R E S E A R C H
Open Access
Positive solutions for a sixth-order boundary
value problem with four parameters
Ravi P Agarwal
1*, B Kovacs
2and D O’Regan
3*Correspondence: [email protected]
1Department of Mathematics, Texas
A&M University-Kingsville, 700 University Blvd., Kingsville, 78363-8202, USA
Full list of author information is available at the end of the article
Abstract
This paper investigates the existence and multiplicity of positive solutions of a sixth-order differential system with four variable parameters using a monotone iterative technique and an operator spectral theorem.
MSC: 34B15; 34B18
Keywords: positive solutions; variable parameters; fixed point theorem; operator spectral theorem
1 Introduction
It is well known that boundary value problems for ordinary differential equations can be used to describe a large number of physical, biological and chemical phenomena. In re-cent years, boundary value problems for sixth-order ordinary differential equations, which arise naturally, for example, in sandwich beam deflection under transverse shear have been studied extensively, see [–] and the references therein. The deformation of the equilib-rium state of an elastic circular ring segment with its two ends simply supported can be described by a boundary value problem involving a sixth-order ordinary differential equa-tion
u()+ u()+u=f(t,u), <t< ,
u() =u() =u() =u() =u()() =u()() = .
()
Liu and Li [] studied the existence and nonexistence of positive solutions of the non-linear fourth-order beam equation
u()(t) +βu(t) –αu(t) =λft,u(t), <t< ,
u() =u() =u() =u() = .
()
They showed that there exists aλ> such that the above boundary value problem has at least two, one, and no positive solutions for <λ<λ,λ=λandλ>λ, respectively.
In this paper, we discuss the existence of positive solutions for the sixth-order boundary value problem
–u()+A(t)u()+B(t)u+C(t)u=D(t) +uϕ+λf(t,u), <t< ,
–ϕ+κϕ=μu, <t< , ()
u() =u() =u() =u() =u()() =u()() = ,
ϕ() =ϕ() = .
For this, we shall assume the following conditions throughout
(H) f(t,u) : [, ]×[,∞)−→[,∞)is continuous;
(H) a,b,c∈R,a=λ+λ+λ> –π,b= –λλ–λλ–λλ> ,c=λλλ<
whereλ≥≥λ≥–π,≤λ< –λ andπ+aπ–bπ+c> , and A,B,C,D∈C[, ]witha=supt∈[,]A(t),b=inft∈[,]B(t)andc=supt∈[,]C(t). LetK=max≤t≤[–A(t) +B(t) –C(t) – (–a+b–c)] and=π+aπ–bπ+c.
Assumption (H) involves a three-parameter nonresonance condition.
More recently Li [] studied the existence and multiplicity of positive solutions for a sixth-order boundary value problem with three variable coefficients. The main difference between our work and [] is that we consider boundary value problem not only with three variable coefficients, but also with two positive parametersλandμ, and the existence of the positive solution depends on these parameters. In this paper, we shall apply the monotone iterative technique [] to boundary value problem () and then obtain several new existence and multiplicity results. In the special case, in [] by using the fixed point theorem and the operator spectral theorem, we establish a theorem on the existence of positive solutions for the sixth-order boundary value problem () withλ= .
2 Preliminaries
LetY=C[, ] andY+={u∈Y:u(t)≥,t∈[, ]}. It is well known thatY is a Banach space equipped with the normu=supt∈[,]|u(t)|. SetX={u∈C[, ] :u() =u() =
u() =u() = }. For givenχ≥ andν≥, we denote the norm · χ,νby
· χ,ν= sup
t∈[,]
u()(t)+χu(t)+νu(t), u∈X.
We also need the spaceX, equipped with the norm
u=max
u,u,u().
In [], it is shown thatXis complete with the norm·χ,νandu, and moreover∀u∈X,
u≤ u≤ u().
Forh∈Y, consider the linear boundary value problem
–u()+au()+bu+cu=h(t), <t< ,
u() =u() =u() =u() =u()() =u()() = ,
()
wherea,b,csatisfy the assumption
π+aπ–bπ+c> , ()
and let=π+aπ–bπ+c. Inequality () follows immediately from the fact that=
> . Since the linel={(a,b,c) :π+aπ–bπ+c= }is the first eigenvalue line of the three-parameter boundary value problem –u()+au()+bu+cu= ,u() =u() = u() =u() =u()() =u()() = , if (a,b,c) lies inl, then by the Fredholm alternative, the existence of a solution of the boundary value problem () cannot be guaranteed.
