Page 208 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 2: Classical methods 195

7.2.5 Fields theories
¡ ¢
N
N
Exercise 2.6.1. Let f ∈ C 2 [a, b] × R × R N , α, β ∈ R . Assume that there
¡ ¢
exists Φ ∈ C 3 [a, b] × R N satisfying Φ (a, α)= Φ (b, β). Suppose also that
f (x, u, ξ)= f (x, u, ξ)+ hΦ u (x, u); ξi + Φ x (x, u)
e
is such that (u, ξ) → f (x, u, ξ) is convex. The claim is then that any solution u
e
of the Euler-Lagrange system
d £ ¤
0 0
(x, u, u ) = f u i (x, u, u ) ,i =1, ..., N
f ξ i
dx
is a minimizer of
( )
Z b
0
(P) inf I (u)= f (x, u (x) ,u (x)) dx = m
u∈X a
© 1 ¡ N ¢ ª
where X = u ∈ C [a, b]; R : u (a)= α, u (b)= β .
The proof is exactly as the one dimensional one and we skip the details.
Exercise 2.6.2. The procedure is very similar to the one of Theorem 2.27. An
exact field Φ = Φ (x, u) covering a domain D ⊂ R N+1 is a map Φ : D → R N so
¢
¡
that there exists S ∈ C 1 D; R N satisfying
(x, u, Φ (x, u)) ,i =1, ..., N
S u i (x, u)= f ξ i
S x (x, u)= f (x, u, Φ (x, u)) − hS u (x, u); Φ (x, u)i .
N
The Weierstrass function is defined, for u, η, ξ ∈ R ,as
E (x, u, η, ξ)= f (x, u, ξ) − f (x, u, η) − hf ξ (x, u, η);(ξ − η)i .
The proof is then identical to the one dimensional case.
Exercise 2.6.3. (i) Wehavebydefinition
⎧
⎨ S u (x, u)= f ξ (x, u, Φ (x, u))

⎩
S x (x, u)= − [S u (x, u) Φ (x, u) − f (x, u, Φ (x, u))] .
We therefore have immediately from Lemma 2.8
S x (x, u)= −H (x, u, S u (x, u)) .

(ii) Using again Lemma 2.8 we obtain
⎧
⎨ H (x, u, S u )= S u (x, u) Φ (x, u) − f (x, u, Φ (x, u))

S u (x, u)= f ξ (x, u, Φ (x, u)) .
⎩
Since S is a solution of Hamilton-Jacobi equation, we get S x = −H (x, u, S u ) as
wished.