Page 119 - Complementarity and Variational Inequalities in Electronics

110 Complementarity and Variational Inequalities in Electronics

that is, also
⎧
⎪ 0 ≤ x 1 ⊥ x 1 − 4x 2 ≥ 0
⎪
⎨
0 ≤ x 2 ⊥ x 1 + x 2 ≥ 0
⎪
⎪
x 3 = 0.
⎩
We deduce easily from this last system that B(M,K) ={0}. Indeed, we first see
that x 1 = 0. If we suppose on the contrary that x 1 > 0, then necessarily x 1 = 4x 2 ,
2
and then 6 x = 0. Therefore x 2 = 0 and x 1 =−4x 2 = 0, a contradiction. Thus
2
3
2
x 1 = 0 and then x = 0. Thus, for all q ∈ R , problem VI(M,q, ) has at least
2
one solution.
4.12 GENERAL RESULTS IN THE NONLINEAR CASE
n
Let : R → R ∪{+∞} be a proper convex lower semicontinuous function.
n
n
n
Let F : R → R and q ∈ R . We consider the following variational inequality
problem:
n
VI(F,q, ):Find u ∈ R such that
n
F(u) + q,v − u + (v) − (u) ≥ 0, ∀v ∈ R . (4.80)
n
Theorem 9. Suppose that ∈ 0 (R ;R∪{+∞}) with 0 ∈ D( ). Assume also
that F is continuous and there exist σ> 0 and α> 0 such that
2
(∀x ∈ D( ), x ≥ σ) : F(x),x ≥ α||x|| .
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Then, for all q ∈ R , problem VI(F,q, ) has at least one solution.
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Proof. Let q ∈ R be given. Using (3.5), we see that problem VI(F,q, ) is
equivalent to the fixed point problem

u = P (u − (F(u) + q)).
n
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Let us now define the continuous homotopy H :[0,1]× R → R by

H(t,u) = P (tu − (tF(u) + q)) = P (u − tF(u) + q + (1 − t)u ).
We claim that there exists R 0 > 0 such that, for all R ≥ R 0 and t ∈[0,1],
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H(t,u)
= u, ∀u ∈ R ,||u|| = R. (4.81)
Indeed, if we suppose the contrary, then we may find sequences {t i } i∈N ⊂[0,1]
n
and {u i } i∈N ⊂ R satisfying ||u i ||→+∞ and u i = H(t i ,u i ). Then
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t i F(u i ) + q + (1 − t i )u i ,v − u i + (v) − (u i ) ≥ 0,∀v ∈ R . (4.82)