Page 190 - Complementarity and Variational Inequalities in Electronics

The Nonregular Dynamical System Chapter | 5 181

and thus

3 L 2
K ={z ∈ R : z 2 ≥ 0 and z 3 ≤ z 2 }.
L 3
Setting M =−RAR −1 , we see that

⎛ 1 ⎞
0 − √ 0
L 3 C 4
⎜ ⎟
1
⎜ R 1 +R 3 R 1 ⎟
M = ⎜ √ − √ ⎟.
L 3 C 4 L 3
⎝ L 2 L 3 ⎠
0 − √ R 1 R 1 +R 2
L 2 L 3 L 2
Example 70. Suppose, for example, that R 1 = R 2 = 0 and R 3 > 0. Then
⎛ 1 ⎞
0 − √ 0
L 3 C 4
⎜ ⎟
M = ⎜ 1 R 3 ⎟ .
⎝ √ 0 ⎠
L 3 C 4 L 3
0 0 0
We have
⎛ ⎞
0 0 0
T ⎜ ⎟
M + M = ⎝ 0 2R 3 0 ⎠ .
L 3
0 0 0
The matrix M is positive semidefinite, and

3
T
ker(M + M ) ={z ∈ R : z 2 = 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 = 0 and z 3 ≤ 0}.
Let us set

L 2
α = .
L 3