Page 109 - Complementarity and Variational Inequalities in Electronics

100 Complementarity and Variational Inequalities in Electronics

Example 51. Let
⎛ ⎞
1 −30
M = ⎝ 1 1 0 ⎠
⎜
⎟
0 0 2
and set
3
(∀x ∈ R ) : (x) = 3 .
R
+
Let us first note 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

T
3
(∀v ∈ R ∩ ker(M )) : q,v ≥ 0.
+
T
We have ker(M ) ={0}, and the necessary condition of solvability is always
satisfied.
The matrix M is positive semidefinite, and
T
3
ker(M + M ) ={x ∈ R : x 1 = x 2 ,x 3 = 0}.
3
We have also D( ) ∞ = R 3 + and K(M, ) ={x ∈ R :−x 2 ≤ x 1 ,x 2 ≤
1 T
3 x 1 ,x 3 ≥ 0}.If x ∈ D( ) ∞ ∩ ker(M + M ) ∩ K(M, ), then 0 ≤ x 1 = x 2
1
and x 2 − x 1 ≤ 0 entail that x 1 = x 2 = 0. Thus
3
T
D( ) ∞ ∩ ker(M + M ) ∩ K(M, ) ={0}.
3
Therefore, for all q ∈ R , problem VI(M,q, ) has at least one solution.
Example 52. Let
⎛ ⎞
1 −30
M = ⎝ 1 1 0 ⎠
⎜
⎟
0 0 2
and set

3
(∀x ∈ R ) : (x) =|x 1 |.
Let us note that a necessary condition of solvability (see Section 4.3)isgiven
by
T
(∀v ∈ ker(M )) : q,v + ∞ (v) ≥ 0.