R E S E A R C H
Open Access
Third order problems with nonlocal
conditions of integral type
Abdelkader Boucherif
1*, Sidi Mohamed Bouguima
2, Zehour Benbouziane
2and Nawal Al-Malki
3*Correspondence: [email protected] 1Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, P.O. Box 5046, Dhahran, 31261, Saudi Arabia Full list of author information is available at the end of the article
Abstract
We discuss the existence of solutions of nonlinear third order ordinary differential equations with integral boundary conditions. We provide sufficient conditions on the nonlinearity and the functions appearing in the boundary conditions that guarantee the existence of at least one solution to our problem. We rely on the method of lower and upper solutions to generate an iterative technique, which is not necessarily monotone.
MSC: 34B10; 34B11; 34B12; 34B13; 34B14; 34B15
Keywords: third order differential equations; integral boundary conditions;a priori
bound on solutions; fixed point theorems; lower and upper solutions; iterative technique
1 Introduction
The purpose of this paper is to establish the existence of solutions for a class of nonlin-ear third order ordinary differential equations with integral boundary conditions. More specifically, we consider the following problem:
u(t) +ft,u(t),u(t),u(t)= , t∈[, ], ()
u() = , ()
u() –au() =
h
u(s),u(s)ds, ()
u() =
h
u(s),u(s)ds, ()
wheref : [, ]×R→R,h
,h:R→Rare continuous functions, andais a
nonnega-tive real number. Several papers have been devoted to the study of third order differential equations with two-point and three-point boundary conditions. See [–], and [] for ref-erences. Problems with integral boundary conditions have been used in the description of many phenomena in the applied sciences. We refer the interested reader to [] and the references therein. Very few papers have dealt with nonlocal conditions for third order differential equations. We can mention [, ] and []. For higher order differential equa-tions with functional boundary condiequa-tions the interested reader can consult []. In this work we use the method of lower and upper solutions to generate a sequence of modified nonlinear problems, each having a unique solution; in this way, we obtain a sequence of
functions, which is uniformly bounded together with their first and second order deriva-tives. We then extract a subsequence converging uniformly to a solution of our original problem ()-(). Contrary to many works in the literature, we develop an iterative tech-nique, which is not necessary monotone. We should point out that our approach is totally different from that of [].
2 Preliminaries
LetI denote the real interval [, ].C(I) is the Banach space of real-valued continuous functions onI, equipped with the normu:=max{|u(t)|;tI}, foru∈C(I). LetDdenote the set of all real-valued functions which are three times continuously differentiable onI. We define the norm ofu∈Dby
uD=u+u+u+u.
LetD={u∈D;u() = }. Then (D, · D) is a Banach space.
Definition A solution of problem ()-() is a functionu∈Dthat satisfies () for every
t∈Iand the conditions () and ().
Definition Letα,β∈Dsatisfyα(t)≤β(t) for everyt∈I. We denote by [α,β] the
set of allv∈Dsuch thatα(t)≤v(t)≤β(t) for everyt∈I.
It is clear that ifu∈[α,β] andu∈D, thenu∈[α,β].
Definition Letα,β∈Dsatisfyα(t)≤β(t) for everyt∈I. LetS(α,β) denote the set
of all functionsu∈Dsuch thatu∈[α,β] andu∈[α,β].
Remark It is clear thatu∈Dandu∈[α,β] imply thatu∈S(α,β).
Definition Letα,β ∈D satisfy α(t)≤β(t) for everyt∈I. Define the operatorp:
D→[α,β] by
(pu)(t) =maxα(t),minu(t),β(t), ∀t∈I,
and the operatorq:D→[α,β] by
(qv)(t) =maxα(t),minv(t),β(t), ∀t∈I.
Remark The operatorspandqare continuous and bounded.
3 Main results
In this section we state and prove our main results. The first result is of independent in-terest and plays a key role in the proof of our second result.
