906
Appendix: All Solutions to the Problems
y(x) = c 1 cos(
√
1 − λx) + c 2 sin(
√
1 − λx)
(4)
According to the boundary conditions y(π) = y(2π) = 0, there are
c 1 cos(π
√
1 − λ) + c 2 sin(π
√
1 − λ) = 0
c 1 cos(2π
√
1 − λ) + c 2 sin(2π
√
1 − λ) = 0
(5)
If Eq. (5) have the nonzero solution, then the coefficient determinant of c 1 and c 2
should be zero, namely
cos(π
√
1 − λ) sin(π
√
1 − λ)
cos(2π
√
1 − λ) sin(2π
√
1 − λ)
= sin(π
√
1 − λ) = 0
( 6 )
As a result π
√
1 − λ = nπ. The characteristic values are λ n = 1 − n
2 , where
n = 1, 2, . . .. Substituting the characteristic values into Eq. (5), we get c 1 = 0.
Substituting these relations into Eq. (4), the characteristic function of the given
problem can be obtained
y n = c 2 sin nx
(7)
where, the constant c 2 can be determined according to the normal characteristic
function namely the isoperimetric condition of the problem, there is
2π
π
c
2
2 sin
2
(nx)dx =
1
2
c
2
2
2π
π
[1 − cos(2nx)]dx =
π
2
c
2
2 = 1
( 8 )
Getting c 2 = ±
2
π
. Therefore the characteristic functions are
y n = ±
2
π
sin(nx) (n = 1, 2, . . .)
(9)
8.21 Find the characteristic value and characteristic function of the functional J [y] =
1
0 [3y
2
− (x + 1)
2 y
2
]dx, such that it conforms to the boundary conditions y(0) =
y(1) = 0 and the isoperimetric condition
1
0 y
2 dx = 1.
Solution The Sturm-Liouville corresponding to the functional is
3y + 2(x + 1)y
+ (x + 1)
2 y
= λy
(1)
or
(x + 1)
2 y
+ 2(x + 1)y
+ (3 − λ)y = 0
( 2 )