Volume 2010, Article ID 598405,11pages doi:10.1155/2010/598405
Research Article
Periodic Boundary Value Problems for
Second-Order Functional Differential Equations
Xiangling Fu and Weibing Wang
Department of Mathematics, Hunan University of Science and Technology, Xiangtan, Hunan 411201, China
Correspondence should be addressed to Xiangling Fu,[email protected] Received 11 June 2009; Accepted 28 January 2010
Academic Editor: Marta Garc´ıa-Huidobro
Copyrightq2010 X. Fu and W. Wang. 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.
This note considers a periodic boundary value problem for a second-order functional differential equation. We extend the concept of lower and upper solutions and obtain the existence of extreme solutions by using upper and lower solution method.
1. Introduction
Upper and lower solution method plays an important role in studying boundary value problems for nonlinear differential equations; see1 and the references therein. Recently, many authors are devoted to extend its applications to boundary value problems of functional differential equations2–5. Supposeαis one upper solution or lower solution of periodic boundary value problems for second-order differential equation; the condition
α0 αTis required. A neutral problem is that whether we can define upper and lower solution without assuming α0 αT. The aim of the present paper is to discuss the following second order functional differential equation
−ut ft, ut, uθt, t∈J,
u0 uT, u0 uT, 1.1
whereJ 0, T, f∈CJ×R2, R,0≤θt≤t, t∈J.
Through this paper, we assume thatM >0, N≥0,and
T2MN≤2, 1.2
Λ1
u∈C2J:ut≥0,1u0 uT, u0≥uT,
Λ2
u∈C2J:ut≥0,1uT u0, u0≤uT.
1.3
Definition 1.1. Functionsα∈C2Jandβ∈C2Jare called lower solution and upper solution
of the boundary value problem1.1, respectively if
−αt≤ft, αt, αθt−Hαt, t∈J,
α0≥αT, 1.4
−βt≥ft, βt, βθthβt, t∈J,
β0≤βT, 1.5
where
Hαt
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
0, α0 αT,
−ct Mct Ncθtα0−αT, α0> αT,
−bt Mbt NbθtαT−α0, α0< αT,
1.6
hβt
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
0, β0 βT,
−ct Mct Ncθtβ0−βT, β0< βT,
−bt Mbt NbθtβT−β0, β0> βT
1.7
andc∈Λ1, b∈Λ2.
Remark 1.2. Clearly,Λ1/∅andΛ2∅. For example,
t
T ∈Λ1, sin πt
2T ∈Λ1, 1− t
T ∈Λ2, sin π
2
1− t
T
2. Comparison Results
We now present the main results of this section.
Lemma 2.1. Assume thatu∈C2Jsatisfies
−ut Mut Nuθt≤0, t∈J,
u0 uT, u0≥uT, 2.1
thenut≤0for allt∈J.
Proof. Suppose, to the contrary, thatut >0 for somet ∈J. We consider the following two cases.
Case 1. ut≥0, ut/0 onJ. It is easy to obtain thatut≥0 onJ. Thusut≡constant K >0 fromu0≥uT. Consequently, we obtain that
MNK−ut Mut Nuθt≤0, 2.2
which contradictsK >0.
Case 2. There existt1, t2∈Jsuch thatut1>0 andut2<0. Hence, two cases are possible.
Subcase 1. u0 uT<0. There exists at3∈0, Tsuch that
ut3 max
t∈J ut>0, u
t
3 0. 2.3
Letut∗ mint∈0,t3ut<0.Then
ut≥MNut∗, t∈0, t3. 2.4
Integrating the above inequality fromst∗≤s≤t3tot3, we obtain
−us≥t3−sMNut∗, t∗≤s≤t3, 2.5
and then integrate fromt∗tot3to obtain
−ut∗< ut3−ut∗≤
t3
t∗
s−t3MNut∗ds
≤ −MN
2 ut∗t3−t∗
2≤ −MN
2 ut∗T
2
2.6
Subcase 2. u0 uT≥0. There exists at3∈Jsuch that
ut3 max
t∈J ut>0. 2.7
Ift3∈0, T, thenut3 0.Ift30 ort3T, thenu0≤0≤uT. Sou0 uT 0.
