EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6128 ISSN 1307-5543 – ejpam.com Published by New York Business Global Group Analysis of a Class of Nonlinear Wave Equations M. Usman1, Akhtar Hussain2,∗, M. Umar Farooq1, Jorge Herrera3 1 College of Electrical and Mechanical Engineering (CEME), National University of Sciences and Technology (NUST), H-12 Islamabad 44000, Pakistan 2 Department of Mathematics, University of Management and Technology, Lahore 54770, Pakistan 3 Facultad de Ciencias Naturales e Ingenieria, Universidad de Bogota Jorge Tadeo Lozano, Bogota 110311, Colombia Abstract. We study the nature of a (2+1)-dimensional nonlinear wave equation using the Lie symmetry analysis method. This problem is reduced to ordinary differential equations (ODEs) using non-similar subalgebras of Lie symmetries. We presented explicit solutions by solving the reduced ODEs. The conserved vectors were constructed using the Lagrange multiplier method. Using these conserved vectors, we also derived the exact solutions of the nonlinear wave equation. Consequently, 3D graphics were used to analyze and illustrate the graphical representations of the solutions. 2020 Mathematics Subject Classifications: 70G65, 35-XX, 70S10, 22E70 Key Words and Phrases: Lie symmetries, conservation laws, invariant solutions, nonlinear wave equations 1. Introduction The study of nonlinear wave phenomena relies extensively on nonlinear partial differ- ential equations (NLPDEs). The interplay between larger-amplitude waves spreading with lower-amplitude waves is the factor that causes the nonlinearity. Finding exact solutions is an essential problem because these equations explain the characteristics and behaviors of nonlinear phenomena. It is important to mention that various methods have been em- ployed to address these nonlinear problems, including the φ6-model expansion method [1], the exp-function method [2], the symmetry method [3], the residual power series approach [4], the homogeneous balanced method [5], the hyperbolic tangent method [6], F-expansion method [7], the unified method [8, 9], the new Jacobi elliptic functions technique [10] and the improved Sardar sub-equation approach [11]. One of the most effective ways to obtain analytical solutions for NLPDEs is to apply Lie symmetry [3]. Any solution to an NLPDE can be converted into a collection of solutions to ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6128 Email address: akhtarhussain21@sms.edu.pk (A. Hussain) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 2 of 29 the same equation via symmetry. Conservation laws play a major role in the development and analysis of various mathematical models. The most efficient technique for determining the conservation laws for NLPDEs is the multiplier method [3, 12]. It is commonly recognized that the main objective of different linear and nonlinear wave equations is to model certain wave phenomena. Ames et al. [13] used the Lie group method to study the general nonlinear form of the (1+1)-dimensional wave equation given by utt − (k(u)ux)x = 0. (1) Bluman et al. [14] constructed the conserved vectors for (1), while the higher-order con- servation laws and the complete categorization of Lie algebra for (1) are presented in [15]. Different types of nonlinear wave equations are presented in [16–22]. The general form of (2+1)-dimensional nonlinear wave equation [23] is given by wtt − (f(w)wx)x − (g(w)wy)y = 0, (2) where the nonlinear functions f(w) and g(w) determine how waves propagate. Usually, this equation simulates the propagation of waves in a nonlinear or nonhomogeneous mediums, where the response of the medium is determined by the amplitude of the wave. Two- dimensional nonlinear acoustic waves, surface water waves, and other waveforms capturing phenomena, such as wave steepening and shock production, can be modeled using (2). The formulation f(w) = αw, g(w) = βw2 in (2) can simulate the nonlinear dispersive waves including solitons in shallow water. By taking f(w) = α, g(w) = βw in (2), the beam dynamics in nonlinear elastic materials can be modeled, where the response is proportional to the deformation. In [23], the double reduction theory is applied on (2) based on the conserved vectors constructed using a partial Lagrangian. In this study, we take the nonlinear wave equation (2) in (2+1) dimensions and use the