International Journal of Analysis and Applications Volume 18, Number 5 (2020), 724-737 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-724 HE’S VARIATIONAL ITERATION METHOD FOR SOLVING MULTI-DIMENSIONAL OF NAVIER STOKES EQUATION MOHAMED ZELLAL1,2,∗, KACEM BELGHABA2 1Department of Common Core in Exact Sciences and Informatics, Faculty of Exact Sciences and Informatics, Hassiba Benbouali University of Chlef, Ouled Fares P.O. Box 78, 02180, Chlef, Algeria 2Laboratory of Mathematics and its Applications (LAMAP), University of Oran1, Oran, Algeria ∗Corresponding author: m.zellal@yahoo.fr Abstract. In this paper, He’s variational iteration method (VIM), established by He in (1999), is adopted to solve two and three dimensional of Navier-Stokes equation in cartesian coordinates. This method is a powerful tool to handle linear and nonlinear models. The main property of the method is its softness and ability to solve nonlinear equations, accurately and easily. Using variational iteration method, it is possible to find the exact solution or a closed approximate solution of a problem. To illustrate the capacity and reliability of this method, some examples and numerical results are provided. 1. Introduction The main aim of this work, is to solve the model of the Navier Stokes equation for an incompressible fluid flow is given as follows [5, 8, 17].  Ut + (U.∇) U = ρ0∇2U − 1ρ∇p, on Ω × (0,T), ∇.U = 0, on Ω × (0,T), U = 0, on ∂Ω × (0,T), (1.1) Received May 18th, 2020; accepted June 8th, 2020; published June 24th, 2020. 2010 Mathematics Subject Classification. 35N05, 35Q30, 65A05, 65N25, 65Q20. Key words and phrases. Navier–Stokes equation; He’s variational iteration method; Lagrange multiplier; correction func- tional; exact solution; approximate solution. ©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 724 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-724 Int. J. Anal. Appl. 18 (5) (2020) 725 where U = (u,v,w), t, p, denote the fluid vector at the point (x; y; z), time and the pressure, respectively, Ω ⊂ R3 and ∂Ω its boundary, ρ is the density, ρ0 denotes the kinematic viscosity of the flow. In Cartesian coordinates, the (3D) Navier-Stokes equation becomes:  ut + uux + vuy + wuz = ρ0 (uxx + uyy + uzz) − 1ρpx, vt + uvx + vvy + wvz = ρ0 (vxx + vyy + vzz) − 1ρpy, wt + uwx + vwy + wwz = ρ0 (wxx + wyy + wzz) − 1ρpz. (1.2) Various kinds of analytical methods and numerical methods were used to solve Navier-Stokes equation, as Adomian decomposition method [2, 17], fractional reduced differential transformation method [5], modified Laplace decomposition method [15], the meshless local Petrov-Galerkin method [22], homotopy perturbation method [23]. The organization of this paper is as follows: In Section 2, we review the procedure of He’s variational iteration method. In Section 3, we show in some examples and numerical results that the present method gives the exact solution or a very good approximation of the exact solution, even for a small n, to illustrate the method, its accuracy, effectiveness and simplicity. A conclusion is given in Section 4. 