310
4 Problems with Variable Boundaries
The general solution is
y =
x
2
4
+ c 1 x + c 2
(2)
Let the abscissas at two tangential points of y = ϕ(x) = 2x − x
2 and the extremal
curve be x 2 and x 3 , from the boundary condition of the left side y(−2) = 1, we get
c 2 = 2c 1 . At the tangential point, there are y(x 2 ) = ϕ(x 2 ), y
(x 2 ) = ϕ
(x 2 ), the
equations are
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
x
2
4
+ c 1 x + c 2 = 2x − x
2
2c 1 = c 2
x
2
+ c 1 = 2 − 2x
(3)
Solve for c 1 = 7 − 3
√
5, c 2 = 14 − 6
√
5, x 2 = −2 +
6
√
5
5
. Thus the equation of
the extremal curve on the left side is
y =
x
2
4
+ (7 − 3
√
5)x + 14 − 6
√
5
( 4 )
At the tangential point, there are
y(x 2 ) = ϕ(x 2 ) = −
76
5
+
36
√
5
5
(5)
y
(x 2 ) = ϕ
(x 2 ) = 6 −
12
√
5
5
(6)
Similarly, the equation of the extremal curve on the right side
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x
2
4
+ c 1 x + c 2 = 2x − x
2
x
2
+ c 1 = 2 − 2x
c 2 = −
5
4
− 3c 1
(7)
Solve for c 1 =
−11+4
√
5
2
, c 2 =
61−24
√
5
4
, x 3 = 3 −
4
√
5
5
. Thus the equation of the
extremal curve on the right side is
y =
x
2
4
−
11 − 4
√
5
2
x +
61 − 24
√
5
4
(8)