Solution of Boundary Layer Problems with Heat Transfer by Optimal Homotopy Asymptotic Method

Abstract

Application of Optimal Homotopy Asymptotic Method (OHAM), a new analytic approximate technique for treatment of Falkner-Skan equations with heat transfer, has been applied in this work. To see the efficiency of the method, we consider Falkner-Skan equations with heat transfer. It provides us with a convenient way to control the convergence of approximate solutions when it is compared with other methods of solution found in the literature as finite difference (N. S. Asaithambi, 1997) and shooting method (Cebeci and Keller, 1971). The obtained solutions show that OHAM is effective, simpler, easier, and explicit.

1. Introduction

Most of the problems in engineering sciences are nonlinear, particularly some of heat transfer problems. Limited analytic methods are presented for the solution of such problems in the literature. Therefore, the researcher’s profound attention is to hunt some new analytic methods for the solution of the problems. Most of the methods like Adomian Decomposition Method (ADM) [1], Variational Iteration Method (VIM) [2], Differential Transform Method (DTM) [3], Radial basis function [4] and Homotopy Perturbation Method (HPM) [5], are used for the solution of weakly nonlinear problems, and limited for strongly nonlinear problems. For the solution of the strongly nonlinear problems the perturbation methods were studied [6–8]. These methods comprise a small parameter which cannot be found easily. To overcome this issue, some new analytic methods such as Artificial Parameters Method [9], Homotopy Analysis Method (HAM) [10], and Homotopy Perturbation Method (HPM) [5] were introduced. These methods pooled the homotopy with the perturbation techniques. Recently, Marinca et al. introduced Optimal Homotopy Asymptotic Method (OHAM) [11–15] for the solution of nonlinear problems which made the perturbation methods independent of the assumption of small parameters.

The Falkner-Skan equation has been considered in the last forty years due to its importance in the boundary layer theory. The boundary layer theory plays a vital role in the diverse area of engineering and scientific applications. The solution of the Falkner-Skan equation has been studied numerically first by Hartree [16]. Smith and Cebeci [17, 18] solved this equation by shooting method. Maksyn [19] solved the Falkner-Skan equation by analytic approximation. Asithambi [20–22] found its solution by finite differences, Liao [23] applied homotopy analysis to solve Falkner-Skan equation, and recently Vera [24] found its solution by Fourier series. An important case is the Blasius equation. This problem was solved by Rosales and Valencia [25] using Fourier series. Boyd [26] found the solution of Falkner-Skan equation by numerical method. An enormous amount of research work has been invested in the study of nonlinear boundary value problems [27–38]. In this paper, we will deal with the Falkner-Skan equations with heat transfer, a nonlinear boundary value problem [24] in different forms.

The motivation of this paper is to enhance OHAM for the solution of Falkner-Skan equation with heat transfer. In [11–15], OHAM has been proved to be useful for obtaining an approximate solution of nonlinear boundary value problems. In this work, we have proved that OHAM is also useful and reliable for the solution of the Falkner-Skan equation with heat transfer, hence showing its validity and great potential for the solution of transient physical phenomenon in science and engineering.

In the succeeding section, the basic idea of OHAM [11–15] is formulated for the solution of boundary value problems arising in heat transfer. In Section 3, the effectiveness of the enhanced formulation of OHAM for Falkner-Skan equation with heat transfer has been studied. Two special cases [24] of Falkner-Skan equation with heat transfer problems have been analyzed.

2. Basic Mathematical Theory of OHAM

Let us consider the following differential equation: ℒ(𝑢(𝑡))+ℎ(𝑡)+𝒩(𝑢(𝑡))=0, (1) along with boundary conditions of the form ℬ𝑢,𝑑𝑢𝑑𝑡=0, (2) where ℒ is the linear operator, 𝑢(𝑡) is an unknown function, ℎ(𝑡) is a known function, 𝒩(𝑢(𝑡)) is a nonlinear differential operator, and ℬ is a boundary operator.

According to OHAM, one can construct an optimal homotopy 𝜙(𝑡,𝑞): Ω×[0,1]→ℜ which satisfies ℒ𝜙(𝑡𝑞ℒ𝜙(𝑡)𝜙,ℬ𝜙,1−𝑞𝑡,𝑞+ℎ)=𝐻𝑡,𝑞+ℎ+𝒩𝑡,𝑞𝑡,𝑞𝜕𝜙𝑡,𝑞𝜕𝑡=0, (3)(4) where 𝑞∈[0,1] is an embedding parameter, 𝜙(𝑡,𝑞) is an unknown function, and 𝐻(𝑞) is a nonzero auxiliary function. The auxiliary function 𝐻(𝑞) is nonzero for 𝑞≠0 and 𝐻(0)=0. Equation (3) is the structure of OHAM homotopy.

