Further solutions of the Falkner-Skan equation · numerical parameter (positive or negative) in the Falkner-Skan equation sets a de-gree of acceleration or deceleration of main stream

FURTHER SOLUTIONS OF THE FALKNER-SKAN EQUATION

LAZHAR BOUGOFFAa, RUBAYYI T. ALQAHTANIb

Department of Mathematics and Statistics, College of Science, Al-Imam Mohammad Ibn SaudIslamic University (IMSIU), Riyadh, Saudi Arabia

E-mail a: [email protected] b: [email protected]

Received April 28, 2017

Abstract. In this paper, we propose a reliable treatment for the Falkner-Skanequation, which can be described as the non-dimensional velocity distribution in thelaminar boundary layer over a flat plate. We propose an algorithm of two steps that willintroduce an exact solution to the equation, followed by a correction to that solution.Three different special cases: Hiemenz flow (β=1), Homann flow (β= 1

2 ), and Blasiusproblem (β = 0) have been considered. When the pressure gradient parameter β takessufficiently large values, we use a transformation of variable that reduces the Falkner-Skan equation into an equivalent boundary value problem in a finite domain, and wesolve this problem by a different technique that allows us to find f ′′(0). Also, variousexact solutions can be obtained in a straightforward manner by using a direct methodwhen β = −1. The new technique, as presented in this paper, has been shown to bevery efficient for solving the Falkner-Skan equation.

Key words: Falkner-Skan equation; Hiemenz flow; Homann flow; Blasiusproblem; approximate solution; exact solution.

1. INTRODUCTION

The Falkner-Skan equation describes the class of the so-called similar laminarflows in boundary layer on a permeable wall, and at varying main - stream velocity [1,2]. This well-known nonlinear third-order differential equation is given by

f ′′′+ff ′′+β(1− (f ′)2

)= 0, 0< η <∞, (1)

subject to the boundary conditions

f(0) = γ, f ′(0) = 0 (2)

andf ′(∞) = 1, (3)

where f is a stream function, η is a distance from the wall, which is called similarityvariable. f ′ defines the velocity component in η-direction, f ′′ defines the shear stressin the boundary layer.The mass-transfer parameter γ in the boundary condition sets the measure for the

Romanian Journal of Physics 63, 102 (2018) v.2.1*2018.2.16#c2fdcf3f

Article no. 102 Lazhar Bougoffa, Rubayyi T. Alqahtani 2

mass flow rate through the wall boundary in either direction. Positive values de-termine flows with suction, negative with blowing through the wall boundary. Thezero value corresponds to flow along impermeable wall with zero mass transfer. Thenumerical parameter β (positive or negative) in the Falkner-Skan equation sets a de-gree of acceleration or deceleration of main stream [3]. The Falkner-Skan equation(1), together with (2)-(3), constitute the theoretical basis of convective heat and masstransfer. The existence and uniqueness question for the problem (1)-(3) was givenby [4] and [5]. Weyl [6] proved that for each value of the parameter β there exists aphysical solution with a positive monotonically decreasing, in [0,1), second deriva-tive that approaches zero as the independent variables goes to infinity. In the caseβ > 1, Craven and Pelietier [7] computed solutions for which df

dη < 0 for some valuesof η. Coppel [8] presented an elegant proof of the existence and uniqueness for β > 0and showed that the wall shear stress f ′′(0) is an increasing function of β. In theworks [9, 10], the extension for β < 0 was discussed and properties of solutions wereinvestigated further. New theoretical results for the solutions of the Falkner-Skanequation were given in Refs. [10] and [11]. Different methods have been proposedto solve this problem: finite difference [9, 12–14], shooting [15–20], finite element[21], group invariance theory [22, 23], Chebyshev spectral method [24, 25], diffe-rential transformation method [26], non-iterative transformation method (ITM) [27],homotopy analysis method [28], and Adomian decomposition method (ADM) [29].Other direct methods and numerical techniques to study nonlinear differential equa-tions of physical relevance have been recently reported [30–32].

