THE METHOD OF SUCCESSIVE APPROXIMATION

From a mathematical point of view, the method of successive approx­imation possesses a number of advantages. It is best to illustrate it by an example.

Consider again the rectangular wing, for which a, e, c, GJ are constants. In this example

Подпись: t—-У for

Подпись: (1)G(x, y) =

THE METHOD OF SUCCESSIVE APPROXIMATION

{s => У ^ x ^ 0)

and the fundamental equation 3 of § 3.2 becomes

THE METHOD OF SUCCESSIVE APPROXIMATION(2)

THE METHOD OF SUCCESSIVE APPROXIMATION THE METHOD OF SUCCESSIVE APPROXIMATION

If we introduce the nondimensional parameters

Подпись: where Подпись: 0(£) = A Подпись: (4) (5)

as the spanwise coordinates, Eq. 2 may be written as

Let 0(1)(f) = f be the first approximation (a guess) of the solution 0(g). Putting 0a) into the integrals on the right-hand side of Eq. 4, we obtain the value of the integrals

/(i>(f) = ^ if drj + f rj dyj =i($ – – j j (6)

Since 8a>(£) is not an exact solution, it does not satisfy Eq. 4. But, if 6a)(£) approximates closely enough to the correct solution, Eq. 4 must be approximately satisfied. Equating A/(1)(f) with 8a)(£) at one point, say £ = 1, we obtain the approximate critical value A = 3.

As a second approximation let us take the result given by Eq. 6, multi­plied by a suitable constant, and assume

*,w(f) = j(f-y) (?)

Then the integral on the right-hand side of Eq. 4 becomes

w 2 12 6 60/ ‘

The second approximate solution of A is again obtained by equating the value of A/<2)(f) to 0<2)(f) at £ = 1, which gives A = 2.5.

The exact solution is A = 7t2/4 = 2.46740. The error at this stage is 1.30 per cent too large. If we take 8m(£) = f%(£ — ff3 + and substitute into the right-hand side of Eq. 4 to obtain /(3)(£), we will have

/(3>(£) = *U(f – III3 -Г &£5 – тЫ7)

and

A = 2.4705

The error of A is now only 0.12 per cent too large.

It is quite evident how to proceed further. We substitute 8(n>(£) into the right-hand side of Eq. 4, and evaluate the integral. Let the result be written as I{n>(£). If Iw(£)/8(n)(£) tends to a constant in the whole range 0 < £ < 1, then 0ln)(£) is an approximate solution and A = 0(я,(1)//(п,(1).

In order to compare the degree of approximation, it is convenient to normalize all 0<n)(f) in such a way that they are equal at a reference

section. In the above example, we have normalized 6(n)(f) so that

Є<п>(1) = 1.

/ 77-

Подпись: • 77 г- ,m2{ Подпись: 7! Подпись: f7 + - Подпись: ■]
THE METHOD OF SUCCESSIVE APPROXIMATION THE METHOD OF SUCCESSIVE APPROXIMATION

Note that, according to § 3.2, the exact solution is/(£) = sin – f. By series expansion,

which may be compared with our successive approximations 0(1)(f), 0<2>(f), 0(S>(£), etc.

The process is actually a repeated integration. It is iterative. So the method of successive approximation is also called the method of iteration.

An integral equation of the form Eq. 3 of § 3.2 is known as a Fredholm integral equation of the second kind. Since the influence function is symmetric, G(x, y) = G(y, x), the method of iteration will converge to the correct eigenfunction corresponding to the smallest (in absolute value) eigenvalue.

The integration process can be very difficult in practice. Often a simple numerical solution can be obtained by replacing the integrals by a finite sum and transforming the governing equation into a matrix equation. The process of iteration can then be applied to the matrix equation in exactly the same manner as described above.

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