Page 109 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
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96                                                      Direct methods

                       Example 3.15 The minimal surface problem has an integrand given by
                                                              q
                                                                      2
                                             f (x, u, ξ)= f (ξ)=  1+ |ξ|
                       that satisfies (H3) with p =1,since

                                                 ⎛               ⎞ 1/2
                                                      ¯         ¯ 2
                                                    n ¯         ¯
                                                   X ¯     ξ    ¯
                                                 ⎜          i    ⎟
                                                      ¯         ¯     ≤ 1 .
                                         |f ξ (ξ)| = ⎝  q        ⎠
                                                      ¯        2  ¯
                                                   i=1  ¯  1+ |ξ|  ¯
                       The equation (E) is the so called minimal surface equation
                                                     ∇u
                                              div q          =0,in Ω
                                                           2
                                                    1+ |∇u|
                       and can be rewritten as
                                                       n
                                      ³        ´      X
                                              2
                                                             u u
                                       1+ |∇u|  ∆u −      u x i x j x i x j  =0,in Ω .
                                                      i,j=1
                                                                    2
                                                                                      0
                       Example 3.16 Let f (x, u, ξ)= f (u, ξ)= g (u) |ξ| ,with 0 ≤ g (u) , |g (u)| ≤
                       g 0 .We then have
                                                    2
                                             0
                                 |f u (u, ξ)| = |g (u)||ξ| , |f ξ (u, ξ)| =2 |g (u)||ξ| ≤ 2g 0 |ξ|
                       We see that if g (u) 6=0,then f does not satisfy (H3) but only the above (H3’).
                                     0
                       We are therefore authorized to write only
                                Z
                                   [f u (u, ∇u) ϕ + hf ξ (u, ∇u); ∇ϕi] dx =0, ∀ϕ ∈ C ∞  (Ω)
                                                                              0
                                  Ω
                                                                          1,2
                       or more generally the equation should hold for any ϕ ∈ W  (Ω) ∩ L ∞  (Ω).
                                                                         0
                          Let us now recall two examples from Section 2.2, showing that, without any
                       hypotheses of convexity of the function f, the converse part of the theorem is
                       false.

                       Example 3.17 (Poincaré-Wirtinger inequality).Let λ>π, n =1 and
                                                              1 ¡  2  2 2 ¢
                                          f (x, u, ξ)= f (u, ξ)=  ξ − λ u
                                                              2
                                                1
                                     ½        Z                                ¾
                                                                        1,2
                             (P)   inf I (u)=    f (u (x) ,u (x)) dx : u ∈ W  (0, 1)  = m.
                                                          0
                                                                        0
                                               0
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