Letut∗ mint∈0,Tut<0.Then
ut≥MNut∗, t∈J. 2.8
Whent∗< t3, same as Subcase 1, we obtain that 1< T2MN/2.
whent∗> t3, integrating the inequality2.8fromt3tost3≤s≤t∗, we obtain
us≥s−t3MNut∗, 2.9
and then integrate fromt3tot∗to obtain
ut∗> ut∗−ut3≥
t∗
t3
s−t3MNut∗ds≥ M2Nut∗t∗−t32≥ MN
2 ut∗T
2
2.10
that implies 1<T2/2MN.This is a contradiction. The proof is complete.
Lemma 2.2. Assume thatu∈C2Jsatisfies
−ut Mut Nuθt −ct Mct Ncθtu0−uT≤0, t∈J, u0> uT, u0≥uT.
2.11
Thenut≤0for allt∈J.
Proof. Put
yt ut ctu0−uT, t∈J, 2.12
theny∈Eandut≤ytfor allt∈Jand
We have
−yt Myt Nyθt −ut Mut Nuθt
−ct Mct Ncθtu0−uT≤0,
y0 1c0u0−c0uT yT,
y0 u0 c0u0−uT≥yT.
2.14
Hence by Lemma 2.1,yt ≤ 0 for all t ∈ J, which implies that ut ≤ 0 for t ∈ J. This completes the proof.
Lemma 2.3. Assume thatu∈C2Jsatisfies
−ut Mut Nuθt −bt Mbt NbθtuT−u0≤0, t∈J,
u0< uT, u0≥uT. 2.15
Thenut≤0for allt∈J.
The proof ofLemma 2.3is similar to that ofLemma 2.2, here we omit it.
Lemma 2.4. Assume thatu∈C2Jsatisfies
−ut Mut Nuθt≤0, t∈J,
u0≤0, uT≤0. 2.16
Thenut≤0for allt∈J.
Proof. Suppose thatut > 0 for somet ∈ J. Then from boundary conditions, we have that there exists at∗∈0,1such that
ut∗ max
t∈J ut>0, u
t∗ 0.
2.17
Suppose thatut ≥0 fort∈ J. It is easy to see thatu0 0 andut ≥0 fort ∈J. Fromu0 0 andut≥0 fort∈J, we obtain thatu0≥0. Therefore,ut≥u0≥0. It follows thatuT maxt∈Jut>0, a contradiction.
Suppose that there existt1, t2 ∈Jsuch thatut1 >0 andut2<0. Lett∗ ∈0, t∗be
such thatut∗ mint∈0,t∗ut≤0. From the first inequality of2.16, we have
ut≥MNut∗, t∈0, t∗. 2.18
Integrating the above inequality fromst∗≤s≤t∗tot∗, we obtain
and then integrate fromt∗tot∗to obtain
−ut∗< ut∗−ut∗≤ t∗
t∗
s−t∗MNut∗ds
≤ −MN
2 ut∗t
∗−t
∗2≤ −M2Nut∗T2.
2.20
Hence,ut∗2−MNT2>0,a contradiction. The proof is complete.
3. Linear Problem
In this section, we consider the boundary value problem
−ut Mut Nuθt σt, t∈J,
u0 uT, u0 uT,
3.1
whereσ∈CJ, R.
Theorem 3.1. Assume that there existα, βwhich are lower and upper solutions of 3.1andα≤β
onJ. Then there exists one unique solutionuto problem3.1andα≤u≤βonJ.
Proof. We first show that the solution of3.1is unique. Letu1andu2be solutions of3.1and
setvu1−u2. Thus
−vt Mvt Nvθt 0, t∈J,
v0 vT, v0 vT.
3.2
ByLemma 2.1, we have thatv≤0 fort∈J, that is,u1≤u2onJ. Similarly, one can obtain that
u2≤u1onJ. Henceu1 u2.
