Page 163 - Complementarity and Variational Inequalities in Electronics

154 Complementarity and Variational Inequalities in Electronics

1
Lemma 2. Let T> 0 and a,b ∈ L (t 0 ,t 0 + T ;R) with b(t) ≥ 0,a.e. t ∈
[t 0 ,t 0 + T ]. Let an absolutely continuous function w :[t 0 ,t 0 + T ]→ R + sat-
isfy
α

(1 − α)w (t) ≤ a(t)w(t) + b(t)w (t), a.e. t ∈[t 0 ,t 0 + T ],
where 0 ≤ α< 1. Then
t a(τ)dτ t t
t
(∀t ∈[t 0 ,t 0 + T ]) : w 1−α (t) ≤ w 1−α (t 0 )e 0 + e s a(τ)dτ b(s)ds.
t 0
Theorem 12 (Continuity in the initial condition). Let t ≥ t 0 . The function

n
x(t;t 0 ,.) : D(∂ϕ) → R ;x 0 → x(t;t 0 ,x 0 )
is uniformly continuous.

Proof. Fix τ ≥ t 0 .Let ε> 0 and set
ε
δ = √ .
e 2ω(τ−t 0 )

We claim that if x 0 ,x ∈ D(∂ϕ), ||x 0 − x || ≤ δ, then ||x(τ;t 0 ,x 0 ) −
0 0

x(τ;t 0 ,x )|| ≤ ε. Indeed, let us set x(t) = x(t;t 0 ,x 0 ) and x (t) = x(t;t 0 ,x ).
0 0
We know that
dx n

(t)+F(x(t)),v−x(t) +ϕ(v)−ϕ(x(t)) ≥ 0, ∀v ∈ R , a.e. t ≥ t 0 , (5.25)
dt
and
dx

(t) + F(x (t)),z − x (t)
dt

n
+ ϕ(z) − ϕ(x (t)) ≥ 0, ∀z ∈ R , a.e. t ≥ t 0 . (5.26)

Setting v = x (t) in (5.25) and z = x(t) in (5.26), we obtain the relations
dx
− (t)+F(x(t)),x (t)−x(t) +ϕ(x(t))−ϕ(x (t)) ≤ 0, a.e. t ≥ t 0 , (5.27)
dt
and

dx

(t)+F(x (t)),x (t)−x(t) +ϕ(x (t))−ϕ(x(t)) ≤ 0, a.e. t ≥ t 0 , (5.28)
dt
which results in

d(x − x)

(t),x (t) − x(t) ≤
ωx (t) − ωx(t),x (t) − x(t)
dt