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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.
Citation: Hussain A, Usman M, Zaman F, Zidan AM, Herrera J (2025) Noether and partial Noether approach for the nonlinear (3+1)-dimensional elastic wave equations. PLoS ONE 20(1): e0315505. https://doi.org/10.1371/journal.pone.0315505
Editor: Timon Idema, Delft University of Technology, NETHERLANDS, KINGDOM OF THE
Received: May 23, 2024; Accepted: November 21, 2024; Published: January 3, 2025
Copyright: © 2025 Hussain et al. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
Data Availability: All relevant data are within the paper.
Funding: The author(s) received no specific funding for this work.
Competing interests: The authors have declared that no competing interests exist.
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 is linearly dependent on the displacement vector Ui according to the equation
(1)
The non-linear elastic model, in contrast, is founded on the concept of non-linear strain, which is represented by
(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
, is expressed as follows
(3)
where λ and μ represent Lame’s coefficients and a1, a2 are constants. The stress tensor, denoted by τnm, can be calculated using the partial derivatives of the Murnaghan potential for the displacement components
(4)
The Cauchy equation of motion, in the absence of body forces, is given by
(5)
In this context, Uj symbolizes the displacement vector, τij 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 for this case are given by
(6)
(7)
For the three-dimensional case, the Murnaghan potential (3) takes the form
(8)
This leads to the following expression for
(9)
The stress components can be calculated using the Eq (4) and are given by
(10)
The Cauchy equation of motion (5) takes the following form in the three-dimensional case
(11)
and the components of the potential are given by
(12)
Lastly, by utilizing Eq (10), we derive the (3+1)-dimensional nonlinear elastic wave equation as follows
(13)
In (13),
,
,
,
.
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]
Since the equation under consideration is of second order, we prolong our vector field as follows
where,
and
are given as follows
(14)
where Dα denotes the total derivative of α.
The invariance criterion of PDE is
(15)
The invariance condition (15) which lead to the following after some calculations. After performing some calculations we get,
(16)
So the infinitesimals are
(17)
We obtain the following symmetry algebra
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.
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,
(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 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 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 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 .
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 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 , 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 , 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 , 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, 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, , 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, .
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, .
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, .
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, .
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, .
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, .
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, .
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, .
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, .
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, .
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, .
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, .
Case 23: μ8 = 0, μ7 = 0, μ4 = 0, μ3 = 0, μ1 = 0, μ6 = 0, μ2 = 0. Then, (18) is reduced to Λ = μ5Λ5. Therefore, .
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, .
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, .
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, .
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, .
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, .
Case 29: μ8 = 0, μ7 = 0, μ5 = 0, μ3 = 0, μ1 = 0, μ6 = 0, μ2 = 0. Then, (18) is reduced to Λ = μ4Λ4. Thus, .
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, ,
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, .
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, ,
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, .
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, .
Case 35: μ8 = 0, μ7 = 0, μ6 = 0, μ4 = 0, μ1 = 0, μ3 = 0, μ5 = 0. Then, (18) is reduced to Λ = μ2Λ2. Thus, .
Case 36: μ8 = 0, μ7 = 0, μ6 = 0, μ4 = 0, μ1 = 0, μ2 = 0, μ5 = 0. Then, (18) is reduced to Λ = μ3Λ3. Thus, .
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, .
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, .
Hence, the one-dimensional optimal system for Eq (13) is given by,
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
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ1 is given by
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
(19)
Infinitesimals of (19) are written as
(20)
Case 1: By taking c3 = 1 and all other constants zero, the characteristic equation associated with (20) is
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
(21)
Infinitesimals of Eq (21) becomes
(22)
Case 1a: By taking c2 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (22) is
which implies that −r + s = p and Y(r, s) = Z(p). Using these invariant variables, (21) is transformed into the following ODE
(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
Now, if Z″ ≠ 0, from (23), we have A + B + (−C − 3D)Z′ = 0 which gives
which provides
.