2. He’s variational iteration method Variational iteration method was first proposed by the Chinese mathematician He ( [9]– [14]). This method has been employed to solve a large variety of linear and nonlinear problems which has various applications in science and engineering, with approximations converging rapidly to accurate solutions. This approach is successfully and effectively applied to various equations such as Burger’s and coupled Burger’s equations [1, 4]. This technique is also employed to solve multispecies Lotka–Volterra equations in [3], the Cauchy reaction-diffusion problem and several test examples are given in [6], in [7] the applications of the present method to solve the Fokker–Planck equation is provided. Author of [9] investigated (VIM) for solving delay differential equations. He’s variational iteration method is used in [10] to give the solution of blasius equation, autonomous ordinary differential equations [11], in [12] the variational iteration method is used to construct solitary solutions and compacton-like solutions for nonlinear dispersive equations. Also this procedure is investigated in [16] for solving Helmholtz equation, for solving integro-differential equations [19]. Authors of [20] employed the variational iteration method for solving a parabolic inverse problem. Wazwaz has used this method for handling linear and nonlinear diffusion equations [24], for solving linear and nonlinear systems of PDEs [25], for rational solutions for KdV, K(2, 2) , Burger, and cubic Boussinesq equations which has various applications in science and engineering [26]. In [27], for analytic treatment for linear and nonlinear ODEs, the (VIM) modified using He’s polynomials to solve biological population model [28]. The convergence of (VIM) is studied in [18, 21]. The (VIM) gives rapidly convergent successive approximations of the exact solution if such a solution exists. Int. J. Anal. Appl. 18 (5) (2020) 726 To illustrate the procedure of this approach, we write a system as shown below [3, 25]  Ltu + R1(u,v,w) + N1(u,v,w) = f1, Ltv + R2(u,v,w) + N2(u,v,w) = f2, Ltw + R3(u,v,w) + N3(u,v,w) = f3, (2.1) subject to initial conditions,   u(x, 0) = g1(x), v(x, 0) = g2(x), w(x, 0) = g3(x), (2.2) where Lt is considered a first-order partial differential operator, Rk and Nk, 1 ≤ k ≤ 3 are linear and nonlinear operators respectively, and f1, f2, f3 source terms. The correction functionals for equations of (2.1) can be written