Page 87 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

74 Classical methods

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hold for any u ∈ X = u ∈ C ([a, b]) : u (a)= α, u (b)= β .The first iden-
tity expresses that the integral is invariant, while the second one says that
0
ϕ (x, u, u ) satisfies the Euler-Lagrange equation identically (it is then called a
null Lagrangian).
With the help of the above observations we immediately obtain the result by
applying Theorem 2.1 to f. Indeed we have that (u, ξ) → f (x, u, ξ) is convex,
e
e
Z b Z b
0
I (u)= f (x, u (x) ,u (x)) dx = f (x, u (x) ,u (x)) dx
0
e
a a
for every u ∈ X and any solution u of (E) also satisfies
³ ´ d h i
0
0
E f ξ (x, u, u ) = f u (x, u, u ) ,x ∈ (a, b) .
e
e
e
dx
This concludes the proof.
With the help of the above elementary theorem we can now fully handle the
Poincaré-Wirtinger inequality.
Example 2.23 (Poincaré-Wirtinger inequality). Let λ ≥ 0, f λ (u, ξ)=
¡ 2 2 2 ¢
ξ − λ u /2 and
½ Z 1 ¾
0
(P λ ) inf I λ (u)= f λ (u (x) ,u (x)) dx = m λ
u∈X
0
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1
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 is
2
(E λ ) u + λ u =0,x ∈ (0, 1) .
00
Note that u 0 ≡ 0 is a solution of (E λ ). Define, if λ<π,
∙ µ ¶¸
λ 1 2
Φ (x, u)= tan λ x − u , (x, u) ∈ [0, 1] × R
2 2
and observe that Φ satisfies all the properties of Theorem 2.21. The function f e
is then

∙ µ ¶¸ 2 ∙ µ ¶¸
1 2 1 λ 2 1 2
f (x, u, ξ)= ξ + λ tan λ x − uξ + tan λ x − u .
e
2 2 2 2
It is easy to see that (u, ξ) → f (x, u, ξ) is convex and therefore applying Theorem
e
2.21 we have that, for every 0 ≤ λ<π,
I λ (u) ≥ I λ (0) , ∀u ∈ X.