Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 33, pp. 1–7.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF SECOND-ORDER IMPULSIVE
DIFFERENTIAL EQUATIONS
HAIFENG LIU, QIAOLUAN LI
Abstract. In this article, we study the asymptotic behavior of all solutions
of 2-th order nonlinear delay differential equation with impulses. Our main
tools are impulsive differential inequalities and the Riccati transformation. We
illustrate the results by an example.
1. Introduction
Consider the impulsive differential equation
0
r(t)(x0 (t))α + p(t)(x0 (t))α + f (t, x(t − δ)) = 0,
x(t+
k)
= Jk (x(tk )),
0
x
(t+
k)
0
= Ik (x (tk )),
t ≥ t0 , t 6= tk ,
k = 1, 2, 3 . . . ,
(1.1)
(1.2)
where α is the quotient of positive odd integers.
The theory of impulsive differential/difference equations is emerging as an important area of investigation, since it is much richer than the corresponding theory
of differential/difference equations without impulsive effects. Moreover, such equations may model several real world phenomena [4]. There are many papers devoted
to the oscillation criteria of differential equations with impulses [2, 5, 6] and to the
asymptotic behavior of all solutions of differential equations without impulses [8].
Recently, Tang [7] studied the equation
(r(t)x0 (t))0 + p(t)x0 (t) + f (t, x(t − δ)) = 0,
x(t+
k)
0 +
x (tk )
= Jk (x(tk )),
0
= Ik (x (tk )),
t 6= tk ,
k = 1, 2, 3 . . . ,
k = 1, 2, 3 . . . .
He obtained sufficient conditions of asymptotic behavior of all solutions of the
equation.
Motivated by [7], using impulsive differential inequality and the Riccati transformation, we study the asymptotic behavior of solutions of (1.1), (1.2).
Definition 1.1. For φ ∈ C([t0 − δ, t0 ], R), a function x : [t0 − δ, +∞) → R is called
a solution of (1.1), (1.2) satisfying the initial value condition
x(t) = φ(t),
t ∈ [t0 − δ, t0 ]
2000 Mathematics Subject Classification. 34K25, 34K45.
Key words and phrases. Impulsive differential equation; asymptotic behavior; second-order.
c
2011
Texas State University - San Marcos.
Submitted January 6, 2011. Published February 23, 2011.
Supported by grant L2009Z02 from the Key Foundation of Hebei Normal University.
1
2
H. LIU, Q. LI
EJDE-2011/33
if the following conditions are satisfied:
(i) x(t) = φ(t) for t ∈ [t0 − δ, t0 ],
(ii) x, x0 are continuously differentiable for t > t0 , t 6= tk (k = 1, 2, . . . ) and
satisfy (1.1),
0 −
0
(iii) x(t−
k ) = x(tk ), x (tk ) = x (tk ), k = 1, 2, . . . and satisfy (1.2).
As is customary, a solution of (1.1), (1.2) is said to be non-oscillatory if it is
eventually positive or eventually negative. Otherwise, it will be called oscillatory.
2. Main results
In this paper, we assume that the following conditions hold:
(t, x)
(H1) f is continuous on [t0 , +∞)×R, xf (t, x) > 0 for x 6= 0, and fg(x)
≥ h(t) for
0 0
0
x 6= 0, where g(γx) ≥ γg(x) for γ > 0, x g (x) > 0, and h, r are continuous
on [t0 , +∞), h(t) ≥ 0, r(t) > 0.
(H2) p, Jk , Ik are continuous on R and there exist positive numbers a∗k , ak , b∗k , bk
such that a∗k ≤ Ikx(x) ≤ ak , b∗k ≤ Jkx(x) ≤ bk .
Rs 0
Rt Q
a∗
dσ)ds = +∞.
(H3) limt→∞ tj tj <tk <s bkk exp(− tj r (σ)+p(σ)
αr(σ)
(H4)
Z tj+m
n−1
Z u r0 (s) + p(s) X n−1
Y m−1
Y
∗
ds du
bj+k aj+l
exp −
αr(s)
tj+m−1
tj
m=1 k=m l=0
Z tj+n
n−1
Z u r0 (s) + p(s) Y
ds du → +∞, as n → ∞.
+
a∗j+k
exp −
αr(s)
tj+n−1
tj
k=0
(H5)
Z
t
lim
t→∞
Z s
1
p(σ)
exp(
dσ)h(s)ds = +∞,
c
t0 r(σ)
<s k
Y
t0 t0 <tk
where
(
aα
k,
ck = aαk
b∗ ,
j
tk − δ 6= tj ,
tk − δ = tj .
In the following, we also assume that solutions to (1.1), (1.2) exist on [t0 , +∞).
