# New Approximate Analytical Solutions of the Falkner-Skan Equation

Authors: Beong In Yun
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We propose an iterative method for solving the Falkner-Skan equation. The method provides approximate analytical solutions which consist of coefficients of the previous iterate solution. By some examples, we show that the presented method with a small number of iterations is competitive with the existing method such as Adomian decomposition method. Furthermore, to improve the accuracy of the proposed method, we suggest an efficient correction method. In practice, for some examples one can observe that the correction method results in highly improved approximate solutions.


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## Body

## 1. Introduction

We consider the well-known Falkner-Skan equation(1.1)y′′′(x)+y(x)y′′(x)+β{1-y′(x)2}=0,0≤x<∞,
subjected to boundary conditions
(1.2)y(0)=y′(0)=0,y′(∞)=1.
This boundary value problem arises in the research of viscous flow past a wedge of angle βπ. β>0 corresponds to flow toward the wedge and β<0 does to flow away from the wedge. The special case, β=0, is called the Blasius equation where the wedge reduces to a flat plate. It is well known that a unique smooth solution of the Falkner-Skan equation (1.1) with the condition (1.2) exists for 0≤β≤1 [1, 2]. For -0.1988<β<0, there exists two solutions, that is, one with y′′(0)>0 and the other one with y′′(0)<0, as shown in the literature [3]. For β>1 the solution is unique under the restriction 0<y′(x)<1 [4, 5].In order to obtain approximate solutions of the Falkner-Skan equation, analytical methods such as Adomian decomposition methods [6–10], variational iterative methods [11–15], and homotopy analysis methods [16–18] can be referred. The objective of this work is to present a new analytical method which provides a simple form of iterate solutions and can be a match for existing methods in accuracy.In the next section we develop an iterative method based on a decomposition of the Falkner-Skan equation (1.1) which was recently introduced in [19] for a particular case, β=0. In the result we derive an iterative formula producing approximate analytical solutions in the form of polynomial series without requiring any differentiations or integrations of the previous iterate solutions. The degree of the nth iterate solution increases very rapidly, in fact, like O(2n) as n→∞. In Section 3, for some values of β we compare the presented solutions with the existing iterate solutions. In Section 4, to improve the accuracy of the presented solutions we suggest a correction method which is composed of the successive differences of the iterations.


## 2. Iteration for Approximate Analytical Solutions

To develop a new iterative method generating analytical series solutions, we consider the following one-point boundary conditions instead of the boundary conditions in (1.2). 						(2.1)y(0)=y′(0)=0,y′′(0)=α,
where the curvature α of the solution is assumed to be known. Actually, the value of α can be obtained by numerical evaluation [8, 20, 21]. In this section we derive iterate solutions based on the method which was introduced in [19] for a special case, β=0.First, fory=y(x) the Falkner-Skan equation (1.1) becomes
(2.2)(y′′+yy′)′=(1+β)(y′)2-β.
From the boundary conditions in (2.1), it follows that
(2.3)y′′+yy′=(1+β)∫0xy′(t)2dt-βx+α∶=A(x),
and thus
(2.4)y′=∫0xA(t)dt-12y2∶=B(x).
In the result we have
(2.5)y(x)=∫0xB(t)dt.Denoting byyn(x) the nth iterate solution and substituting it into the right-hand side of (2.5), with a given initial solution y0(x), we have an iteration formula
(2.6)yn+1(x)=∫0xBn(t)dt,n≥0,
where
(2.7)Bn(x)=∫0xAn(t)dt-12yn(x)2,An(x)=(1+β)∫0xyn′(t)2dt-βx+α.Letδn denote a degree of the nth iterate solution yn(x). Then, from the formulas (2.6) and (2.7), it follows that δn+1=2δn+1 for all n≥0. Solving this recurrence equation, we have
(2.8)δn={4⋅2n-1,whenβ≠0,3⋅2n-1,whenβ=0.For an initial solutiony0(x) satisfying the conditions in (2.1), we may set the nth iterate solution as
(2.9)yn(x)=∑k=0δn-2‍an,kxk+2,
for any integer n≥0. Then, by performing integrations in (2.6) and (2.7) directly, we have
(2.10)An(x)=(1+β)∫0x(∑k=0δn-2‍(k+2)an,ktk+1)2dt-βx+α=α-βx+(1+β)∑k=0δn-2∑j=0δn-2(k+2)(j+2)an,kan,jk+j+3xk+j+3,yn+1′(x)=Bn(x)=αx-β2x2+∑k=0δn-2∑j=0δn-2{(1+β)(k+2)(j+2)(k+j+3)(k+j+4)-12}an,kan,jxk+j+4.
Thus, by integrating Bn(x), we have
(2.11)yn+1(x)=αx22-βx36+∑k=0δn-2∑j=0δn-2{(1+β)(k+2)(j+2)(k+j+3)(k+j+4)-12}an,kan,jk+j+5xk+j+5.On the other hand, referring to the boundary conditions in (2.1) and (1.1), we may take an initial solution as
(2.12)y0(x)=α2x2-β6x3.


