Page 180 - Complementarity and Variational Inequalities in Electronics

The Nonregular Dynamical System Chapter | 5 171

Corollary 12. Suppose that the assumptions of Theorem 11 hold together with
conditions (5.36), (5.42), and (5.43). Suppose also that D(∂(ϕ) is closed. Let
1
n
V ∈ C (R ;R) be a function such that the function ϕ(·) − ϕ(·−@ V(·)) is
lower semicontinuous on D(∂ϕ) and
(∀x ∈ D(∂ϕ)) : V(x) →+∞ as ||x|| → +∞. (5.54)

We suppose also that

(∀x ∈ D(∂ϕ)) :
F(x),∇V(x) + ϕ(x) − ϕ(x −@ V(x)) ≥ 0.

Let M be the largest invariant subset of E(F,ϕ,V ). Then, for each x 0 ∈ D(∂ϕ),
the orbit γ(x 0 ) is bounded, and

lim d(x(τ;t 0 ,x 0 ),M) = 0.
τ→+∞
n
Proof. Let x 0 ∈ D(∂ϕ).Weset V(x 0 ) (V ) ={x ∈ R : V(x) ≤ V(x 0 )} and
= V(x 0 ) (V ) ∩ D(∂ϕ).The set V(x 0 ) (V ) is closed. Moreover, D(∂ϕ) ∩
V(x 0 ) (V ) is bounded (because of (5.54)) and closed. Thus is compact.
Lemma 5 ensures that is invariant. Here x 0 ∈ , and thus γ(x 0 ) ⊂ . There-
fore γ(x 0 ) is bounded. Moreover, from Theorem 18, we obtain

lim d(x(τ;t 0 ,x 0 ),M ∗ ) = 0,
τ→+∞
where M ∗ is the largest invariant subset of E (F,ϕ,V ). It is clear that
M ∗ ⊂ M, and the conclusion follows.

Example 67. Let h> 0 and k> 0. Suppose that
2
(∀x ∈ R ) : F(x) = Ax

with

0 −1
A = .
k h
Let us 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