Page 110 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 101

It reduces here to

3
T
(∀v ∈ R ∩ ker(M )) : q,v +|v 1 |≥ 0.
+
T
We have ker(M ) ={0}, and the necessary condition of solvability is always
satisfied.
The matrix M is positive semidefinite, and
3
T
ker(M + M ) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.
3 ∗
We also have D( ) ∞ = R, D( ∞ ) = R , (D( ∞ )) ={0}, and K(M, ) =
{0}. Thus
T
D( ) ∞ ∩ ker(M + M ) ∩ K(M, ) ={0}.
3
Therefore, for all q ∈ R , problem VI(M,q, ) has at least one solution.
If in addition the matrix M is symmetric, then u is a solution of problem
VI(M,q, ) if and only if u is a solution of the optimization problem

1
min{ Mx,x + q,x + (x)}. (4.66)
x∈R 2
n
This last case is also of particular interest.
n
Corollary 8. Let : R → R ∪{+∞} be a proper convex lower semicontinu-
ous function with closed domain, and let M ∈ R n×n be a positive semidefinite
and symmetric matrix.
n
a) If D( ) ∞ ∩ ker(M) ∩ K(M, ) ={0}, then for each q ∈ R , problem
VI(M,q, ) has at least one solution.
b) Suppose that D( ) ∞ ∩ ker(M) ∩ K(M, )
= {0}.If

(∀v ∈ D( ) ∞ ∩ ker(M) ∩ K(M, ), v
= 0) : q,v + ∞ (v) > 0, (4.67)

then problem VI(M,q, ) has at least one solution.
c) If u 1 and u 2 are two solutions of problem VI(M,q, ), then
u 1 − u 2 ∈ ker(M) (4.68)

and

q,u 1 − u 2 = (u 2 ) − (u 1 ). (4.69)
d) If
(∀x ∈ D( ), z ∈ ker(M)) : (x + z) = (x) (4.70)