Применение уравнения Крокко-Ванга к решению задачи Бласиуса Текст научной статьи по специальности «Математика»

Electronic Journal «Technical Acoustics» http://www .ejta.org

2007, 2

Faiz Ahmad

Department of Mathematics, Faculty of Science, King Abdulaziz University P.O. Box 80203, Jeddah 21589, Saudi Arabia. e-mail: [email protected]

Application of Crocco-Wang equation to the Blasius problem

Received 04.12.2006, published22.01.07

Navier-Stokes problem for a boundary layer can be transformed, by a similarity transformation, into the Blasius problem which is governed by a non-linear ordinary differential equation of order three. Since it is simpler to solve an ordinary differential equation, this transformation leads to an easy evaluation of physical parameters such as the drag and the thickness of the boundary layer. Crocco and independently Wang further transformed this problem to one which is governed by a second order differential equation. In this paper this problem is solved by a classical method and the solution is used to derive two sequences, the first an increasing sequence and the second a decreasing sequence, both converging to the unknown second derivative, at the origin, of the solution to the Blasius equation. Also an asymptotic expression for the solution is obtained.

INTRODUCTION

The two dimensional steady-state laminar viscous flow over a semi-infinite flat plate is modeled by the nonlinear differential equation

where n and f (n) are respectively the dimensionless coordinate and the dimensionless stream functions defined in such a manner that the set of two partial differential Navier-Stokes equations reduces to the single ordinary differential equation (1a) [1]. It is relatively easy to find a series or a numerical solution of the above problem and physical parameters of interest such as the shear-stress distribution along the surface, the drag on the surface and the boundary layer thickness are easily evaluated.

The main hurdle in the solution of the above problem, called the Blasius problem, is the absence of the second derivative f "(0). Once this derivative has been correctly evaluated an analytical solution of the boundary value problem may be readily found. Blasius [2] found the following power series solution of the problem with P0 = -2.

(1b)

(1a)

(2)

where

k > 2, a represents the unknown f "(0).

Howarth [3] solved (1), ( with J30 = -2) numerically and found

a = 0.33206.

(3)

Several authors have devised numerical algorithms to find good approximations to f''(0) (see, for example, Asaithambi [4] and references therein). Asaithambi [4] solved (1) (with J30 = 1) and found f "(0) denoted by a to be

transforms Eq. (1a) into F"' + FF" = 0.

Therefore it is sufficient to consider the Blasius problem with J30 = 1. Henceforth we shall

problem and obtained the solution to a high level of accuracy. J. H. He has used an iterative perturbation technique to find an approximate analytical solution of the Blasius problem [7]. Abbasbandy [8] has used a modified version of the Adomian decomposition method to find a

Crocco proposed a further transformation of the Blasius problem in the 1940’s, see [10] and independently Wang [11] made use of the same transformation which helps to approach the Blasius problem from a new perspective. They used an ingenious idea to transform the Blasius problem into a simpler problem governed by a second order differential equation. They used the transformation

a = 0.469600.

(4)

Fang et al. [5] have shown that the substitution

treat the problem with J30 = 1. Liao [6] applied his homotopy analysis method to the Blasius

(5)

dx y

with the boundary conditions

dly + £ = 0, x E [0, 1)

(6a)

y(0) = f "(0), y'(0) = o, lim y(x) = 0. (6b)

Wang [11] used the Adomian decomposition method to solve (6), and found

x3 x6 x9 x12

y (x) = a--------------------------------------------------------------------------------------------------3- - .... (7)

6a 180a3 2160a5 19008a7

To find a the equation y(1) = 0 is solved for a . He solved this equation retaining six terms of the series (7) and found a = 0.453539 Recently Hashim [12] improved this value to

a = 0.466799 by finding terms of the series (7) up to x24 by the Adomian decomposition method (ADM) and then approximating this function by the [12/12] diagonal Pade' approximant.

Croco-Wang equation (6a) can be made a useful tool to elucidate analytical features of the Blasius problem. An interesting application of this equation recently enabled Ahmad [13] to exhibit degeneracy in the Blasius problem. In the present paper we wish to make further use of this equation to derive following further results.