as  un+1(x,t) = un(x,t) + ∫ t 0 λ1(τ)[Lun(τ) + R1(ũn, ṽn, w̃n) + N1(ũn, ṽn, w̃n) −f1(τ)]dτ, vn+1(x,t) = vn(x,t) + ∫ t 0 λ2(τ)[Lvn(τ) + R2(ũn, ṽn, w̃n) + N2(ũn, ṽn, w̃n) −f2(τ)]dτ, wn+1(x,t) = wn(x,t) + ∫ t 0 λ3(τ)[Lwn(τ) + R3(ũn, ṽn, w̃n) + N3(ũn, ṽn, w̃n) −f3(τ)]dτ, (2.3) where λk, 1 ≤ k ≤ 3, are general Lagrange multipliers, which can be identified optimally via the variational theory, and ũn , ṽn , and w̃n are restricted variations which means δũn = 0 , δṽn = 0 and δw̃n = 0. It is required first to determine the Lagrange multipliers λk that will be identified optimally via integration by parts. The successive approximations un+1(x,t) , vn+1(x,t) , wn+1(x,t) , n ≥ 0, of the solutions u(x,t) , v(x,t), and w(x,t) will follow immediately upon using the Lagrange multipliers obtained and by using selected functions u0 , v0 and w0. The initial values are usually used for the selected zeroth approximations. With the Lagrange multipliers λk determined, then several approximations un(x,t), vn(x,t), wn(x,t), n ≥ 1, can be determined. Finally, the solutions are given   u(x,t) = lim n→∞ un (x, t ), v(x,t) = lim n→∞ vn (x, t ), w(x,t) = lim n→∞ wn (x, t ). (2.4) Int. J. Anal. Appl. 18 (5) (2020) 727 3. Test examples Example 3.1. Consider two-dimensional (2D) Navier-Stokes equations: [2, 5].   ut + uux + vuy = ρ0 (uxx + uyy) ,vt + uvx + vvy = ρ0 (vxx + vyy) , (3.1) with the initial conditions:   u(x,y, 0) = −sin(x + y),v(x,y, 0) = sin(x + y). (3.2) The correction functional for (3.1) reads   un+1(x,y,t) = un(x,y,t) + ∫ t 0 λ1(τ) [ ∂un(x,y,τ) ∂τ + ũn ∂ũn ∂x + ṽn ∂ũn ∂y −ρ0 ( ∂2ũn ∂x2 + ∂2ũn ∂y2 )] dτ, vn+1(x,y,t) = vn(x,y,t) + ∫ t 0 λ2(τ) [ ∂vn(x,y,τ) ∂τ + ũn ∂ṽn ∂x + ṽn ∂ṽn ∂y −ρ0 ( ∂2ṽn ∂x2 + ∂2ṽn ∂y2 )] dτ. (3.3) This yields the stationary conditions   1 + λ1 = 0, λ ′ 1(τ = t) = 0, 1 + λ2 = 0, λ ′ 2(τ = t) = 0. (3.4) As a result we find λ1 = λ2 = −1. (3.5) Substituting these values of the Lagrange multipliers into the functionals (3.3) gives the iteration formulas   un+1(x,y,t) = un(x,y,t) − ∫ t 0 [ ∂un(x,y,τ) ∂τ + un ∂un ∂x + vn ∂un ∂y −ρ0 ( ∂2un ∂x2 + ∂2un ∂y2 )] dτ, vn+1(x,y,t) = vn(x,y,t) − ∫ t 0 [ ∂vn(x,y,τ) ∂τ + un ∂vn ∂x + vn ∂vn ∂y −ρ0 ( ∂2vn ∂x2 + ∂2vn ∂y2 )] dτ. (3.6) We can select u0(x,y,t) = −sin(x+y), v0(x,y,t) = sin(x+y), by using the given initial values. Accordingly, we obtain the following successive approximations: Int. J. Anal. Appl. 18 (5) (2020) 728 u1(x,y,t) = −sin(x + y)(1 − 2ρ0t), v1(x,y,t) = sin(x + y)(1 − 2ρ0t), u2(x,y,t) = −sin(x + y)(1 − 2ρ0t + 2ρ20t 2), v2(x,y,t) = sin(x + y)(1 − 2ρ0t + 2ρ20t 2), u3(x,y,t) = −sin(x + y)(1 − 2ρ0t + 2ρ20t 2 − 4 3 ρ30t 3), v3(x,y,t) = sin(x + y)(1 − 2ρ0t + 2ρ20t 2 − 4 3 ρ30t 3), ... un(x,y,t) = −sin(x + y)(1 − 2ρ0t + 2ρ20t 2 + · · ·), vn(x,y,t) = sin(x + y)(1 − 2ρ0t + 2ρ20t 2 + · · ·). The