Page 269 - Handbook Of Integral Equations

b
2. y(x) – λ (x + t)y(t) dt = f(x).
a
The characteristic values of the equation:

2
2
2
2
6(b + a)+4 3(a + ab + b ) 6(b + a) – 4 3(a + ab + b )
λ 1 = , λ 2 = .
(a – b) 3 (a – b) 3
1 . Solution with λ ≠ λ 1,2 :
◦
y(x)= f(x)+ λ(A 1 x + A 2 ),
where
12f 1 – 6λ(f 1 ∆ 2 – 2f 2 ∆ 1 ) 12f 2 – 2λ(3f 2 ∆ 2 – 2f 1 ∆ 3 )
A 1 = , A 2 = ,
2
2
12 – 12λ∆ 2 – λ ∆ 4 12 – 12λ∆ 2 – λ ∆ 4
1 1
b b

n
n
f 1 = f(x) dx, f 2 = xf(x) dx, ∆ n = b – a .
a a
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
◦
y(x)= f(x)+ Cy 1 (x),
where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
to the characteristic value λ 1 :
1 b + a
y 1 (x)= x + – .
λ 1 (b – a) 2
3 . 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.
◦
4 . The equation has no multiple characteristic values.
b
3. y(x) – λ (Ax + Bt)y(t) dt = f(x).
a
The characteristic values of the equation:

2
2
2
2
3(A + B)(b + a) ± 9(A – B) (b + a) +48AB(a + ab + b )
λ 1,2 = .
AB(a – b) 3
1 . Solution with λ ≠ λ 1,2 :
◦
y(x)= f(x)+ λ(A 1 x + A 2 ),
where the constants A 1 and A 2 are given by
12Af 1 – 6ABλ(f 1 ∆ 2 – 2f 2 ∆ 1 ) 12Bf 2 – 2ABλ(3f 2 ∆ 2 – 2f 1 ∆ 3 )
A 1 = , A 2 = ,
2
2
12 – 6(A + B)λ∆ 2 – ABλ ∆ 4 12 – 6(A + B)λ∆ 2 – ABλ ∆ 4
1 1

b b
n
n
f 1 = f(x) dx, f 2 = xf(x) dx, ∆ n = b – a .
a a
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
◦
y(x)= f(x)+ Cy 1 (x),
where C is an arbitrary constant and y 1 (x) is an eigenfunction of the equation corresponding
to the characteristic value λ 1 :
1 b + a
y 1 (x)= x + – .
λ 1 A(b – a) 2
◦
◦
3 . 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.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 248