Computers and Mathematics with Applications 58 (29) 27 26 Contents lists available at ScienceDirect Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa Study on nonlinear Jeffery Hamel flow by He s semi-analytical methods and comparison with numerical results Z.Z. Ganji, D.D. Ganji, M. Esmaeilpour Department of Mechanical Engineering, Babol University of Technology, P.O. Bo 484, Babol, Iran a r t i c l e i n f o a b s t r a c t Keywords: He s variational iteration Method (VIM) Nonlinear ordinary differential equation of Jeffery Hamel flows He s Homotopy perturbation method (HPM) In this article the problem of Jeffery Hamel flow is presented and the variational iteration method and the homotopy perturbation method are employed to compute an approimation to the solution of the system of nonlinear differential equations governing the problem. Comparisons are made between the Numerical solution (NM) and the results of the He s variational iteration method (VIM) and He s homotopy perturbation method (HPM). The results reveal that these methods are very effective and simple and can be applied for other nonlinear problems. 29 Elsevier Ltd. All rights reserved.. Introduction The incompressible viscous fluid flow through convergent divergent channels is one of the most applicable cases in fluid mechanics, civil, environmental, mechanical and bio-mechanical engineering. The mathematical investigations of this problem were pioneered by [,2], (i.e. Jeffery Hamel flows). Jeffery Hamel flows are an eact similarity solution of the Navier Stokes equations in the special case of two-dimensional flow through a channel with inclined plane walls meeting at a verte and with a source or sink at the verte and have been etensively studied by several authors and discussed in many tetbooks e.g., [3 2], etc. Most scientific problems such as Jeffery Hamel flows and other fluid mechanic problems are inherently nonlinear. Ecept a limited number of these problems, most of them do not have analytical solution. Therefore, these nonlinear equations should be solved using other methods. In the analytical perturbation method, we should eert the small parameter in the equation. Therefore, finding the small parameter and eerting it into the equation are difficulties of this method. Since there are some limitations with the common perturbation method, and also because the basis of the common perturbation method is upon the eistence of a small parameter, developing the method for different applications is very difficult. Therefore, many different methods have recently introduced some ways to eliminate the small parameter. The variational iteration method and homotopy perturbation method (HPM) are well-known methods to solve the nonlinear equations. These methods are introduced by He [3 24] for the first time. These methods have been used by many authors such as Ganji in [25 3] and the references therein to handle a wide variety of scientific and engineering applications such as linear and nonlinear, homogeneous and inhomogeneous as well, because these methods continuously deform a difficult problem into a simple one, which is easy to solve. They were shown by many authors that these methods provide improvements over eisting numerical techniques. With the rapid development of nonlinear science, many different methods were proposed to solve various boundary-value problems (BVP) [32,33] and fractional order [34], such as homotopy perturbation method (HPM) and variational iteration method (VIM) [35,36]. These methods give successive approimations of high accuracy of the solution. In this study, we have applied He s VIM and HPM to find the approimate solutions of nonlinear differential equation governing Jeffery Hamel flow, and have made a comparison with the numerical solution. Corresponding author. Tel./Fa: +98 323425. E-mail addresses: ddg_davood@yahoo.com, mirgang@nit.ac.ir (D.D. Ganji). 898-22/$ see front matter 29 Elsevier Ltd. All rights reserved. doi:.6/j.camwa.29.3.44
28 Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 α q θ u(r, ) θ Fig.. Geometry of the problem. 2. Mathematical formulation Consider the steady two-dimensional flow of an incompressible conducting viscous fluid from a source or sink at the intersection between two rigid plane walls that the angle between them is 2α as shown in Fig.. We assume that the velocity is only along radial direction and depends on r and θ, V(u(r, θ), ) [,]. Using continuity and Navier Stokes equations in polar coordinates: (ρru(r, θ)) =, r r u(r, θ) u(r, θ) = p r ρ r + v ρr From Eq. (): p θ + 2v u(r, θ) =. r 2 θ f (θ) ru(r, θ). [ 2 u(r, θ) + u(r, θ) r 2 r r + r 2 2 u(r, θ) θ 2 ] u(r, θ) r 2 () (2) (3) (4) Using dimensionless parameters: F() f (θ) f ma, θ α (5) and by eliminating P between Eqs. (2) and (3), we obtain an ordinary differential equation for the normalized function profile F() []: F () + 2αReF()F () + 4α 2 F () =. (6) Since we have a symmetric geometry, the boundary conditions will be: F() =, F () =, F() =. (7) The Reynolds number: Re f maα v = U marα v ( ) divergent channel : α >, Uma >. (8) convergent channel : α <, U ma < For solving Eq. (6), we apply another boundary condition, which is F () = η. So the boundary conditions which have been used in the solution of Eq. (6) are as follows: F() =, F () =, F () = η. (9) Now we solve the problem by using two methods. 