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3 Sufficient Conditions of Extrema of Functionals
Solution The Euler equation of the functional is
−2y
3
− 4y
3
− 12x y
2 y
+ 6y
3
+ 12yy
y
= 0
Simplifying it, we get
y
= 0
The solution of the Euler equation is y = c 1 x + c 2 , the solution conforming to
the boundary conditions is y = x − 1, on the extremal curve, the slope of the field
is p = 1. The extremal curve only has one zero point, it conforms to the Jacobi
condition. Using the Legendre condition, there is
F y y = 12x y
2
− 12yy
= 12(x − y) = 12 > 0
Thus the functional can get a weak extremum on the extremal curve y = x − 1.
Example 3.5.9 Making use of the Legendre condition to discriminate whether the
functional J [y] =
x 1
0 (1 − e
−y
2 )dx has extremum, where a > 0, the boundary
conditions are y(0) = 0, y(x 1 ) = y 1 .
Solution Because the integrand of the functional is only the function of y
, so the
solution of the Euler equation is y = c 1 x + c 2 , the solution conforming to the
boundary conditions are y =
y 1
x 1
x, on the extremal curve, the slope of the field is
p =
y 1
x 1
. It only has one zero point, conforms to the Jacobi condition. Using the
Legendre condition, there is
F y y = 4y
2 e
−y
2 − 2e
−y
2 = 2e
−y
2 (2y
2
− 1) = 2e
−y
2
2
y
2
1
x
2
1
− 1
If |y 1 | >
√
2x 1
2
, then F y y > 0, the functional gets a weak minimum on the extremal
curve y =
y 1
x 1
x. If |y 1 | <
√
2x 1
2
, then F y y < 0, the functional gets a weak maximum
on the extremal curve y =
y 1
x 1
x. If |y 1 | =
√
2x 1
2
, then F y y = 0, the functional can not
get extremum on the extremal curve y =
y 1
x 1
x.
Example 3.5.10 Let the functional J [y] =
1
0 yy
2 dx, the boundary conditions are
y(0) = p > 0, y(1) = q > 0. Discuss the extremal situation.
Solution Because the integrand does not contain x, therefore the first integral of the
Euler equation is
yy
2
= c
2
or
3 Sufficient Conditions of Extrema of Functionals
Solution The Euler equation of the functional is
−2y
3
− 4y
3
− 12x y
2 y
+ 6y
3
+ 12yy
y
= 0
Simplifying it, we get
y
= 0
The solution of the Euler equation is y = c 1 x + c 2 , the solution conforming to
the boundary conditions is y = x − 1, on the extremal curve, the slope of the field
is p = 1. The extremal curve only has one zero point, it conforms to the Jacobi
condition. Using the Legendre condition, there is
F y y = 12x y
2
− 12yy
= 12(x − y) = 12 > 0
Thus the functional can get a weak extremum on the extremal curve y = x − 1.
Example 3.5.9 Making use of the Legendre condition to discriminate whether the
functional J [y] =
x 1
0 (1 − e
−y
2 )dx has extremum, where a > 0, the boundary
conditions are y(0) = 0, y(x 1 ) = y 1 .
Solution Because the integrand of the functional is only the function of y
, so the
solution of the Euler equation is y = c 1 x + c 2 , the solution conforming to the
boundary conditions are y =
y 1
x 1
x, on the extremal curve, the slope of the field is
p =
y 1
x 1
. It only has one zero point, conforms to the Jacobi condition. Using the
Legendre condition, there is
F y y = 4y
2 e
−y
2 − 2e
−y
2 = 2e
−y
2 (2y
2
− 1) = 2e
−y
2
2
y
2
1
x
2
1
− 1
If |y 1 | >
√
2x 1
2
, then F y y > 0, the functional gets a weak minimum on the extremal
curve y =
y 1
x 1
x. If |y 1 | <
√
2x 1
2
, then F y y < 0, the functional gets a weak maximum
on the extremal curve y =
y 1
x 1
x. If |y 1 | =
√
2x 1
2
, then F y y = 0, the functional can not
get extremum on the extremal curve y =
y 1
x 1
x.
Example 3.5.10 Let the functional J [y] =
1
0 yy
2 dx, the boundary conditions are
y(0) = p > 0, y(1) = q > 0. Discuss the extremal situation.
Solution Because the integrand does not contain x, therefore the first integral of the
Euler equation is
yy
2
= c
2
or