NonlinearBook10pt November 20, 2007
766 CHAPTER 13
V : D → R satisfies
αkxk
p
≤ V (x) ≤ βkxk
p
, x ∈ D, (13.7)
ρV (f (x)) ≤ V (x), x ∈ D, (13.8)
then the zero solution x(k) ≡ 0 to (13.1) is geometrically stable. Finally, if
D = R
n
and V (·) is s uch that
V (x) → ∞ as kxk → ∞ (13.9)
then (13.6) implies (respectively, (13.7) and (13.8) imply) that the zero
solution x(k) ≡ 0 to (13.1) is globally asymptotically (respectively, globally
geometrically) stable.
Proof. Let ε > 0 be such that B
ε
(0) ⊆ D. S ince B
ε
(0) is compact and
f(x), x ∈ D, is continuous, it follows that
η
△
= max
(
ε, max
x∈B
ε
(0)
kf(x)k
)
(13.10)
exists. Next, let α
△
= min
x∈D: ε≤kxk≤η
V (x). Note α > 0 since 0 6∈ ∂B
ε
(0)
and V (x) > 0, x ∈ D, x 6= 0. Next, let β ∈ (0, α) and define D
β
△
= {x ∈
B
ε
(0) : V (x) ≤ β}. Now, for every x ∈ D
β
, it follows from (13.5) that
V (f (x)) ≤ V (x) ≤ β, and hence, it follows from (13.10) that kf (x)k ≤ η,
x ∈ D
β
. Next, suppose, ad absurdum, that there exists x ∈ D
β
such that
kf(x)k ≥ ε. This implies V (x) ≥ α, which is a contradiction. Hence, for
every x ∈ D
β
, it follows that f(x) ∈ B
ε
(0) ⊂ D
β
, which implies that D
β
is
a positively invariant set (see Definition 13.4) with respect to (13.1). Next,
since V (·) is continuous and V (0) = 0, there exists δ = δ(ε) ∈ (0, ε) such
that V (x) < β, x ∈ B
δ
(0). Now, let x(k), k ∈ Z
+
, satisfy (13.1). Since
B
δ
(0) ⊂ D
β
⊂ B
ε
(0) ⊆ D and D
β
is positively invariant with respect to
(13.1) it follows that for all x(0) ∈ B
δ
(0), x(k) ∈ B
ε
(0), k ∈ Z
+
. Hence,
for all ε > 0 there exists δ = δ(ε) > 0 su ch that if kx(0)k < δ, then
kx(k)k < ε, k ∈ Z
+
, which proves Lyapunov stability.
To prove asymptotic stability suppose that V (f(x)) < V (x), x ∈ D,
x 6= 0, and x(0) ∈ B
δ
(0). Then it follows that x(k) ∈ B
ε
(0), k ∈ Z
+
.
However, V (x(k)), k ∈ Z
+
, is decreasing and bounded from below by zero.
Now, suppose, ad absurdum, that x(k), k ∈ Z
+
, does not converge to zero.
This implies that V (x(k)), k ∈ Z
+
, is lower bounded by a positive number,
that is, there exists L > 0 such that V (x(k)) ≥ L > 0, k ∈ Z
+
. Hence,
by continuity of V (x), x ∈ D, there exists δ
′
> 0 such that V (x) < L for
x ∈ B
δ
′
(0), which further implies that x(k) 6∈ B
δ
′
(0), k ∈ Z
+
. Next, let
L
1
△
=
min
δ
′
≤kxk≤ε
[V (x) − V (f(x))]. Now, (13.6) implies V (x) − V (f(x)) ≥ L
1
,