It is defined that 𝑞=0⟹𝜙(𝑡,0)=𝑢₀(𝑡),𝑞=1⟹𝜙(𝑡,1)=𝑢(𝑡), (5) respectively. Thus, as 𝑞 varies from 0 to1, the solution 𝜙(𝑡,𝑞) varies from 𝑢₀(𝑡) to 𝑢(𝑡), where 𝑢₀(𝑡) is obtained from (1) and (2) for 𝑝=0 as follows: ℒ𝑢₀(𝑡)+ℎ(𝑡)=0,ℬ𝑢₀,𝑑𝑢₀𝑑𝑡=0. (6)

Next, we choose auxiliary function 𝐻(𝑞) in the form 𝐻𝑞=𝑞𝐶₁+𝑞²𝐶₂+𝑞³𝐶₃+⋯, (7) where 𝐶₁,𝐶₂,𝐶₃,… are constants and can be found latter.

To obtain an approximate solution, we expand 𝜙(𝑡,𝑞,𝐶_𝑖) by Taylor’s series about 𝑞 in the following form: 𝜙𝑡,𝑞,𝐶₁,𝐶₂,…,𝐶_𝑖=𝑢₀(𝑡)+^∞_𝑘=1𝑢_𝑘𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖𝑞^𝑘,𝑖=1,2,…. (8)

Now substituting (8) into (1) and (2) and equating the coefficient of like powers of 𝑞, we obtain the zeroth-order problem given by (6), the first- and second-order problems given by (9)–(11), respectively, and the general governing equations for 𝑢_𝑘(𝑡) given by (11): ℒ𝑢₁(𝑡)=𝐶₁𝒩₀𝑢₀(𝑡),ℬ𝑢₁,𝑑𝑢₁ℒ𝑢𝑑𝑡=0,₂(𝑡)−ℒ𝑢₁(𝑡)=𝐶₁𝒩₀𝑢₀(𝑡)+𝐶₁ℒ𝑢₁(𝑡)+𝒩₁𝑢₀(𝑡),𝑢₁(𝑡),ℬ𝑢₂,𝑑𝑢₂ℒ𝑢𝑑𝑡=0,_𝑘(𝑡)−ℒ𝑢_𝑘−1(𝑡)=𝐶_𝑘𝒩₀𝑢₀(𝑡+_{𝑘−1𝑖=1})𝐶_𝑖ℒ𝑢_𝑘−𝑖(𝑡)+𝒩_𝑘−𝑖𝑢₀(𝑡),𝑢₁(𝑡),…,𝑢_𝑘−𝑖(𝑡),ℬ𝑢_𝑘,𝑑𝑢_𝑘𝑑𝑡=0,𝑘=2,3,…, (9)(10)(11) where 𝒩_𝑘−𝑖(𝑢₀(𝑡),𝑢₁(𝑡),…,𝑢_𝑘−𝑖(𝑡)) is the coefficient of 𝑞^𝑘−𝑖 in the expansion series of 𝒩(𝜙(𝑡,𝑞)) about the embedding parameter 𝑝 as follows: 𝒩𝜙𝑡,𝑞,𝐶_𝑖=𝒩₀𝑢₀(𝑡)+_𝑘≥1𝒩_𝑘𝑢₀,𝑢₁,𝑢₂,…,𝑢_𝑘𝑞^𝑘,𝑖=1,2,3,…. (12)

It should be underscored that the 𝑢_𝑘 for 𝑘≥0 are governed by the linear equations with linear boundary conditions that come from the original problem, which can be easily solved.

It has been observed that the convergence of the series (8) depends upon the auxiliary constants 𝐶₁,𝐶₂,…. If it is convergent at 𝑞=1, one has 𝑢𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖=𝑢₀(𝑡)+_𝑘≥1𝑢_𝑘𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖. (13)

Substituting (13) into (1), it results in the following expression for residual: 𝑅𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖=ℒ𝑢𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖+ℎ(𝑡)+𝒩𝑢𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖. (14)

If 𝑅(𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖)=0, then 𝑢(𝑡,𝐶₁,𝐶₂,…,𝐶_𝑖) is the exact solution of the problem. Generally it does not happen, especially in nonlinear problems.

