Page 191 - Complementarity and Variational Inequalities in Electronics

182 Complementarity and Variational Inequalities in Electronics

Remark 36. We have

3
∗
K ={w ∈ R : w 1 = 0,w 2 ≥ 0,w 3 ≤ 0 and w 2 + αw 3 ≥ 0}.
Indeed,
3
∗
K ={w ∈ R :
w,z ≥ 0,∀v ∈ K}.
Let us first suppose that

3
w 1 z 1 + w 2 z 2 + w 3 z 3 ≥ 0,∀(z 1 ,z 2 ,z 3 ) ∈ R : z 2 ≥ 0 and z 3 ≤ αz 2 .
Setting z 2 = z 3 = 0, we get

w 1 z 1 ≥ 0,∀z 1 ∈ R,

and thus w 1 = 0. Setting now z 1 = z 3 = 0, we get

w 2 z 2 ≥ 0,∀z 2 ≥ 0,
and thus w 2 ≥ 0. Setting z 1 = z 2 = 0, we get

w 3 z 3 ≥ 0,∀z 3 ≤ 0,
and thus w 3 ≤ 0. Setting z 1 = 0,z 3 = αz 2 , we get

(w 2 + αw 3 )z 2 ≥ 0,∀z 2 ≥ 0,

and thus w 2 + αw 3 ≥ 0. Let us now suppose that w 1 = 0, w 2 ≥ 0, w 3 ≤ 0, and
w 2 + αw 3 ≥ 0. Then, for all z 1 ∈ R, z 2 ≥ 0, and z 3 ≤ αz 2 ,wehave

w 1 z 1 + w 2 z 2 + w 3 z 3 = w 2 z 2 + w 3 z 3 ,

and from z 3 ≤ αz 2 and w 3 ≤ 0 it follows that w 3 z 3 ≥ αw 3 z 2 . Thus

w 1 z 1 + w 2 z 2 + w 3 z 3 ≥ w 2 z 2 + αw 3 z 2 = (w 2 + αw 3 )z 2 ≥ 0.

A stationary solution z has to satisfy
∗
∗
∗
∗
K z ⊥ Mz ∈ K .
We have
∗
∗
∗
∗
z ∈ K ⇔ z ≥ 0,z ≤ αz ,
2 3 2
∗
∗
∗
z ⊥ Mz ⇔ z = 0.
2