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5 Variational Problems of Conditional Extrema
λ 1 = −
1
2a 2
(5.16)
Substituting λ 1 = −
1
2a 2 , λ 2 =
1
2
back to the Euler Eq. (6), integrating and getting
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
x = A 1 cos
s
a
+ A 2 sin
s
a
y = B 1 cos
s
a
+ B 2 sin
s
a
z = C 1 cos
s
a
+ C 2 sin
s
a
(5.17)
where, A 1 , A 2 , B 1 , B 2 , C 1 , C 2 are all the constants of integration, in order to make
the above equations can hold for arbitrary s value, only needing the determinant
x A 1 A 2
y B 1 B 2
z C 1 C 2
= 0
(5.18)
This is a plane equation through the coordinate origin. This shows that the geodesic
line through two points A, B on the sphere is inevitably on the plane through points
O, A and B, unless AO B happens to constitute diameter, otherwise the arc AB is the
shorter segment of the great circular arc, namely the minor arc.
Example 5.2.3 Find the extremal functions y = y(x) and z = z(x) of the functional
J [y, z] =
1
2
x 1
x 0
(y
2
+ z
2
)dx under the fixed boundary conditions y(x 0 ) = y 0 , z(0) =
z 0 and the constraint condition y
= z.
Solution Let H =
1
2
y
2
+
1
2
z
2
+ λ(y
− z), then the functional without the constraint
condition is
J
∗
[y, z] =
x 1
x 0
Hdx =
x 1
x 0
1
2
y
2
+
1
2
z
2
+ λ(y
− z)
dx
The Euler equations of the functional are
y − λ
= 0
z − λ = 0
or
y = λ
z = λ
,
y
= z
z
= y
It can be obtained from the two equations