Page 337 - Handbook Of Integral Equations

1 . Solution with λ ≠ λ 1,2 :
◦
y(x)= f(x)+ λ[A 1 g 1 (x)+ A 2 g 2 (x)],

where the constants A 1 and A 2 are given by
f 1 – λ(f 1 s 22 – f 2 s 12 ) f 2 – λ(f 2 s 11 – f 1 s 21 )
A 1 = , A 2 = ,
2
2
(s 11 s 22 – s 12 s 21 )λ – (s 11 + s 22 )λ +1 (s 11 s 22 – s 12 s 21 )λ – (s 11 + s 22 )λ +1
b b
f 1 = f(x)h 1 (x) dx, f 2 = f(x)h 2 (x) dx.
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 – λ 1 s 11 λ 1 s 21
y 1 (x)= g 1 (x)+ g 2 (x)= g 1 (x)+ g 2 (x).
λ 1 s 12 1 – λ 1 s 22
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 ∗ =
◦
s 11 + s 22
is double (there is no double characteristic value provided that s 11 = –s 22 ):
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.
1
2
(a) If s 11 ≠ s 22 , then s 12 = – (s 11 – s 22 ) /s 21 , s 21 ≠ 0, and
4
s 11 – s 22
y ∗ (x)= g 1 (x) – g 2 (x).
2s 12
Note that in this case, s 12 and s 21 have opposite signs.
2
1
(b) If s 11 = s 22 , then, in view of s 12 s 21 = – (s 11 – s 22 ) = 0,wehave
4
g 1 (x) for s 12 ≠ 0 and s 21 =0,

y ∗ (x)= g 2 (x) for s 12 = 0 and s 21 ≠ 0,
C 1 g 1 (x)+ C 2 g 2 (x) for s 12 = 0 and s 21 =0,
where C 1 and C 2 are arbitrary constants.

b
19. y(x) – λ [g(x)+ h(t)] m y(t) dt = f(x), m =1, 2, ...
a
k
k
This is a special case of equation 4.9.20, with g k (x)= g (x), h k (t)= C h m–k (t), and
m
k =1, ... , m.
Solution:
m

k
y(x)= f(x)+ λ A k g (x),
k=0
where the A k are constants that can be determined from 4.9.20.

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 316