Page 171 - Handbook Of Integral Equations

x

16. y(x)+ A sin[λ(x – t)]y(t) dt = f(x).
a
This is a special case of equation 2.9.36 with g(t)= 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 sin [λ(x – t)]y(t) dt = f(x).
a
1
3
Using the formula sin β = – sin 3β + 3 sin β, we arrive at an equation of the form 2.5.18:
4 4
x
1 3
y(x)+ – A sin[3λ(x – t)] + A sin[λ(x – t)] y(t) dt = f(x).
4 4
a
x

18. y(x)+ A 1 sin[λ 1 (x – t)] + A 2 sin[λ 2 (x – t)] y(t) dt = f(x).
a
This equation can be solved by the same method as equation 2.3.18, by reducing it to a
fourth-order linear ordinary differential equation with constant coefficients.
Consider the characteristic equation
2
2
2 2
2
2
2
z +(λ + λ + A 1 λ 1 + A 2 λ 2 )z + λ λ + A 1 λ 1 λ + A 2 λ λ 2 = 0, (1)
1 2 1 2 2 1
whose roots, z 1 and z 2 , determine the solution structure of the integral equation.
Assume that the discriminant of equation (1) is positive:
2 2
2
D ≡ (A 1 λ 1 – A 2 λ 2 + λ – λ ) +4A 1 A 2 λ 1 λ 2 >0.
1
2
In this case, the quadratic equation (1) has the real (different) roots
√ √
2
1
2
2
1
2
z 1 = – (λ + λ + A 1 λ 1 + A 2 λ 2 )+ 1 D, z 2 = – (λ + λ + A 1 λ 1 + A 2 λ 2 ) – 1 D.
2 1 2 2 2 1 2 2
Depending on the signs of z 1 and z 2 the following three cases are possible.
Case 1.If z 1 > 0 and z 2 > 0, then the solution of the integral equation has the form
(i = 1, 2):
x

√

y(x)= f(x)+ {B 1 sinh[µ 1 (x – t)] + B 2 sinh µ 2 (x – t) f(t) dt, µ i = z i ,
a
where the coefficients B 1 and B 2 are determined from the following system of linear algebraic
equations:
B 1 µ 1 B 2 µ 2 B 1 µ 1 B 2 µ 2
+ – 1=0, + – 1=0.
2
2
2
2
λ + µ 2 λ + µ 2 λ + µ 2 λ + µ 2
1 1 1 2 2 1 2 2
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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