1.5 Fundamental Lemmas of the Calculus of Variations
49
Arranging the above expression, we obtain Eq. (1.5.12).
Similarly, for η(b) = 0, there is
−η(x) =
b
x
η
(x)dx (a ≤ x ≤ b)
we obtain
b
a
η
2
(x)dx =
b
a
b
x
1 · η
(x)dx
2
dx ≤
b
a
b
x
1
2 dx
b
x
η
2
(x)dx
dx
≤
b
a
(b − x)
b
a
η
2
(x)dx
dx =
1
2
(b − a)
2
b
a
η
2
(x)dx
Arranging the above expression, Eq. (1.5.12) can also be obtained. Quod erat
demonstrandum.
Lemma 1.5.5 can be extended to more general case, which have the following
lemma:
Lemma 1.5.6 Let a function η(x) be n + 1th order continuously differentiable in
the interval [a, b], with η
(n)
(a) = 0 (or η
(n)
(b) = 0), then there is
b
a
[η
(n+1)
(x)]
2 dx ≥
2
(b − a) 2
b
a
[η
(n)
(x)]
2 dx
(1.5.14)
When n = 0, Eq. (1.5.14) reduces to Eq. (1.5.12).
Lemma 1.5.7 Let a function f (x) be continuous in the interval [a, b], arbitrary
function η(x) has first order continuous derivative in the interval [a, b], with η(a) =
η(b) = 0, if the integral
b
a
f (x)η
(x)dx = 0
(1.5.15)
can always hold, there must be in the interval [a, b]
f (x) ≡ C
(1.5.16)
Lemma 1.5.7 is also called the Riemann theorem or Du Bois-Reymond lemma.
Proof 1 With the reduction to absurdity. If f (x) is not a constant in the interval
[a, b], then by the continuity of f (x), there are at least two points ξ , ζ in the interval,
make f (x) has unequal values, might as well suppose f (ξ ) > f (ζ ). Let a 0 and b 0
be a pair of numbers which satisfies the inequality
f (ξ ) > a 0 > b 0 > f (ζ )