LetP(λ) =λ+βλ–α, whereβ< π,α≥. It is easy to see that the equationP(λ) =
has two real rootsλ,λ=
–β±√β+α
withλ≥≥λ> –π . Letλ
be a number such that ≤λ< –λ. In this case, () satisfies the decomposition form
–u()+au()+bu+cu=
–d
dt +λ – d
dt +λ – d
dt +λ u, <t< . ()
Suppose thatGi(t,s) (i= , , ) is the Green’s function associated with
–u+λiu= , u() =u() = . ()
We need the following lemmas.
Lemma [, ] Letωi=
√
|λi|,then Gi(t,s) (i= , , )can be expressed as (i) whenλi> ,
Gi(t,s) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
sinhωitsinhωi( –s) ωisinhωi
, ≤t≤s≤,
sinhωissinhωi( –t) ωisinhωi
, ≤s≤t≤
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
;
(ii) whenλi= ,
Gi(t,s) =
t( –s), ≤t≤s≤,
s( –t), ≤s≤t≤
;
(iii) when–π<λi< ,
Gi(t,s) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
sinωitsinωi( –s) ωisinωi
, ≤t≤s≤,
sinωissinωi( –t) ωisinωi
, ≤s≤t≤
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
.
Lemma [] Gi(t,s) (i= , , )has the following properties (i) Gi(t,s) > ,∀t,s∈(, );
(ii) Gi(t,s)≤CiGi(s,s),∀t,s∈[, ]; (iii) Gi(t,s)≥δiGi(t,t)Gi(s,s),∀t,s∈[, ],
where Ci = , δi = sinhωiωi, if λi> ; Ci = , δi = , if λi = ; Ci=
sinωi, δi =ωisinωi, if –π<λi< .
In what follows, we letDi=maxt∈[,]Gi(t,s)ds.
Lemma [] Let X be a Banach space,K a cone anda bounded open subset of X.Let
θ ∈and T:K∩ ¯→K be condensing.Suppose that Tx=υx for all x∈K∩∂and
Lemma [] Let X be a Banach space,let K be a cone of X.Assume that T:K¯r→K
(here Kr={x∈K| x<r},r> )is a compact map such that Tx=x for all x∈∂Kr.If
x ≤ Txfor x∈∂Kr,then i(T,Kr,K) = .
Now, since
–u()+au()+bu+cu=
–d
dt +λ – d
dt +λ – d
dt +λ u
=
–d
dt +λ – d
dt +λ – d
dt +λ u=h(t), ()
the solution of boundary value problem () can be expressed as
u(t) =
G(t,v)G(v,s)G(s,τ)h(τ)dτds dv, t∈[, ]. ()
Thus, for every givenh∈Y, the boundary value problem () has a unique solutionu∈ C[, ], which is given by ().
We now define a mappingT:C[, ]→C[, ] by
(Th)(t) =
G(t,v)G(v,s)G(s,τ)h(τ)dτds dv, t∈[, ]. ()
Throughout this article, we shall denoteTh=uthe unique solution of the linear bound-ary value problem ().
Lemma [] T:Y−→(X, · χ,ν)is linear completely continuous, whereχ=λ+λ,
ν=λλandT ≤D.Moreover,∀h∈Y+,if u=Th,then u∈X∩Y+,and u≤,u()≥.
We list the following conditions for convenience
(H) f(t,u)is nondecreasing inufort∈[, ]; (H) f(t, ) >c> for allt∈[, ];
(H) f∞=limu→∞f(ut,u) =∞uniformly fort∈[, ];
(H) f(t,ρu)≥ραf(t,u)forρ∈(, )andt∈[, ], whereα∈(, )is independent ofρ andu.
Suppose thatG(t,s) is the Green’s function of the linear boundary value problem
–u+κu= , u() =u() = . ()
Then, the boundary value problem
–ϕ+κϕ=μu, ϕ() =ϕ() = ,
can be solved by using Green’s function, namely,
ϕ(t) =μ
whereκ> –π. Thus, inserting () into the first equation in (), yields
–u()+A(t)u()+B(t)u+C(t)u=μD(t) +u(t)
G(t,s)u(s)ds+λf(t,u),
u() =u() =u() =u() =u()() =u()() = .
