108
2 Variational Problems with Fixed Boundaries
Quod erat demonstrandum.
Let F, F 1 and F 2 be the differentiable functions of x, y, y
, …, then the total
variational symbol has the following fundamental operation properties:
(1) (F 1 + F 2 ) = F 1 + F 2
(2) (F 1 F 2 ) = F 1 F 2 + F 2 F 1
(3) (F
n
) = n F
n−1
F
(4)
F 1
F 2
=
F 2 F 1 −F 1 F 2
F
2
2
(5) [F
(n)
]
= [F
(n+1)
] + F
(n+1)
((x)
The property (5) shows that the operational sequence of the derivation and total
variation can not be interchanged.
(6)
x 1
x 0
Fdx = δ
x 1
x 0
Fdx + (Fx)|
x 1
x 0
=
x 1
x 0
F + F
d
dx
x
dx
The property (6) shows that when
d
dx
x = 0, the operational sequence of the
integral and total variation can not be interchanged too.
(7) d((x) = (dx)
The property (7) shows that for the independent variable, the operational sequence
of the total variation and differential has commutativity.
Proof From the definition of total variation, there is F = F
x + δ F, thus
(F 1 + F 2 ) = (F 1 + F 2 )
x + δ(F 1 + F 2 ) = (F
1 + F
2 ))x + δ(F 1 + F 2 )
= F
1 x + F
2 x + δ F 1 + δ F 2 = (F
1 x + δ F 1 ) + (F
2 x + δ F 2 )
= F 1 + F 2
(F 1 F 2 ) = (F 1 F 2 )
x + δ(F 1 F 2 ) = (F
1 F 2 + F 1 F
2 ))x + F 1 δ F 2 + F 2 δ F 1
= F 1 (F
2 x + δ F 2 ) + F 2 (F
1 x + δ F 1 ) = F 1 F 2 + F 2 F 1
(F
n ) = (F
n )
x + δ(F
n ) = n F
n−1 F
x + n F
n−1 δ F = n F
n−1 (F
x + δ F) = n F
n−1 F
F 1
F 2
=
d
dx
F 1
F 2
x + δ
F 1
F 2
=
F 2 F
1 − F 1 F
2
F
2
2
x +
F 2 δ F 1 − F 1 δ F 2
F
2
2
=
F 2 (F
1 x + δ F 1 ) − F 1 (F
2 x + δ F 2 )
F
2
2
=
F 2 F 1 − F 1 F 2
F
2
2
According to the definition of total variation, there is [F
(n)
] = F
(n+1)
x +
δ[F
(n)
], derive the expression with respect to x, we obtain
[F
(n)
]
= {F
(n+1)
x + δ[F
(n)
]}
= F
(n+2)
x + F
(n+1)
((x)
+ δ[F
(n+1)
]
(2.3.21)