١

Chapter 1 Introduction

1.1 Blasius Equation

If a fluid flows past a solid, a fluid layer is formed adjacent to the

boundary of the solid. This layer is called a boundary layer and strong

viscous effects exist within this layer.

Consider a uniform flow over a flat surface, y=0, 0, .x z≥ − ∞ < < ∞

Equations of the flow in the boundary layer are the continuity equation

0 (1.1)u v

x y

∂ ∂+ =∂ ∂

and the reduced Navier- Stokes equation 2

2(1.2)

u v uu v

x y yν∂ ∂ ∂+ =

∂ ∂ ∂

where u and v are respectively the components of the velocity vector and ν

represents the viscosity of the fluid.

Boundary conditions are

( ,0) 0 0 (1.3 )

( ,0) 0 0 (1.3 )

( , ) as (1.3 )

u x x a

v x y b

u x y U y c

= ≥= ≥→ → ∞

where U is the constant speed of the flow outside the boundary Layer.

Define a stream function ( , )x yψ such that

, (1.4)u vy x

ψ ψ∂ ∂= = −∂ ∂

then equation (1.1) is satisfied identically and equation (1.2) becomes 2 2 3

2 3(1.5)v

y x y x y y

ψ ψ ψ ψ ψ∂ ∂ ∂ ∂ ∂− =∂ ∂ ∂ ∂ ∂ ∂

Blasius used a similarity transformation to reduce (1.5) to an ordinary

differential equation.

٢

A similarity transformation is based on the symmetry analysis of a

differential equation [11,23]. When a symmetry property of a differential

equation is identified it can be exploited to achieve a simplification. If it is

an ordinary differential equation then usually the order of the equation can

be reduced. If it is a partial differential equation then usually the dependent

and independent variables can be combined to achieve a reduction of order

or a reduction of the partial differential equation to an ordinary differential

equation.

In the case of (1.5) symmetry analysis leads to the following

transformation [11]

( , ) ( )

ya

x

x y b x f

η

ψ η

=

=

where a and b are constants and are chosen to make

and ( )fη η dimensionless. They are taken as

,U

a

b U

νν

=

=

With this choice, η is called the dimensionless similarity variable and ( )f η

is called the dimensionless stream function. Now

2

2

2

32

3 2

3

1 ( )( )

2 2

( )

( )

( ) ( )2 2

( ).

U y ff U

x x x

Ufy

aUf

y x

U U y Uf f

x y xx

Uf

y x

ψ ηη ν

ψ η

ψ η

ψ η η ην

ψ ην

∂ ′= − +∂∂ ′=∂∂ ′′=∂

∂ ′′ ′′= − = −∂ ∂

∂ ′′′=∂

٣

A substitution of the above derivatives in equation (1.5) reduces it to 3 2

3 2

1( ) 0 (1.6)

2

d f d ff

d dη

η η+ =

Equation (1.6) is known as the Blasius equation. The boundary condition

(1.3a) transforms to

(0) 0 (1.7 )f a′ =

while (1.3b) becomes

(0) 0 (1.7 )f b=

and (1.3c) reduces to

( ) 1 (1.7 )f as cη η′ → → ∞

Equation (1.6) together with the boundary conditions (1.7a), (1.7b) and

(1.7c) is called the Blasius problem.

1.2 Numerical solution

A numerical solution of the Blasius problem usually uses the shooting

method. In this method it is assumed that

(0) (1.8)f σ′′ =

and the problem is solved with different values of .σ Such values of σ will

lead to different values of as .df

dη

η→ ∞ We seek that value of σ which

will yield an f which satisfies

lim 1.df

dη η→∞=

First accurate numerical solution was obtained by Howarth [17]. More

recently Asaithambi [8], and Cortell[12] have also solved the Blasius

problem by the shooting method.

٤

In practice it is impossible to carry out calculations up to infinity. Hence

an η η∞= is arbitrarily fixed and we demand that

1 when .df

dη η

η ∞=≃

We solve the Blasius equation by the shooting method. We start with

0.1α = and find that ( )f η′ between 12 and13η = is practically a constant

=0.449287. With 0.6α = . We find the slope for 12 13η< < equal to 1.48352.

This means the actual value should lie between 0.1 and 0.6 . Therefore our

next choice is 0.1 0.60.35

2

+ = and so on. We present the results in Table. 1

and 2.

٥

α (12.5)f ′

0.1 0.449287

0.6 1.48352

0.35 1.03571

0.225 0.771459

0.2875 0.908413

0.31875 0.973101

0.334375 1.00465

0.326562 0.988937

0.330468 0.996808

0.3324218 1.00073

0.3314453 0.998771

0.3319335 0.999751

0.3321771 1.00024

0.3320553 0.999996

0.3321162 1.00012

0.3320857 1.00006

0.3320705 1.00003

0.3320629 1.00001

0.3320591 1

Table 1: Sequence of values of α converging to σ

٦

η ( )f η ( )f η′ ( )f η′′

0.0 0 0 0.332059

0.2 0.00664105 0.0664081 0.331985

0.4 0.02656 0.132765 0.331468

0.6 0.059735 0.198938 0.330081

0.8 0.106109 0.264711 0.327391

1.0 0.165573 0.329782 0.323009

1.2 0.23795 0.393778 0.31659

1.4 0.322983 0.456264 0.307867

1.6 0.420323 0.516759 0.296665

1.8 0.529521 0.574761 0.282932

2.0 0.650028 0.629769 0.266753

2.2 0.781197 0.681314 0.248352

2.4 0.922295 0.728985 0.228092

2.6 1.07251 0.772459 0.206455

2.8 1.23098 0.811513 0.184007

3.0 1.39682 0.846048 0.161359

3.2 1.5691 0.876085 0.139129

3.4 1.74696 0.901765 0.117876

3.6 1.92953 0.923334 0.0980867

3.8 2.11604 0.941122 0.0801258

4.0 2.30576 0.955522 0.0642341

4.2 2.49805 0.966961 0.0505193

4.4 2.69237 0.975875 0.0389731

4.6 2.88826 0.982687 0.0294829

4.8 3.08533 0.987793 0.0218711

5.0 3.28329 0.991546 0.0159068

٧

5.2 3.48188 0.994249 0.0113414

5.4 3.68094 0.996159 0.00792786

5.6 3.88031 0.997481 0.00543169

5.8 4.0799 0.998379 0.00364835

6.0 4.27964 0.998976 0.00240198

6.2 4.47948 0.999366 0.00155022

6.4 4.67938 0.999615 0.000980419

6.6 4.87932 0.999711 0.000608191

6.8 5.07928 0.999867 0.000369599

7.0 5.27926 0.999925 0.000220207

7.2 5.47925 0.999959 0.000128431

7.4 5.67924 0.999979 0.0000737176

7.6 5.87924 0.99999 0.0000413441

7.8 6.07924 0.999996 0.0000227099

8.0 6.27924 1 0.0000122503

8.2 6.47924 1 66.4618 10−×

8.4 6.67924 1 63.28575 10−×

Table2: Numerical values of ( ), ( ) and ( )f f fη η η′ ′′

٨

0 2 4 6 8 10 12 14h

0

2

4

6

8

10

12

fHhL

Fig1: ( )f η as a function of η

0 2 4 6 8h

0.5

0.6

0.7

0.8

0.9

1

f'HhL

Fig2: ( )f η′ as a function of η

٩

0 2 4 6 8h

0

0.05

0.1

0.15

0.2

0.25

0.3

f''HhL

Fig3: ( )f η′′ as a function of η

١٠

1.3 Analytical methods

Blasius [10] found a series solution of Eq.(1.6) 1

3 2

0

1( ) ( ) (1.9)

2 (3 2)!

kk kk

k

Af

k

ση η+∞

+

=

= −+∑

where 0 1 1A A= = and

1

1,0

3 12 (1.10)

3

k

k r k rr

kA A A k

r

−

− −=

− = ≥

∑

However the series (1.9) converges slowly and its radius of convergence is

approximately 5.90σ , therefore it cannot be used to find an accurate value

of σ by using the condition

( ) 1f η∞′ =

for sufficiently large η∞ . Blasius found an asymptotic expression and

estimated σ by matching the two expressions for a suitable value of η .

