5.1 Variational Problems with Holonomic Constraints
321
∂ F
∂ y j
+
m
i=1
λ i (x)
∂ϕ i
∂ y j
−
d
dx
∂ F
∂ y
j
= 0 ( j = 1, 2, . . . , n)
(5.1.6)
Proof Begin by finding the variation of the functional (5.1.1), from the condition of
fixed endpoints, then perform integration by parts to the terms with δy
j , and note
that
(δy j )
= δy
j , δy j
x=x 0
= 0, δy j
x=x 1
= 0
It can be got
δ J =
x 1
x 0
n
j=1
∂ F
∂ y j
−
d
dx
∂ F
∂ y
j
δy j dx
(5.1.7)
Because the variables y j are associated with the constraint condition (5.1.2), δy j
are not all independent of each other. Multiplying the expression (5.1.2) by λ i (x)
and integrating in the interval [x 0 , x 1 ], we give
J i =
x 1
x 0
λ i (x)ϕ i (x, y 1 , y 2 , . . . , y n )dx (i = 1, 2, . . . , m)
(5.1.8)
Taking the variation to the above functional, we obtain
δ J i =
x 1
x 0
n
j=1
λ i (x)
∂ϕ i
∂ y j
δy j
dx = 0 (i = 1, 2, . . . , m)
(5.1.9)
Add the expression (5.1.7) and the expressions (5.1.9), and let J
∗
= J +
m
i=1
J i ,
we give
δ J
∗
= δ J +
m
i=1
δ J i =
x 1
x 0
n
j=1
∂ F
∂ y j
−
d
dx
∂ F
∂ y
j
+
m
i=1
λ i (x)
∂ϕ i
∂ y j
δy j dx = 0
(5.1.10)
Because λ i (x) (i = 1, 2, . . . , m) are the m undetermined functions, assuming that
it can be determined by the following m linear equations
∂ F
∂ y j
−
d
dx
∂ F
∂ y
j
+
m
i=1
λ i (x)
∂ϕ i
∂ y j
= 0 ( j = 1, 2, . . . , m)
(5.1.11)