Page 174 - Complementarity and Variational Inequalities in Electronics
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The Nonregular Dynamical System Chapter | 5 165


                           with

                                                         01 −1
                                                  C =
                                                         01     0
                           and
                                                 2
                                           (∀X ∈ R ) :  (X) = ϕ D (X 1 ) + ϕ Z (X 2 ).
                           We have seen that we may set

                                                      z(t) = Rx(t)

                           and
                                                     3              −1
                                              (∀z ∈ R ) :  (z) =  (CR  z)
                           so as to reduce the study of the circuit to the variational inequality
                                         dz         −1
                                        
  (t) − RAR  z(t),v − z(t)
                                         dt
                                                                   n
                                           +  (v) −  (z(t)) ≥ 0,∀v ∈ R , a.e. t ≥ 0.  (5.48)
                           Setting M =−RAR  −1 ,wehave

                                               ⎛            1            ⎞
                                                   0     − √        0
                                                            L 3 C 4
                                               ⎜                         ⎟
                                               ⎜    1     R 1 +R 3   R 1  ⎟
                                          M = ⎜ √                − √     ⎟.
                                                   L 3 C 4  L 3
                                               ⎝                    L 2 L 3 ⎠
                                                   0     − √  R 1  R 1 +R 2
                                                            L 2 L 3  L 2
                           The matrix M is positive semidefinite. Indeed, we have
                                                    ⎛                     ⎞
                                                      0     0        0
                                                T   ⎜     2(R 1 +R 3 )  2R 1  ⎟
                                          M + M =   ⎜ 0           − √     ⎟
                                                            L 3
                                                    ⎝                 L 2 L 3 ⎠
                                                      0  − √ 2R 1  2(R 1 +R 2 )
                                                            L 2 L 3  L 2
                           with
                                                                   2(R 1 + R 3 )
                                               T              T
                                       1 (M + M ) = 0,  2 (M + M ) =         > 0,
                                                                       L 3
                                           2(R 1 + R 2 )
                                      T                            T               T
                               3 (M + M ) =          > 0,  12 (M + M ) = 0,  13 (M + M ) = 0,
                                              L 2
                           and
                                             4(R 1 R 2 + R 1 R 3 + R 2 R 3 )
                                         T                                       T
                                23 (M + M ) =                      > 0,  123 (M + M ) = 0.
                                                     L 2 L 3
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