Hindawi Publishing Corporation
Boundary Value Problems
Volume 2008, Article ID 864297, 18 pages
doi:10.1155/2008/864297
Research Article
On the Solvability of Second-Order Impulsive
Differential Equations with Antiperiodic Boundary
Value Conditions
Yepeng Xing1 and Valery Romanovski2
1
2
Department of Applied Mathematics, Shanghai Normal University, Shanghai 200234, China
Center for Applied Mathematics and Theoretical Physics, University of Maribor, SI-2000 Maribor, Slovenia
Correspondence should be addressed to Yepeng Xing, ypxing@shnu.edu.cn
Received 3 July 2008; Accepted 10 November 2008
Recommended by Colin Rogers
We prove existence results for second-order impulsive differential equations with antiperiodic
boundary value conditions in the presence of classical fixed point theorems. We also obtain the
expression of Green’s function of related linear operator in the space of piecewise continuous
functions.
Copyright q 2008 Y. Xing and V. Romanovski. 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.
In this paper, we mainly study the following second-order impulsive differential
equations with antiperiodic boundary value conditions:
u ft, u, u , t ∈ 0, T \ Ω,
u tk u tk Ik u tk , u tk u tk Jk u tk ,
u0 −uT ,
u 0 −u T ,
k 1, 2, . . . , m,
1.1
2
Boundary Value Problems
n
n
→ Rn is continuous on 0, T \ Ω × Rn × Rn ,
where Ω : m
i1 ti and f : 0, T × R × R
n
n
I, J : R → R are continuous functions.
In 4–12, the authors studied the existence of antiperiodic solutions for first-order,
second-order, or high-order differential equations without impulses, and in 3, 13–16 the
authors were concerned with the antiperiodic solutions of first-order impulsive differential
equations. Also we should mention the work by Cabada et al. in 17 which is concerned
with a certain nth order linear differential equation with constant impulses at fixed times and
nonhomogeneous periodic boundary conditions. So far, to the best of our knowledge, this is
the first work to deal with the antiperiodic solutions to second-order differential equations
with nonconstant impulses. Our method to prove the existence of antiperiodic solutions is
based on the works in 13, 18, 19. We should point out that it is Christopher C. Tisdell who
started with this method.
The article is organized as follows. In Section 2, we 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 to 1.1.
There, differential inequalities are developed and applied to prove the existence of at least
one solution to 1.1. In Section 4, a couple of examples are given to illustrate how the main
results work.
To understand the notation used above and the ideas in the remainder of the
paper, we now briefly introduce some appropriate concepts connected with impulsive
differential equations. Most of the following notation can be found in 1, 2, 4, 5. We assume
that ftk , x, y : limt → tk ft, x, y, ft−k , x, y : limt → t−k ft, x, y exist and ft−k , x, y ftk , x, y, k 1, 2, . . . , m. We introduce and denote the Banach space P C0, T , Rn by
P C 0, T ; Rn : u : 0, T −→ Rn , u ∈ C0, T \ Ω, Rn ,
u is left continuous at t tk , the right-hand limit u tk exists
1.2
with the norm ||u||P C : supt∈0,T ||ut||, 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 space P C1 0, T , Rn by
P C1 0, T ; Rn : u ∈ P C 0, T ; Rn : u ∈ C1 0, T \ Ω, Rn ,
the limits u t−k , u tk exist with u t−k u tk
1.3
with the norm ||u||P C1 : supt∈0,T {||ut||P C , ||u t||P C }.
The following fixed point theorem is our main tool to prove the existence of at least
one solution to 1.1.
Schaefer’s fixed point theorem [19]
Let X be a Banach space and let A : X → X be a completely continuous operator. Then, either
i the operator equation x λAx has a solution for λ 1, or
ii the set S : {x ∈ X, x λAx, λ ∈0, 1} is unbounded.
Y. Xing and V. Romanovski
3
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. Assume p ≥ 0 and q > 0 are two constants. Let α p p2 4q/2, β p −
p2 4q/2. Then for any ht ∈ P C0, T , Rn , ut solves
u − pu − qu ht, t ∈ 0, T , t /
tk , k 1, 2, . . . , m,
u tk u tk Ik u tk , u tk u tk Jk u tk , k 1, 2, . . . , m,
u0 −uT ,
2.1
u 0 −u T if and only if ut is the solution of integral equation
ut T
Gt, shsds 0
m
m
H t, ti Ii u ti G t, ti Ji u ti ,
i1
2.2
i1
where
⎧ αT t−s
eβT t−s
e
⎪
⎪
,
−
⎪
⎨
αT
1e
1 eβT
1
Gt, s α−β ⎪
αt−s
⎪
eβt−s
⎪
⎩e
−
,
αT
1e
1 eβT
⎧ αT t−s
βe
αeβT t−s
⎪
⎪
−
,
⎪
⎨
1 eαT
1 eβT
1
Ht, s α−β ⎪
αt−s
⎪
αeβt−s
⎪
⎩− βe
,
1 eαT
1 eβT
0 ≤ t ≤ s ≤ T,
G
0 ≤ s < t ≤ T,
0 ≤ t ≤ s ≤ T,
H
0 ≤ s < t ≤ T.
Proof. Assume ut is a solution of 2.1 and let vt u t − βut for t / tk , k 1, 2, . . . , m.
We have
v t − αvt u t − α βu t αβut u t − pu t − qut ht.
2.3
Then for t ∈ 0, t1 ,
vt e
αt
e v0 0
e
0
t
−αs
hsds e
αt
v0 t
e
0
−αs
hsds .
2.4
4
Boundary Value Problems
This implies vt1 eαt1 v0 2.1 we get that
t1
0
e−αs hsds. Consequently, from the impulsive condition in
v t1 v t1 J1 u t1 − βI1 u t1 eαt1
v0 t1
e
−αs
hsds
Δ1 ,
2.5
0
where Δi : Ji uti − βIi uti , i 1, 2, . . . , m. Now we integrate 2.3 from t1 to t ∈ t1 , t2 and use 2.5 to obtain
vt e
αt
e
−αt1
v t1 t
e
−αs
hsds eαt v0 eαt−t1 Δ1 t1
t
eαt−s hsds.
2.6
0
It follows that
t2
v t2 v t2 Δ2 eαt2 v0 eαt2 −t1 Δ1 eαt2 −s hsds Δ2 .
2.7
0
Similarly, we have for t ∈ t2 , t3 that
vt eαt v0 eαt−t1 Δ1 eαt−t2 Δ2 t
eαt−s hsds.
2.8
0
To sum up, we have for t ∈ 0, T that
vt eαt v0 eαt−ti Δi ti ∈0,t
t
eαt−s hsds.
2.9
0
Since vt u t − βut, we can deduce in a similar way as to deal with ht v t − αvt
to obtain
ut eβt u0 eβt−ti Ii u ti ti ∈0,t
t
eβt−s vsds.
2.10
0
Now we are in position to show the expression of ut for t ∈ 0, T . To do that, we need
t
to compute 0 eβt−s vsds in 2.10. In what follows we present the expression of ut for
t ∈ t1 , t2 , t2 , t3 step by step and then obtain the general form of ut for t ∈ 0, T .
Y. Xing and V. Romanovski
5
First of all, for t ∈ t1 , t2 , we have
t
e
βt−s
vsds t1
e
0
βt−s
vsds t
2.11
eβt−s vsds.
t1
0
See that
t1
e
βt−s
vsds e
0
0
t
t
eβt−s vsds t1
t1
e v0 βt−s
αs
s
e
αs−τ
hτdτ ds,
0
eβt−s eαs v0 Δ1 eαs−t1 t1
s
2.12
eαs−τ hτdτ ds.
0
Consequently,
t
e
βt−s
vsds 0
t
e
βt−s
e v0 αs
e
0
Integrate
t
0
s
αs−τ
hτdτ ds Δ1
t
0
eβt−s eαs−t1 ds.
2.13
t1
s
eβt−s 0 eαs−τ hτdτds by parts to get
eβt
α−β
t
hse−αs eα−βt − eα−βs ds.
2.14
0
Thus,
t
e
0
βt−s
eβt
v0 eα−βt − 1 e−αt1 Δ1 eα−βt − eα−βt1
vsds α−β
t
−αs α−βt
α−βs
hse
−e
e
ds .
2.15
0
Similarly, we have for t ∈ t2 , t3 that
t
e
0
βt−s
eβt
v0 eα−βt − 1 e−αt1 Δ1 eα−βt − eα−βt1
vsds α−β
e
−αt2
Δ2 e
α−βt
−e
α−βt2
t
hse
−αs
α−βt
α−βs
−e
e
ds .
0
2.16
6
Boundary Value Problems
Now we consider ut for t ∈ t1 , t2 . Clearly,
ut eβt u0 eβt−t1 I1 u t1
t
α−βt
α−βt
eβt
−αt1
α−βt1
−αs α−βt
α−βs
− 1 e Δ1 e
−e
−e
hse
e
ds .
v0 e
α−β
0
2.17
Noting that v0 u 0 − βu0, we have
1
u 0 − βu0eαt αu0 − u 0eβt
ut α−β
t
αt−s
βt−s
αt−t1 βt−t1 hs e
−e
Δ1 − e
e
Δ1 ,
2.18
0
i is denoted by Δ
i J1 ut1 − αI1 ut1 , i 1, 2, . . . , m. Similarly, for t ∈ t2 , t3 where Δ
there holds
1
ut u 0 − βu0eαt αu0 − u 0eβt
α−β
t
αt−s
βt−s
αt−t1 αt−t2 βt−t1 βt−t2 hs e
−e
Δ1 e
Δ2 − e
e
Δ1 − e
Δ2 .
0
2.19
Thus, for t ∈ 0, T ,
1
ut u 0 − βu0eαt αu0 − u 0eβt
α−β
t
αt−s
βt−s
αt−ti βt−ti hs e
−e
e
Δi −
e
Δi .
0
ti ∈0,t
2.20
ti ∈0,t
By the boundary condition of 2.1, we have
T
m
1
αT −s
αT −ti u 0 − βu0 −
,
e
hsds Δi e
1 eαT
0
i1
T
m
1
βT −s
βT −ti .
αu0 − u 0 e
hsds Δi e
1 eβT
0
i1
2.21
Y. Xing and V. Romanovski
7
Substituting 2.21 into 2.20, and also noting that for t ∈ tk , tk1 ,
−
m
m
k
Δi
eαT t−ti Δi eαt−ti Δi
αT t−ti αt−ti e
e
Δ
−
,
i
αT
αT
αT
1e
i1 1 e
i1 1 e
ik1
ti ∈0,t
m
m
k
i
i i
eαT t−ti Δ
eβt−ti Δ
Δ
βT t−ti βt−ti e
−
e
−
,
Δ
i
αT
βT
βT
1e
i1 1 e
i1 1 e
ik1
ti ∈0,t
2.22
we see that ut is the solution of 2.2.
Now assume ut is a solution of 2.2. Then for t /
tk , k 1, 2, . . . , m.
u t T
Gt t, shsds 0
u t T
m
m
Ht t, ti Ii u ti Gt t, ti Ji u ti ,
i1
Gtt t, shsds ht 0
2.23
i1
m
m
Htt t, ti Ii u ti Gtt t, ti Ji u ti .
i1
2.24
i1
It is easy to verify
u t − pu t − qt ht.
2.25
For t tk , k 1, 2, . . . , m, we compute straightforwardly to get
H tk , tk − H tk , tk 1,
Gt tk , tk − Gt tk , tk 1,
G tk , tk − G tk , tk 0,
Ht tk , tk − Ht tk , tk 0,
2.26
Δu tk Jk u tk .
2.27
which implies
Δu tk Ik u tk ,
Now, we prove ut is a solution of 2.1. Then the proof is completed.
For later use, we present the following estimations:
max
|Gt, s| ≤
t,s∈0,T ×0,T 1
α−β
α
|Ht, s| ≤
max
α−β
t,s∈0,T ×0,T eβT
eαT
1 eαT 1 eβT
eβT
eαT
1 eαT 1 eβT
: G0 ,
αG0 ,
8
Boundary Value Problems
max
t,s∈0,T ×0,T |Gt, s| ≤
|Ht t, s| ≤
max
t,s∈0,T ×0,T α
α−β
eβT
eαT
αT
1e
1 eβT
αG0 ,
αT
|α| · |β|
e
eβT
≤ α2 G0 .
