Page 73 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
P. 73

60                                                    Classical methods

                                                                  2
                          Proof 1. Observe firstthatfor any u ∈ C ([a, b]) we have, by straight
                       differentiation,
                                        d
                                          [f (x, u, u ) − u f ξ (x, u, u )]
                                                               0
                                                       0
                                                  0
                                       dx
                                                     ∙                          ¸
                                                                   d
                                               0    0         0               0
                                   = f x (x, u, u )+ u f u (x, u, u ) −  [f ξ (x, u, u )] .
                                                                  dx
                       By Theorem 2.1 we know that any solution u of (P) satisfies the Euler-Lagrange
                       equation
                                       d
                                                                         0
                                                     0
                                         [f ξ (x, u (x) ,u (x))] = f u (x, u (x) ,u (x))
                                       dx
                       hence combining the two identities we have the result.
                          Proof 2. We will use a technique known as variations of the independent
                       variables and that we will encounter again in Chapter 5; the classical deriva-
                       tion of Euler-Lagrange equation can be seen as a technique of variations of the
                       dependent variables.
                                                              −1
                          Let   ∈ R, ϕ ∈ C  ∞  (a, b), λ =(2 kϕ k L ∞)  and
                                                         0
                                         0
                                              ξ (x,  )= x +  λϕ (x)= y.
                       Observe that for | | ≤ 1,then ξ (.,  ):[a, b] → [a, b] is a diffeomorphism with
                       ξ (a,  )= a, ξ (b,  )= b and ξ (x,  ) > 0.Let η (.,  ):[a, b] → [a, b] be its inverse,
                                                x
                       i.e.
                                                  ξ (η (y,  ) , )= y.
                       Since ξ (η (y,  ) , ) η (y,  )=1 and ξ (η (y,  ) , ) η (y,  )+ ξ (η (y,  ) , )=0,
                             x           y              x
                       we find (O (t) stands for a function f so that |f (t) /t| is bounded in a neighbor-
                       hood of t =0)
                                                                     ¡ ¢
                                                                       2
                                                             0
                                          η (y,  )= 1 −  λϕ (y)+ O
                                           y
                                          η (y,  )= −λϕ (y)+ O ( ) .

                       Set for u a minimizer of (P)

                                                 u (x)= u (ξ (x,  )) .
                       Note that, performing also a change of variables y = ξ (x,  ),
                                          Z  b  ¡              ¢


                                                           0
                               I (u )=       f x, u (x) , (u ) (x) dx
                                           a
                                          Z  b
                                                            0
                                      =      f (x, u (ξ (x,  )) ,u (ξ (x,  )) ξ (x,  )) dx
                                                                      x
                                           a
                                          Z  b
                                              ¡                         ¢
                                      =      f η (y,  ) ,u (y) ,u (y) /η (y,  ) η (y,  ) dy .
                                                             0
                                                                   y      y
                                           a
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