Page 103 - Handbook Of Integral Equations

x

18. (x – t) ν+1 J ν [λ(x – t)]y(t) dt = f(x).
a
Solution:
x

y(x)= g(t) dt,
a
where

2 n t
d 2 n–ν–2
g(t)= A + λ (t – τ) J n–ν–2 [λ(t – τ)] f(τ) dτ,
dt 2 a
2 Γ(ν +1) Γ(n – ν – 1)

n–2
A = ,
λ Γ(2ν +2) Γ(2n – 2ν – 3)
where –1< ν < n – 1 and n =1, 2, ...
2
If the right-hand side of the equation is differentiable sufficiently many times and the
conditions f(a)= f (a)= ··· = f x (n–1) (a) = 0 are satisfied, then the function g(t)defining the

x
solution can be written in the form
t d 2 n
g(t)= A (t – τ) n–ν–2 J n–ν–2 [λ(t – τ)]F(τ) dτ, F(τ)= 2 + λ 2 f(τ).
a dτ
•
Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

x

√
19. J 0 λ x – t y(t) dt = f(x).
a
Solution:
d 2 x √
y(x)= I 0 λ x – t f(t) dt.
dx 2
a
x

√
√
20. AJ ν λ x + BJ ν λ t y(t) dt = f(x).
a
√
We consider the interval [a, x] in which J ν λ x does not change its sign.
Solution with B ≠ –A:
1 d √
– A x √
– B

y(x)= ± J ν λ x A+B J ν λ t A+B f (t) dt .
t
A + B dx a
√
Here the sign J ν λ x should be taken.
x
√
√
21. AJ ν λ x + BJ µ β t y(t) dt = f(x).
a
√
√
This is a special case of equation 1.9.6 with g(x)= AJ ν λ x and h(t)= BJ µ β t .
x √ √

22. x – tJ 1 λ x – t y(t) dt = f(x).
a
Solution:
2 d 3 x √
y(x)= I 0 λ x – t f(t) dt.
λ dx 3
a

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 81