This implies,
(24)
Hence, the solution of (13) in original variables is
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ4 is given by
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
(25)
Infinitesimals of (25) are written as
(26)
Case 2: By taking c4 = 1, c5 = 1 and all other constants zero, the characteristic equation associated with (26) is
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
(27)
Infinitesimals of Eq (27) becomes
(28)
Case 2a: By taking c2 = 1, c4 = 1, c6 = 1 and all other constants zero, the characteristic equation associated with (28) is
which implies that −r + s = p and Y(r, s) = r + Z(p). Using these invariant variables, (27) is transformed into the following ODE
(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
Now, if Z″ ≠ 0, then from (29), we have CZ′ + A − 1 = 0 which gives
which provides
.
This implies,
(30)
Hence, the solution of (13) in original variables is
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ5 is given by
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
(31)
Infinitesimals of (31) are written as
(32)
Case 3: By taking c2 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (32) is
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
(33)
Infinitesimals of Eq (33) becomes
(34)
Case 3a: By taking c4 = 1 and all other constants zero, the characteristic equation associated with (34) is
which implies that s = p and Y(r, s) = Z(p). Using these invariant variables, (33) is transformed into the following ODE
(35)
This implies Z(p) = c1p + c2, this gives Y(r, s) = c1s + c2,
This implies,
(36)
So, the solution of (13) in original variables is
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ1 + κΛ3 is given by
From the characteristic equation, we deduce that
. Using these invariant variables, (13) is transformed into the equation given by
(37)
Infinitesimals of (37) are written as
(38)
Case 4: By taking c3 = 1, c6 = 1 and all other constants zero, the characteristic equation associated with (38) is
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
(39)
Infinitesimals of Eq (39) becomes
(40)
Case 4a: By taking c2 = 1, c3 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (40) is
which implies that −r + s = p, Y(r, s) = r + Z(p). Using these invariant variables, (39) is transformed into the following ODE
(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
Now, if Z″ ≠ 0, then from (41), we have (6Dκ2 + C)Z′ + (2B − 1)κ3 − 2Dκ2 + κA − C = 0, which yields
which provides
.
This implies,
(42)
Hence, the solution of (13) in original variables is
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ3 + Λ6 is given by
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
(43)
Infinitesimals of (43) are written as
(44)
Case 5: By taking c13 = 1 and all other constants zero, the characteristic equation associated with (44) is
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
(45)
Infinitesimals of Eq (45) becomes
(46)
Case 5a: By taking c7 = 1 and all other constants zero, the characteristic equation associated with (46) is
which implies that
and Y(r, s) = Z(p). Using these invariant variables, (45) is transformed into the following ODE
(47)
This implies,
(48)
which gives,
(49)
this gives,
(50)
So, the solution of (13) in original variables is
(51)
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ1 + κΛ3 + Λ6 is given by
From the characteristic equation, we deduce that
. Using these invariant variables, (13) is transformed into the equation given by
(52)
Infinitesimals of (52) are written as
(53)
Case 6: By taking c3 = 1 and all other constants zero, the characteristic equation associated with (53) is
which implies that α = r, β = s, and
. Using these invariant variables, (52) is transformed into the following equation
(54)
Infinitesimals of Eq (54) becomes
(55)
Case 6a: By taking c3 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (55) is
which implies that −r + s = p, Y(r, s) = Z(p). Using these invariant variables, (54) is transformed into the following ODE
(56)
This implies,
(57)
which yields,
(58)
this gives,
(59)
So, the solution of (13) in original variables is
(60)
4 Noether’s approach
Let us contemplate a differential system of m-th order
(61)
where xi with i = 1, 2, ⋯, n represents independent variables, and Uα with α = 1, 2, ⋯, m symbolizes the dependent variables. Additionally, the notation Um signifies the m th-order partial derivative.
The Lie-Bäcklund operator is defined as follows
(62)
where
is defined by
In this context, Di denotes the total derivative operator.
The Lie-Bäcklund operator (62), when expressed in characteristic form, takes on the following representation
(63)
where,
constitute Lie characteristic functions.