1. We shall solve (6a), by an alternative method which can easily extend the series (7) to as many terms as we wish. From the results in [11, 12] it seemed that one could determine a to arbitrary accuracy once a large number of terms of the series (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 to a. The convergence is hopelessly slow. However the two sequences enable us to check accuracy of the approximation at each step.

pro / d 2 f \2

2. We shall establish that the integral J0 \nr) dr is bounded below by -4-a and above by

a. If we use the value of a = 0.469606 then we get the exact result J0 )dr =0.37118.

3. We use the above result to find an approximate asymptotic expression for f (r).

The power series solution (2) is known to be valid for 0 < r < 5.69 [6]. This solution combined with an asymptotic solution will represent the function f (n) analytically on the entire domain 0 <r .

1. SOLUTION OF THE CROCCO-WANG EQUATION Re-write Eq. (6a) in the form

yy" = -x. (8)

We assume the solution in the form

y(x) = Y,anx". (9)

n=0

From the boundary conditions (6b), it is clear that

ao = f "(0), a = 0. (10)

We define a = /"(0). Differentiate (9) twice and substitute y and y" in (8). We get

TjTjakann(n -1) x"+k - 2=- x. (ii)

k=0 n = 2

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

2aa2 + (2a1a2 + 6aa3) x + (12aa4 + 6a1a3 + 2a^) x2 + (20aa5 + 12a1a4 + 8a2 a3) x3 +... = - x (12)

from which we get

a2 = 0, a3 =——, a4 = 0, a5 = 0. (13)

6a

Define a new summation index m = n + k - 2 and change the order of summation in (11). We get

» m

TjYPkam-k+2(m - k + 2)(m - k + 1)xm =-x . (14)

m=0 k = 0

On the left side of (14) coefficient of each power of x other than unity must vanish. This gives for m > 2

m

Yjakam-k+2(m - k + 2)(m - k + 1) = °, (15)

k=0

which can be re-written as

m

(m + 2)(m + 1)am+2a0 = akam-k+2 (m - k + 2)(m - k + 1).

k=1

We have obtained the recurrence relation

1m

(m + 2)(m + 1)am+ 2 =-------Yuakam-k+2(m - k + 2)(m - k + 1), m > 2. (16)

a k=1

If 3 is not a divisor of m + 2 it will not divide both k and m - k + 2 From (10) and (13) we note that, with the exception of a0 and a3 other coefficients upto and including a5 are all zero. Using this in (16) we find that a6 ^0 but a7 = a8 = 0. 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 (16) by

1 n -1

3n(3n - 1)a3n =-------V(3n - 3k)(3n - 3k - 1)a3ka3n-3k, n > 2. (17)

a k=1

Starting with a0 =a and a3 = -a Eq. (17) quickly produces

1 1 1 7.01125 X10-

a6 =----3 , a9 =--------------- , a12 =------- , a15 =------------------ -----

6 180a3 9 2160a5 19008a7 a9

1.03002 x 10-6 1.61614 x 10-7 2.65906 x 10-8

a,„ =----------- -, a, =----------------- -------, a, = —

•*18 11 J 21 13 ’ 24 15

a a a

4.53471 x 10-9 7.95335 x 10-10

27 a17 30 a19

Hashim found the first nine coefficients (up to a 24) of (9), in complete agreement with members of the above list. Equation (17) can be easily used in a computer algorithm to extend the series (7) to an arbitrary number of terms with very little effort. As an example we give below coefficient of the x1500 term in the series (9).

4.1809830 x 10-335

a1500 = 999 .

a

2. AN INCREASING SEQUENCE

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

y(x) = Tja3nx3". (18)

n=0

This gives

0 a

a + Z^Jnrr = °. (19)

n=1 &

Hashim [12] solved (19) after truncating the series after 7 terms and found a « 0.463662. He

improved this value to 0.466799 by approximating the series for y(x) up to x21 by the [12/12]

Pade' approximant. Since we can find an arbitrary number of terms in the series (9), we can attempt to solve (19) with a large number of terms. Let F (a) denote the left side of Eq. (19) and let

Fk(a) = ^^0^ k = 11 2-. (20)

n=1 a

and let ak be a root of the equation Fk (a) = °. (21)

We shall solve (21) by a slightly modified Newton's method. We first find ax by solving

Fx(a) = 0 To find ak,k > 2 , we use the iterative scheme

6

*1 = a,

X,.+1 = * - F(*) , i = 1, 2,... (22)

i +1 i T-’f/ \ 5 5

Fk (ak-1)

and we approximate the derivative at a,-1 by

F,(ati) = F,a-1 + h) - F ^-1) (23)

h

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.

