Appendix: All Solutions to the Problems
803
Therefore the extremal curve y = 1 −
x
2
4
taking the strong minimum.
3.4 J [y] =
x 1
0 (y
2
+ 2yy
− 16y
2
)dx, x 1 > 0, y(0) = 0, y(x 1 ) = 0.
Solution It can be seen from the problem 2.47, the solution is y = c 1 cos 4x +
c 2 sin 4x. From the boundary condition y(0) = 0, we get c 1 = 0. Moreover from
y(x 1 ) = 0, we get x 1 =
π
4
or c 2 = 0, we get the extremal curve y = 0. Therefore,
y = c 2 sin 4x forms a extremal curve field taking coordinate origin (0,0) as the
center in the interval
0,
π
4
, and it is around the extremal curve y = 0. furthermore
F y y = 2 > 0 identically holds for an arbitrary y
and the neighborhood of the
extremal curve y = 0, thus it can be seen from the sufficient condition of extremal
value that if 0 < x 1 <
π
4
, then y = c 2 sin 4x reaches the strong minimum. If x 1 >
π
4
,
then there is no the minimum.
3.5 J [y] =
2
−1 y
(1 + x
2 y
)dx, y(−1) = 1, y(2) = 4.
Solution Let the integrand F = y
(1 + x
2 y
), it follows from the problem 2.28 that
the general is y =
c 1
x
+ c 2 , from the boundary conditions y(−1) = 1, y(2) = 4, we
get c 1 = 2, c 2 = 3, y =
2
x
+ 3. Because y
= −
2
x 2 does not exist at point x = 0, it
can not reach the extremal value in the interval [−1, 2].
3.6 J [y] =
2
1 y
(1 + x
2 y
)dx, y(1) = 3, y(2) = 5.
Solution It can be seen from the problem 2.28 or problem 3.5 that the general
solution is y =
c 1
x
+ c 2 , from the boundary conditions y(1) = 3, y(2) = 5, we solve
and get c 1 = −4, c 2 = 7, thus y = 7 −
4
x
, 1 ≤ x ≤ 2. y =
c 1
x
+ c 2 forms the extremal
curve field in the interval [1, 2], and y = 7 −
4
x
is in the extremal curve field.
The equation of the Jacobi equation is
u =
d 1
x
+ d 2
From the Boundary conditions u(1) = 0, u
(1) = 1, solve for d 1 = 1, d 2 = −1.
So the solution of the Jacobi equation is
u =
1
x
− 1
It follows that because u has not other zero except at point x 0 = 1, the extremal
function the functional satisfies the Jacobi strong condition.
Furthermore F y = 1 + 2x
2 y
, the Weierstrass function is
E = (y
+ u)[1 + x
2
(y
+ u)] − y
(1 + x
2 y
) − u(1 + 2x
2 y
) = x
2 u
2
> 0
For an arbitrary y
, there is E > 0, thus the extremal curve y = 7 −
4
x
can reach
the strong minimum. Of course it can also get the weak minimum.
803
Therefore the extremal curve y = 1 −
x
2
4
taking the strong minimum.
3.4 J [y] =
x 1
0 (y
2
+ 2yy
− 16y
2
)dx, x 1 > 0, y(0) = 0, y(x 1 ) = 0.
Solution It can be seen from the problem 2.47, the solution is y = c 1 cos 4x +
c 2 sin 4x. From the boundary condition y(0) = 0, we get c 1 = 0. Moreover from
y(x 1 ) = 0, we get x 1 =
π
4
or c 2 = 0, we get the extremal curve y = 0. Therefore,
y = c 2 sin 4x forms a extremal curve field taking coordinate origin (0,0) as the
center in the interval
0,
π
4
, and it is around the extremal curve y = 0. furthermore
F y y = 2 > 0 identically holds for an arbitrary y
and the neighborhood of the
extremal curve y = 0, thus it can be seen from the sufficient condition of extremal
value that if 0 < x 1 <
π
4
, then y = c 2 sin 4x reaches the strong minimum. If x 1 >
π
4
,
then there is no the minimum.
3.5 J [y] =
2
−1 y
(1 + x
2 y
)dx, y(−1) = 1, y(2) = 4.
Solution Let the integrand F = y
(1 + x
2 y
), it follows from the problem 2.28 that
the general is y =
c 1
x
+ c 2 , from the boundary conditions y(−1) = 1, y(2) = 4, we
get c 1 = 2, c 2 = 3, y =
2
x
+ 3. Because y
= −
2
x 2 does not exist at point x = 0, it
can not reach the extremal value in the interval [−1, 2].
3.6 J [y] =
2
1 y
(1 + x
2 y
)dx, y(1) = 3, y(2) = 5.
Solution It can be seen from the problem 2.28 or problem 3.5 that the general
solution is y =
c 1
x
+ c 2 , from the boundary conditions y(1) = 3, y(2) = 5, we solve
and get c 1 = −4, c 2 = 7, thus y = 7 −
4
x
, 1 ≤ x ≤ 2. y =
c 1
x
+ c 2 forms the extremal
curve field in the interval [1, 2], and y = 7 −
4
x
is in the extremal curve field.
The equation of the Jacobi equation is
u =
d 1
x
+ d 2
From the Boundary conditions u(1) = 0, u
(1) = 1, solve for d 1 = 1, d 2 = −1.
So the solution of the Jacobi equation is
u =
1
x
− 1
It follows that because u has not other zero except at point x 0 = 1, the extremal
function the functional satisfies the Jacobi strong condition.
Furthermore F y = 1 + 2x
2 y
, the Weierstrass function is
E = (y
+ u)[1 + x
2
(y
+ u)] − y
(1 + x
2 y
) − u(1 + 2x
2 y
) = x
2 u
2
> 0
For an arbitrary y
, there is E > 0, thus the extremal curve y = 7 −
4
x
can reach
the strong minimum. Of course it can also get the weak minimum.