Page 192 - Complementarity and Variational Inequalities in Electronics

The Nonregular Dynamical System Chapter | 5 183

∗
∗
Setting w = Mz , we note that
1 1
∗ ∗ ∗ ∗ R 3 ∗ ∗
w =−√ z ,w = √ z + z ,w = 0.
1 2 2 1 2 3
L 3 C 4 L 3 C 4 L 3
Thus
∗
∗
∗
∗
w ∈ K ⇔ z = 0,z ≥ 0.
1
2
The set of stationary solutions is thus given by
3
S(M, ) ={z ∈ R : z 1 ≥ 0,z 2 = 0,z 3 ≤ 0}.
The set S(M, ) is an invariant subset of E(M, ,V ). We claim that it is
∗
the largest one. Indeed, let us study the dynamics in E(M, ,V ).Let z ∈
∗
∗
E(M, ,V ), that is, z = 0 and z ≤ 0. We have
2 3
∗
∗
z 1 (0) = z ,z 2 (0) = 0,z 3 (0) = z ≤ 0,
3
1
(∀t ≥ 0) : z 2 (t) = 0 and z 3 (t) ≤ 0,
and
1

z (t)(v 1 − z 1 (t)) + √ z 1 (t)v 2 + z (t)(v 3 − z 3 (t)) ≥ 0,
1 3
L 3 C 4
∀v 1 ∈ R,v 2 ≥ 0,v 3 ≤ αv 2 , a.e. t ≥ 0.
Setting v 1 = z 1 (t) and v 3 = z 3 (t) ≤ 0(z 3 (t) ≤ 0 ≤ αv 2 for v 2 ≥ 0), we get
z 1 (t)v 2 ,∀v 2 ≥ 0, a.e. t ≥ 0.
∗
It follows by continuity that (∀t ≥ 0) : z 1 (t) ≥ 0. In particular, we have z ≥ 0.
1
Setting v 2 = 0 and v 3 = z 3 (t), we get

z (t) = 0, a.e. t ≥ 0.
1
Thus

∗
(∀t ≥ 0) : z 1 (t) = z ≥ 0.
1
Setting v 1 = z 1 (t) and v 2 = 0, we get

z (t)(v 3 − z 3 (t)) ≥ 0,∀v 3 ≤ 0, a.e. t ≥ 0.
3

Setting v 3 = 0, we get z z 3 (t) ≤ 0. Setting v 3 = 2z 3 (t), we obtain z z 3 (t) ≥ 0.
3 3
Thus
1 d 2
| z 3 (t) | = 0, a.e. t ≥ 0.
2 dt