Electronic Journal of Differential Equations, Vol. 2010(2010), No. 151, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS TO N-TH ORDER NEUTRAL
DYNAMIC EQUATIONS ON TIME SCALES
QIAOLUAN LI, ZHENGUO ZHANG
Abstract. In this article, we study n-th order neutral nonlinear dynamic
equation on time scales. We obtain sufficient conditions for the existence of
non-oscillatory solutions by using fixed point theory.
1. Introduction
This article concerns the n-th order neutral dynamic equation
n
(x(t) + p(t)x(τ (t)))∆ + f1 (t, x(τ1 (t))) − f2 (t, x(τ2 (t))) = 0,
(1.1)
for t ≥ t0 , where t ∈ T, n ∈ N. We assume p ∈ Crd (T, R), τ, τi ∈ Crd (T, T),
τ is strictly increasing, τ (t) < t, τ (t) → ∞, τi (t) → ∞, fi ∈ Crd (T × R, R),
f1 (t, u)f2 (t, u) > 0, and fi is non-decreasing in u. In the sequel, without loss of
generality, we assume that fi (t, u) > 0, i = 1, 2.
In 1988, Stephan Hilger [7] introduced the theory of time scales as a means
of unifying discrete and continuous calculus. Several authors have expounded on
various aspects of this new theory, see [1, 6, 12] and references therein. Recently,
much attention is concerned with questions of existence of non-oscillatory solutions
for dynamic equations on time scales. For significant works along this line, see
[5, 8, 9, 10]. Many results have been obtained for first and second order dynamic
equations, however, few results are available for higher order dynamic equations.
Motivated by these works, we investigate the existence of non-oscillatory solutions
of (1.1).
In Section 2, we present some preliminary material that we will need to show
the existence of solutions of (1.1). We present our main results in Section 3.
2. Preliminaries
We assume the reader is familiar with the notation and basic results for dynamic
equations on time scales. For a review of this topic we direct the reader to the
monographs [2, 3].
We recall x is a solution of (1.1) provided that x(t) + p(t)x(τ (t)) is n times
differentiable, and x satisfies (1.1). A solution x of (1.1) is called non-oscillatory if
x is of one sign when t ≥ T .
2000 Mathematics Subject Classification. 34K40, 34N99, 39A10.
Key words and phrases. Time scales; dynamic equations; non-oscillatory solution.
c
2010
Texas State University - San Marcos.
Submitted June 6, 2010. Published October 21, 2010.
1
2
Q. LI, Z. ZHANG
EJDE-2010/151
We define a sequence of functions gk (s, t), k = 1, 2, . . . as follows.
g0 (s, t) ≡ 1, s, t ∈ Tκ ,
Z s
gk (σ(u), t)∆u, s, t ∈ Tκ .
gk+1 (s, t) =
(2.1)
t
For gk (s, t), we have the following Lemma.
Lemma 2.1 ([11]). Assume s is fixed, and let gk∆ (s, t) be the derivative of gk (s, t)
with respect to t. Then
gk∆ (s, t) = −gk−1 (s, t),
k ∈ N, t ∈ Tκ .
(2.2)
Lemma 2.2 ([4]). Let X be a Banach space, Ω be a bounded closed convex subset
of X and let A, B be maps from Ω to X such that Ax + By ∈ Ω for every pair
x, y ∈ Ω. If A is a contraction and B is completely continuous, then the equation
Ax + Bx = x has a solution in Ω.
Lemma 2.3 ([4]). Let X be a locally convex linear space, S be a compact convex
subset of X, and T : S → S be a continuous mapping with T (S) compact. Then T
has a fixed point in S.
3. Main Results
Theorem 3.1. Assume that 0 < p(t) ≤ p < 1, and there exists b > 0 such that
Z ∞
gn−1 (σ(s), 0)fi (s, b)∆s < ∞, i = 1, 2.
(3.1)
t0
Then (1.1) has a bounded non-oscillatory solution which is bounded away from zero.
Proof. Let BC be the set of bounded functions on [t0 , ∞) with sup norm kxk =
supt≥t0 |x(t)|, t ∈ T. Let Ω ⊂ BC, Ω = {x ∈ BC, 0 < M1 ≤ x(t) ≤ M2 < b, t ≥
t0 , t ∈ T}, where M1 < (1 − p)M2 , then Ω is a closed bounded and convex subset
of BC.
Choose α such that pM2 + M1 < α < M2 , and c = min{M2 − α, α − pM2 −
M
R ∞1 }. We choose t1 > t0 , such that τ (t) ≥ t0 , τi (t) ≥ t0 , i = 1, 2 t ≥ t1 and
g
(σ(s), 0)fi (s, b)∆s ≤ c, i = 1, 2. Define a mapping Γ on Ω as follows.
t1 n−1
(Γx)(t) = (Γ1 x)(t) + (Γ2 x)(t),
where
(
α − p(t)x(τ (t)), t ≥ t1 , t ∈ T,
(Γ1 x)(t) =
(Γ1 x)(t1 ),
t0 ≤ t ≤ t1 , t ∈ T.

