Page 181 - Complementarity and Variational Inequalities in Electronics

172 Complementarity and Variational Inequalities in Electronics

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

Ax,∇V(x) + ψ K (x) − ψ K (x −@ V(x)) = x ≥ 0.
2
k
Here
h 2
E(F,ψ K ,V ) ={x ∈ K : x = 0}={(x 1 ,0);x 1 ≥ 0}.
2
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 . (5.55)
1
Setting v 1 = x 1 (t) in (5.55), we get

kx 1 (t)v 2 ≥ 0,∀v 2 ∈ R, a.e. t ≥ t 0 . (5.56)

Setting v 2 = 0in (5.55),wealsoget

0 ≤ x 1 (t) ⊥ x (t) ≥ 0, a.e. t ≥ t 0 . (5.57)
1
2
From relation (5.57) 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 (5.56) 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 68. Let h> 0 and k> 0. Suppose that
2
(∀x ∈ R ) : F(x) = Ax +∇ (x)