Page 581 - Handbook Of Integral Equations

We assume that Eq. (11) defines a function y(x) for all values of the variable x, positive and
+
–
negative. Let us introduce the functions y (x) and y (x) by formulas (3). Obviously, we have
+
–
y(x)= y (x)+ y (x), and Eq. (11) can be rewritten in the form
∞

+
+
y (x)= K(x – t)y (t) dt, x > 0 (12)
0

∞
+
–
y (x)= K(x – t)y (t) dt, x < 0. (13)
0
+
That is, the function y (x) can be determined by the solution of the integral equation (12) and the
–
+
function y (x) can be expressed via the functions y (x) and K(x) by means of formulas (13). In
this case, we have the relation

∞
+
–
+
y (x)+ y (x)= K(x – t)y (t) dt, (14)
–∞
which is equivalent to the original equation (11).
Let the function K(x) satisfy the condition
|K(x)| < Me v – x as x →∞,
(15)
v + x
|K(x)| < Me as x → –∞,
where v – < 0 and v + > 0. In this case, the function

1 ∞ izx
K(z)= √ K(x)e dx, (16)
2π –∞
is analytic in the strip v – <Im z < v + .
Let us seek the solution of Eq. (11) satisfying the condition
+
|y (x)| < M 1 e µx as x →∞, (17)
where µ < v + (such a solution exists). In this case we can readily verify that the integrals on the
–
right-hand sides in (12) and (13) are convergent, and the function y (x) satisfies the estimate
–
|y (x)| < M 2 e v + x as x → –∞. (18)
It follows from conditions (17) and (18) that the transforms Y + (z) and Y – (z) of the functions
–
+
y (x) and y (x) are analytic functions of the complex variable z for Im z > µ and Im z < v + ,
respectively.
Let us pass to the solution of the integral equation (11) or of Eq. (14), which is equivalent
to (11). To this end, we apply the (alternative) Fourier transform. By the convolution theorem (see
Subsection 7.4-4), it follows from (14) that
√
Y + (z)+ Y – (z)= 2π K(z)Y + (z),

or
W(z)Y + (z)+ Y – (z) = 0, (19)

where
√
W(z)=1 – 2π K(z) ≠ 0. (20)

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 564