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

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 *y*_{n}(*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* - *x*_{0 }*|*?
*a*, *|* *y* - *y*_{0}*| *?
*b*}

satisfies the **Lipschitz
Condition**** **with respect to *y*

(4.4)

then, irrespective of the choice of the initial function, the consecutive approximations *y*_{n}(*x*)
converge on some interval [*x*_{o}*,x*_{o}*+ 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 *y*_{n}(*x*) on the interval [*x*_{0}*, x*_{0}* + 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 *y*_{0}(*x*). At times, it may be advantageous to use in the capacity of *y*_{0}(*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* y*_{0}(*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*)
*= x*^{2}* + y*^{2} 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 *y*_{n-1},* y*_{n }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 *y*_{0}(*x*)
in the form

Using (2.8) and the initial condition, one finds

Thus, *y*_{0}(*x*) =1 + 0.1*x*+ 0.51*x*^{2} 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 *y*_{2}(*x*) - *y*_{1}(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 *y*_{1}(*x*) and *z*_{1}(*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 *x*^{4}, *x*^{5}
and *x*^{6} being small on the interval [0,0.3]. Hence the required approximate solution is

Find three successive approximations to the solutions of the following initial value problems,
using *y*_{0}(*x*) = *y*_{0},* z*_{0}(*x*) = *z*_{0}*.*

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