In recent years a few authors have tried to find alternate methods to

find σ . J. H. He has proposed a perturbation approach to solve this

problem [16] and Abbasbandy has combined this approach with Adomian

decomposition method to find an improved solution of the problem [3].

Liao [19] has used the homotopy analysis method to find an accurate value

of σ .

Recently Wang has used a transformation to transform the Blasius

equation into a second order differential equation in which the condition at

infinity is transformed into

1lim ( ) 0 (1.11)x

y x→

=

[24]. Several authors have used Eq. (1.11) to find a good estimate of

σ ,[9,15,24].

١١

1.4 Iteration perturbation method of He

Following J. H. He [16] we construct an iteration formula for

Blasius equation

1 1

10, (1.12)

2n n nf f f+ +′′′ ′′+ =

where we denote by nf the nth approximate solution. We will use the

perturbation method to find approximately 1.nf +

We begin with

0 2 , (1.13)f b=

Substituting 0f into (1.12) results in

1 1 0. (1.14)f bf′′′ ′′+ =

The solution of Eq. (1.14) is

1

1( ) (1 ), (1.15)bf x e

bηη −= − −

where b is an unknown constant.

Substituting 1f into (1.12), we obtain

2 2

1 1(1 ) 0. (1.16)

2bf e f

bηη − ′′′ ′′+ − − =

We re-write Eq. (1.16) in the form

2 2 2

1 1(1 ) 2 0, (1.17)

2bf bf e b f

bηη − ′′′ ′′ ′′+ + − − − =

and embed an artificial parameter ε in Eq.(1.17):

2 2 2

1 1(1 ) 2 0. (1.18)

2bf bf e b f

bηε η − ′′′ ′′ ′′+ + − − − =

Supposing that the solution of Eq. (1.18) can be expressed as (0) (1)

2 2 2 ..., (1.19)f f fε= + +

we have the following linear equations:

١٢

(0) (0) (0) (0) (0)2 2 2 2 2

(1) (1) (0)2 2 2

(1) (1) (1)2 2 2

[ ] [ ] 0, (0) [ ] (0) 0, and [ ] ( ) 1, (1.20)

1 1 1[ ] [ ] 2 [ ] 0,

2

(0) [ ] 0, and [ ] ( ) 0. (1.21)

b

f b f f f f

f b f e b fb b

f f f

ηη −

′′′ ′′ ′ ′+ = = = +∞ =

′′′ ′′ ′′+ + + − − =

′ ′= = +∞ =

The solution of (1.20) is (0)2 ( ) (1 ) / ,bf e bηη η −= − − substituting it into (1.21)

results in

(1) (1) 22 2

1[ ] [ ] ( 2 1) (1.22)

2b bf b f b e b eη ηη − −′′′ ′′+ = − + − −

The constant b can be identified by the following expression:

2

0

( 2 1) 0. (1.23)b b be b e b e dη η ηη η+∞

− − −+ − − =∫

So we have 21 1 1 2

(2) 0, (1.24)4 3 2

b

b b b

+Γ + − =

which leads to the result

10.28867. (1.25)

12b = =

The exact solution of Eq. (1.22) is not required, the expression, Eq. (1.23),

requires no terms of ne βηη (n1,2,3,…) in (1)2f . So we can assume that the

approximate solution of Eq. (1.22) can be expressed as

(1) 22

1( ) , (1.26)

8b bf Ae e Bη ηη − −= + +

where 1 1and ,

4 8A B

−= = which are identified from the initial conditions

(1) (1)2 2(0) [ ] (0) 0.f f ′= =

By setting 1,ε = we obtain first-order approximate solution, i.e.,

(0) (1)2 2 2( ) ( ) ( ) (1.27)f f fη η η= +

١٣

A highly accurate numerical solution of Blasius equation has been provided

by Howarth [17], who gives the initial slope (0) 0.332057exf ′′ = [17].

Comparing the approximate initial slope (0) (1)

2 2 2(0) [ ] (0) [ ] (0) 0.3095,f f f′′ ′′ ′′= + =

we find that the relative error is 6.8%.

We can improve the accuracy of approximate solution to an high-order

by iteration technique. Substituting 2f into (1.12), and by parallel

operation, the constant b can be identified by the following relation:

2 2

0

1 1 1( 2 1 ) 0, (1.28)

4 8 8b b b b be b e b e e e dη η η η ηη η

+∞− − − − −+ − − − + + =∫

which leads to the result

0.3062. (1.29)b =

It is obvious that 20(0) 0.25 0.3296f b b′′ = + = reaches a very high accuracy, the

0.73% accuracy is remarkably good.

1.5 Homotopy analysis method

Recently Liao has developed a method for solving nonlinear

problems which does not depend on the existence of a small parameter ,

also it generalizes the ideas involved in the decomposition method of

Adomian or the artificial small parameter method of Lyapunov [21]. This

method has been used by Liao and other authors to find analytical

expressions for several nonlinear problems.

Suppose the problem to be solved is

[ ( )] ( ) (1.30)N u r f r=

Introduce a new variable q and consider the equation

0(1 ) [ ( , ) ( )] { [ ( , )] ( )} (1.31 )q L r q u r hq N r q f r− Φ − = Φ −

١٤

where L is a suitably chosen linear operator , 0( )u r is an initial approximate

solution and h is a parameter to be chosen later. In (1.31) put q=0. We get

0( ,0) ( ) (1.32)r u rΦ =

If we let q =1 in (1.31) , we get

[ ( ,1)] ( ) (1.33)N r f rΦ =

Whose solution is the required function u(r). Thus as q varies continuously

from 0 to 1 , the initial approximate solution 0( )u r evolves to the desired

solution u(r) of the problem Liao gives this reason for calling the method as

the homotopy analysis method.

Define

0,

1( ) ( , ) (1.34)

!

n

n n qu r r q

n q =

∂= Φ∂

then

01

( , ) ( ) ( ) . (1.35)nn

n

r q u r u r q∞

=Φ = +∑

If the series converges at q =1 , then

01

( ) ( ,1) ( ) . (1.36)n

u r r u r u∞

== Φ = +∑

The functions ( )nu r are found successively from the equations

1 0[ ( )] { [ ( )] ( )} (1.37)L u r h N u r f r= −

when n=1 and

1[ ( ) ( )] ( ), (1.38)n n nL u r u r hR r−− =

when 2,wheren ≥

1

1 0

1( ) [ ( , ) ] (1.39)

( 1)!

n

n n qR r N r q

n q

−

− =

∂= Φ− ∂

It has been shown by Liao [19] that if we choose h=-1 and L = 0L , the

linear operator chosen for the Adomian method , then the solution

١٥

produced by the homotopy analysis method is identical to the one given by

the Adomian method.

