CC BY
36Electronic Journal «Technical Acoustics» http://www .ejta.org
2007, 21
W. A. Albarakati
Department of Mathematics, Faculty of Science, King Abdulaziz University PO Box 15905, Jeddah 21454, Saudi Arabia, wbarakati@kau.edu.sa
An analytic solution of the stagnation point flow problem
Received 15.10.2007, published 23.11.2007
The Falkner-Skan equation models the laminar flow of an incompressible fluid for several physical situations. A specially interesting case is that of a flow with a stagnation point. This problem is solved analytically in the form of a power series with a finite radius of convergence. By modifying a Pade approximant corresponding to the power series a simple expression is obtained which describes the solution uniformly over the whole domain [0,i).
Keywords: Falkner-Skan equation, stagnation point flow, Pade approximation, Wang transformation
INTRODUCTION
The Falkner-Skan problem
f "(n) + Pof (n)f" (n) + f[1 - [f"(n)]2 ] = 0, n - 0, (1.1a)
f (0) = ° f' (0) = 0,ninif'n) =1 (1.1b)
describes the laminar flow of an incompressible fluid under a variety of circumstances.
When f0 = 1, the above problem corresponds to decelerating, constant and accelerating
flows for respectively f < 0, f = 0 and f > 1. If f0 = j and f = 0, we get the Blasius problem
f(n) + 2 f (n)f " (n) = ° n- 0 (1.2)
with the boundary conditions (1.1b).
If f0 = 1 and p = 1 then the equation
f"(n)+f(n)f" n) +1 - [f ,(n)]2 = 0, n- 0 (13)
with boundary conditions (1.1b) represents the stagnation point flow [1].
The above problems are two-point nonlinear boundary value problems and it is difficult to find analytical solutions which are valid outside a finite interval. A numerical solution of (1.2) was first obtained by Howarth [2] and the Falkner-Skan problem was first solved numerically
by Hartree [3]. Subsequently several authors have devised numerical techniques to efficiently deal with the two problems, see Asaithambi [4, 5] and references therein.
Blasius [6] found an analytical solution of (1.2) in the form of a power series, however the series converges only for 0 < n < 0.5690. Liao [7], by using his homotopy analysis method, found an analytical solution valid on the whole domain [0,o). A power series solution of
(1.3) may be found by using, for example, Adomian decomposition method, but the series converges only in [0,n0) where n0 ~3.2 [8]. To our knowledge no analytical solution of
(1.3) exists which represents the function on the whole domain [0, o). In this paper we shall present a simple expression which fits the function f'(n) on [0, o) and a simple quadrature gives f (n) on the whole domain with remarkable accuracy.
1. ANALYTICAL SOLUTION OF FALKNER-SKAN EQUATION
Asghar [8] found a power series solution by using the Adomian decomposition method. Here we find this solution by a direct method which produces the coefficients in the series with less effort.
Denote the unknown f "(0) by a and let a solution of (1.3) be f (n) = 2 anl". (2.1)
n=0
Substitution of f, f", fin (1.3) yields
n . X'™' X'™' z' . I 1\~ n+k
2 (n + 3)(n + 2)(n + 1)an+3nn +ZZ (n + 2)(n + 1)an+2akin+k + 1
n=0 k=0 n=0
- 22(k+1)(n+1)ak+1an+1nn+k =0.
k=0 n=0
(2.2)
Consider the second summation in (2.2) i.e.
22(n + 2)(n + 1)an+2 ann+k (2.3)
k=0 n=0
and define a new index m = n + k, hence n = m - k , which indicates 0 < n < m . With the
new indices n and m the summation in (2.3) becomes
o m
22(n + 2)(n + 1)a«+2am-nnm . (2.4)
m=0 n=0
Similarly the last summation in (2.2) can be replaced by
o m
22 (m - n +1)(n + 1)am-„+1a„+1nm . (2.5)
m=0 n=0
Thus (2.2) can be written as
2 (m + 3)(m + 2)(m + 1)am+3nm +22 (n + 2)(n + 1)an+2am-nim + 1
m=0 m=0 n=0
o m
- 22 (m - n + 1)(n + 1)am-„+1a„+1nm = 0 .
(2.6)
From the boundary conditions (1.1b) we have a0 = 0, aj = 0. (2.7)
While the assumption f "(0) = a implies a
a2 = -. (2.8)
By equating the constant term in (2.6) to zero we get
a3 =- ^. (29)
Higher coefficients am, m > 1 can be found from the following recurrence relation which
we get by setting the coefficient of nm in (2.6) to zero
am+3 =- m , 3Wm1. ____ [2(n + 2)(n + 1)a«+2a»-»
n=0
- (m - n + 1)(n + 1)am-„+1a„+1], m > 1.