To our knowledge, a general analytic solution of Eq. (1) satisfying the bound-ary conditions (2)-(3) does not exist. The purpose of this work is to present an al-gorithm of two steps that will introduce approximate solutions to the Falkner-Skanequation followed by a correction to that solution. Furthermore, we will considerthree special cases: Hiemenz flow (β = 1), Homann flow (β = 1

2 ) and Blasius pro-blem (β = 0). We use a transformation of variables that reduces (1)-(3) into anequivalent boundary value problem (BVP) in a finite domain, Finally, we solve thisproblem by a different technique that allows us to find f ′′(0), when the pressure gra-dient parameter β takes sufficiently large values. Also, various exact solutions canbe obtained in a straightforward manner by using a direct method when β =−1. Theresults lead to very accurate approximate solutions. Listed below are some specialcases when Eqs. (1)-(3) are solvable.

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3 Further solutions of the Falkner-Skan equation Article no. 102

2. CASE 1: LARGE VALUES OF β

The Falkner-Skan can be written in an equivalent form utilizing m as theFalkner-Skan pressure parameter instead of β :

f ′′′+ff ′′+2m

m+1

(1− (f ′)2

)= 0, (4)

where 0 ≤ m ≤ ∞. The authors have considered only the case of 0 ≤ β ≤ 2. Forlarge values of m→∞, Eq. (1) becomes

f ′′′+ff ′′+2(1− (f ′)2

)= 0. (5)

This case has been studied in Ref. [5]. One of the main problem appears whenβ →∞.

In this section, we will present a direct approach to solve this problem whenthe pressure gradient parameter β takes sufficiently large values, i.e. β →∞.Let us consider the following transformation [33]:

z(x) = f ′′(η) and x= f ′(η). (6)

Differentiating Eq. (1) with respect to η, we get

f ′′′′+ff ′′′+(1−2β)f ′f ′′ = 0 (7)

and from Eq. (6), we have

z(x) =df ′

dη=

dx

dη, (8)

z′(x) =dz

dx=

f ′′′

f ′′ =−f −β1−f ′2

f ′′ (9)

and

z′′(x) =d2z

dx2=

f ′′′′

f ′′2 −f ′′′2

f ′′3 . (10)

Substituting these equations into Eq. (7), we obtain

z′′(x)−β(1−x2)z′(x)

z2(x)+(1−2β)x

1

z(x)= 0. (11)

Since f ′(∞) = 1 and f(0) = 0, dealing, for example, with a simple case γ = 0, wehave z(1) = f ′′(∞) = 0 and z′(0) = −β

σ and z(0) = σ, where f ′′(0) = σ. Hence,the initial-boundary value problem for the Falkner-Skan equation can be transformedinto the new problem{

z′′(x)−β(1−x2) z′(x)

z2(x)+(1−2β)x 1

z(x) = 0, 0< x < 1,

z(0) = σ,z′(0) =−βσ ,z(1) = 0.

(12)



Multiplying both sides of the first-equation of (12) by ξ(x) = e∫ (1−2β)x

β(1−x2)dx

and takinginto account that ξ′(x) = (1−2β)x

β(1−x2)ξ(x), we obtain

ξ(x)z′′(x)+β(1−x2)

(ξ(x)

1

z(x)

)′= 0. (13)

We have

ξ(x) = e(1−2β)

β

∫x

1−x2dx

= (1−x2)1− 1

2β . (14)Thus

z′′(x)+β(1−x2)12β

(ξ(x)

1

z(x)

)′= 0. (15)

The function (1−x2)12β can be approximated by 1 for sufficiently large values of β,

and then ξ(x)≈ 1−x2.More precisely, we obtain the approximated equation

z′′(x)+β

(ξ(x)

1

z(x)

)′= 0. (16)

