Page 126 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
P. 126

The one dimensional case                                          113

                Proposition 4.1 Let g ∈ C ∞  ([a, b] × R) satisfy

                   (H2) there exist 2 >q ≥ 1 and α 2 ,α 3 ∈ R such that
                                             q
                               g (x, u) ≥ α 2 |u| + α 3 , ∀ (x, u) ∈ [a, b] × R .
                Let
                                                 1  2
                                      f (x, u, ξ)=  ξ + g (x, u) .
                                                 2
                Then there exists u ∈ C  ∞  ([a, b]), a minimizer of (P). If, in addition, u →
                g (x, u) is convex for every x ∈ [a, b], then the minimizer is unique.

                   Proof. The existence (and uniqueness, if g is convex) of a solution u ∈
                W 1,2  (a, b) follows from Theorem 3.3. We also know from Theorem 3.11 that it
                satisfies the weak form of the Euler-Lagrange equation
                             Z  b         Z  b
                                 0 0                            ∞
                                u v dx = −    g u (x, u) vdx, ∀v ∈ C  (a, b) .   (4.1)
                                                                0
                              a            a
                To prove further regularity of u,westart by showingthat u ∈ W 2,2  (a, b).This
                follows immediately from (4.1) and from the definition of weak derivative. Indeed
                                                                    2
                since u ∈ W  1,2 ,we havethat u ∈ L ∞  and thus g u (x, u) ∈ L ,leading to
                            ¯        ¯
                            Z  b
                            ¯        ¯
                            ¯    0 0  ¯                          ∞
                               u v dx¯ ≤ kg u (x, u)k  L 2 , ∀v ∈ C  (a, b) .    (4.2)
                            ¯                     L 2 kvk        0
                            ¯  a     ¯
                Theorem 1.36 implies then that u ∈ W 2,2 . We can then integrate by parts (4.1),
                bearinginmindthat v (a)= v (b)= 0, and using the fundamental lemma of the
                calculus of variations (cf. Theorem 1.24), we deduce that
                                  u (x)= g u (x, u (x)) , a.e. x ∈ (a, b) .      (4.3)
                                   00
                We are now in a position to start an iteration process. Since u ∈ W  2,2  (a, b)
                                                     1
                we deduce that (cf. Theorem 1.42) u ∈ C ([a, b]) and hence the function x →
                              1
                g u (x, u (x)) is C ([a, b]), g being C . Returning to (4.3) we deduce that u ∈ C 1
                                                                               00
                                             ∞
                                                                               3
                               3
                and hence u ∈ C . From there we can infer that x → g u (x, u (x)) is C ,and
                                                   3
                                                                    5
                thus from (4.3) we obtain that u ∈ C and hence u ∈ C .Continuing this
                                             00
                process we have indeed established that u ∈ C  ∞  ([a, b]).
                   We will now generalize the argument of the proposition and we start with a
                lemma.
                                     1
                Lemma 4.2 Let f ∈ C ([a, b] × R × R) satisfy (H1), (H2) and (H3’). Then
                any minimizer u ∈ W 1,p  (a, b) of (P) is in fact in W  1,∞  (a, b) and the Euler-
                Lagrange equation holds almost everywhere, i.e.
                               d
                                 [f ξ (x, u, u )] = f u (x, u, u ) ,a.e. x ∈ (a, b) .
                                          0
                                                      0
                              dx
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