Page 272 - Handbook Of Integral Equations

b
2
2
8. y(x) – λ (x – t )y(t) dt = f(x).
a
The characteristic values of the equation:
1
.
λ 1,2 = ±
1
1 (b – a ) – (b – a )(b – a)
5
3 2
5
3
9 5
◦
1 . Solution with λ ≠ λ 1,2 :
2
y(x)= f(x)+ λ(A 1 x + A 2 ),
where the constants A 1 and A 2 are given by

f 1 + λ 1 f –f 2 + λ 1 f 1
5
A 1 = 3 1 ∆ 3 – f 2 ∆ 1 , A 2 = 3 2 ∆ 3 – f 1 ∆ 5 ,

1
1
λ 2 1 ∆ 1 ∆ 5 – ∆ 2 +1 λ 2 1 ∆ 1 ∆ 5 – ∆ 2 +1
5 9 2 5 9 2

b b
2
n
n
f 1 = f(x) dx, f 2 = x f(x) dx, ∆ n = b – a .
a a
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
◦
3
3
3 – λ 1 (b – a )
2
y(x)= f(x)+ Cy 1 (x), y 1 (x)= x + ,
3λ 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.
4 . The equation has no multiple characteristic values.
◦
b
2
2
9. y(x) – λ (Ax + Bt )y(t) dt = f(x).
a
The characteristic values of the equation:

1 1 2 2 4
3
3 (A + B)∆ 3 ± 9 (A – B) ∆ + AB∆ 1 ∆ 5 n n
5
λ 1,2 = 1 1 , ∆ n = b – a .
2
2AB ∆ – ∆ 1 ∆ 5
9 3 5
1 . Solution with λ ≠ λ 1,2 :
◦
2
y(x)= f(x)+ λ(A 1 x + A 2 ),
where the constants A 1 and A 2 are given by
1
f
Af 1 – ABλ 3 1 ∆ 3 – f 2 ∆ 1
A 1 = 1 2 1 1 ,
ABλ 2 ∆ – ∆ 1 ∆ 5 – (A + B)λ∆ 3 +1
9 3 5 3
1 1
Bf 2 – ABλ 3 2 ∆ 3 – f 1 ∆ 5
f
5
A 2 = 1 2 1 1 ,
ABλ 2 ∆ – ∆ 1 ∆ 5 – (A + B)λ∆ 3 +1
9 3 5 3
b b
2
f 1 = f(x) dx, f 2 = x f(x) dx.
a a
◦
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
3
3
3 – λ 1 A(b – a )
2
y(x)= f(x)+ Cy 1 (x), y 1 (x)= x + ,
3λ 1 A(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 .
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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