Page 225 - Handbook Of Integral Equations

1
k
m
27. |x – t | y(t) dt = f(x), k >0, m >0.
0
The transformation
1–m
k
m
z = x , τ = t , w(τ)= τ m y(t)
leads to an equation of the form 3.1.1:
1

|z – τ|w(τ) dτ = F(z), F(z)= mf(z 1/k ).
0
b

28. |x – t| 1+λ y(t) dt = f(x), 0 ≤ λ <1.
a
For λ = 0, see equation 3.1.2. Assume that 0 < λ <1.
◦
1 . Let us remove the modulus in the integrand:
x b

(x – t) 1+λ y(t) dt + (t – x) 1+λ y(t) dt = f(x). (1)
a x
Let us differentiate (1) with respect to x twice and then divide both the sides by λ(λ + 1). As
a result, we obtain
x b
1

(x – t) λ–1 y(t) dt + (t – x) λ–1 y(t) dt = f (x). (2)
xx
a x λ(λ +1)
Rewrite equation (2) in the form
b
y(t) dt 1

|x – t| k = f (x), k =1 – λ. (3)
xx
a λ(λ +1)
See 3.1.29 and 3.1.30 for the solutions of equation (3) for various a and b.
2 . The right-hand side f(x) of the integral equation must satisfy certain relations. By setting
◦
x = a and x = b in (1), we obtain two corollaries
b b
(t – a) 1+λ y(t) dt = f(a), (b – t) 1+λ y(t) dt = f(b). (4)
a a
On substituting the solution y(x) of (3) into (4) and then integrating by parts, we obtain the
desired constraints for f(x).
a
y(t)

29. k dt = f(x), 0 < k <1, 0 < a ≤ ∞.
0 |x – t|
◦
1 . Solution:
1–2k

k–1 d a t 2 dt t f(s) ds
y(x)= –Ax 2 ,
dx x (t – x) 2 0 s 2 (t – s) 2
1–k
1–k
1–k
1 πk 1+ k
–2
A = cos Γ(k) Γ ,
2π 2 2
where Γ(k) is the gamma function.
2
2
◦
2 . The transformation x = z , t = ξ , w(ξ)=2ξy(t) leads to an equation of the form 3.1.31:
√
a
w(ξ)

2
|z – ξ | dξ = f z .
2 k
2
0
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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