Page 182 - Complementarity and Variational Inequalities in Electronics

The Nonregular Dynamical System Chapter | 5 173

with

0 −1
A =
k h
and
2
4
(∀(x 1 ,x 2 ) ∈ R ) : (x 1 ,x 2 ) = x .
2
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
h 2 4 4

F(x),∇V(x) + ψ K (x) − ψ K (x −@ V(x)) = x + x ≥ 0.
2
2
k k
Here
h 2 4 4
E(F,ψ K ,V ) ={x ∈ K : x + x = 0}={(x 1 ,0);x 1 ≥ 0}.
2
2
k k
Let z = (z 1 ,0) ∈ E(F,ψ K ,V ). We claim that if γ(z) ⊂ E(F,ψ K ,V ), then
necessarily z 1 = 0. Indeed, suppose that γ(z) ⊂ E(F,ψ K ,V ) and set x(·) =
dx
x(·,t 0 ,x 0 ), x (·) = (·,t 0 ,x 0 ). From the dynamics in E(F,ψ K ,V ) we have
dt
x (t)(v 1 − x 1 (t)) + kx 1 (t)v 2 ≥ 0, ∀v 1 ≥ 0,v 2 ∈ R, a.e. t ≥ t 0 . (5.58)

1
Setting v 1 = x 1 (t) in (5.58), we get
kx 1 (t)v 2 ≥ 0,∀v 2 ∈ R, a.e. t ≥ t 0 . (5.59)