NonlinearBook10pt November 20, 2007
DISCRETE-TIME NONLINEAR CONTROL 875
Finally, if D = R
n
, w(k) ≡ 0, and
V (x) → ∞ as kxk → ∞, (14.164)
then the solution x(k) = 0, k ∈ Z
+
, of (14.143) is globally asymptotically
stable.
Proof. Let x(k), k ∈ Z
+
, satisfy (14.143). Then
∆V (x(k))
△
= V (x(k + 1)) − V (x(k))
= V (f(x(k))) − V (x(k)) + P
1w
(x(k))w(k)
+w
T
(k)P
2w
(x(k))w(k), k ∈ Z
+
. (14.165)
Hence, with w(k) ≡ 0, it follows from (14.158) that
∆V (x(k)) < 0, k ∈ Z
+
, x(k) 6= 0. (14.166)
Thus , from (14.154), (14.156), and (14.166) it follows that V (·) is a Lyapunov
function for (14.143), which proves local asymptotic stability of the solution
x(k) ≡ 0 with w(k) ≡ 0. Consequently, x(k) → 0 as k → ∞ for all initial
conditions x
0
∈ D
0
for some neighborhood D
0
⊆ D of the origin.
Next, if Γ(x) ≥ 0, x ∈ D, and w(k) ≡ 0, (14.160) implies
L(x(k)) = − ∆V (x(k)) + L(x(k)) + V (f(x(k))) − V (x(k))
≤ −∆V (x(k)) + L(x(k)) + V (f (x(k))) − V (x(k)) + Γ(x(k))
= −∆V (x(k)).
Now, summing over [0, k] yields
k
X
i=0
L(x(i)) ≤ −V (x(k + 1)) + V (x
0
).
Letting k → ∞ and noting that V (x(k)) → 0 as k → ∞ for all x
0
∈ D
0
yields J(x
0
) ≤ V (x
0
).
Next, let x(k), k ∈ Z
+
, satisfy (14.143) w ith w(k) ≡ 0. Then it follows
from (14.160) that
L(x(k)) + Γ(x(k))
= −∆V (x(k)) + L(x(k)) + V (f (x(k))) − V (x(k)) + Γ(x(k))
= −∆V (x(k)).
Summing over [0, k] yields
k
X
i=0
[L(x(i)) + Γ(x(i))] = −V (x(k + 1)) + V (x
0
).