3. Variational iteration method [3] To illustrate the basic concepts of variational iteration method, we consider the following differential equation: Lu + Nu = g() ()
Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 29 where L is a linear operator, N a nonlinear operator, and g() an inhomogeneous term. According to VIM, we can construct a correction function as follows: u n+ () = u n () + λ{l u n (τ) + N ũ n (τ) g(τ)}dτ () where λ is a general Lagrangian multiplier [3], which can be identified optimally via the variational theory [3], the subscript n indicates the nth order approimation, ũ n which is considered as a restricted variation, i.e. δ ũ n =. 4. Basic concept of He s homotopy perturbation method To illustrate the basic ideas of this method, we consider the following equation: A (F) f (r) =, r Ω, with the boundary condition of: ( B u, F ) =, r Γ, n where A is a general differential operator, B a boundary operator, f (r) a known analytical function and Γ is the boundary of the domain Ω. A can be divided into two parts, which are L and N, where L is linear and N is nonlinear. Eq. (6) can therefore be rewritten as follows: L (F) + N (F) f (r) =, r Ω. Homotopy perturbation structure is shown as follows: H (ν, p) = ( p) [L (ν) L (u )] + p [A (ν) f (r)] =, (5) (2) (3) (4) where, ν (r, p) : Ω [, ] R. (6) In Eq. (5), p [, ] is an embedding parameter and u is the first approimation that satisfies the boundary condition. We can assume that the solution of Eq. (6) can be written as a power series in p, as following: ν = ν + pν + p 2 ν 2 + = n ν i p i, i= and the best approimation for the solution is: F = lim ν = ν + ν + ν 2 +. p (7) (8) 5. Application of VIM to Jeffery Hamel flow In this section, we will apply the VIM to nonlinear ordinary differential equation (6). To solve the above equation using VIM, we have the correction function as: F n+ () = F n () + (λ(t)( Fn (t) + 2αRe Fn (t) Fn (t) + 4α 2 Fn (t)) ) dt, (9) where λ is considered as a restricted variation. Its stationary conditions can be obtained as follows: λ (τ) =, + λ (τ) τ=t =, λ (τ) τ=t =, λ (τ) τ=t =. The Lagrange multiplier can therefore be simply identified as λ = ( 2 t)2, and the following iteration formula can be obtained: ( F n+ () = F n () ( Fn )) 2 ( t)2 (t) + 2αRe Fn (t) Fn (t) + 4α 2 Fn (t) dt. (2) (2a) (2b) (2c) (2d)
2 Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26.9.8.7.6.5.4.3.2..25.5.75 Fig. 2. The comparison between the numerical, HPM and VIM solutions for F(), Re = 5, α = 5. Table For HPM terms approimation is used and for VIM 5-iteration is used. α Re η VIM η HPM [2/2] η NM 5 5 3.5394 3.5394 3.5394 5 5.8684 5.8696 5.8692 5 8 6.879 6.8799 6.882 5 6 9.4775 9.4787 9.4786 5 29 2.7393 2.7438 2.7425 5 6.8.7.7 5 7.5333.5326.5326 5 8.284.278.278 5 29.25.295.295 Beginning with an initial approimation, F () = η 2 2 +, components of the iteration formula can be easily found. Using the above variational formula (2), we can obtain the following result: ( F () = F () ( F )) 2 ( t)2 (t) + 2αRe F (t) F (t) + 4α 2 F (t) dt. (22) Substituting F () into Eq. (22) and after some simplifications, we have: F () = + ( 2 η 2 + 2 αreη ) 6 α2 η 4 2 αreη2 6. (23) In the same way, we obtain F 2 () as: F 2 () = + ( 2 η 2 + 2 αreη ) 6 α2 η 4 ( + 2 α η2 Re + 8 α2 ηre 2 + 45 α4 η + ) ( 45 α3 ηre 6 + 28 α3 η 2 Re + ) 56 α2 η 2 Re 2 8 ( + 324 α4 η 2 Re 2 2 96 α3 η 2 Re 3 324 α5 η 2 Re + ) 8 α2 η 3 Re 2 ( 47 52 α4 η 3 Re 2 + ) 95 4 α3 η 3 Re 3 2 262 8 α3 η 4 Re 3 4, (24) where η = F () is to be determined from the boundary conditions later. 5.. Application of HPM to Jeffery Hamel flow In this section, we will apply the HPM to nonlinear ordinary differential equation (6). According to HPM, we can construct homotopy of Eq. (6) as follows: ( p) ( F () F ()) + p ( F () + 2αReF()F () + 4α 2 F () ) =. (25)
Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 2.9.8.7.6.5.4.3.2..25.5.75 Fig. 3. The comparison between the numerical, HPM and VIM solutions for F(), Re = 8 6, α = 5..9.8.7.6.5.4.3.2.. Fig. 4. The comparison between the numerical, HPM and VIM solutions for F(), Re = 29, α = 5..9.8.7.6.5.4.3.2..25.5.75 Fig. 5. The comparison between the numerical, HPM and VIM solutions for F(), Re = 6 7 8, α = 5.
