2.7 Variational Problems Depending on Higher Order Derivatives
155
Example 2.7.6 Determine the extremal curve of the functional J [y] =
l
−l
1
2
μy
2
+ ρy
dx, it satisfies the boundary conditions y(−l) = 0, y
(−l) = 0,
y(l) = 0, y
(l) = 0, where both μ and ρ are constants.
Solution Because the integrand F =
1
2
μy
2
+ ρy does not contain y
, the EulerPoisson equation is
F y +
d
2
dx 2 F y = ρ + μy
(4)
= 0 or y
(4)
= −
ρ
μ
Integrating the above equation four times, we obtain
y = −
ρ
24μ
x
4
+ c 1 x
3
+ c 2 x
2
+ c 3 x + c 4
Making use of the boundary conditions, the last we obtain
y = −
ρ
24μ
x
4
− 2l
2 x
2
+ l
4
= −
ρ
24μ
x
2
− l
2
2
Example 2.7.7 Prove: If F in the functional J [y] =
x 1
x 0
F(x, y, y
, y
)dx does not
explicitly contain y, then the Euler equation of the functional has the first integral
F y −
d
dx
F y = C (constant)
If F does not explicitly contain x, then the Euler equation of the functional has
the first integral
F − y
F y −
d
dx
F y
− y
F y = C
Proof The Euler equation of the functional is
F y −
d
dx
F y +
d
2
dx 2 F y = 0
If F does not explicitly contain y, then there is F y = 0, so the above equation can
be written as
d
dx
F y −
d
2
dx 2 F y = 0
Integrating the above equation once, obviously there is
F y −
d
dx
F y = C