4.2 Variational Problems of Functionals with Several Functions
267
Example 4.2.5 Find the shortest distance between the circle Γ 0 : x
2
+ y
2
= a
2 , z = 0
and the hyperbola Γ 1 : z
2
− x
2
= b
2 , y = 0.
Solution Choosing z as the independent variable. Let the curve pass through an
arbitrary point A(x 0 , y 0 , 0) of Γ 0 , pass through an arbitrary point B(x 1 , 0, z 1 ) of Γ 1 ,
and the found curve is
Γ :
x = x(z)
y = y(z)
(0 ≤ z ≤ z 1 )
(1)
It must make the functional
J [x, y] =
z 1
0
1 + x 2 + y 2 dz
(2)
obtain the minimum. One boundary point A(x 0 , y 0 , 0) of the functional moves on
the circle Γ 0 : Φ 0 = x
2
+ y
2
− a
2
= 0, and another boundary point B(x 1 , 0, z 1 ) is
undetermined on the hyperbola Γ 1 : Φ 1 = z
2
− x
2
− b
2
= 0. The integrand and the
various partial derivatives are
F =
1 + x 2 + y 2 , F x = 0, F x =
x
1 + x 2 + y 2
, F y = 0, F y =
y
1 + x 2 + y 2
(3)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
F x −
d
dz
F x = −
d
dz
x
1 + x 2 + y 2
= 0
F y −
d
dz
F y = −
d
dz
y
1 + x 2 + y 2
= 0
(4)
or
x
1 + x 2 + y 2
= c 1 ,
y
1 + x 2 + y 2
= c 2
(5)
By the above two equations getting y
= kx
, substituting it into the above two
equations, we give x
= k 1 , y
= k 2 , where k, k 1 , k 2 are all constants. Integrating x
and y
, and note that the curve passes through point A(x 0 , y 0 , 0), we obtain
x = k 1 z + x 0
y = k 2 z + y 0
(6)
Its geometric shape is a space straight line. In addition the curve also passes
through point B(x 1 , 0, z 1 ), so that