Page 100 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 91

Proof. We first remark that here D( ) = D( 1 ) × D( 2 ) × ··· × D( n ).
Moreover, as a consequence of assumptions (4.49) and (4.50), each set D( i )
is a nonempty closed convex cone, and thus

D( ) ∞ = D( ).

Moreover, the function is positively homogeneous, and thus

∞ ≡ , D( ∞ ) = D( ).

We claim that (M, ) ∈ P n . Indeed, if x ∈ D( ) ∞ = D( ), then for all
j j T
j ∈{1,...,n}, we see that x,e e = (0 ... 0 x j 0 ... 0) ∈ D( 1 ) × ··· ×
j j
D( j−1 ) × D( j ) × D( j+1 ) × ··· × D( n ), and thus x,e e ∈ D( ) =
D( ∞ ). The matrix M is a P-matrix, and thus (M, ) ∈ P n . Then the existence
is a direct consequence of Corollary 5.
To prove the uniqueness, suppose by contradiction that problem VI(M,q, )
has two different solutions u and U.Weset
w = Mu + q, W = MU + q.

We have
n
w,v − u + (v) − (u) ≥ 0,∀v ∈ R (4.51)
and

n
W,v − U + (v) − (U) ≥ 0,∀v ∈ R . (4.52)
j
j
We may set v = u + u,e e ∈ D( ) (1 ≤ j ≤ n) in (4.51) to get
n n
j

0 ≤ w j u j + k (u k + u j e ) − k (u k ) = w j u j + j (2u j ) − j (u j ).
k
k=1 k=1
Here j (2u j ) = 2 j (u j ), and thus, for all integers 1 ≤ j ≤ n,
0 ≤ w j u j + j (u j ). (4.53)

We check in the same way that, for all integers 1 ≤ j ≤ n,

0 ≤ W j U j + j (U j ). (4.54)

j
j
Let us now set v = u + U,e e ∈ D( ) (1 ≤ j ≤ n) in (4.52) to get
0 ≤ w j U j + j (u j + U j ) − j (u j ) (4.55)