Page 144 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 135

3
T
3
and if z ∈ R and
z ∈ R , then the equation z
z = 0 reduces to
+ +
⎧
⎪ z 1 (
z) 1 = 0
⎪
⎨
z 2 (
z) 2 = 0
⎪
⎪
z 3 (
z) 3 = 0.
⎩
The equation z 1 (
z) 1 = 0 yields
2 R 1 R 1
R 1 z + (1 − α N )z 1 z 2 + (1 − α I )z 1 z 3 = 0.
1
K K
3
2
Recalling that z ∈ R , we must necessarily have R 1 z = 0, and thus z 1 = 0. The
1
+
equation z 3 (Mz) 3 = 0 then yields
R 1 R 1 2
(1 − α N )z 2 z 3 + (1 − α I )z = 0,
3
K K
2
from which we deduce that necessarily R 1 (1 − α I )z = 0, and thus z 3 = 0. The
K 3
equation z 2 (Mz) 2 = 0 then gives
R 1 R 2 2
( (1 − α N ) + )z ,
2
K K
from which we deduce that z 2 = 0. Thus
D( ) ∞ ∩ N 0 (
) ∩ N(
, ) ={0},
3
and Corollary 6 ensures the solvability of VI( , ,q) for any q ∈ R .
Let x and X be two solutions of problem VI( , ,q). Then setting z =
x − X, we may use Proposition 15 to deduce that

z 1 (
z) 1 ≤ 0,z 2 (
z) 2 ≤ 0,z 3 (
z) 3 ≤ 0,

that is,
⎧
⎪ z 1 ≤ 0
⎪
⎨
z 2 + z 2 (R 2 z 2 − α I R 2 z 3 ) ≤ 0 (4.123)
⎪
⎪
z 3 ≤ 0,
⎩
where
= R 1 Kz 1 + R 1 (1 − α N )z 2 + R 1 (1 − α I )z 3 . (4.124)
Either z 1 ≤ 0or z 1 ≥ 0. Let us first suppose that a) z 1 ≤ 0. Then necessarily
≥ 0, and the third relation in (4.123) entails also that z 3 ≤ 0. From (4.124) we