Page 112 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 103

and set
3
(∀x ∈ R ) : (x) = R + (x 1 ) + R + (x 2 ) +|x 3 |.
The matrix M is positive semidefinite, and

3
ker(M) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.
∗
We also have D( ) ∞ = R + ×R + ×R, D( ∞ ) = R + ×R + ×R, (D( ∞ )) =
3
R + × R + ×{0}, and K(M, ) ={x ∈ R : x 1 = x 2 ,x 3 = 0}. Thus
3
D( ) ∞ ∩ ker(M) ∩ K(M, ) ={x ∈ R : x 1 = x 2 ≥ 0,x 3 = 0}.
The condition

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

reduces here to
(∀v 1 > 0) : (q 1 + q 2 )v 1 > 0,
which is equivalent to

q 1 + q 2 > 0.
3
Therefore, for all q ∈ R such that q 1 + q 2 > 0, problem VI(M,q, ) has at
least one solution.

Example 54. Let
⎛ ⎞
1 −10
⎜
M = ⎝ −1 1 0 ⎠
⎟
0 0 1
and set
3
(∀x ∈ R ) : (x) =|x 1 |+|x 2 |+|x 3 |.
The matrix M is positive semidefinite, and

3
ker(M) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.
3
3
∗
We have also D( ) ∞ = R , D( ∞ ) = R , (D( ∞ )) ={0}, and
3
K(M, ) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.
Thus
3
D( ) ∞ ∩ ker(M) ∩ K(M, ) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.