R
∞
n−1

gn−1 (σ(s), t) f1 (s, x(τ1 (s)))

t
(−1)
−f
(s,
x(τ
(s)))
∆s,
t ≥ t1 ,
2
2
(Γ2 x)(t) =


(Γ x)(t ),
t 0 ≤ t ≤ t1 .
2
1
For any x, y ∈ Ω, t ≥ t0 , t ∈ T, we have
(Γ1 x)(t) + (Γ2 y)(t) ≤ α + c ≤ M2 ,
(Γ1 x)(t) + (Γ2 y)(t) ≥ α − pM2 − c ≥ M1 .
EJDE-2010/151
EXISTENCE OF SOLUTIONS?
3
Hence for t ≥ t0 , t ∈ T, Γ1 x + Γ2 y ∈ Ω. Clearly, Γ1 is a contraction mapping on Ω
and Γ2 is continuous. We shall show that Γ2 is completely continuous. In fact, for
any x ∈ Ω, for t0 ≤ t ≤ t1 , (Γ2 x)(t) = (Γ2 x)(t1 ), and for t ≥ t1 , we have
Z ∞
|(Γ2 x)(t)| ≤
gn−1 (σ(s), t)|f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))|∆s
t
Z ∞
≤
gn−1 (σ(s), t)f1 (s, x(τ1 (s)))∆s
t
Z ∞
gn−1 (σ(s), 0)f1 (s, b)∆s ≤ c.
≤
t
Hence Γ2 Ω is uniformly bounded. For ε > 0, there exists a T , such that for t ≥ T ,
Z ∞
ε
gn−1 (σ(s), 0)fi (s, b)∆s < .
2
t
For t, t0 > T , we have
Z ∞
|(Γ2 x)(t) − (Γ2 x)(t0 )| ≤ 2
gn−1 (σ(s), 0)fi (s, b)∆s < ε.
T
For t, t0 ∈ [t1 , T ], we have
|(Γ2 x)(t) − (Γ2 x)(t0 )|
Z ∞
=|
gn−1 (σ(s), t)[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s
t
Z ∞
−
gn−1 (σ(s), t0 )[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s|
Z
t0
t0
≤|
gn−1 (σ(s), t)[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s|
t
Z
T
+
|gn−1 (σ(s), t) − gn−1 (σ(s), t0 )kf1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))|∆s
0
Zt ∞
+
|gn−1 (σ(s), t) − gn−1 (σ(s), t0 )kf1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))|∆s.
T
There exists a δ, so that when |t − t0 | < δ, |(Γ2 x)(t) − (Γ2 x)(t0 )| < ε, which shows
that the family Γ2 Ω is equicontinuous, Γ2 is completely continuous.
By Lemma 2, there exists a fixed point x ∈ Ω, such that Γx = x. It is easily
to see that x is a bounded non-oscillatory solution which is bounded away from
zero.
Theorem 3.2. Assume that 1 < p1 ≤ p(t) ≤ p2 , and (3.1) holds. Then (1.1) has
a bounded non-oscillatory solution which is bounded away from zero.
Proof. We choose t1 > t0 such that
T0 = min{τ (t1 ), inf (τ1 (t)), inf (τ2 (t))} ≥ t0 .
t≥t1
t≥t1
Let BC be the set of bounded functions on [t0 , ∞) with supremum norm kxk =
supt≥t0 |x(t)|, t ∈ T. Define a set Ω ⊂ BC as follows:
n
Ω = x ∈ BC, x∆ (t) ≤ 0, 0 < M1 ≤ x(t) ≤ p1 M1 < b, t ≥ t1 ,
o
x(t) = x(t1 ), T0 ≤ t ≤ t1 .
4
Q. LI, Z. ZHANG
EJDE-2010/151
Then Ω is a closed bounded and convex subset of BC. Let c = min{α−M1 , p1 M1 −
α}, where M1 < α < p1 M1 . We choose t2 ≥ t1 , such that for t ≥ t2 ,
Z ∞
gn−1 (σ(s), 0)fi (s, b)∆s ≤ c.
t
For x ∈ Ω, define
(P
∞
i=1
ψ(t) =
(−1)i−1 x(τ −i (t))
,
Hi (τ −i (t))
t ≥ t2 ,
T 0 ≤ t ≤ t2 ,
ψ(t2 ),
where τ 0 (t) = t, τ i (t) = τ (τ i−1 (t)), τ −i (t) = τ −1 (τ −(i−1) (t)), H0 (t) = 1, Hi (t) =
Qi−1
j
j=0 p(τ (t)), i = 1, 2, . . . . From M1 ≤ x(t) ≤ p1 M1 , we have
0 < ψ(t) ≤ p1 M1 ,
t ≥ t2 , t ∈ T.
Define a mapping Γ on Ω as follows

