Solution of the Jeffery–Hamel flow problem by optimal homotopy asymptotic method

Computers and Mathematics with Applications 59 (2010) 3405–3411

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Solution of the Jeffery–Hamel flow problem by optimal homotopy asymptotic method M. Esmaeilpour a,∗ , D.D. Ganji b a

Ferdowsi University, Department of Mechanical Engineering, Mashhad, Iran

b

Babol University of Technology, Department of Mechanical Engineering, Babol, Iran

article

info

Article history: Received 20 December 2009 Received in revised form 11 March 2010 Accepted 12 March 2010 Keywords: Jeffery–Hamel flows Optimal homotopy analysis method (OHAM) Nonlinear ordinary differential equation

abstract The present article addresses Jeffery–Hamel flow: fluid flow between two rigid plane walls, where the angle between them is 2α . A new analytical method called the optimal homotopy asymptotic method (OHAM) is briefly introduced, and then employed to solve the governing equation. The validity of the homotopy asymptotic method is ascertained by comparing our results with numerical (Runge–Kutta method) results. The effects of the Reynolds number (Re) and the angle between the two walls (2α ) are highlighted in the proposed work. The results reveal that the proposed analytical method can achieve good results in predicting the solutions of such problems. © 2010 Elsevier Ltd. All rights reserved.

1. Introduction The incompressible viscous fluid flow through convergent–divergent channels is one of the most applicable cases in fluid mechanics, civil, environmental, mechanical and biomechanical engineering. The mathematical investigations of this problem were pioneered by Jeffery [1] and Hamel [2] (i.e. Jeffery–Hamel flows). Jeffery–Hamel flows are an exact 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 vertex and with a source or sink at the vertex and have been extensively studied by several authors and discussed in some textbooks and articles: [3–11], etc. Most scientific problems such as Jeffery–Hamel flows and other fluid mechanic problems are inherently nonlinear. Excepting a limited number of these problems, most do not have analytical solutions. Therefore, these nonlinear equations should be solved using other methods. In recent years, much attention has been devoted to the newly developed methods for constructing an analytic solution of an equation; such methods include the homotopy analysis method (HAM) [12–19], the homotopy perturbation method (HPM) [20–26], and the variational iteration method (VIM) [27–32]. Perturbation techniques are too strongly dependent upon the so-called ‘‘small parameters’’. Thus, it is worthwhile developing some new analytic techniques not dependent upon small parameters. One of the newest analytical methods is the optimal homotopy asymptotic method (OHAM) which is not dependent upon a small parameter. Recently, a few authors used this method to solve nonlinear equations arising in engineering problems [33,34]. The aim of the present work is to solve a nonlinear differential equation governing Jeffery–Hamel flow using a new analytical technique called the optimal homotopy asymptotic method (OHAM). The nonlinear equations of these problems are solved through this method and compared with the results obtained by a numerical method (the fourth-order Runge–Kutta one). Some plots and a table are presented to show the reliability and simplicity of the method.

∗

Corresponding author. E-mail address: [email protected] (M. Esmaeilpour).

0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2010.03.024

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2. Fundamentals of the optimal homotopy asymptotic method To explain the basic idea of the OHAM for solving nonlinear differential equations, we consider the following nonlinear differential equation [33,34]: A(u(τ )) + f (τ ) = 0,

(2.1)

subject to a boundary condition: B(u) = 0,

(2.2)

where A is a general differential operator, B is a boundary operator, and f (τ ) is a known analytical function. The operator A can, generally speaking, be divided into two parts: a linear part L and a nonlinear part N. By means of OHAM one first constructs a family of equations:

(1 − p)[L(φ(τ , p)) + f (τ )] = H (p)[L(φ(τ , p)) + f (τ ) + N (φ(τ , p))],

(2.3)

and the boundary condition is B(φ(τ , p)) = 0.

(2.4)

In Eq. (2.3), φ(τ , p) is an unknown function, p ∈ [0 , 1] is an embedding parameter and H (p) is a nonzero auxiliary function for p 6= 0 and H (0) = 0. Thus, when p increases from 0 to 1 the solution φ(τ , p) changes between the initial guess u0 (τ ) and the solution u(τ ). Obviously, when p = 0 and p = 1 it holds that

φ(τ , 0) = u0 (τ ),

φ(τ , 1) = u(τ ).

