation with harmonic correction. Usman et al. [4, 5] conducted a Lie group analysis of the (1+1) and (2+1) elastic wave equations, adding valuable insights to this domain. In addition, many other authors [6, 7] have also been drawn to explore this area, contributing to the growing body of knowledge in the study of elasticity equations. The application of Lie group methods [8–11] has proven to be a useful approach for understanding the symmetries and properties of these equations, making it an essential tool for researchers in this field.

The linear theory of elasticity posits that the strain tensor ${\varkappa }_{ij}$ is linearly dependent on the displacement vector U_i according to the equation

$$\begin{array}{c}\hfill {\varkappa }_{ij}=\frac{1}{2}(U_{i,j}+U_{j,i})\end{array}$$ (1)

The non-linear elastic model, in contrast, is founded on the concept of non-linear strain, which is represented by

$$\begin{array}{c}\hfill {\varkappa }_{ij}=\frac{1}{2}(U_{i,j}+U_{j,i}+U_{k,i}{U}_{k,j}).\end{array}$$ (2)

In the case of a nonlinear strain, the stress is given by the strain rate of a potential function. Among the choices of such a potential functions, the Murnaghan potential appears suitable [1]. This potential is denoted by $\mathcal{W}\left({\varkappa }_{ik}\right)$, is expressed as follows

$$\begin{array}{c}\hfill \mathcal{W}\left({\varkappa }_{ik}\right)=\frac{1}{2}\mathrm{\lambda }{\left({\varkappa }_{mm}\right)}^2+\mu {\left({\varkappa }_{ik}\right)}^2+\frac{1}{3}{a}_1{\varkappa }_{ik}{\varkappa }_{im}{\varkappa }_{km}+a_2{\left({\varkappa }_{ik}\right)}^2{\varkappa }_{mm}+\frac{1}{3}{a}_1{\left({\varkappa }_{mm}\right)}^3,\end{array}$$ (3)

where λ and μ represent Lame’s coefficients and a₁, a₂ are constants. The stress tensor, denoted by τ_nm, can be calculated using the partial derivatives of the Murnaghan potential for the displacement components

$$\begin{array}{c}\hfill {\tau }_{nm}=\frac{\partial \mathcal{W}}{\partial U_{m,n}}·\end{array}$$ (4)

The Cauchy equation of motion, in the absence of body forces, is given by

$$\begin{array}{c}\hfill {\tau }_{ij,i}=\rho {\ddot{U}}_j.\end{array}$$ (5)

In this context, U_j symbolizes the displacement vector, τ_ij stands for the stress tensor, and ρ represents the density of the body.

These equations play a fundamental role in describing the mechanical behavior of elastic materials under the influence of stress and displacement dynamics. The Murnaghan potential represents the elastic energy associated with strain components, while the stress tensor and Cauchy equation of motion help understand the relationship between the stress, displacement, and acceleration of particles within the elastic body.

Considering three-dimensional motion in (x₁, x₂, x₃) and U₁ = U, the strain tensor components ${\varkappa }_{ij}$ for this case are given by

$$\begin{array}{c}\hfill {\varkappa }_{11}=U_{x_1}+\frac{1}{2}{U}_{x_1}^2,{\varkappa }_{22}=\frac{1}{2}{U}_{x_2}^2,{\varkappa }_{33}=\frac{1}{2}{U}_{x_3}^2,\end{array}$$ (6)

$$\begin{array}{c}\hfill {\varkappa }_{12}={\varkappa }_{21}=\frac{1}{2}(U_{x_2}+U_{x_1}{U}_{x_2}),{\varkappa }_{13}={\varkappa }_{31}=\frac{1}{2}(U_{x_3}+U_{x_1}{U}_{x_3}),{\varkappa }_{23}={\varkappa }_{32}=\frac{1}{2}{U}_{x_2}{U}_{x_3}.\end{array}$$ (7)

For the three-dimensional case, the Murnaghan potential (3) takes the form

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathcal{W}=& \frac{1}{2}\mathrm{\lambda }{({\varkappa }_{11}+{\varkappa }_{22}+{\varkappa }_{33})}^2+\mu ({\varkappa }_{11}^2+{\varkappa }_{22}^2+{\varkappa }_{33}^2+{\varkappa }_{12}^2+{\varkappa }_{21}^2+{\varkappa }_{13}^2+{\varkappa }_{31}^2+{\varkappa }_{23}^2+{\varkappa }_{32}^2)\hfill \\ & +\frac{1}{3}{a}_1({\varkappa }_{11}^3+3{\varkappa }_{11}{\varkappa }_{12}^2+3{\varkappa }_{11}{\varkappa }_{13}^2+3{\varkappa }_{33}{\varkappa }_{13}^2+{\varkappa }_{33}^3+3{\varkappa }_{33}{\varkappa }_{32}^2+6{\varkappa }_{12}{\varkappa }_{13}{\varkappa }_{32}+{\varkappa }_{22}^3\hfill \\ & +{\varkappa }_{22}{\varkappa }_{12}^2+3{\varkappa }_{22}{\varkappa }_{32}^2)+a_2({\varkappa }_{11}^2+{\varkappa }_{22}^2+{\varkappa }_{33}^2+{\varkappa }_{12}^2+{\varkappa }_{21}^2+{\varkappa }_{13}^2+{\varkappa }_{31}^2+{\varkappa }_{23}^2+\hfill \\ & {\varkappa }_{32}^2)({\varkappa }_{11}+{\varkappa }_{22}+{\varkappa }_{33})+\frac{1}{3}{a}_3{({\varkappa }_{11}+{\varkappa }_{22}+{\varkappa }_{33})}^3.\hfill \end{array}\end{array}$$ (8)

This leads to the following expression for $\mathcal{W}$

$$\begin{array}{c}\hfill \mathcal{W}=(\frac{\mathrm{\lambda }}{2}+\mu )U_{x_1}^2+(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{3}+a_1+\frac{a_3}{3})U_{x_1}^3+\frac{\mu }{2}(U_{x_2}^2+U_{x_3}^2)+(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{4}+\frac{a_2}{2})(U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2).\end{array}$$ (9)

The stress components can be calculated using the Eq (4) and are given by

$$\begin{array}{c}\begin{array}{c}{\tau }_{11}=(\mathrm{\lambda }+2\mu )U_{x_1}+3(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{3}+a_1+\frac{a_3}{3})U_{x_1}^2+(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{4}+\frac{a_2}{2})(U_{x_2}^2+U_{x_3}^2),\hfill \\ {\tau }_{21}=\mu U_{x_2}+(\mathrm{\lambda }+2\mu +\frac{a_1}{2}+a_1)U_{x_1}{U}_{x_2},\hfill \\ {\tau }_{31}=\mu U_{x_3}+(\mathrm{\lambda }+2\mu +\frac{a_1}{2}+a_1)U_{x_1}{U}_{x_3}.\hfill \end{array}\end{array}$$ (10)

The Cauchy equation of motion (5) takes the following form in the three-dimensional case

$$\begin{array}{c}\hfill {\tau }_{11,1}+{\tau }_{21,2}+{\tau }_{31,3}=\rho U_{\tau \tau },\end{array}$$ (11)

and the components of the potential are given by

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\tau }_{11,1}& =(\mathrm{\lambda }+2\mu )U_{x_1{x}_1}+6(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{3}+a_1+\frac{a_1}{3})U_{x_1}{U}_{x_1{x}_1}+2(\frac{\mathrm{\lambda }}{2}+\mu +\frac{a_1}{4}\hfill \\ & +\frac{a_1}{2})(U_{x_2}{U}_{x_1{x}_2}+U_{x_3}{U}_{x_1{x}_3}),\hfill \\ \hfill {\tau }_{21,2}& =\mu U_{x_2{x}_2}+(\mathrm{\lambda }+2\mu +\frac{a_1}{2}+a_1)(U_{x_2}{U}_{x_1{x}_2}+U_{x_1}{U}_{x_2{x}_2}),\hfill \\ \hfill {\tau }_{31,3}& =\mu U_{x_3{x}_3}+(\mathrm{\lambda }+2\mu +\frac{a_1}{2}+a_2)(U_{x_1}{U}_{x_3{x}_3}+U_{x_3}{U}_{x_1{x}_3}).\hfill \end{array}\end{array}$$ (12)

Lastly, by utilizing Eq (10), we derive the (3+1)-dimensional nonlinear elastic wave equation as follows

$$\begin{array}{c}\hfill \begin{array}{cc}& U_{\tau \tau }-A{U}_{x_1{x}_1}-B(U_{x_2{x}_2}+U_{x_3{x}_3})-C{U}_{x_1}{U}_{x_1{x}_1}-D(U_{x_1}{U}_{x_2{x}_2}+U_{x_1}{U}_{x_3{x}_3}\hfill \\ & +2U_{x_2}{U}_{x_1{x}_2}+2U_{x_3}{U}_{x_1{x}_3})=0.\hfill \end{array}\end{array}$$ (13)

In (13), $A=\frac{\mathrm{\lambda }+2\mu }{\rho }$, $B=\frac{\mu }{\rho }$, $C=\frac{3(\mathrm{\lambda }+2\mu )+2(a_1+3a_2+a_3)}{\rho }$, $D=\frac{1}{\rho }(\mathrm{\lambda }+2\mu +\frac{a_1}{2}+a_2)$.

In Eq (13), λ and μ are Lame’s coefficients and a₁, a₂, and a₃ are the Murnaghan constants. These expressions provide essential insights into the behavior of the (3+1)-nonlinear elastic wave equation and the underlying physical properties of the elastic medium under consideration. The aim of this study was to investigate the underlying properties of elastic media. The Lie group method [12–15] was used to study these properties.

This paper is organized into several sections. In Section 2, we conduct a group analysis and establish an optimal system for the (3+1)-nonlinear elastic wave equation. Section 3 explores group-invariant solutions using the identified optimal system. In Section 4, the authors introduce Noether’s approach for formulating conservation laws and present Noether’s symmetry generators. In Section 5, the focus is shifted to the Lie group and the optimal system of the damped version of the considered equation. This paper proceeds to Section 6, in which symmetry reductions are discussed. Section 7 discusses the partial Lagrangian approach employed to investigate conservation laws. A graphical analysis of the obtained results is presented in Section 8. Finally, in Section 9, we conclude the paper and highlight potential future research directions.

2 Group analysis of the Eq (13)

This section presents a comprehensive Lie symmetry analysis of (13). For this purpose, we consider the vector field given by [16]

$$\mathrm{\Lambda }={\delta }_1\frac{\partial }{\partial x_1}+{\delta }_2\frac{\partial }{\partial x_2}+{\delta }_3\frac{\partial }{\partial x_3}+{\delta }_4\frac{\partial }{\partial \tau }+\mathrm{\Psi }\frac{\partial }{\partial U}·$$

Since the equation under consideration is of second order, we prolong our vector field as follows

$$\begin{array}{c}{\mathrm{\Lambda }}^{\left[2\right]}={\delta }_1\frac{\partial }{\partial x_1}+{\delta }_2\frac{\partial }{\partial x_2}+{\delta }_3\frac{\partial }{\partial x_3}+{\delta }_4\frac{\partial }{\partial \tau }+\mathrm{\Psi }\frac{\partial }{\partial U}+{\mathrm{\Psi }}^{\left[x_1\right]}\frac{\partial }{\partial U_{x_1}}+{\mathrm{\Psi }}^{\left[x_2\right]}\frac{\partial }{\partial U_{x_2}}+{\mathrm{\Psi }}^{\left[x_3\right]}\frac{\partial }{\partial U_{x_3}}+{\mathrm{\Psi }}^{\left[\tau \right]}\frac{\partial }{\partial U_{\tau }}\\ +{\mathrm{\Psi }}^{\left[x_1{x}_1\right]}\frac{\partial }{\partial U_{x_1{x}_1}}+{\mathrm{\Psi }}^{\left[x_1{x}_2\right]}\frac{\partial }{\partial U_{x_1{x}_2}}+{\mathrm{\Psi }}^{\left[x_2{x}_2\right]}\frac{\partial }{\partial U_{x_2{x}_2}}+{\mathrm{\Psi }}^{\left[x_1{x}_3\right]}\frac{\partial }{\partial U_{x_1{x}_3}}+{\mathrm{\Psi }}^{\left[x_2{x}_3\right]}\frac{\partial }{\partial U_{x_2{x}_3}}+{\mathrm{\Psi }}^{\left[x_3{x}_3\right]}\frac{\partial }{\partial U_{x_3{x}_3}}\\ +{\mathrm{\Psi }}^{\left[x_1\tau \right]}\frac{\partial }{\partial U_{x_1\tau }}+{\mathrm{\Psi }}^{\left[x_2\tau \right]}\frac{\partial }{\partial U_{x_2\tau }}+{\mathrm{\Psi }}^{\left[x_3\tau \right]}\frac{\partial }{\partial U_{x_3\tau }}+{\mathrm{\Psi }}^{\left[\tau \tau \right]}\frac{\partial }{\partial U_{\tau \tau }},\end{array}$$

where, ${\mathrm{\Psi }}^{\left[x_1\right]},{\mathrm{\Psi }}^{\left[x_2\right]},{\mathrm{\Psi }}^{\left[x_3\right]},{\mathrm{\Psi }}^{\left[\tau \right]},{\mathrm{\Psi }}^{\left[x_1{x}_1\right]},{\mathrm{\Psi }}^{\left[x_1{x}_2\right]},{\mathrm{\Psi }}^{\left[x_2{x}_2\right]},{\mathrm{\Psi }}^{\left[x_1{x}_3\right]},{\mathrm{\Psi }}^{\left[x_2{x}_3\right]},{\mathrm{\Psi }}^{\left[x_3{x}_3\right]},{\mathrm{\Psi }}^{\left[x_2\tau \right]},{\mathrm{\Psi }}^{\left[x_2\tau \right]},{\mathrm{\Psi }}^{\left[x_3\tau \right]}$ and ${\mathrm{\Psi }}^{\left[\tau \tau \right]}$ are given as follows

