# Solution of the Boundary Layer Equation of the Power-Law Pseudoplastic Fluid Using Differential Transform Method

Authors: Sobhan Mosayebidorcheh
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

The boundary layer equation of the pseudoplastic fluid over a flat plate is considered. This equation is a boundary value problem (BVP) with the high nonlinearity and a boundary condition at infinity. To solve such problems, powerful numerical techniques are usually used. Here, through using differential transform method (DTM), the BVP is replaced by two initial value problems (IVP) and then multi-step differential transform method (MDTM) is applied to solve them. The differential equation and its boundary conditions are transformed to a set of algebraic equations, and the Taylor series of solution is calculated in every sub domain. In this solution, there is no need for restrictive assumptions or linearization. Finally, DTM results are compared with the numerical solution of the problem, and a good accuracy of the proposed method is observed.


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## Body

## 1. Introduction

Boundary layer equation of the Newtonian fluid over a flat plate is one of the classic issues of mechanical engineering. Because of high level of application of non-Newtonian fluids in industry, the boundary layer equation of the non-Newtonian fluids is being noted by engineers. On the other hand, nonlinearity of these equations attracted the interest of mathematicians to evaluate the power and accuracy of the approximate and numerical methods. One of the most significant non Newtonian models is the power law one. In this model, shear thinning or shear thickening property of the fluid is considered and the shear stress has a nonlinear relation with shear rate, while shear stress has a linear relation with shear rate for the Newtonian fluid.As we pointed out above, boundary layer problems are among the nonlinear ones and most of them do not have an exact analytical solution. So, numerical and approximate methods are used by researchers to solve such equations. The most known numerical method used to solve boundary layer problems is the shooting one. Based on this method, boundary value problem is transformed to an initial value problem with unknown initial values. After that, the problem is replaced with a system of first-order ordinary differential equations and usually is solved through Runge-Kutta method. Shooting method is appropriate to solve many boundary value problems, but this one is not useful for solving some BVPs because of the instability of the solution.Approximate techniques like decomposition method (DM), homotopy analysis method (HAM), homotopy perturbation method (HPM), and variational iteration method (VIM) are good substitutes for numerical methods. During the recent years, boundary layer problems have been solved using some of these methods, such as HAM [1–9], HPM [10–14], VIM [15–17], and DM [18–21]. In most of the researches, some modifications were introduced to overcome the nonlinearity and the boundary condition at infinity.Differential transform method (DTM) is also one of the other approximate methods to solve differential equations. Here, DTM and multi-step DTM are used to solve the boundary layer equation of the pseudoplastic fluid. This method was introduced by Zhou [22] to solve initial value problems in analysis of the electrical circuits. After that, DTM is applied to differential algebraic equation [23, 24], partial differential equation [25–30], integral equation [31–33], ordinary differential equation [34–38], and fractional differential equation [39–42]. The method is an iterative technique to find the Taylor series solution of the problem. There is no need for the high calculation cost to determine the coefficients of Taylor series, which is the reverse of the standard Taylor series method.This paper is organized as follows. In Section2, DTM and multi-step DTM are introduced. In Section 3, the boundary layer equation of the pseudoplastic fluid is extracted from continuity and momentum equations as presented in [43]. In Section 4, the problem is solved and the results are illustrated as some figures and tables.