First few coefficients given by (2.10) are as follows
a2 a 1
a4 = 0, a5 =------, a6 =------, a7 =-------,
4 5 120 6 360 7 2520
a3 a2 a
a8 =-------, a9 =----------, a10 =---------------------------------------------------------------------, (2.11)
8 40320 9 90720 10 226800
= 16 + 27a , =-3.77869 x 10-7a3.
11 3991680 12
[2 (n + 2)(n + 1)an+2 m n (m + 3)(m + 2)(m +1)L ^ (2.10)
2. WANG EQUATION By defining
df d2 f
x = —, y = —. dn dn
Wang [9] transformed equation (1.2) to
d y 1 n y—T ^ x = 0. dx 2
Asghar [8] has shown that the same transformation changes equation (1.3) into
y2y” - (1 -x2)y - xy = o, 0 < x < 1. (3.1)
The boundary conditions transform to
y(0) = a y' (0) = --a, y(1) = °. (3.2)
a
Assume a solution of the form
y = Zbnxn. (3.3)
n=0
It is easy to find b2 = —^ and proceeding as in section 2, we obtain the recurrence
2a
relation
i m-1 m-n
bm+2 = ( + 1)( + 2) 2 (n + 2)(n + 1)bm-l-nblbn+2
(m + 1)(m + 2)a t0£0 (3.4)
+ (m + 1)bm+1 - (m - 2)bm-1]? m ^ 1.
First few coefficients generated by (3.4) are as follows
7 0.166667(-3 + a4K -0.625 + 0.208333a4
b3 =-----------^1, b4 = •
3 _ 5 ■’^4 7
2a a
, -0.875 + 0.333333a4 -1.3125 + 0.569444a4 - 0.0222222a8
b5 =----9------> b6 =--------ii-------------------------.
aa
The unknown parameter a can be found by solving the equation y(1) = 0. (3.5)
In practice one finds an approximation ak to a by replacing (3.5) by
n
n=0
Z bn(a) = °. (3.6)
A few members of the sequence {ak }kk=0 are displayed in Table 1.
Table 1. Approximation values f ” (0) = a
k ak
50 1.2321
90 1.2324
250 1.23254
1000 1.23258
The value of a , correct to seven decimal positions as found numerically by shooting method, is a = 1.232589 [5].
3. UNIFORMLY VALID ANALYTICAL SOLUTION
Using the coefficients evaluated in section 2 and the value of a found in section 3 the power series expression for the solution of the stagnation point flow problem up to the power 20 becomes
f (n) = 0.616294n - 0.166667T + 0.0126606T
-0.00342385^ + 0.000396825^ - 0.0000464444^ +0.0000167468^ -5.43469x 10-6n'° +1.96211x10-6n" - 7.07612 x 10-V2
7 i r\-8 14 (41)
(4.2)
+2.04944 x 10-7 n - 6.11228 x 10-8n +2.05581x10-V5 - 6.27563 x10-9n'6 +1.78703 x 10-9n17 - 5.46273 x 10-1y8 +1.65946 x 10-10 - 4.61405 x 10-yy°.
Differentiate (4.1) to get
f ”(n) = 1.23259n - 0.5 x 10-6n" + 0.063303^ - 0.0205431^
+0.00277778^ - 0.000371555^ + 0.000150721^
-0.0000543469^ + 0.0000215832n'°
-8.49134 x 10-6n" + 2.66427 x10-y2 -8.5572 x 10-7n13 + 3.08372 x 10-y4 -1.0041 x10-7n15 + 3.03796 x10-y6 -9.83291 x 10-9 n17 + 3.15297 x 10-9 n'8 -9.22811 x10-1°n19.
Applying Pade [6/6] approximation to (4.2) we get
f ’(n) = 4T)’ (4.3)
f2(n)
where
fj(n) = 1.23259n + 5.06009n2 + 0.0665943^
-0.204148n4 + 0.17892n5 - 0.0139962n6
and
f2(n) = 1 + 4.51091n + 1.88388n2 + 0.547214T
+0.152132n4 + 0.0265334n5 + 0.00419318n6.
