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Appendix: All Solutions to the Problems
E =
1 + y
2
(y + u) 2 −
1 + y
2
y 2 + 2u
1 + y
2
y 3 =
u
2
(1 + y
2
)(3y
+ 2u)
(y + u) 2 y 3
It follows that when (3y
+ 2u)/y
3 is great than zero and less than zero respectively, E function changes its sign, so the family of extremals can not get the strong
extremum. But F y y =
6(1+y
2 )
y 4
> 0, thus the family of extremals can reach the weak
minimum.
3.11 J [y] =
1
0 (y
2
+ y
2
+ 2ye
2x
)dx, y(0) =
1
3
, y(1) =
1
3
e
2 .
Solution F = y
2
+ y
2
+ 2ye
2x , the Euler equation of the functional is
2y + 2e
2x
− 2y
= 0
or
y
− y = e
2x
The general solution of the equation is y =
1
3
e
2x
+ c 1 e
x
+ c 2 e
−x , from the
boundary conditions y(0) =
1
3
, y(1) =
1
3
e
2 , we get c 1 = c 2 = 0, the extremal curve
is y =
1
3
e
2x .
The general solution of the Jacobi equation is
u = d 1 e
x
+ d 2 e
−x
From the boundary conditions u(0) = 0, u
(0) = 1, we get c 1 =
1
2
= −c 2 , thus
the Jacobi equation is
u =
1
2
(e
x
− e
−x
) = sinh x
Because u has not other zero point except at point x 0 = 0, the extremal function
of the functional satisfies the Jacobi strong condition. The Weierstrass function is
E = (y
+ u)
2
+ y
2
+ 2ye
2x
− (y
2
+ y
2
+ 2ye
2x
) − u2y
= u
2
> 0
Therefore the extremal curve y =
1
3
e
2x reaches the strong minimum.
3.12 J [y] =
π
4
0 (y
2
− y
2
+ 6y sin 2x)dx, y(0) = 0, y
π
4
= 1.
Solution F = y
2
− y
2
+ 6y sin 2x, the Euler equation of the functional is
2y + 6 sin 2x + 2y
= 0