8.7 Eigenvalues and Eigenfunctions of Operator Equations
499
y − y
= λy
(1)
or
y
+ (λ − 1)y = 0
( 2 )
The characteristic equation of Eq. (2) is
r
2
+ λ − 1 = 0
( 3 )
Equation (3) has the two roots r 1,2 = ±
√ −(λ − 1). Consider three cases below.
(1) λ − 1 < 0 namely λ < 1. At this time the general solution of Eq. (2) is
y = c 1 e
r 1 x
+ c 2 e
−r 2 x
= c 1 e
√
1−λx
+ c 2 e
−
√
1−λx
(4)
According to the boundary conditions y(0) = y(1) = 0, there are
c 1 + c 2 = 0
c 1 e
r 1 + c 2 e
−r 2 = 0
(5)
It can be got c 1 = 0, c 2 = 0, and y ≡ 0. This is not the solution to find.
(2) λ − 1 = 0 namely λ = 1. At this time the general solution of Eq. (2) is
y = c 1 + c 2 x
(6)
According to the boundary conditions y(0) = y(1) = 0, there are
c 1 = 0
c 1 + c 2 = 0
(7)
It can be got c 1 = 0, c 2 = 0, and y ≡ 0. This is not the solution to find either.
(3) λ − 1 > 0 namely λ > 1. At this time the general solution of Eq. (2) is
y = c 1 cos(
√ λ − 1x) + c 2 sin(
√ λ − 1x)
(8)
According to the boundary conditions y(0) = y(1) = 0, there are
c 1 + 0 = 0
c 1 cos
√ λ − 1 + c 2 sin
√ λ − 1 = 0
(9)