Page 118 - Complementarity and Variational Inequalities in Electronics
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A Variational Inequality Theory Chapter | 4 109


                           This is a contradiction to condition (4.77) since z 
= 0 and we have already
                           proved that z ∈ B(M,K). The sequence {u i } is thus bounded, and we may con-
                           clude as in the proof of Theorem 5.
                              Part c) is obtained as usual from Proposition 15.


                           Example 57. Let
                                                      ⎛           ⎞
                                                         1  −40
                                                                  ⎟
                                                  M = ⎝ 1    1  0 ⎠
                                                      ⎜
                                                         0   0  2
                           and set
                                                             3
                                                       K = R .
                                                             +
                           Let
                                                       ⎛          ⎞
                                                         1   0  0
                                                  	 α = ⎝ 0  α  0 ⎠ .
                                                       ⎜
                                                                  ⎟
                                                         0   0  1
                           We have

                                                     2                 2    2
                                         Mx,	 α x = x + (α − 4)x 1 x 2 + αx + 2x .
                                                     1                 2    3
                           We may thus choose α> 4to have

                                                      3
                                               (∀x ∈ R ) : Mx,	 α x ≥ 0.
                                                      +
                           We have

                                                      x ∈ B(M,K)
                           if and only if

                                                    ⎧
                                                            x 1 ≥ 0
                                                    ⎪
                                                    ⎪
                                                    ⎪
                                                    ⎪
                                                            x 2 ≥ 0
                                                    ⎪
                                                    ⎪
                                                    ⎪
                                                    ⎨
                                                            x 3 ≥ 0
                                                    ⎪
                                                    ⎪
                                                    ⎪ x 1 − 4x 2 ≥ 0
                                                    ⎪
                                                    ⎪
                                                    ⎪
                                                    ⎪
                                                       x 1 + x 2 ≥ 0
                                                    ⎩
                           and
                                                                      2
                                           x 1 (x 1 − 4x 2 ) + x 2 (x 1 + x 2 ) + 2x = 0,
                                                                      3
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