22 Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26.9.8.7.6.5.4.3.2..25.5.75 Fig. 6. The comparison between the numerical, HPM and VIM solutions for F(), Re = 29, α = 5. 2 3 4.25.5.75 Fig. 7. The comparison between the numerical, HPM and VIM solutions for F (), Re = 29, α = 5. We consider F as follows: F() = F () + F () + F 2 () + F 3 () + = n F i (). (26) i= By substituting F from Eq. (26) into Eq. (25) and after some simplifications and rearranging based on powers of p-terms, we have: p : F () =, F () =, F () =, F () = η, (27) p : F () + 2αReF ()F () + 4α2 F () =, F () =, F () =, F () =, (28) p 2 : F () + 2 2αReF ()F () + 2αReF ()F () + 4α2 F () =, F 2() =, F 2 () =, F 2 () =, (29) p 3 : F () + 3 2αReF ()F () + 2 2αReF 2()F () + 2αReF ()F () + 4α2 F 2 () =, F 3 () =, F 3 () =, F 3 () =. Solving Eqs. (27) (3) with boundary conditions, we have: F () = 2 η 2 +, (3) ( F () = 2 η αre + ) 6 ς α2 4 2 η2 αre 6 (32) (3)
Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 23 Table 2 The comparison between the Numerical, HPM and VIM solutions for F and F, when α = 5 and Re = 5. F F NM HPM VIM NM HPM VIM.... 3.53946 3.53924 3.539369.5.995584.995632.995584 3.5827 3.5628 3.5782..98243.98263.98243 3.3869 3.38678 3.386866.5.96826.9694.96827 3.236 3.22878 3.239.2.93226.93789.93227 2.957792 2.957623 2.957753.25.894242.89496.894243 2.66284 2.6693 2.66248.3.856.8546.8563 2.328574 2.328436 2.328542.35.863.8938.866.97687.97565.9766.4.74679.74745.746794.6789.6683.6767.45.68844.688892.68848.234439.234346.234425.5.626948.62722.626953.879794.8797.87979.55.563278.563368.563284.54795.5472.54722.6.498234.4982.49824.243949.243876.243994.65.432573.432492.43258.24728.248.24642.7.366966.36696.366974.25567.255685.25547.75.399.32.3998.447244.44733.44752.8.23824.238233.2383.59972.59983.599464.85.75749.75968.75755.74243.7436.73982.9.552.5479.557.7934.7934.792767.95.5653.56959.56533.83873.838859.838559.....854369.854544.8544 ( F 2 () = 45 α4 η + 8 α2 ηre 2 + ) ( 45 α3 ηre 6 + 56 α2 η 2 Re 2 + ) 28 α3 η 2 Re 8 + 8 α2 η 3 Re 2, (33) ( F 3 () = 54 α3 ηre 3 + 63 α3 ηre 3 + 42 α5 ηre + ) 84 α4 ηre 2 8 ( + 4 α4 η 2 Re 2 56 α3 η 2 Re 3 ) 4 α5 η 2 Re ( 27 2494 8 α4 η 3 Re 2 + 27 ) 4989 6 α3 η 3 Re 3 2 23 2 α3 η 4 Re 3 4, (34) where η = F () is to be determined from the boundary conditions. Solutions F 4 () to F () were too long to be mentioned here; therefore, they are shown graphically. The solution of this equation, when p, will be as follows: F() = F () + F 2 () + F 3 () + + F (). (35) 6. Numerical method The best approimation that can be used is Runge Kutta method. It is often utilized to solve differential equation systems. Third order differential equations can be usually changed into second order equations and then to first order and then be solved through Runge Kutta method: F + f (, F, F, F ) =, F( ) = α, F (36) ( ) = β, F ( ) = η, with the assumption of F = W and F = w = u F = w = f (, F, u, w), w = u = g(, F, u, w), u = f (, F, u, w) = h(, F, u, w) F( ) = α, w( ) = β, u( ) = η. (37) Consequently, the obtained system can be solved through Runge Kutta method.
