Page 327 - Handbook Of Integral Equations

3 . The solution with λ = λ 2 ≠ λ 1 and f 1 = f 2 = 0 is given by the formulas of item 2 in
◦
◦
which one must replace λ 1 and y 1 (x)by λ 2 and y 2 (x), respectively.
2
◦
4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where the characteristic value λ ∗ =
(A + B)s 1
is double:
y(x)= f(x)+ Cy ∗ (x),
where C is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding
to λ ∗ . Two cases are possible.
2 2
(a) If A ≠ B, then 4ABs 0 s 2 = –(A – B) s and
1
(A – B)s 1
y ∗ (x)= g(x) – .
2As 0
2 2
(b) If A = B, then, in view of 4ABs 0 s 2 = –(A – B) s = 0,wehave
1
g(x) for s 0 ≠ 0 and s 2 =0,

y ∗ (x)= 1 for s 0 = 0 and s 2 ≠ 0,
for s 0 = s 2 =0,
C 1 g(x)+ C 2
where C 1 and C 2 are arbitrary constants.

b

7. y(x) – λ [Ag(x)+ Bg(t)+ C]h(t) y(t) dt = f(x).
a
The characteristic values of the equation:

2 2 2 2
(A + B)s 1 + Cs 0 ± (A – B) s +2(A + B)Cs 1 s 0 + C s +4ABs 0 s 2
0
1
λ 1,2 = ,
2
2AB(s – s 0 s 2 )
1
where
b b b

2
s 0 = h(x) dx, s 1 = g(x)h(x) dx, s 2 = g (x)h(x) dx.
a a a
◦
1 . Solution with λ ≠ λ 1,2 :
y(x)= f(x)+ λ[A 1 g(x)+ A 2 ],
where the constants A 1 and A 2 are given by
Af 1 – ABλ(f 1 s 1 – f 2 s 0 )
A 1 = ,
2
2
AB(s – s 0 s 2 )λ – [(A + B)s 1 + Cs 0 ]λ +1
1
C 1 f 1 + Bf 2 – ABλ(f 2 s 1 – f 1 s 2 )
A 2 = 2 ,
2
AB(s – s 0 s 2 )λ – [(A + B)s 1 + Cs 0 ]λ +1
1
b b
f 1 = f(x)h(x) dx, f 2 = f(x)g(x)h(x) dx.
a a
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
◦
1 – λ 1 As 1
y(x)= f(x)+ Cy 1 (x), y 1 (x)= g(x)+ ,

λ 1 As 0
where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding

to the characteristic value λ 1 .

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 306