α − β 1 eαT 1 eβT
2.28
Corollary 2.2. Assume in 2.1 that p 0 and q M2 > 0. Then for any ht ∈ P C0, T , ut is
the solution of
u − M2 u ht,
utk utk Ik utk ,
t ∈ 0, T , t /
tk , k 1, 2, . . . , m,
u tk u tk Jk utk ,
k 1, 2, . . . , m,
2.29
m
m
H t, ti Ii u ti G t, ti Ji u ti ,
2.30
u0 −uT ,
u 0 −u T if and only if ut is the solution of integral equation
ut T
Gt, shsds 0
i1
i1
where
⎧ MT t−s
e−MT t−s
e
⎪
⎪
, 0 ≤ t ≤ s ≤ T,
−
⎪
⎪
MT
1 e−MT
1 ⎨ 1e
Gt, s 2M ⎪
eMt−s
e−Mt−s
⎪
⎪
0 ≤ s < t ≤ T,
⎪
⎩ 1 eMT − 1 e−MT ,
⎧
−MeMT t−s Me−MT t−s
⎪
⎪
−
,
⎪
1 e−MT
1 ⎨ 1 eMT
Ht, s Gt t, s 2M ⎪
⎪ MeMt−s Me−Mt−s
⎪
⎩
,
1 eMT
1 e−MT
2.31
0 ≤ t ≤ s ≤ T,
0 ≤ s < t ≤ T.
Obviously, there hold
max
|Gt, s| ≤
t,s∈0,T ×0,T eMT
,
M 1 eMT
eMT
|Gt t, s| ≤
,
max
t,s∈0,T ×0,T 1 eMT
max
|Ht, s| ≤
t,s∈0,T ×0,T eMT
,
1 eMT
MeMT
max
|Ht t, s| ≤
.
t,s∈0,T ×0,T 1 eMT
We now give Green’s function of 2.1 for p q 0.
2.32
Y. Xing and V. Romanovski
9
Lemma 2.3. For any ht ∈ P C0, T , ut is the solution of
tk , k 1, 2, . . . , m,
u ht, t ∈ 0, T , t /
u tk u tk Ik u tk ,
u tk u tk Jk u tk ,
2.33
u 0 −u T u0 −uT ,
if and only if ut satisfies the integral equation
ut T
Gt, s∗ hsds 0
m
m
H ∗ t, ti Ii u ti G∗ t, ti Ji u ti ,
i1
2.34
i1
where
G∗ t, s −
⎧T
⎪
⎪ t − s,
1 ⎨2
0 ≤ t ≤ s ≤ T,
2⎪
⎪
⎩ T − t s, 0 ≤ s < t ≤ T,
2
⎧ 1
⎪
⎪
⎨− 2 , 0 ≤ t ≤ s ≤ T,
∗
∗
H t, s G t, st ⎪
⎪
⎩1,
0 ≤ s < t ≤ T.
2
2.35
Since the proof is very similar to that of Lemma 2.1, we omit it here. We can check
easily that ut satisfies 2.34 and hence ut is a solution of 2.33. Also we get by
straightforward computation that
max
∗
G t, s ≤ T ,
4
max
t,s∈0,T ×0,T ∗
G t, s t,s∈0,T ×0,T t
max
∗
H t, s ≤ 1 .
2
t,s∈0,T ×0,T 2.36
Recall that a mapping between Banach spaces is compact if it is continuous and carries
bounded sets into relatively compact sets.
Lemma 2.4. Suppose that f : 0, T × Rn × Rn and I, J : Rn → Rn are continuous. Define an
operator A : P C1 0, T , Rn → P C1 0, T , Rn as
T
Aut :
Gt, s fs, us, u s − pu s − qus ds
0
m
m
H t, ti Ii u ti G t, ti Ji u ti ,
i1
i1
where Gt, s and Ht, s are as given in Lemma 2.1. Then A is a compact map.
2.37
10
Boundary Value Problems
Proof. Noting the continuity of f and Ik , Jk , this follows in a standard step-by-step process
and so it is omitted.
3. Main results
In this section, we prove the existence results for 1.1 in presence of Schaefer’s fixed-point
theorem.
Theorem 3.1. Suppose that f : 0, T × Rn × Rn and I, J : Rn → Rn are continuous. If for some
p ≥ 0 and q > 0, there exist nonnegative constants γ, δk , ζk , Lk , Nk , and M such that
ft, x, y − py − qx
≤ γ x y, ft, x, y y
2 M,
m
∀t, x, y ∈ 0, T \ ti × Rn × Rn ,
3.1
i0
Ik x ≤ δk x
Lk ,
m
Jk x ≤ ζk x
Nk ,
δk k1
∀x ∈ Rn ,
m
1
ζk < ,
H
k1
3.2
3.3
where · is the Euclidean inner product, H max{G0 , αG0 , α2 G0 }. Then 1.1 has at least one
solution.