Theorem Letφ:I×R→Rbe continuous,bounded and satisfy the following condition:
Then for anyδ,ρthe boundary value problem
⎧ ⎪ ⎨ ⎪ ⎩
–u(t) =φ(t,u(t)), t∈I, u() –au() =δ,
u() =ρ
()
has a unique solution u.
Proof Uniqueness. Suppose that problem () has two solutionsxanduinD. Putz=x–u.
Thenz() =x() –u() = . We show thatz() = . Suppose this is not true. Then either z() > orz() < . We consider the casez() > . From the condition att= it follows that <z() =az(). Sinceais nonnegative we havez() > , which implies thatzis increasing to the right oft= . Sincez() = there must existξ∈[, ) such thatz(ξ) =maxt∈Iz(t). Then
<z(ξ), =z(ξ) and z(ξ)≤.
The differential equation in () and (Hφ) imply
≥z(ξ)z(ξ) = –φξ,x(ξ)–φξ,u(ξ)x(ξ) –u(ξ)> .
This is a contradiction. Similarly, if we consider the casez() < we will arrive at a con-tradiction. Hencez() = . Now, we have a functionzcontinuous onIwithz() =z() = . Then there existsτ∈Isuch that
z(τ) =max
t∈I z(t), z
(τ) = and z(τ)≤.
Proceeding as before we show thatz(τ) = . So thatz(t) = for allt∈I. This shows that x(t) =u(t) for allt∈I. Sincex() =u() = it follows thatx(t) =u(t) for allt∈I, which shows the uniqueness of the solution.
Existence. Forλ∈[, ] consider the family of problems
⎧ ⎪ ⎨ ⎪ ⎩
–u(t) =λφ(t,u(t)), t∈I, u() –au() =λδ,
u() =λρ.
()
Forλ= problem () has only the trivial solution. Thus, we consider the caseλ∈(, ]. (i)uis a solution of () if and only if it satisfies, for allt∈I,
u(t) = λt a+
δ+aρ+a
( –s)φs,u(s)ds
+ λt
(a+ )
ρ–δ+
( –s)φs,u(s)ds
–λ
t
(t–s)
φ
Indeed, it is clear that the differential equation in () implies
u(t) =u()t+u()t
–λ
t
(t–s)
φ
s,u(s)ds. ()
Then
u(t) =u() +u()t–λ
t
(t–s)φs,u(s)ds. ()
It follows that
λρ=u() =u() +u() –λ
( –s)φs,u(s)ds.
Butu() =au() +λδ, so that
u() = λ a+
ρ–δ+
( –s)φs,u(s)ds
,
and consequently
u() = λ a+
δ+aρ+a
( –s)φs,u(s)ds
.
Now, substitute the expressions ofu() andu() into () to get ().
(ii) We show that there exists a positive constantL, independent ofλ, such that any
possible solutionuof () satisfies
uD≤L. ()
The boundedness ofφimplies that there existsMφ> such that|φ(t,u(t))| ≤Mφfor all
t∈I, so thatu≤Mφ. Then
u()≤ a+
|δ|+a|ρ|+aMφ
()
and
u()≤ a+
|δ|+|ρ|+Mφ
. ()
Combining relations (), (), and () we see that
u
≤M:=
a+
|δ|+ (a+ )|ρ|+ (a+ )Mφ
. ()
Sinceu(t) =tu(s)dsit follows that
Also,u≤Mφand () imply
u≤M:=
a+
|δ|+|ρ|+(a+ )Mφ
.
LetL=M+ M+Mφ. Then any possible solutionuof () satisfies ().
(iii) Define an operator : D →D by (u)(t) = the right-hand side of (). Let
:={u∈D;uD≤L}. Then it is easily seen that (()) is uniformly bounded and equicontinuous. Ascoli-Arzela theorem implies that the operator is compact. More-over, the set of all solutionsuof the equationu=λuis bounded (see ()). It follows from Schaefer theorem (see []) thatu=uhas at least one solution. Thus, () has at least one solution forλ= , which is, in fact, unique from the previous step. Thus,uis a
solution of (). This completes the proof of the theorem.