Next, we prove that ifuis a solution of3.1, thenα≤u≤β. Letmα−u. Ifα0 αT, thenHαt 0 onJ. So we have
−mt Mmt Nmθt≤0, t∈J,
m0 mT, m0≥mT.
3.3
ByLemma 2.1, we have thatmα−u≤0 onJ.
Ifα0> αT, thenHαt −ct Mct Ncθtα0−αT. Thus
−mt Mmt Nmθt −αt Mαt Nαθt
ut−Mut−Nuθt
≤σt−Hαt−σt −Hαt
−−ct Mct Ncθtm0−mT.
It is easy to see that m0 ≥ mT.By Lemma 2.2, we have that m α−u ≤ 0 on J. Analogously,u≤βonJ.
Ifα0< αT, thenHαt −bt Mbt NbθtαT−α0. Thus
−mt Mmt Nmθt −αt Mαt Nαθt
ut−Mut−Nuθt
≤σt−Hαt−σt −Hαt
−−bt Mbt NbθtmT−m0.
3.5
It is easy to see that m0 ≥ mT.By Lemma 2.2, we have that m α−u ≤ 0 on J. Analogously,u≤βonJ.
Finally, we show that3.1has a solution by several steps.
Step 1. Consider the equation
−ut Mut Nuθt σt, t∈J,
u0 uT λ, 3.6
where M, N and σ are defined in3.1. For any λ ∈ R, we show that3.6 has a unique solutionut, λ.
It is easy to check that3.6is equivalent to the integral equation
ut λ
T
0
Gt, sσs−Mus−Nuθsds, 3.7
where
Gt, s
⎧ ⎪ ⎨ ⎪ ⎩
t
TT−s, 0≤t≤s≤T, s
TT−t, 0≤s≤t≤T.
3.8
Define a mappingΦ:CJ, R → CJ, Rby
Φut λ
T
0
Gt, sσs−Mus−Nuθsds. 3.9
ObviouslyCJ, Ris a Banach space with normu supt∈J|ut|. For anyx, y ∈ CJ, R, we have
Φxt−Φyt
T
0
Noting that maxt∈J
T
0Gt, sds T2/8, condition1.2implies thatΦ :CJ, R → CJ, Ris
a contraction mapping. There exists uniqueu ∈ CJ, Rsuch thatΦu u. Thus3.6has a unique solutionut, λ.Moreover,ut, λ∈C2J, R.
Step 2. We show that for anyt ∈J,ut, λandutt, λare continuous inλ, whereut, λis a
unique solution of the problem3.6. Letut, λi, i1,2,be the solution of
−ut Mut Nuθt σt, t∈J, u0 uT λi, i1,2.
3.11
Then
ut, λi λi
T
0
Gt, sσs−Mus, λi−Nuθs, λids, i1,2. 3.12
From3.12, we have that
ut, λ1−ut, λ2 ≤ |λ1−λ2| MNut, λ1−ut, λ2max
t∈J
T
0
Gt, sds
≤ |λ1−λ2|T 2
8 MNut, λ1−ut, λ2.
3.13
Hence
ut, λ1−ut, λ2 ≤
8
8−T2MN|λ1−λ2|. 3.14
Sinceut, λ∈C2J, R,u
tt, λexists for anyt∈Jandλ∈R. From3.12and3.14, we have
utt, λ1−utt, λ2 ≤ MN
T ut, λ1−ut, λ2maxt∈J
t
0
s ds
T
t
T−sds
≤ MNT
2 ut, λ1−ut, λ2
≤ 4MNT
8−T2MN|λ1−λ2|.
Step 3. We show that there exists oneλsuch thatut0, λ utT, λ, whereut, λis the unique
solution of the problem3.6. Put
αt
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
αt, α0 αT,
αt ctα0−αT, α0> αT,
αt btαT−α0, α0< αT,
3.16
βt
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
βt, β0 βT,
βt−ctβT−β0, β0< βT,
βt−btβ0−βT, β0> βT;
3.17
thenαt≤αt, βt≤βtfor anyt∈Jand
−αt Mαt Nαθt≤δt, t∈J,
α0 αT, α0≥αT,
3.18
−βt Mβt Nβθt≥δt, t∈J,
β0 βT, β0≤βT.
