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84 Direct methods
1,2
Part 4 (Converse). We finally prove that if u ∈ u 0 + W 0 (Ω) satisfies (3.1)
1,2
then it is necessarily a minimizer of (D). Let u ∈ u 0 + W 0 (Ω) be any element
1,2
and set ϕ = u − u.Observe that ϕ ∈ W (Ω) and
0
Z
1 2
I (u)= I (u + ϕ)= |∇u + ∇ϕ| dx
2
Ω
Z
= I (u)+ h∇u; ∇ϕi dx + I (ϕ) ≥ I (u)
Ω
since the second term is 0 according to (3.1) and the last one is non negative.
This achieves the proof of the theorem.
3.2.1 Exercises
2
Exercise 3.2.1 Let Ω be as in the theorem and g ∈ L (Ω). Show that
½ Z ∙ ¸ ¾
1 2 1,2
(P) inf I (u)= |∇u (x)| − g (x) u (x) dx : u ∈ W 0 (Ω) = m
Ω 2
1,2
has a unique solution u ∈ W 0 (Ω) which satisfies in addition
Z Z
1,2
h∇u (x); ∇ϕ (x)i dx = g (x) ϕ (x) dx, ∀ϕ ∈ W (Ω) .
0
Ω Ω
3.3 A general existence theorem
The main theorem of the present chapter is the following.
n
Theorem 3.3 Let Ω ⊂ R be a bounded open set with Lipschitz boundary. Let
¡
¢
n
f ∈ C 0 Ω × R × R , f = f (x, u, ξ), satisfy
(H1) ξ → f (x, u, ξ) is convex for every (x, u) ∈ Ω × R;
(H2) there exist p> q ≥ 1 and α 1 > 0, α 2 ,α 3 ∈ R such that
p q n
f (x, u, ξ) ≥ α 1 |ξ| + α 2 |u| + α 3 , ∀ (x, u, ξ) ∈ Ω × R × R .
Let
½ Z ¾
(P) inf I (u)= f (x, u (x) , ∇u (x)) dx : u ∈ u 0 + W 0 1,p (Ω) = m
Ω
where u 0 ∈ W 1,p (Ω) with I (u 0 ) < ∞.Then there exists u ∈ u 0 + W 0 1,p (Ω) a
minimizer of (P).
Furthermore if (u, ξ) → f (x, u, ξ) is strictly convex for every x ∈ Ω,then the
minimizer is unique.