The Noether operators linked with a Lie-Bäcklund operator Λ are given by
(64)
where
denotes the Euler-Lagrange operator, which is defined as
(65) Lagrangian A Lagrangian for the Eq (61) is described by a function
, which adheres to the Euler-Lagrange equation
(66) Noether Symmetry Generator A Lie-Bäcklund operator Λ is considered a Noether symmetry generator of the Eq (61) with respect to a Lagrangian
if there exist gauge functions
that fulfill the condition
(67) Noether Conserved Vectors Each Noether symmetry generator Λ associated with a given Lagrangian
that relates to the Euler-Lagrange differential equations corresponds to a vector T = (T1, T2, ⋯, Tm). The definition of each component Ti is as follows
(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
(69)
Applying the Euler operator (65) to the Eq (69), we get
(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
The vector field Λ is called a Noether symmetry of Eq (13) corresponding to Lagrangian (69) if it fulfills the condition
(71)
where,
and
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
(72)
The following result follows
(73)
By taking f = g = h = p = 0.
then,
Finally, we get the following Noether symmetry generators
4.1.2 Conserved vectors.
For Λ1, δ1 = 1, δ2 = δ3 = δ4 = Ψ = 0 and . Then by Eq (68) we get
This leads to
simplifying the above equation, we get
this leads to
we obtain
For Λ2, δ2 = 1, δ1 = δ3 = δ4 = Ψ = 0 and . 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)
(74)
For Λ3, δ3 = 1, δ1 = δ2 = δ4 = Ψ = 0 and
. From Eq (68), the conserved vector associated with Λ3 becomes
(75)
For Λ4, we consider δ4 = 1, δ1 = δ2 = δ3 = Ψ = 0 and
. From Eq (68), the conserved vector associated with Λ4 becomes
(76)
For Λ5, we follow Ψ = 1, δ1 = δ2 = δ3 = δ4 = 0 and
. From Eq (68), the conserved vector associated with Λ5 becomes
(77)
For Λ6, we have δ2 = x3, δ3 = −x2, δ1 = δ4 = 0 and
. From Eq (68), the conserved vector associated with Λ6 becomes
(78)
For Λ7, we have Ψ = τ, δ1 = δ2 = δ3 = δ4 = 0 and
. From Eq (68), the conserved vector associated with Λ7 becomes
(79)
5 Damped elastic wave equation
The addition of the damping term Uτ to the non-linear elastic wave Eq (13) transforms it into a non-linear damped elastic wave equation given by
(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
The invariance condition for the Eq (80) is
(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
(82)
The infinitesimals we obtain here are given by
We follow the following symmetry algebra given by
Table 4 lists the commutator relation for the generators of symmetry algebra (5).
Following a similar procedure, we obtain the optimal system of symmetry algebra of Eq (80) given by,
6 Group invariant solutions
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ1 is given by
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
(83)
Infinitesimals of (83) are written as
(84)
Case 1: By taking c5 = 1, c6 = 1 and all other constants zero, the characteristic equation associated with (84) is
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
(85)
Infinitesimals of Eq (85) becomes
(86)
Case 1a: By taking c3 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (86) is
which implies that r = p and Y(r, s) = s + Z(p). Using these invariant variables, (85) is transformed into the following ODE
(87)
This implies,
(88)
which provides,
(89)
this implies,
(90)
So, the solution of (80) in original variables is
(91)
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ3 is given by
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
(92)
Infinitesimals of (92) are written as
(93)
Case 2: By taking c3 = 1 and all other constants zero, the characteristic equation associated with (93) is
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
(94)
Infinitesimals of Eq (94) becomes
(95)
Case 2a: By taking c5 = 1, c6 = 1 and all other constants zero, the characteristic equation associated with (95) is
which implies that −r + s = p and Y(r, s) = s + Z(p). Using these invariant variables, (94) is transformed into the following ODE
(96)
This implies,
(97)
which provides,
(98)
this implies,
(99)
So, the solution of (80) in original variables is
(100)
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ5 is given by
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
(101)
Infinitesimals of (101) are written as
(102)
Case 3: By taking c2 = 1, c5 = 1 and all other constants zero, the characteristic equation associated with (102) is
