Page 106 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 97

and set
3
(∀x ∈ R ) : (x) =|x 1 |+ 2|x 2 |+|x 3 |.
The necessary condition of solvability (see Section 4.3) is given by
T
(∀v ∈ ker(M )) : q,v +|x 1 |+ 2|x 2 |+|x 3 |≥ 0.
3
T
We have ker(M ) ={v ∈ R : v 1 = v 2 = 0}, and the necessary condition of
solvability reads
(∀v 3 ∈ R) : q 3 v 3 +|v 3 |≥ 0,

which is equivalent to
−1 ≤ q 3 ≤ 1.
Indeed, we must have

(∀v 3 > 0) : q 3 v 3 + v 3 ≥ 0
and
(∀v 3 < 0) : q 3 v 3 − v 3 ≥ 0,

and thus q 3 + 1 ≥ 0 and q 3 − 1 ≤ 0.
The matrix M is positive semidefinite, and

T
3
ker(M + M ) ={x ∈ R : x 1 = x 2 = 0}.
3
3
∗
We have also D( ) ∞ = R , D( ∞ ) = R , (D( ∞ )) ={0}, and
3
K(M, ) ={x ∈ R : x 1 = x 2 = 0}.
Thus
3
T
D( ) ∞ ∩ ker(M + M ) ∩ K(M, ) ={x ∈ R : x 1 = x 2 = 0}.
3
Here 0 ∈ D( ) = R , and the condition
+
T
(∀v ∈ D( ) ∞ ∩ ker(M + M ) ∩ K(M, ), v
= 0) : q,v + ∞ (v) > 0
reduces to
(∀v 3 ∈ R,v 3
= 0) : q 3 v 3 +|v 3 | > 0,
which is equivalent to

−1 <q 3 < 1.
3
Therefore, for all q ∈ R such that |q 3 | < 1, problem VI(M,q, ) has at least
one solution.