Remark We should emphasize that, unlike Theorem in [], our Theorem gives the uniqueness of the solution and this is essentially utilized in the proof of our Theorem below.
For our second main result we introduce the notion of lower and upper solutions of problem (), (), (), ().
Definition (a) We say thatα∈Dis a lower solution of problem (), (), () if
⎧ ⎪ ⎨ ⎪ ⎩
–α(t)≤f(t,α(t),α(t),α(t)) for allt∈I, α() –aα()≤h(α(s),α(s))ds,
α()≤h(α(s),α(s))ds.
(b) We say thatβ∈Dis an upper solution of problem (), (), () if
⎧ ⎪ ⎨ ⎪ ⎩
–β(t)≥f(t,β(t),β(t),β(t)) for allt∈I, β() –aβ()≥h(β(s),β(s))ds,
β()≥h(β(s),β(s))ds.
To state and prove our second main result we introduce the following assumptions.
(Af) f :I×R→Ris continuous and satisfies
() there existsC> such that any solutionuof (), withu∈S(α,β), satisfies |u(t)| ≤C, for allt∈I;
() (f(t,u(t),v,w) –f(t,u(t),v,w))(v–v) < forv,v∈Rsuch thatv≤v,
u∈D∩[α,β],w∈Randt∈I;
() f(t,α(t),v(t),α(t))≤f(t,u(t),v(t),w)≤f(t,β(t),v(t),β(t))for allu∈D∩[α,β], v∈C∩[α,β],w∈R, andt∈I.
(Ah) h,h:R→Rare continuous and nondecreasing with respect to both arguments.
Remark There are several sufficient conditions that imply (Af)(). See for instance [, Lemma ], [, Lemma ].
Theorem Letα,β∈Dbe,respectively,a lower and an upper solution of problem(),
pair(α,β),whereαis a given lower solution andβis a given upper solution.Then problem (), (), ()has at least one solution u∈S(α,β).
Proof We modify problem (), (), (), () as follows. We define two functionsfC,F:I×
R→Rby
fC(t,u,v,w) =
⎧ ⎪ ⎨ ⎪ ⎩
f(t,p(u),v,C), w≥C,
f(t,p(u),v,w), |w| ≤C,
f(t,p(u),v, –C), w≤–C
()
and
F(t,u,v,w) =fC
t,u,q(v),w=
⎧ ⎪ ⎨ ⎪ ⎩
fC(t,u,α,w), v<α,
fC(t,u,v,w), α≤v≤β,
fC(t,u,β,w), v>β,
()
whereCis the constant from condition (Af)(). Consider the modified problem
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
–u(t) =F(t,u(t),u(t),u(t)) fort∈I, u() = ,
u() –au() =h(u(s),u(s))ds,
u() =h(u(s),u(s))ds.
()
Define a sequence (uj)j∈Nof functions inDas follows. Let
u(t) =γt+β(t), t∈I,
whereγ =maxt∈I(α–β)(t), and forj≥,
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
–uj (t) =F(t,uj–(t),uj(t),uj–(t)) fort∈I,
uj() = ,
uj() –auj() =h(uj–(s),uj–(s))ds,
uj() =h(uj–(s),uj–(s))ds.
()
We shall show that the sequence of modified problems () is such that each problem has a unique solution, which is uniformly bounded, together with its first and second order derivatives. Then we rely on Bolzano-Weierstrass theorem to extract a uniformly conver-gent subsequence, whose limit is the solution of our original problem.
. The sequence (uj)j∈N is well defined. Indeed, for anyt∈I and any z∈Rwe have q(z)∈[α,β] andp(uj–(t))∈[α,β]. It follows that the functionφ:I×R→R, defined by
φ(t,z) =Ft,uj–(t),z,uj–(t)
=ft,p(uj–)(t),q(z),uj–(t)
,
is continuous and bounded for allt∈Iandz∈R. Moreover, condition (Af)() shows that φsatisfies condition (Hφ) in Theorem . It follows from this theorem that () has a unique
solutionuj, for eachj= , , . . . .
