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Fields theories                                                    75

                An elementary passage to the limit leads to Poincaré-Wirtinger inequality

                                    Z  1         Z  1
                                                     2
                                        02
                                       u dx ≥ π 2   u dx, ∀u ∈ X.
                                     0            0
                   For a different proof of a slightly more general form of Poincaré-Wirtinger
                inequality see Theorem 6.1.
                   The way of proceeding, in Theorem 2.21, is, in general, too naive and can be
                done only locally; in fact one needs a similar but more subtle theory.

                                          2
                Definition 2.24 Let D ⊂ R be a domain. We say that Φ : D → R, Φ =
                                                                       1
                Φ (x, u),isan exact field for f covering D if there exists S ∈ C (D) satisfying
                        S u (x, u)= f ξ (x, u, Φ (x, u)) = p (x, u)
                        S x (x, u)= f (x, u, Φ (x, u)) − p (x, u) Φ (x, u)= h (x, u) .
                                         2
                Remark 2.25 (i) If f ∈ C , then a necessary condition for Φ to be exact is
                that p x = h u . Conversely if D is simply connected and if p x = h u then such an
                S exists.
                                                 N
                   (ii) In the case where u :[a, b] → R , N> 1, we have to add to the preceding
                                             ,but also (p i )  =(p j ) , for every 1 ≤ i, j ≤
                                      x                  u j     u i
                remark, not only that (p i ) = h u i
                N.
                   We startwithanelementaryresultthatisa first justification for defining
                such a notion.

                                          2
                Proposition 2.26 Let f ∈ C ([a, b] × R × R), f = f (x, u, ξ),and
                                           Z  b
                                    I (u)=    f (x, u (x) ,u (x)) dx .
                                                         0
                                            a
                            2
                                               1
                Let Φ : D → R , Φ = Φ (x, u) be a C exact field for f covering D, [a, b]×R ⊂D.
                                     2
                Then any solution u ∈ C ([a, b]) of
                                           0
                                          u (x)= Φ (x, u (x))                   (2.20)
                solves the Euler-Lagrange associated to the functional I,namely

                           d
                                                            0
                                         0
                     (E)     [f ξ (x, u (x) ,u (x))] = f u (x, u (x) ,u (x)) ,x ∈ [a, b] .  (2.21)
                          dx
                   Proof. By definition of Φ and using the fact that p = f ξ ,wehave, for any
                (x, u) ∈ D,
                     h u = f u (x, u, Φ)+ f ξ (x, u, Φ) Φ u − p u Φ − pΦ u = f u (x, u, Φ) − p u Φ
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