Page 167 - Complementarity and Variational Inequalities in Electronics

158 Complementarity and Variational Inequalities in Electronics

and

||x(T ;t 0 ,x 0 )|| = R.
We set V(·) = V(x(·;t 0 ,x 0 )). The function V is absolutely continuous on
[t 0 ,T ]. Setting x(·) = x(·;t 0 ,x 0 ), we obtain

dV dx
(t) =
∇V(x(t)), (t) , a.e. t ∈[t 0 ,T ].
dt dt
We have
dx
− (t) + F(x(t)),∇V(x(t))
dt
+ ϕ(x(t) −@ V (x(t))) − ϕ(x(t)) ≥ 0, a.e. t ≥ t 0 .
Thus

dV
(t) ≤−[
F(x(t)),∇(x(t))
dt
+ ϕ(x(t)) − ϕ(x(t) −@ V (x(t)))], a.e. t ∈[t 0 ,T ].

Using our second assumption, we get

dV
(t) ≤ 0, a.e. t ∈[t 0 ,T ].
dt
Using Lemma 2, we see that

(∀t ∈[t 0 ,T ]) : V(t) ≤ V(t 0 ),

which results in

V(T ) ≤ V(t 0 ) = V(x 0 )<a.
However, using our first assumption, we obtain

V(T ) = V(x(T ;t 0 ,x 0 )) ≥ a,

which is a contradiction.

Theorem 14 (Stability). Suppose that the assumptions of Theorem 11 hold to-
gether with conditions (5.36), (5.42), and (5.43). Suppose that there exist σ> 0
n
1
and V ∈ C (R ;R) with V(0) = 0 and such that
(∀x ∈ D(∂ϕ) ∩ B σ ) : V(x) ≥ a(||x||),