Page 242 - Handbook Of Integral Equations

3.5-2. Kernels Containing Sine

∞
6. sin(xt)y(t) dt = f(x).
0
2 ∞
Solution: y(x)= sin(xt)f(t) dt.
π 0
Up to constant factors, the function f(x) and the solution y(t) are the Fourier sine transform
pair.

•
References: H. Bateman and A. Erd´ elyi (vol. 1, 1954), V. A. Ditkin and A. P. Prudnikov (1965).

b

7. sin λ|x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
◦
1 . Let us remove the modulus in the integrand:
x b
sin[λ(x – t)]y(t) dt + sin[λ(t – x)]y(t) dt = f(x). (1)
a x

Differentiating (1) with respect to x twice yields

x b

2λy(x) – λ 2 sin[λ(x – t)]y(t) dt – λ 2 sin[λ(t – x)]y(t) dt = f (x). (2)
xx
a x
Eliminating the integral terms from (1) and (2), we obtain the solution

1 2
y(x)= f (x)+ λ f(x) . (3)

xx
2λ
2 . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting
◦
x = a and x = b in (1), we obtain two corollaries
b b
sin[λ(t – a)]y(t) dt = f(a), sin[λ(b – t)]y(t) dt = f(b). (4)
a a

Substituting solution (3) into (4) followed by integrating by parts yields the desired conditions
for f(x):
sin[λ(b – a)]f (b) – λ cos[λ(b – a)]f(b)= λf(a),

x
(5)

sin[λ(b – a)]f (a)+ λ cos[λ(b – a)]f(a)= –λf(b).
x
The general form of the right-hand side of the integral equation is given by
f(x)= F(x)+ Ax + B, (6)

where F(x) is an arbitrary bounded twice differentiable function, and the coefficients A and B
are expressed in terms of F(a), F(b), F (a), and F (b) and can be determined by substituting

x
x
formula (6) into conditions (5).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 221