Page 235 - Handbook Of Integral Equations

b
6. A sinh λ|x – t| + B sinh µ|x – t| y(t) dt = f(x), –∞ < a < b < ∞.
a
Let us remove the modulus in the integrand and differentiate the equation with respect to x
twice to obtain
b

2 2

2(Aλ + Bµ)y(x)+ Aλ sinh λ|x – t| + Bµ sinh µ|x – t| y(t) dt = f (x), (1)
xx
a

Eliminating the integral term with sinh µ|x – t| from (1) yields
b
2
2
2

2(Aλ + Bµ)y(x)+ A(λ – µ ) sinh λ|x – t| y(t) dt = f (x) – µ f(x). (2)
xx
a
For Aλ+Bµ = 0, this is an equation of the form 3.3.5, and for Aλ+Bµ ≠ 0, this is an equation
of the form 4.3.26.
The right-hand side f(x) must satisfy certain relations, which can be obtained by setting
x = a and x = b in the original equation (a similar procedure is used in 3.3.5).
b

7. sinh(λx) – sinh(λt) y(t) dt = f(x).

a
This is a special case of equation 3.8.3 with g(x) = sinh(λx).
Solution:

1 d f (x)
x
y(x)= .
2λ dx cosh(λx)
The right-hand side f(x) of the integral equation must satisfy certain relations (see item 2 ◦
of equation 3.8.3).
a

8. sinh(βx) – sinh(µt) y(t) dt = f(x), β >0, µ >0.

0
This is a special case of equation 3.8.4 with g(x) = sinh(βx) and λ = µ/β.
b

3
9. sinh λ|x – t| y(t) dt = f(x).
a
3 1 3
Using the formula sinh β = sinh 3β – sinh β, we arrive at an equation of the form 3.3.6:
4 4
b
1 3

4
4 A sinh 3λ|x – t| – A sinh λ|x – t| y(t) dt = f(x).
a
n
b

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

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

k
a x
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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