3.6 Higher Order Variations of Functionals
233
Theorem 3.6.1 Let y = y(x) be the extremal curve of the functional (3.6.1) satisfying the boundary condition (3.2.2). If on y = y(x), δ
2 J ≥ 0 (or ≤ 0), then the
functional obtains weak minimum (or weak maximum) on y = y(x).
Proof It can be seen from the expression (3.6.10) of the increment J of the functional that when the absolute values of δy and δy
are sufficiently small, the sign of
J is decided by δ
2 J , therefore, according to the definition of weak extremum, it is
proven. Quod erat demonstrandum.
Example 3.6.2 Let the functional
J [y] =
x 1
x 0
[
n
k=0
p k (x)y
k y
n−k
]dx
(3.6.24)
where, y ∈ C
1
[x 0 , x 1 ], p k (x) ∈ [x 0 , x 1 ] are given functions (k = 1, 2, · · · , n). Prove
J = J [y + δy] − J [y] =
n
k=1
1
k!
δ
k J
(3.6.25)
Proof The increment of the functional (3.6.24) can be written in the form of the
expression (3.6.12), since the integrand F =
n
k=0
p k (x)y
k y
n−k is a polynomial of
degree n about y and y
, and the sum of the index of the product of both is also n,
there can be at most n-th variation, namely ε n = 0, therefore the expression (3.6.25)
holds. Quod erat demonstrandum.
Example 3.6.3 Let the functional J [y] =
1
2
x 1
x 0
[ p(x)y
2
+ 4q(x)yy
+ r (x)y
2
]dx,
where, p(x), q(x), r (x) ∈ C
1
[x 0 , x 1 ], y ∈ C
2
[x 0 , x 1 ], p(x) = 0, p(x)r (x)−q
2
(x) >
0, and on y = y(x), δ J = 0.
(1) Find δ
2 J ;
(2) Prove that when p(x) > 0, J [y] is an absolute minimum; when p(x) < 0, J [y]
is an absolute maximum.
Solution (1) Setting the integrand F =
1
2
[ p(x)y
2
+2q(x)yy
+r (x)y
2
], the second
variation is
δ
2 F =
1
2
[F yy (δy)
2
+ 2F yy δyδy
+ F y y (δy
)
2
] =
1
2
[ p(δy)
2
+ 2qδyδy
+ r (δy
)
2
]
Thus
δ
2 J =
1
2
x 1
x 0
δ
2 Fdx =
1
2
x 1
x 0
[ p(δy)
2
+ 2qδyδy
+ r (δy
)
2
]dx