Page 177 - Complementarity and Variational Inequalities in Electronics

168 Complementarity and Variational Inequalities in Electronics

Proof. Let T> 0. We define the mapping V :[t 0 ;+∞) → R by the formula
∗
∗
V (t) = V(x(t;t 0 ,x 0 )).
The function x(·) ≡ x(·;t 0 ,x 0 ) is absolutely continuous on [t 0 ,t 0 +T ], and thus
∗
V is a.e. strongly differentiable on [t 0 ,t 0 + T ].Wehave
dV ∗ dx
(t) =
∇V (x(t)), (t) , a.e. t ∈[t 0 ,t 0 + T ].
dt dt
We know (by hypothesis) that

(∀t ≥ t 0 ) : x(t) ∈ D(∂ϕ) ∩
and

dx n

(t)+F(x(t)),v−x(t) +ϕ(v)−ϕ(x(t)) ≥ 0, ∀v ∈ R , a.e. t ≥ t 0 . (5.50)
dt
Setting v = x(t) −@ V(x(t)) in (5.50), we obtain
dx

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

and we may use (5.49) to get
dx

(t),∇V(x(t)) ≤ 0, a.e. t ≥ t 0 . (5.51)
dt
Thus
dV ∗
(t) ≤ 0, a.e. t ∈[t 0 ,t 0 + T ].
dt
dx n
0
n
∞
We know that x ∈ C ([t 0 ,t 0 + T ];R ), ∈ L (t 0 ,t 0 + T ;R ), and V ∈
dt
n
1
n
∗
C (R ;R). It follows that V ∈ W 1,1 (t 0 ,t 0 + T ;R ), and applying, for exam-
∗
ple, Lemma 2, we obtain that V is decreasing on [t 0 ,t 0 + T ]. Since T was
∗
∗
arbitrary, V is decreasing on [t 0 ,+∞). Moreover, V is bounded from below
on [t 0 ,+∞) since γ(x 0 ) ⊂ and V is continuous on the compact set . There-
fore
lim V(x(τ;t 0 ,x 0 )) = k
τ→+∞
for some k ∈ R.
Let y ∈ (x 0 ). There exists {τ i }⊂[t 0 ,+∞) such that τ i →+∞ and
x(τ i ;t 0 ,x 0 ) → y.