Lemma 2.1 ([1]). Let the function m ∈ P C 1 (R+ , R) satisfy the inequalities
m0 (t) ≤ p(t)m(t) + q(t),
m(t+
k ) ≤ dk m(tk ) + bk ,
t 6= tk ,
k = 1, 2, . . . ,
where p, q ∈ P C(R+ , R) and dk ≥ 0, bk are constants, then
Z t
Z t
Y
X Y
dj exp
p(s)ds bk
m(t) ≤ m(t0 )
dk exp
p(s)ds +
t0
t0 <tk <t
Z
+
t
Y
t0 s<tk <t
dk exp
Z
t
t0 <tk <t
p(σ)dσ q(s)ds,
tk <tj <t
tk
t ≥ t0 .
s
(2.1)
Lemma 2.2. Let x be a solution of (1.1), (1.2). Suppose that there exist some
T ≥ t0 such that x(t) > 0, t ≥ T . If (H1)–(H3) are satisfied, then x0 (tk ) > 0 and
x0 (t) > 0 for t ∈ (tk , tk+1 ], where tk ≥ T , k = 1, 2, . . . .
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SECOND-ORDER IMPULSIVE DE’S
3
Proof. We first prove that x0 (tk ) > 0 for any tk ≥ T . If not, there must exist some
0
∗ 0
j such that x0 (tj ) < 0, tj ≥ T and x0 (t+
j ) = Ij (x (tj )) ≤ aj x (tj ) < 0. Let
Z tj r0 (s) + p(s) ds =: β < 0.
x0 (tj ) exp
αr(s)
t0
From (1.1), it is clear that
Z t r0 (s) + p(s) 0
Z t r0 (s) + p(s) f (t, x(t − δ))
x0 (t) exp
=−
exp
ds
ds .
αr(s)
αr(t)(x0 (t))α−1
αr(s)
t0
t0
Since α is the quotient of positive odd integers, (x0 (t))α−1 > 0, we obtain
Z t r0 (s) + p(s) 0
0
ds
< 0.
(2.2)
x (t) exp
αr(s)
t0
Rt 0
Hence, the function x0 (t) exp( t0 r (s)+p(s)
αr(s) ds) is decreasing on (tj , tj+1 ],
Z tj+1 r0 (s) + p(s) Z tj r0 (s) + p(s) x0 (tj+1 ) exp
ds ≤ x0 (t+
)
exp
ds ;
j
αr(s)
αr(s)
t0
t0
i.e.,
Z
x0 (tj+1 ) exp
tj+1
t0
r0 (s) + p(s) ds ≤ a∗j β
αr(s)
and
0
x (tj+2 ) exp
Z
tj+2
t0
r0 (s) + p(s) ds ≤ x0 (t+
j+1 ) exp
αr(s)
≤ a∗j+1 a∗j β.
Z
tj+1
t0
r0 (s) + p(s) ds
αr(s)
By induction, we obtain
x0 (tj+n ) exp
tj+n
Z
t0
n−1
r0 (s) + p(s) Y ∗
ds ≤
aj+k β,
αr(s)
k=0
while for t ∈ (tj+n , tj+n+1 ], we have
Z
Y
x0 (t) ≤
a∗k β exp −
t
r0 (s) + p(s) ds .
αr(s)
t0
tj ≤tk <t
(2.3)
From the condition x(t+
n ) ≤ bn x(tn ), we have the impulsive differential inequality
Z t r0 (s) + p(s) Y
ds , t 6= tk , k = j + 1, j + 2, . . . ,
x0 (t) ≤
a∗k β exp −
αr(s)
t0
tj ≤tk <t
x(t+
k ) ≤ bk x(tk ),
Applying Lemma 2.1, we have
Z
Y
+
∗
x(t) ≤ x(tj )
bk + aj β
tj <tk <t
≤
Y
tj <tk <t
t
Y
bk
tj s<tk <t
Z
n
∗
bk x(t+
)
+
a
β
j
j
t
Y
a∗i
s
t0
tj <ti <s
Z
exp −
a∗i
exp −
b
<s i
Y
tj tj <ti
t = tk , t ≥ tj .
Z
s
t0
r0 (σ) + p(σ) dσ ds
αr(σ)
r0 (σ) + p(σ) o
dσ ds .
αr(σ)
Since x(tk ) > 0 for tk ≥ T , one can find that the above inequality contradicts (H3)
as t → ∞, therefore, x0 (tk ) ≥ 0(t ≥ T ).
4
H. LIU, Q. LI
EJDE-2011/33
∗ 0
By condition (H2), we have x0 (t+
k ) ≥ ak x (tk ) for any tk ≥ T . Because the
R
0
t
r
(s)+p(s)
function x0 (t) exp( t0 αr(s) ds) is decreasing on (tj+i−1 , tj+i ], we obtain
t
Z
0
r0 (s) + p(s)
ds) > 0
αr(s)
x (t) exp(
t0
for any t ∈ (tj+i−1 , tj+i ], which implies x0 (t) ≥ 0 for t ≥ T . The proof is complete.