final form of the solution will be as follows  u(x,y,t) = −sin(x + y) ( 1 + (−2ρ0t) + (−2ρ0t) 2 2! + (−2ρ0t) 3 3! + · · · ) = −sin(x + y)e−2ρ0t, v(x,y,t) = sin(x + y) ( 1 + (−2ρ0t) + (−2ρ0t) 2 2! + (−2ρ0t) 3 3! + · · · ) = sin(x + y)e−2ρ0t. Which is an exact solution and is same as obtained by [5]. Table 1: Numerical results when ρ0 = 0.5, x = 0.1 and y = 0.5 in Example 3.1. t u u3 |u−u3| v v3 |v −v3| 0 -0.56464247 -0.56464247 0 0.56464247 0.56464247 0 0.05 -0.53710453 -0.53710453 0 0.53710453 0.53710453 0 0.1 -0.51090964 -0.51090968 0.4628E-7 0.51090964 0.51090968 0.4628E-7 0.15 -0.48599228 -0.48599262 0.9719 0.48599228 0.48599262 0.9719 0.2 -0.46229015 -0.46229162 0.9246 0.46229015 0.46229162 0.9246 0.25 -0.43974400 -0.43974841 0.8795 0.43974400 0.43974841 0.8795 0.30 -0.41829743 -0.41830832 0.1089E-4 0.41829743 0.41830832 0.1089E-4 0.35 -0.39789682 -0.39792017 0.2335E-4 0.39789682 0.39792017 0.2335E-4 0.40 -0.37849117 -0.37853631 0.4514E-4 0.37849117 0.37853631 0.4514E-4 Int. J. Anal. Appl. 18 (5) (2020) 729 Figure 1. The behavior of u,u3 and v,v3 in Example 3.1 at t = 1 with the parameters ρ0 = 0, 5. Example 3.2. Consider two-dimensional (2D) Navier-Stokes equations:  ut + uux + vuy = ρ0 (uxx + uyy) ,vt + uvx + vvy = ρ0 (vxx + vyy) , (3.7) with the initial conditions:   u(x,y, 0) = −e x+y, v(x,y, 0) = ex+y. (3.8) We follow the same procedure discussed in Example (3.1), we can select u0(x,y,t) = −ex+y, v0(x,y,t) = ex+y in the iteration formulas(3.6), by using the given initial values. Accordingly, we obtain the following successive approximations: Int. J. Anal. Appl. 18 (5) (2020) 730 u1(x,y,t) = −ex+y(1 + 2ρ0t), v1(x,y,t) = e x+y(1 + 2ρ0t), u2(x,y,t) = −ex+y(1 + 2ρ0t + 2ρ20t 2), v2(x,y,t) = e x+y(1 + 2ρ0t + 2ρ 2 0t 2), u3(x,y,t) = −ex+y(1 + 2ρ0t + 2ρ20t 2 + 4 3 ρ30t 3), v3(x,y,t) = e x+y(1 + 2ρ0t + 2ρ 2 0t 2 + 4 3 ρ30t 3), u4(x,y,t) = −ex+y(1 + 2ρ0t + 2ρ20t 2 + 4 3 ρ30t 3 + 2 3 ρ40t 4), v4(x,y,t) = e x+y(1 + 2ρ0t + 2ρ 2 0t 2 + 4 3 ρ30t 3 + 2 3 ρ40t 4), ... un(x,y,t) = −ex+y(1 + (2ρ0t) + (2ρ0t) 2 2! + · · ·), vn(x,y,t) = e x+y(1 + (2ρ0t) + (2ρ0t) 2 2! + · · ·). Finally, the exact solution may be obtained as follows  u(x,y,t) = −e x+ye2ρ0t = −ex+y+2ρ0t, v(x,y,t) = ex+ye2ρ0t = ex+y+2ρ0t, which are exact solutions [5]. Consider the following tables 2 and 3 with the observation that v = −u. Table 2: Numerical results when ρ0 = 1,x = 0.1 and y = 0.5 in Example 3.2. t u u4 |u−u4| = |v −v4| 0 -1.8221188 -1.8221188 0 0.05 -2.0137527 -2.0137526 1.5441055E-7 0.1 -2.2255409 -2.2255359 5.0256955E-6 0.15 -2.4596031 -2.4595643 3.8824935E-5 0.2 -2.7182818 -2.7181155 1.6647662E-4 0.25 -3.0041660 -3.0036490 5.1706393E-4 0.30 -3.3201169 -3.3188072 1.3097397E-3 0.35 -3.6692967 -3.6664144 8.8236261E-3 0.40 -4.0552000 -4.0494768 5.7231449E-3 Int. J. Anal. Appl. 18 (5) (2020) 731 Table 3: Absolute error= |u−u4| when ρ0 = 1 in Example 3.2. t t = 0.1 t = 0.3 t = 0.5 (x,y) |u−u4| |u−u4| |u−u4| (0.1,0.1) 3.36882444 E-6 8.7794477954 E-4 1.2151119386 E-2 0.3,0.1) 4.11469146 E-6 1.0723241752 E-3 1.4841410733 E-2 (0.5,0.2) 5.55425251 E-6 1.4474862325 E-3 2.0033808995 E-2 (0.7,0.2) 6.78397933 E-6 1.7679636768 E-3 2.4469349563 E-2 (0.3,0.3) 5.02569550 E-6 1.3097397053 E-3 1.8127340004 E-2 (0.5,0.3) 6.13839835 E-6 1.5997196885 E-3 2.2140783079 E-2 (0.7,0.5) 9.15741425 E-6 2.3865013406 E-3 3.3030167023 E-2 (0.9,0.5) 1.11848910 E-5 2.9148793198 E-3 4.0343137104 E-2 Example 3.3. Consider three-dimensional (3D) Navier-Stokes equations:  ut + uux + vuy + wuz = ρ0 (uxx + uyy + uzz) , vt + uvx + vvy + wvz = ρ0 (vxx + vyy + vzz) , wt + uwx + vwy + wwz = ρ0 (wxx + wyy + wzz) , (3.9) with the initial condition:   u(x,y,z, 0) = −1 2 x + y + z, v(x,y,z, 0) = x− 1 2 y + z, w(x,y,z, 0) = x + y − 1 2 z. (3.10) The correction functionals for (3.9) read  un+1(x,y,z,t) = un + ∫ t 0 λ1(τ) [ ∂un ∂τ + ũn ∂ũn ∂x + ṽn ∂ũn ∂y + w̃n ∂ũn ∂z −ρ0 ( ∂2ũn ∂x2 + ∂2ũn ∂y2 + ∂2ũn ∂z2 )] dτ, vn+1(x,y,z,t) = vn + ∫ t 0 λ2(τ) [ ∂vn ∂τ + ũn ∂ṽn ∂x + ṽn ∂ṽn ∂y + w̃n ∂ṽn ∂z −ρ0 ( ∂2ṽn ∂x2 + ∂2ṽn ∂y2 + ∂2ṽn ∂z2 )] dτ, wn+1(x,y,z,t) = wn + ∫ t 0 λ3(τ) [ ∂wn ∂τ + ũn ∂w̃n ∂x + ṽn ∂w̃n ∂y + w̃n ∂w̃n ∂z −ρ0 ( ∂2w̃n ∂x2 + ∂2w̃n ∂y2 + ∂2w̃n ∂z2 )] dτ. (3.11) The stationary conditions are thus given by  1 + λ1 = 0, λ ′ 1(τ = t) = 0, 1 + λ2 = 0, λ ′ 2(τ = t) = 0, 1 + λ3 = 0, λ ′ 3(τ = t) = 0, (3.12) so that, the Lagrange multipliers can be identified as follows: λ1 = λ2 = λ3 = −1. (3.13) Int. J. Anal. Appl. 18 (5) (2020) 732 He’s variational iteration method consists of the following scheme:  un+1(x,y,z,t) = un − ∫ t 0 [ ∂un ∂τ + un ∂un ∂x + vn ∂un ∂y + wn ∂un ∂z −ρ0 ( ∂2un ∂x2 + ∂2un ∂y2 + ∂2un ∂z2 )] dτ, vn+1(x,y,z,t) = vn − ∫ t 0 [ ∂vn ∂τ + un ∂vn ∂x + vn ∂vn ∂y + wn ∂vn ∂z −ρ0 ( ∂2vn ∂x2 + ∂2vn ∂y2 + ∂2vn ∂z2 )] dτ, wn+1(x,y,z,t) = wn − ∫ t 0 [ ∂wn ∂τ + un ∂wn ∂x + vn ∂wn ∂y + wn ∂wn ∂z −ρ0 ( ∂2wn ∂x2 + ∂2wn ∂y2 + ∂2wn ∂z2 )] dτ. (3.14) Starting with initial approximations: u0(x,y,z,t) = −12x+y+z,v0(x,y,z,t) = x− 1 2 y+z, and w0(x,y,z,t) = x + y − 1 2 z, from (3.14), other terms of the sequence are computed as follows:  u1(x,y,z,t) = − 1 2 x + y + z − 9 4 xt, v1(x,y,z,t) = x− 1 2 y + z − 9 4 yt, w1(x,y,z,t) = x + y − 1 2 z − 9 4 zt,   u2(x,y,z,t) = − 1 2 x + y + z + (− 1 2 x + y + z) 9 4 t2 − 9 4 xt− 27 16 xt3, v2(x,y,z,t) = x− 1 2 y + z + (x− 1 2 y + z) 9 4 t2 − 9 4 yt− 27 16 yt3, w2(x,y,z,t) = x + y − 1 2 z + (x + y − 1 2 z) 9 4 t2 − 9 4 zt− 27 16 zt3,   u3(x,y,z,t) = (− 1 2 x + y + z) + (− 1 2 x + y + z) 9 4 t2 + (− 1 2 x + y + z) 27 8 t4 + (− 1 2 x + y + z) 81 64 t6 − 9 4 xt− 81 16 xt3 − 243 64 xt5 − 729 1792 xt7, v3(x,y,z,t) = (x− 1 2 y + z) + (x− 1 2 y + z) 9 4 t2 + (x− 1 2 y + z) 27 8 t4 + (x− 1 2 y + z) 81 64 t6 − 9 4 yt− 81 16 yt3 − 243 64 yt5 − 729 1792 yt7, w3(x,y,z,t) = (x + y − 1 2 z) + (x + y − 1 2 z) 9 4 t2 + (x + y − 1 2 z) 27 8 t4 + (x + y − 1 2 z) 81 64 t6 − 9 4 zt− 81 16 zt3 − 243 64 zt5 − 729 1792 zt7, Int. J. Anal. Appl. 18 (5) (2020) 733   u4(x,y,z,t) = (− 1 2 x + y + z) + (− 1 2 + y + z) 9 4 t2 + (− 1 2 x + y + z) 81 16 t4 + (− 1 2 + y + z) 243 32 t6 + (− 1 2 x + y + z) 51759 7168 t8 + (− 1 2 x + y + z) 72171 17920 t10 + (− 1 2 x + y + z) 59049 57344 t12 + (− 1 2 x + y + z) 59049 802816 t14 − 9 4 xt− 81 16 xt3 − 3159 320 xt5 − 21141 1792 xt7 − 31347 3584 xt9 − 98415 28672 xt11 − 59049 114688 xt13 − 177147 16056320 xt15, v4(x,y,z,t) = (x− 1 2 y + z) + (x− 1 2 y + z) 9 4 t2 + (x− 1 2 y + z) 81 16 t4 + (x− 1 2 y + z) 243 32 t6 + (x− 1 2 y + z) 51759 7168 t8 + (x− 1 2 y + z) 72171 17920 t10 + (x− 1 2 y + z) 59049 57344 t12 + (x− 1 2 y + z) 59049 802816 t14 − 9 4 yt− 81 16 yt3 − 3159 320 yt5 − 21141 1792 yt7 − 31347 3584 yt9 − 98415 28672 yt11 − 59049 114688 yt13 − 177147 16056320 yt15, w4(x,y,z,t) = (x + y − 1 2 z) + (x + y − 1 2 z) 9 4 t2 + (x + y − 1 2 z) 81 16 t4 + (x + y − 1 2 z) 243 32 t6 + (x + y − 1 2 z) 51759 7168 t8 + (x + y − 1 2 z) 72171 17920 t10 + (x + y − 1 2 z) 59049 57344 t12 + (x + y − 1 2 z) 59049 802816 t14 − 9 4 zt− 81 16 zt3 − 3159 320 zt5 − 21141 1792 zt7 − 31347 3584 zt9 − 98415 28672 zt11 − 59049 114688 zt13 − 177147 16056320 zt15. Therefore,  u(x,y,z,t) = ( − 1 2 x + y + z )( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) − 9 4 xt ( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) = −1 2 x + y + z − 9 4 xt 1 − 9 4 t2 , v(x,y,z,t) = ( x− 1 2 y + z )( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) − 9 4 yt ( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) = x− 1 2 y + z − 9 4 yt 1 − 9 4 t2 , w(x,y,z,t) = ( x + y − 1 2 z )( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) − 9 4 zt ( 1 + ( 9 4 ) t2 + ( 9 4 )2 t4 + · · · ) = x + y − 1 2 z − 9 4 zt 1 − 9 4 t2 . The same result as [5]. Int. J. Anal. Appl. 18 (5) (2020) 734 Table 4: The (VIM) results for u and u4 approximations in Example 3.3 t t = 0.01 t = 0.3 t = 0.5 (x,y,z) u4 |u−u4| u4 |u−u4| u4 |u−u4| (0.1,0.1,0.2) 0.2478057563 1.44e-11 0.2285415892 2.99e-4 0.2897586783 2.45e-2 (0.3,0.1,0.2) 0.1432822385 4.50e-11 -0.0640993439 1.73e-3 -0.3486963088 7.99e-2 (0.5,0.2,0.3) 0.2388037309 7.51e-11 -0.1068322397 2.89e-3 -0.5811605148 1.33e-1 (0.7,0.2,0.3) 0.1342802131 1.06e-10 -0.3994731729 4.92e-3 -1.2196155019 2.36e-1 (0.9,0.3,0.4) 0.2298017055 1.36e-10 -0.4422060687 6.07e-3 -1.4520797078 2.91e-1 (0.6,0.3,0.4) 0.3865869822 8.97e-11 -0.0032446693 3.03e-3 -0.4943972270 1.34e-1 (0.8,0.4,0.5) 0.4821084745 1.20e-10 -0.0459775653 4.18e-3 -0.7268614329 1.87e-1 (0.9,0.4,0.5) 