Table 1. Roots of Eq. (21)

k a, k a,

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 0.469066 4000 0.469597

The above sequence increases because for larger k more terms in the series (19) are retained and a larger value of a 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 to a . The two sequences together will determine precisely lower and upper bounds for a at each level of approximation.

3. A DECREASING SEQUENCE

Wang [11] and Hashim [12] have found the following expression for — :

y

1 1 x3 x6 x9 2099 x12 1 “ x3n

— = —+--------3 +---------------r +-7 +-o' +.... = —+ / c3n —ttt , (24)

y a 6a3 30a5 144a7 1425600 a9 a a2n+1

where c3n is defined to be the coefficient of x3n /a2n+1 in (24). The solution y(x) of (8) is related to (24) in the following manner

y(x)=a-\0\0ydx. (25)

If we use (24) in (25), we obtain

y(X) =a- -----Xc3n (3 + 3)(3 + 2 2n+1 =a-Z (3 + 3w3 + 2 n+1 (26a)

6a n=1 (3n + 3)(3n + 2)a n=0 (3n + 3)(3n + 2)a

= a-g3„(3n - 1)g2-1 ■ (26b)

Comparing (26b) with (18), we find

c3n-3 =-3n(3n -1)«3n, n (27)

Let x = 1 in (26b). We shall truncate the series after k terms and find an approximate sum of the remainder. We have

0 = y(1) = a z 3n(3n-1)a2n-1 (3k + 3)(3£ + 2)a2k+1 "" (28)

If k is large, we can approximately set

C3k+3 C3k+ 6 _

=a , ^^=a

C3k C3k

and so on. Eq. (28) becomes

k c c 1

0 = a-Z-----3n-3 2 1-----3-Y----------------. (29)

n=13n(3n - 1)a2n-1 a2 +1 ^ (3n + 3)(3n + 2) V ;

It is easy to show that

“ 1 1

z—1— =——

tl (3n + 3)(3n + 2) 3k + 2

and (29) becomes

k c c

0 = a-Z 63n-3

13n(3n - 1)a2n-1 (3k + 2)a2k+1 '

In the notation of Section 2, the above equation may be written as 3k + 3

Fk (a) + —— a3k+3 = 0. (30)

a

Since the approximation used to obtain (29) replaces a larger number by a smaller one, we have F (a) > G(a), where F(a) and G(a) denote respectively the left sides of Eqs. (19) and

(30). The above inequality ensures that if ae is a zero of F(a) then a zero of G(a) greater

than ae exists. We solve (30) for k = 1,2... by the modified Newton's method, as in Section 2.

Let a(k) denote a root of (30). We display a few members of this sequence in Table 2.

Table 2. Roots of Eq. (30)

k a(k) k a(k) k a(k) k a(k)

5 0.482463 100 0.470006 900 0.469634 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 0.4696064

Using the last entries of Tables 1 and 2 we have the following inequality for a:

0.469597 <a< 0.4696064. (31)

4. THE INTEGRAL

Integrate (1.7) with respect to x on [0,1). We get 1 1 1

\\ydx

= a-

4.6a 7.180a3 10.2160a5'"

< a.

Also we have from (32) Pa. K1 1

ydx > a—(----------------1------

J 0 A fir/ 180

1

4 6a 180a 2160a

1

- +...)

= a—a 4

3

= — a.

4

(32)

(33)

(34)

(35)

Eq. (34) follows from the line preceding it because a satisfies the equation y(1) = 0 which becomes, on letting x = 1 in (7),

a = ■

1

1

6a 180a

d 7

- +... .

Now y =

/•1

J 0ydx = J 0

df

2 and x = ^— . Hence dr/ dn

d 2f drf

dn.

(36)

From (33), (35) and (36) we obtain the inequality

3 c“

—a < I

0

'd2 f dr)2

dn < a.