()
Let us consider the boundary value problem
–u()+A(t)u()+B(t)u+C(t)u=h(t), <t< ,
u() =u() =u() =u() =u()() =u()() = .
()
Now, we consider the existence of a positive solution of (). The functionu∈C(, )∩ C[, ] is a positive solution of (), ifu≥,t∈[, ], andu= .
Let us rewrite equation () in the following form
–u()+au()+bu+cu= –A(t) –au()–B(t) –bu–C(t) –cu
+μD(t) +u(t)
G(t,s)u(s)ds+h(t). ()
For anyu∈X, let
Gu= –A(t) –au()–B(t) –bu–C(t) –cu+μD(t)
G(t,s)u(s)ds.
The operatorG:X→Yis linear. By Lemmas and in [],∀u∈X,t∈[, ], we have
(Gu)(t)≤–A(t) +B(t) –C(t) – (–a+b–c)u+μCdu ≤(K+μCd)u≤(K+μCd)uχ,ν,
where C = maxt∈[,]D(t), K = maxt∈[,][–A(t) + B(t) – C(t) – (–a + b – c)], d =
maxt∈[,]G(t,s)ds,χ=λ+λ≥,ν=λλ≥. Hence Gu≤(K+μCd)uχ,ν, and soG ≤(K+μCd). Alsou∈C[, ]∩C(, ) is a solution of () ifu∈Xsatisfies u=T(Gu+h), whereh(t) =μu(t)G(t,s)u(s)ds+h(t),i.e.,
u∈X, (I–TG)u=Th. ()
The operatorI–TGmapsXintoX. FromT ≤Dtogether withG ≤(K+μCd) and the conditionD(K+μCd) < , and applying the operator spectral theorem, we find that (I–TG)–exists and is bounded. Letμ∈(,–DK
DCd), where –DK> , then the condition
D(K+μCd) < is fulfilled. Let L=D(K+μCd), and letμ∗∗=–DDCdK. LetH= (I–TG)–T. Then () is equivalent tou=Hh
. By the Neumann expansion formula,Hcan be expressed by
H=I+TG+· · ·+ (TG)n+· · ·T=T+ (TG)T+· · ·+ (TG)nT+· · ·. ()
Now∀h∈Y+, letu=Th, thenu∈X∩Y+, andu≤,u()≥. Thus, we have
(Gu)(t) = –A(t) –au()–B(t) –bu–C(t) –cu
+μD(t)
G(t,s)u(s)ds≥, t∈[, ].
Hence
∀h∈Y+, (GTh)(t)≥, t∈[, ], ()
and so, (TG)(Th)(t) =T(GTh)(t)≥,t∈[, ].
It is easy to see [] that the following inequalities hold:∀h∈Y+,
–L(Th)(t)≥(Hh)(t)≥(Th)(t), t∈[, ], ()
and, moreover,
(Hh)≤
–L(Th). ()
Lemma [] H:Y −→(X, · κ,ν)is completely continuous, whereχ =λ +λ, ν=
λλ and ∀h∈Y+, –L(Th)(t)≥(Hh)(t)≥(Th)(t), t ∈[, ], and, moreover, Th ≥ ( –L)Hh.
For anyu∈Y+, defineFu=μu(t)G(t,s)u(s)ds+λf(t,u). From (H), we have thatF: Y+→Y+is continuous. It is easy to see thatu∈C[, ]∩C(, ), being a positive solution of (), is equivalent tou∈Y+, being a nonzero solution of
u=HFu. ()
Let us introduce the following notations
Tλ,μu(t) :=TFu(t) =
G(t,v)G(v,s)G(s,τ)
×
μu(τ)
G(τ,s)u(s)ds+λfτ,u(τ) dτds dv,
Qλ,μu:=HFu=TFu+ (TG)TFu+ (TG)TFu+· · ·+ (TG)nTFu+· · ·
=Tλ,μu+ (TG)Tλ,μu+ (TG)Tλ,μu+· · ·+ (TG)nTλ,μu+· · ·,
i.e.,Qλ,μu=HFu. Obviously,Qλ,μ:Y+→Y+is completely continuous. We next show that the operatorQλ,μhas a nonzero fixed point inY+.
LetP={u∈Y+:u(t)≥δ( –L)g(t)u(t),t∈[,]}, whereg(t) =C
G(t,t). It is easy
to see thatPis a cone inY, and now, we showQλ,μ(P)⊂P.
Proof It is clear thatQλ,μ:P→Pis completely continuous. