106
2 Variational Problems with Fixed Boundaries
Example 2.3.4 Let J [y] =
x 1
x 0
(y
2
+ y
2
)dx, find δ J .
Solution F = y
2
+ y
2 , F y = 2y, F y = 2y
, the variation of the functional is
δ J =
x 1
x 0
(2yδy + 2y
δy
)dx = 2
x 1
x 0
(yδy + y
δy
)dx
Example 2.3.5 Let J [y] =
x 1
x 0
(y
3
+ y
2 y
+ y
2
)dx, find δ J .
Solution F = y
3
+ y
2 y
+ y
2 , F y = 3y
2
+ 2yy
, F y = y
2
+ 2y
, the variation of
the functional is
δ J =
x 1
x 0
[(3y
2
+ 2yy
)δy + (y
2
+ 2y
)δy
]dx
Example 2.3.6 Find the variation of the functional J [y] =
x 1
x 0
(x y
2
+ y
e
y
)dx.
Solution F = x y
2
+ y
e
y , F y = 2x y + y
e
y , F y = e
y , the variation of the functional
is
δ J =
x 1
x 0
[(2x y + y
e
y
)δy + e
y
δy
]dx
Let a functional F = F(x, y, y
) be continuous with respect to x, y, y
, and has
enough differentiability, calculate the increment of F
F = F(x, y + δy, y
+ δy
) − F(x, y, y
) = F y δy + F y δy
+ · · · (2.3.16)
then
δ F = F y δy + F y δy
(2.3.17)
is called the variation of a function F. At the moment, the variational expression
(2.3.15) of the functional can be written as
δ J = δ
x 1
x 0
F(x, y, y
)dx =
x 1
x 0
δ F(x, y, y
)dx
(2.3.18)
Equation (2.3.18) shows that under the condition of δ F is a linear function about
δy and δy
, the variational symbol δ and the definite integral symbol
x 1
x 0
can exchange
order. Under certain conditions, this operation can be generalized. For example, let
the functional
J [y 1 , y 2 , · · · , y n ] =
x 1
x 0
F(x, y 1 , y 2 , · · · , y n , y
1 , y
2 , · · · , y
n )dx
(2.3.19)