Page 67 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

54 Classical methods

we deduce that

n 2 Z 1/n 2 1 Z 1 dx
0 ≤ I (u n )= 2 x (1 − 2nx) dx + 2 → 0,as n →∞.
log n 0 log n 1/n x
This indeed shows that m =0.
Case 2.5 f (x, u, ξ)= f (u, ξ).

Although this case is a lot harder to treat than the preceding ones it has
an important property that is not present in the most general case when f =
f (x, u, ξ). The Euler-Lagrange equation is
d
[f ξ (u (x) ,u (x))] = f u (u (x) ,u (x)) ,x ∈ (a, b)
0
0
dx
and according to Theorem 2.7 below, it has a first integral that is given by
f (u (x) ,u (x)) − u (x) f ξ (u (x) ,u (x)) = constant, x ∈ (a, b) .
0
0
0
1. Poincaré-Wirtinger inequality.
We will show, in several steps, that
2
Z µ ¶ Z
b b
02 π 2
u dx ≥ u dx
a b − a a
for every u satisfying u (a)= u (b)= 0. By a change of variable we
immediately reduce the study to the case a =0 and b =1. We will also
prove in Theorem 6.1 a slightly more general inequality known as Wirtinger
inequality which states that
Z Z
1 1
02
2
u dx ≥ π 2 u dx
−1 −1
R 1
among all u satisfying u (−1) = u (1) and udx =0.
−1
We start by writing the problem under the above formalism and we let
¡ 2 ¢
2 2
λ ≥ 0, f λ (u, ξ)= ξ − λ u /2 and
1
½ Z ¾
0
(P λ ) inf I λ (u)= f λ (u (x) ,u (x)) dx = m λ
u∈X
0
© 1 ª
where X = u ∈ C ([0, 1]) : u (0) = u (1) = 0 .Observe that ξ → f λ (u, ξ)
is convex while (u, ξ) → f λ (u, ξ) is not. The Euler-Lagrange equation and
its first integral are
2
02
2 2
u + λ u =0 and u + λ u = constant.
00