2.3 Variations of the Simplest Functionals and Necessary Conditions …
111
J [y] =
1
0
1
x
sin ydx =
∞
1
sin kt
t
dt
Its integral values both exist, taking y =
1
x
, δy =
1
nx
, there is
Φ(ε) = J [y + εδy] =
1
0
1
x
sin
n + ε
nx
dx (n ≥ 2)
Let τ =
n+ε
nx
, dτ = −
n+ε
nx 2 dx, we obtain
Φ(ε) =
∞
n+ε
n
sin τ
τ
dτ =
π
2
−
n+ε
n
0
sin τ
τ
dτ
Supplement definition
sin τ
τ
τ =0 = 1, then
sin τ
τ
is continuous in the interval
0, 1 +
ε
n
, thus
Φ
(ε) = −
1
n + ε
sin
n + ε
n
, Φ
(0) = −
1
n
sin 1
But at this time, the integral
δ J =
1
0
(F y δy + F y δy
)dx =
1
0
1
x
cos
1
x
·
1
nx
dx =
∞
1
cos t
n
dt
does not exist.
Example 2.3.8 Let a functional J [y] = y
2
(x 0 )+
x 1
x 0
(x y + y
2
)dx, find the variation
of the functional defined by Lagrange.
Solution According to the variation of a functional defined by Lagrange, there is
J [y + εδy] = [y(x 0 ) + εδy(x 0 )]
2
+
x 1
x 0
[x(y + εδy) + (y
+ εδy
)
2
]dx
Thus
∂ J [y + εy]
∂ε
= 2[y(x 0 ) + εδy(x 0 )]δy(x 0 ) +
x 1
x 0
[xδy + 2(y
+ εδy
)δy
]dx
Therefore there is
δ J =
∂ J [y + εy]
∂ε
ε=0
= 2y(x 0 )δy(x 0 ) +
x 1
x 0
(xδy + 2y
δy
)dx