(2.5)

We choose the auxiliary function H (p) in the form H (p) = pC1 + p2 C2 + · · ·

(2.6)

where C1 , C2 , . . . are constants which can be determined later. Expanding φ(τ , p) in a series with respect to p, one has

φ(τ , p, Ci ) = u0 (τ ) +

+∞ X

um (τ , Ci )pm ,

i = 1, 2, 3, . . . .

(2.7)

m=1

Substituting Eq. (2.7) into Eq. (2.3), collecting the same powers of p, and equating each coefficient of p to zero, we obtain L(u0 (τ )) + f (τ ) = 0,

B(u0 ) = 0

L(u1 (τ )) = C1 N0 (u0 (τ )),

(2.8)

B(u1 ) = 0,

(2.9)

m−1

L(um (τ ) − um−1 (τ )) = Cm N0 (u0 (τ )) +

X

Ci [L(um−i (τ )) + Nm−i (u0 (τ ), u1 (τ ), . . . , um−i (τ ))],

m = 2, 3, . . .

i=1

(2.10) B(um ) = 0,

Cm = K (t )

where Ni , i ≥ 0, are the coefficients of pi in the nonlinear operator N: N (u(τ )) = N0 (u0 (τ )) + pN1 (u0 (τ ), u1 (τ )) + p2 N2 (u0 (τ ), u1 (τ ), u2 (τ )) + · · ·

(2.11)

and K (t ) is a function which will be defined later. It should be emphasized that the uk for k ≥ 0 are governed by the linear Eqs. (2.8)–(2.10) with the linear boundary conditions that come from original problem, which can be easily solved. The convergence of the series (2.7) depends upon the auxiliary constants C1 , C2 , . . .. If it is convergent at p = 1, one has u(τ , Ci ) = u0 (τ ) +

+∞ X

um (τ , Ci ).

(2.12)

m=1

Generally speaking, the solution of Eq. (2.1) can be determined approximately in the form u˜ (k) = u0 (τ ) +

k X

um (τ , Ci ).

(2.13)

m=1

We note that the last coefficient Ck can be a function of τ . Substituting Eq. (2.13) into Eq. (2.1), there results the following residual: R(τ , Ci ) = L(˜u(k) (τ , Ci )) + f (τ ) + N (˜u(k) (τ , Ci )),

i = 1, 2, . . . .

(2.14)

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α θ

q

u(r,θ )

Fig. 1. Geometry of the problem.

If R(τ , Ci ) = 0 then u˜ (m) (τ , Ci ) happens to be the exact solution. Generally such a case will not arise for nonlinear problems, but we can minimize the functional: J (C1 , C2 , . . . , Cn ) =

Z

b

R2 (τ , C1 , C2 , . . . , Ck )dτ

(2.15)

a

where a and b are two values, depending on the given problem. The unknown constants Ci (i = 1, 2, . . . , k) can be identified from the conditions

∂J ∂J = = · · · = 0. ∂ C1 ∂ C2

(2.16)

With these constants known, the approximate solution (of order k) (2.13) is well determined. The constants Ci can be determined in other forms; for example, if ki ∈ (a, b), i = 1, 2, . . ., and we substitute ki into Eq. (2.14), we obtain the equation [3–34]: R(k1 , Ci ) = R(k2 , Ci ) = · · · = R(km , Ci ) = 0,

i = 1 , 2 , . . . , m.

