Page 113 - Complementarity and Variational Inequalities in Electronics

104 Complementarity and Variational Inequalities in Electronics

The condition

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

reduces here to

(∀v 1 ∈ R,v 1
= 0) : (q 1 + q 2 )v 1 + 2|v 1 | > 0,
which is equivalent to

|q 1 + q 2 | < 2.
3
Therefore, for all q ∈ R such that |q 1 + q 2 | < 2, problem VI(M,q, ) has at
least one solution.
Example 55. Let
⎛ ⎞
1 −10
M = ⎝ −1 1 0 ⎠
⎟
⎜
0 0 1
and set
3
(∀x ∈ R ) : (x) =|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 K(M, ) =
3
{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}.
The condition

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

reduces here to
(∀v 1 ∈ R,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. Moreover, we have
3
(∀x ∈ R ,z ∈ ker(M)) : (x + z) =|x 3 |= (x).