Page 121 - Complementarity and Variational Inequalities in Electronics

112 Complementarity and Variational Inequalities in Electronics

= deg(id R − H(0,.),D R ,0) = deg(id R − P (−q),D R ,0) = 1.
n
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The result now follows from the solution property of Brouwer degree.

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Corollary 11. Suppose that ∈ 0 (R ;R ∪{+∞}) with 0 ∈ D( ). Assume
also that F is continuous and strongly monotone on D( ), that is, there exists
α> 0 such that

2
(∀x,y ∈ D( )) : F(x) − F(y),x − y ≥ α||x − y|| .
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Then, for all q ∈ R , problem VI(F,q, ) has a unique solution.
Proof. Indeed, let us set
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(∀x ∈ R ) : G(x) = F(x) − F(0).

We have
2
(∀x ∈ D( )) : G(x),x ≥ α||x|| .
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Using Theorem 9, we see that for all q ∈ R , the variational inequality
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VI(G,q + F(0), ) has at least one solution. Therefore, for all q ∈ R ,the vari-
ational inequality VI(F,q, ) has at least one solution. Let us now consider two
solutions u 1 and u 2 .Wehave
F(u 1 ) + q,u 2 − u 1 + (u 2 ) − (u 1 ) ≥ 0

and

F(u 2 ) + q,u 1 − u 2 + (u 1 ) − (u 2 ) ≥ 0,
from which we deduce that

2
α||u 1 − u 2 || ≤ 0
and thus the uniqueness of the solution.

3
Example 58. Let us set, for all x = (x 1 ,x 2 ,x 3 ) ∈ R ,
⎛ 2 ⎞
x 1 + x
2
⎜ ⎟
F(x) = ⎝ −x 1 x 2 + x 2 ⎠
2x 3
and
(x) =|x 1 |+|x 3 |.