Page 339 - Handbook Of Integral Equations

∞
22. y(x)+ K(x – t)y(t) dt = Ae λx .
–∞
Solution:
A λx ∞ –λx
y(x)= e , q = K(x)e dx.
1+ q
–∞

∞
23. y(x)+ K(x – t)y(t) dt = A cos(λx)+ B sin(λx).
–∞
Solution:
AI c + BI s BI c – AI s
y(x)= cos(λx)+ sin(λx),
2
2
I + I 2 I + I 2
c s c s

∞ ∞
I c =1 + K(z) cos(λz) dz, I s = K(z) sin(λz) dz.
–∞ –∞

∞
24. y(x) – K(x – t)y(t) dt = f(x).
–∞
Here –∞ < x < ∞, f(x) ∈ L 1 (–∞, ∞), and K(x) ∈ L 1 (–∞, ∞).
For the integral equation to be solvable (in L 1 ), it is necessary and sufficient that
√
1 – 2π K(u) ≠ 0, –∞ < u < ∞, (1)

∞
1 –iux
where K(u)= √ K(x)e dx is the Fourier transform of K(x). In this case, the

2π
–∞
equation has a unique solution, which is given by

∞
y(x)= f(x)+ R(x – t)f(t) dt,
–∞

1 ∞ iux K(u)

R(x)= √ R(u)e du, R(u)= √ .

2π –∞ 1 – 2π K(u)

•
Reference: V. A. Ditkin and A. P. Prudnikov (1965).

∞
25. y(x) – K(x – t)y(t) dt = f(x).
0
The Wiener–Hopf equation of the second kind.*
Here 0 ≤ x < ∞, K(x) ∈ L 1 (–∞, ∞), f(x) ∈ L 1 (0, ∞), and y(x) ∈ L 1 (0, ∞).
For the integral equation to be solvable, it is necessary and sufficient that
ˇ
Ω(u)=1 – K(u) ≠ 0, –∞ < u < ∞, (1)
∞
ˇ
where K(u)= K(x)e iux dx is the Fourier transform (in the asymmetric form) of K(x).
–∞
In this case, the index of the equation can be introduced,
1
∞
ν = –ind Ω(u)= – arg Ω(u) .
2π –∞
1 . Solution with ν =0:
◦

∞
y(x)= f(x)+ R(x, t)f(t) dt,
0
* A comprehensive discussion of this equation is given in Subsection 11.9-1, Section 11.10, and Section 11.11.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 318