Page 156 - Complementarity and Variational Inequalities in Electronics

The Nonregular Dynamical System Chapter | 5 147

dx −1 −1
R ∈ RAR Rx − RB∂ (CR Rx) + RDu
dt
⇔
dz −1 −1 2 −1
∈ RAR z − R R B∂ (CR z) + RDu
dt
⇔
dz −1 −1 T −1
∈ RAR z − R C ∂ (CR z) + RDu.
dt
We set

n
(∀z ∈ R ) : (z) = (CR −1 z). (5.11)
Then
n
D( ) ={z ∈ R : CR −1 z ∈ D( )},
and with Assumption (G1),wehave

T
n
(∀z ∈ R ) : ∂ (z) = R −1 C ∂ (CR −1 z).
n
This allows us to consider, for x 0 ∈ R , problem Q(Rx 0 ,RAR −1 ,RDu, ):
n
Find a function z :[0,+∞[ → R ; t → z(t) such that:
0 n
z ∈ C ([0,+∞[;R ), (5.12)
dz 1 n
∈ L (0,+∞;R ), (5.13)
loc
dt
z(0) = Rx 0 , (5.14)
dz −1
(t) ∈ RAR z(t) + RDu(t) − ∂ (z(t)), a.e. t ≥ 0. (5.15)
dt
Note that this last differential inclusion is equivalent to the variational inequality

dz −1

(t) − RAR z(t) − RDu(t),v − z(t)
dt
n
+ (v) − (z(t)) ≥ 0,∀v ∈ R , a.e. t ≥ 0.
Proposition 23. Suppose that assumptions (G 1 ) and (G 2 ) are satisfied. If
(x,y L ) is a solution of problem P(x 0 ,A,B,C,D,u, ), then z = Rx is a so-
lution of problem Q(Rx 0 ,RAR −1 ,RDu, ).
Reciprocally, if z is solution of problem Q(Rx 0 ,RAR −1 ,RDu, ), then
there exists a function y L such that (R −1 z,y L ) is a solution of problem
P(x 0 ,A,B,C,D,u, ).