Page 102 - Handbook Of Integral Equations

x
2n–1
13. (x – t) 2 J 2n–1 [λ(x – t)]y(t) dt = f(x), n =2, 3, ...
a 2
Solution:
√ 2 n x
π d 2
y(x)= + λ sin[λ(x – t)] f(t) dt.
√ 2n+1 dx 2
2 λ 2 (2n – 2)!! a
x

14. [J ν (λx) – J ν (λt)]y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x)= J ν (λx).

d xf (x)
x
Solution: y(x)= .
dx νJ ν (λx) – λxJ ν+1 (λx)
x

15. [AJ ν (λx)+ BJ ν (λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.8.14. We consider the interval [a, x] in which J ν (λx) does not
change its sign.
Solution with B ≠ –A:

1 d – A x – B

y(x)= ± J ν (λx) A+B J ν (λt) A+B f (t) dt .
t
A + B dx a
Here the sign of J ν (λx) should be taken.
x
16. [AJ ν (λx)+ BJ µ (βt)]y(t) dt = f(x).
a
This is a special case of equation 1.9.6 with g(x)= AJ ν (λx) and h(t)= BJ µ (βt).

x

ν
17. (x – t) J ν [λ(x – t)]y(t) dt = f(x).
a
Solution:
d 2 n–ν–1
2 n x
y(x)= A + λ (x – t) J n–ν–1 [λ(x – t)] f(t) dt,
dx 2 a
2 Γ(ν +1) Γ(n – ν)

n–1
A = ,
λ Γ(2ν +1) Γ(2n – 2ν – 1)
1
where – < ν < n–1 and n =1, 2, ...
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 solution of the integral

x
equation can be written in the form
x n
d 2
y(x)= A (x – t) n–ν–1 J n–ν–1 [λ(x – t)]F(t) dt, F(t)= 2 + λ 2 f(t).
a dt
•
Reference: S. G. Samko, A. A. Kilbas, and O. I. Marichev (1993).

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 80