For the determinations of auxiliary constants, 𝐶_𝑖,𝑖=1,2,…,𝑚, there are different methods like Galerkin’s method, Ritz method, least squares method, and collocation method. One can apply the method of  least squares as follows: 𝐽𝐶₁,𝐶₂,…,𝐶_𝑚=^𝑏_𝑎𝑅²𝑡,𝐶₁,𝐶₂,𝐶₃…,𝐶_𝑚𝑑𝑡, (15) where 𝑎 and 𝑏 are two values, depending on the nature of the given problem.

The auxiliary constants 𝐶_𝑖,𝑖=1,2,…,𝑚 can be optimally found from 𝜕𝐽𝜕𝐶₁=𝜕𝐽𝜕𝐶₂=⋯=𝜕𝐽𝜕𝐶_𝑚=0. (16)

The 𝑚th order approximate solution can be obtained bythese constants. The constants 𝐶_𝑖 can also be determined by another method as follows: 𝑅𝑘₁,𝐶₁,𝐶₂,…,𝐶_𝑚=𝑅𝑘₂,𝐶₁,𝐶₂,…,𝐶_𝑚=⋯=𝑅𝑘_𝑚,𝐶₁,𝐶₂,…,𝐶_𝑚=0,𝑖=1,2,…,𝑚. (17)

The convergence of OHAM is directly proportional to the number of optimal constants 𝐶₁,𝐶₂,… which is determined by (16).

It is easy to observe [13] that the Homotopy Perturbation Method (HPM) proposed by He [4] is a special case of (3) when 𝐻(𝑞)=−𝑞, and on the other hand, the Homotopy Analysis Method (HAM) proposed by Liao [11] is another special case of (3) when 𝐻(𝑞)=𝑞ℏ where ℏ is chosen from “ℏ-curves” [12].

3. Application of OHAM to Falkner-Skan Equations with Heat Transfer

To demonstrate the effectiveness of OHAM formulation, two models are studied.

