Page 331 - 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 , then

A(b – a)+2Bh 1
y ∗ (x)= A + Bxh(x) – h(x).
2h 0
2
2
(b) If A(b – a)= –2Bh 1 , then, in view of h 0 [A(b – a )+2Bh 2 ] = 0,wehave

2
2
 A + Bxh(x) for h 0 ≠ 0 and A(b – a )= –2Bh 2 ,
2
2
y ∗ (x)= h(x) for h 0 = 0 and A(b – a ) ≠ –2Bh 2 ,
 2 2
C 1 [A + Bxh(x)] + C 2 h(x) for h 0 = 0 and A(b – a )= –2Bh 2 ,
where C 1 and C 2 are arbitrary constants.

b
11. y(x) – λ [A +(Bx + Ct)h(x)]y(t) dt = f(x).
a
The characteristic values of the equation:
√
A(b – a)+(B + C)h 1 ± D
λ 1,2 = ,
C{2h 1 [A(b – a)+ Bh 1 ] – h 0 [A(b – a )+2Bh 2 ]}
2
2
2
2
2
D =[A(b – a)+(B – C)h 1 ] +2Ch 0 [A(b – a )+2Bh 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
–1
E 1 = ∆ [f 1 – λC(f 1 h 1 – f 2 h 0 )],

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

1 2 2 2
∆ = C h 1 [A(b – a)+ Bh 1 ] – h 0 2 A(b – a )+ Bh 2 λ – [A(b – a)+(B + C)h 1 ]λ +1,
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 310