$$\begin{array}{c}\begin{array}{c}{\mathrm{\Psi }}^{\left[x_1\right]}={\mathbf{D}}_{x_1}\mathrm{\Psi }-U_{x_1}{\mathbf{D}}_{x_1}{\delta }_1-U_{x_2}{\mathbf{D}}_{x_1}{\delta }_2-U_{x_3}{\mathbf{D}}_{x_1}{\delta }_3-U_{\tau }{\mathbf{D}}_{x_1}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_2\right]}={\mathbf{D}}_{x_2}\mathrm{\Psi }-U_{x_1}{\mathbf{D}}_{x_2}{\delta }_1-U_{x_2}{\mathbf{D}}_{x_2}{\delta }_2-U_{x_3}{\mathbf{D}}_{x_2}{\delta }_3-U_{\tau }{\mathbf{D}}_{x_2}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_3\right]}={\mathbf{D}}_{x_3}\mathrm{\Psi }-U_{x_1}{\mathbf{D}}_{x_3}{\delta }_1-U_{x_2}{\mathbf{D}}_{{\delta }_2}-U_{x_3}{\mathbf{D}}_{x_3}{\delta }_3-U_{\tau }{\mathbf{D}}_{x_3}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[\tau \right]}={\mathbf{D}}_{\tau }\mathrm{\Psi }-U_{x_1}{\mathbf{D}}_{\tau }{\delta }_1-U_{x_2}{\mathbf{D}}_{\tau }{\delta }_2-U_{x_3}{\mathbf{D}}_{\tau }{\delta }_3-U_{\tau }{\mathbf{D}}_{\tau }{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_1{x}_1\right]}={\mathbf{D}}_{x_1}{\mathrm{\Psi }}^{\left[x_1\right]}-U_{x_1{x}_1}{\mathbf{D}}_{x_1}{\delta }_1-U_{x_1{x}_2}{\mathbf{D}}_{x_1}{\delta }_2-U_{x_1{x}_3}{\mathbf{D}}_{x_1}{\delta }_3-U_{x_1\tau }{\mathbf{D}}_{x_1}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_2{x}_2\right]}={\mathbf{D}}_{x_2}{\mathrm{\Psi }}^{\left[x_2\right]}-U_{x_1{x}_2}{\mathbf{D}}_{x_2}{\delta }_1-U_{x_2{x}_2}{\mathbf{D}}_{x_2}{\delta }_2-U_{x_2{x}_3}{\mathbf{D}}_{x_2}{\delta }_3-U_{x_2\tau }{\mathbf{D}}_{x_2}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_3{x}_3\right]}={\mathbf{D}}_{x_3}{\mathrm{\Psi }}^{\left[x_3\right]}-U_{x_1{x}_3}{\mathbf{D}}_{x_3}{\delta }_1-U_{x_2{x}_3}{\mathbf{D}}_{x_3}{\delta }_2-U_{\tau \tau }{\mathbf{D}}_{x_3}{\delta }_3-U_{x_3\tau }{\mathbf{D}}_{x_3}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[\tau \tau \right]}={\mathbf{D}}_{\tau }{\mathrm{\Psi }}^{\left[\tau \right]}-U_{x_1\tau }{\mathbf{D}}_{\tau }{\delta }_1-U_{x_2\tau }{\mathbf{D}}_{\tau }{\delta }_2-U_{x_3\tau }{\mathbf{D}}_{\tau }{\delta }_3-U_{\tau \tau }{\mathbf{D}}_{\tau }{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_1{x}_2\right]}={\mathbf{D}}_{x_2}{\mathrm{\Psi }}^{\left[x_1\right]}-U_{x_1{x}_1}{\mathbf{D}}_{x_2}{\delta }_1-U_{x_1{x}_2}{\mathbf{D}}_{x_2}{\delta }_2-U_{x_1{x}_3}{\mathbf{D}}_{x_2}{\delta }_3-U_{x_1\tau }{\mathbf{D}}_{x_2}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_1\tau \right]}={\mathbf{D}}_{\tau }{\mathrm{\Psi }}^{\left[x_1\right]}-U_{x_1{x}_1}{\mathbf{D}}_{\tau }{\delta }_1-U_{x_1{x}_2}{\mathbf{D}}_{\tau }{\delta }_2-U_{x_1{x}_3}{\mathbf{D}}_{\tau }{\delta }_3-U_{x_1\tau }{\mathbf{D}}_{\tau }{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_2\tau \right]}={\mathbf{D}}_{\tau }{\mathrm{\Psi }}^{\left[x_2\right]}-U_{x_1{x}_2}{\mathbf{D}}_{\tau }{\delta }_1-U_{x_2{x}_2}{\mathbf{D}}_{\tau }{\delta }_2-U_{x_2{x}_3}{\mathbf{D}}_{\tau }{\delta }_3-U_{x_2\tau }{\mathbf{D}}_{\tau }{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_1{x}_3\right]}={\mathbf{D}}_{x_3}{\mathrm{\Psi }}^{\left[x_1\right]}-U_{x_1{x}_1}{\mathbf{D}}_{x_3}{\delta }_1-U_{x_1{x}_2}{\mathbf{D}}_{x_3}{\delta }_2-U_{x_1{x}_3}{\mathbf{D}}_{x_3}{\delta }_3-U_{x_1\tau }{\mathbf{D}}_{x_3}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_2{x}_3\right]}={\mathbf{D}}_{x_3}{\mathrm{\Psi }}^{\left[x_2\right]}-U_{x_1{x}_2}{\mathbf{D}}_{x_3}{\delta }_1-U_{x_2{x}_2}{\mathbf{D}}_{x_3}{\delta }_2-U_{x_2{x}_3}{\mathbf{D}}_{x_3}{\delta }_3-U_{x_2\tau }{\mathbf{D}}_{x_3}{\delta }_4\hfill \\ {\mathrm{\Psi }}^{\left[x_3\tau \right]}={\mathbf{D}}_{\tau }{\mathrm{\Psi }}^{\left[x_3\right]}-U_{x_1{x}_3}{\mathbf{D}}_{\tau }{\delta }_1-U_{x_2{x}_3}{\mathbf{D}}_{\tau }{\delta }_2-U_{\tau \tau }{\mathbf{D}}_{\tau }{\delta }_3-U_{n{u}_3\tau }{\mathbf{D}}_{\tau }{\delta }_4,\hfill \end{array}\hfill \end{array}$$ (14)

where D_α denotes the total derivative of α.

The invariance criterion of PDE is

$$\begin{array}{c}\hfill \begin{array}{cc}& {\mathrm{\Lambda }}^{\left[2\right]}(U_{\tau \tau }-A{U}_{x_1{x}_1}-B(U_{x_2{x}_2}+U_{x_3{x}_3})-C{U}_{x_1}{U}_{x_1{x}_1}-D(U_{x_1}{U}_{x_2{x}_2}+U_{x_1}{U}_{x_3{x}_3})\hfill \\ & -2D(U_{x_2}{U}_{x_1{x}_2}+U_{x_3}{U}_{x_1{x}_3}){\left)\right|}_{\left(13\right)}=0.\hfill \end{array}\end{array}$$ (15)

The invariance condition (15) which lead to the following after some calculations. After performing some calculations we get,

$$\begin{array}{c}\hfill \begin{array}{cc}& {\delta }_{1_{x_1}}={\delta }_{2_{x_2}}={\delta }_{3_{x_3}}={\delta }_{4_{\tau }}={\mathrm{\Psi }}_U,\hfill \\ & {\delta }_{1_{x_2}}={\delta }_{1_{x_3}}={\delta }_{1_{\tau }}={\delta }_{1_U}=0,\hfill \\ & {\delta }_{4_{x_1}}={\delta }_{4_{x_2}}={\delta }_{4_{x_3}}={\delta }_{4_U}=0,\hfill \\ & {\delta }_{2_{x_1}}={\delta }_{2_{\tau }}={\delta }_{2_U}=0,\hfill \\ & {\delta }_{3_{x_1}}={\delta }_{3_{\tau }}={\delta }_{3_U}=0,\hfill \\ & {\mathrm{\Psi }}_{x_1}={\mathrm{\Psi }}_{x_2}={\mathrm{\Psi }}_{x_3}=0,\hfill \\ & {\delta }_{4_{\tau \tau }}=0,\hfill \\ & {\delta }_{2_{x_3{x}_3}}=0,\hfill \\ & {\mathrm{\Psi }}_{\tau \tau }=0,\hfill \\ & {\delta }_{3_{x_2}}+{\delta }_{2_{x_3}}=0.\hfill \end{array}\end{array}$$ (16)

So the infinitesimals are

$$\begin{array}{c}\hfill \begin{array}{cc}& {\delta }_1=C_1{x}_1+C_2,\hfill \\ & {\delta }_2=C_1{x}_2+C_3{x}_3+C_4,\hfill \\ & {\delta }_3=C_1{x}_3-C_3{x}_2+C_5,\hfill \\ & {\delta }_4=C_1\tau +C_6,\hfill \\ & \mathrm{\Psi }=C_{1}U+C_7\tau +C_8.\hfill \end{array}\end{array}$$ (17)

We obtain the following symmetry algebra

$${\mathrm{\Lambda }}_1=\frac{\partial }{\partial \tau },{\mathrm{\Lambda }}_2=\frac{\partial }{\partial U},{\mathrm{\Lambda }}_3=\frac{\partial }{\partial x_1},{\mathrm{\Lambda }}_4=\frac{\partial }{\partial x_2},$$

$${\mathrm{\Lambda }}_5=\frac{\partial }{\partial x_3},{\mathrm{\Lambda }}_6=\tau \frac{\partial }{\partial U},{\mathrm{\Lambda }}_7=x_3\frac{\partial }{\partial x_2}-x_2\frac{\partial }{\partial x_3},$$

$${\mathrm{\Lambda }}_8=x_1\frac{\partial }{\partial x_1}+x_2\frac{\partial }{\partial x_2}+x_3\frac{\partial }{\partial x_3}+\tau \frac{\partial }{\partial \tau }+U\frac{\partial }{\partial U}·$$

Table 1 shows the commutator relation for the symmetry algebra (2). The adjoint representations of the symmetry algebra (15) are presented in Tables 2 and 3.

Table 1. Commutator table.

[Λi, Λj]	Λ1	Λ2	Λ3	Λ4	Λ5	Λ6	Λ7	Λ8
Λ1	0	0	0	0	0	Λ2	0	Λ1
Λ2	0	0	0	0	0	0	0	Λ2
Λ3	0	0	0	0	0	0	0	Λ3
Λ4	0	0	0	0	0	0	−Λ5	Λ4
Λ5	0	0	0	0	0	0	Λ4	Λ5
Λ6	−Λ2	0	0	0	0	0	0	0
Λ7	0	0	0	Λ5	−Λ4	0	0	0
Λ8	−Λ1	−Λ2	−Λ3	−Λ4	-Λ5	0	0	0

Table 2. Adjoint table.

Ad(eϵ)	Λ1	Λ2	Λ3	Λ4
Λ1	Λ1	Λ2	Λ3	Λ4
Λ2	Λ1	Λ2	Λ3	Λ4
Λ3	Λ1	Λ2	Λ3	Λ4
Λ4	Λ1	Λ2	Λ3	Λ4
Λ5	Λ1	Λ2	Λ3	Λ4
Λ6	Λ1 + ϵΛ2	Λ2	Λ3	Λ4
Λ7	Λ1	Λ2	Λ3	cos(ϵ)Λ4 − sin(ϵ)Λ5
Λ8	eϵΛ1	eϵΛ2	eϵΛ3	eϵΛ4

Table 3. Adjoint table.

Ad(eϵ)	Λ5	Λ6	Λ7	Λ8
Λ1	Λ5	Λ6 − ϵΛ2	Λ7	Λ8 − ϵΛ1
Λ2	Λ5	Λ6	Λ7	Λ8 − ϵΛ2
Λ3	Λ5	Λ6	Λ7	Λ8 − ϵΛ3
Λ4	Λ5	Λ6	Λ7 + ϵΛ5	Λ8 − ϵΛ4
Λ5	Λ5	Λ6	Λ7 − ϵΛ4	Λ8 − ϵΛ5
Λ6	Λ5	Λ6	Λ7	Λ8
Λ7	sin(ϵ)Λ4 + cos(ϵ)Λ5	Λ6	Λ7	Λ8
Λ8	eϵΛ5	Λ6	Λ7	Λ8

2.1 Optimal system

The optimal system of subalgebras is a concept in the context of Lie symmetry analysis. This was first introduced by Ibragimov [17] and then carried out by Olver [18]. When analyzing the symmetries of a differential equation using the Lie group method, we often encounter Lie algebra consisting of several infinitesimal generators. However, not all these generators are linearly independent. The optimal system of subalgebras is the maximal set of linearly independent generators that provide a complete description of the symmetries of the differential equation. It represents the minimal set of generators needed to construct all symmetries and invariant solutions of the equation. Here, we discuss the optimal system for (13):

Consider a general element Λ of Lie algebra L⁸ given by,

$$\begin{array}{c}\hfill \mathrm{\Lambda }={\mu }_1{\mathrm{\Lambda }}_1+{\mu }_2{\mathrm{\Lambda }}_2+{\mu }_3{\mathrm{\Lambda }}_3+{\mu }_4{\mathrm{\Lambda }}_4+{\mu }_5{\mathrm{\Lambda }}_5+{\mu }_6{\mathrm{\Lambda }}_6+{\mu }_7{\mathrm{\Lambda }}_7+{\mu }_8{\mathrm{\Lambda }}_8,\end{array}$$ (18)

where μ denotes arbitrary constants.

Case 1: μ₈ ≠ 0, μ₇ ≠ 0. By the adjoint action of Λ₁, Λ₃ and Λ₄, (18) simplifies to Λ = μ₆Λ₆ + μ₇Λ₇ + μ₈Λ₈. Therefore, the corresponding subalgebra is ${\mathcal{R}}_1={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_7+\sigma {\mathrm{\Lambda }}_8$ κ, σ ≠ 0.

Case 2: μ₈ ≠ 0, μ₇ = 0. By the adjoint action of Λ₁, Λ₃, Λ₄ and Λ₅, (18) simplifies to Λ = μ₆Λ₆ + μ₈Λ₈. Hence, the corresponding subalgebra is ${\mathcal{R}}_2={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_8$ κ ≠ 0.

Case 3: μ₈ ≠ 0, μ₇ ≠ 0, μ₆ = 0. By the adjoint action of Λ₁, Λ₂, Λ₃ and Λ₄, (18) simplifies to Λ = μ₇Λ₇ + μ₈Λ₈. Hence, the corresponding subalgebra is ${\mathcal{R}}_3={\mathrm{\Lambda }}_7+\kappa {\mathrm{\Lambda }}_8$ κ ≠ 0.

Case 4: μ₈ ≠ 0, μ₇ = 0, μ₆ = 0. By the adjoint action of Λ₁, Λ₂, Λ₃, Λ₄ and Λ₅, (18) simplifies to Λ = μ₈Λ₈. Therefore, the corresponding subalgebra is ${\mathcal{R}}_4={\mathrm{\Lambda }}_8$.

Case 5: μ₈ = 0, μ₇ ≠ 0, μ₃ ≠ 0. From the adjoint action of Λ₁, Λ₄, Λ₅ and Λ₈, Eq (18) simplifies to Λ = μ₁Λ₁ + μ₃Λ₃ + e^−ϵμ₆Λ₆ + e^−ϵμ₇Λ₇. Therefore, the corresponding subalgebra is ${\mathcal{R}}_5={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6\pm {\mathrm{\Lambda }}_7$ κ ≠ 0.

Case 6: μ₈ = 0, μ₇ ≠ 0, μ₆ = 0, μ₃ ≠ 0. By the adjoint action of Λ₄, Λ₅, Λ₆ and Λ₈, (18) simplifies to Λ = μ₁Λ₁ + μ₃Λ₃ + e^−ϵμ₇Λ₇. Therefore, the corresponding subalgebra is ${\mathcal{R}}_6={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7$, and κ ≠ 0.

Case 7: μ₈ = 0, μ₇ = 0, μ₆ ≠ 0, μ₃ ≠ 0. By the adjoint action of Λ₆, Λ₇ and Λ₈, (18) simplifies to Λ = μ₁Λ₁ + μ₃Λ₃ + e^−ϵμ₆Λ₆. Therefore, the corresponding subalgebra is ${\mathcal{R}}_7={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6$, and κ ≠ 0.

Case 8: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₃ ≠ 0. By the adjoint action of Λ₆ and Λ₇, (18) simplifies to Λ = μ₁Λ₁ + μ₃Λ₃. Therefore, the corresponding subalgebra is ${\mathcal{R}}_8={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3$, and κ ≠ 0.

Case 9: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₅ = 0, μ₄ ≠ 0, μ₃ ≠ 0. By the adjoint action of Λ₆, (18) simplifies to Λ = μ₁Λ₁ + μ₃Λ₃ + μ₄Λ₄. Thus, ${\mathcal{R}}_9={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4$ κ, σ ≠ 0.

Case 10: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₁ = 0, μ₅ = 0, μ₄ ≠ 0, μ₃ ≠ 0. Then, (18) is simplified to Λ = μ₂Λ₂ + μ₃Λ₃ + μ₄Λ₄. Thus, ${\mathcal{R}}_{10}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4$, κ, σ ≠ 0.

Case 11: μ₈ = 0, μ₆ = 0, μ₃ = 0. By the adjoint action of Λ₄, Λ₅, Λ₆ and Λ₈, (18) reduces to Λ = μ₁Λ₁ + e^−ϵμ₇Λ₇. Therefore, ${\mathcal{R}}_{11}={\mathrm{\Lambda }}_1\pm {\mathrm{\Lambda }}_7$.

