Page 643 - Handbook Of Integral Equations

12.4-3. An Equation of the First Kind on a Finite Interval

Consider the singular integral equation of the first kind

1 b ϕ(t)
dt = f(x), a ≤ x ≤ b, (7)
π a t – x
on a finite interval. Its solutions can be constructed by using the theory of the Riemann boundary
value problem for a nonclosed contour (see Subsection 12.3-11). Let us present the final results.
1 . A solution that is unbounded at both endpoints:
◦
b √

1 1 (t – a)(b – t)
ϕ(x)= – √ f(t) dt + C , (8)
π (x – a)(b – x) a t – x
where C is an arbitrary constant and
b
ϕ(t) dt = C. (9)
a
2 . A solution bounded at the endpoint a and unbounded at the endpoint b:
◦
b
1 x – a b – t f(t)
ϕ(x)= – dt. (10)
π b – x t – a t – x
a
3 . A solution bounded at both endpoints:
◦
b

1 f(t) dt
ϕ(x)= – (x – a)(b – x) √ , (11)
π a (t – a)(b – t) t – x
under the condition that
b
f(t) dt

√ = 0. (12)
a (t – a)(b – t)
Solutions that have a singularity point s inside the interval [a, b] can also be constructed. These
solutions have the following form:
4 . A singular solution that is unbounded at both endpoints:
◦
√
b

1 1 (t – a)(b – t) C 2
ϕ(x)= – √ f(t) dt + C 1 + , (13)
π (x – a)(b – x) a t – x x – s
where C 1 and C 2 are arbitrary constants.
5 . A singular solution bounded at one endpoint:
◦
1 b b – t f(t) C
ϕ(x)= – (x – a)(b – x) dt + , (14)
π a t – a t – x x – s

where C is an arbitrary constants.
◦
6 . A singular solution bounded at both endpoints:
b 1
1 f(t) dt A f(t) dt

ϕ(x)= – (x–a)(b–x) √ + , A = √ . (15)
π a (t–a)(b–t) t–x x–s –1 (t–a)(b–t)

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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