Page 290 - Handbook Of Integral Equations

n
b
10. y(x) – λ A k e γ k (x–t) y(t) dt = f(x).
a
k=1
This is a special case of equation 4.9.20 with g k (x)= e γ k x and h k (t)= A k e –γ k t .
1 ∞
11. y(x) – e –|x–t| y(t) dt = Ae µx , 0 < µ <1.
2 0
Solution:
2
y(x)= C(1 + x)+ Aµ –2 (µ – 1)e µx – µ +1 ,
where C is an arbitrary constant.
•
Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).

∞
12. y(x)+ λ e –|x–t| y(t) dt = f(x).
0
Solution:
λ ∞ √
y(x)= f(x) – √ exp – 1+2λ |x – t| f(t) dt
1+2λ 0

λ +1 ∞
√
+ 1 – √ exp – 1+2λ (x + t) f(t) dt,
1+2λ 0
1
where λ > – .
2
•
Reference: F. D. Gakhov and Yu. I. Cherskii (1978).

∞
13. y(x) – λ e –|x–t| y(t) dt =0, λ >0.
–∞
The Lalesco–Picard equation.
Solution:
√ √
 1
 C 1 exp x 1 – 2λ + C 2 exp –x 1 – 2λ for 0 < λ < ,
2

1
y(x)= C 1 + C 2 x for λ = ,
2
 √ √
 1
C 1 cos x 2λ – 1 + C 2 sin x 2λ – 1 for λ > ,
2
where C 1 and C 2 are arbitrary constants.
•
Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971).

∞
14. y(x)+ λ e –|x–t| y(t) dt = f(x).
–∞
1
◦
1 . Solution with λ > – :
2
λ ∞ √
y(x)= f(x) – √ exp – 1+2λ |x – t| f(t) dt.
1+2λ –∞
1
2 .If λ ≤ – , for the equation to be solvable the conditions
◦
2

∞ ∞
f(x) cos(ax) dx =0, f(x) sin(ax) dx =0,
–∞ –∞
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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