5.2 Variational Problems with Differential Constraints
329
Because the functional (5) is similar to the function (4.2.1), the boundary condition
is similar to the expression (4.2.7), the boundary condition can be written as
[(H − x
H x − y
H y − z
H z )δs + H x δx
+ H y δy
+ H z δz
]
s B
s A
= 0
Integrating Eq. (1) twice, we give
x x
+ yy
+ zz
= 0
(5.7)
x x
+ yy
+ zz
+ x
2
+ y
2
+ z
2
= x x
+ yy
+ zz
+ 1 = 0
(5.8)
Integrating Eq. (2), we give
x
x
+ y
y
+ z
z
= 0
(5.9)
Multiplying respectively the first, second, third equation of Eq. (6) by x
, y
, z
and adding them, using Eq. (2), Eqs. (7) and (9), we get
dλ 2
ds
= 0
(5.10)
namely λ 2 is a constant. Again, multiplying respectively the first, second and third
equation of Eq. (6) by x, y, z and adding them, then using Eq. (8), we get
λ 2 + a
2
λ 1 = 0
(5.11)
Because the two ends are fixed, δx = δy = δz = 0, δx
= δy
= δz
= 0, and
since δs is arbitrary, thus the above mentioned boundary condition is simplified to
(H − x
H x − y
H y − z
H z )
s B
s A
= 0
(5.12)
Which needs
H − x
H x − y
H y − z
H z = 0
(5.13)
that is
1 − 2λ 2 = 0
(5.14)
Solving for λ 2
λ 2 =
1
2
(5.15)
From the expression (15) and expression (11), solving for λ 1