Integrating Eq. (16) from 0 to x and taking into account that z′(0) =−βσ and z(0) =

σ, we get

z′(x)+βξ(x)1

z(x)= 0, (17)

which can be written as follows

(z2(x))′ =−2βξ(x). (18)

Integrating Eq. (18) from 0 to x and taking into account that z(0) = σ, we get

z2(x) =−2β

∫ x

0ξ(x)dx+σ2. (19)

Consequently,

z2(x) =−2β(x− x3

3)+σ2. (20)

Thus, it is required to find σ by the given condition z(1) = 0. Then we have

σ2 =4

3β. (21)

Thus

f ′′(0) = σ =±2

√β

3. (22)

It follows that

z(x) =±√

−2β(x− x3

3)+

4

3β. (23)



3. CASE 2: HOMANN FLOW (β = 12 )

For β = 12 , Eq. (12) becomes{

z′′(x)− 12(1−x2) z

′(x)z2(x)

= 0,

z(0) = σ,z′(0) =−βσ ,z(1) = 0.

(24)

We start first by noting that the assumption

z(x) = k, (25)

where k is a parameter to be determined, satisfies the first equation of (24). Usingthe boundary condition z(0) = σ, we obtain k = σ. However, this solution does notsatisfy both boundary conditions z′(0) = −β

σ and z(1) = 0. In view of this, we willemploy the given conditions together with (25) to achieve the second goal of ourapproach by making a correction to (25). We do this by writing the first equation of(24) in an equivalent form as

z′′(x) =1

2(1−x2)

z′(x)

z2(x). (26)

Inserting the obtained solution (25) in the right hand side (RHS) of Eq. (26), weobtain

z(x) = ax+ b, (27)where a and b are two constants of integration. Taking into account the boundaryconditions z(0) = σ, z′(0) =−β

σ =− 12σ , z(1) = 0, we immediately find that

a=− 1

2σ, b= σ (28)

with σ = 1√2= 0.7071. The value of the skin friction for Homann stagnation was

determined to be σ = 0.927682 [15].Returning to the original transformation, we get

f ′′(η) = af ′(η)+ b, (29)

so that

f ′(η) = ceaη− b

a(30)

and

f(η) =c

aeaη− b

aη+d, (31)

where c and d are two constants of integration. Using the boundary conditions (2)-(3), we obtain c=−2σ2 and d=−4σ3.Consequently, the approximate solution of the Homann flow problem is given as

fp(η) = 4σ3e−12σ

η−2σ2η−4σ3, σ = 0.7071. (32)



3.1. THE ERROR REMAINDER

An error analysis ER(η) for fp(η) is considered in Table 1, where the residualER(η) or error remainder is calculated for different values of η by substituting fp(η)for f(η) in Eq. (1) when β = 1

2 :

ER1(η) =−1

2e−

12σ

η+σe−12σ

η(4σ3e−

12σ

η+2σ2η−4σ3)

+1

2

[1−4σ4

(1−e−

12σ

η)2

]. (33)

This demonstrates that the approximate solution is a good approximation of the actualsolution. The complete error analysis is given in Tables 1-4.

4. CASE 3: HIEMENZ STAGNATION FLOW (β = 1)

We note that the solution of the form

f(η) = βη+λ, (34)

where λ is a parameter to be determined, satisfies Eq. (1). Then we have

f ′(η) = β,f ′′(η) = f ′′′(η) = 0. (35)

Inserting this solution into Eq. (1) to get

β(1−β2) = 0. (36)

Thus β =−1, 1, 0. For the Hiemenz flow problem (β = 1), we have

f(η) = η+λ. (37)

Inserting now the boundary condition f(0) = γ we get λ= γ.We write Eq. (1) in an equivalent form as

f ′′′+ff ′′ =−(1− (f ′)2

). (38)

Inserting the obtained solution Eq. (37) in the RHS of Eq. (38), we obtain

f ′′′

f ′′ =−f (39)

Integrating Eq. (39) twice, we obtain

f ′(η) = c

√π

2e

γ2

2 erf

(η+γ√

2

)+d, (40)

where erf is the error function given by

erf(x) =2√π

∫ x

0e−t2dt. (41)



Consequently,

fp(η) = c

√π

2e

γ2

2

[(η+γ)erf

(η+γ√

2

)+

√2

πe−

(η+γ)2

2

]+dη+e, (42)

where c,d, and e are constants of integration.Using the boundary conditions f ′(0) = 0 and f ′(∞) = 1, we obtain

c

√π

2e

γ2

2 erf

(γ√2

)+d= 0 (43)

and

c

√π

2e

γ2

2 +d= 1. (44)

Solving this system for c and d, we obtain

c=1√

π2 e

γ2

2 −√

π2 e

γ2

2 erf(

γ√2

) (45)

and

d=

√π2 e

γ2

2 erf(

γ√2

)√

π2 e

γ2

2 erf(

γ√2

)−√

π2 e

γ2

2

. (46)

The constant e can be determined from the condition f(0) = γ. Thus

e= γ− c

√π

2e

γ2

2

[γerf

(γ√2

)+

√2

πe−

γ2

2

]. (47)

Hence the approximate solution of the Hiemenz flow problem (β = 1) is given byEq. (42). Figures 1 and 2 show the approximate solution.


An error analysis ER(η) for

fp(η) = ηerf

(η√2

)+

√2

πe−

η2

2 −√

2

π(48)

when γ = 0 is considered in Table 3, where the residual ER2(η) or error remainderis calculated for different values of 1≤ η <∞.

ER2(η)=

√2

πe−

η2

2

[−√

2

πη+ηerf(

η√2)+

√2

πe−

η2

2 −√

2

π

]+1−

(erf(

η√2)

)2

.

(49)



0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

fp(

η)

η

Fig. 1 – The approximate solution of the Hiemenz flow problem fp(η) when γ = 0.

Table 1

The error remainder ER4(η) for the value of γ = 0.

η ER4(η)

0 -0.1951342281 -0.07102780912 0.04203181485 0.123399024010 0.0390240252420 0.00111218682550 0.5342136800×10−8

100 0.2810156498×10−17

200 0.3725913644×10−36

500 0.2625174347×10−93

1000 0.6291376971×10−189

2000 0.1799367855×10−380

5000 0.1311231850×10−955



(a) β = 0

0

5

10

15

20

25

0 5 10 15 20 25

fp(

η)

η

Approximation solution Numerical solution

(b) β = 0.5

0

5

10

15

20

25

0 5 10 15 20 25

fp(

η)

η


(c) β = 1

0

5

10

15

20

25

0 5 10 15 20 25

fp(

η)

η


Fig. 2 – The approximate solution compared to numerical solution, γ = 0.



(a) 0≤ η ≤ 20

0

5

10

15

20

0 5 10 15 20

fp(

η)

η

α=0.0α=0.5α=1.0α=1.5α=2.0

(b) 0≤ η ≤ 5

0

1

2

3

4

5

0 1 2 3 4 5

fp(

η)

η

α=0.0α=0.5α=1.0α=1.5α=2.0

Fig. 3 – The approximate solution of the Blasius equation fp(η) when γ = 0 and α= 0, 0.5, 1, 1.5, 2.



(a) 0≤ η ≤ 30

0

5

10

15

20

25

30

0 5 10 15 20 25 30

fp(

η)

η

α=0.0α=-0.5α=-1.0α=-1.5α=-2.0

(b) 0≤ η ≤ 5

-4

-3

-2

-1

0

1

2

3

4

0 1 2 3 4 5

fp(

η)

η

α=0.0α=-0.5α=-1.0α=-1.5α=-2.0

Fig. 4 – The approximate solution of the Blasius equation fp(η) when γ = 0 and α =0, −0.5, −1, −1.5, −2.