It is clear thatuj∈Dforj= , , . . . . Sinceα≤βit follows thatγ ≤, so thatu=
γ +β≤β. On the other hand, we haveγ ≥–mint∈I(β–α)(t)≥–β(t) +α(t); so that
u≥α. It follows thatu∈[α,β], and consequentlyu∈S(α,β). Also, sinceu=β
thenu≤ β. Suppose, by induction, that we haveu–∈S(α,β) and there exists
K≤Csuch thatu–≤K. Let
Mf:=maxf(t,u,v,w);t∈I,u∈[α,β],v∈
α,β,|w| ≤C
,
h=max
h
u(s),u(s)–h
u(s),u(s)ds;u∈S(α,β)
.
Claim. There existsKdepending only onK,Mf,h,α, andβsuch thatu≤ Kandu∈S(α,β).
To prove the claim we start with
u(t) =u() –
t
Fs,u–(s),u(s),u–(s)
ds
=u() –
t
fs,pu–(s)
,qu(s)
,u–(s)ds,
which leads to
u(t)≤u()+Mf. ()
The boundary conditions imply
u() –u() = –au() +
h
u–(s),u–(s)
–h
u–(s),u–(s)
ds.
On the other hand
u() –u() =
u(s)ds=u() –
( –s)fs,pu–(s)
,qu(s)
,u–(s)ds.
It is readily seen that
(a+ )u()≤
Mf +h.
Since|u–()| ≤K(by the induction hypothesis) it follows that
u()≤ a+
Mf
+h
=C. ()
It follows from () and () that
u(t)≤K:=max
C+Mf,α,β
fort∈I. ()
Claim.u∈S(α,β). Sinceu() = it suffices to show thatu∈[α,β],i.e.α≤u≤β.
that W(t)≥ for allt∈I. Suppose by contradiction, that there existsξ∈Isuch that
W(ξ) < . Since W is continuous, it follows that there exists η∈I such that W(η) =
min{W(t);t∈I}< . Hence we haveW(η) < ,W(η) = andW(η) > . Thus,
W(η) =u(η) –α(η) = –F
η,u(η),u–(η),u(η)
–α(η)
= –fη,pu–(η)
,qu(η),u–(η)–α(η).
ButW(η) =u(η) –α(η) < andW(η) =u(η) –α(η) = . Therefore,q(u(η)) =α(η) and u(η) =α(η). Hence,
W(η) = –fη,p(u–)
(η),qu(η)
,u–(η)–α(η) = –fη,u–(η),α(η),u–(η)
–α(η) > .
It follows that
fη,u–(η),α(η),u–(η)
+α(η) < .
Since
α(η)≥–fη,α(η),α(η),α(η),
we infer that
fη,u–(η),α(η),u–(η)
–fη,α(η),α(η),α(η)< .
The above inequality is not possible by (Af)(). Now, ifη= , thenW() < ,W()≥, andW()≥. It follows that
W() =u() –α()
=au() +
h
u–(s),u–(s)
ds–α() < .
Since
W() =u() –α()≥,
we get
aα() –α() +
h
u–(s),u–(s)
ds< .
