Page 251 - Handbook Of Integral Equations

2 2
a β β +1 4x t
y(t) dt
9. F , , µ; 2 β = f(x).
2
2 2
2
0 2 2 (x + t ) (x + t )
Here 0 < a ≤ ∞,0 < β < µ < β + 1, and F(a, b, c; z) is the hypergeometric function.
Solution: a
x 2µ–2 d tg(t) dt
y(x)= ,
2 µ–β
2
Γ(1 + β – µ) dx (t – x )
x
2 Γ(β) sin[(β – µ)π] 1–2β d t s 2µ–1 f(s) ds
g(t)= t .
2 µ–β
2
πΓ(µ) dt 0 (t – s )
If a = ∞ and f(x) is a differentiable function, then the solution can be represented in the
form
2 2
2µ
d ∞ (xt) f (t) β 1 – β 4x t
t
y(x)= A F µ – , µ + , µ +1; dt,
2 2µ–β
2 2
2
2
dt 0 (x + t ) 2 2 (x + t )
Γ(β) Γ(2µ – β) sin[(β – µ)π]
where A = .
πΓ(µ) Γ(1 + µ)
•
Reference: P. P. Zabreyko, A. I. Koshelev, et al. (1975).
3.8. Equations Whose Kernels Contain Arbitrary
Functions
3.8-1. Equations With Degenerate Kernel
b

1. g 1 (x)h 1 (t)+ g 2 (x)h 2 (t) y(t) dt = f(x).
a
This integral equation has solutions only if its right-hand side is representable in the form
f(x)= A 1 g 1 (x)+ A 2 g 2 (x), A 1 = const, A 2 = const . (1)
In this case, any function y = y(x) satisfying the normalization type conditions
b b

h 1 (t)y(t) dt = A 1 , h 2 (t)y(t) dt = A 2 (2)
a a
is a solution of the integral equation. Otherwise, the equation has no solutions.
n
b
2. g k (x)h k (t) y(t) dt = f(x).
a
k=0
This integral equation has solutions only if its right-hand side is representable in the form
n

f(x)= A k g k (x), (1)
k=0
where the A k are some constants. In this case, any function y = y(x) satisfying the normal-
ization type conditions
b

h k (t)y(t) dt = A k (k =1, ... , n) (2)
a
is a solution of the integral equation. Otherwise, the equation has no solutions.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 230