Page 20 - Complementarity and Variational Inequalities in Electronics

10 Complementarity and Variational Inequalities in Electronics

2
FIGURE 2.5 (x) = max{0,x − 1} and ∂ (x).

∂ (x) = 0. If x = 1, then

2
max{0,v − 1}≥ w(v − 1), ∀v ∈ R,
2
if and only if w ∈[0,2]. Indeed, let us first suppose that max{0,v − 1}≥
2
w(v − 1), ∀v ∈ R. Then, for v> 1, we get v − 1 = (v − 1)(v + 1)> 0 and
thus (v − 1)(v + 1) ≥ w(v − 1). It results that (∀v> 1) : (v + 1) ≥ w and thus
2
w ≤ 2. We have also that, for 0 <v < 1, v − 1 = (v − 1)(v + 1)< 0 and thus
0 ≥ w(v − 1), from which we deduce that w ≥ 0 since v − 1 < 0. Let us now
2
suppose that w ∈[0,2]. We check that max{0,v − 1}≥ w(v − 1), ∀v ∈ R.If
2
v> 1, then w(v − 1) ≤ 2(v − 1) ≤ (v + 1)(v − 1) = max{0,v − 1}.If v = 1,
2
then the result is trivial. If v< 1, then w(v − 1) ≤ 0 ≤ max{0,v − 1}.Using
similar computations, we check that if x =−1, then
2
max{0,v − 1}≥ w(v + 1), ∀v ∈ R,
if and only if w ∈[−2,0] (see Fig. 2.5).
n
Example 4. Let : R → R be defined by
n

n
(∀x ∈ R ) : (x) = ϕ i (x i )
i=1
n
with ϕ i ∈ 0 (R ;R ∪{+∞})(1 ≤ i ≤ n). Then
n
w ∈ ∂ (x) ⇔ (v) − (x) ≥
w,v − x ,∀v ∈ R
⇔
n n n

ϕ i (v i ) − ϕ i (x i ) ≥ w i (v i − x i ),∀v 1 ,v 2 ,...,v n ∈ R
i=1 i=1 i=1
⇔