Page 334 - Handbook Of Integral Equations

where the constants A 1 and A 2 are given by
Af 1 – λAB(f 1 s 3 – f 2 s 2 ) Bf 2 – λAB(f 2 s 1 – f 1 s 2 )
A 1 = 2 , A 2 = 2 ,
2
2
AB(s 1 s 3 – s )λ – (As 1 + Bs 3 )λ +1 AB(s 1 s 3 – s )λ – (As 1 + Bs 3 )λ +1
2 2
b b
f 1 = f(x)g(x) dx, f 2 = f(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)+ h(x),
λ 1 As 2
where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
to the characteristic value λ 1 .
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 ∗ =
As 1 + Bs 3
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 As 1 ≠ Bs 3 , then 4ABs = –(As 1 – Bs 3 ) , AB < 0, and
2
As 1 – Bs 3
y ∗ (x)= g(x) – h(x).
2As 2
(b) If As 1 = Bs 3 , then, in view of s 2 = 0,wehave
y ∗ (x)= C 1 g(x)+ C 2 h(x),
where C 1 and C 2 are arbitrary constants.

b
15. y(x) – λ [g(x)h(t)+ h(x)g(t)]y(t) dt = f(x).
a
The characteristic values of the equation:
1 1
λ 1 = √ , λ 2 = √ ,
s 1 + s 2 s 3 s 1 – s 2 s 3
where
b b b

2
2
s 1 = h(x)g(x) dx, s 2 = h (x) dx, s 3 = g (x) dx.
a a a
1 . Solution with λ ≠ λ 1,2 :
◦
y(x)= f(x)+ λ[A 1 g(x)+ A 2 h(x)],
where the constants A 1 and A 2 are given by
f 1 – λ(f 1 s 1 – f 2 s 2 ) f 2 – λ(f 2 s 1 – f 1 s 3 )
A 1 = , A 2 = ,
2
2
2
2
(s – s 2 s 3 )λ – 2s 1 λ +1 (s – s 2 s 3 )λ – 2s 1 λ +1
1 1
b b

f 1 = f(x)h(x) dx, f 2 = f(x)g(x) dx.
a a
◦
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:

s 3
y(x)= f(x)+ Cy 1 (x), y 1 (x)= g(x)+ h(x),
s 2
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 313