Model 1 (see [23]). When an incompressible fluid passes in the vicinity of solid boundaries, the Navier-Stokes equations may be reduced drastically into the boundary layer equations: 𝑢𝜕𝑢𝜕𝑥+𝑣𝜕𝑣𝜕𝑦=𝑈_𝑒𝑑𝑈_𝑒𝜕𝑑𝑥+𝑣²𝑢𝜕²𝑦,𝜕𝑢+𝜕𝑥𝜕𝑣𝜕𝑦=0, (18) where 𝑈_𝑒(𝑥) is the free stream velocity, 𝑢 and 𝑣 are velocity components in 𝑥- and 𝑦-directions, and 𝜐 is the kinematic viscosity. In case of two-dimensional flow, the incompressible boundary layer flow over a wedge, when the free stream velocity is of the form 𝑈_𝑒(𝑥)=𝐾𝑥^𝑚, is the following similarity transformation: 𝑢𝑥,𝑦=𝑈_𝑒(𝑥)𝑓^′𝜂,𝜂=𝑦(𝑚+1)𝐾2𝑥^(𝑚−1)/2,𝜐𝜂=𝑇−𝑇_∞𝑇_𝑤−𝑇_∞. (19)
Using (19) into (18), we obtained the Falkner-Skan equation 𝑑³𝑓𝑑𝜂³𝑑+𝑓²𝑓𝑑𝜂²+𝛽1−𝑑𝑓𝑑𝜂²=0, (20) along with boundary conditions 𝑓(0)=0,𝑓^′(0)=0,𝑓^′(5)=1. (21)
According to (1), we have ℒ𝑓𝜂=𝑓^′′′𝜂𝜂𝒩𝑓𝜂𝜂𝑓+𝛽,ℎ=0,=𝑓^′′𝜂−𝛽𝑓^′𝜂². (22)
The boundary conditions are 𝑓(0)=0,𝑓^′(0)=0,𝑓^′(5)=1. (23)
Applying the method formulation mentioned in Section 2 leads to the following.
Zeroth-Order Problem. Consider 𝑓₀^′′′𝜂𝑓+𝛽=0,₀(0)=0,𝑓^′₀(0)=0,𝑓^′₀(5)=1, (24)(25) from which we obtain 𝑓₀𝜂=1𝜂606+75𝛽²−1060𝛽𝜂³. (26)
First-Order Problem. Consider 𝑓₁^′′′𝜂=𝛽+𝐶₁𝛽−𝛽𝑓^′₀²+𝑓₀𝑓₀^′′+𝑓₀^′′′+𝑓₀^′′′,𝑓₁(0)=0,𝑓^′₁(0)=0,𝑓^′₁(5)=0. (27)(28)
Its solution is 𝑓₁𝜂=−5𝐶₁−485𝛽𝐶₁+161825𝛽²𝐶₁+576625𝛽³𝐶₁𝜂96²+𝐶₁+300023𝛽𝐶₁+300017𝛽²𝐶₁−4805𝛽³𝐶₁𝜂48⁵+−𝛽𝐶₁−90011𝛽²𝐶₁+𝛽900³𝐶₁𝜂48⁶+𝛽²𝐶₁−𝛽1260³𝐶₁𝜂840⁷. (29)
Second-Order Problem. Consider 𝑓₂^′′′𝜂=1+𝐶₁𝑓₁^′′′+𝐶₂𝛽+𝑓₀𝑓₀^′′−𝛽𝑓^′₀+𝑓₀^′′′+𝐶₁𝑓₁𝑓₀^′′−2𝛽𝑓^′₀𝑓^′₁+𝑓₀𝑓₁^′′, (30) with BC 𝑓₂(0)=0,𝑓^′₂(0)=0,𝑓^′₂(5)=0, (31) whose solution is 𝑓₂𝜂=−5𝐶₁−485𝛽𝐶₁+161825𝛽²𝐶₁+576625𝛽³𝐶₁−9695𝐶²₁−4032395𝛽𝐶²₁−504538225𝛽²𝐶²₁[phone]𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁[phone]𝛽⁵𝐶²₁−483845𝐶₂−485𝛽𝐶₂+161825𝛽²𝐶₂+576625𝛽³𝐶₂𝜂96²+𝐶₁+300023𝛽𝐶₁+300017𝛽²𝐶₁−4805𝛽³𝐶₁−4813𝐶²₁−3600041𝛽𝐶²₁[phone]𝛽²𝐶²₁[phone]𝛽³𝐶²₁−2304125𝛽⁴𝐶²₁−1728625𝛽576⁵𝐶²₁+𝐶₂+300023𝛽𝐶₂+300017𝛽²𝐶₂−4805𝛽³𝐶₂𝜂48⁵+−𝛽𝐶₁−90011𝛽²𝐶₁+𝛽900³𝐶₁+48𝛽𝐶²₁−21600151𝛽²𝐶²₁−14400203𝛽³𝐶²₁−510368𝛽256⁴𝐶²₁+125𝛽⁵𝐶²₁−1152𝛽𝐶₂−90011𝛽²𝐶₂+𝛽900³𝐶₂𝜂48⁶+𝛽²𝐶₁−𝛽1260³𝐶₁+𝛽840²𝐶²₁−11260𝛽840³𝐶²₁+𝛽²𝐶₂−𝛽1260³𝐶₂𝜂840⁷+11𝐶²₁[phone]𝛽𝐶²₁[phone]𝛽²𝐶²₁[phone]𝛽³𝐶²₁−32256085𝛽⁴𝐶²₁+8064125𝛽⁵𝐶²₁𝜂16128⁸+−𝛽𝐶²₁−100800253𝛽²𝐶²₁−113400017089𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁−72576055𝛽⁵𝐶²₁𝜂24192⁹+𝛽²𝐶²₁+𝛽70875³𝐶²₁−7000227𝛽⁴𝐶²₁+𝛽567000⁵𝐶²₁𝜂4320¹⁰+−𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁−𝛽1247400⁵𝐶²₁𝜂118800¹¹. (32)
