Page 155 - Complementarity and Variational Inequalities in Electronics

146 Complementarity and Variational Inequalities in Electronics

and

y L (t) ∈ ∂ (y(t)). (5.3)
p
Note that if (∀t ≥ 0) : u(t) = u for some given u ∈ R , then the stationary
solutions of (5.1)–(5.3) are given by the solutions of problem NRM(A,B,C,
D,u, ) discussed in the previous chapter.
1
n
p
We suppose that u ∈ L (0,+∞;R ) and for x 0 ∈ R , we consider problem
loc
n
P(x 0 ,A,B,C,D,u, ): Find a function x :[0,+∞[ → R ; t → x(t), and a
m
function y L :[0,+∞[ → R ; t → y L (t) such that:
0 n
x ∈ C ([0,+∞[;R ), (5.4)
1 n
By L ∈ L (0,+∞;R ), (5.5)
loc
dx 1 n
∈ L (0,+∞;R ), (5.6)
loc
dt
x(0) = x 0 , (5.7)
dx
(t) = Ax(t) − By L (t) + Du(t), a.e. t ≥ 0 (5.8)
dt
y(t) = Cx(t), ∀ t ≥ 0, (5.9)
and
y L (t) ∈ ∂ (y(t)), a.e. t ≥ 0. (5.10)

Let us now make the following two assumptions.

Assumption (G1): There exists a symmetric and invertible matrix R ∈ R n×n
such that
T
R −2 C = B.
n
Assumption (G2): There exists z 0 ∈ R such that is finite and continuous at
y 0 = CR −1 z 0 .

) .Using (5.8), (5.9), and (5.10), we may consider the
Note that R −2 = (R −1 2
differential inclusion
dx
∈ Ax − B∂ (Cx) + Du.
dt
Setting z = Rx, we remark that

dx
∈ Ax − B∂ (Cx) + Du
dt
⇔