doi:10.1155/2009/567329
Research Article
Antiperiodic Boundary Value Problem for
Second-Order Impulsive Differential Equations
on Time Scales
Yepeng Xing, Qiong Wang, and De Chen
Department of Applied Mathematics, Shanghai Normal University, Shanghai 200234, China
Correspondence should be addressed to Yepeng Xing,ypxing-jason@hotmail.com
Received 3 April 2009; Accepted 20 July 2009
Recommended by Alberto Cabada
We prove the existence results for second-order impulsive differential equations on time scales with antiperiodic boundary value conditions in the presence of classical fixed point theorems.
Copyrightq2009 Yepeng Xing et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and Preliminaries
Many evolution processes are characterized by the fact that at certain moments of time they experience a change of state abruptly. Consequently, it is natural to assume that these perturbations act instantaneously, that is, in the form of impulses. It is known that many biological phenomena involving threshold, bursting rhythm models in medicine and biology, optimal control models in economics, pharmacokinetics, and frequency modulated systems do exhibit impulse effects. The branch of modern applied analysis known as “impulsive” differential equations provides a natural framework to mathematically describe the aforementioned jumping processes. The reader is referred to monographs 1–4 and references therein for some nice examples and applications to the above areas.
differential equations with antiperiodic boundary value conditions on time scales. Firstly, we use Schauder’s fixed point theorem to study the existence results of the following boundary value problem:
−xΔΔt ft, xt, xΔt, t∈0, T\Ω:{t
1, t2, . . . , tm},
xt
k
xtk Ikxtk, xΔtkxΔtk Jk
xtk, xΔtk
, k1,2, . . . , m,
x0 −xσT, xΔ0 −xΔσT.
1.1
Secondly, we employ Schaefer’s fixed point to investigate the following antiperiodic boundary value problem on time scales:
xΔΔt ft, xσt, xΔt, t∈0, T\ {t
1, t2, . . . , tm},
xt
k
xtk Ikxtk, xΔtk
xΔt
k Jk
xtk, xΔtk
, k1,2, . . . , m,
x0 −xσT, xΔ0 −xΔσT.
1.2
We assume throughout this paper thatf:0, T×Rn×Rn → Rnis continuous andIk:Rn →
Rn,Jk:Rn×Rn → Rnare also continuousk1,2, . . . , m.
We should mention that1.1, 1.2 are more general than equations considered in 13, 18since f, J both contain xand its derivative. So our result is still new even when 1.1,1.2 reduce to equations studied in 13, 18. The paper is organized as follows. In
Section 2we present the expression of Green’s functions of related linear operator in the space of piecewise continuous functions.Section 3 contains the main results of the paper and is devoted to the existence of solutions to1.1and1.2.
To understand the concept of time scales and the above notation, some definitions are useful.
A time scale is an arbitrary nonempty closed subset of the real numbers. Throughout this paper, we will denote a time scale by the symbolT. And the forward and backward jump operatorsσ, ρ:T → Tare defined by
σt inf{s∈T:s > t}, ρt sup{s∈T:s < t}, 1.3
respectively. The pointt∈Tis called left dense, left scattered, right dense, or right-scattered ifρt t, ρt< t, σt torσt> t, respectively. Points that are right dense and left dense at the same time are called dense. We denoteσσtbyσ2t.
IfThas a left scattered maximumm, defineTk:T− {m}; otherwise, setTkT. The symbolsa, b,a, b, and so on, denote time scales intervals, for example,
a, b {t∈T:a≤t≤b}, 1.4
Definition 1.1. A vector functionf :T → Rn is rd-continuous provided that it is continuous at each right dense point inTand has a left-sided limit at each left dense point inT. The set of rd-continuous functionsf:T → Rnwill be denoted in this paper byCrdT CrdT,Rn.
Definition 1.2. Assumef : T → Ris a function and let t ∈ Tk. Then we define fΔt to be the number provided it exists with the property that given anyε > 0 there exists a neighborhoodUofti.e.,U t−δ, tδ∩Tfor someδ >0such that
fσt−fs−fΔtσt−s≤ε|σt−s|, ∀s∈U. 1.5
We callfΔtthe deltaor Higherderivative offatt.
Definition 1.3. A functionF:T → Ris called an antiderivative off:T → Rprovided
FΔt ft holds∀t∈T. 1.6
Theorem 1.4existence of antiderivatives. Every rd-continuous function has an antiderivative. In particular ift0∈T, thenFdefined by
Ft: t
t0
ftΔt fort∈T 1.7
is an antiderivative off.
We assume that forftk, x, y : limt→t
kft, x, yand ft
−
k, x, y : limt→t−kft, x, y
both exist withft−k, x, y ftk, x, y, k 1,2, . . . , m. In order to define a solution of1.1
and1.2, we introduce and denote the Banach spacePC0, T,Rnby
PC0, T;Rn:{u:0, T−→Rn, u∈C0, T\Ω,Rn,
uis left continuous atttk, the right-hand limitutkexists
1.8
with the normuPC:supt∈0,Tut where · is the usual Euclidean norm and·,· will be the Euclidean inner product.
