Appendix: All Solutions to the Problems
905
If Eq. (7) have the nonzero solution, then the coefficient determinant should be
zero, namely
1
0
cos(
1
2
√
2λ−25)
√
2
sin(
1
2
√
2λ−25)
√
2
=
sin
1
2
√
2λ − 25
√
2
= 0
( 8 )
Consequently
1
2
√
2λ − 25 = nπ. The characteristic values are λ n =
25
2
+
2n
2
π
2
=
25+4n
2 π
2
2
, where n = 1, 2, . . .. Obviously, it follows from the first equation
of Eq. (7) that c 1 = 0. Substituting these relations into Eq. (6), the characteristic
functions of the given problem can be obtained
y n =
c 2 sin(nπ ln x)
√
x
(9)
where, the constant c 2 can be determined according to the normal characteristic
function namely the isoperimetric condition of the problem, there is
e
1
c 2 sin(nπ ln x)
√
x
2
dx =
1
2
c
2
2
e
1
1 − cos(nπ ln x)
x
dx =
1
2
c
2
2 = 1 (10)
Getting c 2 = ±
√
2. Thus the characteristic functions are
y n = ±
√
2 sin(nπ ln x)
√
x
(n = 1, 2, . . .)
(11)
8.20 Find the characteristic value and characteristic function of the functional J [y] =
2π
π (y
2
− y
2
)dx, such that it conforms to the boundary conditions y(π) = y(2π) =
0 and the isoperimetric condition
2π
π y
2 dx = 1.
Solution The Sturm-Liouville corresponding to the functional is
y + y
= λy
(1)
or
y
+ (1 − λ)y = 0
( 2 )
The characteristic equation of Eq. (2) is
r
2
+ 1 − λ = 0
( 3 )
The two roots of Eq. (3) are r 1,2 = ±
√ −(1 − λ). When 1 − λ > 0 namely λ < 1,
the general solution of Eq. (2) is