Page 608 - Handbook Of Integral Equations

11.18-3. Specific Features of the Application of Quadrature Formulas

The accuracy of the resulting solutions essentially depends on the smoothness of the kernel and
the constant term. When choosing the quadrature formula, it is necessary to take into account that
the more accurate an applied formula is, the more serious requirements must be imposed on the
smoothness of the kernel, the solution, and the right-hand side.
If the right-hand side or the kernel have singularities, then it is reasonable to perform a preliminary
transform of the original equation to obtain a more accurate approximate solution. Here the following
methods can be applied.
If the right-hand side f(x) has singularities and the kernel is smooth, then we can introduce the
new unknown function z(x)= y(x) – f(x) instead of y(x), and the substitution of z(x) in the original
equation leads to the equation

b b
z(x) – λ K(x, t)z(t) dt = λ K(x, t)f(t) dt,
a a
in which the right-hand side is smoothed, and hence a solution z(x) is smoother. From the func-
tion z(x) thus obtained we can readily find the desired solution y(x).
For the cases in which the kernel K(x, t) or its derivatives with respect to t have discontinuities
on the diagonal x = t, it is useful to rewrite the equation under consideration in the equivalent form

b b
y(x) 1 – λ K(x, t) dt – λ K(x, t)[y(t) – y(x)] dt = f(x),
a a
where the integrand in the second integral has no singularities because the difference y(t) – y(x)
b

vanishes on the diagonal x = t, and the calculation of the integral K(x, t) dt is performed without
a
unknown functions and is possible in the explicit form.
Example. Consider the equation
1

y(x) – 1 xty(t) dt = 5 x.
2 6
0
Let us choose the nodes x 1 =0, x 2 = 1 , x 3 = 1 and calculate the values of the right-hand side f(x)= 5 x and of the
2 6
kernel K(x, t)= xt at these nodes:

f(0)=0, f 1 = 5 , f(1) = 5 ,
2 12 6
1 1 1 1 1
K(0, 0) = 0, K 0, =0, K(0, 1) = 0, K ,0 =0, K , = ,
2 2 2 2 4

K 1 ,1 = 1 , K(1, 0) = 0, K 1, 1 = 1 , K(1, 1) = 1.
2 2 2 2
On applying Simpson’s rule (see Subsection 8.7-1)
1

F(x) dx ≈ 1 F(0)+4F 1 + F(1)
0 6 2
to determine the approximate values y i (i = 1, 2, 3) of the solution y(x) at the nodes x i we obtain the system
y 1 =0,
11 y 2 – 1 y 3 = 5 ,
12 24 12
– 2 y 2 + 11 y 3 = 5 ,
12 12 6
whose solution is y 1 =0, y 2 = 1 , y 3 = 1. In accordance with the expression (6), the approximate solution can be presented
2
in the form

˜ y(x)= 5 x + 1 × 1 0+4 × 1 × 1 x +1 × 1 × x = x.
6 2 6 2 2
We can readily verify that it coincides with the exact solution.
•
References for Section 11.18: N. S. Bakhvalov (1973), V. I. Krylov, V. V. Bobkov, and P. I. Monastyrnyi (1984),
A. F. Verlan’ and V. S. Sizikov (1986).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 591