24 Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 Table 3 The comparison between the Numerical, HPM and VIM solutions for F and F, when α = 5 and Re = 8. F F NM HPM VIM NM HPM VIM.....2785.2782.28438.5.99965.99965.999647.28898.288998.295..998573.998573.99856.322752.32277.325577.5.996684.996684.996655.3892.38925.385243.2.993834.993834.99378.47928.47956.47545.25.989799.989798.9897.59653.596566.6735.3.984263.984262.98426.7682.76866.774692.35.976795.976794.976592.997839.997898.6427.4.96687.96685.966528.3268.32244.3355.45.953564.95356.9536.7552.765.75342.5.93632.93628.935479 2.2243 2.2253 2.23857.55.9295.929.9272 2.89292 2.892254 2.9826.6.882529.882522.88549 3.74832 3.748338 3.766492.65.842726.84277.84466 4.825574 4.82583 4.838293.7.7987.79795.789235 6.5623 6.5652 6.549.75.72344.723425.72566 7.75658 7.756853 7.72573.8.63663.6366.634533 9.68492 9.6873 9.532967.85.525762.525737.523633.629437.62938.49857.9.385823.385794.38399 3.62898 3.626844 3.45287.95.2887.286.2779 5.256735 5.252 5.92442.... 5.95556 5.94274 5.93967 Table 4 The comparison between the Numerical, HPM and VIM solutions for F and F, when α = 5 and Re = 29. F F NM HPM VIM NM HPM VIM........5.98492.984635.98496.627525.684.627366..93878.939695.93895.99529.883.99224.5.865937.868553.865972.67256.6652.66983.2.773337.776496.773397 2.594 2.228 2.74.25.667229.6723.66738 2.243 2.258 2.29846.3.554528.556558.554648 2.27663 2.27762 2.275987.35.44484.44258.44637 2.227996 2.222496 2.22736.4.333226.33379.33342 2.89568 2.9747 2.88959.45.23359.23386.23387.887234.89563.886698.5.4579.45273.4542.64443.653375.64379.55.69546.69565.6989.37875.38682.378347.6.746.744.7744.3997.336.363.65.4836.494.4526.8287.832742.827475.7.75398.75543.7543.554979.55857.553866.75.96395.96554.95965.285455.28739.28366.8.3984.47.3442.8333.93.58.85.9823.98282.97543.248622.24933.25694.9.7986.78973.78245.577.5969.5245.95.464.45996.45478.79337.794755.78977.....66.74636.367 7. Result and discussion For finding value of F () = η, we solve F() =, and Re = 5 and α = 5 we get the root η = 3.5394. Table shows the value of F () = η for various Reynolds numbers and angles. For finding the root of η by using HPM we apply Padé approimation. As indicated in Figs. 2 7, the comparison between numerical results and VIM and HPM were illustrated in various Reynolds number. And in Tables 2 5, the comparison between numerical result and HPM and VIM have been illustrated via various Reynolds number. 8. Conclusions Results clearly show that both methods of VIM and HPM, which were applied to the Jeffery Hamel flow problem, were capable of solving them with successive rapidly convergent approimations without any restrictive assumptions or transformations causing changes in the physical definition of the problem. Among these two methods, VIM is very user friendly because it reduces the size of calculations and also its iterations are direct and straightforward. But as shown in Figs. 8 and 9, HPM led to more appropriate results when compared with that of VIM. The results
Z.Z. Ganji et al. / Computers and Mathematics with Applications 58 (29) 27 26 25 Table 5 The comparison between the Numerical, HPM and VIM solutions for F, when α = 5 and Re = 7. NM HPM VIM Error HPM Error VIM.......5.9993344.9993345.9993353..92..9972957.9972958.99728785..372.5.9937767.9937768.99376763..854.2.988667.988669.9885954.2.563.25.985937.98594.98494.3.2535.3.972586.9725866.972244.5.387.35.96568.965625.9646.7.5472.4.94464.944642.94453837..7575.45.925854.925867.9249784.3.23.5.95224.95223.938737.7.3477.55.86979373.86979394.8696936.22.7436.6.835656.835684.8328562.28.2294.65.78399899.78399934.7837259.36.2739.7.72535.7253555.72498827.45.32684.75.653254.6532.65282636.56.3748.8.5652948.56538.56472748.7.42.85.45838224.458383.45798998.86.39227.9.339957.33258.329874.2.32555.95.78448.7845.7785374.2.934.......35.3.25 Error.2.5..5.25.5.75 Fig. 8. The comparison between error of HPM and VIM solutions for F(), Re = 29, α = 5. 5.5E-5 5E-5 4.5E-5 4E-5 3.5E-5 Error 3E-5 2.5E-5 2E-5.5E-5 E-5 5E-6.25.5.75 Fig. 9. The comparison between error of HPM and VIM solutions for F (), Re = 7, α = 5.
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