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Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2010, Article ID 908960, 11 pages
doi:10.1155/2010/908960
Research Article
Integral BVPs for a Class of First-Order Impulsive
Functional Differential Equations
Xiaofei He,1 Jingli Xie,1 Guoping Chen,1 and Jianhua Shen2
1
2
Department of Mathematics, College of Zhangjiajie, Jishou University, Jishou, Hunan 416000, China
Department of Mathematics, Hangzhou Normal University, Hangzhou, Zhejiang 310036, China
Correspondence should be addressed to Jianhua Shen, jhshen2ca@yahoo.com
Received 11 November 2009; Accepted 25 January 2010
Academic Editor: Fengde Chen
Copyright q 2010 Xiaofei He 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.
The methods of lower and upper solutions and monotone iterative technique are employed to
the study of integral boundary value problems for a class of first-order impulsive functional
differential equations. Sufficient conditions are obtained for the existence of extreme solutions.
1. Introduction and Preliminaries
In this paper, we study the following integral boundary value problems BVPs for short of
the impulsive functional differential equation
x t btxt ft, xt, Kxt,
t/
tk , t ∈ J 0, T ,
Δxtk Ik xtk , k 1, 2, . . . , m,
T
x0 μ xsds xT , μ ≤ 0,
1.1
0
where f ∈ CJ × R2 , R, Ik ∈ CR, R, 1 ≤ k ≤ m, bt ∈ CR, bt ≤ 0, J 0, T , 0 t0 < t1 < t2 < · · · < tm < tm1 T . K : P CJ → P CJ, where P CJ {u : J → R, u is
−
−
continuous for t ∈ J, t / tk , uti , uti exist, and uti uti , i 1, 2, . . . , m}. Furthermore,
we will assume that K is continuous and monotone nondecreasing, and for any bounded
set A ⊆ P CJ, KA is bounded. Δxtk xtk − xt−k denotes the jump of xt at t tk ;
xtk and xt−k represent the right and left limits of xt at t tk , respectively. Denote J J \ {t1 , t2 , . . . , tm }.
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tk }. P CJ
Let P C1 J {u ∈ P CJ : u be continuously differentiable for t ∈ J, t /
and P C1 J are Banach spaces with the norms
uP C1 J max uP CJ , u P CJ .
uP CJ sup{|ut| : t ∈ J},
1.2
By a solution of 1.1 we mean a u ∈ P C1 J for which problem 1.1 is satisfied.
Note that 1.1 has a very general form, as special instances resulting from 1.1,
one can have impulsive differential equations with deviating arguments and impulsive
differential equations with the Volterra or Fredholm operators. When μ 0, Ik ≡ 0, 1.1
reduces to
x t btxt ft, xt, Kxt,
t ∈ J 0, T ,
x0 xT .
1.3
In 1, Cao and Li. studied and understood existence and stability of solution of this equation
by using fixed theorem and monotone iteration techniques.
When μ 0, bt ≡ 0, 1.1 reduces to
x t ft, xt, Kxt,
Δxtk Ik xtk ,
t/
tk , t ∈ J 0, T ,
k 1, 2, . . . , m,
1.4
x0 xT .
In 2, Li discussed and built the existence theorem of solutions of this equation by using
fixed theorem, upper and lower solutions methods and monotone iterative techniques.
When μ 0, bt ≡ 0, Kxt xt, the equation 1.1 reduces to the periodic
boundary value problem of the impulsive differential equation
x t ft, xt,
t
/ tk , t ∈ J 0, T ,
Δxtk Ik xtk ,
k 1, 2, . . . , m,
1.5
x0 xT .
There are plenty of results on studying the periodic boundary value problem of impulsive
differential equations see 3–8. According to author’s know, there are no dependent
references for studying the 1.1 yet. To fill in this void, we try to find the conditions on f
and Ik , so that make sure that the 1.1 exists extremal solution.