## 3. Examples

### 3.1. A Case ofβ=0 (Blasius Problem)

A case ofβ=0 is called the Blasius problem, and a well-known power series for the solution is
(3.1)S(x)=∑k=0∞(-1)kckαk+1(3k+2)!x3k+2,x∼0,
where the coefficients ck can be computed from the recurrence [1]:
(3.2)ck={1,k=0,1,∑j=0k-1‍(3k-13j)cjck-j-1,k≥2.
In fact, α≈0.46959999 and the series becomes
(3.3)S(x)=α2x2-α2120x5+11α340320x8-73α47983360x11+111α5358758400x14+⋯.
The Blasius series, however, converges only for |x|<ρ≈4.02346 [16, 22, 23]. In this paper we denote by Sn(x) a truncated Blasius series to the nth term of S(x).For the Blasius problem the first four iterate solutions generated by the formula (2.11), with y0(x)=αx2/2, are as follows:
(3.4)y1(x)=αx22-α2x5120,y2(x)=αx22-α2x5120+11α3x840320-α4x11712800,y3(x)=αx22-α2x5120+11α3x840320-5α4x11532224+10033α5x1487178291200-5449α6x173908653056000+83α7x208935557120000-α8x2349080898944000,y4(x)=αx22-α2x5120+11α3x840320-5α4x11532224+9299α5x1429059430400-2173649α6x17355687428096000+13722337α7x20115852476579840000-27184438601α8x2312926008369442488320000+12320831753849α9x26403291461126605635584000000+⋯-α16x47463172433275878342410240000000.
It should be noted that the degree of the presented solution yn(x) increases like O(2n) as shown in (2.8) while that of the truncated Blasius series Sn(x) is 3n-1, or it increases like O(n) as n→∞.For comparison betweenyn(x) and Sn(x), Figure 1 includes graphs of the approximations to the numerical solution y*(x), which is taken for an exact solution, and their errors over an interval [0,ρ0]. We set ρ0=4.02346 which is close to the radius of convergence, ρ of the Blasius series solution mentioned above. Figure 1 shows that the presented solutions yn(x) approximate the exact solution better than the truncated Blasius series Sn(x).Truncated Blasius seriesSn(x) and their errors in (a) and presented solutions yn(x) and their errors in (b), n=2,3,4,5,6, for the Blasius equation (β=0).
(a)
S
n
(
x
)
(b)
y
n
(
x
)