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
(103)
Infinitesimals of Eq (94) becomes
(104)
Case 3a: By taking c2 = 1, c5 = 1 and all other constants zero, the characteristic equation associated with (104) is
which implies that r = p and Y(r, s) = se−γr + Z(p). Using these invariant variables, (103) is transformed into the following ODE
(105)
This implies,
(106)
which provides,
(107)
this implies,
(108)
So, the solution of (80) in original variables is
(109)
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ2 + κΛ4 is given by
From the characteristic equation, we can deduce that
. Using these invariant variables, (80) is transformed into the equation given by
(110)
Infinitesimals of (110) are written as
(111)
Case 4: By taking c3 = 1, c4 = 1 and all other constants zero, the characteristic equation associated with (111) is
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
(112)
Infinitesimals of Eq (112) becomes
(113)
Case 4a: By taking c1 = 1, c3 = 1 and all other constants zero, the characteristic equation associated with (113) is
which implies that r = p, Y(r, s) = s + Z(p). Using these invariant variables, (112) is transformed into the following ODE
(114)
This implies,
(115)
which provides,
(116)
this implies,
(117)
So, the solution of (80) in original variables is
(118)
Invariant Solutions by Using
First, let us consider the infinitesimal generator
The characteristic equation associated with Λ1 + κΛ2 + σΛ3 is given by
From the characteristic equation, we can deduce that
. Using these invariant variables, (80) is transformed into the equation given by
(119)
Infinitesimals of (119) are written as
(120)
Case 5: By taking c3 = 1, c5 = 1 and all other constants zero, the characteristic equation associated with (120) is
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
(121)
Infinitesimals of Eq (121) becomes
(122)
Case 5a: By taking c1 = 1, c3 = 1 and all other constants zero, the characteristic equation associated with (113) is
which implies that r = p, Y(r, s) = s + Z(p). Using these invariant variables, (121) is transformed into the following ODE
(123)
This implies,
(124)
which provides,
(125)
this implies,
(126)
So, the solution of (80) in original variables is
(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 mth order differential system (61) can be expressed as follows
(128)
A function
, is referred to as a partial Lagrangian of the system (61) if the system (61) can be represented in the form
subject to the condition that
for a certain β. Here,
denotes an invertible matrix.
Partial Noether Symmetries The operator Λ as defined in (62), which fulfills the condition
(129)
is classified as a partial Noether symmetry generator, corresponding to the partial Lagrangian
.
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 , 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) (130)
By using the Euler operator (65), we get
7.2 Partial Noether symmetries
The vector field
is referred to as a partial Noether symmetry of Eq (80) concerning the partial Lagrangian (130) if it fulfills the condition
(131)
where,
and
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
(132)
The solution of the above system follows
By taking f = g = h = p = 0,
we get,
Finally, we had the following partial Noether symmetry generators
7.3 Conserved vectors
For Λ1, we have δ1 = eγτ, δ2 = δ3 = δ4 = Ψ = 0 and , we follow from Eq (68)
This lead to the component of the conserved vector given by,
we get
this implies
we obtain the result
For Λ2, δ2 = eγτ, δ1 = δ3 = δ4 = Ψ = 0 and
. From Eq (68), the conserved vector associated with Λ2 becomes
(133)
For Λ3, δ3 = eγτ, δ1 = δ2 = δ4 = Ψ = 0 and
. From Eq (68), the conserved vector associated with Λ3 becomes
(134)
For Λ4, we have δ2 = x3eγτ, δ3 = −x2eγτ, δ1 = δ4 = Ψ = 0 and
. From Eq (68), the conserved vector associated with Λ4 becomes
(135)
For Λ5, Ψ = eγτ, δ1 = δ2 = δ3 = δ4 = 0 and
. From Eq (68), the conserved vector associated with Λ5 becomes
(136)
For Λ6, we have Ψ = 1, δ1 = δ2 = δ3 = δ4 = 0 and
. From Eq (68), the conserved vector associated with Λ6 becomes
(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.
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.
Acknowledgments
The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through large group Research Project under grant number RGP.2/16/45.
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