It is well known that the monotone iterative technique offers an approach for
obtaining approximate solutions of nonlinear differential equations, for details, see 4 and
the references therein. There also exist several works devoted to the applications of this
technique to boundary value problems of impulsive differential equations, see, for example,
1–3, 5–14. In this paper, we consider 1.1 by using the method of upper and lower
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3
solutions combined with monotone iterative technique. This technique plays an important
role in constructing monotone sequences which converge to the solutions of our problems. In
presence of a lower solution α and an upper solution β with α ≤ β, we show under suitable
conditions the sequences converge to the solutions of 1.1 by using the method of upper and
lower solutions and monotone iterative technique.
Definition 1.1. The functions α, β ∈ P C1 J are called lower solution and upper solution of
1.1, respectively, if
α t btαt ≤ ft, αt, Kαt,
t
/ tk , t ∈ J,
Δαtk ≤ Ik αtk , k 1, 2, . . . , m,
T
α0 μ αsds ≤ αT , μ ≤ 0.
0
β t btβt ≥ f t, βt, Kβ t , t / tk , t ∈ J,
Δβtk ≥ Ik βtk , k 1, 2, . . . , m,
T
β0 μ βsds ≥ βT , μ ≤ 0.
1.6
0
In what follows we define the set
α, β w ∈ P CJ, R : αt ≤ wt ≤ βt, t ∈ J
1.7
for α, β ∈ P CJ, R and α ≤ β.
We list the following conditions.
H1 αt, βt are lower and upper solutions of 1.1 such that αt ≤ βt.
H2 There exists M ≥ 0 such that
f t, x, y − f t, x, y ≥ −Mx − x,
1.8
for αt ≤ x ≤ x ≤ βt, T αt ≤ y ≤ y ≤ T βt, t ∈ J.
H3 There exist 0 ≤ Lk < 1, k 1, 2, . . . , m such that
Ik x − Ik y ≥ −Lk x − y ,
for αt ≤ y ≤ x ≤ βt, t ∈ J.
k 1, 2, . . . , m,
1.9
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2. Main Results
To obtain our main results, we need the following lemmas.
Lemma 2.1 see 9. Suppose that the following conditions are satisfied.
A0 Sequence {tk } satisfies 0 ≤ t0 < t1 < t2 · · · , and limn → ∞ tn ∞.
A1 m ∈ P C1 J, R and mt is left continuous at tk , k 1, 2, . . ..
A2 For k 1, 2, . . . , t ≥ t0 ,
m t ≤ ptmt qt, t /
tk , t ∈ J,
m tk ≤ dk mtk bk , k 1, 2, . . . , m,
2.1
where q, p ∈ CR , R, bk , dk ≥ 0 are constants, then
mt ≤ mt0 t
dk exp
psds
t0 <tk ≤t
⎛
⎝
t0 <tk ≤t
t0
t
dj exp
tk <tj ≤t
t ⎞
psds ⎠bk
2.2
tk
t
dk exp
pσdσ qsds.
t0 s<tk <t
s
Lemma 2.2 see 12. If m ∈ P C1 J and
m t ≤ −Mmt,
t/
tk , t ∈ J 0, T ,
Δmtk ≤ −Lk mtk ,
k 1, 2, . . . , m,
2.3
m0 ≤ mT ,
where M > 0, 0 < Lk ≤ 1, then mt ≤ 0, t ∈ J.
Lemma 2.3. If x ∈ P CJ, M > 0, 0 < Lk ≤ 1, k 1, 2, . . . , m, and 1/1 − e−MT then the equation
x t Mxt σt,
k0
Lk < 1,
t/
tk , t ∈ J 0, T ,
Δxtk −Lk xtk dk ,
x0 d xT ,
has one unique solution.
m
k 1, 2, . . . , m,
d∈R
2.4
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5
Proof. Firstly, we prove that 2.4 is equivalent to the integral equation
xt −
e−Mt
d
1 − e−MT
T
Gt, sσsds 0
m
Gt, tk −Lk xtk dk ,
2.5
k0
where
⎧ −Mt−s
e
⎪
⎪
⎪
,
⎨
1 − e−MT
Gt, s −MT t−s
e
⎪
⎪
⎪
,
⎩
1 − e−MT
0 ≤ s < t ≤ T,
2.6
0 ≤ t ≤ s ≤ T.