### 3.2. A Case ofβ≠0

We refer to another analytical solution obtained by the Adomian decomposition method as follows:(3.5)ynA(x)=y0(x)+∑k=1nuk(x),n=1,2,…,
with
(3.6)uk+1(x)=-L-1(Ak(x)),k=0,1,2,…,u0(x)=y0(x), and Adomian polynomial 𝒜k(x) generated by the recursive formula [8]:(3.7)Ak(x)=1k!dkdλk[N(∑j=0kλjuj(x))]λ=0,
where L-1 is an inverse operator of L=d3/dx3 and N is a nonlinear operator defined as Nz(x)=z(x)z′′(x)-βz′(x)2.Presented solutionsyn(x), n=2,3,4,5,6, given in (2.11) and their errors are compared with those of Adomian's solutions ynA(x) for the Falkner-Skan equation with β=0.5 and β=1 in Figures 2 and 3, respectively. We chose y0(x) in (2.12) as an initial solution and took α=y′′(0)=0.9277 for β=0.5 and α=1.2326 for β=1, as given in [1]. One can see that the presented solutions are competitive with the Adomian solutions. Moreover, it should be noticed that unlike the Adomian solutions, the presented solutions do not require any integration or differentiation.The Adomian solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=0.5.
(a)
ynA(x)
(b)
y
n
(
x
)The Adomian solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=1.
(a)
ynA(x)
(b)
y
n
(
x
)For the so-called decelerated flow (orβ<0), it is well known that a unique solution with α>0 exists when -0.1988<β<0 [4]. So we take an example of β=-0.14 with α=0.2365 as given in [8]. Comparison between the presented solutions yn(x) and the Adomian solutions ynA(x) for this case is given in Figure 4. Therein we can find superiority of the presented solutions in accuracy.Adomian's solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=-0.14.
(a)
ynA(x)
(b)
y
n
(
x
)


## 3.1. A Case ofβ=0 (Blasius Problem)

A case ofβ=0 is called the Blasius problem, and a well-known power series for the solution is
(3.1)S(x)=∑k=0∞(-1)kckαk+1(3k+2)!x3k+2,x∼0,
where the coefficients ck can be computed from the recurrence [1]:
(3.2)ck={1,k=0,1,∑j=0k-1‍(3k-13j)cjck-j-1,k≥2.
In fact, α≈0.46959999 and the series becomes
(3.3)S(x)=α2x2-α2120x5+11α340320x8-73α47983360x11+111α5358758400x14+⋯.
The Blasius series, however, converges only for |x|<ρ≈4.02346 [16, 22, 23]. In this paper we denote by Sn(x) a truncated Blasius series to the nth term of S(x).For the Blasius problem the first four iterate solutions generated by the formula (2.11), with y0(x)=αx2/2, are as follows:
(3.4)y1(x)=αx22-α2x5120,y2(x)=αx22-α2x5120+11α3x840320-α4x11712800,y3(x)=αx22-α2x5120+11α3x840320-5α4x11532224+10033α5x1487178291200-5449α6x173908653056000+83α7x208935557120000-α8x2349080898944000,y4(x)=αx22-α2x5120+11α3x840320-5α4x11532224+9299α5x1429059430400-2173649α6x17355687428096000+13722337α7x20115852476579840000-27184438601α8x2312926008369442488320000+12320831753849α9x26403291461126605635584000000+⋯-α16x47463172433275878342410240000000.
It should be noted that the degree of the presented solution yn(x) increases like O(2n) as shown in (2.8) while that of the truncated Blasius series Sn(x) is 3n-1, or it increases like O(n) as n→∞.For comparison betweenyn(x) and Sn(x), Figure 1 includes graphs of the approximations to the numerical solution y*(x), which is taken for an exact solution, and their errors over an interval [0,ρ0]. We set ρ0=4.02346 which is close to the radius of convergence, ρ of the Blasius series solution mentioned above. Figure 1 shows that the presented solutions yn(x) approximate the exact solution better than the truncated Blasius series Sn(x).Truncated Blasius seriesSn(x) and their errors in (a) and presented solutions yn(x) and their errors in (b), n=2,3,4,5,6, for the Blasius equation (β=0).
(a)
S
n
(
x
)
(b)
y
n
(
x
)


