314
4 Problems with Variable Boundaries
x + c 1 = 2 −
2x
5
(7)
Solve for x = 10 −
√
10, c 1 =
7
√
10
5
− 12, c 2 = 7(11 − 2
√
10). Thus the extremal
curve is
y =
x
2
2
+
7
√
10
5
− 12
x + 7(11 − 2
√
10) (10 −
√
10 ≤ x ≤ 10)
(8)
Coupled with the parabolic part of inequality, there are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
y =
x
2
2
+
2 −
7
√
10
5
x + 7
(0 ≤ x ≤
√
10)
y = x
2 −
x
5
(
√
10 ≤ x ≤ 10 −
√
10)
y =
x
2
2
+
7
√
10
5
− 12
x + 7(11 − 2
√
10) (10 −
√
10 ≤ x ≤ 10)
(9)
Example 4.6.5 Under the condition y ≤ x
2 , find the shortest curve through two
points A(−2, 3) and B(2, 3).
Solution The functional is J [y] =
2
−2
1 + y 2 dx. Because the integrand function
is only the function of y
, the integral of the Euler equation is the straight line
y = c 1 x + c 2
(1)
From the boundary condition x = −2, y = 3, we obtain
−2c 1 + c 2 = 3
( 2 )
Let the tangential point of the extremal curve and the inequality be x 1 , then
c 1 = 2x 1
(3)
c 1 x 1 + c 2 = x
2
1
(4)
Combining the above Eqs. (2), (3) and (4), we solve for x 1 = −3 (casting out),
x 1 = −1, c 1 = −2, c 2 = −1. Therefore the extremal curve is y = −2x − 1.
From the boundary condition x = 2, y = 3, we give 2c 1 + c 2 = 3. Let the
tangential point of the extremal curve and the inequality be x 2 , Using the above
mentioned same method, we solve for x 2 = 3(casting out), x 2 = 1, c 1 = 1, c 2 = 1.
Or according to the symmetry of functional graph, the result can also be obtained.
Thus the extremal curve is y = 2x + 1. On the whole, there are