Page 158 - Handbook Of Integral Equations

x

15. y(x) – A sinh(λt)y(t) dt = f(x).
a
This is a special case of equation 2.9.2 with g(x)= A and h(t) = sinh(λt).
Solution:

x
A
y(x)= f(x)+ A sinh(λt)exp cosh(λx) – cosh(λt) f(t) dt.
a λ
x
16. y(x)+ A sinh[λ(x – t)]y(t) dt = f(x).
a
This is a special case of equation 2.9.30 with g(x)= A.
◦
1 . Solution with λ(A – λ)>0:
Aλ x
y(x)= f(x) – sin[k(x – t)]f(t) dt, where k = λ(A – λ).
k a
2 . Solution with λ(A – λ)<0:
◦
Aλ x
y(x)= f(x) – sinh[k(x – t)]f(t) dt, where k = λ(λ – A).
k a

3 . Solution with A = λ:
◦
x

y(x)= f(x) – λ 2 (x – t)f(t) dt.
a
x
3
17. y(x)+ A 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 2.3.18:
4 4
x
1 3
y(x)+ A sinh 3λ(x – t) – A sinh[λ(x – t)] y(t) dt = f(x).
4 4
a
x

18. y(x)+ A 1 sinh[λ 1 (x – t)] + A 2 sinh[λ 2 (x – t)] y(t) dt = f(x).
a
1 . Introduce the notation
◦
x x
I 1 = sinh[λ 1 (x – t)]y(t) dt, I 2 = sinh[λ 2 (x – t)]y(t) dt,
a a
x x
J 1 = cosh[λ 1 (x – t)]y(t) dt, J 2 = cosh[λ 2 (x – t)]y(t) dt.
a a
Successively differentiating the integral equation four times yields (the first line is the original
equation)

y + A 1 I 1 + A 2 I 2 = f, f = f(x), (1)
y + A 1 λ 1 J 1 + A 2 λ 2 J 2 = f , (2)

x
x
2
2

y xx +(A 1 λ 1 + A 2 λ 2 )y + A 1 λ I 1 + A 2 λ I 2 = f , (3)
2
1
xx
3
3
y xxx +(A 1 λ 1 + A 2 λ 2 )y + A 1 λ J 1 + A 2 λ J 2 = f xxx , (4)

2
1
x
3
3
4
4

y xxxx +(A 1 λ 1 + A 2 λ 2 )y +(A 1 λ + A 2 λ )y + A 1 λ I 1 + A 2 λ I 2 = f xxxx . (5)
1
2
1
xx
2
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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