## 2. Differential Transform Method

The differential transform is defined as follows:(1)X(k)=1k![dkx(t)dtk]t=t0,
wherex(t)is an arbitrary function andX(k)is the transformed function. The inverse transformation is as follows:
(2)x(t)=∑k=0∞X(k)(t-t0)k.
Substituting (1) into (2), we have
(3)x(t)=∑k=0∞(t-t0)kk![dkx(t)dtk]t=t0.
The functionx(t)is usually considered as a series with limited terms and (2) can be rewritten as
(4)x(t)≈∑k=0mX(k)(t-t0)k,
wheremrepresents the number of Taylor series’ components. Usually, through elevating this value, we can increase the accuracy of the solution.Although the DTM series solution is a good approximate of the exact solution, but this series is diverged for greater areas. There are two ways to overcome this problem. One is the using pade approximate technique. Usually, Pade approximate gives us more exact information about the behavior of the solution (see [44]). Another one is to use multi-step DTM. Based on this one, solution domain is divided to some sub domains.To solve a differential equation in the domain[0,T]using multi-step DTM, this domain is divided toNsections. We suppose the sub domains are equal and length of each sub domain isH=T/N. So, there is a separate function for every sub domain as follows:
(5)x(t)={x1(t),t∈[t1,t2]⋮xi(t),t∈[ti,ti+1]⋮xN(t),t∈[tN,tN+1],
whereti=(i-1)H. Multi-step DTM for every sub domain is defined as
(6)Xi(k)=Hkk![dkxi(t)dtk]t=ti.
The inverse multi-step DTM is
(7)xi(t)=∑k=0∞Xi(k)(t-tiH)k.
Some of the properties of DTM and multi-step DTM are shown in Table 1. These properties are extracted from (1) and (6).Table 1
The properties of the DTM and multi-step DTM.
Original function
DTM
Multi-step DTM
f
(
t
)
=
g
(
t
)
±
s
(
t
)
F
(
k
)
=
G
(
k
)
±
S
(
k
)
F
(
k
)
=
G
(
k
)
±
S
(
k
)
f
(
t
)
=
c
g
(
t
)
F
(
k
)
=
c
G
(
k
)
F
(
k
)
=
c
G
(
k
)
f
(
t
)
=
d
n
g
(
t
)
d
t
n
F
(
k
)
=
(
k
+
n
)
!
k
!
G
(
k
+
n
)
F
(
k
)
=
(
k
+
n
)
!
H
n
k
!
G
(
k
+
n
)
f
(
t
)
=
g
(
t
)
s
(
t
)
F
(
k
)
=
∑
r
=
0
k
G
(
r
)
S
(
k
-
r
)
F
(
k
)
=
∑
r
=
0
k
G
(
r
)
S
(
k
-
r
)


## 3. Mathematical Formulation

Two-dimensional boundary layer equations for an incompressible fluid are(8)U∂U∂X+V∂U∂Y=1ρ∂τ∂Y(9)∂U∂X+∂V∂Y=0,
whereρis the fluid density,τis the shear stress, andUandVare the velocities in direction ofxandy, respectively. For the non-Newtonian power-law fluid, the shear stress is calculated through the following relation:
(10)τ=m|∂U∂Y|n-1∂U∂Y,
wheremandnare the flow consistency and power-law index, respectively.Here, the following dimensionless parameters are used:(11)Re=ρU∞2-nLnm,x=XL,y=YLRe1/(n+1),u=UU∞,v=VU∞Re1/(n+1),
whereLis the plate length,U∞is the far field velocity, andReis generalized Reynolds number.Combining (8), (10), and (11), the dimensionless momentum equation can be obtained:
(12)u∂u∂x+v∂u∂y=∂∂y(|∂u∂y|n-1∂u∂y)
with the boundary conditions
(13)u(x,0)=v(x,0)=0,u(x,∞)=1,u(0,y)=1.
Similarity variable and stream function for the problem is defined as [45]
(14)η=y(n(n+1)x)1/(n+1)Ψ=(n(n+1)x)1/(n+1)f(η).
Regarding (14), the velocities can be obtained as a function of the similarity variable:
(15)u=∂Ψ∂y=f′(η)v=-∂Ψ∂x=1n+1[n(n+1)xn]1/(n+1)(ηf′(η)-f(η)).
Substituting (15) in (12), we have
(16)(|f′′|n-1f′′)′+nff′′=0.
Equation (16) is the boundary layer equation of the power-law fluid as a function of indexn. For pseudoplastic fluid (n<1) the value of thef′′(η)is more than zero [43, 45–48]. So, (16) can be simplified as
(17)f′′′(η)+f(η)(f′′(η))2-n=0forn≤1,f(0)=0,f′(0)=0,f′(∞)=1.


## 4. Solution of the Problem

In this section, we try to solve (17) using DTM and multi-step DTM. The solution consists of two stages; first through mathematical relations and applying DTM, the value off′′(0)is found. After that, boundary layer equation of the pseudoplastic fluid is solved as an initial value problem (IVP) using multi-step DTM.


### 4.1. Applying DTM (Obtainingf′′(0))

The BVP (17) can be transformed to an initial value problem with the replacement of the unknown initial conditionf′′(0)=λinstead of the boundary condition at infinityf′(∞)=1:
(18)f′′′(η)+f(η)s(η)=0forn≤1f(0)=0,f′(0)=0,f′′(0)=λ,
where
(19)s(η)=(f′′(η))2-n.
By applying the DTM on (18) atη=0, the recursive relation is derived:
(20)F(k+3)=-1(k+1)(k+2)(k+3)∑r=0kF(r)S(k-r)F(0)=0,F(1)=0,F(2)=λ2.
Regarding (19),
(21)s′(η)f′′(η)=(2-n)f′′′(η)s(η).
Taking differential transform on (21), we have
(22)∑r=0k(r+1)S(r+1)(k-r+1)(k-r+2)F(k-r+2)=(2-n)∑r=0k(r+1)(r+2)(r+3)F(r+3)S(k-r).
From (19), the value ofS(0)can be obtained:
(23)S(0)=s(0)=(f′′(0))2-n=λ2-n.
Combining (20), (22), and (23), the coefficients of the Taylor series solution of (18) can be calculated:
(24)F(0)=0,F(1)=0,F(2)=λ2,F(3)=0,F(4)=0,F(5)=-1120λ3-n,F(6)=0,F(7)=0,F(8)=11920λ5-2n-14032λ5-2nn,F(9)=0,F(10)=0,F(11)=-891900800λ7-3n+102719958400λ7-3nn-171280λ7-3nn2,⋮
Now, consider a similar equation (18) in whichλ=1:
(25)g′′′(η)+g(η)(g′′(η))2-n=0forn≤1g(0)=0,g′(0)=0,g′′(0)=1.
In the same manner, the coefficientsG(k)are
(26)G(0)=0,G(1)=0,G(2)=12,G(3)=0,G(4)=0,G(5)=-1120,G(6)=0,G(7)=0,G(8)=11920-14032n,G(9)=0,G(10)=0,G(11)=-891900800+102719958400n-171280n2,⋮
Substituting (24) and (26) in (2), thef(η)andg(η)are obtained. These functions have a relation as follows:
(27)f(η)=λqg(λpη),p=2-n3,q=2n-13.
Forn=1, both values ofpandqare 1/3 for the Blasius equation. From (27), we have
(28)f′(η)=λq+pg′(λpη)⟶f′(∞)=λq+pg′(∞).
Regarding (28) and the boundary condition at infinity in (17), the value ofλcan be obtained:
(29)λ=(1g′(∞))3/(1+n).


### 4.2. Applying Multi-Step DTM (Solution of IVPs)

As was noted, multi-step DTM usually is used for solving problems in which Taylor series is diverged in the solution domain. This technique was already used to solve engineering and computational problems [49–57].To solve boundary layer problems, the domain[0,∞)is replaced by[0,η∞). Butη∞should be great enough that the solution is not to be dependent on. The solution domain should be divided toNequal parts (H=η∞/N). So, we have
(30)gi′′′(ηi)+gi(ηi)si(ηi)=0,(i-1)H≤ηi<iH,for1≤i≤N,
where
(31)si(η)=(gi′′(η))2-n.
Applying multi-step DTM on (30) and (31), we have
(32)Gi(k+3)=-H3(k+1)(k+2)(k+3)∑r=0kGi(r)Si(k-r),for1≤i≤N∑r=0k(r+1)HSi(r+1)(k-r+1)(k-r+2)H2Gi(k-r+2)=(2-n)∑r=0k(r+1)(r+2)(r+3)H3Gi(r+3)Si(k-r),for1≤i≤N.
So
(33)Si(1)=2-nGi(2)[3Gi(3)Si(0)].
And fork≥1(34)Si(k+1)=12(k+1)Gi(2)[∑r=0k-1-∑r=0k-1(r+1)Si(r+1)(k-r+1)×(k-r+2)Gi(k-r+2)+(2-n)∑r=0k(r+1)(r+2)(r+3)Gi×(r+3)Si(k-r)∑r=0k-1].
The initial conditions of the problem are considered for first sub domain (i=1). Regarding (6), (25), and (31), initial conditions transformed as follows:
(35)G1(0)=g1(0)=0G1(1)=Hg1′(0)=0G1(2)=H22g1′′(0)=H22S1(0)=s1(0)=(g1′′(0))2-n=1.
The boundary conditions of each sub domain are the continuity of theg(ηi),g′(ηi), andg′′(ηi). These boundary conditions can be obtained from (7):(36)gi(ηi+1)=∑k=0mGi(k)gi+1(ηi+1)=Gi+1(0)⟶Gi+1(0)=∑k=0mGi(k)(37)gi′(ηi+1)=∑k=1mkHGi(k)gi+1′(ηi+1)=Gi+1(1)H⟶Gi+1(1)=∑k=1mkGi(k)gi′′(ηi+1)=∑k=2mk(k-1)H2Gi(k)gi+1′′(ηi+1)=2H2Gi+1(2)⟶Gi+1(2)=12∑k=2mk(k-1)Gi(k).
And we have forS(38)Si+1(0)=si+1(ηi+1)=(gi+1′′(ηi+1))2-n=(∑k=2mk(k-1)H2Gi(k))2-n.
The value of theg′(η∞)can be calculated by differentiating from (7):
(39)g′(∞)≃g′(η∞)=gN′(ηN+1)=∑k=1mkHGN(k).
The unknown parameter (λ) is calculated from (29). Now, (18) is solved with a similar process like (25) using multi-step DTM. The only difference is that the conditionf′′(0)=λis replaced by the conditiong′′(0)=1.In Figure1  f′(η)is illustrated for different values ofnand compared with the numerical solution and results of Lemieux et al. [43]. The numerical solution of the problem is done by the MATLAB software. Figures 2 and 3 show the variation off(η)andf′′(η), respectively. The most important step of this scheme is to choose the appreciate finite value ofη∞. Thus to estimate this value, we start with an initial guess forη∞and solve the problem. The solution process is repeated with another larger value ofη∞until the values off′′(0)andg′′(0)differ regarding the desired accuracy.Figure 1
Variation off′(η)and comparison with the numerical solution and results of Lemieux et al. [43].Figure 2
Variation off(η)for different values ofn.Figure 3
Variation off′′(η)for different values ofn.In Table2 the values of thef′′(0)=λare compared with the numerical solution and results of Lemieux et al. [43]. The approximate solution of the problem is presented in Table 3 for a special case (n=0.7).Table 2
Comparison of the DTM results, numerical solution, and results of Lemieux et al. [43].
DTM
Numerical
Lemieux et al. [43]
f
′
′
(
0
)
f
′
′
(
0
)
f
′
′
(
0
)
n
=
0.1
0.1091
0.1140
0.1085
n
=
0.2
0.1472
0.1493
0.1470
n
=
0.3
0.1881
0.1889
0.1882
n
=
0.4
0.2306
0.2308
0.2307
n
=
0.5
0.2734
0.2734
0.2734
n
=
0.6
0.3156
0.3157
0.3156
n
=
0.7
0.3567
0.3567
0.3567
n
=
0.8
0.3962
0.3962
0.3962
n
=
0.9
0.4339
0.4339
0.4339
n
=
1
0.4696
0.4696
0.4696Table 3
The solution off(η) for n=0.7 when η∞=10 and N=10.
Approximate solution
0
≤
η
<
1
f
(
η
)
=
0.1793
η
2
-
0.0008
η
5
+
0.0000
η
8
+
⋯,
1
≤
η
<
2
f
(
η
)
=
0.1784
+
0.3546
(
η
-
1
)
+
0.1716
(
η
-
1
)
2
-
0.0074
(
η
-
1
)
3
-
0.0033
(
η
-
1
)
4
+
⋯,
2
≤
η
<
3
f
(
η
)
=
0.6937
+
0.6617
(
η
-
2
)
+
0.1290
(
η
-
2
)
2
-
0.0197
(
η
-
2
)
3
-
0.0018
(
η
-
2
)
4
+
⋯,
3
≤
η
<
4
f
(
η
)
=
1.4637
+
0.8577
(
η
-
3
)
+
0.0679
(
η
-
3
)
2
-
0.0182
(
η
-
3
)
3
-
0.0021
(
η
-
3
)
4
+
⋯,
4
≤
η
<
5
f
(
η
)
=
2.3733
+
0.9481
(
η
-
4
)
+
0.0272
(
η
-
4
)
2
-
0.0090
(
η
-
4
)
3
-
0.0020
(
η
-
4
)
4
+
⋯,
5
≤
η
<
6
f
(
η
)
=
3.3416
+
0.9824
(
η
-
5
)
+
0.0098
(
η
-
5
)
2
-
0.0033
(
η
-
5
)
3
+
0.0009
(
η
-
5
)
4
+
⋯,
6
≤
η
<
7
f
(
η
)
=
4.3312
+
0.9948
(
η
-
6
)
+
0.0036
(
η
-
6
)
2
-
0.0012
(
η
-
6
)
3
+
0.0003
(
η
-
6
)
4
+
⋯,
7
≤
η
<
8
f
(
η
)
=
5.3286
+
0.9993
(
η
-
7
)
+
0.0014
(
η
-
7
)
2
-
0.0004
(
η
-
7
)
3
+
0.0001
(
η
-
7
)
4
+
⋯,
8
≤
η
<
9
f
(
η
)
=
6.2390
+
1.0012
(
η
-
8
)
+
0.0006
(
η
-
8
)
2
-
0.0002
(
η
-
8
)
3
+
0.0000
(
η
-
8
)
4
+
⋯,
9
≤
η
≤
10
f
(
η
)
=
7.3307
+
1.0020
(
η
-
9
)
+
0.0003
(
η
-
9
)
2
-
0.0001
(
η
-
9
)
3
+
0.0000
(
η
-
9
)
4
+
⋯.


