Page 127 - Complementarity and Variational Inequalities in Electronics

118 Complementarity and Variational Inequalities in Electronics

Proposition 21. Suppose that assumptions (H1) and (H2) are satisfied
and let be defined as in (4.90).i)If (x,y L ) is a solution of problem
n
NRM(A,B,C,D,u, ), then x ∈ R is a solution of the variational inequality
VI(−PA,−PDu, ), that is,

n
−PAx − PDu,v − x + (v) − (x) ≥ 0,∀v ∈ R . (4.92)
n
ii) If x ∈ R is a solution of the variational inequality VI(−PA,−PDu, ),
m
then there exists y L ∈ R such that (x,y L ) is a solution of problem NRM(A,B,
C,D,u, ).
Proof. Let (x,y L ) be a solution of problem (4.87)–(4.89). Then

0 ∈ Ax − B∂ (Cx) + Du,

which is equivalent to

0 ∈ PAx − PB∂ (Cx) + PDu

since P is invertible. Thus

T
0 ∈ PAx − C ∂ (Cx) + RDu.
The existence of a vector ¯y 0 = C ¯x 0 at which is finite and continuous ensures
that (see Proposition 3)
n T
(∀z ∈ R ) : C ∂ (Cz) = ∂ (z).

Thus
0 ∈ PAx + PDu − ∂ (x),

that is,
n
−PAx − PDu,v − x + (v) − (x) ≥ 0,∀v ∈ R .
Suppose now that x is solution of problem (4.92). As before, we see that

0 ∈ Ax − B∂ (Cx) + Du.
Therefore there exists y L ∈ ∂ (Cx) such that

0 = Ax − By L + Du.
Then we obtain relations (4.87)–(4.89) by setting y = Cx.