Page 326 - Handbook Of Integral Equations

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
f 1 – λ[f 1 s 3 – (b – a)f 2 ] f 2 – λ(f 2 s 1 – f 1 s 2 )
A 1 = , A 2 = ,
2
2
[s 1 s 3 – (b – a)s 2 ]λ – (s 1 + s 3 )λ +1 [s 1 s 3 – (b – a)s 2 ]λ – (s 1 + s 3 )λ +1
b b
f 1 = f(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 s 1
y(x)= f(x)+ Cy 1 (x), y 1 (x)= g(x)+ ,
λ 1 (b – a)
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 ∗ =
s 1 + s 3
is double:
s 1 – s 3
y(x)= f(x)+ Cy ∗ (x), y ∗ (x)= g(x) – .
2(b – a)
Here C is an arbitrary constant and y ∗ (x) is an eigenfunction of the equation corresponding
to λ ∗ .
b
6. y(x) – λ [Ag(x)+ Bg(t)]h(t) y(t) dt = f(x).
a
The characteristic values of the equation:

2 2
(A + B)s 1 ± (A – B) s +4ABs 0 s 2
λ 1,2 = 2 1 ,
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 ) Bf 2 – ABλ(f 2 s 1 – f 1 s 2 )
A 1 = , A 2 = ,
2
2
2
2
AB(s – s 0 s 2 )λ – (A + B)s 1 λ +1 AB(s – s 0 s 2 )λ – (A + B)s 1 λ +1
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 305