(a) 0≤ η ≤ 30

0

5

10

15

20

25

30

0 5 10 15 20 25 30

fp(

η)

η

γ=0.0γ=0.5γ=1.0γ=1.5γ=2.0

(b) 0≤ η ≤ 5

0

1

2

3

4

5

0 1 2 3 4 5

fp(

η)

η

γ=0.0γ=0.5γ=1.0γ=1.5γ=2.0

Fig. 5 – The approximate solution of the Blasius equation fp(η) when α= 0 and γ = 0, 0.5, 1, 1.5, 2.



Table 2

The error remainder ER3(η) for different value of γ.

η ER3(η), γ=0.5 ER3(η), γ=1 ER3(η), γ=1.5 ER3(η), γ=20 0 0 0 01 0.3678794412 0.1353352832 0.04978706837 0.018315638892 0.5032147244 0.1536509221 0.05226582055 0.018651101525 0.3350779415 0.02699718793 0.002212643382 0.1816017802×10−3

10 0.06068692292 0.0004086014290 0.2753120978×10−5 0.1855038260×10−7

20 0.0008626007266 0.3916191882×10−7 0.1777948364×10−11 0.8071873084×10−16

50 0.6805092491×10−9 0.9450874255×10−20 0.1312532111×10−30 0.1822837228×10−41

100 0.1909462350×10−19 0.3682875216×10−41 0.7103345013×10−63 0.1370057562×10−84

200 0.1909462350×10−19 0.2753954089×10−84 0.1024491844×10−127 0.3811187498×10−171

500 0.1331925918×10−105 0.3555163627×10−214 0.9489407965×10−323 0.2532903490×10−431

1000 0.7117451831×10−214 0.5070882939×10−431 0.3612789294×10−648 0.2573959337×10−865

2000 0.1014684184×10−430 0.5150495210×10−865 0.2614370199×10−1299 0.1327043567×10−1733

5000 0.9176527672×10−1082 0.1684510105×10−2167 0.3092209160×10−3253 0.5676283841×10−4337

Table 3


η ER2(η)

0 11 0.3262573048

2 0.0482564534

5 0.1779034859×10−5

10 0.2187308194×10−21

20 0.3692604684×10−86

50 0.1010532795×10−541

100 0.5224985463×10−2170

200 0.4077397794×10−8684

500 0.1238358721×10−54284

1000 0.9223046463×10−217145

2000 05000 0

5. CASE 4: BLASIUS FLOW (β = 0)

5.1. γ = 0

For β = 0, we havef(η) = γ. (50)

Proceeding as before, inserting the obtained solution Eq. (50) in the RHS of Eq. (39),we obtain

f(η) =c

γ2e−γη+dη+e, γ = 0. (51)



Table 4


η ER1(η)

0 01 0.2236783820

2 0.2518178278

5 0.889029848×10−1

10 0.55999659×10−2

20 0.29018999482×10−4

50 0.19179816079×10−4

100 0.19179816064×10−4

200 0.19179816064×10−4

500 0.19179816064×10−4

1000 0.19179816064×10−4

2000 0.19179816064×10−4

5000 0.19179816064×10−4

Using the boundary conditions (2)-(3), we obtain

c= γ, d= 1, e= γ− 1

γ. (52)

Consequently, the approximate solution of Eqs. (1)-(3) when β = 0 is given as

fp(η) =1

γe−γη+η+γ− 1

γ, γ = 0. (53)

If f ′(0)=α, where α is a constant, then c= γ(1−α). The initial condition f ′(0)=αindicates the slip condition at the wall. The case where α= 0 represents no-slip.