Case 12: μ₈ = 0, μ₆ = 0, μ₁ = 0, μ₃ ≠ 0. By the adjoint action of Λ₄, Λ₅ and Λ₈ Eq (18) reduces to Λ = μ₂Λ₂ + μ₃Λ₃ + e^−ϵμ₇Λ₇. Therefore, ${\mathcal{R}}_{12}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7,\kappa \ne 0$.

Case 13: μ₈ = 0, μ₆ = 0, μ₁ = 0, μ₂ = 0, μ₃ ≠ 0. By the adjoint action of Λ₄, Λ₅ and Λ₈ (18), reduces to Λ = μ₃Λ₃ + e^−ϵμ₇Λ₇. Therefore, ${\mathcal{R}}_{13}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7$.

Case 14: μ₈ = 0, μ₆ = 0, μ₁ = 0, μ₂ ≠ 0, μ₃ = 0. By the adjoint action of Λ₄, Λ₅ and Λ₈, (18) reduces to Λ = μ₂Λ₂ + e^−ϵμ₇Λ₇. Therefore, ${\mathcal{R}}_{14}={\mathrm{\Lambda }}_2\pm {\mathrm{\Lambda }}_7$.

Case 15: μ₈ = 0, μ₁ = 0, μ₃ ≠ 0. By the adjoint action of Λ₁, Λ₄, Λ₅ and Λ₈ Eq (18) is reduced to Λ = μ₃Λ₃ + e^−ϵμ₆Λ₆ + e^−ϵμ₇Λ₇. Thus, ${\mathcal{R}}_{15}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6\pm {\mathrm{\Lambda }}_7$.

Case 16: μ₈ = 0, μ₇ = 0, μ₁ = 0, μ₃ ≠ 0. By the adjoint action of Λ₁, Λ₇ and Λ₈, (18) reduces to Λ = μ₃Λ₃ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{16}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6$.

Case 17: μ₈ = 0, μ₇ = 0, μ₄ = 0, μ₁ = 0, μ₅ ≠ 0. By the adjoint action of Λ₁ and Λ₈ Eq (18) reduces to Λ = μ₃Λ₃μ₅Λ₅ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{17}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5\pm {\mathrm{\Lambda }}_6,\kappa \ne 0$.

Case 18: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₁ = 0, μ₄ = 0, μ₂ = 0, μ₅ ≠ 0. Then, (18) is reduced to Λ = μ₃Λ₃ + μ₅Λ₅. Therefore, ${\mathcal{R}}_{18}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0$.

Case 19: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₁ = 0, μ₅ = 0, μ₂ = 0, μ₄ ≠ 0. Then, (18) is reduced to Λ = μ₃Λ₃ + μ₄Λ₄. Therefore, ${\mathcal{R}}_{19}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0$.

Case 20: μ₈ = 0, μ₇ = 0, μ₅ = 0, μ₁ = 0, μ₄ ≠ 0. By the adjoint action of Λ₁ and Λ₈ Eq (18) reduces to Λ = μ₃Λ₃ + μ₄Λ₄ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{20}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_4\pm {\mathrm{\Lambda }}_6,\kappa \ne 0$.

Case 21: μ₈ = 0, μ₇ = 0, μ₄ = 0, μ₃ = 0, μ₁ = 0, μ₅ ≠ 0. By the adjoint action of Λ₁ and Λ₈, (18) reduces to Λ = μ₅Λ₅ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{21}={\mathrm{\Lambda }}_5\pm {\mathrm{\Lambda }}_6$.

Case 22: μ₈ = 0, μ₇ = 0, μ₄ = 0, μ₃ = 0, μ₁ = 0, μ₅ = 0. By the adjoint action of Λ₁, (18) is reduced to Λ = μ₆Λ₆. Thus, ${\mathcal{R}}_{22}={\mathrm{\Lambda }}_6$.

Case 23: μ₈ = 0, μ₇ = 0, μ₄ = 0, μ₃ = 0, μ₁ = 0, μ₆ = 0, μ₂ = 0. Then, (18) is reduced to Λ = μ₅Λ₅. Therefore, ${\mathcal{R}}_{23}={\mathrm{\Lambda }}_5$.

Case 24: μ₈ = 0, μ₇ ≠ 0, μ₃ = 0, μ₁ = 0. By the adjoint action of Λ₁, Λ₄ and Λ₅, (18) reduces to Λ = μ₆Λ₆ + μ₇Λ₇. Therefore, ${\mathcal{R}}_{24}={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_7,\kappa \ne 0$.

Case 25: μ₈ = 0, μ₇ ≠ 0, μ₆ = 0, μ₃ = 0μ₂ = 0, μ₁ = 0. By the adjoint action of Λ₄ and Λ₅, Eq (18) is reduced to Λ = μ₇Λ₇. Thus, ${\mathcal{R}}_{25}={\mathrm{\Lambda }}_7$.

Case 26: μ₈ = 0, μ₇ = 0, μ₅ = 0, μ₄ = 0, μ₃ = 0. By the adjoint action of Λ₁ and Λ₈, Eq (18) is reduced to Λ = μ₁Λ₁ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{26}={\mathrm{\Lambda }}_1\pm {\mathrm{\Lambda }}_6$.

Case 27: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₅ = 0, μ₄ = 0, μ₃ = 0. By the adjoint action of Λ₆, (18) reduces to Λ = μ₁Λ₁. Thus, ${\mathcal{R}}_{27}={\mathrm{\Lambda }}_1$.

Case 28: μ₈ = 0, μ₇ = 0, μ₅ = 0, μ₃ = 0, μ₁ = 0, μ₆ ≠ 0. By the adjoint action of Λ₁ and Λ₈, Eq (18) is reduced to Λ = μ₄Λ₄ + e^−ϵμ₆Λ₆. Therefore, ${\mathcal{R}}_{28}={\mathrm{\Lambda }}_4\pm {\mathrm{\Lambda }}_6$.

Case 29: μ₈ = 0, μ₇ = 0, μ₅ = 0, μ₃ = 0, μ₁ = 0, μ₆ = 0, μ₂ = 0. Then, (18) is reduced to Λ = μ₄Λ₄. Thus, ${\mathcal{R}}_{29}={\mathrm{\Lambda }}_4$.

Case 30: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₃ ≠ 0, μ₅ ≠ 0. From the adjoint action of Λ₆ and Λ₈, Eq (18) reduces to Λ = μ₁Λ₁ + μ₃Λ₃ + μ₅Λ₅. Thus, ${\mathcal{R}}_{30}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0$,

Case 31: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₃ = 0, μ₅ ≠ 0. By the adjoint action of Λ₆, Eq (18) is reduced to Λ = μ₁Λ₁ + μ₅Λ₅. Therefore, ${\mathcal{R}}_{31}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0$.

Case 32: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₁ = 0, μ₃ ≠ 0, μ₅ ≠ 0. Then, (18) is reduced to Λ = μ₂Λ₂ + μ₃Λ₃ + μ₅Λ₅. Thus, ${\mathcal{R}}_{32}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0$,

Case 33: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₁ = 0, μ₃ ≠ 0, μ₅ = 0. Then, (18) is reduced to Λ = μ₂Λ₂ + μ₃Λ₃. Therefore, ${\mathcal{R}}_{33}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3,\kappa \ne 0$.

Case 34: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₁ = 0, μ₃ = 0, μ₅ ≠ 0. Then, (18) is reduced to Λ = μ₂Λ₂ + μ₅Λ₅. Therefore, ${\mathcal{R}}_{34}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0$.

Case 35: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₁ = 0, μ₃ = 0, μ₅ = 0. Then, (18) is reduced to Λ = μ₂Λ₂. Thus, ${\mathcal{R}}_{35}={\mathrm{\Lambda }}_2$.

Case 36: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₄ = 0, μ₁ = 0, μ₂ = 0, μ₅ = 0. Then, (18) is reduced to Λ = μ₃Λ₃. Thus, ${\mathcal{R}}_{36}={\mathrm{\Lambda }}_3$.

Case 37: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₅ = 0, μ₃ = 0, μ₄ ≠ 0. By the adjoint action of Λ₆, Eq (18) is reduced to Λ = μ₁Λ₁ + μ₄Λ₄. Therefore, ${\mathcal{R}}_{37}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0$.

Case 38: μ₈ = 0, μ₇ = 0, μ₆ = 0, μ₅ = 0, μ₃ = 0, μ₁ = 0, μ₄ ≠ 0. Then, (18) is reduced to Λ = μ₂Λ₂ + μ₄Λ₄. Therefore, ${\mathcal{R}}_{38}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0$.

Hence, the one-dimensional optimal system for Eq (13) is given by,

$$\begin{array}{cc}\begin{array}{l}\begin{array}{c}{\mathcal{R}}_1={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_7+\sigma {\mathrm{\Lambda }}_8,\kappa ,\sigma \ne 0\hfill \end{array}\hfill \\ \begin{array}{c}{\mathcal{R}}_2={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_8,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_3={\mathrm{\Lambda }}_7+\kappa {\mathrm{\Lambda }}_8,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_4={\mathrm{\Lambda }}_8\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_5={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6\pm {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_6={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_7={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_8={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_9={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{10}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{11}={\mathrm{\Lambda }}_1\pm {\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{12}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{13}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{14}={\mathrm{\Lambda }}_2\pm {\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{15}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6\pm {\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{16}={\mathrm{\Lambda }}_3\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{17}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5\pm {\mathrm{\Lambda }}_6,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{18}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{19}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\end{array}& \begin{array}{l}\begin{array}{c}{\mathcal{R}}_{20}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_4\pm {\mathrm{\Lambda }}_6,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{21}={\mathrm{\Lambda }}_5\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{22}={\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{23}={\mathrm{\Lambda }}_5\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{24}={\mathrm{\Lambda }}_6+\kappa {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{25}={\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{26}={\mathrm{\Lambda }}_1\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{27}={\mathrm{\Lambda }}_1\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{28}={\mathrm{\Lambda }}_4\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{29}={\mathrm{\Lambda }}_4\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{30}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{31}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{32}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{33}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{34}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{35}={\mathrm{\Lambda }}_2\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{36}={\mathrm{\Lambda }}_3\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{37}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{38}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\end{array}\end{array}$$

3 Group invariant solutions

Invariant solutions [19] are the solutions to differential equations that remain unchanged under the action of a given symmetry group. In the context of the Lie group method, invariant solutions are those that are preserved by the symmetry transformations generated by the infinitesimal generators of the group. In this section, we focus on investigating the invariant solutions of (13).

Invariant Solutions by Using ${\mathcal{R}}_{27}={\mathrm{\Lambda }}_1.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_1=\frac{\partial }{\partial \tau }·$

The characteristic equation associated with Λ₁ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{1}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that x₁ = α, x₂ = β, x₃ = λ, U = X(α, β, λ). Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\alpha }-A)X_{\alpha \alpha }+(-D{X}_{\alpha }-B)X_{\beta \beta }+(-D{X}_{\alpha }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D(X_{\beta }{X}_{\alpha \beta }+X_{\mathrm{\lambda }}{X}_{\alpha \mathrm{\lambda }})=0,\end{array}$$ (19)

Infinitesimals of (19) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_2\alpha +c_5,\hfill \\ \hfill {\delta }_{\beta }& =-c_1\mathrm{\lambda }+c_2\beta +c_6,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_1\beta +c_2\mathrm{\lambda }+c_3,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_{2}X+c_4.\hfill \end{array}\end{array}$$ (20)

Case 1: By taking c₃ = 1 and all other constants zero, the characteristic equation associated with (20) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{0},$$

which implies that α = r, β = s, and X(α, β, λ) = Y(r, s). Using these invariant variables, (19) is transformed into the following equation

$$\begin{array}{c}\hfill (-D{Y}_r-B)Y_{ss}+(-C{Y}_r-A)Y_{rr}-2D{Y}_s{Y}_{rs}=0.\end{array}$$ (21)

Infinitesimals of Eq (21) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_{1}r+c_4,\hfill \\ \hfill {\delta }_s& =c_{1}s+c_2,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_{1}Y+c_3.\hfill \end{array}\end{array}$$ (22)

Case 1a: By taking c₂ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (22) is

$$\frac{dr}{1}=\frac{ds}{1}=\frac{dY}{0},$$

which implies that −r + s = p and Y(r, s) = Z(p). Using these invariant variables, (21) is transformed into the following ODE

$$\begin{array}{c}\hfill -Z^{″}(A+B+(-C-3D)Z^{\prime })=0.\end{array}$$ (23)

If Z″ = 0, then Z(p) = c₁p + c₂. This implies Y(r, s) = c₁(s − r) + c₂, this gives X(α, β, λ) = c₁(β − α) + c₂.

So, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1(x_2-x_1)+c_2.$$

Now, if Z″ ≠ 0, from (23), we have A + B + (−C − 3D)Z′ = 0 which gives $Z\left(p\right)=c_1+\frac{(A+B)}{(C+3D)}p,$ which provides $Y(r,s)=c_1+\frac{(A+B)}{(C+3D)}(s-r)$.

This implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1+\frac{(A+B)}{(C+3D)}(\beta -\alpha ).\end{array}$$ (24)

Hence, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1+\frac{(A+B)}{(C+3D)}(x_2-x_1).$$

Invariant Solutions by Using ${\mathcal{R}}_{29}={\mathrm{\Lambda }}_4.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_4=\frac{\partial }{\partial x_2}·$

The characteristic equation associated with Λ₄ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{1}=\frac{d{x}_3}{0}=\frac{d\tau }{0}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that τ = α, x₁ = β, x₃ = λ, U = X(α, β, λ). Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\beta }-A)X_{\beta \beta }+(-D{X}_{\beta }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D{X}_{\mathrm{\lambda }}{X}_{\beta \mathrm{\lambda }}+X_{\alpha \alpha }=0.\end{array}$$ (25)

Infinitesimals of (25) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_1\alpha +c_2,\hfill \\ \hfill {\delta }_{\beta }& =c_1\beta +c_6,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_1\mathrm{\lambda }+c_5,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_{1}X+c_3\alpha +c_4.\hfill \end{array}\end{array}$$ (26)

Case 2: By taking c₄ = 1, c₅ = 1 and all other constants zero, the characteristic equation associated with (26) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{1},$$

which implies that α = r, β = s, and X(α, β, λ) = λ + Y(r, s). Using these invariant variables, (25) is transformed into the following equation

$$\begin{array}{c}\hfill (-C{Y}_s-A)Y_{ss}+Y_{rr}=0.\end{array}$$ (27)

Infinitesimals of Eq (27) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_{1}r+c_2,\hfill \\ \hfill {\delta }_s& =c_{3}s+c_4,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =(3c_3-2c_1)Y-\frac{2A}{C}(c_1-c_3)s+c_{5}r+c_6.\hfill \end{array}\end{array}$$ (28)

Case 2a: By taking c₂ = 1, c₄ = 1, c₆ = 1 and all other constants zero, the characteristic equation associated with (28) is

$$\frac{dr}{1}=\frac{ds}{1}=\frac{dY}{1},$$

which implies that −r + s = p and Y(r, s) = r + Z(p). Using these invariant variables, (27) is transformed into the following ODE

$$\begin{array}{c}\hfill -Z^{″}(C{Z}^{\prime }+A-1)=0.\end{array}$$ (29)

If Z″ = 0, then Z(p) = c₁p + c₂. This implies Y(r, s) = c₁(s − r) + c₂ + r, this gives X(α, β, λ) = c₁(β − α) + c₂ + α + λ.

So, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1(x_1-\tau )+c_2+\tau +x_3.$$

Now, if Z″ ≠ 0, then from (29), we have CZ′ + A − 1 = 0 which gives $Z\left(p\right)=c_1-\frac{(A-1)}{C}p,$ which provides $Y(r,s)=c_1+r-\frac{(A-1)}{C}(s-r)$.

This implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1+\alpha +\mathrm{\lambda }-\frac{(A-1)}{C}(\beta -\alpha ).\end{array}$$ (30)

Hence, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1+\tau +x_3-\frac{(A-1)}{C}(x_1-\tau ).$$

Invariant Solutions by Using ${\mathcal{R}}_{23}={\mathrm{\Lambda }}_5.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_5=\frac{\partial }{\partial x_3}·$

The characteristic equation associated with Λ₅ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{0}=\frac{d{x}_3}{1}=\frac{d\tau }{0}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that τ = α, x₁ = β, x₂ = λ, U = X(α, β, λ). Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\beta }-A)X_{\beta \beta }+(-D{X}_{\beta }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D{X}_{\mathrm{\lambda }}{X}_{\beta \mathrm{\lambda }}+X_{\alpha \alpha }=0.\end{array}$$ (31)

Infinitesimals of (31) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_1\alpha +c_2,\hfill \\ \hfill {\delta }_{\beta }& =c_1\beta +c_6,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_1\mathrm{\lambda }+c_5,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_{1}X+c_3\alpha +c_4.\hfill \end{array}\end{array}$$ (32)

Case 3: By taking c₂ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (32) is

$$\frac{d\alpha }{1}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{0}=\frac{dX}{1},$$

which implies that β = r, λ = s, and X(α, β, λ) = α + Y(r, s). Using these invariant variables, (31) is transformed into the following equation

$$\begin{array}{c}\hfill (-D{Y}_r-B)Y_{ss}+(-C{Y}_r-A)Y_{rr}-2D{Y}_s{Y}_{rs}=0.\end{array}$$ (33)

Infinitesimals of Eq (33) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_{1}r+c_4,\hfill \\ \hfill {\delta }_s& =c_{1}s+c_2,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_{1}Y+c_3.\hfill \end{array}\end{array}$$ (34)

Case 3a: By taking c₄ = 1 and all other constants zero, the characteristic equation associated with (34) is

$$\frac{dr}{1}=\frac{ds}{0}=\frac{dY}{0},$$

which implies that s = p and Y(r, s) = Z(p). Using these invariant variables, (33) is transformed into the following ODE

$$\begin{array}{c}\hfill -B{Z}^{″}=0.\end{array}$$ (35)

This implies Z(p) = c₁p + c₂, this gives Y(r, s) = c₁s + c₂,

This implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1\mathrm{\lambda }+c_2+\alpha ,\end{array}$$ (36)

So, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1{x}_2+c_2+\tau .$$

Invariant Solutions by Using ${\mathcal{R}}_8={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3=\frac{\partial }{\partial \tau }+\kappa \frac{\partial }{\partial x_1}·$

The characteristic equation associated with Λ₁ + κΛ₃ is given by

$$\frac{d{x}_1}{\kappa }=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{1}=\frac{dU}{0}·$$

From the characteristic equation, we deduce that $x_2=\alpha ,x_3=\beta ,\tau -\frac{x_1}{\kappa }=\mathrm{\lambda },U=X(\alpha ,\beta ,\mathrm{\lambda })$. Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill \begin{array}{c}\hfill -B{\kappa }^3{X}_{\alpha \alpha }-B{\kappa }^3{X}_{\beta \beta }+{\kappa }^{2}D{X}_{\mathrm{\lambda }}{X}_{\alpha \alpha }+{\kappa }^{2}D{X}_{\mathrm{\lambda }}{X}_{\beta \beta }+2{\kappa }^{2}D{X}_{\beta }{X}_{\beta \mathrm{\lambda }}+2{\kappa }^{2}D{X}_{\alpha }{X}_{\alpha \mathrm{\lambda }}\\ \hfill +{\kappa }^3{X}_{\mathrm{\lambda }\mathrm{\lambda }}-\kappa A{X}_{\mathrm{\lambda }\mathrm{\lambda }}+C{X}_{\mathrm{\lambda }}{X}_{\mathrm{\lambda }\mathrm{\lambda }}=0,\end{array}\end{array}$$ (37)

Infinitesimals of (37) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_1\beta +c_2\alpha +c_3,\hfill \\ \hfill {\delta }_{\beta }& =-c_1\alpha +c_2\beta +c_6,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_2\mathrm{\lambda }+c_5,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_{2}X+c_4.\hfill \end{array}\end{array}$$ (38)

Case 4: By taking c₃ = 1, c₆ = 1 and all other constants zero, the characteristic equation associated with (38) is

$$\frac{d\alpha }{1}=\frac{d\beta }{1}=\frac{d\mathrm{\lambda }}{0}=\frac{dX}{0},$$

which implies that λ = r, β − α = s and X(α, β, λ) = λ + Y(r, s). Using these invariant variables, (37) is transformed into the following equation

$$\begin{array}{c}\hfill ({\kappa }^3-\kappa A+C{Y}_r)Y_{rr}+(-2B{\kappa }^3+2D{\kappa }^2{Y}_r)Y_{ss}+4D{\kappa }^2{Y}_s{Y}_{rs}=0,\end{array}$$ (39)

Infinitesimals of Eq (39) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_{1}r+c_4,\hfill \\ \hfill {\delta }_s& =c_{1}s+c_2,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_{1}Y+c_3.\hfill \end{array}\end{array}$$ (40)

Case 4a: By taking c₂ = 1, c₃ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (40) is

$$\frac{dr}{1}=\frac{ds}{1}=\frac{dY}{1},$$

which implies that −r + s = p, Y(r, s) = r + Z(p). Using these invariant variables, (39) is transformed into the following ODE

$$\begin{array}{c}\hfill -Z^{″}((6D{\kappa }^2+C)Z^{\prime }+(2B-1){\kappa }^3-2D{\kappa }^2+\kappa A-C)=0,\end{array}$$ (41)

If Z″ = 0, then Z(p) = c₁p + c₂. This implies Y(r, s) = c₁(s − r) + c₂ + r, this gives X(α, β, λ) = c₁(β − α − λ) + c₂ + 2λ.

So, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=c_1(x_3-x_2-(\tau -\frac{x_1}{\kappa }\left)\right)+c_2+2(\tau -\frac{x_1}{\kappa }).$$

Now, if Z″ ≠ 0, then from (41), we have (6Dκ² + C)Z′ + (2B − 1)κ³ − 2Dκ² + κA − C = 0, which yields $Z\left(p\right)=\frac{(2D{\kappa }^2-\kappa A+C-(2B-1){\kappa }^3)p}{6D{\kappa }^2+C}+c_1,$ which provides $Y(r,s)=\frac{(2D{\kappa }^2-\kappa A+C-(2B-1){\kappa }^3)(s-r)}{6D{\kappa }^2+C}+c_1+r$.

This implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=\frac{(2D{\kappa }^2-\kappa A+C-(2B-1){\kappa }^3)(\beta -\alpha -\mathrm{\lambda })}{6D{\kappa }^2+C}+c_1+2\mathrm{\lambda }.\end{array}$$ (42)

Hence, the solution of (13) in original variables is

$$U(x_1,x_2,x_3,\tau )=\frac{(2D{\kappa }^2-\kappa A+C-(2B-1){\kappa }^3)(x_3-x_2-(\tau -\frac{x_1}{\kappa }\left)\right)}{6D{\kappa }^2+C}+c_1+2(\tau -\frac{x_1}{\kappa }).$$

Invariant Solutions by Using ${\mathcal{R}}_{16}={\mathrm{\Lambda }}_3+{\mathrm{\Lambda }}_6.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_3+{\mathrm{\Lambda }}_6=\frac{\partial }{\partial x_1}+\tau \frac{\partial }{\partial U}·$

The characteristic equation associated with Λ₃ + Λ₆ is given by

$$\frac{d{x}_1}{1}=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{0}=\frac{dU}{\tau }·$$

From the characteristic equation, we can deduce that τ = α, x₂ = β, x₃ = λ, U = x₁τ + X(α, β, λ). Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill (-\alpha D-B)X_{\beta \beta }+(-\alpha D-B)X_{\mathrm{\lambda }\mathrm{\lambda }}+X_{\alpha \alpha }=0,\end{array}$$ (43)

Infinitesimals of (43) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =\frac{2(\alpha D+B)((c_2\beta +c_4)D+\frac{9}{2}{c}_1\mathrm{\lambda })}{3D^2},\hfill \\ \hfill {\delta }_{\beta }& =\frac{2}{9D^2}{c}_2{(\alpha D+B)}^3+\frac{c_2}{2}{\beta }^2+c_4\beta +\frac{9}{2D}{c}_1\beta \mathrm{\lambda }-\frac{c_2}{2}{\mathrm{\lambda }}^2-c_3\mathrm{\lambda }+c_{13},\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =-\frac{9}{4D}{c}_1{\beta }^2+(c_2\mathrm{\lambda }+c_3)\beta +\frac{9}{4D}{c}_1{\mathrm{\lambda }}^2+c_4\mathrm{\lambda }+c_5+c_1{(\alpha +\frac{B}{D})}^3,\hfill \\ \hfill {\mathrm{\Psi }}_X& =\frac{1}{3D{e}^{\sqrt{a_3}\mathrm{\lambda }}{e}^{\sqrt{a_2}\beta }}\left(3D{c}_7\right(c_{11}{\left(e^{\sqrt{a_3}\mathrm{\lambda }}\right)}^2+c_{12}\left)\right({\left(e^{\sqrt{a_2}\beta }\right)}^2{c}_9+c_{10}\left)AiryAi\right(-\hfill \\ & \frac{(\alpha D+B){(-D(a_2+a_3))}^{\frac{1}{3}}}{D})+3D{c}_8(c_{11}{\left(e^{\sqrt{a_3}\mathrm{\lambda }}\right)}^2+c_{12}\left)\right({\left(e^{\sqrt{a_2}\beta }\right)}^2{c}_9+c_{10})AiryBi\hfill \\ & (-\frac{(\alpha D+B){(-D(a_2+a_3))}^{\frac{1}{3}}}{D})-2e^{\sqrt{a_2}\beta }{e}^{\sqrt{a_3}\mathrm{\lambda }}X((c_2\beta -\frac{3}{2}{c}_6)D+\frac{9}{2}{c}_1\mathrm{\lambda })).\hfill \end{array}\end{array}$$ (44)

Case 5: By taking c₁₃ = 1 and all other constants zero, the characteristic equation associated with (44) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{0},$$

which implies that α = r, β = s, X(α, β, λ) = Y(r, s). Using these invariant variables, (43) is transformed into the following equation

$$\begin{array}{c}\hfill (-rD-B)Y_{ss}+Y_{rr}=0.\end{array}$$ (45)

Infinitesimals of Eq (45) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =\frac{(rD+B)(c_{7}D-4c_{1}s)}{D},\hfill \\ \hfill {\delta }_s& =-\frac{4}{3D^2}{c}_1{(rD+B)}^3-3c_1{s}^2+\frac{3}{2}D{c}_{7}s+c_8,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =\frac{1}{e^{\sqrt{a_2}s}}\left(c_3\right(c_5{\left(e^{\sqrt{a_2}s}\right)}^2+c_6\left)AiryAi\right(-\frac{{(-a_{2}D)}^{\frac{1}{3}}(rD+B)}{D})+c_4(c_5{\left(e^{\sqrt{a_2}s}\right)}^2+c_6)AiryBi\hfill \\ & (-\frac{{(-a_{2}D)}^{\frac{1}{3}}(rD+B)}{D})+e^{\sqrt{a_2}s}Y(c_{1}s+c_2)).\hfill \end{array}\end{array}$$ (46)

Case 5a: By taking c₇ = 1 and all other constants zero, the characteristic equation associated with (46) is

$$\frac{dr}{(rD+B)}=\frac{ds}{\frac{3}{2}sD}=\frac{dY}{0},$$

which implies that $\frac{s}{{(rD+B)}^{\frac{3}{2}}}=p$ and Y(r, s) = Z(p). Using these invariant variables, (45) is transformed into the following ODE

$$\begin{array}{c}\hfill 9D^2{p}^2{Z}^{″}+15D^{2}p{Z}^{\prime }-4Z^{″}=0.\end{array}$$ (47)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=c_1+hypergeom\left(\right[\frac{1}{2},\frac{5}{6}],[\frac{3}{2}],\frac{9}{4}{D}^2{p}^2)c_{2}p,\end{array}$$ (48)

which gives,

$$\begin{array}{c}\hfill Y(r,s)=c_1+hypergeom\left(\right[\frac{1}{2},\frac{5}{6}],[\frac{3}{2}],\frac{9}{4}{D}^2\frac{s^2}{{(rD+B)}^3})c_2\frac{s}{{(rD+B)}^{\frac{3}{2}}},\end{array}$$ (49)

this gives,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1+hypergeom\left(\right[\frac{1}{2},\frac{5}{6}],[\frac{3}{2}],\frac{9}{4}{D}^2\frac{{\beta }^2}{{(\alpha D+B)}^3})c_2\frac{\beta }{{(\alpha D+B)}^{\frac{3}{2}}},\end{array}$$ (50)

So, the solution of (13) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=c_1+x_1\tau +hypergeom\left(\right[\frac{1}{2},\frac{5}{6}],[\frac{3}{2}],\frac{9}{4}{D}^2\frac{x_2^2}{{(\tau D+B)}^3})c_2\frac{x_2}{{(\tau D+B)}^{\frac{3}{2}}}.\end{array}$$ (51)

Invariant Solutions by Using ${\mathcal{R}}_7={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+{\mathrm{\Lambda }}_6.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+{\mathrm{\Lambda }}_6=\frac{\partial }{\partial \tau }+\kappa \frac{\partial }{\partial x_1}+\tau \frac{\partial }{\partial U}·$

The characteristic equation associated with Λ₁ + κΛ₃ + Λ₆ is given by

$$\frac{d{x}_1}{\kappa }=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{1}=\frac{dU}{\tau }·$$

From the characteristic equation, we deduce that $x_2=\alpha ,x_3=\beta ,\tau -\frac{x_1}{\kappa }=\mathrm{\lambda },U=\frac{2\kappa x_1\tau -x_1^2}{2{\kappa }^2}+X(\alpha ,\beta ,\mathrm{\lambda })$. Using these invariant variables, (13) is transformed into the equation given by

$$\begin{array}{c}\hfill \begin{array}{c}\hfill ({\kappa }^3-A\kappa +C{X}_{\mathrm{\lambda }}-C\mathrm{\lambda })X_{\mathrm{\lambda }\mathrm{\lambda }}+{\kappa }^2(D{X}_{\mathrm{\lambda }}-B\kappa -D\mathrm{\lambda })X_{\beta \beta }+{\kappa }^2(D{X}_{\mathrm{\lambda }}-B\kappa -D\mathrm{\lambda })X_{\alpha \alpha }\\ \hfill +2D{\kappa }^2{X}_{\alpha }{X}_{\alpha \mathrm{\lambda }}+2D{\kappa }^2{X}_{\beta }{X}_{\beta \mathrm{\lambda }}+A\kappa -C{X}_{\mathrm{\lambda }}+C\mathrm{\lambda }=0,\end{array}\end{array}$$ (52)

Infinitesimals of (52) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_1\beta +c_2,\hfill \\ \hfill {\delta }_{\beta }& =-c_1\alpha +c_5,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_3,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_3\mathrm{\lambda }+c_4.\hfill \end{array}\end{array}$$ (53)

Case 6: By taking c₃ = 1 and all other constants zero, the characteristic equation associated with (53) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{\mathrm{\lambda }},$$

which implies that α = r, β = s, and $X(\alpha ,\beta ,\mathrm{\lambda })=\frac{{\mathrm{\lambda }}^2}{2}+Y(r,s)$. Using these invariant variables, (52) is transformed into the following equation

$$\begin{array}{c}\hfill -{\kappa }^3(B{Y}_{ss}+B{Y}_{rr}-1)=0.\end{array}$$ (54)

Infinitesimals of Eq (54) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =B(F_4(s-Ir)+F_6(s+Ir))-c_{2}s+\frac{c_1}{2}r+c_4,\hfill \\ \hfill {\delta }_s& =-BI(-F_6(s+Ir)+F_4(s-Ir))+\frac{c_1}{2}s+c_{2}r+c_3,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_{1}Y+F_3(s-Ir)+r{F}_4(s-Ir)+F_5(s+Ir)+r{F}_6(s+Ir).\hfill \end{array}\end{array}$$ (55)

Case 6a: By taking c₃ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (55) is

$$\frac{dr}{1}=\frac{ds}{1}=\frac{dY}{0},$$

which implies that −r + s = p, Y(r, s) = Z(p). Using these invariant variables, (54) is transformed into the following ODE

$$\begin{array}{c}\hfill -2B{\kappa }^3{Z}^{″}+{\kappa }^3=0.\end{array}$$ (56)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=\frac{1}{4B}{p}^2+c_{1}p+c_2,\end{array}$$ (57)

which yields,

$$\begin{array}{c}\hfill Y(r,s)=\frac{1}{4B}{(-r+s)}^2+c_1(-r+s)+c_2,\end{array}$$ (58)

this gives,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=\frac{{\mathrm{\lambda }}^2}{2}+\frac{1}{4B}{(-\alpha +\beta )}^2+c_1(-\alpha +\beta )+c_2.\end{array}$$ (59)

So, the solution of (13) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=\frac{2\kappa x_1\tau -x_1^2}{2{\kappa }^2}+\frac{{(\tau -\frac{x_1}{\kappa })}^2}{2}+\frac{1}{4B}{(-x_2+x_3)}^2+c_1(-x_2+x_3)+c_2.\end{array}$$ (60)

4 Noether’s approach

Let us contemplate a differential system of m-th order

$$\begin{array}{c}\hfill G(x,U,U^{\tau },U^{\left(1\right)},U^{\left(2\right)}\cdots ,U^{\left(m\right)})=0,\end{array}$$ (61)

where xⁱ with i = 1, 2, ⋯, n represents independent variables, and U^α with α = 1, 2, ⋯, m symbolizes the dependent variables. Additionally, the notation U^m signifies the m th-order partial derivative.

