Page 132 - Complementarity and Variational Inequalities in Electronics

A Variational Inequality Theory Chapter | 4 123

Here the matrix M is positive semidefinite and symmetric, D( ) ∞ = R − ×R + ,

2
ker{M}={v ∈ R : v 2 =−v 1 },
2
and K(M, ) ={v ∈ R : v 2 =−v 1 }. Thus
2
D( ) ∞ ∩ ker{M}∩ K(M, ) ={v ∈ R : v 1 ≤ 0,v 2 =−v 1 }.
Then, for all v ∈ D( ) ∞ ∩ ker{M}∩ K(M, ), v
= 0, we have −v 1 = v 2 > 0,
and thus

q,v = (E 1 − u)v 1 + (E 2 − u)v 2 = v 2 (E 2 − E 1 )> 0. (4.108)
We may apply Corollary 8, which ensures that system (4.107) has at least one
solution.
Using the first relation in part (c) of Corollary 8, we first note that if I and I ¯
¯
are two solutions of (4.103), then I −I ∈ ker{M}, and thus i 1 −i 1 =−(i 2 −i 2 ),
¯
¯
that is,
i 1 + i 2 = i 1 + i 2 . (4.109)
¯
¯
∗ ∗ T
∗
Therefore, if I = (i i ) is a solution of system (4.107), then the current
1 2
∗
∗
∗
through the resistor R, that is, i = i + i , is uniquely determined.
1 2
Using now the second relation of part (c) of Corollary 8, we also obtain that
(E 1 − u)i 1 + (E 2 − u)i 2 = (E 1 − u)i 1 + (E 2 − u)i 2 . (4.110)
¯
¯
System (4.109)–(4.110) may be written as
¯
(i 1 − i 1 ) + (i 2 − i 2 ) = 0
¯
and
¯
¯
(E 1 − u)(i 1 − i 1 ) + (E 2 − u)(i 2 − i 2 ) = 0,
that is,
A

¯
1 1 i 1 − i 1 0
= .
E 1 − uE 2 − u ¯ 0
i 2 − i 2
¯
Here E 2
= E 1 , and thus det(A) = E 2 − E 1
= 0. Thus i 1 = i 1 and i 2 = i 2 .The
¯
solution of (4.107) is thus unique.
Using relations (4.103)–(4.105), we see that
u − E 1 u − E 2
∗
∗
∗
∗
i + i = min{i , }= max{i , }.
2
1
1
2
R R