312
4 Problems with Variable Boundaries
on the left side y(−2) =
π
3
+
√
3
2
− 2, we give c 2 = 2c 1 +
π
3
+
√
3
2
− 2. At tangential
point M 1 (x 1 , y 1 ), we give y(x 1 ) = ϕ(x 1 ), y
(x 1 ) = ϕ
(x 1 ). Thus, the equations are
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
c 1 x + c 2 = −2 sin 2x
c 2 = 2c 1 +
π
3
+
√
3
2
− 2
c 1 = −2 cos 2x
(2)
Solve for c 1 = 1, c 2 =
π
3
+
√
3
2
, x 1 = −
π
3
. Thus, the equation of the extremal
curve on the left is
y = x +
π
3
+
√
3
2
(3)
At the tangential point, there are
y(x 1 ) = ϕ(x 1 ) =
√
3
2
(4)
y
(x 1 ) = ϕ
(x 1 ) = 1
( 5 )
According to the symmetry of the graphics, the equation of the extremal curve on
the right side is
y = −x +
π
3
+
√
3
2
(6)
At the tangential point, there are
y(x 4 ) = ϕ(x 4 ) =
√
3
2
(7)
y
(x 4 ) = ϕ
(x 4 ) = −1
( 8 )
Point M 2 and point M 3 should be the tangential point of common tangent of the
sine curve y = − sin 2x(−2 ≤ x ≤ 0) and sine curve y = sin 2x(0 ≤ x ≤ 2),
obviously, the tangential points are respectively
M 2
−
π
4
, 1
, M 3
π
4
, 1
Thus, the extremal curve to be found is