If xt ∈ P C1 J is solution of 2.4, then, by directly integrating we obtain
e−Mt
d
xt −
1 − e−MT
T
Gt, sσsds 0
m
Gt, tk −Lk xtk dk .
2.7
k0
If xt ∈ P C1 J is solution of the above-mentioned integral equation, then
e−Mt
x t −M −
d
1 − e−MT
−Mxt σt,
T
Gt, sσsds 0
m
Gt, tk −Lk xtk dk σt
k0
t/
tk ,
Δxtk −Lk xtk dk , k 1, 2, . . . , m,
T −MT −s
m −MT −tk e
e
1
d
σsds
x0 −
−Lk xtk dk ,
−MT
−MT
1−e
1 − e−MT
01−e
k0
xT −
e−MT
d
1 − e−MT
2.8
T
m −MT −tk e−MT −s
e
σsds
−Lk xtk dk .
−MT
1 − e−MT
01−e
k0
This yields x0 d xT . So 2.4 is equivalent to the integral equation
xt −
e−Mt
d
1 − e−MT
T
Gt, sσsds 0
m
Gt, tk −Lk xtk dk .
2.9
k0
Now, we define operator A : P CJ → P CJ as
e−Mt
d
Axt −
1 − e−MT
T
0
Gt, sσsds m
k0
Gt, tk −Lk xtk dk .
2.10
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International Journal of Differential Equations
For each x, y ∈ P CJ,
Axt − Ay t ≤
m
m
1
1
≤
x
−
y
L
Lk xtk − ytk ,
k
−MT
−MT
1−e
1−e
k0
k0
2.11
and so
Axt − Ay t ≤
m
1
Lk xtk − ytk .
−MT
1−e
k0
2.12
This indicates that A : P CJ → P CJ is a contraction mapping. Then there is one
unique x ∈ P CJ such that Ax x, that is, 2.4 has an unique solution xt. The proof is
complete.
Theorem 2.4. If the conditions H1 , H2 , H3 are all satisfied, and, in addition, if there exist
M > 0, 0 < Lk ≤ 1, k 1, 2, . . . , m, such that 1/1 − e−MT m
k0 Lk < 1, then the
impulsive equation 1.1 has minimal and maximal solutions ρt, rt ∈ P C1 J in α, β, and
there are monotone sequences {αn }, {βn } convergeing uniformly to ρt, rt in J, respectively, where
α0 α, β0 β, and αn t, βn t are lower and upper solutions of 1.1, respectively.
Proof. For each ψ ∈ α, β, we consider the equation
x t f t, ψt, Kψ t − btψt − M xt − ψt , t /
tk , t ∈ J,
Δxtk Ik ψtk − Lk xtk − ψtk , k 1, 2, . . . , m,
T
x0 μ ψsds xT , μ ≤ 0.
2.13
0
By Lemma 2.3, we know that 2.13 has a unique solution xt ∈ P C1 J. Now, we define
operator A : P C1 J → P C1 J as Aψ x.
We will prove that {αn }, {βn } have the following properties.
a α0 ≤ Aα0 , Aβ0 ≤ β0 .
b A is monotone nondecreasing on α0 , β0 .
Proofs of properties a, b are divided into three steps to proceed.
Step 1. Suppose that p α0 − α1 , then
p α0 − α1 ≤ −btαt ft, αt, Kαt − ft, αt, Kαtt
btαt Mα1 t − αt −Mpt,
t
/ tk ,
Δptk Δα0 − Δα1 ≤ Ik αtk − Ik αtk Lk α1 tk − αtk −Lk ptk ,
p0 α0 0 − α1 0 ≤ pT .
By Lemma 2.2, we obtain pt ≤ 0, t ∈ J, so α0 t ≤ α1 t.
t tk ,
2.14
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Step 2. Suppose that p β1 − β0 , then
p β1 − β0 ≤ btβt − f t, βt, Kβ t f t, βt, Kβt t
− btβt − M β1 t − βt −Mpt, t tk ,
Δptk Δβ1 − Δβ0 ≤ −Ik βtk Ik βtk − Lk β1 tk − βtk −Lk ptk ,
t tk ,
2.15
p0 β1 0 − β0 0 ≤ pT .