## 4.1. Applying DTM (Obtainingf′′(0))

The BVP (17) can be transformed to an initial value problem with the replacement of the unknown initial conditionf′′(0)=λinstead of the boundary condition at infinityf′(∞)=1:
(18)f′′′(η)+f(η)s(η)=0forn≤1f(0)=0,f′(0)=0,f′′(0)=λ,
where
(19)s(η)=(f′′(η))2-n.
By applying the DTM on (18) atη=0, the recursive relation is derived:
(20)F(k+3)=-1(k+1)(k+2)(k+3)∑r=0kF(r)S(k-r)F(0)=0,F(1)=0,F(2)=λ2.
Regarding (19),
(21)s′(η)f′′(η)=(2-n)f′′′(η)s(η).
Taking differential transform on (21), we have
(22)∑r=0k(r+1)S(r+1)(k-r+1)(k-r+2)F(k-r+2)=(2-n)∑r=0k(r+1)(r+2)(r+3)F(r+3)S(k-r).
From (19), the value ofS(0)can be obtained:
(23)S(0)=s(0)=(f′′(0))2-n=λ2-n.
Combining (20), (22), and (23), the coefficients of the Taylor series solution of (18) can be calculated:
(24)F(0)=0,F(1)=0,F(2)=λ2,F(3)=0,F(4)=0,F(5)=-1120λ3-n,F(6)=0,F(7)=0,F(8)=11920λ5-2n-14032λ5-2nn,F(9)=0,F(10)=0,F(11)=-891900800λ7-3n+102719958400λ7-3nn-171280λ7-3nn2,⋮
Now, consider a similar equation (18) in whichλ=1:
(25)g′′′(η)+g(η)(g′′(η))2-n=0forn≤1g(0)=0,g′(0)=0,g′′(0)=1.
In the same manner, the coefficientsG(k)are
(26)G(0)=0,G(1)=0,G(2)=12,G(3)=0,G(4)=0,G(5)=-1120,G(6)=0,G(7)=0,G(8)=11920-14032n,G(9)=0,G(10)=0,G(11)=-891900800+102719958400n-171280n2,⋮
Substituting (24) and (26) in (2), thef(η)andg(η)are obtained. These functions have a relation as follows:
(27)f(η)=λqg(λpη),p=2-n3,q=2n-13.
Forn=1, both values ofpandqare 1/3 for the Blasius equation. From (27), we have
(28)f′(η)=λq+pg′(λpη)⟶f′(∞)=λq+pg′(∞).
Regarding (28) and the boundary condition at infinity in (17), the value ofλcan be obtained:
(29)λ=(1g′(∞))3/(1+n).


