Page 111 - Complementarity and Variational Inequalities in Electronics

102 Complementarity and Variational Inequalities in Electronics

and

(∀e ∈ ker(M),e
= 0) : q,e
= 0, (4.71)
then problem VI(M,q, ) has at most one solution.
e) If the function is strictly convex, then problem VI(M,q, ) has at most
one solution.

T
Proof. Let us first remark that ker(M + M ) = ker(M) and that part a) and the
first relation in part c) are direct consequences of parts a) and c) in Corollary 7.
Using the equivalence of problem VI(M,q, ) with problem (4.66),wealsoget
1 1
Mu 1 ,u 1 + q,u 1 + (u 1 ) = Mu 2 ,u 2 + q,u 2 + (u 2 ),
2 2

from which we deduce the second relation in part c). We have indeed M(u 1 −
u 2 ) = 0, and thus

Mu 1 ,u 1 −ÐMu 2 ,u 2 = Mu 1 ,u 1 −ÐMu 2 ,u 1 + Mu 2 ,u 1 −ÐMu 2 ,u 2
= M(u 1 − u 2 ),u 1 + Mu 2 ,u 1 − u 2
= M(u 1 − u 2 ),u 1 + u 2 ,M(u 1 − u 2 ) = 0.

It is easy to check that part d) is a direct consequence of part c). It suffices to
set e = u 2 − u1 and to see that e = 0. Suppose on the contrary that e
= 0. Then
from part c) we get

−Ðq,e = (u 1 + e) − (u 1 ),
and then using (4.70), we get q,e = 0 and a contradiction to (4.71).
Moreover, let x 0 ∈ D( ). Then, for all v ∈ ker(M),wehave

T
M x 0 ,v = x 0 ,Mv = 0.
Therefore, condition (4.63) in this case is equivalent to condition (4.67).Partb)
is then a direct consequences of part b) in Corollary 7.
1
Finally, if is strictly convex, then the function x à Mx,x + q,x +
2
(x) also is strictly convex, and thus problem (4.66) has at most one solu-
tion. The result in part e) follows since problem (4.66) is equivalent to problem
VI(M,q, ).
Example 53. Let
⎛ ⎞
1 −10
M = ⎝ −1 1 0 ⎠
⎜
⎟
0 0 1