Page 258 - Handbook Of Integral Equations

∞

15. K(x – t)y(t) dt = f(x).
–∞
The Fourier transform is used to solve this equation.
1 . Solution:
◦
f(u)
1 ∞ ˜ iux
y(x)= e du,
˜
2π –∞ K(u)
1 ∞ 1 ∞
˜ –iux ˜ –iux
f(u)= √ f(x)e dx, K(u)= √ K(x)e dx.
2π –∞ 2π –∞
The following statement is valid. Let f(x) ∈ L 2 (–∞, ∞) and K(x) ∈ L 1 (–∞, ∞). Then
for a solution y(x) ∈ L 2 (–∞, ∞) of the integral equation to exist, it is necessary and sufficient
˜
˜
that f(u)/K(u) ∈ L 2 (–∞, ∞).
2 . Let the function P(s)defined by the formula
◦
1 ∞ –st
= e K(t) dt
P(s)
–∞
be a polynomial of degree n with real roots of the form
s s s

P(s)= 1 – 1 – ... 1 – .
a 1 a 2 a n
Then the solution of the integral equation is given by

d
y(x)= P(D)f(x), D = .
dx
•
References: I. I. Hirschman and D. V. Widder (1955), V. A. Ditkin and A. P. Prudnikov (1965).
∞

16. K(x – t)y(t) dt = f(x).
0
The Wiener–Hopf equation of the first kind. This equation is discussed in Subsection 10.5-1
in detail.

b
3.8-4. Other Equations of the Form K(x, t)y(t) dt = F (x)
a
∞

17. K(ax – t)y(t) dt = Ae λx .
–∞
Solution:
A λ ∞ λ
y(x)= exp x , B = K(z)exp – z dz.
B a a
–∞
∞

18. K(ax – t)y(t) dt = f(x).
–∞
The substitution z = ax leads to an equation of the form 3.8.15:
∞

K(z – t)y(t) dt = f(z/a).
–∞

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 237