1.6 Liao's solution of the Blasius problem

Liao [18] wrote the Blasius equation in the form

0

3 2

3 2

(1 ) [ ( , , , ) ( )]

( , , , ) 1 ( , , , )[ ( , , , ) ] (1.40)

2

q L F h q f

F h q F h qqh F h q

η β ηη β η βη β

η η

− − =

∂ ∂+∂ ∂

where [0, ), 0, 0, [0,1]h qη β∈ ∞ ≠ > ∈

with the boundary conditions

(0, , , ) (0, , , ) 0, ( , , , ) 1 (1.41)F h q F h q F h qβ β β′ ′= = ∞ =

Here h and β are free parameters to be chosen later. Note that at q=0, we

have

0( , , ,0) ( )F h fη β η=

and at q=1

( , , ,1) ( )F h fη β η=

where the initial approximation 0( )f η can be chosen freely Liao chooses the

linear operator L as 3 2

3 2, 0L β β

η η∂ ∂= + >

∂ ∂

and

0

1 exp( )( )f

βηη ηβ

− −= −

as the initial approximation which satisfies the boundary conditions

f(0)= (0)f ′ =0 and ( ) 1f ′ ∞ = . Successive elements of the series

1 2( , , ) , ( , , ) ,......f h f hη β η β are calculated recursively and the solution is

formally written as

١٦

01

( , , )( , , , ) ( ) , (1.42)

!kk

k

f hf h q f q

k

η βη β η∞

== +∑

Liao also finds a sequence of approximations for (0, , )f h β′′ as

1

22

2 3

2 3 33

3 3 5 3

(1 )4

1 5(1 )

2 24

3 5 275 5(1 )

4 8 576 24

hh

h hh

h h h hh

σ ββ

σ ββ β

σ ββ β β β

= + −

= + − −

= + − − − +

⋮

He finds that with 30 35

93and , 0.33205and 0.33206

10hβ σ σ= = − = = . Thus

93,

10hβ = = − are the appropriate values of these parameters since they

yield the correct value of (0)f ′′ found by Howarth [17]. When we set

93,

10hβ = = − and q=1 in (1.42) we find that the solution with the series

truncated after 35 terms gives numerical values on 0 100η≤ ≤ which agree

very well with the solution of Howarth [17].

1.7 Generalized Blasius problem

Cortell numerically studied the problem

0 (1.43 )

(0) (0) 0, ( ) 1 (1.43 )

af ff a

f f f b

′′′ ′′+ =′ ′= = ∞ =

for various values of a and observed that an increase in a decreases

( )and ( )also ( )f f fη η η′ ′′ near 0η = decreases at first with an increase in a but

further away it increases with a. Fang et al. [13] explained the above

behavior analytically by making the following transformation

( ) ( )

a

f aF

ηζ

η ζ

=

=

١٧

Then the problem(1.43) transforms to 3 2

3 2

0

0 (1.44)

(0) 0, 1

d F d FF

d d

dF dFF

d dζ ζ

ζ ζ

ζ ζ= →∞

+ =

= = =

whose solution can be easily obtained. They showed that

[ ]

2

3 22

2

3 22

( , ) 1( ) (1.45)

2

( , )(1.46)

2

( , ) 1( ) 1 (1.47)

2

f a dFF

a da

f a d F

a da

f a d FF

a da

η ζ ζζ

η ηζ

η ζ ζζ

∂ = − ∂

′∂ = −∂

′′∂ = −∂

Now from (1.45)and (1.46) we see that 0, 0,f f

a a

′∂ ∂≤ <∂ ∂

therefore andf f ′

decrease with an increase in a . From (1.47) we see that the right side is

negative to start with but becomes positive later on. This explains the

behavior of f ′′ noted by Cortell [12].

١٨

Chapter 2 Wang's Equation

2.1 Introduction

The two dimensional constant laminar viscous flow over a semi-infinite

flat plate is modeled by the nonlinear differential equation

0( ) ( ) ( ) 0, [0, ), (2.1 )f f f aη β η η η′′′ ′′+ = ∈ ∞

With the boundary conditions

(0) (0) 0, ( ) 1 (2.1 )f f f b′ ′= = ∞ =

The above boundary value problem is called the Blasius problem [10] and

it successfully describes the velocity distribution and other physical

parameters associated with the fluid motion in the boundary layer along the

plane [22].The main hurdle in the solution of the above problem is the

absence of the second derivative (0)f ′′ .

Once this derivative has been correctly evaluated an analytical solution

of the boundary value problem may be readily found. Blasius [10] found

the following power series solution of the problem with 0

1

2β =

13 2

0

1( ) ( ) (2.2)

2 (3 2)!

kk kk

k

Af

k

ση η+∞

+

=

= −+∑

where 0 1 1A A= = and

1

1,0

3 12

3

k

k r k rr

kA A A k

r

−

− −=

− = ≥

∑

Where σ represents the unknown (0).f ′′ Howarth [17] solved (1.1), (with

0

1

2β = ), numerically and found

0.33206 (2.3)σ =

Several authors have devised numerical algorithms to find good

approximations to (0)f ′′ see, for example, Asaithambi [8] and references

١٩

therein . Asithambi [8] solved (2.1) , (with 0 1β = ), and found (0)f ′′ ,

denoted by α , to be

0.46900 (2.4)α =

Fang et al. [13] have shown that the substitution

0

0

1( ) ( )f Fη β η

β=

transforms Eq. (2.1a) into

0F FF′′′ ′′+ =

Therefore it is sufficient to consider the Blasius problem with 0 1β = .

Henceforth we shall treat the problem with 0 1.β = Liao [18] applied his

homotopy analysis method to the Blasius problem and obtained the

solution to a high level of accuracy. J. H. He [16] has used an iterative

perturbation technique to find an approximate analytical solution of the

Blasius problem. Abbasbandy [3] has used a modified version of the

Adomian decomposition method to find a numerical solution while Cortell

[12] has studied the dependence of the solution on the parameter0β .

Recently Wang [24] has introduced a fresh approach to the Blasius problem

. He used an ingenious idea to transform the Blasius problem into a simpler

problem governed by a second order differential equation. He used the

transformation

( ) , ( ) (2.5)x f y fη η′ ′′= =

to transform (2.1) to 2

20, [0,1) (2.6 )

d y xx a

dx y+ = ∈

with the boundary conditions

1(0) (0), (0) 0, lim ( ) 0 (2.6 )

xy f y y x b

→′′ ′= = =

٢٠

Wang [24] used the Adomian decomposition method to solve (2.6), and

found 3 6 9 12

3 5 7( ) ...... (2.7)

6 180 2160 19008

x x x xy x α

α α α α= − − − −

To find α the equation y(1)=0 is solved for .α He solved this equation

retaining six terms of the series (2.7) and found 0.453539.α = Recently

Hashim [15] improved this value to 0.466799α = by finding terms of the

series (2.7) up to 24x by the Adomian decomposition method (ADM) and

then approximating this function by the [12/12] diagonal Pade'

approximating .

2.2 The Adomian decomposition method

Adomian's decomposition method is used for solving operator

equations of the from

(2.8)u Nuν= +

where N is a nonlinear differential operator and ν is known. Since the

method does not resort to linearization or assumptions of week

nonlinearity, the nonlinearities which can be handled are quite general and

the solutions generated may be more realistic than those achieved by

simplifying the model of the physical problem to achieve conditions

required for other techniques [14].

Adomian's method assumes that the solution u can be expanded as an

infinite series:

0

(2.9)nn

u u∞

==∑

With , .nu X n∈ ∀ The image Nu has an expansion

0

(2.10)nn

Nu A∞

==∑

٢١

With , .nA X n∈ ∀ Substituting (2.9) and (2.10) into (2.8) gives

0 0

(2.11)n nn n

u Aν∞ ∞

= == +∑ ∑

which is satisfied formally if we set

0

1

(2.12)

(2.13)n n

u

u A

ν

+

==

Thus, we need to determine the so-called Adomian polynomials

, 1,2,3,...nA n = To determine the nA s′ , a scalar parameter λ is introduced to

set

0

( ) (2.14)nn

n

u uλ λ∞

==∑

And

0

( ) (2.15)nn

n

Nu Aλ λ∞

==∑

As in [4,5], we can deduce that the coefficients , 1,2,3,...nA n = are given by

0

1( ) (2.16)

!

n

n n

dA Nu

n d λλλ =

=

Thus,

0 0

1

1 0 11

0

2 22 0 2 0 1

( ) ,

( ) ( )( ) ,

1( )( ) ( )( ) ,

2!

