102
2 Variational Problems with Fixed Boundaries
y (x) − y
(x)
< d 1 [y 1 (x), y(x)]
(2.3.6)
For arbitrary ε 1 > 0, ε 2 > 0, when d 1 [y 1 (x), y(x)] is sufficiently small, there
must be
¯
F y − F y
< ε 1 ,
¯
F y − F y
< ε 2
(2.3.7)
therefore, there is
J =
x 1
x 0
( ¯
F y δy + ¯
F y δy
)dx
=
x 1
x 0
(F y δy + F y δy
)dx +
x 1
x 0
[( ¯
F y − F y )δy + ( ¯
F y − F y )δy
]dx
=
x 1
x 0
(F y δy + F y δy
)dx + εd 1 [y 1 , y]
(2.3.8)
where
εd 1 [y 1 , y] =
x 1
x 0
[( ¯
F y − F y )δy + ( ¯
F y − F y )δy
]dx
(2.3.9)
and ε approaches zero with d 1 [y 1 , y] approaching zero. This is because there is
x1
x0
[( ¯
F y − F y )δy + ( ¯
F y − F y )δy
]dx
≤
x1
x0
¯
F y − F y
|δy|dx +
x1
x0
¯
F y − F y
δy
dx
< (ε 1 + ε 2 )d 1 [y 1 , y](x 1 − x 0 ) = ε
d 1 [y 1 , y]
(2.3.10)
and the ε
approaches zero with d 1 [y 1 , y] approaching zero.
The Taylor mean value theorem of binary function can be written as the following
form
F(x, y + δy, y
+ δy
) − F(x, y, y
) = F y δy + F y δy
+ R 1
(2.3.11)
where, R 1 is the first order Lagrange’s remainder, it can be expressed as
R 1 =
1
2
∂ F(x, y + θδy, y
+ θδy
)
∂ y
δy +
∂ F(x, y + θδy, y
+ θδy
)
∂ y
δy
(2.3.12)
where, 0 ≤ θ ≤ 1.