R∞

(−1)n−1 t gn−1 (σ(s), t)

α +
P2
i+1
(Γx)(t) = × i=1 (−1) fi (s, ψ(τi (s)))∆s, t ≥ t2 ,


(Γx)(t ),
T 0 ≤ t ≤ t2 .
2
Then Γ satisfies the following conditions:
(a) ΓΩ ⊆ Ω. In fact, for any x ∈ Ω, (Γx)(t) ≥ α − c ≥ M1 , (Γx)(t) ≤ α + c ≤ p1 M1 .
(b) Γ is continuous which is easy to show.
(c) Similar to Theorem 1, Γ is equicontinuous.
By Lemma 3, there exists x ∈ Ω, such that x = Γx; i.e.,
Z ∞
n−1
x(t) = α + (−1)
gn−1 (σ(s), t)[f1 (s, ψ(τ1 (s))) − f2 (s, ψ(τ2 (s)))]∆s.
t
Since ψ(t) + p(t)ψ(τ (t)) = x(t), we obtain
ψ(t) + p(t)ψ(τ (t))
Z ∞
n−1
= α + (−1)
gn−1 (σ(s), t)[f1 (s, ψ(τ1 (s))) − f2 (s, ψ(τ2 (s)))]∆s.
t
So ψ(t) satisfies (1.1) for t ≥ t0 , t ∈ T, and
proof is complete.
p1 −1
−1
(t))
p1 p2 x(τ
≤ ψ(t) ≤ x(t). The
Theorem 3.3. Assume that −1 < p1 ≤ p(t) ≤ 0, and (3.1) holds. Then (1.1) has
a bounded non-oscillatory solution which is bounded away from zero.
Proof. Let BC be the set of bounded functions on [t0 , ∞) with sup norm kxk =
supt≥t0 |x(t)|. We choose M1 , M2 < b such that 0 < M1 < α < (1 + p1 )M2 . Let
Ω = {x ∈ BC, M1 ≤ x(t) ≤ M2 < b, t ≥ t0 }. Then Ω is a closed bounded and
convex subset of BC. Let c = min{α−M1 ,R(1+p1 )M2 −α}. We choose t1 ≥ t0 , such
∞
that τ (t) ≥ t0 , τi (t) ≥ t0 , for t ≥ t1 and t1 gn−1 (σ(s), 0)fi (s, b)∆s ≤ c, i = 1, 2.
Define a mapping Γ on Ω as follows:
(Γx)(t) = (Γ1 x)(t) + (Γ2 x)(t),
EJDE-2010/151
EXISTENCE OF SOLUTIONS?
5
where
(
α − p(t)x(τ (t)), t ≥ t1 ,
(Γ1 x)(t) =
(Γ1 x)(t1 ),
t 0 ≤ t ≤ t1 ,