A Nu

dA Nu DN u u

d

A DN u u D N u u

λ

λ

λ

λλ

=

=

=

= =

= +

⋮

And so on, where [ ]0( )rD N u denotes the r-th derivative of N at 0u X∈ . We

should note that while nA is a polynomial in 0 1 2, , ,...,u u u its dependence on

0u is decided by the form of N and its derivatives. However generally, nA

depends only on 0 1, ,..., nu u u so that (2.16) allows successive calculation of

٢٢

, 0,1,2,...nu n = at any stage the solution can be approximated by the partial

sum 1

0

N

N jj

s u−

=

=∑

2.3 Wang's transformation

Define ( ), ( ),x f y fη η′ ′′= = then

( )( ) .

( ) ( ),

( )

dy d df

dx d dx

f fdx fd

ηηη

η ηη

η

′′=

′′′ ′′′= =

′′

Using this in (2.1a) we get

0( ) ( ) ( ) 0 (2.17)dy

f f fdx

η β η η′′ ′′+ =

Also assuming ( ) 0,f η′′ ≠ we can write (2.1a) as

0

( )( ) (2.18)

( )

ff

f

η β ηη

′′′= −

′′

Integrate from 0 to η .

[ ] 000

0

0

0

0

ln( ( )) ( )

( )ln ( )

(0)

( ) (0) exp ( ) (2.19)

f f s ds

ff s ds

f

f f f s ds

ηη

η

η

η β

η β

η β

′′ = −

′′ = − ′′

′′ ′′= −

∫

∫

∫

Therefore ( ) 0f η′′ > if (0) 0.f ′′ > Our assumption that ( ) 0f η′′ ≠ is justified.

We can cancel ( )f η′′ in Eq.(2.17). We get 0 ( ) 0dy

fdx

β η+ =

Differentiate again with respect to x

٢٣

2

02

2

02

2

02

( ) 0

0

0

d y df

dx dx

d y xor

dxdxd

d y xor

dx y

ηβ η

β

η

β

′+ =

+ =

+ =

The above equation was obtained by Wang [24]. The new boundary

condition are as follows

x=0 corresponds to 0df

dη= which occurs at 0η = . At this point

2

2

d fy

dα

η= = . Thus (0)y α= .

Also 0 ( )dy

fdx

β η= − . At 0,η = the right side will be zero giving us (0) 0y ′ = .

The condition lim 1df

dη η→∞= indicates that the domain over which Wang's

equation holds is [0,1). From (2.19) we conclude that ,

lim ( ) 0fη

η→∞

′′ =

This condition transforms to

1lim ( ) 0x

y x→

=

2.4 S0lution of Wang's equation by Adomian

decomposition method

0, 1, (2.20)x

y o xy

′′ + = < <

Subject to the conditions

(0) 0, (1) 0, (2.21)y y′ = =

Along with the condition

(0) (2.22)y α=

٢٤

Following the standard procedure of the ADM [4], we write Eq. (2.20) in

the operator form

, (2.23)Ly Ny=

Where the linear operator 2

2

dL

dx= is easily invertible x

Nyy

= − represents

the nonlinear term. Applying the inverse operator 0 0

(.)x x

dxdx∫ ∫ to both sides of

Eq.(2.23) and incorporating the first condition of (2.21) and condition

(2.22) yield 1( ) ( ) (2.24)y x L Nyα −= +

The ADM defines the solution y(x) by the series: 0

nn

y y∞

==∑

The nonlinear term Ny is defined by:

0n

n

Ny A∞

==∑

where the Adomian polynomials nA can be generated for all types of

nonlinearities according to algorithm set by Adomian [4],

0 0

1, 0,1,2,... (2.25)

!

ni

n ini

dA N y n

n d λ

λλ

∞

= =

= = ∑

In the ADM, the components ny are determined by the recursive algorithm:

10 1, ( ), 0k ky y L A kα −

+= = ≤

Some of the Adomian polynomials for xNy

y

−= calculated form (10) are

٢٥

00

11 2

0

21 2

2 3 20 0

33 1 2 1

3 2 3 40 0 0

1,

,

2,

A xy

yA x

y

y yA x

y y

y y y yA x

y y y

= −

= − −

= − −

= − − + −

⋮

For numerical comparison purposes, we construct the solution

0 1 2 3( ) ( ) ( ) ( ) ( ) ... asy x y x y x y x y x= + + + +

lim ( ) ( ), (2.26)nnx y xφ

→∞=

Where ( )n xφ is the n-term approximate solution. The theoretical treatment

of the convergence of the ADM has been considered and discussed in [1,2].

Below we present first few members of the set of Adomian polynomials

associated with this problem as well as the functions 0 1 2, , ...y y y calculated

from them.

0 0

3 4

1 1 3

6 7

2 23 5

9 10

3 35 7

12 13

4 47 9

15 16

5 59 11

, ,

, ,6 6

, ,180 30

, ,2160 144

2099, ,

19008 1425600

2099 31453, ,

2993766000 99792000

xy A

x xy A

x xy A

x xy A

x xy A

x xy A

αα

α α

α α

α α

α α

α α

= = −

= − = −

= − = −

= − = −

= − = −

= − = −

⋮ ⋯

The value ofα can be found by using the last condition in (2.21), i.e.

y(1)=0.

٢٦

By solving (1) 0, 1,2..,6k kφ = = , we find

1

2

3

4

5

6

0.408248

0.441743

0.452576

0,457674

0.460566

0.462404

αααααα

======

2.5 Pade approximation

Hashim [15] improved the above values by approximating the function y

into a rational function with respect to the variable x using diagonal Pade

approximation. He employed the Maple built-in diagonal Pade

approximants together with the condition (1) 0nφ = to get the approximate

value for α .

He presented the calculated values for α using the nφ for both the ADM

and ADM-Pade approximation. The ADM with Pade approximation gives

more accurate results compared with the standard ADM without Pade

approximation. The accuracy of the solution will be further improved if

more terms in the series solution are used. Hashim's results are reproduced

in Table 3.

Level of Pade approximation Value of α

[3/3] 0.447214

[6/6] 0.463257

[9/9] 0.466791

[12/12] 0.466799

Table 3: Evaluation of α using Pade approximation

٢٧

Chapter 3 Analytical Solution of Wang's Equation

3.1 Introduction

In this chapter we shall find an analytical solution of Wang's equation

by a method independent of Adomian decomposition method. This method

can easily extend the series (2.7) to as many terms as we wish. From the

results in [15,24] it seemed that one could determineα to arbitrary

accuracy once a large number of terms of the series (2.7) was available.

Unfortunately this expectation turns out to be false. We shall derive two

sequences one of them increasing the other decreasing, both of them

converging toα . The convergence is hopelessly slow. However the two

sequences enable us to check accuracy of the approximation at each step.

We shall establish that the integral22

20

d fd

dη

η

∞ ∫ is bounded below by 3

4α

and above by α . If we use the value of 0.469606α = then we get the exact

result22

20

0.37118d f

dd

ηη

∞ =

∫

We use the above result to find an approximate asymptotic expression for

( )f η . The power series solution (2.2) is known to be valid for 0 5.69η≤ ≤

[18]. This solution combined with an asymptotic solution will represent the

function ( )f η analytically on the entire domain 0 η≤ ≤ ∞ .

3.2 Blasius solution

First we obtain Blasius solution give by (2.2) of the last chapter.

