Page 231 - Handbook Of Integral Equations

b
λt
3. |e λx – e | y(t) dt = f(x), λ >0.
a
λx
This is a special case of equation 3.8.3 with g(x)= e .
Solution:
1 d
f (x) .
y(x)= e –λx x
2λ dx
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
a
µt
4. |e βx – e | y(t) dt = f(x), β >0, µ >0.
0
This is a special case of equation 3.8.4 with g(x)= e βx and λ = µ/β.
n
b

5. A k exp λ k |x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
k=1
1 . Let us remove the modulus in the kth summand of the integrand:
◦
b x b

I k (x)= exp λ k |x – t| y(t) dt = exp[λ k (x – t)]y(t) dt + exp[λ k (t – x)]y(t) dt. (1)
a a x
Differentiating (1) with respect to x twice yields
x b

I = λ k exp[λ k (x – t)]y(t) dt – λ k exp[λ k (t – x)]y(t) dt,
k
a x
x b (2)

I =2λ k y(x)+ λ 2 exp[λ k (x – t)]y(t) dt + λ 2 exp[λ k (t – x)]y(t) dt,

k k k
a x
where the primes denote the derivatives with respect to x. By comparing formulas (1) and (2),
we find the relation between I and I k :

k
2

I =2λ k y(x)+ λ I k , I k = I k (x). (3)
k
k
◦
2 . With the aid of (1), the integral equation can be rewritten in the form
n

A k I k = f(x). (4)
k=1
Differentiating (4) with respect to x twice and taking into account (3), we obtain
n n
2

σ 1 y(x)+ A k λ I k = f (x), σ 1 =2 A k λ k . (5)
xx
k
k=1 k=1
Eliminating the integral I n from (4) and (5) yields
n–1
2 2 2

σ 1 y(x)+ A k (λ – λ )I k = f (x) – λ f(x). (6)
n
n
xx
k
k=1
Differentiating (6) with respect to x twice and eliminating I n–1 from the resulting equation
with the aid of (6), we obtain a similar equation whose right-hand side is a second-order
n–2
linear differential operator (acting on y) with constant coefficients plus the sum B k I k .If
k=1
we successively eliminate I n–2 , I n–3 , ... , I 1 with the aid of double differentiation, then we
finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n–1) with
constant coefficients.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 210