In a similar fashion to the above, define and denote the Banach spacePC10, T,Rn
by
PC10, T;Rn: u∈PC0, T;Rn:uΔt∈C0, T\Ω,Rn,
the limitsuΔt−k, uΔtkexist withuΔt−kuΔtk
1.9
with the normuPC1 :supt∈0,T{utPC,uΔtPC}.
Theorem 1.5Schaefer’s fixed point theorem19. LetXbe a Banach space and letA:X → X
be a completely continuous operator. Then either:
ithe operator equationxλAxhas a solution forλ1.or iithe setS:{x∈X, xλAx, λ∈0,1}is unbounded.
2. Expression of Green’s Function
In this part, we present the expression of Green’s functions for second-order impulsive equations with antiperiodic conditions.
Lemma 2.1. For anyht∈PC0, T,Rn,xtsolves
if and only ifxtis the solution of integral equation
Similarly, we Integrate2.3from 0 totstep by step to get Theorem 1.117, page 46that
whereGΔ, fΔdenotes the derivatives ofG, Hwith respect to the first variable. That is
Compute straightforwardly to get
xΔΔt −ht. 2.11
Thus, the proof is completed.
Therefore we have the upper bounds
max
t,s∈0,σ2T×0,σT|Gt, σs| ≤ 1 4
σT 2σ2T:G0,
max
t,s∈0,σ2T×0,σT|Ht, s| 1 2,
max
t,s∈0,σ2T×0,σT
GΔt, s 1
2.
2.17
Recall that a mapping between Banach spaces is compact if it is continuous and carries bounded sets into relatively compact sets.
Lemma 2.2. Suppose thatf :0, T×Rn×Rn → Rn,Ik:Rn → Rn, andJk:Rn×Rn → Rnare continuous. Define an operatorA:PC10, σ2T,Rn → PC10, σ2T,Rnas
Axt: σT
0
Gt, σsfs, xs, xΔsΔsm
k1
Ht, tkIkxtk
−m k1
Gt, tkJk
xtk, xΔtk
,
2.18
whereGt, sandHt, sare as given inLemma 2.1. ThenAis a compact map.
Proof. From2.10we know
AxΔt −1
2 t
0
fs, xs, xΔsΔs1
2 σT
t f
s, xs, xΔsΔsm
k1 Jk
xtk, xΔtk
.
2.19
Then the continuality off, Ik, JkimpliesAis a continuous map fromPCJ,RntoPCJ,Rn.
On the other hand, for any bounded subsetS⊂PCJ,Rn.2.19implies{AxΔt|xt∈S} is also a bounded subset ofPCJ,Rn. Deducing in a similar way as proving20, Lemma 2.4, we haveAis a compact map.
Lemma 2.3. x∈PC1J,Rn∩PC2J,Rnis a solution of 1.1if and only ifxt∈PC1J,Rnis
a fixed point ofA.
Proof. It can be easily obtain fromLemma 2.1, and we omit the proof here.
3. Main Results
We first set
α lim
xy→ ∞
max
t∈0,T
f
t, x, y
xy
,
βk lim
x → ∞
I
kx x
k1,2, . . . , m,
γk lim
xy→ ∞
Jk
x, y
xy
k1,2, . . . , m.
3.1
Theorem 3.1. Suppose thatf :0, T×Rn×Rn → Rn,Ik:Rn → RnandJk:Rn×Rn → Rnare
continuousk1,2, . . . , m. If
ηmaxη1, η2
<1, 3.2
where
η1 1
2
ασT m
k1 γk
σT 2σ2T 1 2
m
k1 βk,
η2ασT
m
k1 γk.
3.3
then1.1has at least one solution inPC1J,Rn∩PC2J,Rn.
Proof. By Lemma 2.2it is sufficient to show that A has at least one solution inPC1J,Rn. First, we can choose by3.2α> α, βk> βk, γk > γk, k1,2, . . . , msuch that
η 1
1 2
ασT m
k1 γ
k
σT 2σ2T1 2
m
k1 β
k<1,
η
2ασT
m
k1 γ
k<1.
3.4
By3.1we can choose a positive numberNsuch that
f
t, x, y< αxy, ∀t∈0, σT, xy≥N. 3.5
Then
f
where
M max
t∈0,σT,xy≤Nf
t, x, y<∞. 3.7
Similarly, there exist positive constantsMk, Mk,k1,2, . . . , msuch that
IkxβkxMk, ∀x∈Rn, 3.8
Jk
x, y≤γ
k
xyMk, ∀x, y∈Rn. 3.9
It then follows by2.17,2.18,3.6–3.9that
Axt ≤ 1
4σT
σT 2σ2TαxtxΔtM
1 2
m
k1 β
kx m
k1 Mk
1 4σT
σT 2σ2T
m
k1 γ
k
xtxΔtm
k1 Mk
≤η
1xPC1M1, ∀t∈
0, σ2T,
3.10
where
M1 1
4σT
σT 2σ2TM
m
k1 Mk
1
2 m
k1 Mk
3.11
is a constant.
