Page 28 - Complementarity and Variational Inequalities in Electronics

18 Complementarity and Variational Inequalities in Electronics

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FIGURE 2.10 A cone K and the corresponding polar cone K =−K .
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Example 14. Let K = R . Then
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K = R .
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Example 15. Let a 1 ,a 2 ,...,a m ∈ R be given vectors. Set
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K ={x ∈ R :
a 1 ,x ≥ 0,
a 2 ,x ≥ 0,...,
a m ,x ≥ 0}.
Then using Farkas’ lemma (see e.g. [83]), we obtain

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K ={λ 1 a 1 + λ 2 a 2 + ··· + λ m a m : λ 1 ≥ 0,λ 2 ≥ 0,...,λ m ≥ 0}.
A more general form of the complementarity relation is

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U ∈ K,V ∈ K , and
U,V = 0,
which may be also written as

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K U ⊥ V ∈ K .
We have:
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K U ⊥ V ∈ K ⇔−V ∈ ∂
K (U)
⇔ V ∈ K &
V,h − U ≥ 0, ∀h ∈ K.
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Indeed, let U,V ∈ R satisfy the complementarity relation K U ⊥ V ∈ K .
Then (∀h ∈ K) :
V,h ≥ 0, and since
V,U = 0, we see that
(∀h ∈ K) :
V,h − U ≥ 0,

meaning that −V ∈ ∂
K (U). Reciprocally, if −V ∈ ∂
K (U), then U ∈ K and
(∀h ∈ K) :
V,h − U ≥ 0. We have 2U ∈ K, and we may thus set h = 2U to