R
∞
n−1

gn−1 (σ(s), t) f1 (s, x(τ1 (s)))

t
(−1)
−f
(s,
x(τ
(s)))
∆s,
t ≥ t1 ,
2
2
(Γ2 x)(t) =


(Γ x)(t ),
t 0 ≤ t ≤ t1 .
2
1
For x, y ∈ Ω, t ≥ t0 , we have
(Γ1 x)(t) + (Γ2 y)(t) ≤ α − p1 M2 + c ≤ M2 ,
(Γ1 x)(t) + (Γ2 y)(t) ≥ α − c ≥ M1 .
Hence for t ≥ t0 , Γ1 x + Γ2 y ∈ Ω. Clearly, Γ1 is a contraction mapping on Ω and Γ2
is continuous. Similar to Theorem 1, we can prove that Γ2 is completely continuous.
So that there exists x ∈ Ω such that x = Γx. The proof is complete.
Theorem 3.4. Assume that p1 ≤ p(t) ≤ p2 < −1 and (3.1) holds. Then (1.1) has
a bounded non-oscillatory solution which is bounded away from zero.
Proof. Let BC be the bounded functions on [t0 , ∞). We choose 0 < M1 < M2 < b,
such that −p1 M1 < α < (−p2 − 1)M2 . Let Ω = {x ∈ BC, M1 ≤ x(t) ≤ M2 , t ≥ t0 },
c = min{ (α+Mp11p1 )p2 , (−p2 − 1)M2 − α}. Choose t1 ≥ t0 such that for t ≥ t1 ,
Z ∞
τ −1 (τi (t)) ≥ t0 ,
gn−1 (σ(s), 0)fi (s, b)∆s ≤ c, i = 1, 2.
τ −1 (t)
Define two maps Γ1 , Γ2 on Ω as follows:
(
x(τ −1 (t))
α
− p(τ −1
(t)) − p(τ −1 (t)) , t ≥ t1 ,
(Γ1 x)(t) =
(Γ1 x)(t1 ),
t 0 ≤ t ≤ t1 .

n−1 R ∞
(−1)


 p(τ −1 (t)) τ −1 (t) gn−1 (σ(s), t) f1 (s, x(τ1 (s)))
t ≥ t1 ,
(Γ2 x)(t) = −f2 (s, x(τ2 (s))) ∆s,



