Page 89 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

76 Classical methods

and hence
f u (x, u, Φ)= h u (x, u)+ p u (x, u) Φ (x, u) .
We therefore get for every x ∈ [a, b]
d d
[f ξ (x, u, u )] − f u (x, u, u )= [p (x, u)] − [h u (x, u)+ p u (x, u) Φ (x, u)]
0
0
dx dx
= p x + p u u − h u − p u Φ = p x − h u =0
0
sincewehavethat u = Φ and p x = h u , Φ beingexact. Thuswehavereached
0
the claim.
The next theorem is the main result of this section and was established by
Weierstrass and Hilbert.
2
Theorem 2.27 (Hilbert Theorem).Let f ∈ C ([a, b] × R × R) with ξ →
2
f (x, u, ξ) convex for every (x, u) ∈ [a, b] × R.Let D ⊂ R be a domain and
2
Φ : D → R , Φ = Φ (x, u),be an exact field for f covering D. Assume that there
1
exists u 0 ∈ C ([a, b]) satisfying
(x, u 0 (x)) ∈ D, ∀x ∈ [a, b]
u (x)= Φ (x, u 0 (x)) , ∀x ∈ [a, b]
0
0
then u 0 is a minimizer for I, i.e.
Z b
I (u)= f (x, u (x) ,u (x)) dx ≥ I (u 0 ) , ∀u ∈ X
0
a
where ½ ¾
1
u ∈ C ([a, b]) : u (a)= u 0 (a) ,u (b)= u 0 (b)
X = .
with (x, u (x)) ∈ D, ∀x ∈ [a, b]
Remark 2.28 (i) Observe that according to the preceding proposition we have
that such a u 0 is necessarily a solution of the Euler-Lagrange equation.
(ii) As already mentioned it might be very difficult to construct such exact
fields. Moreover, in general, D does not contain the whole of [a, b] × R and,
consequently, the theorem will provide only local minima. The construction of
such fields is intimately linked with the so called Jacobi condition concerning
conjugate points (see the bibliography for more details).

Proof. Denote by E the Weierstrass function defined by
E (x, u, η, ξ)= f (x, u, ξ) − f (x, u, η) − f ξ (x, u, η)(ξ − η)

or in other words

f (x, u, ξ)= E (x, u, η, ξ)+ f (x, u, η)+ f ξ (x, u, η)(ξ − η) .