8.9 Section 2 Background .1 Cosmology .11 Math Section 8.5

#### Book 9 Computational Mathematics Chapter 8 Ordinary Differential Equation Initial Value Problems

Section 4 Successive approximations

Consider the Cauchy Problem of the first-order differential equation

(4.1)

with the initial condition

(4.2)

The method of successive approximations obtains the solution y(x) as the limit of a sequence of functions yn(x) which are determined by the recurrence formula

(4.3)

It has been shown ([36]) that, if the right-hand term in the domain R{| x - x0 |? a, | y - y0| ? b}

satisfies the Lipschitz Condition with respect to y

(4.4)

then, irrespective of the choice of the initial function, the consecutive approximations yn(x) converge on some interval [xo,xo+ h] to the solution of this Cauchy Problem.

If f (x, y) is continuous in a rectangle R, then the error of the approximate solution yn(x) on the interval [x0, x0 + h] is estimated by the inequality:

(4.5)

where and h is determined by

Any function which is sufficiently close to the exact solution may be taken as the initial approximation y0(x). At times, it may be advantageous to use in the capacity of y0(x) an approximate solution of (4.1) in the form of a partial sum of a power series (Example 2).

Note: The method of successive approximations can also be used for solving systems of differential equations ( Example 3) and differential equations of order n, if the latter are written in the form of a system of first order equations. It has been pointed out that for the solution of a differential equation to be expandable in a power series the right-hand term of the equation must be analytic, whereas this is not required for the method of successive approximations. Therefore, generally speaking, the method of successive approximations is more widely used. It can also be used when the solution of a differential equation cannot be expanded in a power series. But this method, nevertheless, has the shortcoming that it involves evaluation of more and more cumbersome integrals.

Example 1. Find three successive approximations to the solution of the equation

(4.7)

for the initial condition y(0) = 0.

Solution: Taking into consideration the initial condition, replace Equation (4.7) by the integral equation:

Set the initial approximation y0(x) = 0. Then the first approximation is given by

Similarly, the second and third approximations are

Next, use (4.5) to estimate the error involved in the last approximation. Since the function f (x,y) = x2 + y2 is defined and continuous throughout the entire plane, any numbers may be used for a and b. For the sake of definiteness, let a = 1, b = 0.5, whence

By (4.6), let h=0.4, when on the interval [0,0.4]

whence

Note: Formula (4.5) tends to overestimates the error. When using this method, one chooses for a a value such that yn-1, yn coincide within the limits of permissible accuracy (cf. Example 2).

Example 2: Find for the equation

with the initial condition y(0) = 1 an approximate solution on the interval [0, 0.2] with an accuracy of 10-6.

Solution: Choose the initial approximation y0(x) in the form

Using (2.8) and the initial condition, one finds

Thus, y0(x) =1 + 0.1x+ 0.51x2 and

and the first approximation is

Consider the difference

Its maximum value occurs at x = 0.2:

whence the required accuracy has not yet been achieved. Note that in the expression for y, the sum of the last two terms does not exceed 10-5, whence one may set

to obtain

and the second approximation

The difference y2(x) - y1(x) on the interval [0, 0.2] is

Thus,

Example 3: Given the system of equations

(4.9)

and the initial conditions y(0) = 1, z(0) = 1/2, employ successive approximations to find the solution on the interval [0; 0.3] to an accuracy of 5?10-3.

Solution. Write (4.9) as integral equations

Use the initial values to find y'(0) =1?, z'(0)= -l, choose the initial approximations

and find y1(x) and z1(x):

In the same manner, one finds

The error estimate on the interval [0,0.3] is:

These differences lie now within the required limits, the terms x4, x5 and x6 being small on the interval [0,0.3]. Hence the required approximate solution is

Exercises

Find three successive approximations to the solutions of the following initial value problems, using y0(x) = y0, z0(x) = z0.

Employing successive approximations, find approximate solutions of the following differential equations on the interval [0,1] with an accuracy of 10-3, setting y0(x) = y0: