Page 307 - Handbook Of Integral Equations

2
2 .For λ(2A + λ)= –k < 0, the general solution of equation (1) is given by
◦
2Aλ x
y(x)= C 1 cosh(kx)+ C 2 sinh(kx)+ f(x) – sinh[k(x – t)] f(t) dt, (3)
k a

where C 1 and C 2 are arbitrary constants.
2
For λ(2A + λ)= k > 0, the general solution of equation (1) is given by
x
2Aλ
y(x)= C 1 cos(kx)+ C 2 sin(kx)+ f(x) – sin[k(x – t)] f(t) dt. (4)
k a
For λ =2A, the general solution of equation (1) is given by

x
y(x)= C 1 + C 2 x + f(x)+4A 2 (x – t)f(t) dt. (5)
a
The constants C 1 and C 2 in solutions (3)–(5) are determined by conditions (2).

b
30. y(x)+ A t sin(λ|x – t|)y(t) dt = f(x).
a
This is a special case of equation 4.9.39 with g(t)= At. The solution of the integral equation
can be written via the Bessel functions (or modified Bessel functions) of order 1/3.

b
3
31. y(x)+ A sin (λ|x – t|)y(t) dt = f(x).
a
3
1
Using the formula sin β = – sin 3β + 3 sin β, we arrive at an equation of the form 4.5.32
4 4
with n =2:
b

1
3
y(x)+ – A sin(3λ|x – t|)+ A sin(λ|x – t|) y(t) dt = f(x).
4 4
a
n
b
32. y(x)+ A k sin(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)= sin(λ k |x – t|)y(t) dt = sin[λ k (x – t)]y(t) dt + sin[λ k (t – x)]y(t) dt. (1)
a a x
Differentiating (1) with respect to x twice yields
x b

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

I =2λ k y(x) – λ 2 k sin[λ k (x – t)]y(t) dt – λ 2 k sin[λ k (t – x)]y(t) dt,
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
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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