OPTICAL SOLITONS AND LIE SYMMETRY ANALYSIS OF KAUP–NEWELL EQUATION Arshdeep Kaur1, Anupma Bansal2, Rajeev Budhiraja3 1,3Department of Mathematics MMEC, Maharishi Markandeshwar (Deemed to be University) Mullana, Ambala-133203 (Haryana), India E-mail: dhimanarshdeep500@gmail.com 2Department of Mathematics D.A.V. College for Women, Ferozepur-152001(Punjab), India. E-mail: anupma2512@yahoo.co.in Abstract This manuscript is based on the Kaup-Newell equation, specifically contemplated to be a kind of nonlinear Schrödinger equation. One of the frameworks derived from the well-known nonlinear Schrödinger equation is the Kaup–Newell equation which is one of the three derivative forms of the nonlinear Schrödinger equation that is frequently explored in nonlinear Optics. A systematic investigation of Lie symmetry method is pertained to derive the symmetry reductions of the given equation. By utilizing these symmetry reductions, the nonlinear Partial differential equation are transformed into the Ordinary differential equations. Consequently, the solutions are obtained with arbi- trary parameters through various methods which are in the form of hyperbolic functions, rational functions and trigonometric functions. Keywords: Lie Classical method, Kaup–Newell equation, G ′ G –Expansion Method, Symme- try reductions, Optical Solitons. 1 Introduction The Schrödinger equation is a core concept in Quantum mechanics with widespread appli- cations in different areas. This equation is referred as a partial differential equation that describes wave function of physical systems like fluid dynamics, waves in water, etc. Op- tical outcomes [1-5] of nonlinear Schrödinger equation are fundamental solutions to design the high-performance communication systems, understanding wave interactions in nonlin- ear media and pushing forward technologies such as super continuum generation and pulse compression. In this direction, Exact traveling wave solutions are key to grasping wave dynamics in nonlinear media, influencing fields such as Optics, Fluid dynamics and more. They provide insights into pulse behavior in optical fibers and other nonlinear systems. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 89 Article History: Received: 12-01-2025, Revised: 15-02-2025, Accepted: 01-03-2025 Thereby, in this supervision, multivarious adequate methods for finding out accurate and exact solutions to the Schrödinger equation are determined and evolved in the course of the numerous former decades. Besides the assorted distinct methods, Lie group method, also stated as Lie symmetry method [11-20], is apparently a significant method to calculate the solutions of nonlinear partial differential equations. The Lie symmetry methodology is smoothly appertained to many physical and engineering methods. Under the Lie Group of point transformations when particular differential equations are invariant then reduction exists. By utilizing these symmetry reductions, the nonlinear partial differential equations are transformed into ordinary differential equations. Then employing various methods and techniques exact traveling wave solutions are accumulated. 1.1 Governing Model The Kaup–Newell equation is given by: qt + ιaqxx + b(|q|2q)x = 0, (1.1) where ‘q(x, t)′ is a complex-valued function, ‘a′ demonstrates velocity dispersion parameter, ‘b′ the nonlinearity coefficient. The Kaup-Newell equation is a specific type of nonlinear partial differential equation derived from the nonlinear Schrödinger equation. It describes wave phenomena and is particularly relevant in field such as plasma physics and optics. The Kaup-Newell equation can model various physical situations involving stability, interactions and soliton solutions. 