Page 470 - Handbook Of Integral Equations

or, in the detailed notation,

1 f(a) x f (t)

t
y(x)= + dt . (25)
Γ(µ)Γ(1 – µ) (x – a) 1–µ (x – t) 1–µ
a
Taking into account the relation
1 sin(πµ)
= ,
Γ(µ)Γ(1 – µ) π
we now arrive at the solution of the generalized Abel equation in the form
sin(πµ) f(a) x f (t) dt

t
y(x)= + , (26)
π (x – a) 1–µ a (x – t) 1–µ
which coincides with that obtained above in Subsection 8.4-4.
•
References for Section 8.5: K. B. Oldham and J. Spanier (1974), Yu. I. Babenko (1986), S. G. Samko, A. A. Kilbas, and
O. I. Marichev (1993).

8.6. Equations With Weakly Singular Kernel

8.6-1. A Method of Transformation of the Kernel
Consider the Volterra integral equation of the first kind with polar kernel
L(x, t)
K(x, t)= , 0 < α < 1. (1)
(x – t) α

The integral equation in question can be represented in the form
x
L(x, t)

y(t) dt = f(x), (2)
(x – t) α
0
where we assume that the functions L(x, t) and ∂L(x, t)/∂x are continuous and bounded. To solve
Eq. (2), we multiply it by dx/(ξ – x) 1–α and integrate from 0 to ξ, thus obtaining
ξ
L(x, t) dx f(x) dx
x ξ
y(t) dt = .
(x – t) α (ξ – x) 1–α (ξ – x) 1–α
0 0 0
By setting
ξ
L(x, t) dx
K (ξ, t)= 1–α α ,
∗
t (ξ – x) (x – t)
ξ
f(x) dx
ϕ(ξ)= 1–α , ϕ(0) = 0,
0 (ξ – x)
we obtain another integral equation of the first kind with the unknown function y(t):
ξ
∗
K (ξ, t)y(t) dt = ϕ(ξ), (3)
0
in which the kernel K (ξ, t) has no singularities.
∗
It can be shown that any solution of Eq. (3) is a solution of Eq. (2). Thus, after transforming
Eq. (2) to the form (3), we can apply any methods available for continuous kernels to the latter
equation.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 452