We want to solve

0( ) ( ) ( ) 0 (3.1)

(0) 0, (0) 0, (0)

f f f

f f f

η β η ησ

′′′ ′′+ =′ ′′= = =

٢٨

Let 0

( ) ,nn

n

f aη η∞

==∑

Successivederivativesgive us

1

1

2

2

3

3

( ) ,

( ) ( 1) ,

( ) ( 1)( 2) ,

nn

n

nn

n

nn

n

f na

f n n a

f n n n a

η η

η η

η η

∞−

=∞

−

=∞

−

=

′ =

′′ = −

′′′ = − −

∑

∑

∑

Put them in (3.1)

3 20

3 2 0

( 1)( 2) ( 1) 0n k nn k n

n k n

n n n a k k a aη β η η∞ ∞ ∞

− −

= = =

− − + − =∑ ∑ ∑

The above expression is equivalent to

3 0 20 0 0

( 3)( 2)( 1) ( 2)( 1) 0n k nn k n

n k n

n n n a k k a aη β η η∞ ∞ ∞

+ += = =

+ + + + + + =∑ ∑ ∑

Or

3 0 20 0 0

( 3)( 2)( 1) ( 2)( 1) 0 (3.2)m k nm k n

m k n

m m m a k k a aη β η∞ ∞ ∞

++ +

= = =

+ + + + + + =∑ ∑∑

Define . Therefore .m k n n m k= + = −

Since 0, this implies0 .n k m≥ ≤ ≤ . Eq. (3.2) becomes

3 0 20 0 0

3 0 20 0

3 0

( 3)( 2)( 1) ( 2)( 1) 0

( 3)( 2)( 1) ( 2)( 1) 0

Sinceeach coefficient in theaboveseries must vanish, we get

( 3)( 2)( 1) ( 2)

mm m

m k m km m k

mm

m k m km k

m

m m m a k k a a

m m m a k k a a

m m m a k

η β η

β η

β

∞ ∞

+ + −= = =

∞

+ + −= =

+

+ + + + + + =

+ + + + + + =

+ + + + +

∑ ∑∑

∑ ∑

20

03 2

0

( 1) 0

or

( 2)( 1) (3.3)( 3)( 2)( 1)

m

k m kk

m

m k m kk

k a a

a k k a am m m

β

+ −=

+ + −=

+ =

= − + ++ + +

∑

∑

Since 1 20, 0, ,2!oa a aσ= = = we get from (3.3)

٢٩

20

3 4 5

2 30

6 7 8

3 40

9 10 11

0, 0,5!

110, 0,

8!

3750, 0,

11!

a a a

a a a

a a a

β σ

β σ

β σ

= = = −

= = =

= = = −

It is clear that we shall get non zero ma for m=2,5,8,11,…

Let 3 1, 1,2,3,...m = − =ℓ ℓ

3 10

3 2 2 3 10

( 2)( 1) , 0,3,6,...,3 3(3 )(3 1)(3 2) k k

k

a k k a a kβ −

+ + − −=

= − + + = −+ + ∑

ℓ

ℓ ℓℓ

ℓ ℓ ℓ

We shall get non zero coefficients if

2 2,5,8,..., for this reason wedefine 2 3 2, 0,1,2,...

or 3 , 0 1

k k r r

k r k

+ = + = + == ≤ ≤ −ℓ

Now Eq. (3.3) transforms to 1

03 2 3 2 3 3 1

0

(3 2)(3 1) , (3.4)(3 )(3 1)(3 2) r r

r

a r r a aβ −

+ + − −=

= − + ++ + ∑

ℓ

ℓ ℓℓ ℓ ℓ

In order to get the solution in the form (2.2) we define 1

3 23 2 (3.5)

(3 2)!

ba

σ ++

+ =+

ℓ

ℓ

ℓℓ

and replace 3 2 3 2, ra a+ +ℓ etc. in (3.4). We get

1 113 2 0 3 2 3 3 1

0

(3 2)(3 1) .(3 2)! 3 (3 1)(3 2) (3 2)! (3 3 1)!

r rr r

r

b b br r

r r

σ β σ σ+ + −−+ + − −

=

= − + ++ + + + − −∑ℓ ℓℓ

ℓ ℓ

ℓ ℓ ℓ ℓ ℓ

Simplification gives 1

3 2 3 3 13 2 0

0

1

0 3 2 3 3 10

(3 1)!(3 )!(3 3 1)!

3 1(3.6)

3

r r

r

r rr

b bb

r r

b br

β

β

−+ − −

+=

−

+ − −=

= − −− −

− = −

∑

∑

ℓ

ℓ

ℓ

ℓ

ℓ

ℓℓ

ℓ

From (3.5) we find

22 2!

ba σ=

٣٠

Since 2 ,2

aσ= we get 2 1b =

Formula (3.6) with 2 1b = produces the solution given by Blasius [10] in

1908.

3.3 Solution of Wang's equation

Wang's equation can be made a useful tool to elucidate analytical

features of the Blasius problem. He solved this equation by Adomian

decomposition method. However this method requires a lot of computation

including double integration. In this section we shall solve Wang's equation

by the same approach as was used to find Blasius solution. Re-write Eq.

(2.6a) in the form

(3.7)yy x′′ = −

We assume the solution in the form

0

( ) (3.8)nn

n

y x a x∞

=

=∑

From the boundary conditions (2.6b), it is clear that

0 1(0), 0 (3.9)a f a′′= =

We define (0)fα ′′= . We are following the same notation as used by Wang

[24] and Hashim [15], since in (2.1a) has now been put equal to unity.

Differentiate y twice and substitute y and y ′′ in (3.7). We get

2

0 2

( 1) (3.10)n kk n

k n

a a n n x x∞ ∞

+ −

= =

− = −∑∑

Writing first few terms of the series on the left we obtain 2 2

2 1 2 3 4 1 3 2

35 1 4 2 3

2 (2 6 ) (12 6 2 )

(20 12 8 ) ... (3.11)

a a a a x a a a a x

a a a a a x x

α α αα

+ + + + + +

+ + + = −

٣١

from which we get

2 3 4 5

10, , 0, 0 (3.12)

6a a a a

α= = − = =

Define new summation index 2m n k= + − and change the order of

summation in (3.10). We get

20 0

( 2)( 1) (3.13)m

mk m k

m k

a a m k m k x x∞

− += =

− + − + = −∑∑

On the left side of (3.13) coefficient of each power of x other than unity

must vanish. This gives for 2m ≥

20

( 2)( 1) 0 (3.14)m

k m kk

a a m k m k− +=

− + − + =∑

which can be re-written as

2 0 21

( 2)( 1) ( 2)( 1)m

m k m kk

m m a a a a m k m k+ − +=

+ + = − − + − +∑

We have obtained the recurrence relation

2 21

1( 2)( 1) ( 2)( 1), 2 (3.15)

m

m k m kk

m m a a a m k m k mα+ − +

=

+ + = − − + − + ≥∑

If 3 is not a divisor of 2m + it will not divide both k and 2m k− + . From

(3.9) and (3.12) we note that, with the exception of 0a and 3a , other

coefficients upto and including 5a are all zero. Using this in (3.15) we find

that 6 0a ≠ but 7 8 0a a= = . Applying the principle of mathematical induction

we can easily establish that only those coefficients whose suffixes are

multiples of 3 fail to vanish. We can replace (3.15) by 1

3 3 3 31

13 (3 1) (3 3 )(3 3 1) , 2 (3.16)

n

n k n kk

n n a n k n k a a nα

−

−=

− = − − − − ≥∑

Define

33 2 1

(3.17)(3 )!

nn n

ba

n α −= −

then (3.16) becomes

٣٢

13 3 3 3

2 1 2 1 2( ) 11

( ) ( ) ( )13 (3 1) (3 3 )(3 3 1) .

(3 )! (3 )! (3 3 )!

nn k n k

n k n kk

b b bn n n k n k

n k n kα α α α

−−

− − − −=

− − −− = − − − −−∑

Or 1

3 3 3 31

(3 2)!

(3 )!(3 3 2)!

n

n k n kk

nb b b

k n k

−

−=

−=− −∑

Or

1

3 3 3 31

3 2(3.18)

3

n

n k n kk

nb b b

k

−

−=

− =

∑

From (3.17) we get

00 01

33 6

ba b

ba

αα

α

−= − = −

= −

therefore 0 31, 1b b= − = .

