Page 683 - Handbook Of Integral Equations

The equation for the correction has the form
x 1

2
ϕ 1 (x)= –2 t ϕ 1 (t) dt + x 4
0 4
and can be solved by any of the known methods for Volterra linear equations of the second kind. In the case under
consideration, we apply the successive approximation method, which leads to the following results (the number of the step
is indicated in the superscript):
4
ϕ (0) = 1 4 x ,
1
x
7
4
1
4
ϕ (1) = 1 4 x – 2 1 6 1 4 x – 14 x ,
t dt =
1
4
0
x

(2) 1 4 2 1 4 1 7 1 4 1 7 1 10
ϕ = x – 2 t t – x dt = x – x + x .
1 4 4 14 4 14 70
0
We restrict ourselves to the second approximation and obtain
4
7
y 2 (x)= –x + 1 x – 1 x + 1 x 10
4 14 70
and then pass to the third iteration step of the Newton–Kantorovich method:
y 3 (x)= y 2 (x)+ ϕ 2 (x),
22
ε 2 (x)= 1 x 10 – 1 x 13 – 1 x 16 + 1 x 19 + 1 x ,
160 1820 7840 9340 107800
x
7
ϕ 2 (x)= ε 2 (x)+2 t –t + 1 4 1 t + 1 t 10 ϕ 2 (t) dt.
t –
4 14 70
0
When solving the last equation, we restrict ourselves to the zero approximation and obtain
7
22
4
y 3 (x)= –x + 1 x – 1 x + 23 x 10 – 1 x 13 – 1 x 16 + 1 x 19 + 1 x .
4 14 112 1820 7840 9340 107800
The application of the successive approximation method to the original equation leads to the same result at the fourth step.
As usual, in the numerical solution the integral is replaced by a quadrature formula. The main
difficulty of the implementation of the method in this case is in evaluating the derivative of the
kernel. The problem can be simplified if the kernel is given as an analytic expression that can be
differentiated in the analytic form. However, if the kernel is given by a table, then the evaluation
must be performed numerically.
14.2-5. The Collocation Method
When applied to the solution of a Volterra equation of the first kind in the Urysohn form
x

K x, t, y(t) dt = f(x), a ≤ x ≤ b, (21)
a
the collocation method is as follows. The interval [a, b] is divided into N parts on each of which the
desired solution can be presented by a function of a certain form
˜ y(x)= Φ(x, A 1 , ... , A m ), (22)
involving free parameters A i , i =1, ... , m.
On the (k + 1)st part x k ≤ x ≤ x k+1 , where k =0, 1, ... , N – 1, the solution can be written in the
form
x

K x, t, ˜y(t) dt = f(x) – Ψ k (x), (23)
x k
where the integral

x k

Ψ k (x)= K x, t, ˜y(t) dt, (24)
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 666