The Lie-Bäcklund operator is defined as follows

$$\begin{array}{c}\hfill {\mathrm{\Lambda }}_1={\delta }^i\frac{\partial }{\partial x^i}+{\mathrm{\Psi }}^{\varpi }\frac{\partial }{\partial U^{\varpi }}+\sum _{s⩾1}{\mathrm{\Psi }}_{i_1\cdots i_s}^{\varpi }\frac{\partial }{\partial U_{i_1\cdots i_s}^{\varpi }},\end{array}$$ (62)

where ${\mathrm{\Psi }}_{i_1\cdots i_s}^{\varpi }$ is defined by

$$\begin{array}{c}{\mathrm{\Psi }}_i^{\varpi }={\mathbf{D}}_i\left({\mathrm{\Psi }}^{\varpi }\right)-U_j^{\varpi }{\mathbf{D}}_i\left({\delta }^j\right),\hfill \\ {\mathrm{\Psi }}_{i_1\cdots i_s}^{\varpi }={\mathbf{D}}_{i_s}\left({\mathrm{\Psi }}_{i_1\cdots i_{s-1}}^{\varpi }\right)-U_{j{i}_1\cdots i_{s-1}}^{\varpi }{\mathbf{D}}_{i_s}\left({\delta }^j\right),s>1,\hfill \end{array}$$

In this context, D_i denotes the total derivative operator.

The Lie-Bäcklund operator (62), when expressed in characteristic form, takes on the following representation

$$\begin{array}{c}\hfill \mathrm{\Lambda }={\delta }^i\frac{\partial }{\partial x^i}+W^{\varpi }\frac{\partial }{\partial U^{\varpi }}+{\mathbf{D}}_i\left(W^{\varpi }\right)\frac{\partial }{\partial U_i^{\varpi }}+{\mathbf{D}}_i{\mathbf{D}}_j\left(W^{\varpi }\right)\frac{\partial }{\partial U_{ij}^{\varpi }}+\cdots ,\end{array}$$ (63)

where,

$$W^{\varpi }={\mathrm{\Psi }}^{\delta }-{\delta }^j{U}_j^{\varpi },\varpi =1,2,\dots ,m,$$

constitute Lie characteristic functions.

The Noether operators linked with a Lie-Bäcklund operator Λ are given by

$$\begin{array}{c}\hfill N^i={\mathrm{\Psi }}^i+W^{\varpi }\frac{\delta }{\delta U_i^{\varpi }}+\sum _{s⩾1}{\mathbf{D}}_{i_1}\cdots {\mathbf{D}}_{i_s}\left(W^{\varpi }\right)\frac{\delta }{\delta U_{i{j}_1\dots j_s}^{\varpi }},i=1,2,\dots ,n,\end{array}$$ (64)

where $\frac{\delta }{\delta U_i^{\varpi }}$ denotes the Euler-Lagrange operator, which is defined as

$$\begin{array}{c}\hfill \frac{\delta }{\delta U_i^{\varpi }}=\frac{\partial }{\partial U_i^{\varpi }}+\sum _{s⩾1}{(-1)}^s{\mathbf{D}}_{j_1}\cdots {\mathbf{D}}_{j_s}\frac{\partial }{\partial U_{i{j}_1\dots j_s}^{\varpi }},i=1,2,\dots ,n,\varpi =1,2,\cdots ,m.\end{array}$$ (65)

Lagrangian A Lagrangian for the Eq (61) is described by a function $\mathcal{L}=\mathcal{L}(x,U,U^{\left(1\right)},\cdots ,U^{(m-1)})$, which adheres to the Euler-Lagrange equation

$$\begin{array}{c}\hfill \frac{\delta \mathcal{L}}{\delta U^{\alpha }}=0.\end{array}$$ (66)

Noether Symmetry Generator A Lie-Bäcklund operator Λ is considered a Noether symmetry generator of the Eq (61) with respect to a Lagrangian $\mathcal{L}$ if there exist gauge functions ${\mathcal{G}}^i=({\mathcal{G}}^1,\cdots ,{\mathcal{G}}^m)$ that fulfill the condition

$$\begin{array}{c}\hfill \mathrm{\Lambda }\left(\mathcal{L}\right)+\mathcal{L}{\mathbf{D}}_i\left({\mathrm{\Psi }}^i\right)={\mathbf{D}}_i\left({\mathbf{G}}^i\right).\end{array}$$ (67)

Noether Conserved Vectors Each Noether symmetry generator Λ associated with a given Lagrangian $\mathcal{L}$ that relates to the Euler-Lagrange differential equations corresponds to a vector T = (T¹, T², ⋯, T^m). The definition of each component Tⁱ is as follows

$$\begin{array}{c}\hfill T^i={\mathcal{G}}^i-N^i\mathcal{L}={\mathcal{G}}^i-{\delta }^i\mathcal{L}-W^{\varpi }\frac{\delta \mathcal{L}}{\delta U_i^{\varpi }}-\sum _{s⩾1}{\mathbf{D}}_{i_1\cdots i_s}\left(W^{\varpi }\right)\frac{\delta \mathcal{L}}{\delta U_{i_1\cdots i_s}^{\varpi }},\end{array}$$ (68)

Vector Tⁱ serves as a conserved quantity for Eq (66). Eq (67) is employed to identify the Noether symmetries, whereas (68) provides the associated Noether conserved vectors.

4.1 Noether’s approach to Eq (13)

In this subsection, we utilize Noether’s approach [20] to determine the conservation laws of (13). We begin by determining the Noether symmetries, and then establish the corresponding conservation laws using the Noether theorem.

The Lagrangian for Eq (13) given by

$$\begin{array}{c}\hfill \mathcal{L}=\frac{A{U}_{x_1}^2}{2}+\frac{B}{2}(U_{x_2}^2+U_{x_3}^2)+\frac{C{U}_{x_1}^3}{6}-\frac{U_{\tau }^2}{2}+\frac{D}{2}(U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2).\end{array}$$ (69)

Applying the Euler operator (65) to the Eq (69), we get

$$\begin{array}{c}\hfill \frac{\delta \mathcal{L}}{\delta u}=U_{\tau \tau }-A{U}_{x_1{x}_1}-B(U_{x_2{x}_2}+U_{x_3{x}_3})-C{U}_{x_1}{U}_{x_1{x}_1}-D(U_{x_1}{U}_{x_2{x}_2}+U_{x_1}{U}_{x_3{x}_3})\\ \hfill -2D(U_{x_2}{U}_{x_1{x}_2}+U_{x_3}{U}_{x_1{x}_3})=0.\end{array}$$ (70)

Hence, we can deduce Noether symmetries by employing the calculated Lagrangian. These symmetries, identified through Noether’s theorem, enable us to establish the conservation laws associated with Eq (13). These conservation laws offer valuable insights into the underlying physics and behavior of the system.

4.1.1 Computation of the noether symmetries

We consider the following vector field

$$\mathrm{\Lambda }={\delta }_1\frac{\partial }{\partial x_1}+{\delta }_2\frac{\partial }{\partial x_2}+{\delta }_3\frac{\partial }{\partial x_3}+{\delta }_4\frac{\partial }{\partial \tau }+\mathrm{\Psi }\frac{\partial }{\partial U}·$$

The vector field Λ is called a Noether symmetry of Eq (13) corresponding to Lagrangian (69) if it fulfills the condition

$$\begin{array}{c}\hfill {\mathrm{\Lambda }}^{\left[1\right]}\mathcal{L}+\mathcal{L}({\mathbf{D}}_{\tau }{\delta }_4+{\mathbf{D}}_{x_1}{\delta }_1+{\mathbf{D}}_{x_2}{\delta }_2+{\mathbf{D}}_{x_3}{\delta }_3)={\mathbf{D}}_{\tau }{\mathcal{G}}_1+{\mathbf{D}}_{x_1}{\mathcal{G}}_2+{\mathbf{D}}_{x_2}{\mathcal{G}}_3+{\mathbf{D}}_{x_3}{\mathcal{G}}_4,\end{array}$$ (71)

where,

$${\mathrm{\Lambda }}^{\left[1\right]}=\mathrm{\Lambda }+{\mathrm{\Psi }}^{x_1}\frac{\partial }{\partial U_{x_1}}+{\mathrm{\Psi }}^{x_2}\frac{\partial }{\partial U_{x_2}}+{\mathrm{\Psi }}^{x_3}\frac{\partial }{\partial U_{x_3}}+{\mathrm{\Psi }}^{\tau }\frac{\partial }{\partial U_{\tau }},$$

and ${\mathcal{G}}_i(x_1,x_2,x_3,\tau ,U),i=1,2,3,4$ are gauge terms. We compare the coefficients of derivatives of U in the above equation to obtain the determining system. After performing some steps we get

$$\begin{array}{c}\hfill \begin{array}{cc}& A{\delta }_{2_{x_2}}-A{\delta }_{1_{x_1}}+A{\delta }_{3_{x_3}}+A{\delta }_{4_{\tau }}+2A{\mathrm{\Psi }}_U+C{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & B{\delta }_{1_{x_1}}-B{\delta }_{2_{x_2}}+B{\delta }_{3_{x_3}}+B{\delta }_{4_{\tau }}+2B{\mathrm{\Psi }}_U+D{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & B{\delta }_{1_{x_1}}+B{\delta }_{2_{x_2}}-B{\delta }_{3_{x_3}}+B{\delta }_{4_{\tau }}+2B{\mathrm{\Psi }}_U+D{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & 2{\mathrm{\Psi }}_U+{\delta }_{1_{x_1}}+{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}-{\delta }_{4_{\tau }}=0,\hfill \\ & 3{\mathrm{\Psi }}_U-2{\delta }_{1_{x_1}}+{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & 3{\mathrm{\Psi }}_U-{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & 3{\mathrm{\Psi }}_U+{\delta }_{2_{x_2}}-{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & {\delta }_{1_{x_2}}={\delta }_{1_{x_3}}={\delta }_{1_{\tau }}={\delta }_{1_U}=0,\hfill \\ & {\delta }_{2_{x_1}}={\delta }_{2_{\tau }}={\delta }_{2_U}=0,\hfill \\ & {\delta }_{2_{x_3}}+{\delta }_{3_{x_2}}=0,\hfill \\ & {\delta }_{3_{x_1}}={\delta }_{3_{\tau }}={\delta }_{3_U}=0,\hfill \\ & {\delta }_{4_{x_1}}={\delta }_{4_{x_2}}={\delta }_{4_{x_3}}={\delta }_{4_U}=0,\hfill \\ & {\mathrm{\Psi }}_{x_2}={\mathrm{\Psi }}_{x_3}={\mathcal{G}}_U^3={\mathcal{G}}_U^4=0,\hfill \\ & A{\mathrm{\Psi }}_{x_1}-{\mathcal{G}}_U^2=0,\hfill \\ & -{\mathrm{\Psi }}_{\tau }-{\mathcal{G}}_U^1=0,\hfill \\ & {\mathcal{G}}_{\tau }^1+{\mathcal{G}}_{x_1}^2+{\mathcal{G}}_{x_2}^3+{\mathcal{G}}_{x_3}^4=0.\hfill \end{array}\end{array}$$ (72)

The following result follows

$$\begin{array}{c}\hfill \begin{array}{cc}& {\delta }_1=C_1,{\delta }_2=C_3{x}_3+C_4,{\delta }_3=-C_3{x}_2+C_5,{\delta }_4=C_2,\mathrm{\Psi }=C_6\tau +C_7,\hfill \\ & {\mathcal{G}}^1=-C_{6}U+X_{x_1},{\mathcal{G}}^2=-X_{\tau }+\int (-{\mathcal{G}}_{x_2}-h_{x_3})d{x}_1+p(x_2,x_3,\tau ),\hfill \\ & {\mathcal{G}}^3=g(x_1,x_2,x_3,\tau ),{\mathcal{G}}^4=h(x_1,x_2,x_3,\tau ).\hfill \end{array}\end{array}$$ (73)

By taking f = g = h = p = 0.

then,

$${\mathcal{G}}^1=-C_{6}U,{\mathcal{G}}^2={\mathcal{G}}^3={\mathcal{G}}^4=0.$$

Finally, we get the following Noether symmetry generators

$${\mathrm{\Lambda }}_1=\frac{\partial }{\partial x_1},{\mathrm{\Lambda }}_2=\frac{\partial }{\partial x_2},{\mathrm{\Lambda }}_3=\frac{\partial }{\partial x_3},{\mathrm{\Lambda }}_4=\frac{\partial }{\partial \tau },{\mathrm{\Lambda }}_5=\frac{\partial }{\partial U},$$

$${\mathrm{\Lambda }}_6=x_3\frac{\partial }{\partial x_2}-x_2\frac{\partial }{\partial x_3},{\mathrm{\Lambda }}_7=\tau \frac{\partial }{\partial U},and{\mathcal{G}}^1=-U.$$

4.1.2 Conserved vectors

For Λ₁, δ₁ = 1, δ₂ = δ₃ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. Then by Eq (68) we get

$$T^{\tau }={\mathcal{G}}_1-\mathcal{L}{\delta }_4-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{\tau }}·$$

This leads to

$$T^{\tau }=-U_{x_1}{U}_{\tau }$$

$$T^{x_1}={\mathcal{G}}_2-\mathcal{L}{\delta }_1-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_1}},$$

simplifying the above equation, we get

$$T^{x_1}=\frac{A}{2}{U}_{x_1}^2+\frac{C}{3}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}-\frac{B}{2}{U}_{x_2}^2-\frac{B}{2}{U}_{x_3}^2,$$

$$T^{x_2}={\mathcal{G}}_3-\mathcal{L}{\delta }_2-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_2}},$$

this leads to

$$T^{x_2}=B{U}_{x_1}{U}_{x_2}+D{U}_{x_1}^2{U}_{x_2}$$

$$T^{x_3}={\mathcal{G}}_4-\mathcal{L}{\delta }_3-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_3}},$$

we obtain

$$T^{x_3}=B{U}_{x_1}{U}_{x_3}+D{U}_{x_1}^2{U}_{x_3}·$$

For Λ₂, δ₂ = 1, δ₁ = δ₃ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. The conserved vector associated with Λ₂ represents the conserved quantities associated with ether symmetry Λ₂ of the equation. These conserved quantities remain constant over time due to the symmetry provided by Λ₂ and follow from Eq (68)

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-U_{x_2}{U}_{\tau },\hfill \\ & T^{x_1}=A{U}_{x_1}{U}_{x_2}+\frac{C}{2}{U}_{x_1}^2{U}_{x_2}+\frac{D}{2}(U_{x_2}^3+U_{x_2}{U}_{x_3}^2),\hfill \\ & T^{x_2}=-\frac{A}{2}{U}_{x_1}^2-\frac{C}{6}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}+\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)+\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2),\hfill \\ & T^{x_3}=U_{x_2}(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (74)

For Λ₃, δ₃ = 1, δ₁ = δ₂ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₃ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-U_{x_3}{U}_{\tau },\hfill \\ & T^{x_1}=A{U}_{x_1}{U}_{x_3}+\frac{C}{2}{U}_{x_1}^2{U}_{x_3}+\frac{D}{2}(U_{x_2}^2{U}_{x_3}+U_{x_3}^3),\hfill \\ & T^{x_2}=U_{x_2}(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}),\hfill \\ & T^{x_3}=-\frac{A}{2}{U}_{x_1}^2-\frac{C}{6}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}+\frac{B}{2}(-U_{x_2}^2+U_{x_3}^2)+\frac{D}{2}(-U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2).\hfill \end{array}\end{array}$$ (75)

For Λ₄, we consider δ₄ = 1, δ₁ = δ₂ = δ₃ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₄ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-\frac{A}{2}{U}_{x_1}^2-\frac{C}{6}{U}_{x_1}^3-\frac{U_{\tau }^2}{2}-\frac{B}{2}(U_{x_2}^2+U_{x_3}^2)-\frac{D}{2}(U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2),\hfill \\ & T^{x_1}=A{U}_{x_1}{U}_{\tau }+\frac{C}{2}{U}_{x_1}^2{U}_{\tau }+\frac{D}{2}(U_{x_2}^2{U}_{\tau }+U_{x_3}^2{U}_{\tau }),\hfill \\ & T^{x_2}=U_{x_2}(B{U}_{\tau }+D{U}_{x_1}{U}_{\tau }),\hfill \\ & T^{x_3}=U_{x_3}(B{U}_{\tau }+D{U}_{x_1}{U}_{\tau }).\hfill \end{array}\end{array}$$ (76)

For Λ₅, we follow Ψ = 1, δ₁ = δ₂ = δ₃ = δ₄ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₅ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=U_{\tau },\hfill \\ & T^{x_1}=-A{U}_{x_1}-\frac{C}{2}{U}_{x_1}^2-\frac{D}{2}(U_{x_2}^2+U_{x_3}^2),\hfill \\ & T^{x_2}=-(B{U}_{x_2}+D{U}_{x_1}{U}_{x_2}),\hfill \\ & T^{x_3}=-(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (77)

For Λ₆, we have δ₂ = x₃, δ₃ = −x₂, δ₁ = δ₄ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₆ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=(-x_3{U}_{x_2}+x_2{U}_{x_3})U_{\tau },\hfill \\ & T^{x_1}=(A{U}_{x_1}+\frac{C}{2}{U}_{x_1}^2+\frac{D}{2}(U_{x_2}^2+U_{x_3}^2))(x_3{U}_{x_2}-x_2{U}_{x_3}),\hfill \\ & T^{x_2}=-x_3(\frac{A}{2}{U}_{x_1}^2+\frac{C}{6}{U}_{x_1}^3-\frac{U_{\tau }^2}{2}-\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)-\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2))-y{U}_{x_2}{U}_{x_3}(B+D{U}_{x_1}),\hfill \\ & T^{x_3}=x_2(\frac{A}{2}{U}_{x_1}^2+\frac{C}{6}{U}_{x_1}^3-\frac{U_{\tau }^2}{2}+\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)+\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2))+x_3{U}_{x_2}{U}_{x_3}(B+D{U}_{x_1}).\hfill \end{array}\end{array}$$ (78)

For Λ₇, we have Ψ = τ, δ₁ = δ₂ = δ₃ = δ₄ = 0 and ${\mathcal{G}}_1=-U,{\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₇ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-U+\tau U_{\tau },\hfill \\ & T^{x_1}=-\tau (A{U}_{x_1}+\frac{C}{2}{U}_{x_1}^2+\frac{D}{2}(U_{x_2}^2+U_{x_3}^2)),\hfill \\ & T^{x_2}=-\tau (B{U}_{x_2}+D{U}_{x_1}{U}_{x_2}),\hfill \\ & T^{x_3}=-\tau (B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (79)

5 Damped elastic wave equation

The addition of the damping term U_τ to the non-linear elastic wave Eq (13) transforms it into a non-linear damped elastic wave equation given by

$$\begin{array}{c}\hfill \begin{array}{c}\hfill U_{\tau \tau }-A{U}_{x_1{x}_1}-B(U_{x_2{x}_2}+U_{x_3{x}_3})-C{U}_{x_1}{U}_{x_1{x}_1}-D(U_{x_1}{U}_{x_2{x}_2}+U_{x_1}{U}_{x_3{x}_3})\\ \hfill -2D(U_{x_2}{U}_{x_1{x}_2}+U_{x_3}{U}_{x_1{x}_3})+\gamma U_{\tau }=0,\end{array}\end{array}$$ (80)

where γ is the damping parameter.

In this section, we present a comprehensive Lie symmetry analysis of the nonlinear damped elastic wave Eq (80).

We assume the following form of the vector field

$$\mathrm{\Lambda }={\delta }_1\frac{\partial }{\partial x_1}+{\delta }_2\frac{\partial }{\partial x_2}+{\delta }_3\frac{\partial }{\partial x_3}+{\delta }_4\frac{\partial }{\partial \tau }+\mathrm{\Psi }\frac{\partial }{\partial U}·$$

The invariance condition for the Eq (80) is

$$\begin{array}{c}\hfill \begin{array}{cc}& {\mathrm{\Lambda }}^{\left[2\right]}(U_{\tau \tau }-A{U}_{x_1{x}_1}-B(U_{x_2{x}_2}+U_{x_3{x}_3})-C{U}_{x_1}{U}_{x_1{x}_1}-D(U_{x_1}{U}_{x_2{x}_2}+U_{x_1}{U}_{x_3{x}_3})\hfill \\ & -2D(U_{x_2}{U}_{x_1{x}_2}+U_{x_3}{U}_{x_1{x}_3})+\gamma U_{\tau }{\left)\right|}_{(6.1)}=0.\hfill \end{array}\end{array}$$ (81)

Upon comparing the coefficients of derivatives of U in the given expressions, we get a set of determining system and solving it, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}& {\delta }_{1_{x_1}}={\delta }_{1_{x_2}}={\delta }_{1_{x_3}}={\delta }_{1_{\tau }}={\delta }_{1_U}=0,\hfill \\ & {\delta }_{4_{x_1}}={\delta }_{4_{x_2}}={\delta }_{4_{x_3}}={\delta }_{4_{\tau }}={\delta }_{4_U}=0,\hfill \\ & {\delta }_{2_{x_1}}={\delta }_{2_{x_2}}={\delta }_{2_{\tau }}={\delta }_{2_U}=0,\hfill \\ & {\delta }_{3_{x_1}}={\delta }_{3_{x_3}}={\delta }_{3_{\tau }}={\delta }_{3_U}=0,\hfill \\ & {\mathrm{\Psi }}_{x_1}={\mathrm{\Psi }}_{x_2}={\mathrm{\Psi }}_{x_3}={\mathrm{\Psi }}_U=0,\hfill \\ & {\delta }_{3_{x_2}}+{\delta }_{2_{x_3}}=0,\hfill \\ & {\mathrm{\Psi }}_{\tau \tau }+\gamma {\mathrm{\Psi }}_{\tau }=0,\hfill \\ & {\delta }_{2_{x_3{x}_3}}=0,\hfill \end{array}\end{array}$$ (82)

The infinitesimals we obtain here are given by

$${\delta }_1=C_1,{\delta }_2=C_2{x}_3+C_3,{\delta }_3=-C_2{x}_2+C_4,{\delta }_4=C_5,\mathrm{\Psi }=C_6+C_7{e}^{-\gamma \tau }.$$

We follow the following symmetry algebra given by

$$\begin{array}{c}{\mathrm{\Lambda }}_1=\frac{\partial }{\partial \tau },{\mathrm{\Lambda }}_2=\frac{\partial }{\partial U},{\mathrm{\Lambda }}_3=\frac{\partial }{\partial x_1},{\mathrm{\Lambda }}_4=\frac{\partial }{\partial x_2},{\mathrm{\Lambda }}_5=\frac{\partial }{\partial x_3},\\ {\mathrm{\Lambda }}_6=e^{-\gamma \tau }\frac{\partial }{\partial U},{\mathrm{\Lambda }}_7=x_3\frac{\partial }{\partial x_2}-x_2\frac{\partial }{\partial x_3}·\end{array}$$

Table 4 lists the commutator relation for the generators of symmetry algebra (5).

Table 4. Commutator table.

[Λi, Λj]	Λ1	Λ2	Λ3	Λ4	Λ5	Λ6	Λ7
Λ1	0	0	0	0	0	−γΛ6	0
Λ2	0	0	0	0	0	0	0
Λ3	0	0	0	0	0	0	0
Λ4	0	0	0	0	0	0	- Λ5
Λ5	0	0	0	0	0	0	Λ4
Λ6	γΛ6	0	0	0	0	0	0
Λ7	0	0	0	Λ5	−Λ4	0	0

Following a similar procedure, we obtain the optimal system of symmetry algebra of Eq (80) given by,

$$\begin{array}{cc}\begin{array}{l}\begin{array}{c}{\mathcal{R}}_1={\mathrm{\Lambda }}_1\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_2={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_3={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_4={\mathrm{\Lambda }}_2\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_5={\mathrm{\Lambda }}_3\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_6={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_3,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_7={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_4,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_8={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_9={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{10}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{11}={\mathrm{\Lambda }}_4\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{12}={\mathrm{\Lambda }}_5\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{13}={\mathrm{\Lambda }}_1+{\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{14}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{15}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{16}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{17}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_6,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{18}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\end{array}& \begin{array}{l}\begin{array}{c}{\mathcal{R}}_{19}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_4,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{20}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{21}=\mathrm{\Lambda }\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{22}={\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{23}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_4,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{24}={\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{25}={\mathrm{\Lambda }}_1+{\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{26}={\mathrm{\Lambda }}_1+{\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_7,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{27}={\mathrm{\Lambda }}_4\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{28}={\mathrm{\Lambda }}_2+{\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_5+\sigma {\mathrm{\Lambda }}_7,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{29}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{30}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_5,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{31}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_7,\kappa ,\sigma \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{32}={\mathrm{\Lambda }}_3+\kappa {\mathrm{\Lambda }}_7,\kappa \ne 0\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{33}={\mathrm{\Lambda }}_5\pm {\mathrm{\Lambda }}_6\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{34}={\mathrm{\Lambda }}_6\pm {\mathrm{\Lambda }}_7\hfill \end{array}\\ \begin{array}{c}{\mathcal{R}}_{35}={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_3+\sigma {\mathrm{\Lambda }}_7,\kappa ,\sigma \ne 0,\hfill \end{array}\end{array}\end{array}$$

6 Group invariant solutions

Invariant Solutions by Using ${\mathcal{R}}_1={\mathrm{\Lambda }}_1.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_1=\frac{\partial }{\partial \tau }·$

The characteristic equation associated with Λ₁ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{1}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that x₁ = α, x₂ = β, x₃ = λ, U = X(α, β, λ). Using these invariant variables, (80) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\alpha }-A)X_{\alpha \alpha }+(-D{X}_{\alpha }-B)X_{\beta \beta }+(-D{X}_{\alpha }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D(X_{\beta }{X}_{\alpha \beta }+X_{\mathrm{\lambda }}{X}_{\alpha \mathrm{\lambda }})=0.\end{array}$$ (83)

Infinitesimals of (83) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_2\alpha +c_5,\hfill \\ \hfill {\delta }_{\beta }& =-c_1\mathrm{\lambda }+c_2\beta +c_6,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_1\beta +c_2\mathrm{\lambda }+c_3,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_{2}X+c_4.\hfill \end{array}\end{array}$$ (84)

Case 1: By taking c₅ = 1, c₆ = 1 and all other constants zero, the characteristic equation associated with (84) is

$$\frac{d\alpha }{1}=\frac{d\beta }{1}=\frac{d\mathrm{\lambda }}{0}=\frac{dX}{0},$$

which implies that λ = r, β − α = s and X(α, β, λ) = Y(r, s). Using these invariant variables, (83) is transformed into the following equation

$$\begin{array}{c}\hfill ((C+3D)Y_s-A-B)Y_{ss}+(D{Y}_s-B)Y_{rr}+2D{Y}_r{Y}_{rs}=0.\end{array}$$ (85)

Infinitesimals of Eq (85) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_{1}r+c_2,\hfill \\ \hfill {\delta }_s& =c_{1}s+c_4,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_{1}Y+c_3.\hfill \end{array}\end{array}$$ (86)

Case 1a: By taking c₃ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (86) is

$$\frac{dr}{0}=\frac{ds}{1}=\frac{dY}{1},$$

which implies that r = p and Y(r, s) = s + Z(p). Using these invariant variables, (85) is transformed into the following ODE

$$\begin{array}{c}\hfill -Z^{″}(B-D)=0.\end{array}$$ (87)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=c_{1}p+c_2,\end{array}$$ (88)

which provides,

$$\begin{array}{c}\hfill Y(r,s)=c_{1}r+c_2+s,\end{array}$$ (89)

this implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1\mathrm{\lambda }+c_2+(\beta -\alpha ).\end{array}$$ (90)

So, the solution of (80) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=c_1{x}_3+c_2+(x_2-x_1).\end{array}$$ (91)

Invariant Solutions by Using ${\mathcal{R}}_5={\mathrm{\Lambda }}_3.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_3=\frac{\partial }{\partial x_1}·$

The characteristic equation associated with Λ₃ is given by

$$\frac{d{x}_1}{1}=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{0}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that τ = α, x₂ = β, x₃ = λ, U = X(α, β, λ). Using these invariant variables, (80) is transformed into the equation given by

$$\begin{array}{c}\hfill X_{\alpha \alpha }-B{X}_{\beta \beta }-B{X}_{\mathrm{\lambda }\mathrm{\lambda }}+\gamma X_{\alpha }=0.\end{array}$$ (92)

Infinitesimals of (92) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =\frac{c_2}{B}\mathrm{\lambda }-\frac{2}{\gamma }{c}_4\beta +c_{13},\hfill \\ \hfill {\delta }_{\beta }& =-c_1\mathrm{\lambda }-\frac{2}{\gamma }{c}_{4}B\alpha +c_{12},\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_1\beta +c_2\alpha +c_3,\hfill \\ \hfill {\mathrm{\Psi }}_X& =\frac{1}{e^{\sqrt{a_2}\beta }}\left(2\right(c_{10}\text{sin}\left(\frac{\sqrt{a_{2}B-a_1}\mathrm{\lambda }}{\sqrt{B}}\right)+c_{11}\text{cos}\left(\frac{\sqrt{a_{2}B-a_1}\mathrm{\lambda }}{\sqrt{B}}\right)\left)\right(e^{-\frac{\alpha \sqrt{{\gamma }^2+4a_1}}{2}}{c}_7\hfill \\ & +e^{\frac{\alpha \sqrt{{\gamma }^2+4a_1}}{2}}{c}_6\left)B\right({\left(e^{\sqrt{a_2}\beta }\right)}^2{c}_8+c_9)e^{-\frac{\alpha \gamma }{2}}+2e^{\sqrt{a_2}\beta }((c_4\beta +c_5)B-\frac{c_2}{2}\gamma \mathrm{\lambda }\left)X\right).\hfill \end{array}\end{array}$$ (93)