2 Lie Symmetry Analysis Recently, symmetry analysis has become popular for studying differential equations. Lie Symmetry analysis is the most effective method for evaluating the precise and exact solu- tions of nonlinear partial differential equations. In this document, we procure symmetry reductions, the symmetries and the solutions of Kaup-Newell equation [6-10] by emphasiz- ing Lie classical method. To determine the symmetries of the Kaup-Newell equation, firstly we consider the complex function q(x,t) as q(x, t) = u(x, t) + ιv(x, t). (2.1) The equation can be expressed in terms of its real and imaginary components as- ut − avxx + 2bu2ux + 2buvvx + bu2ux + bv2ux = 0, (2.2) vt + auxx + 2buvux + 2bv2vx + bu2vx + bv2vx = 0. (2.3) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 90 The one parameter Lie group of infinitesimal transformations constitutes as aforementioned- ũ = u+ εη(x, t, u, v) +O(ε2), ṽ = v + εφ(x, t, u, v) +O(ε2), x̃ = x+ εξ(x, t, u, v) +O(ε2), t̃ = t+ ετ(x, t, u, v) +O(ε2), (2.4) where ε is group parameter. Applying these one-parameter Lie group of infinitesimal trans- formations, we obtain the invariance condition of the equation (1.1): ηt − aφxx + 2bu2ηx + 4buuxη + 2buvxφ+ 2buvφx + 2bvvxη + bu2ηx + 2buuxη +bv2ηx + 2bvuxφ = 0, (2.5) φt + aηxx + 2buvηx + 2buuxφ+ 2bvuxη + 2bv2φx + 4bvvxφ+ bu2φx + 2buvxη +bv2φx + 2bvvxφ = 0, (2.6) The mentioned transformation abandons the entire set of resolutions of the equation (1.1) invariant and brings about an over determined linear system of equations for the infinitesi- mals ξ(x, t, u, v), τ(x, t, u, v), η(x, t, u, v). By using these values of the infinitesimal ηt,φt, ηx, φx, ηxx and φxx into equation (2.5) and (2.6), we equalize the alike powers of differentials to 0. After this, we will be able to pertain the system of determining equations as under: (i) τx = 0, τu = 0, τv = 0 (ii) ξu = 0, ξv = 0 (iii) − aφv + aηu + 2aξx − aτt = 0 (iv) φuu = 0, φuv = 0, φvv = 0 (v) − 3u2ξx + 3u2τtb+ 2uvbφu − 2uvbηv − v2bξx + v2bτt + 6buη + 2bvφ− 2aφxu − ξt = 0 (vi) 2u2bηv + 2uvbφv − 2uvbηu − 2uvbξx + 2uvbτt − 2bv2ηv + 2buφ+ 2bvη + aξxx − 2aφxv = 0 (vii) φu = −ηv (viii) v2bηx + 2uvbφx − aφxx + ηt + 3u2bηx = 0 (ix) ηvv = 0, ηuu = 0, ηuv = 0 (x) − bu2ξx + bu2τt − 2buvφu + 2buvηv − 3bv2ξx + 3bv2τt + 2buη + 6bvφ+ 2aηxv − ξx = 0 (xi) − 2bu2φu − 2buvξx + 2buvηu + 2buvτt − 2buvφv + 2bv2φu + 2buφ+ 2bvη + 2aηxu − aξxx = 0 (xii) − 2aξx + aηu + aτt − aφv = 0 (xiii) φu + ηv = 0 (xiv) aηxx + 2buvηx + 3bv2φx + bu2φx + φt = 0 (2.7) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 91 By working through these determining equations, the below-mentioned solutions will be obtained- η = uC2 − vC1, φ = uC1 + vC2, ξ = −2xC2 + C3, τ = −4tC2 + C4, (2.8) where C1, C2, C3, C4 illustrates the arbitrary constants. From the above obtained Lie symmetries the vector fields are as follows: V1 = u ∂ ∂v − v ∂ ∂u , V2 = v ∂ ∂v − u ∂ ∂u − 2x ∂ ∂x − 4t ∂∂t , V3 = ∂ ∂x , V4 = ∂ ∂t . (2.9) By utilizing the various linear combinations of the above mentioned vector fields i.e., V1, V2, V3, V4 we avail the similarity reductions of the Kaup-Newell equation. 