4.6 One-Sided Variational Problems
313
y =
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x +
π
3
+
√
3
2
−2 ≤ x ≤ −
π
3
− sin 2x
−
π
3
≤ x ≤ −
π
4
1
−
π
4
≤ x ≤
π
4
sin 2x
π
4
≤ x ≤
π
3
−x +
π
3
+
√
3
2
π
3
≤ x ≤ 2
(9)
Example 4.6.4 Find the extremal curve of the functional J [y]
=
10
0 (2x y
− y
2
)dx, Let the admissible curve pass through fixed points (0,7)
and (10,7), and satisfy the inequality y ≥ x
2 −
x
5
.
Solution The Euler equation of the functional is
d
dx
(2x − 2y
) = 0
( 1 )
The first integral is y
= x + c 1 . Integrating again, we obtain y =
x
2
2
+ c 1 x + c 2 .
From the boundary condition (0,7), we give c 2 = 7, thus y =
x
2
2
+ c 1 x + 7. Let the
extremal curve and inequality be tangent, there is
x
2
2
+ c 1 x + 7 = 2x −
x
2
5
(2)
x + c 1 = 2 −
2x
5
(3)
Solve for x =
√
10 and c 1 = 2 −
7
√
10
5
from Eqs. (2) and (3). Therefore the
extremal curve is
y =
x
2
2
+
2 −
7
√
10
5
x + 7 (0 ≤ x ≤
√
10)
(4)
In addition from the boundary condition (10, 7), c 2 = −43 − 10c 1 can be
determined, thus
y =
x
2
2
+ c 1 x − 10c 1 − 43
(5)
Let the extremal curve and the inequality be contingence, there are
x
2
2
+ c 1 x − 10c 1 − 43 = 2x −
x
2
5
(6)