Page 125 - Complementarity and Variational Inequalities in Electronics

116 Complementarity and Variational Inequalities in Electronics

On the other hand,

n
(∀x ∈ R ,t > 0) : (0) − (tx) ≥− F(tx),tx ,
and thus
(tx) (0)
n
(∀x ∈ R ,t > 0) : ≤ + F(tx),x ,
t t
which results in
∞ (x) ≤ r (x).
F
3
Example 59. Let us set, for all x = (x 1 ,x 2 ,x 3 ) ∈ R ,
⎛ ⎞
|x 1 |
⎜ ⎟
F(x) = ⎝ |x 1 + x 2 | ⎠
x 1 + x 2

and
(x) = 3 .
R
+
3
Here ∞ = and D( ∞ ) = D( ) ∞ = R .Wehave
+
3
(∀x ∈ R ) : F(x),x =|x 1 |x 1 +|x 1 + x 2 |x 2 + x 1 x 3 + x 2 x 3 ≥ 0.
+
We also have

3
3
N(F) ∩ R ={x ∈ R : x 1 = x 2 = 0,x 3 ≥ 0}.
+
The mapping F is positively homogeneous, and thus, if x ∈ R 3 and
+
F(x),x > 0, then r (x) =+∞. The condition
F
(∀v ∈ D( ) ∞ ,v
= 0) : r (u) + q,v + ∞ (v) > 0
F
reduces here to
3
(∀v ∈ N(F) ∩ R ,v
= 0) : q,v > 0,
+
which is equivalent to
(∀v 3 > 0) : q 3 v 3 > 0

or also to q 3 > 0. Therefore, for all q ∈ R n such that q 3 > 0, problem
VI(F,q, ) has at least one solution.