(Γ2 x)(t1 ),
t0 ≤ t ≤ t1 .
c
−α
For x, y ∈ Ω, (Γ1 x)(t) + (Γ2 y)(t) ≥ −α
p1 + p2 ≥ M1 , (Γ1 x)(t) + (Γ2 y)(t) ≤ p2 −
M2
c
p2 − p2 ≤ M2 . So (Γ1 x)(t) + (Γ2 y)(t) ∈ Ω. Since p1 ≤ p(t) ≤ p2 ≤ −1, we get
Γ1 is contraction. We shall prove that Γ2 is completely continuous. In fact, for all
x ∈ Ω, t0 ≤ t ≤ t1 , (Γ2 x)(t) = (Γ2 x)(t1 ). For t ≥ t1 ,
Z ∞
c
1
|(Γ2 x)(t)| ≤ −
gn−1 (σ(s), 0)fi (s, x(τi (s)))∆s ≤ − ,
p2 τ −1 (t)
p2
so Γ2 Ω is uniformly bounded. By the conditions, for all ε > 0, there exists a T > t1
such that
Z ∞
−p2 ε
gn−1 (σ(s), 0)fi (s, b)∆s <
.
2
τ −1 (T )
For all x ∈ Ω, t, t0 ≥ T , we have
|(Γ2 x)(t) − (Γ2 x)(t0 )| ≤ −
2
p2
Z
∞
gn−1 (σ(s), 0)fi (s, b)∆s < ε.
τ −1 (T )
6
Q. LI, Z. ZHANG
EJDE-2010/151
1
Since τ −1 (t), p(τ −1
(t)) are continuous on [t1 , T ], they are uniformly continuous on
[t1 , T ]. Let |gn−1 (σ(t), 0)fi (t, b)| ≤ M , when t ∈ [t1 , T ]. Hence for each ε > 0, there
exists a δ > 0 such that for t, t0 ∈ [t1 , T ], |t − t0 | < δ, we have
|
1
1
ε
−p2 ε
−
| < , |τ −1 (t) − τ −1 (t0 )| <
,
p(τ −1 (t)) p(τ −1 (t0 ))
3c
3M
Z ∞
|p2 |ε
|gn−1 (σ(s), t) − gn−1 (σ(s), t0 )|fi (s, b)∆s <
.
3
−1
τ (t1 )
For all x ∈ Ω, when t, t0 ∈ [t1 , T ] and |t − t0 | < δ, we have
|(Γ2 x)(t) − (Γ2 x)(t0 )|
Z ∞
1
gn−1 (σ(s), t)[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s
=|
p(τ −1 (t)) τ −1 (t)
Z ∞
1
−
gn−1 (σ(s), t0 )[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s|
p(τ −1 (t0 )) τ −1 (t0 )
Z ∞
1
1
−
|
≤|
gn−1 (σ(s), t)|f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))|∆s
p(τ −1 (t)) p(τ −1 (t0 )) τ −1 (t)
Z τ −1 (t0 )
1
+|
k
gn−1 (σ(s), t)[f1 (s, x(τ1 (s))) − f2 (s, x(τ2 (s)))]∆s|
p(τ −1 (t0 )) τ −1 (t)
Z ∞
1
|
+|
|gn−1 (σ(s), t) − gn−1 (σ(s), t0 )|
p(τ −1 (t0 )) τ −1 (t0 )
×|
2
X
(−1)i+1 fi (s, x(τi (s)))|∆s
i=1
εc
M |p2 |ε |p2 |ε
<
+
·
+
= ε,
3c |p2 | 3M
|p2 |3
which shows that the family Γ2 Ω is equicontinuous, so Γ2 is completely continuous.
By Lemma 2, there exists a fixed point x ∈ Ω such that Γx = x. It is easily to see
that x is a bounded non-oscillatory solution which is bounded away from zero. Example. On the time scale T = {q n : n ∈ N0 , q > 1}, consider the dynamic
equation
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2 2
4
1
(x(t) − √ x(ρ(t)))∆ + 2
x (ρ (t))
q
q 10 t3 (t + q 2 )2
(3.2)
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2 3
−
x (ρ (t)) = 0,
q 10 t3 (t + q 2 )3
where ρ is the backward operator, ρ2 (t) = ρ(ρ(t)), ρ3 (t) = ρ(ρ2 (t)). In this equation,
n = 4, p(t) = − √1q , τ (t) = ρ(t) = qt , τ1 (t) = ρ2 (t), τ2 (t) = ρ3 (t),
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b ,
q 10 t3 (t + q 2 )2
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
f2 (b) =
b .
q 10 t3 (t + q 2 )3
f1 (t, b) = 2
EJDE-2010/151
EXISTENCE OF SOLUTIONS?
7
By the definition of gk (s, t),
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b
q 10 s3 (s + q 2 )2
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
≤2
b ,
q 10 s2
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b
g4−1 (σ(s), 0) · f2 (s, b) ≤ s3
q 10 s3 (s + q 2 )3
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b ,
≤
q 10 s3
g4−1 (σ(s), 0) · f1 (s, b) ≤ s3 2
and
Z
∞
t0
∞
Z
t0
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b ∆s < ∞,
q 10 s2
√
( q − 1)(q + 1)2 (q 2 + 1)(q 2 + q + 1) 2
b ∆s < ∞.
q 10 s3
It is obviously that (3.2) satisfies all conditions of Theorem 3. Hence (3.2) has a
bounded non-oscillatory solution which is bounded away from zero. In fact x(t) =
1 + 1t is a solution of (3.2).
Acknowledgements. This research was supported by grants L2009Z02 from the
Main Foundation of Hebei Normal University, and L2006B01 from the Doctoral
Foundation of Hebei Normal University. The authors would like to thank the
anonymous referee for his or her careful reading and the comments on improving
the presentation of this article.
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8
Q. LI, Z. ZHANG
EJDE-2010/151
Qiaoluan Li
College of Mathematics and Information Science, Hebei Normal University,
Shijiazhuang, 050016, China
E-mail address: qll71125@163.com
Zhenguo Zhang
College of Mathematics and Information Science, Hebei Normal University,
Shijiazhuang, 050016, China.
Information College, Zhejiang Ocean University, Zhoushan, 316000, China
E-mail address: zhangzhg@mail.hebtu.edu.cn