(2.17)

It is easy to observe that the so-called homotopy perturbation method (HPM) proposed by He [20–22] is only a special case of Eq. (2.3) when H (p) = −p, and the homotopy analysis method (HAM) proposed by Liao [12–15] is also another special case of Eq. (2.3) when H (p) = ph¯ and F (¯r , τ ) = 1, where the parameter h¯ is determined from so-called ‘‘h¯ -curves’’. 3. Applications In this section, we shall illustrate the analytical scheme by means of fluid flow problems, which have been discussed in the literature. 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, where the angle between them is 2α as shown in Fig. 1. We assume that the velocity is only along the radial direction and depends on r and θ , V (u(r , θ ), 0) [10,11]. Using continuity and the Navier–Stokes equations in polar coordinates,

ρ∂ r∂r

(ru(r , θ )) = 0,

u( r , θ )

−

(3.1)

∂ u(r , θ ) 1 ∂p ∂ 2 u(r , θ ) 1 ∂ u(r , θ ) 1 ∂ 2 u( r , θ ) u(r , θ ) =− +v + + − ∂r ρ ∂r ∂r2 r ∂r r2 ∂θ 2 r2 

1 ∂p 2v ∂ u(r , θ ) + 2 = 0. ρ r ∂θ r ∂θ

 (3.2) (3.3)

From Eq. (3.1), f (θ ) ≡ ru(r , θ ).

(3.4)

Using dimensionless parameters, F (x) ≡

f (θ ) fmax

,x ≡

θ α

(3.5)

and eliminating P between Eqs. (3.2) and (3.3), we obtain an ordinary differential equation for the normalized function profile F (x) [10]: F 000 (x) + 2α Re F (x)F 0 (x) + 4α 2 F 0 (x) = 0.

(3.6)

Since we have a symmetric geometry, the boundary conditions will be F (0) = 1,

F 0 (0) = 0,

F (1) = 0.

(3.7)

The Reynolds number is Re ≡

fmax α

v

=

Umax r α

v



divergent channel : α > 0, Umax > 0 . convergent channel : α < 0, Umax < 0



(3.8)

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4. Application of OHAM to the Jeffery–Hamel flow problem Now we apply optimal homotopy asymptotic method to solve Eq. (3.6). We choose the linear operator

∂ 3 φ(x, p) , ∂ x3

L(φ(x, p)) = f (x) = 0,

(4.1)

and we define a nonlinear operator N (φ(x, p)) = 2α Re φ(x, p)

∂φ(x, p) ∂φ(x, p) + 4α 2 , ∂x ∂x

(4.2)

and the boundary conditions are

∂φ(0, p) = 0. ∂x The zeroth-order problem is given by Eq. (2.8) (f (τ ) = 0): φ(0, p) = 1,

d3 F0 dx3

φ(1, p) = 0,

(4.3)

= 0,

(4.4)

F0 (0) = 1,

dF0 (0)

F0 (1) = 0,

dx

= 0.

It is obtained that F0 (x) = 1 − x2 .

(4.5)

The first-order problem is given by Eq. (2.9): d3 F1 dx3

− 4C1 α 2

dF0 dx

F1 (0) = 0,

− C1

d3 F 0

−

dx3

F1 (1) = 0,

d3 F 0

− 2C1 α Re F0 (x)

dx3 dF1 (0) dx

dF0 dx

= 0,

(4.6)

= 0.

The solution of Eq. (4.6) is given by F1 (x) = −4α C1

 −

1 120

Re x6 +

1 24

   1 4 2 C1 α Re + C1 α 2 x2 . (2α + Re) x4 + 2

15

(4.7)

3

The second-order problem is given by Eq. (2.10) for m = 2:

     3   3  ∂ 3 F2 ∂ 3 F1 dF1 (x) dF0 (x) ∂ F1 ∂ F0 2 2 − − 4C α − 4C α − C − C 1 2 1 2 3 ∂ x3 ∂ x3 dx dx ∂ x ∂ x3       ∂ F0 ∂ F1 ∂ F0 − 2C2 α Re F0 (x) 2C1 α Re F0 (x) =0 − 2C1 α Re F1 (x) ∂x ∂x ∂x ∂ F2 (0) = 0. F2 (0) = 0, F2 (1) = 0, ∂x

(4.8)

So, the solution of Eq. (4.8) is given by F2 (x) = −

+

4 15

α

1 120



1 360

α C12 Re2 x10 +

1 336

(−9α C12 Re2 − 18α 2 C12 Re)x8

(−15C1 Re + 20α 3 C12 − 15C12 Re − 15C2 Re + 30α 2 C12 Re

+ 9α C12 Re2 )x6 +

1

(30C12 α + 15C2 Re − 10α 3 C12 + 30C2 α + 30C1 α   1 163 2 2 2 − 2α C12 Re2 + 15C1 Re + 15C12 Re − 9α 2 C12 Re)x4 + − α C1 Re 24