The monotonicity ofhleads to
aα() –α() +
h
This is not possible by the properties of the lower solutionα. Finally,W()≥. Indeed,
W() =u() –α() =
h
u–(s),u–(s)
ds–α()
≥
h
α(s),α(s)ds–α()≥,
by definition of the lower solutionα. Similarly we show thatu≤β. Thus, we have shown
thatu∈[α,β], which implies thatu∈S(α,β). Therefore, we have proved that the
se-quences (uj)j∈N, (uj)j∈N, and (uj)j∈N are uniformly bounded on the intervalI. Bolzano-Weierstrass theorem implies that we can extract subsequences (ujm)jm∈N, (ujm)jm∈N and
(ujm)jm∈Nthat are uniformly convergent onI. Using the diagonalization process, if
neces-sary, we shall assume thatlimm→∞ujm=limm→∞u
jm–=w,limm→∞u
jm=limm→∞u
jm–=
vandlimm→∞ujm=limm→∞ujm–=u. To complete the proof of our second main result we prove thatu=v,u=wanduis the desired solution of our original problem. Since ujm(t) =
t
ujm(s)ds, it follows from the uniform convergence of the two subsequences
that u(t) =tv(s)ds, and this equality implies thatu() = andu=v. Also, we have ujm(t) =ujm() +ujm()t+
t
(t–s)ujm(s)ds=u
jm()t+
t
(t–s)ujm(s)ds, which implies that
u(t) =u()t+t(t–s)w(s)ds, from which we readily getu=w. It is clear that
u∈S(α,β) and u≤K. ()
The differential equation –ujm(t) =F(t,ujm–(t),ujm(t),u
jm–(t)), for t∈ I, implies that
–ujm(t) = –ujm() +tF(s,ujm–(s),ujm(s),u
jm–(s))ds. The continuity ofFand the uniform
convergence of the respective subsequences imply that –u(t) = –u() +tF(s,u(s),u(s), u(s))ds, so that
–u(t) =Ft,u(t),u(t),u(t) fort∈I. The definition ofFand () show that
–u(t) =ft,u(t),u(t),u(t) fort∈I. ()
Similarly we can show that
u() –au() =
h
u(s),u(s)ds
and
u() =
h
u(s),u(s)ds.
We see thatuis a solution of (), (), (), (). Moreover,u∈S(α,β). This completes the
proof of our main result.
Competing interests
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Author details
1Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, P.O. Box 5046, Dhahran, 31261, Saudi Arabia. 2Department of Mathematics, University of Tlemcen, P.O. Box 119, Tlemcen, 13000, Algeria. 3Department of Mathematics, Dammam University, P.O. Box 838, Dammam, Saudi Arabia.
Acknowledgements
A Boucherif is grateful to King Fahd University of Petroleum and Minerals for its constant support. The authors are grateful to the anonymous referees and the handling editor for comments that led to the improvement of the presentation of the manuscript.
Received: 23 January 2014 Accepted: 23 May 2014 References
1. Boucherif, A, Al-Malki, N: Nonlinear three-point third order boundary value problems. Appl. Math. Comput.190, 1168-1177 (2007)
2. Minhós, FM: On some third order nonlinear boundary value problems: existence, location and multiplicity results. J. Math. Anal. Appl.339, 1342-1353 (2008)
3. Grossinho, MR, Minhós, FM: Existence results for some third order separated boundary value problems. Nonlinear Anal.47, 2407-2418 (2001)
4. Klokov, YA: Upper and lower functions in boundary value problems for third-order ordinary differential equations. Differ. Equ.36, 1762-1769 (2000)
5. Boucherif, A, Bouguima, SM, Al-Malki, N, Benbouziane, ZN: Third order differential equations with integral boundary conditions. Nonlinear Anal.71, e1736-e1743 (2009)
6. Benbouziane, ZN, Boucherif, A, Bouguima, SM: Third order nonlocal multipoint boundary value problems. Dyn. Syst. Appl.13, 41-48 (2004)
7. Gray, M: Uniqueness implies uniqueness for nonlocal boundary value problems for third order ordinary differential equations. Dyn. Syst. Appl.16, 277-284 (2007)
8. Wang, Y, Ge, W: Existence of solutions for a third order differential equation with integral boundary conditions. Comput. Math. Appl.53, 144-154 (2007)
9. Graef, J, Kong, L, Minhós, F: Higher order functional boundary value problems: existence and location results. Acta Sci. Math.77, 87-100 (2011)
10. Smart, DR: Fixed Point Theorems. Cambridge University Press, Cambridge (1974)
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