238
3 Sufficient Conditions of Extrema of Functionals
satisfies the fixed boundary condition u| Γ = g(x, y), g ∈ C(Γ ), where Γ
is the closed boundary curve of D, D = D + Γ ; p ∈ C
1
(D), f ∈ C(D),
and p > 0, u ∈ C
2
(D). Prove: u = u(x, y) makes J [u] obtain an absolute
minimum.
3.23 Find the extremal curve of the functional J [y] =
1
2
3
1 (x
2 y
2
+ 4yy
)dx, and
discuss the extremal property, the boundary conditions are y(1) = 0, y(3) = 1.
3.24 Discuss the extremal case of the functional J [y] =
x 1
0 (1 + x)y
2 dx, where
x 1 > 0, the boundary conditions are y(0) = 0, y(x 1 ) = y 1 .
3.25 Discuss the extremal case of the functional J [y] =
π
2
0 (y
2
− y
2
)dx, the
boundary conditions are y(0) = 1, y
π
2
= 1.
3.26 Discuss whether the functional J [y] =
1
0 y
3 dx can obtain a strong
extremum? The boundary conditions are y(0) = 0, y(1) = 1.
3.27 Judge whether the functional J [y] =
1
0 (εy
2
+ y
2
+ x
2
)dx has extremum
for various different parameters ε, the boundary conditions are y(0) = 0,
y(1) = 1.
3.28 Using the Legendre condition to judge whether the functional J [y] =
1
0 (y
2
+ x
2
)dx has extremum, the boundary conditions are y(0) = −1,
y(1) = 1.
3.29 Using the Legendre condition to judge whether the functional J [y] =
x 1
0 (1 − e
−y
4 )dx has extremum, the boundary conditions are y(0) = 0,
y(x 1 ) = y 1 , where, x 1 > 0, y 1 > 0.
3.30 Verify the extremum of the functional J [y, z] =
x 1
0
1 + y 2 + z 2 dx, the
boundary conditions are y(0) = 0, y(x 1 ) = y 1 , z(0) = 0, z(x 1 ) = z 1 , where,
x 1 > 0, y 1 > 0, z 1 > 0.