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 u u v x y y ν ∂ ∂ ∂ + = ∂ ∂ ∂ 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 ) ux x a vx y b uxy U y c = ≥ = ≥ → →∞ where U is the constant speed of the flow outside the boundary Layer. Define a stream function ( , ) xy ψ such that , (1.4) u v y 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.
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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].
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