Theorem 2.3. If (H1)-(H3), (H5) are satisfied, then every solution x of (1.1),
(1.2) satisfies lim inf t→∞ |x(t)| = 0.
Proof. Let x be a solution of (1.1)-(1.2), and by contradiction assume that
lim inf |x(t)| > 0.
t→∞
Without loss of generality, we may assume that x(t) > 0 on (t0 , +∞). By Lemma
2.2, x0 (t) > 0 for all t ≥ t0 . We use a Riccati transformation of the form
V (t) =
r(t)(x0 (t))α
.
g(x(t − δ))
(2.4)
Differentiating V (t), we obtain
(r(t)(x0 (t))α )0 g(x(t − δ)) − r(t)(x0 (t))α g 0 (x(t − δ))x0 (t − δ)
g 2 (x(t − δ))
0
α
−p(t)(x (t)) − f (t, x(t − δ)) x0 (t − δ)g 0 (x(t − δ)) 2
=
V (t)
−
g(x(t − δ))
r(t)(x0 (t))α
V (t)
≤ −p(t)
− h(t).
r(t)
V 0 (t) =
From (2.4) and (H1), it is clear that
V (t+
k)=
≤
0 + α
r(t+
k )(x (tk ))
g(x(t+
k − δ))

r(tk )(x0 (tk ))α aα
 g(x(t −δ)) k = aα
k V (tk ) = ck V (tk ),
k
0

r(tk )(x (tk ))α aα
k
g(x(t+
j ))
≤
aα
k
b∗
j
V (tk ) = ck V (tk ),
tk − δ 6= tj ,
tk − δ = tj ,
where ck ’s are defined in (H5). Applying Lemma 2.1, we have
Z t p(s) Y
V (t) ≤
ck exp −
ds
t0 r(s)
t0 <tk <t
Z t Y
Z s p(σ) i
h
1
× V (t0 ) −
exp
dσ h(s)ds .
t0 r(σ)
t0 t <t <s ck
0
k
By (H5), the above inequality is impossible. The proof is complete.
Lemma 2.4. Let x be a solution of (1.1), (1.2). Suppose that there exist some
T ≥ t0 such that x(t) > 0, t ≥ T . If (H1), (H2), (H4) are satisfied, then x0 (tk ) > 0
and x0 (t) > 0 for t ∈ (tk , tk+1 ], where tk ≥ T, k = 1, 2, . . . .
EJDE-2011/33
SECOND-ORDER IMPULSIVE DE’S
5
Proof. Firstly, for x(t) > 0, t ≥ T , we will prove that x0 (tk ) > 0, for any tk ≥ T ,
T ≥ t0 . If not, there exist some j such that x0 (tj ) < 0, tj ≥ T and x0 (t+
j ) =
Ij (x0 (tj )) ≤ a∗j x0 (tj ) < 0. From (1.1), it is clear that
Z t r0 (s) + p(s) Z t r0 (s) + p(s) 0
f (t, x(t − δ))
ds
=−
exp
ds .
x0 (t) exp
αr(s)
αr(t)(x0 (t))α−1
αr(s)
t0
t0
Since α is the quotient of positive odd integers, (x0 (t))α−1 > 0, we obtain
Z t r0 (s) + p(s) 0
x0 (t) exp
ds
< 0.
αr(s)
t0
R 0
t
Hence, the function x0 (t) exp t0 r (s)+p(s)
ds
is decreasing on (tj , tj+1 ],
αr(s)
Z
tj+1 r0 (s) + p(s) Z tj r0 (s) + p(s) x0 (tj+1 ) exp
ds ≤ x0 (t+
)
exp
ds ),
j
αr(s)
αr(s)
t0
t0
i.e.,
0
x (tj+1 ) ≤
a∗j x0 (tj ) exp
Z
tj+1
−
tj
and
r0 (s) + p(s) ds
αr(s)
tj+2
Z
x0 (tj+2 ) ≤ a∗j+1 a∗j x0 (tj ) exp −
tj
r0 (s) + p(s) ds .
αr(s)
By induction, we obtain
0
x (tj+n ) ≤
n−1
Y
a∗j+k x0 (tj ) exp
Z
tj+n
−
tj
k=0
Because the function x0 (t) exp(
Rt
t0
0
r0 (s) + p(s) ds .
αr(s)
r (s)+p(s)
αr(s) ds)
x0 (t) ≤ a∗j x0 (tj ) exp −
Z
t
tj
is decreasing on (tj , tj+1 ], we have
r0 (s) + p(s) ds , t ∈ (tj , tj+1 ].