Formula (3.18) with 0 31, 1b b= − = gives a solution of Wang's equation

similar to Blasius solution found in the last section.

In (3.18) let n=2. We get

6 3 3

44

3b b b

= =

Also n=3 give,

2

9 3 9 3 3 6 6 31

7 7 7168

3 3 6k kk

b b b b b b bk −

=

= = + =

∑

Then (3.17) gives,

6 3 3

9 5 5

4 1

6! 180168 1

9! 2160

a

a

α α

α α

= − = −

= − = −

We calculate first few coefficients in the above manner and list them in the

following lines.

٣٣

0 0

3 3

6 6 3

9 9 5

12 12 7

6

15 15 9

6

18 18 11

7

21 21 13

24

1,

11,

61

4,180

1168,

21601

25200,19008

7.01125 109168432,

1.03002 106594561792,

1.61614 108257020125920,

16498091239838

b a

b a

b a

b a

b a

b a

b a

b a

b

α

α

α

α

α

α

α

α

−

−

−

= − =

= = −

= = −

= = −

= = −

×= = −

×= = −

×= = −

=8

24 15

9

27 27 17

10

30 30 19

2.65906 10208,

4.53471 1049377815204079398400,

7.95335 10210964851697107227166720,

a

b a

b a

α

α

α

−

−

−

×= −

×= = −

×= = −

⋮ ⋱

Hashim [15] found the first nine coefficients (up to 24a ) of (3.8), in

complete agreement with members of the above list. Equation (3.16) can be

easily used in a computer algorithm to extend the series (2.7) to an arbitrary

number of terms with very little effort. As an example we give below

coefficient of the 1500x term in the series (3.8). 335

1500 999

4.1809830 10a

α

−×= −

٣٤

3.4 An increasing sequence

To find α we have to solve the equation (1) 0,y = where

33

0

( ) (3.19)nn

n

y x a x∞

==∑

This gives

32 1

1

0 (3.20)nn

n

aαα

∞

−=

+ =∑

Hashim [15] solved (3.20) after truncating the series after 7 terms and

found 0.463662α ≃ . He improved this value to 0.466799 by approximating

the series for y(x) up to 21x by the [12/12] Pade approximant. Since we can

find an arbitrary number of terms in the series (3.20), we can attempt to

solve (3.23)with a large number of terms. Let ( )F α denote the left side of

Eq. (3.20) and let

32 1

1

( ) , 1,2,3,.... (3.21)k

nk n

n

aF kα α

α −=

= + =∑

and let kα be a root of the equation

( ) 0 (3.22)kF α =

We shall solve (3.8) by a slightly modified Newton's method. We first find

1 1bysolving ( ) 0Fα α = . To find , 2,k kα ≥ we use the iterative scheme

1 1

11

(3.23)

( ), 1,2,...

( )

k

k ii i

k k

x

F xx x i

F

α

α

−

+−

=

= − =′

and we approximate the derivative at 1kα − by

1 11

( ) ( )( ) (3.24)k k k k

k k

F h FF

h

α αα − −−

+ −′ =

and choose h a sufficiently small number. The above scheme works

efficiently and produces a sequence some of whose members are given in

the following table.

٣٥

1 0.408248 100 0.469365

2 0.441743 200 0.469496

5 0.460566 300 0.469535

7 0.463662 1000 0.469583

15 0.467293 2000 0.469592

25 0.468366 3000 0.469595

50 0.468066 4000 0.469597

k kk kα α

Table 4

Roots of Eq. (3.22)

The above sequence increases because for larger k more terms in the series

(3.23) are retained and a larger value of α will be needed to satisfy the

equation. The sequence converges very slowly and no estimate about the

accuracy of any term is available. In the next section we shall obtain a

decreasing sequence also converging α . The two sequences together will

determine precisely lower and upper bounds for α at each level of

approximation.

3.5 A decreasing sequence for α .

Hashim [15] and Wang [24] have found the following expression for 1y

3 6 9 12

3 5 7 9

3

3 2 11

1 1 2099... (3.25)

6 30 144 1425600

1 n

n nn

x x x x

y

xc

α α α α α

α α

∞

+=

= + + + + +

= +∑

where 3nc is defined to be the coefficient of 3

2 1

n

n

x

α + in (3.25). The solution

y(x) of (3.7) is related to (3.25) in the following manner

0 0

( ) (3.26)x x x

y x dxy

α= − ∫ ∫

If we use (3.25) in (3.26), we obtain

٣٦

3 3 3

3 2 11

3 33

2 10

33 3

2 11

( ) (3.27. )6 (3 3)(3 2)

(3 3)(3 2)

(3.27. )3 (3 1)

n

n nn

nn

nn

nn

nn

x xy x c a

n n

c x

n n

c xb

n n

αα α

αα

αα

+∞

+=

+∞

+=

∞−

−=

= − −+ +

= −+ +

= −−

∑

∑

∑

Compare (3.27.b) with (3.19), we find

3 3 33 (3 1) , 1 (3.28)n nc n n a n− = − − ≥

Let x=1 in (3.27.b) . We shall truncate the series after k terms and find an

approximate sum of the remainder. We have

3 3 32 1 2 1

1

0 (1) ...... (3.29)3 (3 1) (3 3)(3 2)

kn k

n kn

c cy

n n k kα

α α−

− +=

= = − − −− + +∑

If k is large, we can approximately set

23 3

3

43 6

3

k

k

k

k

c

c

c

c

α

α

+

+

=

=

and so on. Eq. (3.29) becomes

3 3 32 1 2 1

1

10 (3.30)

3 (3 1) (3 3)(3 2)

kn k

n kn n k

c c

n n n nα

α α

∞−

− += =

= − −− + +∑ ∑

It is easy to show that

1 1

(3 3)(3 2) 3 2n k n n k

∞

=

=+ + +∑

and (3.30) becomes

3 3 32 1 2 1

1

03 (3 1) (3 2)

kn k

n kn

c c

n n kα

α α−

− +=

= − −− +∑

In the notation of Section 3, the above equation may be written as

3 32 1

3 3( ) 0 (3.31)k kk

kF aα

α ++

++ =

Since the approximation used to obtain (3.30) replaces a larger number by a

smaller one, we have

٣٧

( ) ( )F Gα α≥

Where ( )and ( )F Gα α denote respectively the left sides of Eqs. (3.20) and

(3.31). The above inequality ensures that if eα is a zero of ( )F α then a zero

of ( )G α greater than eα exists. We solve (5.3) for k=1,2,… by the modified

Newton's method.

Let ( )kα denote a root of (5.3). We display a few members of this sequence

in Table 5. ( ) ( ) ( ) ( )

5 0.482463 100 0.470006 900 0.469633 2100 0.4696130

7 0.478413 200 0.469784 1000 0.4696298 3000 0.4696088

15 0.473268 300 0.469716 1500 0.4696190 3500 0.4696074

50 0.470505 500 0.469665 1800 0.4696155 4000 0.469

k k k kk k k kα α α α

6064

Table 5

Roots of Eq. (3.31)

Using the last entries of Tables 4 and 5 we have the following inequality

for α

0.469597 0.4696064 (3.32)α< <

3.6 Pade approximation

In chapter 2 we discussed Hashim's results which were found by

approximating the power series solution of Wang by a Pade approximant.

letting x=1 and putting the numerator equal to zero. From the [12/12]

approximant he found 0.466799.α ≃

We have shown above that the exact value of α lies between 0.4695975

and 0.4696064. Thus Hashim's result is correct to only two decimal

positions and, in order to achieve a more accurate result by this method,

Pade' approximants of very high order must be used. However the amount

of work involved in ADM becomes prohibitive after a few terms which

٣٨

precludes the use of this approach to high orders. We can use our solution

to carry this method to higher orders to generate a sequence of numbers

which should converge to .α We find that, up to a point, this sequence

increases and appears to be converging to the exact value of ,α each term

being a better approximation as compared with the corresponding number

given by a straightforward solution of the equation y(1)=0 after truncating

the series (2.7) and retaining an appropriate number of terms. However

after this stage the Pade' approximations start yielding numbers which,

instead of giving an improved approximation, produce errors of increasing

magnitude. Thus this method appears to be of limited utility.

