Page 143 - Complementarity and Variational Inequalities in Electronics

134 Complementarity and Variational Inequalities in Electronics

4
x 4 = 0 since x ∈ R .From Corollary 6 it follows that for each u ∈ R, problem
+
VI(M, ,Fu) has at least one solution.
Example 65. Suppose that diodes D 1 and D 4 are ideal and for diodes D 2
and D 3 , consider the practical model with electrical superpotential

V 1 x if x ≥ 0
(∀x ∈ R) : ϕ(x) =
V 2 x if x< 0
with V 2 < 0 <V 1 .Here

4 ∗
(∀x ∈ R ) : (x) = R + (x 1 ) + ϕ (−x 2 ) + R + (x 3 ) + ϕ(x 4 )
∗
and ϕ ≡ [V 2 ,V 1 ] . Thus D( ) = R + ×[V 2 ,V 1 ]× R + × R, D( ) ∞ = R + ×
{0}× R + × R, and D( ∞ ) = R + ×{0}× R + × R. We check that

T
D( ) ∞ ∩ ker{M + M }∩ K(M, ) ={0}.
T
Indeed, here x 2 = 0, and x ∈ ker{M + M } yields x 1 =−x 3 and thus x 1 =
x 3 = 0 since x 1 ≥ 0 and x 2 ≥ 0. Using then x ∈ K(M, ), we also get x 4 = 0.
From Corollary 6 it follows that for each u ∈ R, problem VI(M, ,Fu) has at
least one solution.

Let us now consider the second variational inequality of the model described
3
in Section 3.4 of Chapter 3, that is, VI( , ,q) with q ∈ R , = 3 , and
R
+
⎛ ⎞
R 1 K R 1 (1 − α N ) R 1 (1 − α I )
⎝ R 1 K
⎟

= 1 ⎜ R 1 (1 − α N ) + R 2 R 1 (1 − α I ) − α I R 2 ⎠ .
K
R 1 K R 1 (1 − α N ) R 1 (1 − α I )
The matrix
is of class P 0 . Indeed,
1
1 (
) = R 1 > 0, 2 (
) = (R 1 (1 − α N ) + R 2 )> 0,
K
1
3 (
) = (R 1 (1 − α I )) > 0, 12 (
) = R 1 R 2 > 0,
K
and

13 (
) = 0, 23 (
) = R 1 R 2 > 0, 123 (
) = 0.
We have
T
z
z = z 1 (
z) 1 + z 2 (
z) 2 + z 3 (
z) 3 ,