Page 222 - INTRODUCTION TO THE CALCULUS OF VARIATIONS
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Chapter 4: Regularity                                             209

                while
                                  3
                                 x x 2        x 1 x 2  ¡  2  2 ¢
                                  1
                                                               0
                                       00
                              =      V (|x|)+       2x +3x    V (|x|)
                         u x 1 x 1                    1     2
                                   2             3
                                 |x|           |x|
                                  3
                                 x x 1        x 1 x 2  ¡  2  2 ¢
                                  2
                                                               0
                              =      V (|x|)+       2x +3x    V (|x|)
                                       00
                                   2             3
                         u x 2 x 2                    2     1
                                 |x|           |x|
                                                4
                                  2 2
                                                    2 2
                                 x x          x + x x + x  4 2
                                                    1 2
                                                1
                                  1 2
                                       00
                              =      V (|x|)+               V (|x|)+ V (|x|) .
                                                             0
                                   2                  3
                         u x 1 x 2
                                 |x|               |x|
                We therefore get that
                                               0              ∞
                                 u x 1 x 1  ,u x 2 x 2  ∈ C (Ω) ,u x 1 x 2  / ∈ L  (Ω) .
                Exercise 4.3.4. Let V (r)= log |log r|. A direct computation shows that
                                      x 1                  x 2
                                                               0
                                    =     0              =    V (|x|)
                                 u x 1   V (|x|) and u x 2
                                      |x|                  |x|
                and therefore
                                          x 2 1        x 2 2
                                                           0
                                       =     V (|x|)+    V (|x|)
                                              00
                                           2            3
                                  u x 1x 1
                                         |x|          |x|
                                          x 2 2        x 2 1
                                              00
                                                           0
                                       =     V (|x|)+    V (|x|)
                                           2            3
                                  u x 2 x 2
                                         |x|          |x|
                                         x 1 x 2       x 1 x 2
                                                             0
                                       =       00          V (|x|) .
                                            2             3
                                  u x 1 x 2   V (|x|) −
                                          |x|          |x|
                This leads to
                                             0
                                            V (|x|)      −1         1
                             ∆u = V (|x|)+         =   2       2  ∈ L (Ω)
                                    00
                                              |x|    |x| |log |x||
                                        1
                                     / ∈ L (Ω). Summarizing the results we indeed have that
                while u x 1 x 1  ,u x 1 x 2  ,u x 2x 2
                                        1
                u/∈ W  2,1  (Ω) while ∆u ∈ L (Ω). We also observe (compare with Example 1.33
                (ii)) that, trivially, u/∈ L ∞  (Ω) while u ∈ W  1,2  (Ω),since
                                ZZ              Z  1/2  dr       2π
                                        2
                                    |∇u| dx =2π              =      .
                                                            2
                                  Ω              0   r |log r|  log 2
                A much more involved example due to Ornstein [79] produces a u such that
                                               1              1
                                 u x 1 x 1  ,u x 2 x 2  ∈ L (Ω) ,u x 1 x 2  / ∈ L (Ω) .
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