Case 2: By taking c₃ = 1 and all other constants zero, the characteristic equation associated with (93) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{0},$$

which implies that α = r, β = s, and X(α, β, λ) = Y(r, s). Using these invariant variables, (92) is transformed into the following equation

$$\begin{array}{c}\hfill Y_{rr}-B{Y}_{ss}+\gamma Y_r=0.\end{array}$$ (94)

Infinitesimals of Eq (94) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =-\frac{2}{\gamma }{c}_{1}s+c_5,\hfill \\ \hfill {\delta }_s& =-\frac{2}{\gamma }B{c}_{1}r+c_6,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =(c_{1}s+c_2)Y+c_3\frac{{\left(e^{\frac{-a_1^2\sqrt{B}(\sqrt{B}r+s)}{4\sqrt{B}{a}_1+\gamma }}\right)}^4{e}^{\frac{a_1\gamma (\sqrt{B}r+s)}{4\sqrt{B}{a}_1+\gamma }}}{{\left(e^{\frac{\gamma a_1\sqrt{B}(-\frac{s}{\sqrt{B}}+r)}{8\sqrt{B}{a}_1+2\gamma }}\right)}^2}{c}_4.\hfill \end{array}\end{array}$$ (95)

Case 2a: By taking c₅ = 1, c₆ = 1 and all other constants zero, the characteristic equation associated with (95) is

$$\frac{dr}{1}=\frac{ds}{1}=\frac{dY}{0},$$

which implies that −r + s = p and Y(r, s) = s + Z(p). Using these invariant variables, (94) is transformed into the following ODE

$$\begin{array}{c}\hfill Z^{″}-B{Z}^{″}-\gamma Z^{\prime }=0.\end{array}$$ (96)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=c_1+c_2{e}^{\frac{-\gamma p}{B-1}},\end{array}$$ (97)

which provides,

$$\begin{array}{c}\hfill Y(r,s)=c_1+c_2{e}^{\frac{-\gamma (s-r)}{B-1}},\end{array}$$ (98)

this implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1+c_2{e}^{\frac{-\gamma (\beta -\alpha )}{B-1}}.\end{array}$$ (99)

So, the solution of (80) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=c_1+c_2{e}^{\frac{-\gamma (x_2-\tau )}{B-1}}.\end{array}$$ (100)

Invariant Solutions by Using ${\mathcal{R}}_{12}={\mathrm{\Lambda }}_5.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_5=\frac{\partial }{\partial x_3}·$

The characteristic equation associated with Λ₅ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{0}=\frac{d{x}_3}{1}=\frac{d\tau }{0}=\frac{dU}{0}·$$

From the characteristic equation, we can deduce that τ = α, x₁ = β, x₂ = λ, U = X(α, β, λ). Using these invariant variables, (80) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\beta }-A)X_{\beta \beta }+(-D{X}_{\beta }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D{X}_{\mathrm{\lambda }}{X}_{\beta \mathrm{\lambda }}+\gamma X_{\alpha }+X_{\alpha \alpha }=0.\end{array}$$ (101)

Infinitesimals of (101) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_3,\hfill \\ \hfill {\delta }_{\beta }& =c_4,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_5,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_1+c_2{e}^{-\alpha \gamma }.\hfill \end{array}\end{array}$$ (102)

Case 3: By taking c₂ = 1, c₅ = 1 and all other constants zero, the characteristic equation associated with (102) is

$$\frac{d\alpha }{0}=\frac{d\beta }{0}=\frac{d\mathrm{\lambda }}{1}=\frac{dX}{e^{-\alpha \gamma }},$$

which implies that α = r, β = s, and X(α, β, λ) = λe^−αγ + Y(r, s). Using these invariant variables, (101) is transformed into the following equation

$$\begin{array}{c}\hfill (-C{Y}_s-A)Y_{ss}+\gamma Y_r+Y_{rr}=0.\end{array}$$ (103)

Infinitesimals of Eq (94) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_3,\hfill \\ \hfill {\delta }_s& =c_{1}s+c_2,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =3c_{1}Y+\frac{2A}{C}{c}_{1}s+c_4+c_5{e}^{-\gamma r}.\hfill \end{array}\end{array}$$ (104)

Case 3a: By taking c₂ = 1, c₅ = 1 and all other constants zero, the characteristic equation associated with (104) is

$$\frac{dr}{0}=\frac{ds}{1}=\frac{dY}{e^{-\gamma r}},$$

which implies that r = p and Y(r, s) = se^−γr + Z(p). Using these invariant variables, (103) is transformed into the following ODE

$$\begin{array}{c}\hfill Z^{″}+\gamma Z^{\prime }=0.\end{array}$$ (105)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=c_1+c_2{e}^{-\gamma p},\end{array}$$ (106)

which provides,

$$\begin{array}{c}\hfill Y(r,s)=c_1+(c_2+s)e^{-\gamma r},\end{array}$$ (107)

this implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=c_1+(c_2+\beta )e^{-\gamma \alpha }+\mathrm{\lambda }{e}^{-\alpha \gamma },\end{array}$$ (108)

So, the solution of (80) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=c_1+(c_2+x_1)e^{-\gamma \tau }+x_2{e}^{-x_1\gamma }.\end{array}$$ (109)

Invariant Solutions by Using ${\mathcal{R}}_{15}={\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_4.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_2+\kappa {\mathrm{\Lambda }}_4=\frac{\partial }{\partial U}+\kappa \frac{\partial }{\partial x_2}·$

The characteristic equation associated with Λ₂ + κΛ₄ is given by

$$\frac{d{x}_1}{0}=\frac{d{x}_2}{\kappa }=\frac{d{x}_3}{0}=\frac{d\tau }{0}=\frac{dU}{1}·$$

From the characteristic equation, we can deduce that $\tau =\alpha ,x_1=\beta ,x_3=\mathrm{\lambda },U=\frac{y}{\kappa }+X(\alpha ,\beta ,\mathrm{\lambda })$. Using these invariant variables, (80) is transformed into the equation given by

$$\begin{array}{c}\hfill (-C{X}_{\beta }-A)X_{\beta \beta }+(-D{X}_{\beta }-B)X_{\mathrm{\lambda }\mathrm{\lambda }}-2D{X}_{\mathrm{\lambda }}{X}_{\beta \mathrm{\lambda }}+\gamma X_{\alpha }+X_{\alpha \alpha }=0.\end{array}$$ (110)

Infinitesimals of (110) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_3,\hfill \\ \hfill {\delta }_{\beta }& =c_4,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_5,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_1+c_2{e}^{-\alpha \gamma }.\hfill \end{array}\end{array}$$ (111)

Case 4: By taking c₃ = 1, c₄ = 1 and all other constants zero, the characteristic equation associated with (111) is

$$\frac{d\alpha }{1}=\frac{d\beta }{1}=\frac{d\mathrm{\lambda }}{0}=\frac{dX}{0},$$

which implies that λ = r, −α + β = s and X(α, β, λ) = Y(r, s). Using these invariant variables, (110) is transformed into the following equation

$$\begin{array}{c}\hfill (-C{Y}_s-A+1)Y_{ss}+(-D{Y}_s-B)Y_{rr}-2D{Y}_r{Y}_{rs}-\gamma Y_s=0.\end{array}$$ (112)

Infinitesimals of Eq (112) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_2,\hfill \\ \hfill {\delta }_s& =c_3,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_1.\hfill \end{array}\end{array}$$ (113)

Case 4a: By taking c₁ = 1, c₃ = 1 and all other constants zero, the characteristic equation associated with (113) is

$$\frac{dr}{0}=\frac{ds}{1}=\frac{dY}{1},$$

which implies that r = p, Y(r, s) = s + Z(p). Using these invariant variables, (112) is transformed into the following ODE

$$\begin{array}{c}\hfill (-B-D)Z^{″}-\gamma =0.\end{array}$$ (114)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=\frac{-\gamma }{2(B+D)}{p}^2+c_{1}p+c_2,\end{array}$$ (115)

which provides,

$$\begin{array}{c}\hfill Y(r,s)=\frac{-\gamma }{2(B+D)}{r}^2+c_{1}r+c_2+s,\end{array}$$ (116)

this implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=\frac{-\gamma }{2(B+D)}{\mathrm{\lambda }}^2+c_1\mathrm{\lambda }+c_2+(\beta -\alpha ),\end{array}$$ (117)

So, the solution of (80) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=\frac{y}{\kappa }+\frac{-\gamma }{2(B+D)}{x}_3^2+c_1{x}_3+c_2+(x_1-\tau ).\end{array}$$ (118)

Invariant Solutions by Using ${\mathcal{R}}_6={\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_3.$

First, let us consider the infinitesimal generator ${\mathrm{\Lambda }}_1+\kappa {\mathrm{\Lambda }}_2+\sigma {\mathrm{\Lambda }}_3=\frac{\partial }{\partial \tau }+\kappa \frac{\partial }{\partial U}+\sigma \frac{\partial }{\partial x_1}·$

The characteristic equation associated with Λ₁ + κΛ₂ + σΛ₃ is given by

$$\frac{d{x}_1}{\sigma }=\frac{d{x}_2}{0}=\frac{d{x}_3}{0}=\frac{d\tau }{1}=\frac{dU}{\kappa }·$$

From the characteristic equation, we can deduce that $x_2=\alpha ,x_3=\beta ,\tau -\frac{x_1}{\sigma }=\mathrm{\lambda },U=\frac{\kappa }{\sigma }{x}_1+X(\alpha ,\beta ,\mathrm{\lambda })$. Using these invariant variables, (80) is transformed into the equation given by

$$\begin{array}{c}\hfill \begin{array}{c}\hfill ({\sigma }^3-A\sigma +C{X}_{\mathrm{\lambda }}-C\kappa )X_{\mathrm{\lambda }\mathrm{\lambda }}+{\sigma }^2((-B\sigma +D{X}_{\mathrm{\lambda }}-\kappa D)X_{\beta \beta }+(-B\sigma +D{X}_{\mathrm{\lambda }}-\kappa D)X_{\alpha \alpha }\\ \hfill +\sigma \gamma X_{\mathrm{\lambda }}+2D{X}_{\alpha }{X}_{\alpha \mathrm{\lambda }}+2D{X}_{\beta }{X}_{\beta \mathrm{\lambda }})=0,\end{array}\end{array}$$ (119)

Infinitesimals of (119) are written as

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_{\alpha }& =c_1\beta +c_2,\hfill \\ \hfill {\delta }_{\beta }& =-c_1\alpha +c_5,\hfill \\ \hfill {\delta }_{\mathrm{\lambda }}& =c_4,\hfill \\ \hfill {\mathrm{\Psi }}_X& =c_3.\hfill \end{array}\end{array}$$ (120)

Case 5: By taking c₃ = 1, c₅ = 1 and all other constants zero, the characteristic equation associated with (120) is

$$\frac{d\alpha }{0}=\frac{d\beta }{1}=\frac{d\mathrm{\lambda }}{0}=\frac{dX}{1},$$

which implies that α = r, λ = s, and X(α, β, λ) = β + Y(r, s). Using these invariant variables, (119) is transformed into the following equation

$$\begin{array}{c}\hfill ({\sigma }^3-A\sigma +C{Y}_s-\kappa C)Y_{ss}+2\left(\right(\frac{D}{2}{Y}_s-\frac{B}{2}\sigma -\frac{D}{2}\kappa )Y_{rr}+D{Y}_r{Y}_{rs}+\frac{\sigma }{2}\gamma Y_s){\sigma }^2=0.\end{array}$$ (121)

Infinitesimals of Eq (121) becomes

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\delta }_r& =c_2,\hfill \\ \hfill {\delta }_s& =c_3,\hfill \\ \hfill {\mathrm{\Psi }}_Y& =c_1.\hfill \end{array}\end{array}$$ (122)

Case 5a: By taking c₁ = 1, c₃ = 1 and all other constants zero, the characteristic equation associated with (113) is

$$\frac{dr}{0}=\frac{ds}{1}=\frac{dY}{1},$$

which implies that r = p, Y(r, s) = s + Z(p). Using these invariant variables, (121) is transformed into the following ODE

$$\begin{array}{c}\hfill -((\sigma B+D(\kappa -1))Z^{″}-\gamma \sigma ){\sigma }^2=0,\end{array}$$ (123)

This implies,

$$\begin{array}{c}\hfill Z\left(p\right)=\frac{-\gamma \sigma }{2(\sigma B+D(\kappa -1\left)\right)}{p}^2+c_{1}p+c_2,\end{array}$$ (124)

which provides,

$$\begin{array}{c}\hfill Y(r,s)=\frac{-\gamma \sigma }{2(\sigma B+D(\kappa -1\left)\right)}{r}^2+c_{1}r+c_2+s,\end{array}$$ (125)

this implies,

$$\begin{array}{c}\hfill X(\alpha ,\beta ,\mathrm{\lambda })=\frac{-\gamma \sigma }{2(\sigma B+D(\kappa -1\left)\right)}{\alpha }^2+c_1\alpha +c_2+\mathrm{\lambda }+\beta ,\end{array}$$ (126)

So, the solution of (80) in original variables is

$$\begin{array}{c}\hfill U(x_1,x_2,x_3,\tau )=\frac{\kappa }{\sigma }{x}_1+\frac{-\gamma \sigma }{2(\sigma B+D(\kappa -1\left)\right)}{x}_2^2+c_1{x}_2+c_2+\tau -\frac{x_1}{\sigma }+x_3.\end{array}$$ (127)

7 Partial noether approach

In cases in which the standard Lagrangian is either unavailable or challenging to ascertain, an alternative approach involves the use of a partial Lagrangian. Conservation laws are then derived using the partial Noether approach, which was introduced by Kara and Mahomed [21].

Partial Lagrangian The m^th order differential system (61) can be expressed as follows

$$\begin{array}{c}\hfill H_{\alpha }=H_{\alpha }^0+H_{\alpha }^1=0,\end{array}$$ (128)

A function $\mathcal{L}=\mathcal{L}(x,U,U^{\left(1\right)},\cdots ,U^{\left(n\right)}),n\le m$, is referred to as a partial Lagrangian of the system (61) if the system (61) can be represented in the form

$$\frac{\delta \mathcal{L}}{\delta U^{\alpha }}=f_{\alpha }^{\beta }{H}_{\beta }^1$$

subject to the condition that $H_{\beta }^1\ne 0$ for a certain β. Here, $\left(f_{\alpha }^{\beta }\right)$ denotes an invertible matrix.

Partial Noether Symmetries The operator Λ as defined in (62), which fulfills the condition

$$\begin{array}{c}\hfill \mathrm{\Lambda }\left(\mathcal{L}\right)+\mathcal{L}\left({\mathbf{D}}_i{\mathrm{\Psi }}^i\right)={\mathbf{D}}_i\left({\mathcal{G}}^i\right)+({\mathrm{\Psi }}^{\varpi }-{\mathrm{\Psi }}^j{U}_j^{\varpi })\frac{\delta \mathcal{L}}{\delta U^{\varpi }},i=1,2,\cdots ,n,\varpi =1,2,\cdots ,m,\end{array}$$ (129)

is classified as a partial Noether symmetry generator, corresponding to the partial Lagrangian $\mathcal{L}$.

Conserved Vectors To find the conserved vector related to the system described by Eq (61), which is associated with a symmetry generator (partial Noether) Λ corresponding to the partial Lagrangian $\mathcal{L}$, we utilize Eq (68).

