Page 100 - Handbook Of Integral Equations

x

3. [AJ 0 (λx)+ BJ 0 (λt)]y(t) dt = f(x).
a
For B = –A, see equation 1.8.2. We consider the interval [a, x] in which J 0 (λx) does not
change its sign.
Solution with B ≠ –A:
1 d – A x – B

y(x)= ± J 0 (λx) A+B J 0 (λt) A+B f (t) dt .
t
A + B dx a
Here the sign of J 0 (λx) should be taken.
x
4. (x – t) J 0 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x
y(x)= g(t) dt,
a
where
1 d 2 2 3 t
g(t)= + λ (t – τ) J 1 [λ(t – τ)] f(τ) dτ.
λ dt 2 a
x

5. (x – t)J 1 [λ(x – t)]y(t) dt = f(x).
a
Solution:
1 d 2 2 4 x 2
y(x)= + λ (x – t) J 2 [λ(x – t)] f(t) dt.
3λ 3 dx 2 a
x

2
6. x – t J 1 [λ(x – t)]y(t) dt = f(x).
a
Solution:
x
y(x)= g(t) dt,
a
where
1 d 2 2 5 t
2
g(t)= + λ t – τ J 2 [λ(t – τ)] f(τ) dτ.
9λ 3 dt 2 a
x

n
7. x – t J n [λ(x – t)]y(t) dt = f(x), n =0, 1, 2, ...
a
Solution:
2 2n+2 x
d 2 n+1
y(x)= A + λ (x – t) J n+1 [λ(x – t)] f(t) dt,
dx 2 a
2 n!(n + 1)!

2n+1
A = .
λ (2n)! (2n + 2)!
If the right-hand side of the equation is differentiable sufficiently many times and the
conditions f(a)= f (a)= ··· = f (2n+1) (a) = 0 are satisfied, then the solution of the integral

x x
equation can be written in the form
x 2n+2
d 2
y(x)= A (x – t) 2n+1 J 2n+1 [λ(x – t)]F(t) dt, F(t)= 2 + λ 2 f(t) dt.
a dt

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 78