We illustrate our method by an example. Let us truncate the series (2.7)

after five terms and let x=1. We get an approximate value of α by solving

the equation

3 5 7

1 1 1 10 (3.33)

6 180 2160 19008α

α α α α− − − − =

If we divide through with 2

1and letpα

α= , the equation becomes

2 3 4

1 0 (3.34)6 180 2160 19008

p p p p− − − − =

We replace the left side of (3.34) by its [2/2] Pade' approximant. A Pade'

approximant can be easily found by using algebraic softwares like

MATLAB or Mathematica. This gives

2

2

119 731

396 3960 0 (3.35)53 1

1396 594

p p

p p

− +=

− +

By equating the numerator of the fraction on the left side of (3.35) to zero

we find p =4.65968. An approximate value of α is given by

10.463257

p= . This agrees with the entry which appears in the table of

٣٩

Hashim[15] corresponding to his Pade [6/6] approximant. In Table 6 we

present approximate values of α calculated by using successive Pade'

approximants starting with the [2/2] approximant and going up to the

[23/23] approximant. For comparison, we also include values of this

parameter obtained by solving an equation of the from(3.34) after steadily

increasing the number of terms. Note that our [23/23] approximant

corresponds to the [69/69] approximant of Hashim assuming that one could

go as far as that by adopting the ADM approach.

2 0.457674 0.463257 14 0.468525 0.469075

3 0.462404 0.466791 15 0.468611 0.469086

4 0.464572 0.468061 16 0.468686 0.469097

5 0.465791 0.468645 17 0.468750 0.469107

6 0.466561 0.468956 18 0.468807 0.469118

7 0.467088 0.4689

i ii ii iα α α α

80 19 0.468857 0.469123

8 0.467469 0.468997 20 0.468901 0.469124

9 0.467756 0.469011 21 0.468941 0.469053

10 0.467980 0.469025 22 0.468977 0.469977

11 0.468158 0.469038 23 0.469009 0.474672

12 0.468303 0.469051 24 0.469039no value

13 0.468424 0.469063 25 0.469066no value

Table 6. Values of iα found by solving the equation y(1)=0, when we

retain 2i+1 terms in (2.7) and iα given by the [i/i] Pade' approximant.

A glance at the Table indicates that successive terms of the sequence iα give better approximation to the exact value of α for i =2,3,…,22 but 23α is much worse compared with 23α . Also the sequence ends abruptly at

23α because the Pade' [24/24] approximant equated to zero does not yield a

real root leading to a suitable entry in the Table. We are faced with a

dilemma. On the one hand the sequence iα continues beyond i =23 and the

٤٠

subsequent terms yield an increasing sequence which converges to α , on

the other hand the corresponding Pade' approximant either produces

an iα which is worthless or it fails to generate any value of iα at all. This is

strange since it is well known that an [m/m] Pade' approximant gives a

better representation of a function than its Taylor series truncated after 2m

terms. The first 22 terms of Table 4 mirror this fact. The anomalous

behavior after i=23 can be explained in only one way that the Pade'

approximants from the stage [23/23] onwards no longer adequately

represent the corresponding Taylor polynomial on an interval containing

the unique zero of the polynomial. Probably this occurs due to ill-

conditioned matrices which are involved in the calculation of coefficients

in the Pade' approximants of sufficiently high orders.

3.7 The integral 2

0

[ ( )] .f dη η∞

′′∫

Integrate (2.7) with respect to x on [0.1). We get 1

3 50

1 1 1.... (3.36)

4.6 7.180 10.2160

(3.37)

ydx αα α α

α

= − − −

<

∫

Also we have from (3.36) 1

3 50

1 1 1 1( ....)

4 6 180 2160

1(3.38)

43

(3.39)4

ydx αα α α

α α

α

> − + + +

= −

=

∫

Eq. (3.38) follows from the line preceding it because α satisfies the

equation y(1)=0, which becomes, on letting x=1 in (2.7),

3

1 1....

6 180α

α α= + +

٤١

Now 2

2and .

d f dfy x

d dη η= = Hence

1 22

20 0

[ ] (3.40)d f

ydx dd

ηη

∞

=∫ ∫

From (3.37), (3.39) and (3.40) we obtain the inequality, 22

20

3(3.41)

4

d fd

dα η α

η

∞ < <

∫

If we use 0.469606α = in (3.36) then we can replace (3.41) by the equality 22

20

0.37118 (3.42)d f

dη

∞ =

∫

Write the Blasius equation 0f ff′′′ ′′+ = in the form

ff

f

′′′= −

′′

An integration gives

0

( ) exp ( ) (3.43)f f u duη

η α

′′ = − ∫

It is clear from (3.43) that for 0 , ( ) 0.fη η′′≤ < ∞ ≥ As a consequence ( )f η′ is

nonnegative and monotonically increasing and ( )f η is also increasing and

unbounded. Since ( )f η′ is bounded above by unity, the curve ( )y f η= does

not intersect the line ,y η= because otherwise ( )f η being a convex

function, its slope, at the point of intersection must exceed unity. This

means the graph of ( )y f η= is flanked on the left by the line .y η= The

initial condition lim ( ) 1fη

η→∞

′ = indicates that the function ( )f η has an

asymptote with slope unity but lying below the line .y η= Let the equation

of this asymptote be

0y η η= −

Since 0

0

0

( ) ( ) ,f d dη η

η

η η η η η≥ −∫ ∫ an exponentiation leads to the inequality

٤٢

20

0

1( ) exp ( ) exp ( ) (3.44)

2f f u du

η

η α α η η ′′ = − ≤ − − ∫

Square both sides of (3.44) and integrate on [0, )∞ . We find

2 20

0

0.37118 exp{ ( ) } (3.45)α η η∞

≤ − −∫

where we have used (3.42). Now

2

0

02 2

02

20

0

0 0

0

exp{ ( ) }

(3.46)2

u

u u

u

e du

e du e du

e du

η

η

η

η η

π

∞ ∞−

−

∞− −

−

− − =

= +

= +

∫ ∫

∫ ∫

∫

Combining (3.45) and (3.46) we obtain

02

20

0.37118

2

0.79695 (3.47)

ue duη π

α− ≥ −

=

∫

The above inequality is satisfied if

0 1.16 (3.48)η ≥

The above result indicates that the line

1.16 (3.49)y η= −

is approximately an asymptote to the curve ( ).y f η= Since this curve

eventually behaves like a straight line of slope unity, it follows that, for

large ,η

( ) 1.16f η η≈ −

A numerical solution of the Blasius problem indicates that 0 1.23.η = The

above result gives this number with an error less than 6 percent.

We were able to apply a classical method to solve Wang' equation which

gave the solution to an arbitrary number of terms. In principle a modern

٤٣

method such as the Adomian decomposition method also gives the solution

to as many terms as we wish but the computational effort involved to

calculate the 3000th term, for example, would be tremendous. However our

advantage in easy calculation of the two sequences (k){ }and{ }kα α , is

somewhat neutralized by their extremely slow convergence to α .

٤٤

Chapter 4 Limitations of Adomian Decomposition Method

4.1 Introduction

In the last two chapters we solved Blasius equation and its transformed

version, Wang equation, by Adomian decomposition method as well as by

a direct method.

In both cases the solution comes out in the form of an infinite series.

However the series converges within a finite interval and it is impossible to

get any information about the solution for large values of η . To evaluate

(0)f ′′ , correct to six decimal places, we need to find a series solution of the

Wang equation up to several thousand terms.

