Page 473 - Handbook Of Integral Equations

Example. For the interval [–1, 1], the parameters in formula (1) acquire the following values:
Chebyshev’s formula (n = 6):
2 1
A 1 = A 2 = ··· = = , x 1 = –x 6 = –0.8662468181,
n 3 (5)
x 2 = –x 5 = –0.4225186538, x 3 = –x 4 = –0.2666354015.
Gauss’ formula (n = 7):
A 1 = A 7 = 0.1294849662, A 2 = A 6 = 0.2797053915,
A 3 = A 5 = 0.3818300505, A 4 = 0.4179591837,
(6)
x 1 = –x 7 = –0.9491079123, x 2 = –x 6 = –0.7415311856,
x 3 = –x 5 = –0.4058451514, x 4 =0.
Note that a vast literature is devoted to quadrature formulas, and the reader can find books of
interest (e.g., see G. A. Korn and T. M. Korn (1968), N. S. Bakhvalov (1973), S. M. Nikol’skii
(1979)).

8.7-2. The General Scheme of the Method

Let us solve the Volterra integral equation of the first kind
x
K(x, t)y(t) dt = f(x), f(a)=0, (7)
a
on an interval a ≤ x ≤ b by the method of quadratures. The procedure of constructing the solution
involves two stages:
◦
1 . First, we determine the initial value y(a). To this end, we differentiate Eq. (7) with respect to x,
thus obtaining
x
K(x, x)y(x)+ K (x, t)y(t) dt = f (x).

x
x
a
By setting x = a,we find that
f (a) f (a)

x
x
y 1 = y(a)= = .
K(a, a) K 11
◦
2 . Let us choose a constant integration step h and consider the discrete set of points x i = a+h(i–1),
i =1, ... , n.For x = x i , Eq. (7) acquires the form

x i
K(x i , t)y(t) dt = f(x i ), i =2, ... , n, (8)
a
Applying the quadrature formula (1) to the integral in (8) and choosing x j (j =1, ... , i)tobethe
nodes in t, we arrive at the system of equations
i

A ij K(x i , x j )y(x j )= f(x i )+ ε i [y], i =2, ... , n, (9)
j=1
where the A ij are the coefficients of the quadrature formula on the interval [a, x i ] and ε i [y]isthe
truncation error. Assume that the ε i [y] are small and neglect them; then we obtain a system of linear
algebraic equations in the form

i

A ij K ij y j = f i , i =2, ... , n, (10)
j=1

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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