Page 564 - Handbook Of Integral Equations

representation of the solution of the nonhomogeneous integral equation via the resolvent of the
original equation is more convenient. To obtain the desired representation, we note that formula (3)
can be transformed to the expression

K(u)
Y(u)=[1 – R(u)]F(u), R(u)= . (5)
1+ K(u)
On the basis of (5), by applying the Fourier inversion formula and the convolution theorem (for
transforms) we obtain

∞
1
y(x)= f(x) – √ R(x – t)f(t) dt, (6)
2π –∞
where the resolvent R(x – t) of the integral equation (1) is given by the relation

1 ∞ K(u) –iux
R(x)= √ e du, (7)
2π –∞ 1+ K(u)
Thus, to determine the solution of the original integral equation (1), it suffices to find the func-
tion R(x) by formula (7).
The function R(x) is a solution of Eq. (1) for a special form of the function f(x). Indeed, it
follows from formulas (3) and (5) that for Y(u)= R(u) the function F(u) is equal to K(u). This
means that, for f(x) ≡ K(x), the function y(x) ≡ R(x) is a solution of Eq. (1), i.e., the resolvent of
Eq. (1) satisfies the integral equation

1 ∞
R(x)+ √ K(x – t)R(t) dt = K(x), –∞ < x < ∞. (8)
2π –∞
Note that to calculate direct and inverse Fourier transforms, one can use the corresponding tables
from Supplements 6 and 7 and the books by H. Bateman and A. Erd´ elyi (1954) and by V. A. Ditkin
and A. P. Prudnikov (1965).
Example. Let us solve the integral equation

∞

y(x) – λ exp α|x – t| y(t) dt = f(x), –∞ < x < ∞, (9)
–∞
which is a special case of Eq. (1) with kernel K(x – t) given by the expression
√
K(x)= – 2πλe –α|x| , α >0. (10)
Let us find the function R(x). To this end, we calculate the integral

∞ 2αλ
e
K(u)= – λe –α|x| iux dx = – 2 2 . (11)
–∞ u + α
In this case, formula (5) implies
K(u) 2αλ
R(u)= = – , (12)
2
2
1+ K(u) u + α – 2αλ
and hence

1 ∞ 2 ∞ αλ

R(x)= √ R(u)e –iux du = – e –iux du. (13)
2
2
2π –∞ π –∞ u + α – 2αλ
Assume that λ < 1 2 α. In this case the integral (13) makes sense and can be calculated by means of the theory of residues on
applying the Jordan lemma (see Subsections 7.1-4 and 7.1-5). After some algebraic manipulations, we obtain
√ αλ √
2
R(x)= – 2π √ exp –|x| α – 2αλ (14)
2
α – 2αλ
and finally, in accordance with (6), we obtain

αλ ∞ √
2
y(x)= f(x)+ √ exp –|x – t| α – 2αλ f(t) dt, –∞ < x < ∞. (15)
2
α – 2αλ –∞
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 547