Page 145 - Complementarity and Variational Inequalities in Electronics

136 Complementarity and Variational Inequalities in Electronics

now deduce that necessarily z 2 ≥ 0. The second relation in (4.123) then gives

+ (R 2 z 2 − α I R 2 z 3 ) ≤ 0.

We have

(R 2 z 2 − α I R 2 z 3 ) ≤ + (R 2 z 2 − α I R 2 z 3 ) ≤ 0.

However, R 2 z 2 ≥ 0 and −α I R 2 z 3 ≥ 0, so that

(R 2 z 2 − α I R 2 z 3 ) = 0.

Let us now suppose that b) z 1 ≥ 0. Then necessarily ≤ 0, and the third relation
in (4.123) entails also that z 3 ≥ 0. From (4.124) we now deduce that necessarily
z 2 ≤ 0. We have z 2 ≥ 0, and the second relation in (4.123) then gives

z 2 (R 2 z 2 − α I R 2 z 3 ) ≤ z 2 + z 2 (R 2 z 2 − α I R 2 z 3 ) ≤ 0,

and since z 2 ≤ 0, we get

R 2 z 2 − α I R 2 z 3 ≥ 0.
However, z 2 ≤ 0 and z 3 ≥ 0 yields R 2 z 2 − α I R 2 z 3 ≤ 0, so that

R 2 z 2 − α I R 2 z 3 = 0.

So, in both cases a) and b), we have proved that R 2 z 2 − α I R 2 z 3 = 0, and thus

1
(R 2 z 2 − α I R 2 z 3 ) = 0.
K
This property implies

1 1
(R 2 x 2 − α I R 2 x 3 ) = (R 2 X 2 − α I R 2 X 3 ),
K K
that is, the uniqueness of the output signal

R 2

V o = R 2 I E = (I − α I I ).
K
For the last two examples, we get the existence of at least one solution x of
VI(M, ,Fu). Using now Proposition 15, we see that if x is another solution,
then
T
x − x ∈ ker{M + M }, (4.125)