Page 156 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
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The Douglas-Courant-Tonelli method                                143


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                Integrating this identity and changing the variables, letting (x ,y )= ϕ (x, y),
                                                                       0
                in the right hand side we find (recalling (5.11)) that
                     h                        i            h                    i
                 RR             2  ¯ ¯       ¯ 2        RR          2          2
                   Ω   |u 0 (x ,y )| + u 0 (x ,y )  dx dy =  Ω  |v x (x, y)| + |v y (x, y)|  dxdy
                                                    0
                            0
                              0
                                         0
                                                  0
                                           0 ¯
                        x
                                     y
                           h³           ´                            i
                        RR      2      2  ¡      ¢
                     −       |v x | − |v y |  λ x − µ  +2 hv x ; v y i (λ y + µ ) dxdy + o ( ) .
                          Ω                     y                  x
                                                                                (5.12)
                   Step 3.2. We now use Riemann theorem to find a conformal mapping

                                             α : Ω → Ω

                which is also a homeomorphism from Ω onto Ω . We can also impose that the
                mapping verifies


                                           α (w i )= ϕ (w i )
                where w i are the points that enter in the definition of S.
                   We finally let





                                v (x, y)= u ◦ α (x, y)= v ◦ ψ ◦ α (x, y)

                where u is as in Step 3.1.

                   Since v ∈ S,we deduce that v ∈ S. Therefore using the conformal invariance
                of the Dirichlet integral (see Exercise 5.3.1), we find that
                                      ZZ
                                     1    h        2   ¯      ¯ 2 i

                                                       ¯
                          D (v )=          |v (x, y)| + v (x, y) ¯  dxdy
                                             x
                                                        y
                                     2   Ω
                                      ZZ   h                        i
                                     1                2  ¯        ¯ 2
                                                         ¯
                                                 0
                                 =          |u 0 (x ,y )| + u 0 (x ,y )  dx dy 0
                                                                        0
                                                               0
                                                    0
                                                                 0 ¯
                                              x
                                                           y
                                     2
                                         Ω
                which combined with (5.12) leads to
                                            ZZ
                                                h³          ´  ¡      ¢ i
                                                     2      2
                        D (v )= D (v) −           |v x | − |v y |  λ x − µ y  dxdy
                                          2
                                              Ω
                                      ZZ
                                   −      [hv x ; v y i (λ y + µ )] dxdy + o ( ) .
                                                        x
                                        Ω

                Since v , v ∈ S and v is a minimizer of the Dirichlet integral, we find that
                    ZZ
                        h³          ´                            i
                             2      2  ¡      ¢
                          |v x | − |v y |  λ x − µ  +2 hv x ; v y i (λ y + µ ) dxdy =0.  (5.13)
                                             y                  x
                      Ω
                   Step 3.3. We finally choose in an appropriate way the functions λ, µ ∈
                   ¡ ¢
                C ∞  Ω that appeared in the previous steps. We let σ, τ ∈ C  ∞  (Ω) be arbitrary,
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