3.19
Using3.18and3.19, one easily obtains thatαt≤βtfor anyt∈J.
Put λ ∈ α0, β0, then α0 αT ≤ u0, λ uT, λ ≤ β0 βT.Using
Lemma 2.3, we easily obtain thatαt≤ut, λ≤βtonJ. Hence
ut0, α0≥α0, utT, αT≤αT, 3.20
ut
0, β0≤β0, ut
T, βT≥βT. 3.21
Define a function
Pλ ut0, λ−utT, λ, 3.22
whereut, λis the unique solution of the problem3.6. SincePis continuous and
Pα0Pβ0≤0, 3.23
there exists oneλ0 ∈α0, β0such thatPλ0 0,that is,ut0, λ0 utT, λ0. Obviously,
4. Main Result
Theorem 4.1. Let the following conditions hold.
H1The functionsα,βare lower and upper solutions of1.1, respectively, andα≤βonJ.
H2The constantsM, Nin definition of upper and lower solutions satisfy
ft, x, y−ft, x, y≥ −Mx−x−Ny−y 4.1
forαt≤x≤x≤βt, αθt≤y≤y≤βθt, t∈J.
Then, there exist monotone sequences {αn}, {βn} with α0 α, β0 β such that
limn→ ∞αnt ρt,limn→ ∞βnt rtuniformly onJ, andρ, rare the minimal and the maximal
solutions of 1.1, respectively, such that
α0≤α1≤α2≤ · · · ≤αn≤ρ≤x≤r≤βn≤ · · · ≤β2≤β1≤β0 4.2
onJ, wherexis any solution of 1.1such thatαt≤xt≤βtonJ.
Proof. Letα, β {u∈C2J: αt≤ ut≤βt, t ∈J}. For anyγ ∈α, β, we consider the
equation
−ut Mut Nuθt ft, γt, γθtMγt Nγθt, t∈J,
u0 uT, u0 uT. 4.3
Theorem 3.1implies that the problem4.3has a unique solutionu ∈ C2J. We define an
operatorAbyuAγ, thenAis an operator fromα, βtoα, β. We shall show that
aα≤Aα, Aβ≤β;
bAis nondecreasing inα, β.
FromAα∈α, βandAβ∈α, β, we have thataholds. To proveb, we show that
Aμ1≤Aμ2ifα≤μ1≤μ2≤β.
Letρ∗1Aμ1, ρ2∗Aμ2,andpρ∗1−ρ∗2; then byH2, we have
−pt Mpt Npθtft, μ1t, μ1θt
Mμ1t Nμ1θt
−ft, μ2t, μ2θt
−Mμ2t−Nμ2θt
≤0,
4.4
andp0 pT, p0 pT.ByLemma 2.1,pt≤0, which impliesAμ1≤Aμ2.
Define the sequence{αn}, {βn}withα0 α, β0 βsuch thatαn1 Aαn, βn1 Aβn
forn0,1,2, . . . .Fromaandb, we have
on t ∈ J. Therefore, there exist ρ, r such that limn→ ∞αnt ρt, limn→ ∞βnt rt
uniformly onJ. Clearly,ρ, rare solutions of1.1.
Finally, we prove that ifx∈α0, β0is one solution of1.1, thenρt≤xt≤rton
J. Sinceα0t≤xt≤β0tandAxx, by propertyb, we obtain thatα1t≤xt≤β1t
fort∈J. Using propertybrepeatedly, we have
αnt≤xt≤βnt 4.6
for alln. Passing to the limit asn → ∞, we obtainρt ≤xt≤ rt, t ∈J.This completes the proof.
Acknowledgment
This work is supported by Scientific Research Fund of Hunan Provincial Education Department09B033and the NNSF of China10871062.
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