Appendix: All Solutions to the Problems
817
In the found problem, there are
F y =
y
1 + y 2 + z 2
, F z =
z
1 + y 2 + z 2
F y y =
1 + z 2
(1 + y 2 + z 2 )
3
2
, F y z =
−y z
(1 + y 2 + z 2 )
3
2
= F z y , F z z =
1 + y 2
(1 + y 2 + z 2 )
3
2
where, y
=
y 1
x 1
, z
=
z 1
x 1
.
The Legendre condition are
F y y =
1 + z
2
(1 + y 2 + z 2 )
3
2
> 0,
F y y F y z
F z y F z z
=
(1 + y
2
)(1 + z
2
) − y
2 z
2
(1 + y 2 + z 2 ) 3
=
1
(1 + y 2 + z 2 ) 2 > 0
It follows that the functional can obtain a weak minimum on the extremal curves
y =
y 1
x 1
x and z =
z 1
x 1
x.
Chapter 4: Solutions to the Problems in Problems
with Variable Boundaries
4.1 Find the first variation of the functional J [y] =
1
0 y
3 y
2 dx, y(0) = 1.
Solution The first variation of the functional is
δ J =
1
0
(3y
2 y
2
δy + 2y
3 y
δy
)dx
=
1
0
[3y
2 y
2
δy − (2y
3 y
)
δy]dx + 2y
3 y
δy
1
0
=
1
0
[3y
2 y
2
− (2y
3 y
)
]δydx + 2y
3
(1)y
(1)δy(1)
4.2 Let the functional J [y] =
l
0
1
2
E I y
2
− qy
dx, where, E I = c, q is the given
function of x, y(0) = y(l) = 0. Derive the Euler equation and the natural boundary
condition by the variational equation.
Solution The first variation of the functional is
δ J =
l
0
(E I y
δy
− qδy)dx = 0