Page 96 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
P. 96

The model case: Dirichlet integral                                 83

                Therefore returning to the above inequality we have indeed obtained that

                                         lim infI (u ν ) ≥ I (u) .
                                          ν→∞
                   Step 3. Wenow combinethe twosteps. Since {u ν } was a minimizing se-
                quence (i.e. I (u ν ) → inf {I (u)} = m) and for such a sequence we have lower
                semicontinuity (i.e. lim inf I (u ν ) ≥ I (u)) we deduce that I (u)= m,i.e. u is a
                minimizer of (D).
                                                                        1,2
                   Part 2 (Uniqueness). Assume that there exist u, v ∈ u 0 + W  (Ω) so that
                                                                       0
                                           I (u)= I (v)= m
                and let us show that this implies u = v.Denote by w =(u + v) /2 and observe
                               1,2                      2
                that w ∈ u 0 + W  (Ω). The function ξ → |ξ| being convex, we can infer that
                              0
                w is also a minimizer since
                                               1       1
                                   m ≤ I (w) ≤  I (u)+ I (v)= m,
                                               2       2
                which readily implies that
                                "                             #
                              Z                     ¯        ¯ 2
                                  1    2   1    2   ¯ ∇u + ∇v ¯
                                   |∇u| +   |∇v| −  ¯        ¯  dx =0 .
                                  2        2        ¯   2    ¯
                               Ω
                                                         2
                Appealing once more to the convexity of ξ → |ξ| , we deduce that the integrand
                is non negative, while the integral is 0. This is possible only if
                                               ¯         ¯ 2
                              1    2   1    2  ¯ ∇u + ∇v ¯
                               |∇u| +   |∇v| −  ¯        ¯  =0 a.e. in Ω .
                              2        2       ¯    2    ¯
                                                     2
                We now use the strict convexity of ξ → |ξ| to obtain that ∇u = ∇v a.e. in Ω,
                which combined with the fact that the two functions agree on the boundary of
                                    1,2
                Ω (since u, v ∈ u 0 + W  (Ω)) leads to the claimed uniqueness u = v a.e. in Ω.
                                   0
                   Part 3 (Euler-Lagrange equation). Let us now establish (3.1). Let   ∈ R and
                      1,2                                       1,2
                ϕ ∈ W   (Ω) be arbitrary. Note that u +  ϕ ∈ u 0 + W  (Ω), which combined
                     0                                          0
                with the fact that u is the minimizer of (D) leads to
                                                   Z
                                                      1          2
                              I (u) ≤ I (u +  ϕ)=      |∇u +  ∇ϕ| dx
                                                      2
                                                    Ω
                                               Z
                                                                2
                                    = I (u)+      h∇u; ∇ϕi dx +   I (ϕ) .
                                                Ω
                The fact that   is arbitrary leads immediately to (3.1), which expresses nothing
                else than                           ¯
                                          d         ¯
                                           I (u +  ϕ) ¯  =0 .
                                         d          ¯
                                                      =0
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