Proof. Define an integral operator A as
Au T
m
m
Gt, s fs, us, u s − pu s − qus ds H t, ti Ii u ti G t, ti Ji u ti ,
0
i1
i1
3.4
where Gt, s and Ht, s follow the forms of G and H in Lemma 2.1. By Lemma 2.4, A
is a compact mapping. Also, it follows from Lemma 2.1 that ut is a fixed point of A if and
only if ut satisfies
u t − pu t − qut ft, ut, u t − pu t − qut, t / tk , k 1, 2, . . . , m,
u tk u tk Ik u tk , u tk u tk Jk u tk , k 1, 2, . . . , m,
u0 −uT ,
3.5
u 0 −u T ,
which is equivalent to 1.1. Consequently, all that we need to do is to verify that A has at
least one fixed point. With this in mind, we assume ut is a solution of
u λAu,
λ ∈ 0, 1.
3.6
Y. Xing and V. Romanovski
11
That is,
ut T
Gt, sλ fs, us, u s − pu s − qus ds
0
m
m
λ H t, ti Ii u ti λ G t, ti Ji u ti .
i1
3.7
i1
It is equivalent to say that ut satisfies
u t − pu t − qut λ ft, ut, u t − pu t − qut ,
u tk u tk λIk u tk ,
t
/ tk , k 1, 2, . . . , m,
u tk u tk λJk u tk ,
u0 −uT ,
k 1, 2, . . . , m,
3.8
u 0 −u T .
Firstly, we see that for λ ∈ 0, 1,
λ
ft, ut, u t − pu t − qut
≤ λ γ ut u t, ft, ut, u t u t
2 M
γ ut u t, λft, ut, u t λ
u t
2 λM
γ ut u t, u t − 1 − λpu t qut λ
u t
2 λM
γ ut u t, u t − 1 − λp q ut, u t
3.9
− q1 − λ
ut
2 − p1 − λ
u t
2 λ
u t
2 λM
≤ γ ut u t, u t − 1 − λp q ut, u t u t
2 M
γ ut u t, u t u t − 1 − λp q ut, u t − ut, u t M.
Further more, by the antiperiodic boundary condition we have
T
1
ut, u tdt 2
0
T
T
1
d
ut
2 uT 2 − u0
2 0,
2
0 dt
1
ut u t, u t u tdt uT u T 2 − u0 u 0
2 0.
2
0
3.10
12
Boundary Value Problems
As a result,
T
λft, ut, u t − pu t − qutdt ≤ MT.
3.11
0
Now we show that any potential solution of 3.6 is bounded a priori. By 3.2 and 3.11, we
obtain
ut
λ
Aut
T
Gt, sλ ft, ut, u t − pu t − qut dt
0
m
m
λ H t, ti Ii u ti λ G t, ti Ji u ti i1
i1
≤ G0
T
m m λft, ut, u t − pu t − qutdt λαG0 Ii u ti λG0 Ji u ti 0
i1
m
m
MT α Lk Nk
≤ G0
k1
m
m
MT Lk Nk
k1
G0
m m
α ζ i u t i δi u t i i1
k1
≤ G1
i1
m m
δi u t i
.
G1
ζi u ti
i1
k1
i1
i1
3.12
Taking the supremum and rearranging, we get by 3.3 that
m
G1 T M m
k1 Lk k1 Nk
sup ut
≤
.
m
1 − G1 m
t∈0,T k1 δk k1 ζk
3.13
Differentiating both sides of 3.7 and noting 2.23, we obtain
m
G2 T M m
k1 Lk k1 Nk
sup u t
≤
,
m
1 − G2 m
t∈0,T k1 δk k1 ζk
3.14
G2 max αG0 , α2 G0 .
3.15
where
Y. Xing and V. Romanovski
13
Thus,
ut
P C1 max
!
m
m
G1 T M m
G2 T M m
k1 Lk k1 Nk
k1 Lk k1 Nk
: R.
,
m
m
1 − G1 m
1 − G2 m
k1 δk k1 ζk
k1 δk k1 ζk
3.16
Now we have shown that any possible solution of 3.6 is bounded by R which is independent
of λ. By Scheafer’s fixed theorem we know that A has at least one fixed point. Therefore, the
proof is completed.
Suppose both p 0 and q M2 in Theorem 3.1. We obtain the following theorem.
