4.1 Variational Problems of the Simplest Functional
257
When ϕ = 0, the functional (1) has the maximum, namely the time needed for
light passing the person to the highest point of the light path is
J =
n 0 π
2ac
(16)
When ϕ =
π
2
, the functional (1) has the minimum
J =
n 0 (3π + 2)
8ac
(17)
Example 4.1.7 Find the shortest distance between the parabola y = x
2 and the
straight line y = x − 5.
Solution This problem boils down to finding the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx. The general solution of the Euler equation of the functional is the
straight line y = c 1 x + c 2 , the constraint condition is that the left endpoint of the
extremal curve is on the parabola y = ϕ(x) = x
2 , while the right endpoint is on
the straight line y = ψ(x) = x − 5, and ϕ(x 0 ) = 2x 0 , ψ(x 1 ) = 1. According to
Eq. (4.1.38), the system of equations about the four unknown numbers x 0 , x 1 , c 1 and
c 2 is
c 1 x 0 + c 2 = x
2
0 , c 1 x 1 + c 2 = x 1 − 5
1 + 2x 0 c 1 = 0, 1 + c 1 = 0
Solving the system of equations, we get
x 0 =
1
2
, x 1 =
23
8
, c 1 = −1, c 2 =
3
4
Thus the equation of the extremal curve is y =
3
4
− x, and the shortest distance
between two given curves is
J [y] =
23
8
1
2
1 + (−1) 2 dx =
√
2x
23
8
1
2
=
19
√
2
8
Example 4.1.8 Find the shortest distance between point A(1, 0) and the ellipse
4x
2
+ 9y
2
= 36.
Solution This problem boilds down to finding the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx. The extremal curve of the functional is straight line y = c 1 x + c 2 ,
the constraint condition is that the left endpoint of the extremal curve passes through
point A(1, 0), at the moment x 0 = 1, y 0 = 0, the right endpoint is undetermined on
the ellipse 4x
2
+ 9y
2
= 36. The elliptic equation can be written as y = ψ(x) =