Lie group approach to obtain its solutions. Several interesting similarity reductions were performed using similar conjugacy classes of Lie algebra, which were computed using the matrix method. Invariant solutions were obtained for these symmetry reductions. The local conserved vectors for (2) were constructed and utilized to obtain the exact solutions of (2). For certain parameter values, 3D images of the solutions in various structures were created to obtain a better analysis of the solutions. The remainder of this paper is structured as follows: Section 2 is devoted to the clas- sification of the Lie symmetries and the construction of the associated optimal set for (2). Section 3 addresses symmetry reduction and invariant solutions. The local conservation laws are presented in Section 4, which produce the exact solutions stated in Section 5. Section 6 presents a geometric analysis of the obtained results. 2. Lie point symmetries and optimal system This section explains the fundamental terms required to calculate the infinitesimals of (2). We take a Lie group of infinitesimal transformations with one parameter that acts on M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 3 of 29 the dependent variable w and the independent variables x, y, and t of Eq. (2) x̃ → x+ ςζ1(x, y, t, w) +O(ς2), ỹ → y + ςζ2(x, y, t, w) +O(ς2), t̃ → t+ ςζ3(x, y, t, w) +O(ς2), w̃ → w + ςΦ(x, y, t, w) +O(ς2), (3) with ς � 1 being a parameter of the group and ζ1, ζ2, ζ3 and Φ are the infinitesimal functions for the variables x, y, t and w respectively, which should be computed later. The vector field associated with the Lie group of transformations (3) takes the form Y = ζ1 ∂ ∂x + ζ2 ∂ ∂y + ζ3 ∂ ∂t +Φ ∂ ∂w · (4) The operator Y can be identified as the Lie point symmetry generator of (2) if it meets the invariance criterion [3] Y [2](wtt − (f(w)wx)x − (g(w)wy)y)|(2) = 0, (5) where Y [2] denotes the prolongation of Y up to order two. We obtain a system of linear coupled PDEs for infinitesimals by setting the coefficients of the dependent variable and its differential to zero. These coupled PDEs also referred to as determining equations, are solved to produce the infinitesimals, we then substitute these infinitesimals in Eq. (4) to extract the Lie point symmetries for (2), which are stated as follows: Case-1: f(w),g(w) are arbitrary [23]. In this case, the solution of (5) provides a four-dimensional Lie algebra of (2) spanned by the following vector fields Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + y ∂ ∂y + t ∂ ∂t . (6) The Lie algebra for (2) is extended for the cases presented as follows: Case-2: f(w) = αw, g(w) = βw2. Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + 2t ∂ ∂t − 2w ∂ ∂w , Y5 = y ∂ ∂y − t ∂ ∂t + 2w ∂ ∂w . (7) Case-3: f(w) = α, g(w) = βw: Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = t ∂ ∂x + x α ∂ ∂t , Y5 = y ∂ ∂y + 2w ∂ ∂w , Y6 = x ∂ ∂x + t ∂ ∂t − 2w ∂ ∂w (8) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 4 of 29 Case-4: f(w) = eαw, g(w) = eβw: Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + βt (β − α) ∂ ∂t + 2 (α− β) ∂ ∂w , Y5 = y ∂ ∂y + αt (−β + α) ∂ ∂t + 2 (β − α) ∂ ∂w (9) Case-5: f(w) = wα, g(w) = wβ: Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + βt (β − α) ∂ ∂t − 2w1+βw1+α wβ (β − 1)w1+α − w1+βwα (α− 1) ∂ ∂w , Y5 = y ∂ ∂y + αt (−β + α) ∂ ∂t + 2w1+βw1+α wβ (β − 1)w1+α − w1+βwα (α− 1) ∂ ∂w (10) Case-6: f(w) = α lnw, g(w) = β lnw: Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + y ∂ ∂y + t ∂ ∂t , Y5 = y ∂ ∂x − βx α ∂ ∂y (11) Case-7: f(w) = α, g(w) = β: [22] Y1 = ∂ ∂x , Y2 = ∂ ∂y , Y3 = ∂ ∂t , Y4 = x ∂ ∂x + y ∂ ∂y + t ∂ ∂t , Y5 = w ∂ ∂w , Y6 = t ∂ ∂y + y β ∂ ∂t , Y7 = t ∂ ∂x + x α ∂ ∂t , Y8 = y ∂ ∂x − βx α ∂ ∂y , Y9 = xt ∂ ∂x + yt ∂ ∂y + ( y2 2β + x2 2α + t2 2 ) ∂ ∂t − tw 2 ∂ ∂w , Y10 = xy ∂ ∂x + ( β t2 + y2 ) α− β x2 2α ∂ ∂y + yt ∂ ∂t − yw 2 ∂ ∂w , Y11 = ( −β t2 + y2 ) α− β x2 2α ∂ ∂x − βxy α ∂ ∂y − xβt α ∂ ∂t + xβw 2α ∂ ∂w (12) 2.1. Optimal System The optimal systems for the symmetry generators of (2) are derived using the algorithm described in [3]. The adjoint action representation [15] is written as follows Ad(exp(ςYi)Yj) = Yj − ς[Yi,Yj ] + ς2 2! [Yi, [Yi,Yj ]]− . . . . (13) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 5 of 29 Computation of Basic Invariants Let f be a function on Lie algebra L such that f(Ad(exp(ςY)X )) = f(X ) ∀ Y ∈ L, ς ∈ R. (14) Now, if we write f(y1Y1 + . . .