Adding (25), (27), and (29), we obtain 𝑓𝜂=𝐴₂𝜂²+𝐴₃𝜂³+𝐴₅𝜂⁵+𝐴₆𝜂⁶+𝐴₇𝜂⁷+𝐴₈𝜂⁸+𝐴₉𝜂⁹+𝐴₁₀𝜂¹⁰+𝐴₁₁𝜂¹¹, (33) where 𝐴₂=1+105𝛽4−5𝐶₁−245𝛽𝐶₁8+1825𝛽²𝐶₁+288625𝛽³𝐶₁−4895𝐶²₁−4032395𝛽𝐶²₁−504538225𝛽²𝐶²₁[phone]𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁[phone]𝛽⁵𝐶²₁−483845𝐶₂−485𝛽𝐶₂+161825𝛽²𝐶₂+576625𝛽³𝐶₂,𝐴96₃=−𝛽6,𝐴₅=𝐶₁+150023𝛽𝐶₁+150017𝛽²𝐶₁−2405𝛽³𝐶₁−2413𝐶²₁−3600041𝛽𝐶²₁[phone]𝛽²𝐶²₁[phone]𝛽³𝐶²₁−2304125𝛽⁴𝐶²₁−1728625𝛽576⁵𝐶²₁+𝐶₂+300023𝛽𝐶₂+300017𝛽²𝐶₂−4805𝛽³𝐶₂,𝐴48₆=−𝛽𝐶₁−45011𝛽²𝐶₁+𝛽450³𝐶₁+24𝛽𝐶²₁−21600151𝛽²𝐶²₁−14400203𝛽³𝐶²₁−510368𝛽256⁴𝐶²₁+125𝛽⁵𝐶²₁−1152𝛽𝐶₂−90011𝛽²𝐶₂+𝛽900³𝐶₂,𝐴48₇=𝛽²𝐶₁−𝛽630³𝐶₁+𝛽420²𝐶²₁−11260𝛽840³𝐶²₁+𝛽²𝐶₂−𝛽1260³𝐶₂,𝐴840₈=11𝐶²₁[phone]𝛽𝐶²₁[phone]𝛽²𝐶²₁[phone]𝛽³𝐶²₁−32256085𝛽⁴𝐶²₁+8064125𝛽⁵𝐶²₁,𝐴16128₉=−𝛽𝐶²₁−100800253𝛽²𝐶²₁−113400017089𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁−72576055𝛽⁵𝐶²₁,𝐴24192₁₀=𝛽²𝐶²₁+𝛽70875³𝐶²₁−7000227𝛽⁴𝐶²₁+𝛽567000⁵𝐶²₁,𝐴4320₁₁=−𝛽³𝐶²₁[phone]𝛽⁴𝐶²₁−𝛽1247400⁵𝐶²₁.118800 (34)
For the computation of the constants 𝐶₁ and 𝐶₂ applying the method of least square mentioned in (14)–(16), we get 𝐶₁𝐶=−0.08144083042145557,₂=−0.05004674822211045,for𝛽=1. (35)
Putting these values in (30), we obtained the approximate solution of the form 𝑓𝜂=6.67768×10⁻¹𝜂²−1.677×10⁻¹𝜂³+7.00357×10⁻³𝜂⁵−1.20593×10⁻⁴𝜂⁶+8.18634×10⁻⁵𝜂⁷−3.23784×10⁻⁶𝜂⁸+5.32977×10⁻⁷𝜂⁹−7.89596×10⁻⁸𝜂¹⁰+2.65857×10⁻⁹𝜂¹¹. (36)
Now consider the energy equation of an incompressible fluid which passes through the vicinity of the solid boundaries: 𝑢𝜕𝑇𝜕𝑥+𝑣𝜕𝑇𝜕𝜕𝑦=𝑎²𝑇𝜕𝑦². (37)
Upon using the transformation in (19) into (37), we obtained 𝑑²𝜐𝑑𝜂²+𝑃𝑟𝑓𝑑𝜐𝑑𝜂=0, (38) with boundary conditions 𝜐(0)=1,𝜐(5)=0, (39) where “𝑎” is the thermal diffusivity and 𝑃𝑟 is the prandtl number.
Applying the method formulated in Section 2 leads to the following.
Zeroth-Order Problem. Consider 𝜐₀^′′𝜂=0,𝜐(0)=1,𝜐(5)=0. (40)
Its solution is 𝜐₀𝜂=5−𝜂5. (41)
First-Order Problem. Consider 𝜐₁^′′𝜂=𝑃𝑟𝐶₁𝑓₀𝜂𝜐^′₀𝜂+1+𝐶₁𝜐₀^′′𝜂,𝜐(0)=0,𝜐(5)=0, (42)
whose solution is