Even after using this value of (0)f ′′ to solve the Blasius problem, the series

solution converges only for 5.90η ≤ . To get a complete picture we have to

solve the Blasius problem numerically such as Runge-Kutta method.

Liao has developed a method, called the homotopy analysis method,

which attempts to accelerate the converge of the series solution. In chapter

1 we described his method of solution. However there are a number of

arbitrary parameters in the method and one has to depend on guess work to

estimate a suitable value of them.

In general it is useful to make an asymptotic analysis of a problem

before tackling it by the Adomian or the homotopy analysis methods. Such

methods produce series solutions which, for nonlinear problems, only

converge within a small interval.

٤٥

4.2 A simple problem

To highlight the deficiencies of the Adomian and the homotopy

analysis method, we consider a very simple nonlinear problem 2

22

(4.50)

(0) 0, (0) 0 (4.51)

d yy x

dxy y

+ =

′= =

First we find an approximate analytical solution for large x. Consider a

transformation

(4.52)y u x= +

Then 1

2

3

2

2 2

1

2

1(4.53)

4

and 2 (4.54)

y u x

y u x

y u x u x

−

−

′ ′= +

′′ ′′= −

= + +

For 1x >> , the second term on the right of (4.53) may be dropped. Also we

assume u x<< , an assumption justified later. Therefore we drop 2u in

(4.54) and Eq. (4.50) is transformed to a linear equation

2 0 (4.55)u x u′′ + =

Now it is well-known that the equation 2

2 2 2 2 2 2 22

(1 2 ) [( ) ] 0 (4.56)rd y dyx s x s r a r x y

dx dxα+ − + − + =

has the general solution

1 2[ ( ) ( )] (4.57)s r ry x c J ax c Y axα α= +

Multiply (4.55) with 2x . We get 5

2 22 0 (4.58)x u x u′′ + =

٤٦

Compare (4.58) with (4.56). We get

2 2 2

2 2

1 2 0

0

52

22

s

s r

r

a r

α− =

− =

=

=

Therefore 1 5 2 4 2, , ,

2 4 5 5s r aα= = = = and the general solution of

(4.55) is

2

5

5 5

4 41 2 2

5

4 2 4 2[ ( ) ( )]

5 5u x c J x c Y x= +

Since the Bessel functions are oscillatory with decreasing amplitude such

that

( ) 0 as

and ( ) 0 as

J x x

Y x xα

α

→ → ∞→ → ∞

our assumption that u x<< is justified for large x.

From (4.52) we see that the solution of Eq. (4.50), for large x, will be small

oscillations superimposed on the parabola y x= .

In Fig.4 we present the numerical solution of the problem which confirms

our qualitative analysis.

٤٧

0 10 20 30 40 50

0

1

2

3

4

5

6

7

Fig.4 : Numerical solution of Eq. (4.50).

In Fig.5 we present the solution along with the curve y x= ,

shown dashed .

0 10 20 30 40 50

0

1

2

3

4

5

6

7

Fig. 5 : The solution of Eq. (4.50) together with y x= shown dashed.

٤٨

4.3 Solution by Adomian method If we solve the problem by Adomian decomposition method, we

get the following solution 3 8 13 18 23

28 33

38 43

x x x 95 31x

6 2016 943488 48502831104 9203412201984

74849x 189251x

13507369511417413632 21395673306085183193088

38700523x 1061168785x

2801579230592089310166515712 50011106613036187601787024

xy = − + − +

− +

− +

48

53

58

310272

1527994261379x

47494608245909822416547577359008530432

986872159994333x

20444184729624559311214550180022959197913088

3659317869389202813517x

50960628520373191763798652378101228449208971561009

−

+

−

63

152

63328266525722085749107x

597156645001733061088192608566590194967830728751905243136+

In Fig. 6 we plot the above solution. It is clear that the series diverges

when 3.5x > and the oscillatory behavior depicted in Figs. 4 and 5 is

nowhere to be seen.

0 0.5 1 1.5 2 2.5 3 3.5

0

1

2

3

4

5

6

Fig. 6

٤٩

If we add one more term to the solution it becomes

3 8 6 13

-9 18 -12 23 -15 28

-18 33 -20 38 -23 43

-26 48 -29 53 -32 58

y = 0.166667x 0.000496032x 1.0599 10

1.95865 10 3.36832 10 5.54135 10

8.84529 10 1.38138 10 2.12187 10

3.2172 10 4.82715 10 7.18068 10

1.0605

x

x x x

x x x

x x x

−− + ×− × + × − ×

+ × − × + ×

− × + × − ×

+ -34 63 -37 6810 1.55666 10 .x x× − ×

The graph of shown in Fig.7. Again the divergence of the series is clear

beyond x =3.3

0 0.5 1 1.5 2 2.5 3 3.5

-3

-2

-1

0

1

2

Fig. 7

٥٠

4.4 Homotopy Analysis Method

It has been shown by Liao [19] that the Adomian solution up to m

terms

0

mn

nn

y a x=

=∑

Can be improved by the homotopy analysis method to

,0

( ) (4.59)m

nm n n

n

y h a xµ=

=∑

where , ( )m n hµ is called an approaching function and is defined as

( ),0

1( ) ( ) 1 (4.60)

m nkn

m nk

n kh h h

kµ

−

=

− + = − +

∑

where h is arbitrary and should be chosen so that the solution (4.59) agrees

with the exact solution over as wide an interval as possible.

If we choose m =7 , the homotopy solution becomes

2 3 4 5 6 3

2 2 3 4 5 8 3 2 3 4 13

4 2 3 18 5 2 23

5 2

1(1 )(1 )

6

(1 ) (1 2 3 4 5 6 ) (1 ) (1 3 6 10 15 )

2016 943488

95(1 ) (1 4 10 20 ) 31(1 ) (1 5 15 )

48502831104 9203412201984

74849(1 ) (1 5 15 )

y h h h h h h h x

h h h h h h x h h h h h x

h h h h x h h h x

h h h

= − + + + + + +

− + + + + + − + + + +− +

− + + + − + +− +

− + +−28 6 33189251(1-h) (1 6 )

13507369511417413632 21395673306085183193088

x h x++

We plot the above solution for various values of h and observe that the

homotopy solution also suffers from the convergence problem. In any case

it fails to produce the oscillatory behavior of the solution shown in Fig. 4

and 5 .

٥١

0 0.5 1 1.5 2 2.5 3 3.5

0

1

2

3

4

5

6

7

Fig. 8 : Homotopy solution with h = 0.1

0 0.5 1 1.5 2 2.5 3 3.5

0

1

2

3

4

5

6

7

Fig. 9 : Homotopy solution with h = -0.1

٥٢

0 0.5 1 1.5 2 2.5 3 3.5

0

1

2

3

4

5

6

Fig. 10 : Homotopy solution with h = 0.2

0 0.5 1 1.5 2 2.5 3 3.5

-3

-2

-1

0

1

2

Fig. 11 : Homotopy solution with h = -0.2

٥٣

0 1 2 3 4

-3

-2

-1

0

1

2

Fig. 12 : Homotopy solution with h = 0.5

0 0.5 1 1.5 2 2.5 3 3.5-40

-30

-20

-10

0

10

20

Fig. 13 : Homotopy solution with h =- 0.5

We note that the homotopy solution in Figs. 11-13 produces a zero but the

exact solution is free of zeros. This shows that a bad choice of the

parameter h can lead to a solution which is totally wrong even for small

values of x.

٥٤

4.5 Conclusion

By applying the Adomian and homotopy analysis method to our

simple nonlinear problem we have shown that these methods produce

series solutions which converge over a relatively short interval and it is

impossible to get any information for large x. For this one should turn to

asymptotic analysis of the problem.

٥٥

References

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٥٦

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