816
Appendix: All Solutions to the Problems
Solution The Euler equation of the functional is
y
= 0
Integrating two times, we get y = c 1 x + c 2 , the solution conforming to the
boundary conditions is y = 2x − 1, it is included in the extremal curve field y =
c 1 x + c 2 . For an arbitrary y
, the Legendre condition is
F y y = 2 > 0
Therefore the functional obtains the strong minimum on the extremal curve y =
2x − 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.
Solution Because the functional is only the function of y
, the solution the Euler
equation of the functional is y = c 1 x + c 2 , the solution conforming to the boundary
conditions is y =
y 1
x 1
x, it is included in the central field of extremal curve y = cx
taking the coordinate origin (0,0) as the center, and y
=
y 1
x 1
. The Legendre condition
is
F y y = 4y
2 e
−y
4 (3 − 4y
4
) = 4
y
2
1
x
2
1
e
−
y 4
1
x 4
1
3 − 4
y
4
1
x
4
1
It follows that the sign of F y y depends on the sign of
3 − 4
y
4
1
x
4
1
. On the extremal
curve y =
y 1
x 1
x, if
y 1
x 1
<
4
3
4
, then F y y > 0, the functional can obtain the weak
minimum; If
y 1
x 1
=
4
3
4
, then F y y = 0, the functional can not obtain the weak
extremum; If
y 1
x 1
>
4
3
4
, then F y y < 0, the functional can obtain the weak maximum.
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.
Solution Because the functional is only the function of y
and z
, the solutions of
the Euler equations are
y = c 1 x + c 2
z = c 3 x + c 4
The solutions conforming to the boundary conditions are y =
y 1
x 1
x, z =
z 1
x 1
x, they
are the straight lines through the origin.
Appendix: All Solutions to the Problems
Solution The Euler equation of the functional is
y
= 0
Integrating two times, we get y = c 1 x + c 2 , the solution conforming to the
boundary conditions is y = 2x − 1, it is included in the extremal curve field y =
c 1 x + c 2 . For an arbitrary y
, the Legendre condition is
F y y = 2 > 0
Therefore the functional obtains the strong minimum on the extremal curve y =
2x − 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.
Solution Because the functional is only the function of y
, the solution the Euler
equation of the functional is y = c 1 x + c 2 , the solution conforming to the boundary
conditions is y =
y 1
x 1
x, it is included in the central field of extremal curve y = cx
taking the coordinate origin (0,0) as the center, and y
=
y 1
x 1
. The Legendre condition
is
F y y = 4y
2 e
−y
4 (3 − 4y
4
) = 4
y
2
1
x
2
1
e
−
y 4
1
x 4
1
3 − 4
y
4
1
x
4
1
It follows that the sign of F y y depends on the sign of
3 − 4
y
4
1
x
4
1
. On the extremal
curve y =
y 1
x 1
x, if
y 1
x 1
<
4
3
4
, then F y y > 0, the functional can obtain the weak
minimum; If
y 1
x 1
=
4
3
4
, then F y y = 0, the functional can not obtain the weak
extremum; If
y 1
x 1
>
4
3
4
, then F y y < 0, the functional can obtain the weak maximum.
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.
Solution Because the functional is only the function of y
and z
, the solutions of
the Euler equations are
y = c 1 x + c 2
z = c 3 x + c 4
The solutions conforming to the boundary conditions are y =
y 1
x 1
x, z =
z 1
x 1
x, they
are the straight lines through the origin.