5.2. γ = 0

Proceeding as before, we obtain

f(η) =c

k2e−kη+dη+e. (54)

Using the boundary conditions (2)-(3), we obtain

c= k, d= 1, e=−1

k. (55)

Consequently, the approximate solution of Eqs. (1)-(3) when β = 0 is given as

fp(η) =1

ke−kη+η− 1

k. (56)



To find the value of k, we use the transformation [33]:{z(η)z′′(η)+η = 0,z′(0) = 0,z(1) = 0

(57)

with z(0) = σ.Proceeding as in [34], we define the solution z(η) by an infinite series in the form

z(η) =

∞∑n=0

Cnηn. (58)

Thus

z′(η) =

∞∑n=0

(n+1)Cn+1ηn, (59)

z2(η) =

∞∑n=0

ηnn∑

k=0

CkCn−k (60)

and

z′2(η) =

∞∑n=0

ηnn∑

k=0

(k+1)(n+1−k)Ck+1Cn+1−k. (61)

The substitution into Eq. (57) yields

∞∑n=0

ηnn∑

k=0

CkCn−k = σ2− 1

3η3

−2

∞∑n=2

ηn

n(n−1)

n−2∑k=0

(k+1)(n−1−k)Ck+1Cn−1−k. (62)



We equate the coefficients of the like powers of η on the left side and on the rightside to arrive at recurrence relations for the coefficients. Thus

C20 = σ2,∑1

k=0CkC1−k = 0,∑2k=0CkC2−k = −C2

1 ,∑3k=0CkC3−k = −1

3 +13

∑1k=0(k+1)(2−k)Ck+1C2−k,∑4

k=0CkC4−k = +13

∑2k=0(k+1)(3−k)Ck+1C3−k,∑5

k=0CkC5−k = +16

∑3k=0(k+1)(4−k)Ck+1C4−k,∑6

k=0CkC6−k = +16

∑4k=0(k+1)(5−k)Ck+1C5−k,

...

(63)

The coefficients C0,C1, ... are

C0 = σ,

C1 = C2 = 0,

C3 = − 16σ ,

C4 = C5 = 0,

C6 = − 1180σ3 ,

...

(64)

We can now compute the truncated decomposition series z(η) = z0(η) + z1(η) +z2(η)+ z3(η)+z4(η)+z5(η)+ z6(η). Consequently, the solution is given by

z(η) = σ− η3

6σ− η6

180σ3. (65)

Thus, it is required to find σ by the given condition z(1) = 0; then we get

σ− 1

6σ− 1

180σ3= 0. (66)



Then, we get σ = 0.4417, that is f ′′p (0) = 0.44174. The numerical value of f ′′(0)

by the Runge-Kutta method is given as f ′′(0) = 0.46960. Finally, substituting thisvalue into Eq. (56), we obtain k = 0.44174. Consequently, the approximate solutionof the initial-boundary value problem for the Blasius equation when γ = 0, which ischaracterized by k = 0.44174, is given by Eq. (56).


The error remainder ER(η) is calculated for different values of η :

5.3.1. Case 1: γ = 0

ER3(η) = e−2γη+e−γη(η−1). (67)

5.3.2. Case 2: γ = 0

ER4(η) = e−2kη+e−kη(kη−1−k2), k = 0.44174. (68)where the residuals ERi(x), i = 3,4 or errors remainders are calculated for smalland large values of η, which show that the present solutions are highly accurate.The figures 3-5 have been drawn to show the approximate solutions of the Blasiusequation.

6. EXACT SOLUTIONS OF THE FALKNER-SKAN EQUATION WHEN β =−1

The solutions for the value of β < 0 represent decelerating flows. We integrateEq. (1) from 0 to η, using the fact that

ff ′′ = (ff ′)′−f ′2. (69)

The substitution of Eq. (69) into Eq. (1) leads to

f ′′′+(ff ′)′ = 1. (70)

Integrating Eq. (70) from 0 to η and taking into account that f(0) = γ and f ′(0) = 0,we obtain

f ′′+ff ′ = η+f ′′(0). (71)Consequently Eq. (71) can also be integrated analytically, and we obtain the nonli-near Riccati equation

f ′+1

2f2 =

1

2η2+f ′′(0)η+

γ2

2. (72)