By Lemma 2.2, we obtain pt ≤ 0, t ∈ J, so β1 t ≤ β0 t.
Similary we can show that α1 t ≤ β1 t, hence α0 t ≤ α1 t ≤ β1 t ≤ β0 t.
Step 3. If n m, αm−1 ≤ αm ≤ βm ≤ βm−1 , then when n m 1, let p αm − αm1 . Then
p αm − αm1 ≤ −btαm−1 t ft, αm−1 t, Kαm−1 t
− Mαm t − αm−1 t − ft, αm t, Kαm tt
btαm t Mαm1 t − αm t
−btαm−1 − αm − Mpt,
2.16
t
/ tk .
Furthermore, bt ≤ 0, αm−1 − αm ≤ 0, thus p α m − α m1 ≤ −Mpt,
Δptk Δαm − Δαm1 Ik αm−1 tk − Lk αm tk − αm−1 tk − Ik αm tk Lk αm1 tk − αm tk ≤ −Lk ptk , t tk ,
T
p0 αm 0 − αm1 0 μ αm s − αm−1 sds pT ≤ pT .
2.17
0
By Lemma 2.2, we obtain pt ≤ 0, t ∈ J, so αm t ≤ αm1 t.
tk ,
Similarly, we can assume that p βm1 − βm . When t /
− βm
≤ −btβm t f t, βm t, Kβm t − M βm1 t − βm t
p βm1
− f t, βm−1 t, Kβm−1 t t btβm−1 t M βm t − βm−1 t
−bt βm − βm−1 − Mpt.
2.18
Furthermore, bt ≤ 0, βm − βm−1 ≤ 0, thus p βm1
− βm
≤ −Mpt, when t tk ,
Δptk Δβm1 − Δβm Ik βm tk − Lk βm1 tk − βm tk − Ik βm−1 tk Lk βm tk − βm−1 tk ≤ −Lk ptk ,
T
βm−1 s − βm s ds pT ≤ pT ,
p0 βm1 0 − βm 0 μ
0
hence by Lemma 2.2, we obtain pt ≤ 0, t ∈ J, so βm1 t ≤ βm t.
2.19
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International Journal of Differential Equations
In the same way we can prove that αm1 t ≤ βm1 t.
Thus by mathematical induction we can know that
αn−1 ≤ αn ≤ βn ≤ βn−1 ,
n 0, 1, 2, . . . , t ∈ J.
2.20
So far, we finish the proof of the properties a, b.
Now we prove that αn , βn , n 0, 1, 2, . . . , are lower and upper solutions of 1.1.
Similarly, we can use mathematical induction to prove this.
When n 0, α0 , β0 are already lower and upper solutions of 1.1.
When n 1,
α1 t ft, αt, Kαt − btαt − Mα1 t − αt
− ft, α1 t, Kα1 t ft, α1 t, Kα1 t
≤ ft, α1 t, Kα1 t − btα1 t,
t
/ tk , t ∈ J,
Δα1 tk Ik αtk − Lk α1 tk − αtk Ik α1 tk − Ik α1 tk ≤ Ik α1 tk , k 1, 2, . . . , m,
T
T
α1 0 μ α1 sds ≤ α1 0 μ αsds α1 T ,
0
2.21
μ ≤ 0.
0
Thus α1 is lower solution of 1.1.
Suppose that αn is lower solution of 1.1 when n m.
Then when n m 1,
αm1 t ft, αm t, Kαm t − btαm t − Mαm1 t − αm t
− ft, αm1 t, Kαm1 t ft, αm1 t, Kαm1 t
≤ ft, αm1 t, Kαm1 t − btαm1 t,
t/
tk , t ∈ J,
Δαm1 tk Ik αm tk − Lk αm1 tk − αm tk Ik αm1 tk 2.22
− Ik αm1 tk ≤ Ik αm1 tk , k 1, 2, . . . , m,
T
T
αm1 0 μ αm1 sds ≤ αm1 0 μ αm sds αm1 T .
0
0
Thus by mathematical induction we can know that αn is lower solution of 1.1. In the
same way we can prove that βn is upper solution of 1.1.