## 4.2. Applying Multi-Step DTM (Solution of IVPs)

As was noted, multi-step DTM usually is used for solving problems in which Taylor series is diverged in the solution domain. This technique was already used to solve engineering and computational problems [49–57].To solve boundary layer problems, the domain[0,∞)is replaced by[0,η∞). Butη∞should be great enough that the solution is not to be dependent on. The solution domain should be divided toNequal parts (H=η∞/N). So, we have
(30)gi′′′(ηi)+gi(ηi)si(ηi)=0,(i-1)H≤ηi<iH,for1≤i≤N,
where
(31)si(η)=(gi′′(η))2-n.
Applying multi-step DTM on (30) and (31), we have
(32)Gi(k+3)=-H3(k+1)(k+2)(k+3)∑r=0kGi(r)Si(k-r),for1≤i≤N∑r=0k(r+1)HSi(r+1)(k-r+1)(k-r+2)H2Gi(k-r+2)=(2-n)∑r=0k(r+1)(r+2)(r+3)H3Gi(r+3)Si(k-r),for1≤i≤N.
So
(33)Si(1)=2-nGi(2)[3Gi(3)Si(0)].
And fork≥1(34)Si(k+1)=12(k+1)Gi(2)[∑r=0k-1-∑r=0k-1(r+1)Si(r+1)(k-r+1)×(k-r+2)Gi(k-r+2)+(2-n)∑r=0k(r+1)(r+2)(r+3)Gi×(r+3)Si(k-r)∑r=0k-1].
The initial conditions of the problem are considered for first sub domain (i=1). Regarding (6), (25), and (31), initial conditions transformed as follows:
(35)G1(0)=g1(0)=0G1(1)=Hg1′(0)=0G1(2)=H22g1′′(0)=H22S1(0)=s1(0)=(g1′′(0))2-n=1.
The boundary conditions of each sub domain are the continuity of theg(ηi),g′(ηi), andg′′(ηi). These boundary conditions can be obtained from (7):(36)gi(ηi+1)=∑k=0mGi(k)gi+1(ηi+1)=Gi+1(0)⟶Gi+1(0)=∑k=0mGi(k)(37)gi′(ηi+1)=∑k=1mkHGi(k)gi+1′(ηi+1)=Gi+1(1)H⟶Gi+1(1)=∑k=1mkGi(k)gi′′(ηi+1)=∑k=2mk(k-1)H2Gi(k)gi+1′′(ηi+1)=2H2Gi+1(2)⟶Gi+1(2)=12∑k=2mk(k-1)Gi(k).
And we have forS(38)Si+1(0)=si+1(ηi+1)=(gi+1′′(ηi+1))2-n=(∑k=2mk(k-1)H2Gi(k))2-n.
The value of theg′(η∞)can be calculated by differentiating from (7):
(39)g′(∞)≃g′(η∞)=gN′(ηN+1)=∑k=1mkHGN(k).
The unknown parameter (λ) is calculated from (29). Now, (18) is solved with a similar process like (25) using multi-step DTM. The only difference is that the conditionf′′(0)=λis replaced by the conditiong′′(0)=1.In Figure1  f′(η)is illustrated for different values ofnand compared with the numerical solution and results of Lemieux et al. [43]. The numerical solution of the problem is done by the MATLAB software. Figures 2 and 3 show the variation off(η)andf′′(η), respectively. The most important step of this scheme is to choose the appreciate finite value ofη∞. Thus to estimate this value, we start with an initial guess forη∞and solve the problem. The solution process is repeated with another larger value ofη∞until the values off′′(0)andg′′(0)differ regarding the desired accuracy.Figure 1
Variation off′(η)and comparison with the numerical solution and results of Lemieux et al. [43].Figure 2
Variation off(η)for different values ofn.Figure 3