0.4298467157 1.35e-10 -0.1922980316 5.19e-3 -1.0460889265 2.40e-1 Table 5: The (VIM) results for v and v4 approximations in Example 3.3. t t = 0.01 t = 0.3 t = 0.5 (x,y,z) v4 |v −v4| v4 |v −v4| v4 |v −v4| (0.1,0.1,0.2) 0.2478057563 1.43e-11 0.2285415892 2.99e-4 0.2897586783 2.45e-2 (0.3,0.1,0.2) 0.4478507665 1.35e-11 0.4784496263 1.17e-3 0.6957494596 7.57e-2 (0.5,0.2,0.3) 0.6956565227 2.77e-11 0.7069912153 1.47e-3 0.9855081381 1.00e-1 (0.7,0.2,0.3) 0.8957015329 2.70e-11 0.9568992524 2.35e-3 1.3914989194 1.51e-1 (0.9,0.3,0.4) 1.1435072892 4.12e-11 1.1854408415 2.65e-3 1.6812575977 1.76e-1 (0.6,0.3,0.4) 0.8434397739 4.24e-11 0.8105787860 1.33e-3 1.0722714258 9.92e-2 (0.8,0.4,0.5) 1.0912455303 5.66e-11 1.0391203750 1.63e-3 1.3620301041 1.24e-1 (0.9,0.4,0.5) 1.1912680354 5.63e-11 1.1640743935 2.07e-3 1.5650254948 1.49e-1 Int. J. Anal. Appl. 18 (5) (2020) 735 Table 6: The (VIM) results for w and w4 approximations in Example 3.3. t t = 0.01 t = 0.3 t = 0.5 (x,y,z) w4 |w −w4| w4 |w −w4| w4 |w −w4| (0.1,0.1,0.2) 0.0955214924 3.00e-11 -0.0427328960 1.15e-3 -0.2324642059 5.33e-2 (0.3,0.1,0.2) 0.2955665025 2.93e-11 0.2071751411 2.79e-4 0.1735265755 2.09e-3 (0.5,0.2,0.3) 0.5433722588 4.35e-11 0.4357167302 1.95e-5 0.4632852538 2.24e-2 (0.7,0.2,0.3) 0.7434172689 4.28e-11 0.6856247672 8.96e-4 0.8692760351 7.36e-2 (0.9,0.3,0.4) 0.9912230253 5.70e-11 0.9141663564 1.19e-3 1.1590347135 9.81e-2 (0.6,0.3,0.4) 0.6911555100 5.82e-11 0.5393043008 1.19e-4 0.5500485415 2.19e-2 (0.8,0.4,0.5) 0.9389612664 7.24e-11 0.7678458899 1.79e-4 0.8398072198 4.59e-2 (0.9,0.4,0.5) 1.0389837714 7.20e-11 0.8927999085 6.17e-4 1.0428026105 7.15e-2 It can be observed through tables [1-6] that this method is efficient and accurate for different values of time and place. Conclusion. There are two main objectives for this work. The first presents an alternative approach to variation iteration method to handle non-linear problems. The second is to use this method to solve the two- dimensional (2D) and three-dimensional (3D) Navier-Stokes equations in Cartesian coordinates. It is obvious that the method gives several successive approximations through determining the Lagrange multipliers and using the iteration. (VIM) reduces the size of calculations and facilitates the computational work when compared with (ADM) or (HPM) techniques. He’s variational iteration method is suitable as an alternative approach to current techniques being employed to a wide variety of problems in physics. The Navier-Stokes equations were examined. The desired solutions were obtained rapidly and in a direct way. The two goals were achieved. Acknowledgements: The two authors, sincerely thank the editor and the referees for their guidance and suggestions. 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