Page 343 - Handbook Of Integral Equations

∞ 1 x
b
34. y(x) – K y(t) dt = Ax .
0 t t
A solution:
∞
1
A b b–1
y(x)= x , B =1 – K ξ dξ.
B 0 ξ
It is assumed that the improper integral is convergent and B ≠ 0. The general solution of
the integral equations is the sum of the above solution and the solution of the homogeneous
equation 4.9.33.

∞ 1 x
35. y(x) – K y(t) dt = f(x).
0 t t
The solution can be obtained with the aid of the inverse Mellin transform:

c+i∞
1 f(s) –s

y(x)= x ds,
2πi c–i∞ 1 – K(s)

where f and K stand for the Mellin transforms of the right-hand side and the kernel of the

integral equation,

∞ ∞
f(s)= f(x)x s–1 dx, K(s)= K(x)x s–1 dx.

0 0
Example. For f(x)= Ae –λx and K(x)= 1 –x
e , the solution of the integral equation has the form
2

4A
 for λx >1,

 3
 (3 – 2C)(λx)
y(x)= ∞
1
 –2A for λx <1.


s
 (λx) k ψ(s k )
k=1
Here C = 0.5772 ... is the Euler constant, ψ(z) = [ln Γ(z)] z is the logarithmic derivative of the gamma function,

and the s k are the negative roots of the transcendental equation Γ(s k ) = 2, where Γ(z) is the gamma function.
•
Reference: M. L. Krasnov, A. I. Kisilev, and G. I. Makarenko (1971).
b
36. y(x)+ |x – t|g(t)y(t) dt = f(x), a ≤ x ≤ b.
a
1 . Let us remove the modulus in the integrand,
◦
x b
y(x)+ (x – t)g(t)y(t) dt + (t – x)g(t)y(t) dt = f(x). (1)
a x
Differentiating (1) with respect to x yields

x b

y (x)+ g(t)y(t) dt – g(t)y(t) dt = f (x). (2)
x
x
a x
Differentiating (2), we arrive at a second-order ordinary differential equation for y = y(x),

y +2g(x)y = f (x). (3)
xx
xx
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 322