Page 328 - 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)s 1 + Cs 0
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.
2
(a) If As 1 – (Bs 1 + Cs 0 ) ≠ 0, then 4As 0 (Bs 2 + Cs 1 )= –[(A – B)s 1 – Cs 0 ] and
(A – B)s 1 – Cs 0
y ∗ (x)= g(x) – .
2As 0
2
(b) If (A – B)s 1 = Cs 0 , then, in view of 4As 0 (Bs 2 + Cs 1 )= –[(A – B)s 1 – Cs 0 ] =0, we
have
g(x) for s 0 ≠ 0 and Bs 2 = –Cs 1 ,

y ∗ (x)= 1 for s 0 = 0 and Bs 2 ≠ –Cs 1 ,
for s 0 = 0 and Bs 2 = –Cs 1 ,
C 1 g(x)+ C 2
where C 1 and C 2 are arbitrary constants.

b

8. y(x) – λ [A + B(x – t)h(t)]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 = ,
2
B A(b – a)[2h 1 – (b + a)h 0 ] – 2B(h – h 0 h 2 )
1
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

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

f 1 = A f(x) dx – B 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) – Bh 1 ]
y(x)= f(x)+ Cy 1 (x), y 1 (x)=1 + x,
2
2
λ 1 [A(b – a ) – 2Bh 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 307