+ ynYn) = f(y1 + . . .+ yn), then for the basis {Y1, . . . ,Yn}, we get [24] ∑ 1≤i,i≤N yiΘj([Yk,Yi]) ∂f ∂yi = 0. (15) The basic invariants in the adjoint representations are obtained from the solution of the linear system of PDEs (15). The adjoint transformation matrix [24] is given by A = n∏ i=1 Ai, Ai = Ad(eςiYi). (16) Let Y = n∑ i=1 kiYi, Ỹ = n∑ i=1 k̃iYi ∈ L. For simplicity, we write Y = (k1 k2 · · · kn), Ỹ = (k̃1 k̃2 · · · k̃n). Then, Y and Ỹ are in the same conjugacy class, if we have [24] Ỹ = YA. (17) The solution of (17) gives the values of ςi in terms of ki, which can be used to simplify k̃i. 2.1.1. Optimal system for Case 1 The non-zero commutators of Lie algebra L4 presented in (6) are given by [Y1,Y4] = Y1, [Y2,Y4] = Y2, [Y3,Y4] = Y3. (18) The adjoint action representations of Lie symmetries (6) are obtained using (13) and are presented in Table 1. Table 1: Adjoint Table Ad(eς) Y1 Y2 Y3 Y4 Y1 Y1 Y2 Y3 Y4 − ςY1 Y2 Y1 Y2 Y3 Y4 − ςY2 Y3 Y1 Y2 Y3 Y4 − ςY3 Y4 eςY1 eςY2 eςY3 Y4 The adjoint transformation matrix of L4 is constructed using (16) and is shown below M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 6 of 29 A =  eς4 0 0 0 0 eς4 0 0 0 0 eς4 0 −ς1e ς4 −ς2e ς4 −ς3e ς4 1  Using (17), we obtain the following system of equations, which is used to obtain ςi: k̃1 = k1e ς4 − k4ς1e ς4 , k̃2 = k2e ς4 − k4ς2e ς4 , k̃3 = k3e ς4 − k4ς3e ς4 , k̃4 = k4. (19) The basic invariants for (6) can be calculated by finding a solution of the following system of linear PDEs, which is derived using (15). k4 ∂ϑ ∂k1 = 0, k4 ∂ϑ ∂k2 = 0, k4 ∂ϑ ∂k3 = 0, k1 ∂ϑ ∂k1 + k2 ∂ϑ ∂k2 + k3 ∂ϑ ∂k3 = 0. (20) The solution of (20) gives ϑ(k1, k2, k3, k4) = F (k4). So, the first basic invariant is k4, which will be the first vertix of the tree. k4 k4 = 0 k3 = 0, k1 6= 0 k2 = 0 Case-5 k2 6= 0 Case-4 k3 6= 0, k1 6= 0 k2 = 0 Case-3 k2 6= 0 Case-2 k4 6= 0 Case-1 Case-1: For k4 6= 0, the representative element is S1 = Y4. By substituting k̃4 = 1 in (19), we get ς1 = k1 k4 , ς2 = k2 k4 , ς3 = k3 k4 . When k4 = 0, the solution of (20) provides ϑ(k1, k2, k3, k4) = F (k2k1 , k3 k1 , k4). This implies M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 7 of 29 that the new invariants are k2 k1 , k3k1 . Since k1 6= 0, so the next two vertices will be k3 and k2. Case-2: For k4 = 0, k3 6= 0, k2 6= 0, k1 6= 0, the corresponding element of the optimal system is S2 = Y1 + Y2 + Y3. Case-3: For k4 = 0, k3 6= 0, k1 6= 0, k2 = 0, the associated element is S3 = Y1 + Y3. Case-4: For k4 = 0, k3 = 0, k1 6= 0, k2 6= 0, the conjugacy class is of the form S4 = Y1+Y2. Case-5: For k4 = 0, k3 = 0, k1 6= 0, k2 = 0, the representative element is S5 = Y1. Hence, the one-dimensional optimal system for Lie algebra (6) is given by S1 = Y4, S2 = Y1 + Y2 + Y3, S3 = Y1 + Y3, S4 = Y1 + Y2, S5 = Y1. (21) 2.1.2. Optimal system for Case 2 For Lie algebra L5 given in (7), we have the following non-zero commutators [Y1,Y4] = Y1, [Y2,Y5] = Y2, [Y3,Y4] = 2Y3, [Y3,Y5] = −Y3. (22) The adjoint representations of (7) are given in Table 2. Table 2: Adjoint Table Ad(eς) Y1 Y2 Y3 Y4 Y5 Y1 Y1 Y2 Y3 Y4 − ςY1 Y5 Y2 Y1 Y2 Y3 Y4 Y5 − ςY2 Y3 Y1 Y2 Y3 Y4 − 2ςY3 Y5 + ςY3 Y4 eςY1 Y2 e2ςY3 Y4 Y5 Y5 Y1 eςY2 e−ςY3 Y4 Y5 The adjoint transformation matrix of L5 is given as A =  eς4 0 0 0 0 0 eς5 0 0 0 0 0 e2ς4−ς5 0 0 −ς1e ς4 0 −2ς3e 2ς4−ς5 1 0 0 −ς2e ς5 ς3e 2ς4−ς5 0 1  Eq. (17) yields the following system of equations k̃1 = k1e ς4 − k4ς1e ς4 , k̃2 = k2e ς5 − k5ς2e ς5 , k̃3 = k3e 2ς4−ς5 − 2k4ς3e 2ς4−ς5 + k5ς3e 2ς4−ς5 , k̃4 = k4, k̃5 = k5. (23) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 8 of 29 By using the formula (15), we obtain the following system of linear PDEs k4 ∂ϑ ∂k1 = 0, k5 ∂ϑ ∂k2 = 0, k5 ∂ϑ ∂k3 − 2k4 ∂ϑ ∂k3 = 0, k1 ∂ϑ ∂k1 + 2k3 ∂ϑ ∂k3 = 0, k2 ∂ϑ ∂k2 + k3 ∂ϑ ∂k3 = 0. (24) By solving equation (24), we get Φ(k1, k2, k3, k4, k5) = F (k4, k5). So, the first two vertices of the tree are k4 and k5. k5 k5 = 0 k4 = 0 k2 = 0, k1 6= 0 k3 = 0 Case-7 k3 6= 0 Case-6 k2 6= 0, k1 6= 0 k3 = 0 Case-5 k3 6= 0 Case-4 k4 6= 0 Case-3 k5 6= 0 k4 = 0 Case-2 k4 6= 0 Case-1 Case-1: For k5 6= 0, k4 6= 0, we have M1 = Y4 + Y5. By substituting k̃4 = k̃5 = 1 in (23), we obtain ς1 = k1 k4 , ς2 = k2 k5 , ς3 = k3 2k4−k5 . Case-2: For k5 6= 0, k4 = 0, we have M2 = Y5 + aY1, a ∈ R. By substituting k̃1 = k̃5 = 1 in (23), we obtain ς2 = k2 k5 , ς3 = k3 2k4−k5 . Case-3: For k5 = 0, k4 6= 0, we have M3 = Y4 + aY2, a ∈ R. By substituting k̃2 = k̃4 = 1 in (23), we obtain ς1 = k1 k4 , ς3 = k3 2k4−k5 . When k4 = k5 = 0, the solution of (24) gives ϑ(k1, k2, k3, k4, k5) = F (k2k3 k21 , k4, k5). This means that the new invariant is k2k3 k21 . Since k1 6= 0, so the next two vertices will be k2 and k3. Case-4: For k4 = k5 = 0, k3 6= 0, k2 6= 0, k1 6= 0, we have M4 = Y1 + Y2 + Y3. Case-5: For k4 = k5 = 0, k3 = 0, k2 6= 0, k1 6= 0, we have M5 = Y1 + Y2. Case-6: For k4 = k5 = 0, k3 6= 0, k2 = 0, k1 6= 0, we have M6 = Y1 + Y3. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 9 of 29 Case-7: For k4 = k5 = 0, k2 = k3 = 0, k1 6= 0, we have M7 = Y1. So, the one-dimensional optimal system of subalgebras of Lie algebra (7) is presented as M1 = Y4 + Y5, M2 = Y5 + aY1, a ∈ R, M3 = Y4 + aY2, a ∈ R, M4 = Y1 + Y2 + Y3, M5 = Y1 + Y2, M6 = Y1 + Y3, M7 = Y1. (25) 2.1.3. Optimal system for Case 3 For Lie algebra L6 presented in (8), the non-zero commutators are provided by [Y1,Y4] = Y3 α , [Y1,Y6] = Y1, [Y2,Y5] = Y2, [Y3,Y4] = Y1, [Y3,Y6] = Y3. (26) In Table 3, the adjoint actions of (8) are presented. Table 3: Adjoint Table Ad(eς) Y1 Y2 Y3 Y4 Y5 Y6 Y1 Y1 Y2 Y3 Y4 − ς α Y3 Y5 Y6 − ςY1 Y2 Y1 Y2 Y3 Y4 Y5 − ςY2 Y6 Y3 Y1 Y2 Y3 Y4 − ςY1 Y5 Y6 − ςY3 Y4 (e ς√ α −e −ς√ α ) 2 √ α Y3 + (e ς√ α +e −ς√ α ) 2 Y1 Y2 (e ς√ α +e −ς√ α ) 2 Y3 − (e −ς√ α −e ς√ α ) √ α 2 Y1 Y4 Y5 Y6 Y5 Y1 eςY2 Y3 Y4 Y5 Y6 Y6 eςY1 Y2 eςY3 Y4 Y5 Y6 For Lie algebra (8), the adjoint transformation matrix is presented as follows A =  (e ς4√ α+e −ς4√ α )eς6 2 0 (e ς4√ α−e −ς4√ α )eς6 2 0 0 0 0 eς5 0 0 0 0 − √ α(e −ς4√ α −e ς4√ α )eς6 2 0 (e ς4√ α+e −ς4√ α )eς6 2 0 0 0 −ς3(e ς4√ α+e −ς4√ α )eς6 2 + ς1(e −ς4√ α −e ς4√ α )eς6 2 √ α 0 −ς3(e ς4√ α−e −ς4√ α )eς6 2 √ α − ς1(e ς4√ α+e −ς4√ α )eς6 2α 1 0 0 0 −ς2e ς5 0 0 1 0 −ς1(e ς4√ α+e −ς4√ α )eς6 2 + ς3 √ α(e −ς4√ α −e ς4√ α )eς6 2 0 −ς1(e ς4√ α−e −ς4√ α )eς6 2 √ α − ς3 √ α(e ς4√ α+e − ς4√ α )eς6 2 0 0 1  By using adjoint transformation matrix in (17), we obtain the following system of equations M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 10 of 29 k̃1 = k1 (e ς4√ α + e −ς4√ α )eς6 2 − k3 √ α(e −ς4√ α − e ς4√ α )eς6 2 + k4 (−ς3(e ς4√ α + e −ς4√ α )eς6 2 + ς1(e −ς4√ α − e ς4√ α )eς6 2 √ α ) + k6 (−ς1(e ς4√ α + e −ς4√ α )eς6 2 + ς3 √ α(e −ς4√ α − e ς4√ α )eς6 2 ) , k̃2 = k2e ς5 − k5ς2e ς5 , k̃3 = k1 (e ς4√ α − e −ς4√ α )eς6 2 + k3 (e ς4√ α + e −ς4√ α )eς6 2 + k4 (−ς3(e ς4√ α − e −ς4√ α )eς6 2 √ α − ς1(e ς4√ α + e −ς4√ α )eς6 2α ) + k6 (−ς1(e ς4√ α − e −ς4√ α )eς6 2 √ α − ς3 √ α(e ς4√ α + e − ς4√ α )eς6 2 ) , k̃4 = k4, k̃5 = k5, k̃6 = k6. (27) For the computation of basic invariants, we have the following system of linear PDEs derived from (15). k6 ∂ϑ ∂k1 + k4 α ∂ϑ ∂k3 = 0, k5 ∂ϑ ∂k2 = 0, k4 ∂ϑ ∂k1 + k6 ∂ϑ ∂k3 = 0, k3 ∂ϑ ∂k1 + k1 α ∂ϑ ∂k3 = 0, k2 ∂ϑ ∂k2 = 0, k1 ∂ϑ ∂k1 + k3 ∂ϑ ∂k3 = 0. (28) This implies Φ(k1, k2, k3, k4, k5, k6) = F (k4, k5, k6). So, the invariants k4, k5 and k6 are the vertices of the tree. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 11 of 29 k6 k6 = 0 k5 = 0 k4 = 0 Case-8 k4 6= 0 Case-7 k5 6= 0 k4 = 0 Case-6 k4 6= 0 Case-5 k6 6= 0 k5 = 0 k4 = 0 Case-4 k4 6= 0 Case-3 k5 6= 0 k4 = 0 Case-2 k4 6= 0 Case-1 Case-1: For k6 6= 0, k5 6= 0, k4 6= 0, we have N1 = Y4+Y5+Y6. By substituting k̃4 = k̃5 = k̃6 = 1 in (27), we obtain ς1 = √ αk1k6+k3 ( k4 α + √ α 2 k6 2) , ς2 = k2 k5 , ς3 = k1 k4 − k6 2k4 ( √ αk1k6+k3 ( k4 α + √ α 2 k6 2) ) , ς4 = 0. Case-2: For k6 6= 0, k5 6= 0, k4 = 0, we have N2 = Y5 + Y6. By substituting k̃5 = k̃6 = 1 in (27), we obtain ς1 = 2k1 k6 , ς2 = k2 k5 , ς3 = k3√ αk6 , ς4 = 0. Case-3: For k6 6= 0, k5 = 0, k4 6= 0, we have N3 = Y6 +Y4 + aY2, a ∈ R. By substituting k̃2 = k̃4 = k̃6 = 1 in (27), we obtain ς1 = √ αk1k6+k3 ( k4 α + √ α 2 k6 2) , ς3 = k1 k4 − k6 2k4 ( √ αk1k6+k3 ( k4 α + √ α 2 k6 2) ) , ς4 = 0. Case-4: For k6 6= 0, k5 = 0, k4 = 0, we have N4 = Y6 + aY2, a ∈ R. By substituting k̃2 = k̃6 = 1 in (27), we obtain ς1 = 2k1 k6 , ς3 = k3√ αk6 , ς4 = 0. Case-5: For k6 = 0, k5 6= 0, k4 6= 0, we have N5 = Y4 + Y5. By substituting k̃4 = k̃5 = 1 in (27), we obtain ς1 = αk3 k4 , ς2 = k2 k5 , ς3 = k1 k4 , ς4 = 0. Case-6: k6 = 0, k5 6= 0, k4 = 0. For ς2 = k2 k5 , ς4 = ( ln (k1−k3 k1+k3 ) )√ α 2 , we get N6 = Y5 + aY1, a ∈ R. If we choose ς2 = k2 k5 , ς4 = ( ln ( √ αk3−k1√ αk3+k1 ) )√ α 2 , we get N7 = Y5 + aY3, a ∈ R. Case-7: For k6 = 0, k5 = 0, k4 6= 0, we have N8 = Y4 + aY2, a ∈ R. By substituting k̃2 = k̃4 = 1 in (27), we obtain ς1 = αk3 k4 , ς3 = k1 k4 , ς4 = 0. Case-8: k6 = 0, k5 = 0, k4 = 0. For ς4 = ( ln (k1−k3 k1+k3 ) )√ α 2 , we get N9 = Y2 + aY1, a ∈ R. If we choose ς4 = ( ln ( √ αk3−k1√ αk3+k1 ) )√ α 2 , we get N10 = Y2 + aY3, a ∈ R. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 12 of 29 So, the one-dimensional optimal system of subalgebras of Lie algebra (8) is presented as N1 = Y4 + Y5 + Y6, N2 = Y5 + Y6, N3 = Y6 + Y4 + aY2, a ∈ R, N4 = Y6 + aY2, a ∈ R, N5 = Y4 + Y5, N6 = Y5 + aY1, a ∈ R, N7 = Y5 + aY3, a ∈ R, N8 = Y4 + aY2, a ∈ R, N9 = Y2 + aY1, a ∈ R, N10 = Y2 + aY3, a ∈ R. (29) 3. Group invariant solutions 3.1. Similarity reductions for Case 1 Case-a: Consider Y1 = ∂ ∂x . The solution of the characteristic equation for Y1 provides the symmetry invariants p = y, q = t, w(x, y, t) = H(p, q). (30) By inserting (30) in (2), we get Hqq − g′Hp 2 − gHpp = 0. (31) Infinitesimals for (31) are as follows ζp = c1p+ c2, ζq = c1q + c3, ΦH = 0. (32) Case-a1: In (32), set c3 = 1 and ci = 0 for i = 1, 2. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (33) The substitution of (33) in (31) yields the following ODE gµ′′ + g′µ′2 = 0. (34) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 13 of 29 If we take g = b1µ+ b2, then the solution of (34) takes the form µ(r) = −b2 + √ 2c1b1r + 2c2b1 + b2 2 b1 , (35) which implies H(p, q) = −b2 + √ 2c1b1p+ 2c2b1 + b2 2 b1 . (36) Hence, the solution of (2) is w(x, y, t) = −b2 + √ 2c1b1y + 2c2b1 + b2 2 b1 , (37) where c1 and c2 are constants of integration. This solution is same, as obtained in [22]. Case-b: Consider Y1 + Y2 = ∂ ∂x + ∂ ∂y . The solution of the characteristic equation for Y1 + Y2 provides the symmetry invariants p = t, q = y − x, w(x, y, t) = H(p, q). (38) By inserting (38) in (2), we get Hpp − (f ′ + g′)Hq 2 − (f + g)Hqq = 0. (39) Infinitesimals for (39) are as follows ζp = c1p+ c2, ζq = c1q + c3, ΦH = 0. (40) Case-b1: In (40), set c2 = 1 and ci = 0 for i = 1, 3. Then we obtain the symmetry invariants r = q, H(p, q) = µ(r). (41) The substitution of (41) in (39) yields the following ODE (f + g)µ′′ + (f ′ + g′)µ′2 = 0. (42) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 14 of 29 If we take f = a1µ 2 + a2, g = b1µ, then the solution of (42) takes the form µ(r) = 1 2a1 ( 12c1ra1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2r2a1 2 + 72c1c2ra1 2 + 36c1ra1a2b1 − 6c1rb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1 ) 1 3 − (4a1a2 − b1 2) / ( 2a1(12c1ra1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2r2a1 2 + 72c1c2ra1 2 + 36c1ra1a2b1 − 6c1rb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1) 1 3 ) − b1 2a1 , (43) which implies H(p, q) = 1 2a1 ( 12c1qa1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2q2a1 2 + 72c1c2qa1 2 + 36c1qa1a2b1 − 6c1qb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1 ) 1 3 − (4a1a2 − b1 2) / ( 2a1(12c1qa1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2q2a1 2 + 72c1c2qa1 2 + 36c1qa1a2b1 − 6c1qb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1) 1 3 ) − b1 2a1 . (44) Hence, the solution of (2) is w(x, y, t) = 1 2a1 ( 12c1qa1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2q2a1 2 + 72c1c2qa1 2 + 36c1qa1a2b1 − 6c1qb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1 ) 1 3 − (4a1a2 − b1 2) / ( 2a1(12c1qa1 2 + 12c2a1 2 + 6a1a2b1 − b1 3 + 2(36c1 2q2a1 2 + 72c1c2qa1 2 + 36c1qa1a2b1 − 6c1qb1 3 + 36c2 2a1 2 + 36c2a1a2b1 − 6c2b1 3 + 16a1a2 3 − 3a2 2b1 2) 1 2a1) 1 3 ) − b1 2a1 , (45) where c1 and c2 are constants of integration. 3.2. Similarity reductions for Case 2 Case-a: Consider Y1 + Y3 = ∂ ∂x + ∂ ∂t . The solution of the characteristic equation for Y1 + Y3 provides the symmetry invariants p = y, q = −x+ t, w(x, y, t) = H(p, q). (46) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 15 of 29 By inserting (46) in (2), we get Hqq − αHHqq − αHq 2 − βH2Hpp − 2βHHp 2 = 0. (47) Infinitesimals for (47) are as follows ζp = c1p+ c2, ζq = c1q + c3, ΦH = 0. (48) Case-a1: In (48), set c2 = 1 and ci = 0 for i = 1, 3. Then we obtain the symmetry invariants r = q, H(p, q) = µ(r). (49) The substitution of (49) in (47) yields the following ODE µ′′ − αµµ′′ − αµ′2 = 0. (50) The solution of (50) takes the form µ(r) = 1− √ 1 + (2c1r + 2c2)α α , (51) which implies H(p, q) = 1− √ 1 + (2c1q + 2c2)α α . (52) Hence, the solution of (2) is w(x, y, t) = 1− √ 1 + (2c1(t− x) + 2c2)α α , (53) where c1 and c2 are constants of integration. Case-a2: In (48), set c3 = 1 and ci = 0 for i = 1, 2. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (54) The substitution of (54) in (47) yields the following ODE µ2µ′′ + 2µµ′2 = 0. (55) The solution of (55) takes the form µ(r) = (3c1r + 3c2) 1 3 , (56) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 16 of 29 which implies H(p, q) = (3c1p+ 3c2) 1 3 . (57) Hence, the solution of (2) is w(x, y, t) = (3c1y + 3c2) 1 3 , (58) where c1 and c2 are constants of integration. Case-a3: In (48), set c2 = c3 = 1 and c1 = 0. Then we obtain the symmetry invariants r = q − p, H(p, q) = µ(r). (59) The substitution of (59) in (47) yields the following ODE −βµ2µ′′ − 2βµµ′2 − αµµ′′ − αµ′2 + µ′′ = 0. (60) The solution of (60) takes the form µ(r) = 1 2β ( 12c1β 2r+12c2β 2−α3+2(36c1 2β2r2+72c1c2β 2r−6c1α 3r−36c1αβr+36c2 2β2− 6c2α 3 − 36c2αβ − 3α2 − 16β) 1 2β − 6αβ ) 1 3 + ( α2 + 4β )/( 2β(12c1β 2r + 12c2β 2 − α3 + 2(36c1 2β2r2+72c1c2β 2r−6c1α 3r−36c1αβr+36c2 2β2−6c2α 3−36c2αβ−3α2−16β) 1 2β− 6αβ) 1 3 ) − α 2β , which implies H(p, q) = 1 2β ( 12c1β 2(q − p) + 12c2β 2 − α3 + 2(36c1 2β2(q − p)2 + 72c1c2β 2(q − p) − 6c1α 3(q − p) − 36c1αβ(q − p) + 36c2 2β2 − 6c2α 3 − 36c2αβ − 3α2 − 16β) 1 2β − 6αβ ) 1 3 +( α2 + 4β )/( 2β(12c1β 2(q − p) + 12c2β 2 − α3 + 2(36c1 2β2(q − p)2 + 72c1c2β 2(q − p) − 6c1α 3(q− p)− 36c1αβ(q− p) + 36c2 2β2 − 6c2α 3 − 36c2αβ − 3α2 − 16β) 1 2β − 6αβ) 1 3 ) − α 2β . Hence, the solution of (2) is w(x, y, t) = 1 2β ( 12c1β 2(t−x−y)+12c2β 2−α3+2(36c1 2β2(t−x−y)2+72c1c2β 2(t−x−y)− 6c1α 3(t−x−y)−36c1αβ(t−x−y)+36c2 2β2−6c2α 3−36c2αβ−3α2−16β) 1 2β−6αβ ) 1 3 +( α2 + 4β )/( 2β(12c1β 2(t−x−y)+12c2β 2−α3+2(36c1 2β2(t−x−y)2+72c1c2β 2(t−x−y)− 6c1α 3(t−x−y)−36c1αβ(t−x−y)+36c2 2β2−6c2α 3−36c2αβ−3α2−16β) 1 2β−6αβ) 1 3 ) − α 2β , where c1 and c2 are constants of integration. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 17 of 29 Case-b: Consider Y1 = ∂ ∂x . The solution of the characteristic equation for Y1 provides the symmetry invariants p = t, q = y, w(x, y, t) = H(p, q). (61) By inserting (61) in (2), we get Hpp − βH2Hqq − 2βHHq 2 = 0. (62) Infinitesimals for (62) are as follows ζp = c1p+ c2, ζq = c3q + c4, ΦH = −H(c1 − c3). (63) Case-b1: In (63), set c4 = 1 and ci = 0 for i = 1, 2, 3. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (64) The substitution of (64) in (62) yields the following ODE µ′′ = 0. (65) The solution of (65) takes the form µ(r) = c1r + c2, (66) which implies H(p, q) = c1p+ c2. (67) Hence, the solution of (2) is w(x, y, t) = c1t+ c2, (68) where c1 and c2 are constants of integration. This solution is also obtained in [22]. Case-b2: In (63), set c2 = c4 = 1 and ci = 0 for i = 1, 3. Then we obtain the symmetry invariants r = q − p, H(p, q) = µ(r). (69) The substitution of (69) in (62) yields the following ODE µ′′ − βµ2µ′′ − 2βµµ′2 = 0. (70) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 18 of 29 The solution of (70) takes the form µ(r) = ( 2 1 3 (2 1 3 ( √ −4 + 9(−c1r − c2)2β + (3c1r + 3c2) √ β) 2 3 + 2) )/( 2 √ β( √ −4 + 9(−c1r − c2)2β + (3c1r + 3c2) √ β) 1 3 ) . (71) which implies H(p, q) = ( 2 1 3 (2 1 3 ( √ −4 + 9(−c1(q − p)− c2)2β + (3c1(q − p) + 3c2) √ β) 2 3 + 2) )/( 2 √ β ( √ −4 + 9(−c1(q − p)− c2)2β + (3c1(q − p) + 3c2) √ β) 1 3 ) . (72) Hence, the solution of (2) is w(x, y, t) = ( 2 1 3 (2 1 3 ( √ −4 + 9(−c1(y − t)− c2)2β + (3c1(y − t) + 3c2) √ β) 2 3 + 2) )/( 2 √ β ( √ −4 + 9(−c1(y − t)− c2)2β + (3c1(y − t) + 3c2) √ β) 1 3 ) , (73) where c1 and c2 are constants of integration. Case-b3: In (63), set c3 = 1 and ci = 0 for i = 1, 2, 4. Then we obtain the symmetry invariants r = p, H(p, q) = qµ(r). (74) The substitution of (74) in (62) yields the following ODE µ′′ − 2βµ3 = 0. (75) The solution of (75) takes the form µ(r) = c2 JacobiSN ( (ι √ βr + c1)c2, ι ) , (76) which implies H(p, q) = c2 JacobiSN ( (ι √ βp+ c1)c2, ι ) q. (77) Hence, the solution of (2) is w(x, y, t) = c2 JacobiSN ( (ι √ βt+ c1)c2, ι ) y, (78) where c1 and c2 are constants of integration. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 19 of 29 3.3. Similarity reductions for Case 3 Case-a: Consider Y2 = ∂ ∂y . The solution of the characteristic equation for Y2 provides the symmetry invariants p = t, q = x, w(x, y, t) = H(p, q). (79) By inserting (79) in (2), we get Hpp − αHqq = 0. (80) Infinitesimals for (80) are as follows ζp = f5(q + √ αp) + f6(q − √ αp), ζq = √ αf5(q + √ αp)− √ αf6(q − √ αp) + c2, ΦH = c1H+ f3(q + √ αp) + f4(q − √ αp). (81) Case-a1: In (81), set c2 = 1, c1 = 0 and all arbitrary functions zero. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (82) The substitution of (82) in (80) yields the following ODE µ′′ = 0. (83) The solution of (83) takes the form µ(r) = c1r + c2, (84) which implies H(p, q) = c1p+ c2. (85) Hence, the solution of (2) is w(x, y, t) = c1t+ c2, (86) where c1 and c2 are constants of integration. Case-a2: In (81), set c1 = c2 = 1 and all arbitrary functions zero. Then we obtain the symmetry invariants r = p, H(p, q) = eqµ(r). (87) The substitution of (87) in (80) yields the following ODE µ′′ − αµ = 0. (88) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 20 of 29 The solution of (88) takes the form µ(r) = (c1e 2 √ αr + c2)e − √ αr, (89) which implies H(p, q) = (c1e 2 √ αp + c2)e q− √ αp. (90) Hence, the solution of (2) is w(x, y, t) = (c1e 2 √ αt + c2)e x− √ αt, (91) where c1 and c2 are constants of integration. Case-b: Consider Y5 + aY3 = y ∂ ∂y + 2w ∂ ∂w + a ∂ ∂t . The solution of the characteristic equation for Y5 + aY3 provides the symmetry invariants p = x, q = t− a ln (y), w(x, y, t) = y2H(p, q). (92) By inserting (92) in (2), we get −a2βHHqq − 6βH2 + 7βHHq − βa2Hq 2 − αHpp +Hqq = 0. (93) Infinitesimals for (93) are as follows ζp = c1, ζq = c2, ΦH = 0. (94) Case-b1: In (94), set c2 = 1 and c1 = 0. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (95) The substitution of (95) in (93) yields the following ODE αµ′′ + 6βµ2 = 0. (96) The solution of (96) takes the form µ(r) = −WeierstrassP (r + c1, 0, c2)α β , (97) which implies H(p, q) = −WeierstrassP (p+ c1, 0, c2)α β . (98) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 21 of 29 Hence, the solution of (2) is w(x, y, t) = −WeierstrassP (x+ c1, 0, c2)αy 2 β , (99) where c1 and c2 are constants of integration. Case-c: Consider Y2 + aY3 = ∂ ∂y + a ∂ ∂t . The solution of the characteristic equation for Y2 + aY3 provides the symmetry invariants p = x, q = t− ay, w(x, y, t) = H(p, q). (100) By inserting (100) in (2), we get Hqq − αHpp − a2βHHqq − a2βHq 2 = 0. (101) Infinitesimals for (101) are as follows ζp = c1p+ c2, ζq = c3q + c4, ΦH = −2(c1 − c3)(a 2βH− 1) a2β . (102) Case-c1: In (102), set c2 = 1 and other constants zero. Then we obtain the symmetry invariants r = q, H(p, q) = µ(r). (103) The substitution of (103) in (101) yields the following ODE µ′′ − a2βµµ′′ − a2βµ′2 = 0. (104) The solution of (104) takes the form µ(r) = 1− √ 1 + 2a2β(c1r + c2) a2β , (105) which implies H(p, q) = 1− √ 1 + 2a2β(c1q + c2) a2β . (106) Hence, the solution of (2) is w(x, y, t) = 1− √ 1 + 2a2β(c1(t− ay) + c2) a2β , (107) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 22 of 29 where c1 and c2 are constants of integration. Case-c2: In (102), set c4 = 1 and other constants zero. Then we obtain the symmetry invariants r = p, H(p, q) = µ(r). (108) The substitution of (108) in (101) yields the following ODE µ′′ = 0. (109) The solution of (109) takes the form µ(r) = c1r + c2, (110) which implies H(p, q) = c1p+ c2. (111) Hence, the solution of (2) is w(x, y, t) = c1x+ c2, (112) where c1 and c2 are constants of integration. Case-c3: In (102), set c2 = c4 = 1 and other constants zero. Then we obtain the symmetry invariants r = q − p, H(p, q) = µ(r). (113) The substitution of (113) in (101) yields the following ODE (−a2βµ− α+ 1)µ′′ − a2βµ′2 = 0. (114) The solution of (114) takes the form µ(r) = −α+ 1− √ 2(c1r + c2)a2β + (α− 1)2 a2β , (115) which implies H(p, q) = −α+ 1− √ 2(c1(q − p) + c2)a2β + (α− 1)2 a2β . (116) Hence, the solution of (2) is w(x, y, t) = −α+ 1− √ 2(c1(t− ay − x) + c2)a2β + (α− 1)2 a2β , (117) where c1 and c2 are constants of integration. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 23 of 29 4. Conservation Laws This section demonstrates how to use the multiplier approach [12] to compute conser- vation laws for (2). The multipliers Q(x, y, t, w) for (2) can be computed by solving the equation given by [25] δ δw (Q (wtt − (f(w)wx)x − (g(w)wy)y)) = 0. (118) Equation (118) yields the following multiplier functions Q1 = xyt, Q2 = xt, Q3 = xy, Q4 = x, Q5 = yt, Q6 = t, Q7 = y, Q8 = 1. (119) The conserved vectors for (2) can be derived by substituting the multipliers (119) in the following relation [25] Q(wtt − (f(w)wx)x − (g(w)wy)y) = DtΓ t +DxΓ x +DyΓ y, (120) and are formulated as follows Γ1 =  Γ1 t = xy(twt − w), Γ1 x = t 2(x 2g(w)wy − 2xyf(w)wx + 2y ∫ fdw), Γ1 y = −xt 2 g(w)(xwx + 2ywy). (121) Γ2 =  Γ2 t = x(twt − w), Γ2 x = t(−xf(w)wx + ∫ fdw), Γ2 y = −xtg(w)wy. (122) Γ3 =  Γ3 t = xywt, Γ3 x = y ∫ fdw − xyf(w)wx + x2 2 g(w)wy, Γ3 y = −x 2g(w)(xwx + 2ywy). (123) Γ4 =  Γ4 t = xwt, Γ4 x = −xf(w)wx + ∫ fdw, Γ4 y = −xg(w)wy. (124) Γ5 =  Γ5 t = y(twt − w), Γ5 x = t(xg(w)wy − yf(w)wx), Γ5 y = −t(xwx + ywy)g(w). (125) Γ6 =  Γ6 t = twt − w, Γ6 x = −tf(w)wx, Γ6 y = −tg(w)wy. (126) Γ7 =  Γ7 t = ywt, Γ7 x = xg(w)wy − yf(w)wx, Γ7 y = −g(w)(xwx + ywy). (127) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 24 of 29 Γ8 =  Γ8 t = wt, Γ8 x = −f(w)wx, Γ8 y = −g(w)wy. (128) The conserved vectors Γ2,Γ4,Γ6,Γ8 are same, which are derived in [23] by using a partial Lagrangian. 5. Exact solutions via conservation laws This section deals with the obtention of exact solutions for (2) by using its conserved vectors. By using the technique explained in [26, 27], the conserved vector (Γt,Γx,Γy) of (2) satisfies the condition stated as DtΓ t = 0, DxΓ x = 0, DyΓ y = 0. (129) For f(w) = αw, g(w) = βw2, the conserved vector (122) becomes Γ2 =  Γ2 t = x(twt − w), Γ2 x = αtw 2 (w − 2xwx), Γ2 y = −βxtw2wy. (130) The following system is acquired by the insertion of conserved vector (130) in (129). wtt = 0, wx(−2xwx + w) + w(−wx − 2xwxx) = 0, 2wy 2 + wwyy = 0. (131) The exact solution of (2) is founded by solving the system (131) and is described as follows w(x, y, t) = √√√√2c3(c1y + c2) 2 3x+ 2 ( −(12c1y + 12c2) 1 3 4 + ι √ 3(12c1y + 12c2) 1 3 4 )2 t + 2c4(c1y + c2) 1 3 (312 1 3 − 2c3(18) 1 3x+ 4c3 2x2) 1 4√ − 53073 6692 ιc3x+3ι √ 3+9 6c32(18) 1 3 x2−318 2 3 c3x+27 . (132) For f(w) = α, g(w) = βw, the conserved vector (121) takes the following form Γ1 =  Γ1 t = wt, Γ1 x = −αwx, Γ1 y = −βwwy. (133) M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 25 of 29 By substituting the conserved vector (133) in (129), we get wtt = 0, wxx = 0, wy 2 + wwyy = 0. (134) This provides the solution of (2) given by w(x, y, t) = −xt √ 2c1y + 2c2 + √ c1y + c2(c3x+ c4t+ c5). (135) The solutions obtained in this section are not invariant under the Lie algebra. 6. Graphical representation of solutions In the present section, 3D graphical representations of the obtained findings are given. Geometrical analysis is provided because mathematical expressions are insufficient to de- scribe the physical wave patterns. Figure 1 describes the 3D graphics of (73) in the range of variables 1 ≤ t ≤ 2, 1 ≤ y ≤ 3 and 1 ≤ t ≤ 2, 1 ≤ y ≤ 25 by assuming c1 = c2 = 1 and β = 1. In figure 2, an exponential wave pattern of (91) is shown in the range of variables 1 ≤ t ≤ 2, 1 ≤ x ≤ 30 and 1 ≤ t ≤ 20, 1 ≤ x ≤ 4 by assuming c1 = c2 = 1 and α = 1. Figure 3 describes the 3D graphics of (135) in the range of vari- ables 1 ≤ x ≤ 2, 1 ≤ y ≤ 2, t = 2 and 1 ≤ x ≤ 20, 1 ≤ y ≤ 60, t = 40 by assuming c1 = c2 = c3 = c4 = c5 = 1. (a) (b) Figure 1: 3D graphic depictions of (73), by considering c1 = c2 = 1 and β = 1. M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 26 of 29 (a) (b) Figure 2: 3D graphic depictions of (91) by considering c1 = c2 = 1 and α = 1. (a) t=2 (b) t=40 Figure 3: 3D graphic depictions of (135) by considering c1 = c2 = c3 = c4 = c5 = 1. 7. Conclusions Through the implementation of the Lie group analysis method, we investigated the characteristics of a (2+1)-dimensional nonlinear wave equation. We built an optimal set of subalgebras of Lie algebra, which are one-dimensional, and inspected its representatives in order to investigate invariant solutions and similarity reductions, which provide a mul- M. Usman, A. Hussain, M. U. Farooq, J. Herrera / Eur. J. Pure Appl. Math, 18 (4) (2025), 6128 27 of 29 titude of explicit exact solutions using computerized symbolic computation. We applied the Lagrange multiplier approach to generate conserved vectors. We have also obtained precise solutions to the nonlinear wave equation through the use of these conserved vec- tors. Consequently, graphical representations of the answers are analyzed and illustrated using 3D visuals. Ethics approval and consent to participate: All the authors approve their con- sent. Consent for publication: The authors approve this version for publication. 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