(37)

If we use a = 0.469606 in (32) then we can replace (37) by the equality

d,

101 dn

J

Write the Blasius equation fm + f f" = 0 in the form

rfff

— = - f. f"

An integration gives

f n j

f n = a exp I -Jf (u)du

(39)

It is clear from (39) that for 0 <n <ro , f \rj) > 0. As a consequence f '(r/) is nonnegative and monotonically increasing and f (rf) is also increasing and unbounded. Since f' (n) is bounded above by unity, the curve y = f (rf) does not intersect the line y = n, because otherwise f (rf) being a convex function, its slope, at the point of intersection must exceed unity. This means the graph of y = f (rf) is flanked on the left by the line y = n. The initial condition limn^ro f'(n) = 1 indicates that the function f (rf) has an asymptote with slope unity but lying below the line y = n . Let the equation of this asymptote be

y = n-n

f (n)dn>I (n-n0)dn, an exponentiation leads to the inequality

0 J^0

f "(n = a exp |-Jf (u)du j < a exp f- 2 (n - n)2 J. (40)

Square both sides of (40) and integrate on [0, ro) We find

ro

0.37118 < a2 J exp {-(n-%)2}, (41)

0

where we have used (38). Now

ro ro ro n I V0

J exp {-{ - n0 )2} = J e u du = Je-u du + Je-u du =---------+ Je~u du . (42)

0 -n0 0 0 2 0

Combining (41) and (42) we obtain

]e-'du > 0.79695. (43)

0 a 2

The above inequality is satisfied if

(44)

The above result indicates that the line

y = n-116.

(45)

is approximately an asymptote to the curve y = f (rf) Since this curve eventually behaves like a straight line of slope unity, it follows that, for large n

A numerical solution of the Blasius problem indicates that n0 = 1 23 . The above result gives this number with an error less than 6 percent.

CONCLUSIONS

We were able to apply a classical method to solve Crocco-Wang equation which gave the solution to an arbitrary number of terms. In principle a modern 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 3000-th term, for example, would be tremendous. However our advantage in easy calculation of the two sequences {ak} and {a( k)} is somewhat neutralized by their extremely slow convergence to a .

Results of Section 4 coupled with those in [13] demonstrate that Crocco-Wang equation contains much more information than just an analytical tool to evaluate f "(0). We may hope that, in future, this equation will be helpful in elucidating more features associated with qualitative as well as quantitative theory of the Blasius problem.

REFERENCES

[1] B. J. Cantwell. Introduction to Symmetry Analysis. Cambridge University Press, 2002.

[2] H. Blasius. Grenzschichten in Flussigkeiten mit kleiner Reibung. Z. Math. u. Phys. 56 (1908) 1-37.

[3] L. Howarth. On the solution of the laminar boundary layer equations. Proc. London Math. Soc. A 164 (1938) 547-579.

[4] Asai Asaithambi. Solution of the Falkner-Skan equation by recursive evaluation of Taylor coefficients. J. Comput. Appl. Math. 176 (2005) 203-214.

[5] T. Fang, F. Guo, C. F. Lee. A note on the extended Blasius problem. Applied Mathematics Letters 19 (2006) 613-617.

[6] S. J. Liao. An explicit, totally analytic approximate solution for Blasius' viscous flow problems. International Journal of Non-Linear Mechanics 34 (1999) 759-778.

[7] J. H. He. A simple perturbation approach to Blasius equation. Appl. Math. Comput. 140

(2003) 217-222.

f(if) « r/-n<5.

[8] S. Abbasbandy. A numerical solution of Blasius equation by Adomian decomposition method and comparison with homotopy perturbation method. Chaos, Solitons and Fractals 31 (2007)257-260.

[9] R. Cortell. Numerical solutions of the classical Blasius flat-plate problem. Appl. Math. Comput. 170 (2005) 706-710.

[10] Chia-Shun Yih. Fluid Mechanics - A Concise Introduction to the Theory, McGraw-Hill, New York, 1969.

[11] L. Wang. A new algorithm for solving classical Blasius equation. Appl. Math. Comput. 157

(2004) 1-9.

[12] I. Hashim. Comments on A new algorithm for solving classical Blasius equation by L. Wang. Appl. Math. Comput. 176 (2006), 700-703.

[13] F. Ahmad. Degeneracy in the Blasius problem, submitted for publication to the Electronic Journal of Differential Equations.