Page 107 - Complementarity and Variational Inequalities in Electronics

98 Complementarity and Variational Inequalities in Electronics

Example 50. Let

⎛ ⎞
2 −10
M = ⎝ 0 1 0 ⎠
⎟
⎜
0 0 0
and set
3
(∀x ∈ R ) : (x) = 3 .
R
+
Let us first remark that a necessary condition of solvability (see Section 4.3)is
given by
T
(∀v ∈ ker(M )) : q,v + ∞ (v) ≥ 0.
It reduces here to
3 T
(∀v ∈ R ∩ ker(M )) : q,v ≥ 0.
+
3
T
We have ker(M ) ={v ∈ R : v 1 = v 2 = 0}, and the necessary condition of
solvability reads
(∀v 3 ≥ 0) : q 3 v 3 ≥ 0,

which is equivalent to q 3 ≥ 0.
The matrix M is positive semidefinite, and

3
T
ker(M + M ) ={x ∈ R : x 1 = x 2 = 0}.
3
3
We have also D( ) ∞ = R and K(M, ) ={x ∈ R : 2x 1 ≥ x 2 ≥ 0}. Thus
+
T 3
D( ) ∞ ∩ ker(M + M ) ∩ K(M, ) ={x ∈ R : x 1 = x 2 = 0,x 3 ≥ 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
3
(∀v ∈ R ,v 1 = v 2 = 0,v 3 > 0) : q,v > 0,
that is,
(∀v 3 > 0) : q 3 v 3 > 0,

3
or also q 3 > 0. Thus, for all q ∈ R such that q 3 > 0, problem VI(M,q, ) has
at least one solution.