7.1 Partial Noether approach to Eq (80)

In this subsection, we utilize Noether’s partial approach [21] to derive the conservation laws of Eq (80). Initially, we identify the partial Noether symmetries, which are infinitesimal transformations preserving Eq (80). Subsequently, we established associated conservation laws corresponding to these symmetries. The partial Noether’s approach is a powerful method for uncovering conserved quantities. We assume the partial Lagrangian for Eq (80)

$$\begin{array}{c}\hfill \mathcal{L}=\frac{A{U}_{x_1}^2}{2}+\frac{B}{2}(U_{x_2}^2+U_{x_3}^2)+\frac{C{U}_{x_1}^3}{6}-\frac{U_{\tau }^2}{2}+\frac{D}{2}(U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2).\end{array}$$ (130)

By using the Euler operator (65), we get

$$\frac{\delta \mathcal{L}}{\delta U}=-\gamma U_{\tau }·$$

7.2 Partial Noether symmetries

The vector field

$$\mathrm{\Lambda }={\delta }_1\frac{\partial }{\partial x_1}+{\delta }_2\frac{\partial }{\partial x_2}+{\delta }_3\frac{\partial }{\partial x_3}+{\delta }_4\frac{\partial }{\partial \tau }+\mathrm{\Psi }\frac{\partial }{\partial U},$$

is referred to as a partial Noether symmetry of Eq (80) concerning the partial Lagrangian (130) if it fulfills the condition

$$\begin{array}{c}\hfill \begin{array}{cc}& {\mathrm{\Lambda }}^{\left[\mathbf{1}\right]}\mathcal{L}+\mathcal{L}({\mathbf{D}}_{\tau }{\delta }_4+{\mathbf{D}}_{x_1}{\delta }_1+{\mathbf{D}}_{x_2}{\delta }_2+{\mathbf{D}}_{x_3}{\delta }_3)=(\mathrm{\Psi }-{\delta }_1{U}_{x_1}-{\delta }_2{U}_{x_2}-{\delta }_3{U}_{x_3}-{\delta }_4{U}_{\tau })\frac{\delta \mathcal{L}}{\delta U}\hfill \\ & +{\mathbf{D}}_{\tau }{\mathcal{G}}_1+{\mathbf{D}}_{x_1}{\mathcal{G}}_2+{\mathbf{D}}_{x_2}{\mathcal{G}}_3+{\mathbf{D}}_{x_3}{\mathcal{G}}_4,\hfill \end{array}\end{array}$$ (131)

where,

$${\mathrm{\Lambda }}^{\left[1\right]}=\mathrm{\Lambda }+{\mathrm{\Psi }}^{x_1}\frac{\partial }{\partial U_{x_1}}+{\mathrm{\Psi }}^{x_2}\frac{\partial }{\partial U_{x_2}}+{\mathrm{\Psi }}^{x_3}\frac{\partial }{\partial U_{x_3}}+{\mathrm{\Psi }}^{\tau }\frac{\partial }{\partial U_{\tau }},$$

and ${\mathcal{G}}_i(x_1,x_2{x}_3,\tau ,U),i=1,2,3,4$ are gauge terms. We compare the coefficients of derivatives of U in the above equation, we obtain the determining system, and then solving this system gives

$$\begin{array}{c}\hfill \begin{array}{cc}& A{\delta }_{2_{x_2}}-A{\delta }_{1_{x_1}}+A{\delta }_{3_{x_3}}+A{\delta }_{4_{\tau }}+2A{\mathrm{\Psi }}_U+C{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & B{\delta }_{1_{x_1}}-B{\delta }_{2_{x_2}}+B{\delta }_{3_{x_3}}+B{\delta }_{4_{\tau }}+2B{\mathrm{\Psi }}_U+D{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & B{\delta }_{1_{x_1}}+B{\delta }_{2_{x_2}}-B{\delta }_{3_{x_3}}+B{\delta }_{4_{\tau }}+2B{\mathrm{\Psi }}_U+D{\mathrm{\Psi }}_{x_1}=0,\hfill \\ & 2{\mathrm{\Psi }}_U+{\delta }_{1_{x_1}}+{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}-{\delta }_{4_{\tau }}+2\gamma {\delta }_4=0,\hfill \\ & 3{\mathrm{\Psi }}_U-2{\delta }_{1_{x_1}}+{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & 3{\mathrm{\Psi }}_U-{\delta }_{2_{x_2}}+{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & 3{\mathrm{\Psi }}_U+{\delta }_{2_{x_2}}-{\delta }_{3_{x_3}}+{\delta }_{4_{\tau }}=0,\hfill \\ & {\delta }_{1_{x_2}}={\delta }_{1_{x_3}}={\delta }_{1_U}=0,\hfill \\ & {\delta }_{2_{x_1}}={\delta }_{2_U}=0,\hfill \\ & {\delta }_{2_{x_3}}+{\delta }_{3_{x_2}}=0,\hfill \\ & {\delta }_{3_{x_1}}={\delta }_{3_U}=0,\hfill \\ & {\delta }_{4_{x_1}}={\delta }_{4_{x_2}}={\delta }_{4_{x_3}}={\delta }_{4_U}=0,\hfill \\ & {\mathrm{\Psi }}_{x_2}={\mathrm{\Psi }}_{x_3}={\mathcal{G}}_U^3={\mathcal{G}}_U^4=0,\hfill \\ & A{\mathrm{\Psi }}_{x_1}-{\mathcal{G}}_U^2=0,\hfill \\ & \gamma \mathrm{\Psi }-{\mathrm{\Psi }}_{\tau }-{\mathcal{G}}_U^1=0,\hfill \\ & -\gamma {\delta }_1+{\delta }_{1_{\tau }}=0,\hfill \\ & -\gamma {\delta }_2+{\delta }_{2_{\tau }}=0,\hfill \\ & -\gamma {\delta }_3+{\delta }_{3_{\tau }}=0,\hfill \\ & {\mathcal{G}}_{\tau }^1+{\mathcal{G}}_{x_1}^2+{\mathcal{G}}_{x_2}^3+{\mathcal{G}}_{x_3}^4=0.\hfill \end{array}\end{array}$$ (132)

The solution of the above system follows

$${\delta }_1=C_1{e}^{\gamma \tau },{\delta }_2=(C_2{x}_3+C_3)e^{\gamma \tau },{\delta }_3=(-C_2{x}_2+C_4)e^{\gamma \tau },{\delta }_4=0,\mathrm{\Psi }=C_5+C_6{e}^{\gamma \tau }$$

$${\mathcal{G}}^1=C_5\gamma U-X_{x_2}+\int (-h_{x_1}-{\mathcal{G}}_{x_3})dt+p(x_1,x_2,x_3),{\mathcal{G}}^2=h(x_1,x_2,x_3,\tau ),$$

$${\mathcal{G}}^3=X_{\tau },{\mathcal{G}}^4=g(x_1,x_2,x_3,\tau ).$$

By taking f = g = h = p = 0,

we get,

$${\mathcal{G}}^1=C_5\gamma U,{\mathcal{G}}^2={\mathcal{G}}^3={\mathcal{G}}^4=0.$$

Finally, we had the following partial Noether symmetry generators

$${\mathrm{\Lambda }}_1=e^{\gamma \tau }\frac{\partial }{\partial x_1},{\mathrm{\Lambda }}_2=e^{\gamma \tau }\frac{\partial }{\partial x_2},{\mathrm{\Lambda }}_3=e^{\gamma \tau }\frac{\partial }{\partial x_2},{\mathrm{\Lambda }}_4=e^{\gamma \tau }(x_3\frac{\partial }{\partial x_2}-x_2\frac{\partial }{\partial x_3}),$$

$${\mathrm{\Lambda }}_5=e^{\gamma \tau }\frac{\partial }{\partial U},{\mathrm{\Lambda }}_6=\frac{\partial }{\partial U},and{\mathcal{G}}^1=\gamma U.$$

7.3 Conserved vectors

For Λ₁, we have δ₁ = e^γτ, δ₂ = δ₃ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$, we follow from Eq (68)

$$T^{\tau }={\mathcal{G}}_1-\mathcal{L}{\delta }_4-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{\tau }}·$$

This lead to the component of the conserved vector given by,

$$T^{\tau }=-e^{\gamma t}{U}_{x_1}{U}_{\tau }$$

$$T^{x_1}={\mathcal{G}}_2-\mathcal{L}{\delta }_1-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_1}},$$

we get

$$T^{x_1}=e^{\gamma \tau }(\frac{A}{2}{U}_{x_1}^2+\frac{C}{3}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}-\frac{B}{2}{U}_{x_2}^2-\frac{B}{2}{U}_{x_3}^2),$$

$$T^{x_2}={\mathcal{G}}_3-\mathcal{L}{\delta }_2-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_2}},$$

this implies

$$T^{x_y}=e^{\gamma \tau }(B{U}_{x_1}{U}_{x_2}+D{U}_{x_1}^2{U}_{x_2}),$$

$$T^{x_3}={\mathcal{G}}_4-\mathcal{L}{\delta }_3-(\mathrm{\Psi }-U_{x_1}{\delta }_1-U_{x_2}{\delta }_2-U_{x_3}{\delta }_3-U_{\tau }{\delta }_4)\frac{\partial \mathcal{L}}{\partial U_{x_3}},$$

we obtain the result

$$T^{x_3}=e^{\gamma \tau }(B{U}_{x_1}{U}_{x_3}+D{U}_{x_1}^2{U}_{x_3}).$$

For Λ₂, δ₂ = e^γτ, δ₁ = δ₃ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₂ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-e^{\gamma \tau }{U}_{x_2}{U}_{\tau },\hfill \\ & T^{x_1}=e^{\gamma \tau }(A{U}_{x_1}{U}_{x_2}+\frac{C}{2}{U}_{x_1}^2{U}_{x_2}+\frac{D}{2}(U_{x_2}^3+U_{x_2}{U}_{x_3}^2)),\hfill \\ & T^{x_2}=e^{\gamma \tau }(-\frac{A}{2}{U}_{x_1}^2-\frac{C}{6}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}+\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)+\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2)),\hfill \\ & T^{x_3}=e^{\gamma \tau }{U}_{x_2}(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (133)

For Λ₃, δ₃ = e^γτ, δ₁ = δ₂ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₃ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=-e^{\gamma \tau }{U}_{x_3}{U}_{\tau },\hfill \\ & T^{x_1}=e^{\gamma \tau }(A{U}_{x_1}{U}_{x_3}+\frac{C}{2}{U}_{x_1}^2{U}_{x_3}+\frac{D}{2}(U_{x_2}^2{U}_{x_3}+U_{x_3}^3)),\hfill \\ & T^{x_2}=e^{\gamma \tau }{U}_{x_2}(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}),\hfill \\ & T^{x_3}=e^{\gamma \tau }(-\frac{A}{2}{U}_{x_1}^2-\frac{C}{6}{U}_{x_1}^3+\frac{U_{\tau }^2}{2}+\frac{B}{2}(-U_{x_2}^2+U_{x_3}^2)+\frac{D}{2}(-U_{x_1}{U}_{x_2}^2+U_{x_1}{U}_{x_3}^2)).\hfill \end{array}\end{array}$$ (134)

For Λ₄, we have δ₂ = x₃e^γτ, δ₃ = −x₂e^γτ, δ₁ = δ₄ = Ψ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₄ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=(-x_3{U}_{x_2}+x_2{U}_{x_3})e^{\gamma \tau }{U}_{\tau },\hfill \\ & T^{x_1}=e^{\gamma \tau }(z{U}_{x_2}-x_2{U}_{x_3})(A{U}_{x_1}+\frac{C}{2}{U}_{x_1}^2+\frac{D}{2}(U_{x_2}^2+U_{x_3}^2)),\hfill \\ & T^{x_2}=-x_3{e}^{\gamma \tau }(\frac{A}{2}{U}_{x_1}^2+\frac{C}{6}{U}_{x_1}^3-\frac{U_{\tau }^2}{2}-\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)-\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2))-y{U}_{x_2}{U}_{x_3}{e}^{\gamma \tau }(B+D{U}_{x_1}),\hfill \\ & T^{x_3}=x_2{e}^{\gamma \tau }(\frac{A}{2}{U}_{x_1}^2+\frac{C}{6}{U}_{x_1}^3-\frac{U_{\tau }^2}{2}+\frac{B}{2}(U_{x_2}^2-U_{x_3}^2)+\frac{D}{2}(U_{x_1}{U}_{x_2}^2-U_{x_1}{U}_{x_3}^2))+z{U}_{x_2}{U}_{x_3}{e}^{\gamma \tau }(B+D{U}_{x_1}).\hfill \end{array}\end{array}$$ (135)

For Λ₅, Ψ = e^γτ, δ₁ = δ₂ = δ₃ = δ₄ = 0 and ${\mathcal{G}}_1={\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₅ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=e^{\gamma \tau }{U}_{\tau },\hfill \\ & T^{x_1}=e^{\gamma \tau }(-A{U}_{x_1}-\frac{C}{2}{U}_{x_1}^2-\frac{D}{2}(U_{x_2}^2+U_{x_3}^2)),\hfill \\ & T^{x_2}=-e^{\gamma \tau }(B{U}_{x_2}+D{U}_{x_1}{U}_{x_2}),\hfill \\ & T^{x_3}=-e^{\gamma \tau }(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (136)

For Λ₆, we have Ψ = 1, δ₁ = δ₂ = δ₃ = δ₄ = 0 and ${\mathcal{G}}_1=\gamma U,{\mathcal{G}}_2={\mathcal{G}}_3={\mathcal{G}}_4=0$. From Eq (68), the conserved vector associated with Λ₆ becomes

$$\begin{array}{c}\hfill \begin{array}{cc}& T^{\tau }=\gamma U+U_{\tau },\hfill \\ & T^{x_1}=-(A{U}_{x_1}+\frac{C}{2}{U}_{x_1}^2+\frac{D}{2}(U_{x_2}^2+U_{x_3}^2)),\hfill \\ & T^{x_2}=-(B{U}_{x_2}+D{U}_{x_1}{U}_{x_2}),\hfill \\ & T^{x_3}=-(B{U}_{x_3}+D{U}_{x_1}{U}_{x_3}).\hfill \end{array}\end{array}$$ (137)

8 Graphical analysis

In this section, we explore the graphical analysis of the nonlinear elastic wave equations, which proves crucial when dealing with exact solutions. This analysis helps us grasp the intricate nature of complex physical phenomena. Figures labeled as Figs 1–4 exemplify the interplay of two and three-dimensional motion of elastic waves. Notably, Figs 1 and 2 validate the presence of exponential within the elastic wave equations. These figures are based on specific parameter values, providing valuable insights into the behavior of the elastic waves.

Fig 1. Behaviour of three-dimensional elastic motion by (100) with parameters c₁ = 1, c₂ = 3, B = 2 and γ = 1.

Fig 4. Behaviour of two-dimensional elastic motion by (109) with parameters c₁ = 1, c₂ = 2 and γ = 1.

Fig 2. Behaviour of two-dimensional elastic motion by (100) with parameters c₁ = 1, c₂ = 3, B = 2 and γ = 1.

Fig 3. Behaviour of three-dimensional elastic motion by (109) with parameters c₁ = 1, c₂ = 2 and γ = 1.

9 Conclusions

In this study, our main objective was to investigate the (3+1)-dimensional elasticity wave equations using the widely employed Lie group method. This research is novel because no previous studies have applied the Lie group method to these equations. Our exploration provides unique analytical solutions that are invariant under certain transformations or symmetry generators, highlighting the novelty of our results. Additionally, we derived the conservation laws of linear momentum and energy within the model. To achieve this, we formulated classical and partial Lagrangians for the elasticity wave equation and its damped version, respectively. This exploration of conservation laws using Noether’s theorem is presented for the first time in the literature for the (3+1)-dimensional elasticity wave equations, as is the application of the partial Lagrangian approach to these models. Our results demonstrate the effectiveness of the partial Lagrangian approach when a classical Lagrangian is unavailable, extending the applications of variational principles.

We accomplished this goal by thoroughly discussing both Noether’s approach and partial Noether’s approach to elasticity equations, resulting in the successful derivation of conserved vectors. The soundness of the Lie group method was confirmed through our analysis. Furthermore, we found that our adopted technique was also suitable for exploring group invariant solutions. The promising results have motivated us to further research and study additional classes within the elasticity medium. These findings pave the way for future investigations in the field of elasticity equations.

Data Availability

All relevant data are within the paper.

Funding Statement

The author(s) received no specific funding for this work.