## 3.2. A Case ofβ≠0

We refer to another analytical solution obtained by the Adomian decomposition method as follows:(3.5)ynA(x)=y0(x)+∑k=1nuk(x),n=1,2,…,
with
(3.6)uk+1(x)=-L-1(Ak(x)),k=0,1,2,…,u0(x)=y0(x), and Adomian polynomial 𝒜k(x) generated by the recursive formula [8]:(3.7)Ak(x)=1k!dkdλk[N(∑j=0kλjuj(x))]λ=0,
where L-1 is an inverse operator of L=d3/dx3 and N is a nonlinear operator defined as Nz(x)=z(x)z′′(x)-βz′(x)2.Presented solutionsyn(x), n=2,3,4,5,6, given in (2.11) and their errors are compared with those of Adomian's solutions ynA(x) for the Falkner-Skan equation with β=0.5 and β=1 in Figures 2 and 3, respectively. We chose y0(x) in (2.12) as an initial solution and took α=y′′(0)=0.9277 for β=0.5 and α=1.2326 for β=1, as given in [1]. One can see that the presented solutions are competitive with the Adomian solutions. Moreover, it should be noticed that unlike the Adomian solutions, the presented solutions do not require any integration or differentiation.The Adomian solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=0.5.
(a)
ynA(x)
(b)
y
n
(
x
)The Adomian solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=1.
(a)
ynA(x)
(b)
y
n
(
x
)For the so-called decelerated flow (orβ<0), it is well known that a unique solution with α>0 exists when -0.1988<β<0 [4]. So we take an example of β=-0.14 with α=0.2365 as given in [8]. Comparison between the presented solutions yn(x) and the Adomian solutions ynA(x) for this case is given in Figure 4. Therein we can find superiority of the presented solutions in accuracy.Adomian's solutionsynA(x) and their errors in (a) and presented solutions yn(x) and their errors in (b) for the Falkner-Skan equation with β=-0.14.
(a)
ynA(x)
(b)
y
n
(
x
)


