Page 330 - 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.
4 . Solution with λ = λ 1,2 = λ ∗ and f 1 = f 2 = 0, where the characteristic value λ ∗ =
◦
2
is double:
A(b – a)+(B + C)h 1
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)+(C – B)h 1 ≠ 0, then
A(b – a)+(C – B)h 1
y ∗ (x)=1 – x.
2 2
A(b – a )+2Ch 2
2
2
(b) If A(b – a)+(C – B)h 1 = 0, then, in view of h 0 [A(b – a )+2Ch 2 ] = 0,wehave

2
2
 1 for h 0 ≠ 0 and A(b – a )= –2Ch 2 ,
2
y ∗ (x)= x for h 0 = 0 and A(b – a ) ≠ –2Ch 2 ,
2
 2 2
C 1 + C 2 x for h 0 = 0 and A(b – a )= –2Ch 2 ,

where C 1 and C 2 are arbitrary constants.

b
10. y(x) – λ [A + B(x – t)h(x)]y(t) dt = f(x).
a
The characteristic values of the equation:

2
2
2
A(b – a) ± [A(b – a)+2Bh 1 ] – 2Bh 0 [A(b – a )+2Bh 2 ]
λ 1,2 = ,
B{h 0 [A(b – a )+2Bh 2 ] – 2h 1 [A(b – a)+ Bh 1 ]}
2
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 AE 1 +(BE 1 x + E 2 )h(x) ,
where the constants E 1 and E 2 are given by

E 1 = ∆ –1 f 1 + λB(f 1 h 1 – f 2 h 0 ) ,

2
2
E 2 = B∆ –1 –f 2 + λf 2 A(b – a)+ Bh 1 + λf 1 2 1 A(b – a )+ Bh 2 ,

1 2 2 2
∆ = B h 0 A(b – a )+ Bh 2 – h 1 [A(b – a)+ Bh 1 ] λ – A(b – a)λ +1,
2
b b
f 1 = f(x) dx, f 2 = xf(x) dx.
a a
2 . Solution with λ = λ 1 ≠ λ 2 and f 1 = f 2 =0:
◦
1 – λ 1 [A(b – a)+ Bh 1 ]
y(x)= f(x)+ Cy 1 (x), y 1 (x)= A + Bxh(x)+ h(x),
λ 1 h 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 309