Theorem 3.2. Assume that f : 0, T × Rn × Rn and I, J : Rn → Rn are continuous. If for some
M > 0 there exist nonnegative constants γ, δk , ζk , Lk , N k , and M∗ such that
ft, x, y − M2 x ≤ γ x, ft, x, y y
2 M∗ ,
∀t, x, y ∈
m
0, T \ ti
× Rn × Rn ,
i0
Ik x ≤ δk x
Lk ,
Jk x ≤ ζk x
Nk ,
m
δk k1
∀x ∈ Rn ,
3.17
m
1
ζk < ,
Ȟ
k1
where · is the Euclidean inner product, Ȟ max{eMT /M1 eMT , eMT /1 eMT , MeMT /
M1 eMT }, then 1.1 has at least one solution.
Proof. Consider the mapping
A : P C1 0, T , Rn −→ P C1 0, T , Rn ,
Aut T
0
Gt, shsds m
m
H t, ti Ii u ti G t, ti Ji u ti ,
i1
3.18
3.19
i1
where
⎧ MT t−s
e−MT t−s
e
⎪
⎪
, 0 ≤ t ≤ s ≤ T,
−
⎪
⎪
MT
1 e−MT
1 ⎨ 1e
Gt, s 2M ⎪
eMt−s
e−Mt−s
⎪
⎪
0 ≤ s < t ≤ T,
⎪
⎩ 1 eMT − 1 e−MT ,
⎧
−MeMT t−s Me−MT t−s
⎪
⎪
−
,
⎪
⎨
1 eMT
1 e−MT
1
Ht, s Gt t, s 2M ⎪
Mt−s
⎪
Me−Mt−s
⎪
⎩ Me
,
1 eMT
1 e−MT
3.20
0 ≤ t ≤ s ≤ T,
3.21
0 ≤ s < t ≤ T.
14
Boundary Value Problems
By Lemma 2.4, A is a compact mapping. Consider the equation
u Au.
3.22
To show that A has at least one fixed point, we apply Schaefer’s theorem by showing that all
potential solutions to
u λAu,
λ ∈ 0, 1,
3.23
are bounded a priori, with the bound being independent of λ. With this in mind, let ut be a
solution of 3.23. Note that ut is also a solution to
u t − M2 ut λ ft, ut, u t − M2 ut ,
u tk u tk λIk u tk ,
t
/ tk , k 1, 2, . . . , m,
u tk u tk λJk u tk ,
u0 −uT ,
k 1, 2, . . . , m,
3.24
u 0 −u T .
On one hand, we see that for λ ∈ 0, 1,
λft, ut, u t ≤ λ γ ut, ft, ut, u t u t
2 M∗
γ ut, λft, ut, u t λ
u t
2 λM
γ ut, u t − 1 − λM2 ut λ u t, u t λM∗
γ ut, u t − 1 − λM2 ut
2 λu t, u t λM∗
3.25
≤ γ ut, u u t, u t M∗
γ
d
ut, u t M∗ .
dt
On the other hand, by the antiperiodic boundary condition we have
T
0
ut, u u t, u t dt uT , u T −u0, u 0 −u0, −u 0−u0, u 0 0.
3.26
Y. Xing and V. Romanovski
15
It therefore follows that
T
λft, ut, u tdt ≤ M∗ T.
3.27
0
Consequently,
ut
λ
Aut
m
m
T
Gt, sλ ft, ut, u t dt λ H t, ti Ii u ti λ
t, ti Ji u ti 0
i1
i1
≤ Ǧ0
T
m m λft, ut, u t − pu t − qutdt λMǦ0 Ii u ti λǦ0 u ti 0
i1
≤ Ǧ1
m
m
M T Lk N k
∗
k1
i1
m m
Ǧ1
ζ i u t i δi u t i ,
k1
i1
i1
3.28
where Ǧ0 eMT /M1 eMT , Ǧ1 max{eMT /M1 eMT , eMT /1 eMT }.
We compute directly to get
m
Ǧ1 T M∗ m
k1 Lk k1 Nk
sup ut
≤
.
m
1 − Ǧ1 m
t∈0,T k1 δk k1 ζk
3.29
Differentiating both sides of 3.19, we obtain
m
Ǧ2 T ∗ M m
k1 Lk k1 Nk
sup u t
≤
,
m
1 − Ǧ2 m
t∈0,T k1 δk k1 ζk
3.30
where
Ǧ2 max
!
eMT
MeMT
,
.
