Page 161 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
P. 161

148                                                   Minimal surfaces

                                                               n
                       Theorem 5.28 (Korn-Müntz).Let Ω ⊂ R be a bounded domain with C    2,α ,
                       0 <α < 1, boundary and consider the problem
                                ⎧
                                        ³         ´
                                                 2        n P
                                ⎪
                                ⎪ Mu = 1+ |∇u|     ∆u −      u x i x j  =0   in Ω
                                                               u u x i x j
                                ⎨
                                                        i,j=1
                                ⎪
                                ⎪
                                ⎩
                                                   u = u 0                  on ∂Ω.
                                                                     ¡ ¢
                       Then there exists  > 0 so that, for every u 0 ∈ C 2,α  Ω with ku 0 k  2,α ≤  ,the
                                                                                 C
                                                      ¡ ¢
                       above problem has a (unique) C 2,α  Ω solution.
                       Remark 5.29 It is interesting to compare this theorem with the preceding one.
                       We see that we have not made any assumption on the mean curvature of ∂Ω;
                       but we require that the C 2,α  norm of u 0 to be small. We should also observe
                       that the above mentioned result of Finn and Osserman shows that if Ω is non
                       convex and if ku 0 k C 0 ≤  , this is, in general, not sufficient to get existence of
                       solutions. Therefore we cannot, in general, replace the condition ku 0 k  2,α small,
                                                                                   C
                              C 0 small.
                       by ku 0 k
                          Proof. The proof is divided into three steps. We write
                                                          n
                                                         X                   2
                                                               u u
                                 Mu =0 ⇔ ∆u = N (u) ≡       u x i x j x i x j  − |∇u| ∆u.  (5.14)
                                                        i,j=1
                       From estimates of the linearized equation ∆u = f (Step 1) and estimates of the
                       nonlinear part N (u) (Step 2), we will be able to conclude (Step 3) with the help
                       of Banach fixed point theorem.
                          Step 1. Let us recall the classical Schauder estimates concerning Poisson
                       equation (see Theorem 6.6 and page 103 in Gilbarg-Trudinger [49]). If Ω ⊂ R n
                                              n
                       is a bounded domain of R with C 2,α  boundary and if
                                                 ⎧
                                                 ⎨ ∆u = f    in Ω
                                                                                       (5.15)
                                                 ⎩
                                                    u = ϕ   on ∂Ω
                       we can then find a constant C = C (Ω) > 0 so that the (unique) solution u of
                       (5.15) satisfies
                                           kuk C 2,α ≤ C (kfk C 0,α + kϕk C 2,α) .     (5.16)
                          Step 2. We now estimate the nonlinear term N. We will show that we can
                                                                                         ¡ ¢
                       find a constant, still denoted by C = C (Ω) > 0,so that for every u, v ∈ C 2,α  Ω
                       we have
                                                                    2
                             kN (u) − N (v)k C 0,α ≤ C (kuk C 2,α + kvk C 2,α) ku − vk C 2,α .  (5.17)
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