## 4. Improvement of the Approximate Solution

In Figure1, for the case of β=0, one can see that both the proposed solutions yn(x) and the truncated Blasius series Sn(x) overshoot and undershoot, alternately, as n increases. This tendency, in the proposed solutions, is continued to the cases of other values of β as can be observed in Figures 2–4. Considering this feature, we propose a correction method as follows.Letyj=yj(x), j=n-1,n,n+1, be three successive iterate solutions obtained by the formula (2.11). To improve the accuracy of the iterate solutions, we suggest two corrections of the form
(4.1)y^n+1=θnyn+(1-θn)yn+1,y^n=θnyn-1+(1-θn)yn
for a weight function θn=θn(x) such that 0≤θn(x)<1. Setting a constraint y^n+1=y^n, we can determine θn as
(4.2)θn=yn-yn+12yn-yn-1-yn+1=1-11+rn,
where rn=rn(x) is a ratio of the successive differences defined as
(4.3)rn=-yn+1-ynyn-yn-1.
We may surmise that rn(x)≥0 because, as mentioned above, the proposed solutions yn(x) alternately overshoot and undershoot as n increases. In practice, for the Blasius problem (β=0), Figure 5(a) depicts the graphs of rn(x) on the interval [0,ρ0] for n=3,4,5,6, which shows rn(x)≥0 for all x≥0. Thus from (4.2) we have 0≤θn(x)<1 so that the correction y^n(x) or y^n+1(x) in (4.1) with the weight θn(x) in (4.2) is reasonable. The graph of θn(x) is given in Figure 5(b).Graphs of the ratiosrn(x) and the weight functions θn(x), n=3,4,5,6 (β=0).
(a)
rn(x)
(b)
θ
n
(
x
)Forβ=0, for example, Figure 6 shows the approximations of y^n(x), n=2,4,6, and their errors on an extended interval [0,ρ0+1]. Furthermore, Table 1 includes L2-norm errors of Sn+1(x), yn+1(x), and y^n(x) with respect to the numerical solution y*(x) on the interval [0,ρ0]. The L2-norm error E(y) is defined as
(4.4)E(y)=(∫0ρ0(y(x)-y*(x))2dx)1/2.
From Figure 6 and Table 1, compared with the results of yn+1(x) in Figure 1, we can see that the correction y^n(x) highly improves the accuracy of the original iterate solution yn+1(x).Table 1
Comparison of theL2-norm errors of the truncated Blasius series Sn+1(x), the presented solution yn+1(x), and its correction y^n(x) for the Blasius problem (β=0).
n
E
(
S
n
+
1
)
E
(
y
n
+
1
)
E
(
y
^
n
)
2
3
.
9
×
1
0
-
1
2
.
0
×
1
0
-
1
2
.
4
×
1
0
-
3
3
3
.
8
×
1
0
-
1
1
.
1
×
1
0
-
1
9
.
5
×
1
0
-
3
4
3
.
4
×
1
0
-
1
5
.
6
×
1
0
-
2
1
.
9
×
1
0
-
3
5
3
.
3
×
1
0
-
1
2
.
5
×
1
0
-
2
1
.
7
×
1
0
-
3
6
3
.
0
×
1
0
-
1
1
.
0
×
1
0
-
2
4
.
6
×
1
0
-
4Figure 6
Corrected solutionsy^n(x) of the presented solutions yn(x) and their errors, n=2,4,6 (β=0).In addition, absolute errors ofyn+1(x) and y^n(x), n=2,6, for each β=0.5,1,-0.14 are depicted in Figure 7. Similar to Table 1, Table 2 includes L2-norm errors of  yn+1A(x),  yn+1(x),  and y^n(x) on the interval [0,ρ0]. We can also observe that the correction y^n(x) highly improves the accuracy of the original iterate solution yn+1(x).Table 2
Comparison of theL2-norm errors of the Adomian solution yn+1A(x), the presented solution yn+1(x), and its correction y^n(x) for various values of β.
β
n
E
(
y
n
+
1
A
)
E
(
y
n
+
1
)
E
(
y
^
)
2
9
.
9
×
1
0
-
2
1
.
6
×
1
0
-
1
5
.
7
×
1
0
-
3
0.5
4
2
.
1
×
1
0
-
3
5
.
1
×
1
0
-
2
2
.
4
×
1
0
-
3
6
5
.
4
×
1
0
-
3
1
.
2
×
1
0
-
2
5
.
1
×
1
0
-
4
2
3
.
1
×
1
0
-
1
2
.
0
×
1
0
-
1
4
.
6
×
1
0
-
3
1.0
4
9
.
3
×
1
0
-
3
5
.
5
×
1
0
-
2
1
.
5
×
1
0
-
3
6
5
.
1
×
1
0
-
3
1
.
2
×
1
0
-
2
4
.
1
×
1
0
-
4
2
2
.
8
×
1
0
-
1
1
.
2
×
1
0
-
1
2
.
0
×
1
0
-
3
-
0.14
4
2
.
0
×
1
0
-
1
2
.
0
×
1
0
-
2
9
.
7
×
1
0
-
4
6
1
.
5
×
1
0
-
1
2
.
3
×
1
0
-
3
1
.
3
×
1
0
-
4Absolute errors of the presented solutionsyn+1(x) (thin lines) and those of the corrected solutions y^n(x) (thick lines) for β=0.5,1,-0.14.
(a)
n=2
(b)
n
=
6


## 5. Conclusion

We have developed a new iterative method generating analytical solutions to the Falkner-Skan equation (1.1) with the boundary conditions (2.1). In practice, for several cases of β, it was shown that the presented method with a small number of iterations is available and efficient, compared with a well-known existing method such as the Adomian decomposition method. Moreover, we have proposed a simple correction method which improves the accuracy and the rate of convergence of the presented method.Although in this paper we considered-0.1988<β≤1 with the boundary conditions y(0)=y′(0)=0, the proposed method may be extended to the general case of β with boundary conditions, y(0)=γ and y′(0)=λ, as long as the existence of the solution is assured under some additional restrictions. In addition, we leave theoretical analysis of the error and the radius of convergence for a future work.



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*Source: 170802-2012-07-10.xml*