8.7 Eigenvalues and Eigenfunctions of Operator Equations
501
r
2
+ r + λ = 0
( 4 )
The two roots of Eq. (4) are r 1,2 = −
1
2
±
1
2
√ −(4λ − 1). Consider three cases
below.
(1) 4λ − 1 < 0 namely λ <
1
4
. At this time the general solution of Eq. (3) is
y(t) = c 1 e
r 1 t
+ c 2 e
−r 2 t
(5)
Consequently the general solution of Eq. (2) is
y(x) = c 1 x
r 1 + c 2 x
−r 2
(6)
According to the boundary conditions y(1) = y(2) = 0, there are
c 1 + c 2 = 0
c 1 2
r 1 + c 2 2
−r 2 = 0
(7)
It can be obtained c 1 = 0, c 2 = 0, and y ≡ 0. This is not the solution to find.
(2) 4λ − 1 = 0 namely λ =
1
4
. At this time the general solution of Eq. (3) is
y(t) = (c 1 + c 2 t)e
−
1
2 t
(8)
Consequently the general solution of Eq. (2) is
y(x) =
c 1 + c 2 ln x
√
x
(9)
According to the boundary conditions y(1) = y(2) = 0, there are
⎧
⎨
⎩
c 1 + 0 = 0
1
√
2
(c 1 + c 2 ln 2) = 0
(10)
It can be obtained c 1 = 0, c 2 = 0, and y ≡ 0. This is not the solution to find
either.
(3) 4λ − 1 > 0 namely λ >
1
4
. At this time the general solution of Eq. (3) is
y(t) = e
−
1
2 t
c 1 cos
√
4λ − 1
2
t + c 2 sin
√
4λ − 1
2
t
(11)