Page 183 - Complementarity and Variational Inequalities in Electronics

174 Complementarity and Variational Inequalities in Electronics

Setting v 2 = 0in (5.58),wealsoget

0 ≤ x 1 (t) ⊥ x (t) ≥ 0, a.e. t ≥ t 0 . (5.60)
1
2
From relation (5.60) we obtain d x (t) = 0, a.e. t ≥ t 0 , from which it easily
dt 1
follows that x 1 (t) = z 1 ,a.e. t ≥ t 0 . Relation (5.59) then gives z 1 = 0. Therefore,
M ={0} is the largest invariant subset of E(F,ψ K ,V ), and for any x 0 ∈ K,we
have

lim x(t;t 0 ,x 0 ) = 0.
t→+∞
Example 69. Let h> 0 and k> 0. Suppose that

2
(∀x ∈ R ) : F(x) = Ax +∇ (x)
with

0 −1
A =
k 0
and
4
2
(∀(x 1 ,x 2 ) ∈ R ) : (x 1 ,x 2 ) = x .
1
Let us also consider the closed convex set
2
K ={(x 1 ,x 2 ) ∈ R : x 1 ≥ 0}.
We set
1 1
2 2 2
(∀ (x 1 ,x 2 ) ∈ R ) : V(x 1 ,x 2 ) = x + x .
2
1
2 2k
We check that both assumptions of Corollary 12 are satisfied. Indeed,
1
2
(∀ (x 1 ,x 2 ) ∈ R ) :∇V(x 1 ,x 2 ) = x 1 , x 2 .
k
If x ∈ K, then
k − 1
x −@ V(x) = (0, x 2 ) ∈ K,
k
and thus
∀x ∈ K : ψ K (x) − ψ K (x −@ V(x)) = 0.

We obtain
4

F(x),∇V(x) + ψ K (x) − ψ K (x −@ V(x)) = 4x ≥ 0.
1