Page 159 - Handbook Of Integral Equations

Eliminating I 1 and I 2 from (1), (3), and (5), we arrive at a fourth-order linear ordinary
differential equation with constant coefficients:
2
2
2
2
2 2
y xxxx – (λ + λ – A 1 λ 1 – A 2 λ 2 )y +(λ λ – A 1 λ 1 λ – A 2 λ λ 2 )y =

2
1 2
2
xx
1
1
(6)
2 2
2
2

f xxxx – (λ + λ )f xx + λ λ f.
2
1 2
1
The initial conditions can be obtained by setting x = a in (1)–(4):
y(a)= f(a), y (a)= f (a),

x
x

y (a)= f (a) – (A 1 λ 1 + A 2 λ 2 )f(a), (7)
xx
xx

y xxx (a)= f xxx (a) – (A 1 λ 1 + A 2 λ 2 )f (a).
x
On solving the differential equation (6) under conditions (7), we thus find the solution of the
integral equation.
2 . 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, (8)
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 (8) 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 (8) has the real (different) roots
√ √
2
1
2
1
2
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
2
2
2
2
2
2
2
2
λ 1 (µ – λ ) λ 2 (µ – λ ) λ 1 (µ – λ ) λ 2 (µ – λ )
1 2 1 1 2 2 2 1
B 1 = A 1 2 2 + A 2 2 2 , B 2 = A 1 2 2 + A 2 2 2 .
µ 1 (µ – µ ) µ 1 (µ – µ ) µ 2 (µ – µ ) µ 2 (µ – µ )
2 1 2 1 1 2 1 2
Case 2.If z 1 < 0 and z 2 < 0, then the solution of the integral equation has the form
x

y(x)= f(x)+ {B 1 sin[µ 1 (x – t)] + B 2 sin µ 2 (x – t) f(t) dt, µ i = |z i |,
a
where the coefficients B 1 and B 2 are found by solving 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
Case 3.If z 1 > 0 and z 2 < 0, then the solution of the integral equation has the form
x

y(x)= f(x)+ {B 1 sinh[µ 1 (x – t)] + B 2 sin µ 2 (x – t) f(t) dt, µ i = |z i |,
a
where 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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