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162  Complementarity and Variational Inequalities in Electronics


                           Let us denote respectively by λ 1 > 0 and λ max > 0 the smallest and greatest
                           eigenvalues of the symmetric positive definite matrix V .Wehave
                                      1 d                α
                                         
V x(t),x(t) ≤−    
V x(t),x(t)  a.e. t ≥ t 0 .
                                      2 dt              λ max
                           Using the Gronwall inequality, we get
                                                                       2α  (t−t 0 )
                                                                     −
                                       (∀t ≥ t 0 ) :
V x(t),x(t) ≤
Vx 0 ,x 0  e  λ max  .
                           Thus

                                                           λ max      α  (t−t 0 )
                                        (∀t ≥ t 0 ) :||x(t)|| ≤  ||x 0 ||e −  λ max  .
                                                            λ 1
                           The result follows from the last inequality by setting


                                                      λ max      α
                                                c 1 =     ,c 2 =    .
                                                       λ 1      λ max
                              Inequality (5.46) entails that the trivial solution of (5.37)–(5.40) is asymp-
                           totically stable. It is also globally attractive in the sense that

                                         (∀x 0 ∈ D(∂ϕ)) : lim ||x(t;t 0 ,x 0 )|| = 0.
                                                       t→+∞
                           Remark 33. Let
                                                       n
                                                 (∀x ∈ R ) : F(x) = Mx
                           with M ∈ R n×n .If M is positive stable, then there exists a symmetric positive
                                                               T
                           definite matrix V ∈ R n×n  such that VM + M V is positive definite. Thus
                                                   n
                                             (∀x ∈ R ,x  = 0) :
Mx,V x  > 0.
                           Therefore, there exists α> 0 such that

                                                                      2
                                                    n
                                             (∀x ∈ R ) :
Mx,Gx ≥ α||x|| .
                           Indeed, suppose on the contrary that
                                                       n                  2
                                          (∀α> 0)(∃x ∈ R ) :
Mx,Gx  <α||x|| .
                                                     n
                           We can find a sequence {x n }⊂ R such that
                                                                 1     2
                                            (∀n ∈ N) :
Mx n ,Gx n   < ||x n || .
                                                                 n
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