(2.5)
αr(s)
Integrating (2.5) from m to t, we have
Z t
Z
x(t) ≤ x(m) + a∗j x0 (tj )
exp −
m
u
tj
r0 (s) + p(s) ds du,
αr(s)
tj < m < tj+1 .
Let t → tj+1 , m → t+
j . We have
tj+1
u
r0 (s) + p(s) ds du
αr(s)
tj
tj
Z tj+1
Z u r0 (s) + p(s) ds du,
≤ bj x(tj ) + a∗j x0 (tj )
exp −
αr(s)
tj
tj
∗ 0
x(tj+1 ) ≤ x(t+
j ) + aj x (tj )
Z
Z
exp −
and
tj+2
u
r0 (s) + p(s) ds )du
αr(s)
tj+1
tj+1
Z tj+1
Z u r0 (s) + p(s) ∗
0
≤ bj+1 bj x(tj ) + aj bj+1 x (tj )
exp −
ds du
αr(s)
tj
tj
Z tj+2
Z u r0 (s) + p(s) + a∗j+1 a∗j x0 (tj )
exp −
ds du.
αr(s)
tj+1
tj
∗
0
x(tj+2 ) ≤ x(t+
j+1 ) + aj+1 x (tj+1 )
Z
Z
exp −
6
H. LIU, Q. LI
EJDE-2011/33
By induction, we have
Z
h n−1
X n−1
Y m−1
Y
x(tj+n ) ≤ x0 (tj )
bj+k a∗j+l
+
a∗j+k
Z
tj+n
Z
tj+n−1
u
tj
r0 (s) + p(s) ds du
αr(s)
i n−1
Y
r0 (s) + p(s)
ds du +
bj+k x(tj ).
αr(s)
u
exp −
k=0
Z
exp −
tj+m−1
m=1 k=m l=0
n−1
Y
tj+m
tj
k=0
Since x(tk ) > 0 (tk ≥ T ), we find that the above inequality contradicts condition
(H4), therefore x0 (tk ) ≥ 0 for t ≥ T . Further, for t ∈ (tj , tj+1 ], we obtain
Z t r0 (s) + p(s) Z tj+1 r0 (s) + p(s) 0
0
x (t) exp
ds ≥ x (tj+1 ) exp
ds > 0,
αr(s)
αr(s)
t0
t0
which implies x0 (t) > 0 for t ≥ T . This completes the proof.
Using Lemma 2.4, we have the following Theorem.
Theorem 2.5. If (H1), (H2), (H4), (H5) are satisfied, then every solution x of
(1.1), (1.2) satisfies lim inf t→∞ |x(t)| = 0.
Example. Consider
0
3
1
1
1
t(x0 (t))3 − x0 (t) + 2 x(t − ) = 0, t 6= k, t ≥ ,
t
3
2
k
0 +
0
+
x (k ) =
x (k), x(k ) = x(k), k = 1, 2, . . . .
k+1
Comparing with (1.1), (1.2), we see that r(t) = t, p(t) = −1, α = 3, δ = 1/3,
tk+1 − tk > 1/3 and ak = a∗k = k/(k + 1), bk = b∗k = 1. Obviously (H1), (H2) are
satisfied,
Z t Y
Z s r0 (σ) + p(σ) a∗k
exp −
dσ ds
lim
t→∞ t
bk
3r(σ)
tj
j tj <tk <s
Z t
ds
> (j + 1) lim
= +∞,
t→∞ t s + 1
j
and
Z
t
1
exp
c
<s k
Y
lim
t→∞
t0 t0 <tk
Z t Y
= lim
t→∞
>
(
t0 t0 <tk <s
Z t
1
lim
2 t→∞
Z
s
t0
p(σ) dσ h(s)ds
r(σ)
1 α
1
) exp(− ln s + ln t0 ) 2 ds
ak
s
ds = +∞.
t0
So (H3) and (H5) are satisfied. By Theorem 2.3, it is clear that every solution of
this equation satisfies lim inf t→∞ |x(t)| = 0.
References
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59-62.
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[3] J. Jiao, L. Chen and L. Li; Asymptotic behavior of solutions of second-order nonlinear impulsive differential equations, J. Math. Anal. Appl., 337, 2008, 458-463.
[4] V. Lakshmikantham, D. Bainov, P. Simeonov; Theory of impulsive differential equations,
World Scientific, Singapore, 1989.
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equation with impulses, J. Comp. Appl. Math., 197, 2006, 78-88.
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impulses, Comput. Math. Appl., 53, 2007, 1740-1749.
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Haifeng Liu
Department of Science and Technology, Hebei Normal University,
Shijiazhuang, 050016, China
E-mail address: liuhf@mail.hebtu.edu.cn
Qiaoluan Li
College of Mathematics and Information Science, Hebei Normal University,
Shijiazhuang, 050016, China
E-mail address: qll71125@163.com