NonlinearBook10pt November 20, 2007
148 CHAPTER 3
Proof. Let x(t), t ≥ 0, be a solution to (3.1) with x(0) ∈ D
c
. Since
V
′
(x)f(x) ≤ 0, x ∈ D
c
, it follows that
V (x(t)) − V (x(τ)) =
Z
t
τ
V
′
(x(s))f(x(s))ds ≤ 0, t ≥ τ,
and h en ce, V (x(t)) ≤ V (x(τ)), t ≥ τ, which implies that V (x(t)) is a
nonincreasing function of t. Next, since V (·) is continuous on the compact
set D
c
, there exists β ∈ R such that V (x) ≥ β, x ∈ D
c
. Hence,
γ
x
0
△
= lim
t→∞
V (x(t)) exists. Now, for all p ∈ ω(x
0
) there exists an
increasing unbounded sequence {t
n
}
∞
n=0
, with t
0
= 0, such that x(t
n
) → p
as n → ∞. Since V (x), x ∈ D
c
, is continuous, V (p) = V (lim
n→∞
x(t
n
)) =
lim
n→∞
V (x(t
n
)) = γ
x
0
, and hence, V (x) = γ
x
0
on ω(x
0
). Now, since D
c
is compact and positively invariant it follows that x(t), t ≥ 0, is bounded,
and hence, it follows from Theorem 2.41 that ω(x
0
) is a nonempty, compact
invariant set. Hence, it follows that V
′
(x)f(x) = 0 on ω(x
0
) and thus
ω(x
0
) ⊂ M ⊂ R ⊂ D
c
. Finally, since x(t) → ω(x
0
) as t → ∞, it f ollows
that x(t) → M as t → ∞.
The construction of V (·) in Theorem 3.3 can be used to guarantee
the existence of the compact positively invariant s et D
c
. Specifically, if
D
β
= {x ∈ D : V (x) ≤ β}, where β > 0, is bounded and
˙
V (x) ≤ 0,
x ∈ D
β
, then we can always take D
c
= D
β
. As discussed in Section 3.1,
if V (·) is positive definite, then D
β
is bounded for sufficiently small β > 0.
Alternatively, if V (·) is radially unbounded, then D
β
is bounded for every
β > 0 irrespective of whether V (·) is positive definite or not.
Example 3. 5. The Barbashin-Krasovskii-LaSalle invariant set theo-
rem can be used to examine th e stability of limit cycles. To see this, consider
the nonlinear dynamical system
˙x
1
(t) = 4x
2
1
(t)x
2
(t) − g
1
(x
1
(t))[x
2
1
(t) + 2x
2
2
(t) −4], x
1
(0) = x
10
, t ≥ 0,
(3.38)
˙x
2
(t) = −2x
3
1
(t) −g
2
(x
2
(t))[x
2
1
(t) + 2x
2
2
(t) − 4], x
2
(0) = x
20
, (3.39)
where x
1
g
1
(x
1
) > 0, x
1
6= 0, x
2
g
2
(x
2
) > 0, x
2
6= 0, g
1
(0) = 0, and g
2
(0) = 0.
Now, note th at the s et defined by the ellipse E
△
= {(x
1
, x
2
) ∈ R × R :
x
2
1
+ 2x
2
2
− 4 = 0} is invariant since
d
dt
[x
2
1
(t) + 2x
2
2
(t) −4] = −[2x
1
(t)g
1
(x
1
(t)) + 4x
2
(t)g
2
(x
2
(t))]
·[x
2
1
(t) + 2x
2
2
(t) −4]
= 0, t ≥ 0, (3.40)
and hence, if (x
1
(0), x
2
(0)) ∈ E, th en (x
1
(t), x
2
(t)) ∈ E, t ≥ 0. The motion