Now we replace the functions f(n) and f2(n) by respectively gx(n) and g2(n) so
that
f ”(n) = (4.4)
g2(n)
where
gj(n) = 1.23259n + 5.06009n2 + 0.0665943T -0.204148T
2
+0.17892T - 0.0139962n6 + cirf
and
g2(n) = 1 + 4.51091n + 1.88388n2 + 0.547214n3 + 0.152132n4
+0.0265334n5 + 0.00419318n6 + T
where c is a constant to be determined. It is clear that for every c the expression (4.4) will represent f ”(n), for small n, since it will differ only slightly from the [6/6] Pade approximant for the function. Also for large n the exponential term will dominate other terms and the limiting value will approach unity as required by the boundary condition f ” (w) = 1. An optimum choice of c should ensure that the ratio g1(r)/ g2(n) differs as little as possible from the exact value of f ”(n) on the entire domain [0, w). To achieve this we numerically evaluate the integral
r» g1(r) t
J° g2T)dn (4 5)
and choose c so as to make the above integral as close to f (nw) for some suitable but
arbitrary nw . We choose nw = 8 which forces us to fix c = 0.00051. With this choice of
parameters the integral (4.5) gives 7.35212 which agrees with the exact numerical value to four decimal positions. Also
fr(8) =1
g2(8) ’
which implies that the function g1(n)/ g2(n) has already attained its asymptotic behavior at
8 and any integration beyond this point will yield approximate values for f (n) in agreement with the exact values to at least four decimal positions. Let fc (n) denote an approximate value found from
fcc(n)du. (46)
In Table 2 we compare a few approximate values, as found from (4.6), with the exact values of the solution, obtained numerically.
Table 2. Comparison of approximate values with exact
n Approximate fa (n) from (4.6) f (n) from a numerical solution
0 0 0
0.4 0.0880566 0.0880566
0.8 0.312428 0.312423
1.2 0.622033 0.622028
1.6 0.9798 0.9798
2 1.3620 1.3621
2.4 1.7555 1.7552
2.8 2.1534 2.1530
3.2 2.5526 2.5523
3.6 2.9523 2.9522
4 3.3522 3.3521
4.4 3.7521 3.7521
4.8 4.1521 4.1521
5.2 4.5521 4.5521
5.6 4.9521 4.9521
6 5.3521 5.3521
6.4 5.7521 5.7521
6.8 6.1521 6.1521
It is clear that for 0 < n < ®, the maximum error is less than 2 parts in ten thousand and it steadily decreases as we go down the Table. Even this error could be decreased by considering a Pade approximant of an order higher than [6/ 6] considered in this paper.
CONCLUSIONS
Sometimes it is possible to find an approximate solution of a nonlinear problem in the form of an infinite series. However the series may not converge outside a finite interval whereas the solution is known to exist on an infinite domain. In the case of the Blasius problem, the domain was extended to infinity by Liao [7] by making use of the homotopy analysis method and by Ahmad and Albarakati [10] by employing a technique similar to the one used in the present work. It is possible to use the homotopy technique to solve Falkner-Skan equation, however this will result in lengthy expressions. We have been successful in finding a short expression because, by modifying the Pade approximation, we have incorporated the asymptotic behavior of the function. As a result this expression gains in accuracy as the values of n are increased in contradistinction to solutions in the form of series or rational expressions.
ACKNOWLEDGMENTS
Financial support for this research was provided by the Deanship of Scientific Research of King Abdulaziz University. The author also wishes to express her gratitude to Professors Faiz Ahmad and Hassan Alsereihy for technical assistance.
REFERENCES
[1] I. G. Currie. Fundamental Mechanics of Fluids. McGrow-Hill, New York, 1974.
[2] L. Howarth. On the solution of the laminar boundary layer equations. Proc. Roy. Soc. London A 164 (1938) 547-579.
[3] D. R. Hartree. On an equation occurring in Falkner and Skan's approximate treatment of the equations of the boundary layer, Proc. Camb. Philo. Soc. 33 (1937) 223-239.
[4] A. Asaithambi. A finite difference method for the Falkner-Skan equation. Appl. Math. Comput. 92 (1998) 135-141.
[5] Asai Asaithambi. Solution of the Falkner-Skan equation by recursive evaluation of Taylor coefficients. J. Comput. Appl. Math. 176 (2005) 203-214.
[6] H. Blasius. Grenzschichten in Flussigkeiten mit kleiner Reibung. Z. Math. u. Phys. 56 (1908) 1-37.
[7] S. J. Liao. An explicit, totally analytic approximate solution for Blasius' viscous flow problems. Int. J. Non-Linear Mech. 34 (1999) 759-778.
[8] S. Asghar. Private communication.
[9] L. Wang. A new algorithm for solving classical Blasius equation. Appl. Math. Comput.
157 (2004) 1-9.
[10] F. Ahmad and W. A. Albarakati. A uniformly valid analytic solution of the Blasius problem. Submitted for publication to Communications in Nonlinear Science and Numerical Simulation.