𝜐₁𝜂=5𝑃𝑟𝐶₁+242516𝑃𝑟𝛽𝐶₁𝜂−𝑃𝑟𝐶₁−160048𝑃𝑟𝛽𝐶₁𝜂⁴+1600𝑃𝑟𝛽𝐶₁𝜂⁵. (43)

Second-Order Problem. Consider 𝜐₂^′′𝜂=𝑃𝑟𝐶₂𝑓₀𝜂+𝐶₁𝑓₁𝜂𝜐₀𝜂+1+𝐶₁𝜐₁^′′𝜂+𝑃𝑟𝐶₁𝑓₀𝜂𝜐^′₁𝜂+𝐶₂𝜐₀^′′𝜂,𝜐₂(0)=0,𝜐₂(5)=0. (44)

We obtain the following solution: 𝜐₂𝜂=5𝑃𝑟𝐶₁+242516𝑃𝑟𝛽𝐶₁+65𝑃𝑟𝐶²₁+4032125𝑃²𝑟𝐶²₁[phone]𝑃𝑟𝛽𝐶²₁+625𝑃²𝑟𝛽𝐶²₁[phone]𝑃𝑟𝛽²𝐶²₁[phone]𝑃²𝑟𝛽²𝐶²₁[phone]𝑃𝑟𝛽³𝐶²₁+15125𝑃𝑟𝐶₂+242516𝑃𝑟𝛽𝐶₂𝜂+−𝑃𝑟𝐶₁−160048𝑃𝑟𝛽𝐶₁+𝑃𝑟𝐶²₁+114400𝑃576²𝑟𝐶²₁−164𝑃𝑟𝛽𝐶²₁+5𝑃144²𝑟𝛽𝐶²₁−365𝑃𝑟𝛽²𝐶²₁+6912125𝑃768²𝑟𝛽²𝐶²₁−125𝑃𝑟𝛽³𝐶²₁−1152𝑃𝑟𝐶₂−160048𝑃𝑟𝛽𝐶₂𝜂⁴+1600𝑃𝑟𝛽𝐶₁+1600𝑃𝑟𝛽𝐶²₁−1𝑃576²𝑟𝛽𝐶²₁−5𝑃384²𝑟𝛽²𝐶²₁+1600𝑃𝑟𝛽𝐶₂𝜂⁵+−𝑃𝑟𝐶²₁−𝑃630000²𝑟𝐶²₁−6300023𝑃𝑟𝛽𝐶²₁−𝑃630000²𝑟𝛽𝐶²₁−252017𝑃𝑟𝛽²𝐶²₁−1008005𝑃²𝑟𝛽²𝐶²₁+2016𝑃𝑟𝛽³𝐶²₁𝜂2016⁷+𝑃𝑟𝛽𝐶²₁+𝑃252000²𝑟𝛽𝐶²₁+2880011𝑃𝑟𝛽²𝐶²₁+𝑃252000₂𝑟𝛽²𝐶²₁−2304𝑃𝑟𝛽³𝐶²₁𝜂13440⁸+−𝑃𝑟𝛽²𝐶²₁−𝑃453600²𝑟𝛽²𝐶²₁+51840𝑃𝑟𝛽³𝐶²₁𝜂302400⁹. (45)

Adding (41), (43), and (45), we obtain 𝜐𝜂=1+𝐵₁𝜂+𝐵₄𝜂⁴+𝐵₅𝜂⁵+𝐵₇𝜂⁷+𝐵₈𝜂⁸+𝐵₉𝜂⁹, (46) where 𝐵₁=−15+5𝑃𝑟𝐶₁+12258𝑃𝑟𝛽𝐶₁+65𝑃𝑟𝐶²₁+4032125𝑃²𝑟𝐶²₁[phone]𝑃𝑟𝛽𝐶²₁+625𝑃²𝑟𝛽𝐶²₁[phone]𝑃𝑟𝛽²𝐶²₁[phone]𝑃²𝑟𝛽²𝐶²₁[phone]𝑃𝑟𝛽³𝐶²₁+15125𝑃𝑟𝐶₂+242516𝑃𝑟𝛽𝐶₂,𝐵₄=−𝑃𝑟𝐶₁−130024𝑃𝑟𝛽𝐶₁+𝑃𝑟𝐶²₁+114400𝑃576²𝑟𝐶²₁−164𝑃𝑟𝛽𝐶²₁+5𝑃144²𝑟𝛽𝐶²₁−365𝑃𝑟𝛽²𝐶²₁+6912125𝑃768²𝑟𝛽²𝐶²₁−125𝑃𝑟𝛽³𝐶²₁−1152𝑃𝑟𝐶₂−160048𝑃𝑟𝛽𝐶₂,𝐵₅=1300𝑃𝑟𝛽𝐶₁+1600𝑃𝑟𝛽𝐶²₁−1𝑃576²𝑟𝛽𝐶²₁−5𝑃384²𝑟𝛽²𝐶²₁+1600𝑃𝑟𝛽𝐶₂,𝐵₇=−𝑃𝑟𝐶²₁−𝑃630000²𝑟𝐶²₁−6300023𝑃𝑟𝛽𝐶²₁−𝑃630000²𝑟𝛽𝐶²₁−252017𝑃𝑟𝛽²𝐶²₁−1008005𝑃²𝑟𝛽²𝐶²₁+2016𝑃𝑟𝛽³𝐶²₁,𝐵2016₈=𝑃𝑟𝛽𝐶²₁+𝑃252000²𝑟𝛽𝐶²₁+2880011𝑃𝑟𝛽²𝐶²₁+𝑃252000₂𝑟𝛽²𝐶²₁−2304𝑃𝑟𝛽³𝐶²₁,𝐵13440₉=−𝑃𝑟𝛽²𝐶²₁−𝑃453600²𝑟𝛽²𝐶²₁+51840𝑃𝑟𝛽³𝐶²₁.302400 (47)

Using (46) in (38) and applying the method of least square, we obtain 𝐶₁𝐶=−0.027238050474420454,₂=−0.016903606167477613. (48)

Substituting these values in (46) for 𝛽=1 and 𝑃𝑟=10, we obtain 𝜐𝜂=1−1.29659𝜂+2.95285×10⁻²𝜂⁴−2.27213×10⁻³𝜂⁵−2.12478×10⁻⁴𝜂⁵+3.45784×10⁻⁶𝜂⁸−1.42298×10⁻⁶𝜂⁹. (49)

Model 2 (see [23]). In case of 𝛽=0 in (52), we obtain the Blasius equation which is the famous equation of fluid dynamics and represent the problem of an incompressible fluid that passes through on a semi-infinite flat plate.
One has 𝑑³𝑓𝑑𝜂³𝑑+𝑓²𝑓𝑑𝜂²𝑓(0)=0,=0,𝑓^′(0)=0,𝑓^′(5)=1. (50)(51)
According to (1), we define the operators ℒ𝑓𝜂=𝑓^′′′𝜂𝜂𝒩𝑓𝜂𝜂𝑓,𝑔=0,=𝑓^′′𝜂, (52) where 𝑓^′′(𝜂) and 𝑓^′′′(𝜂) represent the second and third derivatives of 𝑓(𝜂) with respect to 𝜂.

Applying the method formulated in Section 2 leads to the following.

Zeroth-Order Problem. Consider 𝑓₀^′′′𝜂𝑓=0,₀(0)=0,𝑓^′₀(0)=0,𝑓^′₀(5)=1. (53)(54)

Its solution is 𝑓₀𝜂=𝜂².10 (55)

First-Order Problem. Consider 𝑓₁^′′′𝜂=−𝐶₁𝑓₀𝜂𝑓₀^′′𝜂−𝑓₀^′′′𝜂−𝐶₁𝑓₀^′′′𝜂,𝑓₁(0)=0,𝑓^′₁(0)=0,𝑓^′₁(5)=0, (56)

whose solution is 𝑓₁𝜂=−5𝐶₁𝜂48²+𝐶₁𝜂3000⁵. (57)

Second-Order Problem. Consider 𝑓₂^′′′𝜂=−𝐶₂𝑓₀𝑓₀^′′−𝐶₁𝑓₁𝜂𝑓₀^′′𝜂−𝐶₁𝑓₀𝜂𝑓₁^′′𝜂−𝐶₂𝑓₀^′′′𝜂−1+𝐶₁𝑓₁^′′′𝜂,𝑓₂(0)=0,𝑓^′₂(0)=0,𝑓^′₂(5)=0. (58) We obtain the following solution: 𝑓₂𝜂=−5𝐶₁−4895𝐶²₁−40325𝐶₂𝜂48²+𝐶₁−300013𝐶²₁+𝐶36000²𝜂3000⁵+11𝐶²₁𝜂5040000⁸. (59)

