Page 193 - Complementarity and Variational Inequalities in Electronics

184 Complementarity and Variational Inequalities in Electronics

It results in
2
2
∗ 2
(∀t ≥ 0) :|z 3 (t)| =|z 3 (0)| =|z | .
3
Here (∀t ≥ 0) : z 3 (t) ≤ 0, and thus
∗
(∀t ≥ 0) : z 3 (t) = z 3 (0) = z ≤ 0.
3
Thus any invariant subset of E(M, ,V ) is a subset of S(M, ). Therefore, for
any z 0 ∈ K,wehave

lim d(z(t;0,z 0 ),S(M, )) = 0.
t→+∞
Example 71. Suppose, for example, that R 1 > 0, R 2 > 0, and R 3 > 0. We have

0 0 0
⎛ ⎞
T ⎜ 0 2(R 1 +R 3 ) 2R 1 ⎟
⎜
M + M = L 3 − √ ⎟ .
⎝ L 2 L 3 ⎠
0 − √ 2R 1 2(R 1 +R 2 )
L 2 L 3 L 2
The matrix M is positive semidefinite, and

T
3
ker(M + M ) ={z ∈ R : z 2 = z 3 = 0}.
We set
1 2
3
(∀ z ∈ R ) : V(z) = ||z|| .
2
Thus
(∀ z ∈ K) :
Mz,∇V(z) =
Mz,z ≥ 0
and

T
3
E(M, ,V ) = K ∩ ker(M + M ) ={z ∈ R : z 2 = z 3 = 0}.
A stationary solution z has to satisfy
∗
∗
∗
∗
K z ⊥ Mz ∈ K ,
where
∗ 3
K ={w ∈ R : w 1 = 0,w 2 ≥ 0,w 3 ≤ 0 and w 2 + αw 3 ≥ 0}

L 2
with α = .Wehave
L 3
∗
∗
∗
∗
z ∈ K ⇔ z ≥ 0,z ≤ αz ,
2 3 2