832
Appendix: All Solutions to the Problems
Solution Let F = y
3 , the integrand only depends on y
, the general solution of the
Euler equation of the functional is
y = c 1 x + c 2
From y(0) = 0, y(x 1 ) = y 1 , we get c 2 = 0, c 1 =
y 1
x 1
, therefore the extremal curve
is y =
y 1
x 1
x. There is not the corner point.
4.20 For the extremal problem of the functional J [y] =
x 1
0 (y
4
− 6y
2
)dx, is there
any solution with the corner point? The boundary conditions are y(0) = 0, y(x 1 ) =
y 1 .
Solution Because F y y = 12y
2
− 12 can be equal to zero, the extremal curve can
have the corner point. The Euler equation of the functional is
4y
2 y
− 12y
= 4y
(y
+
√
3)(y
−
√
3) = 0
The equation has three general solutions
y = c 1 x + c 2 , y = −
√
3x + c 3 , y =
√
3x + c 4
F y = 4y
3
− 12y
= 4y
(y
+
√
3)(y
−
√
3)
F − y
F y = y
4
− 6y
2
− 4y
4
+ 12y
2
= −3y
2
(y
+
√
2)(y
−
√
2)
Let the abscissa of the broken point be x c , then from the broken conditions there
is
4y
(y
+
√
3)(y
−
√
3)
x=x c −0
= 4y
(y
+
√
3)(y
−
√
3)
x=x c +0
−3y
2
(y
+
√
2)(y
−
√
2)
x=x c −0
= −3y
2
(y
+
√
2)(y
−
√
2)
x=x c +0
In order to make the above two equations simultaneously hold, there should choose
A
y
(x c − 0) =
√
3
y
(x c + 0) = −
√
3
B
y
(x c − 0) = −
√
3
y
(x c + 0) =
√
3
For A, there is
y =
√
3x + c 4 (0 ≤ x ≤ x c )
−
√
3x + c 3 (x c ≤ x ≤ x 1 )