Page 179 - Complementarity and Variational Inequalities in Electronics

170 Complementarity and Variational Inequalities in Electronics

Proof. Since γ(x 0 ) is bounded, by Remark 34 iii) and iv), (x 0 ) is nonempty,
and
lim d(x(τ;t 0 ,x 0 ), (x 0 )) = 0.
τ→+∞
Let us now check that (x 0 ) ⊂ E (F,ϕ,V ). We first note that

(x 0 ) ⊂ γ(x 0 ) ⊂ D(∂ϕ) ∩ = D(∂ϕ) ∩ .

By Lemma 4 there exists k ∈ R such that V(x) = k,∀x ∈ (x 0 ).Let z ∈ (x 0 ).
Using Theorem 17, we see that x(t;t 0 ,z) ∈ (x 0 ),∀t ≥ t 0 , and thus

V(x(t;t 0 ,z)) = k,∀t ≥ t 0 ,

so that
d
V(x(t;t 0 ,z)) = 0, a.e. t ≥ t 0 . (5.52)
dt
Setting x(·) ≡ x(·;t 0 ,z), we check as in the proof of Lemma 4 that

dx

∇V (x(t)), (t) ≤−
F(x(t)),∇V(x(t))
dt
− ϕ(x(t)) + ϕ(x(t) −@ V (x(t))), a.e. t ≥ t 0 . (5.53)

From (5.52) and (5.53) we deduce that

F(x(t)),∇V(x(t)) + ϕ(x(t)) − ϕ(x(t) −@ V (x(t))) ≤ 0, a.e. t ≥ t 0 .
Using our assumption on ϕ, we see that the mapping

t →
F(x(t;t 0 ,z)),∇V(x(t;t 0 ,z)) + ϕ(x(t;t 0 ,z))
− ϕ(x(t;t 0 ,z) −@ V(x(t;t 0 ,z)))

is lower semicontinuous on [t 0 ,+∞), and thus taking the liminf as t → t 0 ,we
obtain

F(z),∇V(z) + ϕ(z) − ϕ(z −@ V(z)) ≤ 0.
We obtain that z ∈ E (F,ϕ,V ). Finally, (x 0 ) ⊂ M since (x 0 ) ⊂ E (F,ϕ,V )
and (x 0 ) is invariant (see Theorem 17). The conclusion follows.

Remark 35. Note that the conditions of Theorem 18 ensure that
S(F,ϕ) ∩ ⊂ M.

Indeed, Proposition 24 yields S(F,ϕ) ∩ ⊂ E (F,ϕ,V ), so that S(F,ϕ) ∩
is invariant.