The first result in the analysis of Riccati equation (72) is: if one particular solution toEq. (72) can be chosen as follows f1 = η+f ′′(0), then f ′′(0) must satisfy

f ′′2(0)+2 = γ2, (73)

that is

f ′′2(0) = γ2−2, γ2−2≥ 0 (74)

and the general solution is obtained as

f(η) = f1(η)+1

w(η), (75)

where w satisfies the corresponding first - order linear equation

w′−f1w =1

2, (76)

and we can obtain its closed form solution

w(η) =

√π2 e

f ′′2(0)2 erf

(η+f ′′(0)√

2

)+2C

2e−η2

2−f ′′(0)η

, (77)

where erf is the error function and C is a constant of integration.A set of solutions to the Riccati equation is then given by

f(η) = η+f ′′(0)+2e−

η2

2−f ′′(0)η√

π2 e

f ′′2(0)2 erf

(η+f ′′(0)√

2

)+2C

, (78)

and therefore f satisfies the boundary condition (3), that is f ′(∞) = 1.Inserting now the boundary condition f(0) = γ into Eq. (78) to get

f ′′(0)+2√

π2 e

f ′′2(0)2 erf

(f ′′(0)√

2

)+2C

= γ. (79)

Consequently, this leads to

C =1

γ−f ′′(0)− 1

2

√π

2e

f ′′2(0)2 erf

(f ′′(0)√

2

). (80)

Thus we have proved.Lemma 1

The solution of the Falkner-Skan equation when β = −1 subject to the boundaryconditions (2)-(3) can be exactly obtained by Eq. (80), where C is determined by Eq.(79) and γ2−2≥ 0.



7. CONCLUSION

In this work, we have considered the Falkner-Skan equation, which can bedescribed as the non-dimensional velocity distribution in the laminar boundary layerover a flat plate. Very good approximate analytic solutions are obtained by a directmethod when β takes large values. Also, three special cases: Hiemenz flow (β = 1),Homann flow (β = 1

2 ), and Blasius problem (β = 0) have been considered. We havealso demonstrated that the exact solution can be obtained in a straightforward mannerby using a direct method when the pressure gradient parameter takes the value β =−1.

REFERENCES

1. V. M. Falkner, S. W. Skan, Some approximations of the boundary layers equations, Philos. Mag.Soc. 12, 865-896 (1931).

2. H. Blasius, Grenrschichten in Flussigkeiten mit kleiner Reibung, Z. Math. u. Phys. 56, 1-37 (1908).3. A. Solodov, V. Ochkov, Differential Models An Introduction with Mathcad, Springer (2005).4. D. R. Hartree, On the equation occurring in Falkner-Skan approximate treatment of the equations

of the boundary layer, Proc. Camb. Phil. Soc. 33, 223-239 (1937).5. K. Stewartson, Further solutions of the Falkner-Skan equation, Proc. Camb. Phil. Soc. 50, 454-465

(1954).6. H. Weyl, On the differential equation of the simplest boundary-layer problems, Ann. Math. 43,

381-407 (1942).7. A. H. Craven, L. A. Pelietier, Reverse flow solutions of the Falkner-Skan equation for λ > 1,

Mathematika. 19, 135-138 (1972).8. W. A. Coppel, On a differential equation of boundary layer theory, Philos. Trans. Roy. Soc. London

Ser. A. 253, 101-136 (1960).9. A. Asaithambi, A finite-difference method for the solution of the Falkner-Skan equation, Appl.

Math. Comput. 92, 135-141 (1998).10. G. C. Yang, New results of Falkner-Skan equation arising in boundary layer theory, Appl. Math.

Comput. 202, 406-412 (2008).11. O. Pade, On the solution of Falkner-Skan equation, J. Math. Anal. Appl. 285, 264-274 (2003).12. A. Asaithambi, A second-order finite-difference method for the Falkner-Skan equation, Appl.

