Page 101 - Complementarity and Variational Inequalities in Electronics

92 Complementarity and Variational Inequalities in Electronics

j
j
and v = U + u,e e ∈ D( ) (1 ≤ j ≤ n) in (4.51) to get
0 ≤ W j u j + j (U j + u j ) − j (U j ). (4.56)

Setting v = 0in (4.51), we also get
0 ≥ w,u + (u),

and thus
n

0 ≥ [w j u j + j (u j )]. (4.57)
j=1
We prove in the same way that
n

0 ≥ [W j U j + j (U j )]. (4.58)
j=1
Using (4.53) and (4.54) together with (4.57) and (4.58), we see that, for all
integers 1 ≤ j ≤ n,

w j u j + j (u j ) = 0,W j U j + j (U j ) = 0.
Then, for all integers 1 ≤ j ≤ n,

(u − U) j (M(u − U)) j = (u j − U j )(w j − W j )
= u j w j + U j W j − u j W j − U j w j
≤− j (u j ) − j (U j ) + j (U j + u j ) − j (U j )
+ j (u j + U j ) − j (u j )
= 2 j (u j + U j ) − 2( j (u j ) + j (U j )).

Note that
1 1 1 1
2 j (u j + U j ) = 2 j (2( u j + U j )) = 4 j ( u j + U j )
2 2 2 2
≤ 2( j (u j ) + j (U j )).

Thus, for all integers 1 ≤ j ≤ n,
(u − U) j (M(u − U)) j ≤ 0.

The matrix M is a P-matrix, and thus

x
= 0 =⇒ ∃ α ∈{1,...,n}: x α (Mx) α > 0. (4.59)
Recalling that u − U
= 0, a contradiction to (4.59) has been obtained.