500
8 Direct Methods of Variational Problems
If Eq. (9) have the nonzero solution, then the determinant of coefficients should
be zero, namely
1
0
cos
√ λ − 1 sin
√ λ − 1
= sin
√ λ − 1 − 0 = sin
√
λ − 1 = 0
(10)
Consequently
√ λ − 1 = nπ. The characteristic values are λ n = 1 + n
2
π
2 , where
n = 1, 2, . . .. Obviously, it follows from the first equation of Eq. (9) that c 1 = 0.
Substituting these relations into Eq. (8), the characteristic function of the given
problem can be obtained
y n = c 2 sin nπx
(11)
where, c 2 can be determined by the formal characteristic function namely the
isoperimetric condition, there is
1
0
(c 2 sin nπx)
2 dx =
1
2
c
2
2
1
0
(1 − cos 2nπx)dx =
1
2
c
2
2 = 1
(12)
It can be obtained c 2 = ±
√
2. Thus the characteristic functions are
y n = ±
√
2 sin nπx (n = 1, 2, . . .)
(13)
Example 8.7.4 Find the characteristic values and characteristic functions of the
functional J [y] =
2
1 x
2 y
2 dx, such that it conforms to the boundary conditions
y(1) = y(2) = 0 and the isoperimetric condition
2
1 y
2 dx = 1.
Solution The Sturm–Liouville equation corresponding to the functional is
−2x y
− x
2 y
= λy
(1)
or
x
2 y
+ 2x y
+ λy = 0
( 2 )
Making substitutions, let x = e
t , then there is dx = e
t dt = xdt,
dt
dx
=
1
x
,
y
=
dy
dx
=
dy
dt
dt
dx
=
1
x
dy
dt
, y
=
d
dx
dy
dx
= −
1
x 2
dy
dt
+
1
x 2
d
2 y
dt 2 . If the derivative of y with
respect to x is expressed by the derivative of y with respect to t, then Eq. (2) can be
transformed into the second order linear equation with constant coefficients
d
2 y
dt 2 +
dy
dt
+ λy = 0
( 3 )
Its characteristic equation is
8 Direct Methods of Variational Problems
If Eq. (9) have the nonzero solution, then the determinant of coefficients should
be zero, namely
1
0
cos
√ λ − 1 sin
√ λ − 1
= sin
√ λ − 1 − 0 = sin
√
λ − 1 = 0
(10)
Consequently
√ λ − 1 = nπ. The characteristic values are λ n = 1 + n
2
π
2 , where
n = 1, 2, . . .. Obviously, it follows from the first equation of Eq. (9) that c 1 = 0.
Substituting these relations into Eq. (8), the characteristic function of the given
problem can be obtained
y n = c 2 sin nπx
(11)
where, c 2 can be determined by the formal characteristic function namely the
isoperimetric condition, there is
1
0
(c 2 sin nπx)
2 dx =
1
2
c
2
2
1
0
(1 − cos 2nπx)dx =
1
2
c
2
2 = 1
(12)
It can be obtained c 2 = ±
√
2. Thus the characteristic functions are
y n = ±
√
2 sin nπx (n = 1, 2, . . .)
(13)
Example 8.7.4 Find the characteristic values and characteristic functions of the
functional J [y] =
2
1 x
2 y
2 dx, such that it conforms to the boundary conditions
y(1) = y(2) = 0 and the isoperimetric condition
2
1 y
2 dx = 1.
Solution The Sturm–Liouville equation corresponding to the functional is
−2x y
− x
2 y
= λy
(1)
or
x
2 y
+ 2x y
+ λy = 0
( 2 )
Making substitutions, let x = e
t , then there is dx = e
t dt = xdt,
dt
dx
=
1
x
,
y
=
dy
dx
=
dy
dt
dt
dx
=
1
x
dy
dt
, y
=
d
dx
dy
dx
= −
1
x 2
dy
dt
+
1
x 2
d
2 y
dt 2 . If the derivative of y with
respect to x is expressed by the derivative of y with respect to t, then Eq. (2) can be
transformed into the second order linear equation with constant coefficients
d
2 y
dt 2 +
dy
dt
+ λy = 0
( 3 )
Its characteristic equation is