By αn−1 ≤ αn ≤ βn ≤ βn−1 , n 0, 1, 2, . . . , t ∈ J, we can know that when
n → ∞, {αn },{βn } have limits ρt, rt, respectively. Since they are independent of t when
n → ∞{αn }, {βn } converge uniformly to ρt, rt and αn ≤ ρt ≤ rt ≤ βn ≤ βn−1 , n 0, 1, 2, . . . , t ∈ J.
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According to αn ,βn satisfying 2.13, that is,
αn t ft, αn−1 t, Kαn−1 t − btαn−1 t − Mαn t − αn−1 t,
t/
tk , t ∈ J,
Δαn tk Ik αn−1 tk − Lk αn tk − αn−1 tk , k 1, 2, . . . , m,
T
αn 0 μ αn−1 sds αn T , μ ≤ 0,
0
βn t f t, βn−1 t, Kβn−1 t − btβn−1 t − M βn t − βn−1 t , t /
tk , t ∈ J,
Δβn tk Ik βn−1 tk − Lk βn tk − βn−1 tk , k 1, 2, . . . , m,
T
βn 0 μ βn−1 sds βn T , μ ≤ 0,
2.23
0
when n → ∞, we have
ρ t f t, ρt, Kρ t − btρt, t / tk , t ∈ J,
Δρtk Ik ρtk , k 1, 2, . . . , m,
T
ρ0 μ ρsds ρT , μ ≤ 0.
0
r t ft, rt, Krt − btrt,
t
/ tk , t ∈ J,
2.24
Δrtk Ik rtk , k 1, 2, . . . , m,
T
r0 μ rsds rT , μ ≤ 0.
0
Equation 2.24 indicates that ρt, rt are solutions of 1.1.
Lastly, we prove that ρt, rt are minimal and maximal solutions of the equation 1.1
in α, β.
Suppose that xt is a solution of the equation and satisfies xt ∈ α, β, t ∈ J,
obviously, we can assume that there is an n such that αn ≤ x ≤ βn .
If pt αn1 − x, then
p αn1 − x ≤ −btαn t ft, αn t, Kαn t
− Mαn1 t − αn t − ft, xt, Kxtt btxt
≤ −btαn − x − Mpt, t /
tk .
2.25
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International Journal of Differential Equations
And since bt ≤ 0, αm − x ≤ 0, p ≤ −Mpt,
Δptk Δαn1 − Δx Ik αn tk − Lk αn1 tk − αn tk − Ik xtk ≤ −Lk ptk ,
T
p0 αn1 0 − x0 μ xs − αn sds pT ≤ pT .
0
t tk ,
2.26
Hence by Lemma 2.2, we can obtain pt ≤ 0, t ∈ J, so αn1 t ≤ xt. Similarly, we can obtain:
xt ≤ βn1 t, t ∈ J. This indicates that αn t ≤ xt ≤ βn1 t, t ∈ J, n 0, 1, 2, . . . . Hence
when n → ∞, we can obtain that ρt ≤ xt ≤ rt, t ∈ J. This ends the proof.
Finally, we give an example to illustrate the efficiency of our results.
Example 2.5. Consider the problem of
x t − xt sin t −xt t t
xsds,
0
0 < t < 1, t /
t1 ,
1
1
Δxt1 − xt1 , t1 ,
8
2
1
x0 − xsds x1,
2.27
0
t
where bt sin t, ft, xt, Kxt −xt t 0 xsds, I1 x −x. Obviously, αt 0, βt 1 − t are the lower solution and upper solution for 2.27 with αt ≤ βt,
t
respectively. ft, x, Kx − ft, y, Ky −x − y − 0 xs − ysds, I1 x − I1 y −x − y.
Let T 1, Lk 1/8, the conditions of Theorem 2.4 are all satisfied, so problem 2.27 has the
maximal and minimal solutions in the segement αt, βt.
Acknowledgments
This work was supported by the NNSF of China 10571050; 10871062, the Hunan Provincial
Natural Science Foundation of China 09JJ6010, and the Zhejiang Provincial Natural Science
Foundation of China Y6090057.
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