Similarly, we can prove that
AxΔt≤η
2xPC1M2, ∀t∈
0, σ2T, 3.12
whereM2 1/2Mm
k1Mkis a constant.
Consequently, by3.10and3.12we have
AxPC1≤ηxPC1M, ∀x∈PC1
0, σ2T,Rn, 3.13
where
ηmaxη 1, η2
Thus we can chooser >0 such that
ABr⊂Br, 3.15
whereBr {x∈PC10, σ2T,Rn| xPC1 ≤r}.On the other hand, fromLemma 2.3Ais a
complete continuous operator. Hence by Schauder’s fixed point theoremAhas fixed point in
Br. This completes the proof.
Remark 3.2. Assumption3.2is true if there hold f
t, x, y
xy ⇒0,
Jk
x, y
xy −→0, aspq−→ ∞,
Ikx
x −→0, asx −→ ∞k1,2, . . . , m.
3.16
We now go on to study the existence results for1.2. We should mention that the idea used in the following theorem is initiated by Tisdell21,22.
Theorem 3.3. Suppose thatf:0, T×Rn×Rn → Rn, Ik:Rn → RnandJk:Rn×Rn → Rnare continuousk1,2, . . . , m. If there exist nonnegative constantsγ, δk, ζk, Lk, NkandMsuch that
f
t, x, y≤γx, ft, x, y y2M, ∀t, x, y∈
0, σT\Ω×Rn×Rn, 3.17
Ikx ≤δkxLk, Jkx, y≤ζkxyNk, ∀x, y∈Rn,
max !
1 2
m
k1
δi2G0
m
i1 ζi,
m
i1 ζi
"
<1, 3.18
where· is the Euclidean inner product. Then1.2has at least one solution.
Proof. Consider the mapping
A:PC10, σ2T,Rn−→PC10, σ2T,Rn 3.19
Axt
σT
0
Gt, σsfs, xσs, xΔsΔsm
k1
Ht, tkIkxtk
m
k1
Gt, tkJk
xtk, xΔtk
,
3.20
where
Gt, s
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
−1
2
1
2σT−ts
0≤s≤t≤σ2T,
−1
2
1
2σT t−s
0≤t < s≤σT,
Ht, s GΔt, s
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1
2, 0≤s≤t≤σ
2T,
−1
2, 0≤t < s≤σT.
H∗
In a similar way as we prove Lemmas2.1,2.2and2.3we obtain
ix∈PC1J,Rn∩PC2J,Rnis a solution of1.2if and only ifxt∈PC1J,Rnis a
fixed point ofA;
iiAis a compact operator. Consider the equation
xAx. 3.21
To show Ahas at least one fixed point, we apply Schaefer’s theorem by showing that all potential solutions to
xλAx, λ∈0,1 3.22
are bounded a priori, with the bound being independent ofλ. With this in mind, letxtbe a solution of3.22. Note thatxtis also a solution to
xΔΔt λft, xσt, xΔt, t /t
k, k1,2, . . . , m,
xt
k
xtk λIkxtk, xΔtkxΔtk λJk
xtk, xΔtk
, k1,2, . . . , m,
x0 −xσT, xΔ0 −xΔσT.
3.23
On one hand, we see that forλ∈0,1
λft, xσt, xΔt≤λ#γ$xσt, ft, xσt, xΔt%xΔt2M&
γ$xσt, λft, xσt, xΔt%λxΔt2λM
γ$xσt, xΔΔt λxΔt, xΔt%λM
≤γ$xσt, xΔΔt xΔt, xΔt%M
γ$xt, xΔt%ΔM.
3.24
On the other hand, by the antiperiodic boundary condition we have
σT
0
$
It therefore follows that
Example 3.4. Consider antiperiodic value problem onT'∞k03k1,3k2'{3k2}
uΔΔt uσt uΔt2cost, t∈0,1, t /1,
u1 u1 1
8u1 4, u
Δ1 uΔ1−1
8u1 10,
u0 −u3, uΔ0 −uΔ3.
3.31
We claim4.2has at least one solution.
Proof. LetT 3 andft, x, y xxy2costinTheorem 3.3. ChoosingM4, we have for t, x, y∈0,1'{2,3} ×R2that
f
t, x, y|x||x|y21,
x, ft, x, y y2x2x2y2y2xcost≥x2x2y2y2− |x|. 3.32
Since minv≥0{v2−2v} ≥ −1, we havex2y2y2− |x|y2y2x2− |x|1>0.Thus, for γ1 andM2
γx, ft, x, y y24≥ft, x, y, ∀t, x, y∈0,1×R2. 3.33
Moreover,G09/4,1/2δ12G0ξ1 1/2×1/8 2×9/4×1/8 5/8<1, Then the
conclusion follows fromTheorem 3.3.
Acknowlegments
The research was Supported by Leading Academic Discipline Project of Shanghai Normal University DZL805, the National Natural Science Foundation of China 10671127, the National Ministry of Education of China 20060270001, and the National Natural Science Foundation of Shanghai08ZR1416000.
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