Page 281 - Handbook Of Integral Equations

b
43. y(x)+ A |x – t| 2n+1 y(t) dt = f(x), n = 0,1,2, ...
a
Let us remove the modulus in the integrand:

x b

y(x)+ A (x – t) 2n+1 y(t) dt + A (t – x) 2n+1 y(t) dt = f(x). (1)
a x
The k-fold differentiation of (1) with respect to x yields

x b

(k) 2n+1–k k 2n+1–k (k)
y (x)+ AB k (x – t) y(t) dt +(–1) AB k (t – x) y(t) dt = f x (x),
x
a x (2)
B k =(2n + 1)(2n) ... (2n +2 – k), k =1, 2, ... ,2n +1.
Differentiating (2) with k =2n + 1, we arrive at the following linear nonhomogeneous
differential equation with constant coefficients for y = y(x):
y (2n+2) + 2(2n + 1)! Ay = f x (2n+2) (x). (3)
x

Equation (3) must satisfy the initial conditions which can be obtained by setting x = a in (1)
and (2):

b
y(a)+ A (t – a) 2n+1 y(t) dt = f(a),
a
b (4)
(k) k 2n+1–k (k)
y (a)+(–1) AB k (t – a) y(t) dt = f (a), k =1, 2, ... ,2n +1.
x x
a
These conditions can be reduced to a more habitual form containing no integrals. To this end,
y must be expressed from equation (3) in terms of y (2n+2) and f (2n+2) and substituted into (4),
x x
and then one must integrate the resulting expressions by parts (sufficiently many times).

4.1-5. Kernels Containing Rational Functions

b 1 1
44. y(x) – λ + y(t) dt = f(x).
a x t
This is a special case of equation 4.9.2 with g(x)=1/x.
Solution:

A 1
y(x)= f(x)+ λ + A 2 ,
x
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.2.

b 1 1

45. y(x) – λ – y(t) dt = f(x).
a x t
This is a special case of equation 4.9.3 with g(x)=1/x.
Solution:
A 1
y(x)= f(x)+ λ + A 2 ,
x
where A 1 and A 2 are the constants determined by the formulas presented in 4.9.3.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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