Variation off′′(η)for different values ofn.In Table2 the values of thef′′(0)=λare compared with the numerical solution and results of Lemieux et al. [43]. The approximate solution of the problem is presented in Table 3 for a special case (n=0.7).Table 2
Comparison of the DTM results, numerical solution, and results of Lemieux et al. [43].
DTM
Numerical
Lemieux et al. [43]
f
′
′
(
0
)
f
′
′
(
0
)
f
′
′
(
0
)
n
=
0.1
0.1091
0.1140
0.1085
n
=
0.2
0.1472
0.1493
0.1470
n
=
0.3
0.1881
0.1889
0.1882
n
=
0.4
0.2306
0.2308
0.2307
n
=
0.5
0.2734
0.2734
0.2734
n
=
0.6
0.3156
0.3157
0.3156
n
=
0.7
0.3567
0.3567
0.3567
n
=
0.8
0.3962
0.3962
0.3962
n
=
0.9
0.4339
0.4339
0.4339
n
=
1
0.4696
0.4696
0.4696Table 3
The solution off(η) for n=0.7 when η∞=10 and N=10.
Approximate solution
0
≤
η
<
1
f
(
η
)
=
0.1793
η
2
-
0.0008
η
5
+
0.0000
η
8
+
⋯,
1
≤
η
<
2
f
(
η
)
=
0.1784
+
0.3546
(
η
-
1
)
+
0.1716
(
η
-
1
)
2
-
0.0074
(
η
-
1
)
3
-
0.0033
(
η
-
1
)
4
+
⋯,
2
≤
η
<
3
f
(
η
)
=
0.6937
+
0.6617
(
η
-
2
)
+
0.1290
(
η
-
2
)
2
-
0.0197
(
η
-
2
)
3
-
0.0018
(
η
-
2
)
4
+
⋯,
3
≤
η
<
4
f
(
η
)
=
1.4637
+
0.8577
(
η
-
3
)
+
0.0679
(
η
-
3
)
2
-
0.0182
(
η
-
3
)
3
-
0.0021
(
η
-
3
)
4
+
⋯,
4
≤
η
<
5
f
(
η
)
=
2.3733
+
0.9481
(
η
-
4
)
+
0.0272
(
η
-
4
)
2
-
0.0090
(
η
-
4
)
3
-
0.0020
(
η
-
4
)
4
+
⋯,
5
≤
η
<
6
f
(
η
)
=
3.3416
+
0.9824
(
η
-
5
)
+
0.0098
(
η
-
5
)
2
-
0.0033
(
η
-
5
)
3
+
0.0009
(
η
-
5
)
4
+
⋯,
6
≤
η
<
7
f
(
η
)
=
4.3312
+
0.9948
(
η
-
6
)
+
0.0036
(
η
-
6
)
2
-
0.0012
(
η
-
6
)
3
+
0.0003
(
η
-
6
)
4
+
⋯,
7
≤
η
<
8
f
(
η
)
=
5.3286
+
0.9993
(
η
-
7
)
+
0.0014
(
η
-
7
)
2
-
0.0004
(
η
-
7
)
3
+
0.0001
(
η
-
7
)
4
+
⋯,
8
≤
η
<
9
f
(
η
)
=
6.2390
+
1.0012
(
η
-
8
)
+
0.0006
(
η
-
8
)
2
-
0.0002
(
η
-
8
)
3
+
0.0000
(
η
-
8
)
4
+
⋯,
9
≤
η
≤
10
f
(
η
)
=
7.3307
+
1.0020
(
η
-
9
)
+
0.0003
(
η
-
9
)
2
-
0.0001
(
η
-
9
)
3
+
0.0000
(
η
-
9
)
4
+
⋯.


## 5. Conclusion

In this paper, the boundary layer equation of the pseudoplastic fluid over a flat plate was extracted from the boundary layer theory as presented in [43]. This equation is a boundary value problem with high nonlinearity and boundary condition at infinity. Using DTM, the BVP transformed to a pair of initial value problems and multi-step DTM is used to solve them.Using the method, the differential equation and boundary conditions are transformed into a recurrence set of equations. Finally, the coefficients of power series are obtained based on the solution of this set of equations. The results validated with the numerical solution, and a good accuracy is observed. The proposed method overcame nonlinearity without using linearization or restrictive assumptions.



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*Source: 685454-2013-11-26.xml*