Page 222 - Handbook Of Integral Equations

b
2n+1
16. x – t y(t) dt = f(x), n =0, 1, 2, ...
a
Solution:
1 (2n+2)
y(x)= f x (x). (1)
2(2n + 1)!
The right-hand side f(x) of the equation must satisfy certain conditions. To obtain these
conditions, one must substitute solution (1) into the relations

b b (–1) k+1
(t – a) 2n+1 y(t) dt = f(a), (t – a) 2n–k y(t) dt = f x (k+1) (a),
a a A k
A k =(2n + 1)(2n) ... (2n +1 – k); k =0, 1, ... ,2n,

and then integrate the resulting equations by parts.

∞
y(t) dt

17. = f(x).
x + t
0
The left-hand side of this equation is the Stieltjes transform.
1 . By setting
◦
τ
z
x = e , t = e , y(t)= e –τ/2 w(τ), f(x)= e –z/2 g(z),
we obtain an integral equation with difference kernel of the form 3.8.15:

∞
w(τ) dτ

= g(z),
1
–∞ 2 cosh 2 (z – τ)
whose solution is given by
1 ∞ iux 1 ∞ –iuz
w(z)= √ cosh(πu) ˜g(u)e du, ˜ g(u)= √ g(z)e dz.
2π 3 –∞ 2π –∞
•
Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).
◦
2 . Under some assumptions, the solution of the original equation can be represented in the
form
(–1) n 2n+1 (n) (n+1)
y(x) = lim x f x (x) , (1)
n→∞ (n – 1)! (n + 1)! x
which is the real inversion of the Stieltjes transform.
An alternative form of the solution is

(–1) n 2n 2n (n) (n)
e
y(x) = lim x f x (x) . (2)
n→∞ 2π n x
To obtain an approximate solution of the integral equation, one restricts oneself to a
specific value of n in (1) or (2) instead of taking the limit.

•
Reference: I. I. Hirschman and D. V. Widder (1955).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 201