Third-Order Problem. Consider 𝑓₃^′′′𝜂=−𝐶₃𝑓₀𝜂𝑓₀^′′𝜂−𝐶₂𝑓₁𝜂𝑓₀^′′𝜂−𝐶₁𝑓₂𝜂𝑓₀^′′𝜂−𝐶₂𝑓₀𝜂𝑓₁^′′𝜂−𝐶₁𝑓₁𝜂𝑓₁^′′𝜂−𝐶₁𝑓₀𝜂𝑓₂^′′𝜂−𝐶₃𝑓₀^′′′𝜂−𝐶₂𝑓₁^′′′𝜂−1+𝐶₁𝑓₂^′′′𝜂,𝑓₃(0)=0,𝑓^′₃(0)=0,𝑓^′₃(5)=0. (60) Its solution is 𝑓₃𝜂=−5𝐶₁−4895𝐶²₁[phone]𝐶³₁−1935365𝐶₂−4895𝐶₁𝐶₂−20165𝐶₃𝜂48²+𝐶₁−300013𝐶²₁−18000631𝐶³₁+𝐶4032000²−300013𝐶₁𝐶₂+𝐶18000₃𝜂3000⁵+11𝐶²₁−252000011𝐶³₁[phone]𝐶₁𝐶₂𝜂2520000⁸+𝐶³₁𝜂66528000⁹. (61) From (53), (57), (59), and (61), we obtain 𝑓𝜂=𝐷₂𝜂²+𝐷₅𝜂⁵+𝐷₉𝜂⁹+𝐷₁₁𝜂¹¹, (62) where 𝐷₂=1−105𝐶₁−1695𝐶²₁[phone]𝐶³₁−1935365𝐶₂−2495𝐶₁𝐶₂−20165𝐶₃,𝐷48₅=𝐶₁−100013𝐶²₁−12000631𝐶³₁+𝐶4032000²−150013𝐶₁𝐶₂+𝐶18000³,𝐷3000₉=11𝐶²₁−168000011𝐶³₁[phone]𝐶₁𝐶₂,𝐷2520000₁₁=𝐶³₁.66528000 (63) Using (62) in (50) and applying the technique as discussed in (14)–(16), we obtain 𝐶₁𝐶=−0.7047085843229471,₂𝐶=0.4265066662188469,₃=0.0092786353498473. (64) Substituting these values in (62), we have 𝑓𝜂=1.95147×10⁻¹𝜂²−6.83434×10⁻⁴𝜂⁵+2.79896×10⁻⁶𝜂⁸−5.26047×10⁻⁹𝜂¹¹. (65)

4. Results and Discussions

The formulation presented in Section 2 provides highly accurate solutions for the problems demonstrated in Section 3. We have used Mathematica 7 for most of our computational work. In Table 1, we have presented the initial slope 𝑓^′′(0) for different values of 𝛽 obtained by OHAM. Table 1 shows that the results obtained by OHAM are in excellent agreement with the results found in the literature. Table 1 shows a benchmark for the initial slope 𝑓^′′(0) found by different authors. It is found that the method present in this work is very good and provides the same values that optimized numerical methods with which it is compared. Figure 1 shows the variation of the function 𝑓^′(𝜂) against 𝜂 for different values of 𝛽 with OHAM, while Figure 2 shows the variation of the function 𝜐^′(𝜂) with respect to 𝜂 for different values of 𝛽. Table 2 shows the solution of the Blasius equation obtained by the present method. In order to verify the accuracy of the present method, we have compared the results obtained by OHAM to the results available in the literature and found an excellent agreement. Figures 2 and 3 show the variation of 𝑓^′(𝜂) with respect to 𝜂 for the Blasius equation which is identical to results in the literature [41].

Table 1 

Initial slope 𝑓^′′(0) obtained by OHAM for different values of 𝛽.

Table 2 

Solution of the Blasius equation by OHAM.

Figure 1 

𝑓′(𝜂) for different values of 𝛽.

Figure 2 

𝜐′(𝜂) for different values of 𝛽.

Figure 3 

Plot of 𝑓′(𝜂) with respect to 𝜂.

5. Conclusion

In this work, we have seen the effectiveness of OHAM [11–15] to Falkner-Skan, Energy and Blasius equations. By applying the basic idea of OHAM to Falken-Skan, Energy and Blasius equations, we found that it is simpler in applicability and, more convenient to control convergence and involved less computational overhead. Therefore, OHAM shows its validity and great potential for the solution Falken-Skan, Energy and Blasius equations, with heat transfer problems arising in science and engineering.