328
5 Variational Problems of Conditional Extrema
From the boundary conditions, the integral constants c 1 =
3
2
, c 2 = −
7
2
, c 3 =
c 4 = 1 are determined, thus there is
⎧
⎪ ⎨
⎪ ⎩
y 1 =
1
2
x
3
−
7
4
x
2
+ x + 1
y 2 =
3
2
x
3
−
7
2
x
2
+ 1
(5.8)
Finally the minimum of the functional is obtained
J [y] =
1
2
2
0
3x −
7
2
2
dx =
13
4
(5.9)
Example 5.2.2 The geodesic line problem. Two points A and B have been given on
the sphere x
2
+ y
2
+ z
2
= a
2 . Find the shortest arc connecting the two points.
Solution The differential of arc is (ds)
2
= (dx)
2
+ (dy)
2
+ (dz)
2 , let x
=
dx
ds
,
y
=
dy
ds
, z
=
dz
ds
, then there are the constraint conditions
ϕ 1 = x
2
+ y
2
+ z
2
− a
2
= 0
(5.1)
ϕ 2 = x
2
+ y
2
+ z
2
− 1 = 0
( 5 . 2 )
The functional expressed by arc length is
J =
s B
s A
ds
(5.3)
Let
H = 1 + λ 1 (s)ϕ + λ 2 (s)ϕ 2 = 1 + λ 1 (x
2
+ y
2
+ z
2
− a
2
) + λ 2 (x
2
+ y
2
+ z
2
− 1)
(5.4)
then the functional without the constraint condition is
J
•
=
s B
s A
H ds =
s B
s A
[1 + λ 1 (x
2
+ y
2
+ z
2
− a
2
) + λ 2 (x
2
+ y
2
+ z
2
− 1)]ds
(5.5)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
d
ds
(λ 2 x
) − λ 1 x = 0
d
ds
(λ 2 y
) − λ 1 y = 0
d
ds
(λ 2 z
) − λ 1 z = 0
(5.6)