Page 329 - 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 – a)
(A ≠ 0) 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.
(a) If A(b – a) – 2Bh 1 ≠ 0, then

A(b – a) – 2Bh 1
y ∗ (x)=1 – x.
2 2
A(b – a ) – 2Bh 2
2
2
(b) If A(b – a) – 2Bh 1 = 0, then, in view of h 0 [A(b – a ) – 2Bh 2 ] = 0, we have

2
2
 1 for h 0 ≠ 0 and A(b – a )=2Bh 2 ,
2
y ∗ (x)= x for h 0 = 0 and A(b – a ) ≠ 2Bh 2 ,
2
 2 2
C 1 + C 2 x for h 0 = 0 and A(b – a )=2Bh 2 ,
where C 1 and C 2 are arbitrary constants.
b
9. y(x) – λ [A +(Bx + Ct)h(t)]y(t) dt = f(x).
a
The characteristic values of the equation:
√
A(b – a)+(C + B)h 1 ± D
λ 1,2 = ,
2
B A(b – a)[2h 1 – (b + a)h 0 ]+2C(h – h 0 h 2 )
1
2
2
2
D =[A(b – a)+(C – B)h 1 ] +2Bh 0 [A(b – a )+2Ch 2 ],
where
b b b
2
h 0 = h(x) dx, h 1 = xh(x) dx, h 2 = x h(x) dx.
a a a
1 . Solution with λ ≠ λ 1,2 :
◦
y(x)= f(x)+ λ(A 1 + A 2 x),
where the constants A 1 and A 2 are given by

2
2
1
A 1 = ∆ –1 f 1 – λ Bf 1 h 1 – Cf 2 h 2 – A(b – a )f 2 ,
2

A 2 = ∆ –1 f 2 – λ A(b – a)f 2 – Bf 1 h 0 + Cf 2 h 1 ,

2 2
1
∆ = B A(b – a) h 1 – (b + a)h 0 + C(h – h 0 h 2 ) λ +[A(b – a)+(B + C)h 1 ]λ +1,
2 1

b b b
f 1 = A f(x) dx + C xf(x)h(x) dx, f 2 = B f(x)h(x) dx.
a a a
◦
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
2 – 2λ 1 [A(b – a)+ Ch 1 ]
y(x)= f(x)+ Cy 1 (x), y 1 (x)=1 + x,

2
2
λ 1 [A(b – a )+2Ch 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 308