Page 128 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

The one dimensional case 115

Theorem 4.3 Let f ∈ C ∞ ([a, b] × R × R) satisfy (H2), (H3’) and

(H1’) f ξξ (x, u, ξ) > 0, ∀ (x, u, ξ) ∈ [a, b] × R × R .

Then any minimizer of (P) is in C ∞ ([a, b]).
Remark 4.4 (i) Note that (H1’) is more restrictive than (H1). This stronger
condition is usually, but not always as will be seen in Theorem 4.5, necessary to
gethigherregularity.
(ii) Proposition 4.1 is, of course, a particular case of the present theorem.
k
(iii) The proof will show that if f ∈ C , k ≥ 2, then the minimizer is also
k
C .
Proof. We will propose a different proof in Exercise 4.2.1. The present one
is more direct and uses Lemma 2.8.
Step 1. We know from Lemma 4.2 that x → ϕ (x)= f ξ (x, u (x) , u (x)) is in
0
W 1,1 (a, b) and hence it is continuous. Appealing to Lemma 2.8 (and the remark
following this lemma), we have that if

H (x, u, v)= sup {vξ − f (x, u, ξ)}
ξ∈R
then H ∈ C ∞ ([a, b] × R × R) and, for every x ∈ [a, b],wehave

ϕ (x)= f ξ (x, u (x) , u (x)) ⇔ u (x)= H v (x, u (x) ,ϕ (x)) .
0
0
Since H v , u and ϕ are continuous, we infer that u is continuous and hence
0
1
u ∈ C ([a, b]). We therefore deduce that x → f u (x, u (x) , u (x)) is continuous,
0
which combined with the fact that (cf. (4.5))
d
0
[ϕ (x)] = f u (x, u (x) , u (x)) ,a.e. x ∈ (a, b)
dx
1
(or equivalently, by Lemma 2.8, ϕ = −H u (x, u, ϕ))leads to ϕ ∈ C ([a, b]).
0
Step 2. Returning to our Hamiltonian system
⎧
⎨ u (x)= H v (x, u (x) ,ϕ (x))
0
ϕ (x)= −H u (x, u (x) ,ϕ (x))
⎩
0
1
∞
we can start our iteration. Indeed since H is C and u and ϕ are C we deduce
2
from our system that, in fact, u and ϕ are C . Returning to the system we get
3
that u and ϕ are C . Finally we get that u is C , as wished.
∞
We conclude the section by giving an example where we can get further
regularity without assuming the non degeneracy condition f ξξ > 0.