3.6 Higher Order Variations of Functionals
231
R ≤ 0, the functional is absolute maximum. This shows that the second variation of
the functional is associated with the sufficient condition of extremum.
For the functional J [y(x)], a function Φ(ε) can be introduced, such that Φ(ε) =
J [y(x) + εδy], if its second derivative
∂
2 J [y(x)+εδy]
∂ε 2
ε=0
with respect to ε exists at
ε = 0, then Φ
(0) is called the quadratic variation or second variation of the
functional J [y(x)] at y = y(x), it is written as δ
2 J , that is
δ
2 J = Φ
(0) =
∂
2 J [y(x) + εδy]
∂ε 2
ε=0
(3.6.17)
The second variation of such a defined functional is equivalent to the second
variation of the proceeding defined functional for the determinate functional on class
of integrable functions, and sometimes it is easier to calculate the variation of a
functional.
Example 3.6.1 Find the second variation of the functional J [y] =
x 1
x 0
(x y
2
+ y
3
)dx.
Solution S and R are respectively
S =
1
2
F yy −
d
dx
F yy
= x, R =
1
2
F y y = 3y
The second variation of the functional is
δ
2 J =
x 1
x 0
[S(δy)
2
+ R(δy
)
2
]dx =
x 1
x 0
[x(δy)
2
+ 3y
(δy
)
2
]dx
For the integrand F in the functional (3.6.1), it can be expanded into the polynomial
of degree n by the Taylor formula, that is
x 1
x 0
F(x, y + δy, y
+ δy
)dx =
x 1
x 0
F(x, y, y
)dx +
x 1
x 0
(F y δy + F y δy
)dx+
1
2!
x 1
x 0
[F yy (δy)
2
+ 2F yy δyδy
+ F y y (δy
)
2
]dx + · · · +
1
n!
x 1
x 0
δy
∂
∂ y
+ δy
∂
∂ y
n
Fdx + ε n
(3.6.18)
where, ε n is the higher order infinitesimal than d
n
1 [y, y + δy]. Thus we get
J = δ J + δ
2 J + · · · + δ
n J + ε n =
n
k=1
δ
k J + ε n
(3.6.19)
231
R ≤ 0, the functional is absolute maximum. This shows that the second variation of
the functional is associated with the sufficient condition of extremum.
For the functional J [y(x)], a function Φ(ε) can be introduced, such that Φ(ε) =
J [y(x) + εδy], if its second derivative
∂
2 J [y(x)+εδy]
∂ε 2
ε=0
with respect to ε exists at
ε = 0, then Φ
(0) is called the quadratic variation or second variation of the
functional J [y(x)] at y = y(x), it is written as δ
2 J , that is
δ
2 J = Φ
(0) =
∂
2 J [y(x) + εδy]
∂ε 2
ε=0
(3.6.17)
The second variation of such a defined functional is equivalent to the second
variation of the proceeding defined functional for the determinate functional on class
of integrable functions, and sometimes it is easier to calculate the variation of a
functional.
Example 3.6.1 Find the second variation of the functional J [y] =
x 1
x 0
(x y
2
+ y
3
)dx.
Solution S and R are respectively
S =
1
2
F yy −
d
dx
F yy
= x, R =
1
2
F y y = 3y
The second variation of the functional is
δ
2 J =
x 1
x 0
[S(δy)
2
+ R(δy
)
2
]dx =
x 1
x 0
[x(δy)
2
+ 3y
(δy
)
2
]dx
For the integrand F in the functional (3.6.1), it can be expanded into the polynomial
of degree n by the Taylor formula, that is
x 1
x 0
F(x, y + δy, y
+ δy
)dx =
x 1
x 0
F(x, y, y
)dx +
x 1
x 0
(F y δy + F y δy
)dx+
1
2!
x 1
x 0
[F yy (δy)
2
+ 2F yy δyδy
+ F y y (δy
)
2
]dx + · · · +
1
n!
x 1
x 0
δy
∂
∂ y
+ δy
∂
∂ y
n
Fdx + ε n
(3.6.18)
where, ε n is the higher order infinitesimal than d
n
1 [y, y + δy]. Thus we get
J = δ J + δ
2 J + · · · + δ
n J + ε n =
n
k=1
δ
k J + ε n
(3.6.19)
