3.5 Sufficient Conditions of Extrema of Functionals
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It is included in the center field of extremal curve of y = 2 ln(cx + 1). The
Weierstrass function is
E = e
y y
2
− e
y p
2
− (y
− p)2e
y p = e
y
(y
− p)
2
That shows that for any y
, there is E ≥ 0. According to Theorem 3.5.2, the
functional gets a strong minimum on the extremal curve y = 2 ln(x + 1).
Example 3.5.4 Let the functional J [y] =
1
0 (y
2
+ yy
3
)dx, the boundary conditions are y(0) = 0, y(1) = 0. Discuss its extremal situation.
Solution Because F = y
2
+ yy
3 does not contain x, so the Euler equation of the
functional has the first integral
y
2
+ yy
3
− y
(2y
+ 3yy
2
) = −c
or
y
2
(1 + 2yy
) = c
Solve for
y = c 1 x + c 2
y
2
= c 3 x + c 4
From the boundary conditions y(0) = 0, y(1) = 0, we get c 1 = c 2 = c 3 = c 4 =
0. so y = 0.
The Weierstrass function is
E = (y
+ u)
2
+ y(y
+ u)
3
− (y
2
+ yy
3
) − u(2y
+ 3yy
2
) = u
2
≥ 0
Although E ≥ 0, but because the extremal curve y = 0 does not satisfy the Jacobi
strong condition, therefore the functional only can get a weak minimum on y = 0.
It is observed from this example that on the extremal curve, for any y
, even if the
Weierstrass function E ≥ 0, it is not necessarily to be able to draw the conclusion
of the strong extremum existing.
Example 3.5.5 Let the functional J [y] =
2
1 (x
2 y
2
+ 6y
2
+ 12y)dx, the boundary
conditions are y(1) = 0, y(2) = 3. Discuss its extremal situation.
Solution F = x
2 y
2
+ 6y
2
+ 12y, the Euler equation of the functional is
12y + 12 − 4x y
− 2x
2 y
= 0
Making a replacement x = e
t , the original equation is changed into