1 eMT 1 eMT
3.31
Thus,
ut
P C1 max
!
m
m
Ǧ1 T M∗ m
Ǧ2 T M∗ m
k1 Lk k1 Nk
k1 Lk k1 Nk
: Ř.
,
m
m
1 − Ǧ1 m
1 − Ǧ2 m
k1 δk k1 ζk
k1 δk k1 ζk
3.32
Then the proof is completed.
16
Boundary Value Problems
Similarly, we can prove the following existence result for M 0 in Theorem 3.2.
Theorem 3.3. Suppose that f : 0, T × Rn × Rn and I, J : Rn → Rn are continuous. If there exist
nonnegative constants γ, δk , ζk , Lk , Nk , and M such that
ft, x, y
≤ γ x, ft, x, y y
2 M,
Ik x ≤ δk x
Lk ,
t, x, y ∈
m 0, T \ ti × Rn × Rn ,
Jk x ≤ ζk x
Nk ,
i0
∀x ∈ Rn ,
3.33
m
m
1
δk ζ k < ∗ ,
G0
k1
k1
where · is the Euclidean inner product, G∗0 max{T/4, 1/2}, then 1.1 has at least one solution.
4. Examples
In this part, we show how our main theorems work by a couple of examples.
Example 4.1. The scalar second-order impulsive equations with antiperiodic boundary value
condition
u ut u t3 2ut u t t,
1 u t1 u t1 u t1 1,
10
u0 −u1,
t ∈ 0, 1, t /
t1 ,
1 u t1 u t1 − u t1 2,
10
4.1
u 0 −u 1,
where t1 ∈ 0, 1, have at least one solution.
Proof. Let T 1 and ft, x, y x y3 2x y t in Theorem 3.1. For p 1, q 2, we have
α 2, β −1, and
|ft, x, y − y − 2x| |x y|3 1,
∀t, x, y ∈ 0, 1 × R2 .
4.2
On the other hand, for t, x, y ∈ 0, 1 × R2 ,
x y, ft, x, y y2 x y4 2x2 3xy 2y2 x yt
≥ x y4 − |x y| 4|x| · |y| 3xy
≥ x y4 − |x y|.
4.3
Y. Xing and V. Romanovski
17
Noting minv≥0 {v4 − v3 − v} > −2, we have for γ 1 and M 3 that
γ x y, ft, x, y y2 M ≥ |ft, x, y − y − 2x|,
∀t, x, y ∈ 0, 1 × R2 .
4.4
Moreover, H α2 G0 4/3e2 /1 e2 e−1 /1 e−1 ≈ 1.53298, δ1 ζ1 0.2 < 1/H. Then
the conclusion follows from Theorem 3.1.
Example 4.2. Consider antiperiodic value problem
u t ut utu t2 cos t,
1 u t1 u t1 u t1 4,
4
u0 −u1,
t ∈ 0, 1, t /
t1 ,
1 u t1 u t1 − u t1 ,
2
4.5
u 0 −u 1.
We claim that 4.5 has at least one solution.
Proof. Let T 1 and ft, x, y x xy2 cos t in Theorem 3.2. Choosing M 1, we have for
t, x, y ∈ 0, 1 × R2 that
|ft, x, y − x| |x|y2 cos t,
x, ft, x, y y2 x2 x2 y2 y2 xcos t ≥ x2 x2 y2 y2 − |x|.
4.6
Since minv≥0 {v2 − v} > −1, we have x2 y2 y2 − |x|y2 y2 x2 − |x| 1 > 0. Thus, for
γ 1 and M∗ 2,
γ x, ft, x, y y2 M∗ ≥ |ft, x, y − x|,
∀t, x, y ∈ 0, 1 × R2 .
4.7
Moreover, Ȟ e/1 e, δ1 ζ1 3/4 < 1/Ȟ. Then the conclusion follows from Theorem 3.2.
Acknowledgments
This research is supported by Ad Futura Scientific and Educational Foundation of the
Republic of Slovenia, the Ministry of Higher Education, Science and Technology of the
Republic of Slovenia; the Nova Kreditna Banka Maribor; TELEKOM Slovenije; National
Natural Science Foundation of China 10671127; National Natural Science Foundation
of Shanghai 08ZR1416000; and Foundation of Science and Technology Commission of
Shanghai Municipality 06XD14034.
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