3.5 Sufficient Conditions of Extrema of Functionals
219
J =
x 1
x 0
E(x, y, y
, p)dx
It is observed from the above expression that if E(x, y, y
, p) ≥ 0 (or ≤ 0), then
J ≥ 0 (or ≤ 0). Thus according to the definition of the strong minimum (or strong
maximum), that is proved. Quod erat demonstrandum.
Note that the three conditions in Theorems 3.5.1 and 3.5.2 are not only the
sufficient conditions of extremum of the functional, but also the necessary conditions.
Theorem 3.5.3 If the curve y = y(x) (x 0 ≤ x ≤ x 1 ) is the extremal curve of the
functional (3.2.1) satisfying the boundary condition (3.2.2), and the extremal curve is
included in the extremal curve field with the domain D, in D for the extremal function
p = p(x, y) and an arbitrary value y
, the Weierstrass condition, namely E-function
does not change its sign, then the functional (3.2.1) satisfying the boundary condition
(3.2.2) obtains an absolute extremum on the extremal curve y = y(x). When E ≥ 0,
it obtains an absolute minimum; When E ≤ 0, it obtains an absolute maximum.
Proof According to the expression (3.3.10), the increment of the functional J [y(x)]
on the extremal curve y = (x) can be expressed as
J =
x 1
x 0
E(x, y, y
, p)dx
That shows that if E(x, y, y
, p) ≥ 0 (or ≤ 0), then J ≥ 0 (or ≤ 0). Therefore
according to the definition of the absolute minimum (or absolute maximum), that is
proved. Quod erat demonstrandum.
Example 3.5.1 Discuss the extremal situation of the functional J [y] =
1
0 (y
3
+ ay
)dx, where, a is an arbitrary real number, the boundary conditions are
y(0) = 0, y(1) = 1.
Solution Since the integrand is only the function of y
, so the general solution of the
Euler equation is y = c 1 x + c 2 . From the boundary conditions y(0) = 0, y(1) = 1,
to solve for c 2 = 0, c 1 = 1, thus the extremal curve is y = x. The extremal curve of
y = cx in the closed interval [0, 1] forms a central field taking the coordinate origin
(0, 0) as the center, and y = x is located in the extremal curve field. On the extremal
curve, the slope of the field is p = 1. The Weierstrass function is
E = y
3
+ ay
− ( p
3
+ ap) − (y
− p)(3 p
2
+ a) = (y
− p)
2
(y
+ 2 p)
If y
gets a value near p, then E ≥ 0, according to Theorem 3.5.1, the functional
can obtain a weak minimum on the extremal curve y = x. If y
is an arbitrary value,
then (y
+ 2 p) can have an arbitrary sign, namely E-function cannot maintain fixed
sign, so the sufficient condition of strong maximum is not satisfied, and the sufficient
condition is also necessary. Therefore,on the extremal curve y = x, the functional
can not get the strong extremum.
219
J =
x 1
x 0
E(x, y, y
, p)dx
It is observed from the above expression that if E(x, y, y
, p) ≥ 0 (or ≤ 0), then
J ≥ 0 (or ≤ 0). Thus according to the definition of the strong minimum (or strong
maximum), that is proved. Quod erat demonstrandum.
Note that the three conditions in Theorems 3.5.1 and 3.5.2 are not only the
sufficient conditions of extremum of the functional, but also the necessary conditions.
Theorem 3.5.3 If the curve y = y(x) (x 0 ≤ x ≤ x 1 ) is the extremal curve of the
functional (3.2.1) satisfying the boundary condition (3.2.2), and the extremal curve is
included in the extremal curve field with the domain D, in D for the extremal function
p = p(x, y) and an arbitrary value y
, the Weierstrass condition, namely E-function
does not change its sign, then the functional (3.2.1) satisfying the boundary condition
(3.2.2) obtains an absolute extremum on the extremal curve y = y(x). When E ≥ 0,
it obtains an absolute minimum; When E ≤ 0, it obtains an absolute maximum.
Proof According to the expression (3.3.10), the increment of the functional J [y(x)]
on the extremal curve y = (x) can be expressed as
J =
x 1
x 0
E(x, y, y
, p)dx
That shows that if E(x, y, y
, p) ≥ 0 (or ≤ 0), then J ≥ 0 (or ≤ 0). Therefore
according to the definition of the absolute minimum (or absolute maximum), that is
proved. Quod erat demonstrandum.
Example 3.5.1 Discuss the extremal situation of the functional J [y] =
1
0 (y
3
+ ay
)dx, where, a is an arbitrary real number, the boundary conditions are
y(0) = 0, y(1) = 1.
Solution Since the integrand is only the function of y
, so the general solution of the
Euler equation is y = c 1 x + c 2 . From the boundary conditions y(0) = 0, y(1) = 1,
to solve for c 2 = 0, c 1 = 1, thus the extremal curve is y = x. The extremal curve of
y = cx in the closed interval [0, 1] forms a central field taking the coordinate origin
(0, 0) as the center, and y = x is located in the extremal curve field. On the extremal
curve, the slope of the field is p = 1. The Weierstrass function is
E = y
3
+ ay
− ( p
3
+ ap) − (y
− p)(3 p
2
+ a) = (y
− p)
2
(y
+ 2 p)
If y
gets a value near p, then E ≥ 0, according to Theorem 3.5.1, the functional
can obtain a weak minimum on the extremal curve y = x. If y
is an arbitrary value,
then (y
+ 2 p) can have an arbitrary sign, namely E-function cannot maintain fixed
sign, so the sufficient condition of strong maximum is not satisfied, and the sufficient
condition is also necessary. Therefore,on the extremal curve y = x, the functional
can not get the strong extremum.