Page 489 - Handbook Of Integral Equations

Example 2. Let us apply the operator method (for n = 2) to solve the generalized Abel equation with exponent 3/4:
x y(t) dt
y(x) – b = f(x). (17)
(x – t) 3/4
0
We first consider the integral operator with difference kernel
x
L [y(x)] ≡ K(x – t)y(t) dt.
0
2
Let us find L :
x t

2

L [y] ≡ L L [y] = K(x – t)K(t – s)y(s) ds dt
0 0
x x x
= y(s) ds K(x – t)K(t – s) dt = K 2 (x – s)y(s) ds, (18)
0 s 0
z
K 2 (z)= K(ξ)K(z – ξ) dξ.
0
In the proof of this formula, we have reversed the order of integration and performed the change of variables ξ = t – s.
For the power-law kernel
µ
K(ξ)= bξ ,
we have
2
Γ (1 + µ)
K 2 (z)= b 2 z 1+2µ . (19)
Γ(2+2µ)
For Eq. (17) we obtain
3 1 b 2
2 1
µ = – , K 2 (z)= A √ , A = √ Γ ( ).
4 z π 4
Therefore, the auxiliary equation (14) corresponding to n = 2 has the form
x y(t) dt

y(x) – A √ = Φ(x), (20)
0 x – t
where
x f(t) dt
Φ(x)= f(x)+ b 3/4 .
0 (x – t)
After the substitution A → –λ and Φ → f, relation (20) coincides with Eq. (6), and the solution of Eq. (20) can be obtained
by formula (12).
Remark. It follows from (19) that the solution of the generalized Abel equation with exponent β
x
y(t) dt
y(x)+ λ β = f(x)
0 (x – t)
can be reduced to the solution of a similar equation with the different exponent β 1 =2β – 1. In
particular, the Abel equation (6), which corresponds to β = 1 , is reduced to the solution of an
2
equation with degenerate kernel for β 1 =0.
9.4-3. A Method for Solving “Quadratic” Operator Equations
Suppose that the solution of the linear (integral, differential, etc.) equation

y(x) – λL [y]= f(x) (21)
is known for an arbitrary right-hand side f(x) and for any λ from the interval (λ min , λ max ). We
denote this solution by
y = Y (f, λ). (22)

Let us construct the solution of the more complicated equation
2
y(x) – aL [y] – bL [y]= f(x), (23)

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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