Page 341 - Handbook Of Integral Equations

b
4.9-3. Other Equations of the Form y(x)+ K(x, t)y(t) dt = F (x)
a

∞
26. y(x) – K(x + t)y(t) dt = f(x).
–∞
The Fourier transform is used to solving this equation.
Solution: √

f(u)+
1 ∞ 2π f(–u)K(u) iux

y(x)= √ √ e du,
2π –∞ 1 – 2π K(u)K(–u)

where
1 ∞ –iux 1 ∞ –iux

f(u)= √ f(x)e dx, K(u)= √ K(x)e dx.
2π –∞ 2π –∞
•
Reference: V. A. Ditkin and A. P. Prudnikov (1965).

∞
βt
27. y(x)+ e K(x + t)y(t) dt = Ae λx .
–∞
Solution:
e λx – k(λ)e –(β+λ)x ∞ (λ+β)x
y(x)= , k(λ)= K(x)e dx.
1 – k(λ)k(–β – λ)
–∞

∞
βt
28. y(x)+ [e K(x + t)+ M(x – t)]y(t) dt = Ae λx .
–∞
Solution:
I k (λ)e px – [1 + I m (p)]e λx
y(x)= A , p = –λ – β,
I k (λ)I k (p) – [1 + I m (λ)][1 + I m (p)]
where

∞ ∞
I k (λ)= K(z)e (β+λ)z dz, I m (λ)= M(z)e –λz dz.
–∞ –∞

∞
29. y(x) – K(xt)y(t) dt = f(x).
0
The solution can be obtained with the aid of the inverse Mellin transform:

1 c+i∞ –s
f(s)+ K(s)f(1 – s)
y(x)= x ds,
2πi 1 – K(s)K(1 – s)

c–i∞
where f and K stand for the Mellin transforms of the right-hand side and of the kernel of the

integral equation,

∞ ∞
f(s)= f(x)x s–1 dx, K(s)= K(x)x s–1 dx.

0 0
•
Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971).

∞
β
λ
30. y(x) – K(xt)t y(t) dt = Ax .
0
Solution:
λ
x + I β+λ x –β–λ–1 ∞ µ
y(x)= A , I µ = K(ξ)ξ dξ.
1 – I β+λ I –λ–1
0
It is assumed that all improper integrals are convergent.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 320