Appendix: All Solutions to the Problems
805
Solution F = x
2 y
2
+ 12y
2 , the Euler equation of the functional is
24y − 4x y
− 2x
2 y
= 0
or
x
2 y
+ 2x y
− 12y = 0
This is the Euler equation, let x = e
T , the original equation is transformed to
D(D − 1)y + 2Dy − 12y = 0
Its characteristic equation is
r (r − 1) + 2r − 12 = 0
Solve for two roots r 1 = 3, r 2 = −4.
The general solution of the equation is
y = c 1 e
3t
+ c 2 e
−4t
= c 1 x
3
+ c 2 x
−4
From the boundary conditions y(1) = 1, y(2) = 0, we get c 1 = 1, c 2 = 0, thus
the extremal curve is y = x
3 . y = c 1 x
3 forms the extremal curve field in the interval
[1, 2], and y = x
3 is located in the extremal curve field.
The general solution of the Jacobi equation is u = d 1 x
3
+ d 2 x
−4 . From the
boundary conditions u(1) = 0, u
(1) = 1, we get d 1 =
1
12
= −d 2 , hence the
solution of the Jacobi equation is u =
1
12
(x
3
− x
−4
). It has zero only at point x 0 = 1,
thus the extremal curve of the functional satisfies the Jacobi strong condition.
The Weierstrass function is
E = x
2
(y
+ u)
2
+ 12y
2
− x
2 y
2
− 12y
2
− u2x
2 y
= x
2 u
2
> 0
Therefore the extremal curve y = x
3 can reach the strong minimum.
3.10 J [y] =
x 1
x 0
1+y
2
y 2 dx, y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution It follows from the problem 2.48 that the family of extremals y =
sinh(c 1 x + c 2 ). From the boundary conditions y(x 0 ) = y 0 , y(x 1 ) = y 1 , we get
c 1 =
arcsinhy 1 −arcsinhy 0
x 1 −x 0
, c 2 =
x 1 arcsinhy 0 −x 0 arcsinhy 1
x 1 −x 0
, so the extremal curve is
y = sinh
arcsinhy 1 − arcsinhy 0
x 1 − x 0
x +
x 1 arcsinhy 0 − x 0 arcsinhy 1
x 1 − x 0
The Weierstrass function is