Appendix: All Solutions to the Problems
831
y =
x + c 2 (0 ≤ x ≤ x c )
−x + c 2
(x c ≤ x ≤ 4)
From y(0) = 0, we get c 2 = 0; From y(4) = 2, we get c 2
= 6. Therefore there
is
y =
x
(0 ≤ x ≤ x c )
−x + 6 (x c ≤ x ≤ 4)
Because the two extremal curves intersect at the broken point, namely x = −x +6,
we get x = x 1 = 3, so the piecewise extremal curve is
y =
x
(0 ≤ x ≤ 3)
−x + 6 (3 ≤ x ≤ 4)
Similarly, for B, the piecewise extremal curve is
y =
−x (0 ≤ x ≤ 1)
x − 2 (1 ≤ x ≤ 4)
The functional reaches the absolute minimum on the whole broken curve.
4.18 For the extremal problem of the functional J [y] =
x 1
x 0
(y
2
+ 2x y − y
2
)dx, is
there any solution with the corner point? The boundary conditions are y(x 0 ) = y 0 ,
y(x 1 ) = y 1 .
Solution Let F = y
2
+ 2x y − y
2 , the Euler equation of the functional is
2x − 2y − 2y
= 0
or
y
+ y = x
The solution of the Euler equation is
y = c 1 cos x + c 2 sin x + x
It follows that y is a smooth continuous function, there is not the corner point.
In fact, since F y y = 2 > 0, in this problem it can only reach the extremum on the
smooth curve.
4.19 For the extremal problem of the functional J [y] =
x 1
0 y
3 dx, is there any
solution with the corner point? The boundary conditions are y(0) = 0, y(x 1 ) = y 1 .