2

−

2 21

α 3 C12 Re +

4 15

C1 α Re −

2 15

α 4 C12 +

4 15

α C12 Re +

9450

4 15

 2 2 2 α C2 Re + α 2 C12 + C2 α 2 + C1 α 2 x2 . 3

3

3

(4.9)

Substitution of Eqs. (4.5), (4.7) and (4.9) into Eq. (2.13) yields the second-order approximate solution (k = 2) for Eqs. (3.6): F˜ (2) (x) = F0 (x) + F1 (x) + F2 (x)

(4.10)

M. Esmaeilpour, D.D. Ganji / Computers and Mathematics with Applications 59 (2010) 3405–3411

a

b

c

d

e

f

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Fig. 2. Sample profiles of the F (x) function for: (a) α = 3 and Re = 50; (b) α = 3 and Re = 210; (c) α = 5 and Re = 50; (d) α = 5 and Re = 100; (e) α = 5 and Re = 118; (f) α = 5 and Re = 160.

and substituting this approximate solution into Eq. (2.14) yields the residual and the functional J: R(x, C1 , C2 ) = J (C1 , C2 ) =

d3 F˜ dx3

dF˜ dF˜ + 2α Re F˜ (x) + 4α 2 , dx

dx

(4.11)

∞

Z

R2 (x, C1 , C2 )dx.

(4.12)

0

For x = 0.1, 0.5 in Eq. (4.10), we obtain the approximate solution of the third order for α = 5 and Re = 50. In this particular case, the constants C1 and C2 are

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M. Esmaeilpour, D.D. Ganji / Computers and Mathematics with Applications 59 (2010) 3405–3411

a

b

c

d

e

f

Fig. 3. Sample profiles of the F (x) function for: (a) α = −3 and Re = 50; (b) α = −3 and Re = 210; (c) α = −5 and Re = 50; (d) α = −5 and Re = 150; (e) α = −5 and Re = 200; (f) α = −5 and Re = 210.

C1 = −1.224613762,

C2 = 0.07286874234,

(4.13)

and the approximate solution of second order will be in the form F (x) = 1 − 1.760914211x2 + 1.279282010x4 − 0.02114898331x10 − 0.7018673114x6 + 0.2046484977x8 .

(4.14)

In a similar manner, we will obtain other solutions for different cases of α and Re. Table 1 shows the errors involved in the differential transformation method, along with the numerical solution obtained by the fourth-order Runge–Kutta method. Figs. 2 and 3 display plots of F (x) for different cases of α and Re.

M. Esmaeilpour, D.D. Ganji / Computers and Mathematics with Applications 59 (2010) 3405–3411

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Table 1 Comparison of the optimal homotopy asymptotic method (OHAM) and Runge–Kutta solutions for α = 5 and Re = 50. x

FOHAM

FNM

Error = |FNM − FOHAM |

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1.000000000 0.982518086 0.931565885 0.851381546 0.748260398 0.629538651 0.502428940 0.372933832 0.245081977 0.120715623 0.000000001

1.00000000 0.98243124 0.93122597 0.85061063 0.74679081 0.62694818 0.49823446 0.36696635 0.23812375 0.11515193 0.00000000

0.000000000 0.000086843 0.000339917 0.000770918 0.001469584 0.002590470 0.004194482 0.005967484 0.006958228 0.005563692 0.000000001

5. Conclusion In this paper, Jeffery–Hamel flow has been considered and a kind of analytical method, called the optimal homotopy asymptotic method (OHAM), is also used to obtain the exact solution of this problem in the form of analytical expressions. Also, this problem is solved by a numerical method (the Runge–Kutta method of order 4) using the software MAPLE, and the results and the corresponding results from the two different methods OHAM and NM are compared in Table 1 and Figs. 2–3. The results obtained in this paper confirm the notion that the OHAM method is a powerful and efficient technique for finding exact solutions for differential equations which have great significance in different fields of science and engineering. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34]

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