Math. Comput. 156, 779-786 (2004).13. P. M. Beckett, Finite difference solution of boundary-layer type equations, Int. J. Comput. Math.

14, 183-190 (1983).14. B. Oskam, A. E. P. Veldman, Branching of the Falkner-Skan solutions for λ < 1, J. Engg. Math.

16, 295-308 (1982).15. A. Asaithambi, A numerical method for the solution of the Falkner-Skan equation, Appl. Math.

Comput. 81, 259-264 (1997).16. A. Asaithambi, Solution of the Falkner-Skan equation by recursive evaluation of Taylor coeffi-

cients, J. of Comput. Appl. Math. 176, 203-214 (2005).17. T. Cebeci, H. B. Keller, Shooting and parallel shooting methods for solving the Falkner-Skan

boundary-layer equation, J. Comp. Phys. 7, 289-300 (1971).



18. C. W. Chang et al., The Lie-group shooting method for boundary layer equations in fluid mechan-ics, J. Hydrodynamics Ser. B, 18, 103-108 (2006).

19. C. S. Liu, J. R. Chang, The Lie-group shooting method for multiple-solutions of Falkner-Skanequation under suctioninjection conditions, Int. J. Non-Linear Mech. 43, 844-851 (2008).

20. I. Sher, A. Yakhot, New approach to the solution of the Falkner-Skan equation, AIAA J. 39, 965-967 (2001).

21. A. Asaithambi, Numerical solution of the Falkner-Skan equation using piecewise linear functions,Appl. Math. Comput. 159, 267-273 (2004).

22. R. Fazio, The Falkner-Skan equation: Numerical solutions within group invariance theory, Calcolo31, 115-124 (1994).

23. R. Fazio, A novel approach to the numerical solution of boundary value problems on infinite inter-vals, SIAM J. Numer. Anal. 33, 1473-1483 (1996).

24. E. M. E. Elbarbary, Chebyshev finite difference method for the solution of boundary-layer equa-tions, Appl. Math. Comp. 160, 487-497 (2005).

25. H. Nasr et al., H.M. Chebyshev solution of laminar boundary layer flow, Int. J. Comput. Math. 33,127-132 (1990).

26. B. L. Kuo, Application of the differential transformation method to the solutions of the Falkner-Skan wedge flow, Acta Mech. 164, 161-174 (2003).

27. R. Fazio, Blasius problem and FalknerSkan model: Topfers algorithm and its extension, Comput.Fluids. 73, 202-209 (2013).

28. S. J. Liao, A non-iterative numerical approach for two-dimensional viscous flow problems gov-erned by the Falkner-Skan equation, Int. J. Numer. Meth. Fluids. 35, 495-518 (2001).

29. N. S. Elgazery, Numerical solution for the Falkner-Skan equation, Chaos Solit. Fract. 35, 738-746(2008).

30. A. M. Wazwaz et al., Reliable treatment for solving boundary value problems of pantograph delaydifferential equations, Rom. Rep. Phys. 69, 102 (2017).

31. L. Bougoffa, Exact solutions of a generalized Bratu equation, Rom. J. Phys. 62, 110 (2017).32. A. M. Wazwaz, The successive differentiation method for solving Bratu equation and Bratu-type

equations, Rom. J. Phys. 61, 774–783 (2016).33. L. L. Wang, A new algorithm for solving classical Blasius equation, Appl. Math. Comput. 157,

1-9 (2004).34. L. Bougoffa, A. M. Wazwaz, New approximate solutions of the Blasius equation, Internat. J. Nu-

mer. Methods Heat Fluid Flow. 25, 1590-1599 (2015).


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Further solutions of the Falkner-Skan equation · numerical parameter (positive or negative) in the Falkner-Skan equation sets a de-gree of acceleration or deceleration of main stream