Page 76 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Hamiltonian formulation 63

Remark 2.9 (i) The lemma remains partially true if we replace the hypothesis
(2.4) by the weaker condition
ξ → f (x, u, ξ) is strictly convex.

1
In general, however the function H is only C , as the following simple example
shows
1 4 3 4/3
f (x, u, ξ)= |ξ| and H (x, u, v)= |v| .
4 4
(See also Example 2.13.)
(ii) The lemma remains also valid if the hypothesis (2.5) does not hold but
then, in general, H is not anymore finite everywhere as the following simple
example suggests. Consider the strictly convex function
q
2
f (x, u, ξ)= f (ξ)= 1+ ξ
and observe that
⎧ √
⎨ − 1 − v 2 if |v| ≤ 1
H (v)=
⎩
+∞ if |v| > 1.
k
k
(iii) The same proof leads to the fact that if f ∈ C , k ≥ 2,then H ∈ C .
Proof. We divide the proof into several steps.
Step 1. Fix (x, u) ∈ [a, b] × R. From the definition of H and from (2.5) we
deduce that there exists ξ = ξ (x, u, v) such that
⎧
⎨ H (x, u, v)= vξ − f (x, u, ξ)
(2.11)
v = f ξ (x, u, ξ) .
⎩
Step 2. The function H is easily seen to be continuous. Indeed let (x, u, v),
(x ,u ,v ) ∈ [a, b] × R × R, using (2.11) we find ξ = ξ (x, u, v) such that
0
0
0
H (x, u, v)= vξ − f (x, u, ξ) .
Appealingtothe definition of H we also have
0
0
0
0
0
0
H (x ,u ,v ) ≥ v ξ − f (x ,u ,ξ) .
Combining the two facts we get
0
0
0
0
0
H (x, u, v) − H (x ,u ,v ) ≤ (v − v ) ξ + f (x ,u ,ξ) − f (x, u, ξ) ,
0
since the reverse inequality is obtained similarly, we deduce the continuity of H
from the one of f (in fact only in the variables (x, u)).