3 Reductions and Exact Solutions of Kaup-Newell Equation A prominent motive to calculate symmetries of differential equations aims to utilize those to generate the exact solutions and obtain symmetry reductions. In this subsection, we use the different combinations of vector fields and reduce the Kaup-Newell equation to ordinary differential equations in each case to further explore the exact solutions. We will consider the following vector fields for the reduction of the system of equations: (1) V1 + αV3 + βV4, (2) V2, (3) V3 + ωV4. (3.1) In each case, by using characteristics equation, as under, we will try to obtain the reductions: dx ξ = dt τ = du η = dv φ . (3.2) 3.1 Vector field V1 + αV3 + βV4 On using the characteristic equations (3.2) we acquire the similarity variables as shown below- q = F (ξ)e ι( t β +G(ξ)) , ξ = βx− αt. (3.3) Using equation (3.3) in equation (2.2) and (2.3), we get ordinary differential equation as: −aβ2F (ξ)G′(ξ) 2 + F (ξ)( 1 β − αG′(ξ)) + aβ2F ′′(ξ) + bβF (ξ)3G′(ξ) = 0, (3.4) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 92 and 2β2F ′(ξ)G′(ξ) + β2F (ξ)G′′(ξ) + 3bβF ′(ξ)F (ξ)2 − αF ′(ξ) = 0. (3.5) In this context, the prime(’) signifies the differentiation w.r.t variable ξ. On Solving these equations we get F (ξ) = K1, (3.6) where K1 is any constant and G(ξ) = ( bc2β − α+ √ b2c4β2 − 2bc2αβ + 4aβ + α2 2aβ2 ) ξ + C1. (3.7) Using these values we get the solution of Kaup-Newell equation as q(x, t) = K1e ι ( t β + ( bc2β−α+ √ b2c4β2−2bc2αβ+4aβ+α2 2aβ2 ) (βx−αt)+C1 ) . (3.8) 3.2 Vector field V2 By Solving the characteristic equation (3.2) aforementioned similarity variable is calculated as u = F (ξ)t −1 4 , v = G(ξ)t −1 4 , ξ = x2 t , (3.9) On using equation (3.9) in equation (2.2) and (2.3), we get ordinary differential equation as: −1 4 F (ξ)−ξF ′(ξ)−4aξG′′(ξ)−2aG′(ξ)+6b √ ξF ′(ξ)+2b √ ξG2(ξ)F ′(ξ)+4b √ ξG(ξ)G′(ξ) = 0, and −1 4 G(ξ)−ξG′(ξ)+2aF ′(ξ)+4aξF ′′(ξ)+4bF (ξ)G(ξ)F ′(ξ) √ ξ+6b √ ξG2(ξ)G′(ξ)+2b √ ξF 2(ξ)G′(ξ) = 0. (3.10) where prime(’) denotes the differentiation w.r.t variable ξ. Due to complexity of the reduced equations, we just obtain a constant solution for the equation. 3.3 Vector field V3 + ωV4 Similarity variables are acquired in the light of the characteristic equation as: ξ = x− t ω , q = F (ξ)eιG(ξ). (3.11) On applying these values we acquire the aforementioned system of Ordinary differential equations as − 1 ω F (ξ)G′(ξ) + aF ′′(ξ)− aF (ξ)(G′(ξ))2 + bF (ξ)3G′(ξ) = 0, (3.12) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 93 and − 1 ω F ′(ξ)− 2aF ′(ξ)G′(ξ)− aF (ξ)G′′(ξ) + 3bF (ξ)2F ′(ξ) = 0. (3.13) Multiply equation(3.13) with F (ξ) and on integrating we obtain −aG′(ξ)F (ξ)2 − F (ξ)2 2ω + 3 4 bF (ξ)4 = 0, (3.14) Above equation leads to: G′(ξ) = −12aω + 3b 4a F (ξ)2 +B, (3.15) where B is referred to as the arbitrary constant. On substituting this value in (3.12) and again multiply with F ′(ξ) and integrating, we get b2 32a F (ξ)6 − b 8aw F (ξ)4 + 1 8aω2 F (ξ)2 + a 2 (F ′(ξ))2 = 0, (3.16) Substitute (F (ξ))2 = P (ξ) into (3.16) equation, we get b2ω2P (ξ)4 − 4bωP (ξ)3 + 4P (ξ)2 + 4a2ω2(P ′(ξ))2 = 0. (3.17) On Solving the equation (3.17), we acquire the solutions which are given underneath: (i)q(x, t) = √ 2 bωe ι(−12aω+ 3ω 2a +B)(x− t ω )+C2 , (ii)q(x, t) = √ 1 bω − tanh ( C1− 1 2 ι(x− t ω ) ωa ) bω × e ι ( −12aω(x− t ω )+ 3(x− t ω ) 4aω + 3ι log(tanh(C1− ι 2 (x− t ω ))−1) 4aω + 3ι log(tanh(C1− ι 2 (x− t ω ))+1) 4aω +B(x− t ω ) ) (3.18) 3.4 Exact traveling wave solutions using (G ′ G )-Expansion Method In this section, we aim to determine wave solutions by using (G ′ G )-Expansion Method for equation (3.17). The progression of wave solutions originates from these procedures: 1. Taking into Consideration, the structure of Ordinary Differential Equation (3.17) is formulated as ( G′ G ) : F̃ (χ) = aq ( G′ G )q + aq−1 ( G′ G )q−1 + ........ (3.19) Here aq are constants to be determine, where q ranges from 0 to infinity. As G = G(χ) persuades the LDE of second order which is of the form written below: G′′ + λ̃G′ + µ̃G = 0. (3.20) where aq(aq 6= 0); where q is called the balance number, aq−1,.....a0, λ̃ and µ̃ are constants to be determine later. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 94 2. Determining the positive integer q in (3.19), by equalizing the higher order nonlinear terms to the higher order derivatives in (3.17) 3. Replacing (3.19) into (3.17) and pertain the ODE (3.20), then gather all the entire terms of ( G′ G ) containing homogeneous power, and equalize each coefficient to zero, resulting in a variety of algebraic equations for anaysis aq, aq−1,..... a0 ,c, λ̃ and µ̃. 4. The general solution of (3.20) is noted, then substituting the values of aq in (3.19) we will get the exact solutions of (1.1) The general solutions of Equation(3.20) can be written as ( G′ G ) =  √ λ̃2−4µ̃ 2 ( C1sinh ( 1 2 √ λ̃2−4µ̃ ) χ+C2cosh ( 1 2 √ λ̃2−4µ̃ ) χ C1cosh ( 1 2 √ λ̃2−4µ̃ ) χ+C2sinh ( 1 2 √ λ̃2−4µ̃ ) χ ) − λ̃ 2 , λ̃ 2 − 4µ̃ > 0 √ 4µ̃−λ̃2 2 ( −C1sin ( 1 2 √ 4µ̃−λ̃2 ) χ+C2cos ( 1 2 √ 4µ̃−λ̃2 ) χ C1cos ( 1 2 √ 4µ̃−λ̃2 ) χ+C2sin ( 1 2 √ 4µ̃−λ̃2 ) χ ) − λ̃ 2 , λ̃ 2 − 4µ̃ < 0 c2 c1+c2χ − λ̃ 2 , λ̃2 − 4µ̃ = 0. (3.21) In this context, C1 and C2 both illustrates the arbitrary constants. Considering equation (3.17), balancing with P ′(ξ)2 and P (ξ)4 bestows q = 1, thus, the result is of the below-stated form P (ξ) = a0 + a1( G′ G ), (3.22) We substitute the equation (3.22) in the equation(3.17) and after this, we equalize the coefficients of the algebraic equation to zero, we will get the following algebraic system of equations:- (G ′ G )4 : b2w2a41 + 4w2a2a21 = 0, (G ′ G )3 : −4ba31 + 8w2a2a21λ+ 4b2w2a0a 3 1 = 0, (G ′ G )2 : −12ba0a 2 1 + 4a21 + 6b2w2a20a 2 1 + 4w2a2a21λ 2 + 8w2a2a21µ = 0, (G ′ G )1 : −12ba20a1 + 8a0a1 + 4b2w2a2a21µλ = 0, (G ′ G )0 : b2w2a40 − 4ba30 + 4w2a2a21µ 2 + 4a20 = 0. (3.23) Resolving the overhead algebraic equations; yields the following cases: Case-I V = V, a = 1 2 ιa1b,− 1 2 ιa1b, b = b, µ = a0(−2 + ba0) ba21 , w = 1, λ = 2(−1 + ba0) ba1 , a0 = a0, a1 = a1. (3.24) Case-II V = V, a = 1 2 ιa1b,− 1 2 ιa1b, b = b, µ = a0(−2 + ba0) ba21 , w = −1, λ = 2(−1 + ba0) ba1 , a0 = a0, a1 = a1. (3.25) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 95 Corresponding to the above cases we will obtain the traveling wave solutions. The general outcomes of the equation (3.24) can be expressed as: Subcase-I when λ2 − 4µ > 0, then solution is: P (ξ) = a0 +a1 − (−1+ba0) a1b + √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21  C1sinh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω ) C1cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2sinh √√√√ (−1+ba20) a21b 2 − a0(−2+ba0) ba21 (x− t ω )   . (3.26) F (ξ) =√√√√√√√√√a0 + a1 − (−1+ba0) a1b + √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21  C1sinh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω ) C1cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2sinh √√√√ (−1+ba20) a21b 2 − a0(−2+ba0) ba21 (x− t ω )  , (3.27) G(ξ) = 6ιa1bξ − 3 2 ι √ (−1+ba0) 2 b2a21 −a0(−2+ba0) ba21 log ( C1 cosh( √ 1 b2a21 ξ)+C2 sinh( √ 1 b2a21 ξ) ) √ 1 b2a21 − 3ιξ 2a1b +Bξ, (3.28) using these values we get solution q(x, t) =√√√√√√√√√a0 + a1 − (−1+ba0) a1b + √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21  C1sinh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω ) C1cosh √ (−1+ba0)2 a21b 2 − a0(−2+ba0) ba21 (x− t ω )+C2sinh √√√√ (−1+ba20) a21b 2 − a0(−2+ba0) ba21 (x− t ω )   e ι(6ιa1b(x− t ω )− 3 2 ι √ (−1+ba0)2 b2a21 − a0(−2+ba0) ba21 log C1 cosh( √ 1 b2a21 (x− t ω ))+C2 sinh( √ 1 b2a21 (x− t ω ))  √ 1 b2a21 − 3ι(x− t ω ) 2a1b +B(x− t ω )) . (3.29) The Graphical representation of this case is represented in Figure 1 and in Figure 2. Subcase-II when λ2 − 4µ < 0, then solution is: P (x, t) = a0 +a1 − (−1+ba0) a1b + √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 −C1sin √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω ) C1cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2 √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )   . (3.30) F (ξ) = √√√√√√√a0 + a1 − (−1+ba0) a1b + √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 −C1sin √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω ) C1cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2 √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )  , (3.31) G(ξ) = −12aωξ + Bξ 3b 8a √√√√√√−2(−2a0C1 cos( √ 4µ−λ2 2 )−2a0C2 sin( √ 4µ−λ2 2 )+ √ 4µ−λ2 2 a1C1 sin( √ 4µ−λ2 2 )− √ 4µ−λ2a1C2 cos( √ 4µ−λ2 2 )+λC1 cos( √ 4µ−λ2 2 )+λC2 sin( √ 4µ−λ2 2 ) C1 cos( √ 4µ−λ2 2 )+C2 sin( √ 4µ−λ2 2 )  ξ. (3.32) using these values we get the solution q(x, t) = √√√√√√√a0 + a1 − (−1+ba0) a1b + √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 −C1sin √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω ) C1cos √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )+C2 √ a0(−2+ba0) ba21 − (−1+ba0)2 a21b 2 (x− t ω )   eι(−12aωξ+Bξ)× e ι  3b 8a √√√√√√√√−2(−2a0C1 cos( √ 4µ−λ2 2 )−2a0C2 sin( √ 4µ−λ2 2 )+ √ 4µ−λ2 2 a1C1 sin( √ 4µ−λ2 2 )− √ 4µ−λ2a1C2 cos( √ 4µ−λ2 2 )+λC1 cos( √ 4µ−λ2 2 )+λC2 sin( √ 4µ−λ2 2 ) C1 cos( √ 4µ−λ2 2 )+C2 sin( √ 4µ−λ2 2 ) ξ . (3.33) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 96 Figure 1: Traveling wave solution of |q(x, t)| of equation (3.29) when λ2− 4µ > 0 for case-I a0 = 2, a1 = 1, C1 = 4, C2 = 5, w = 1, b = 1, ξ = x− t w . Figure 2: Traveling wave solution of |q(x, t)| of equation (3.29) when λ2− 4µ > 0 for case-II a0 = 2, a1 = 1, C1 = 4, C2 = 5, w = −1, b = 1, ξ = x− t w . Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 10s (2025) https://internationalpubls.com 97 where ξ = x− t w . Subcase-III when λ2 − 4µ = 0, then solution is: P (x, t) = ( −(−1 + ba0) a1b + C2 C1 + C2ξ ) . (3.34) In this instance, both C1 and C2 are treated the arbitrary constants. F (x, t) = √ −K2 + ( C2 C1 + C2ξ ) , (3.35) Where K2 = −1+ba0 a1b . is any constant. G(x, t) = −12aξ + 3 8  b √ −(K2C1+K2C2ξ−C2) (C1+C2ξ) (C1+C2ξ)(2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) √ K2C 2 2+C2 2arctan( √ K2C 2 2(2K2C2ξ+2K2C1−C2)) 2K2C2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) ) aC2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) √ K2C 2 2  + Bξ. (3.36) using these values we get solution q(x, t) = √ −K2 + ( C2 C1+C2ξ ) × e ι −12aξ+3 8  b √ −(K2C1+K2C2ξ−C2) (C1+C2ξ) (C1+C2ξ)(2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) √ K2C 2 2+C2 2arctan( √ K2C 2 2(2K2C2ξ+2K2C1−C2)) 2K2C2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) ) aC2 √ −(K2C1+K2C2ξ−C2)(C1+C2ξ) √ K2C 2 2 +Bξ  . (3.37) where ξ = x− t w . 4 Conclusion The goal of this study was to explore the various Lie symmetries and the invariant solutions of Kaup-Newell equations. The first step is to determine the infinitesmal generators of the equation using the Lie group method. We then used the G′ G -expansion procedure to derive the traveling wave solutions. These traveling wave solutions are expressed using hyperbolic, trigonometric, and rational functions that incorporate arbitrary parameters. We derived exact soliton solutions for the Kaup-Newell equation. These solutions may hold importance across various scientific disciplines. 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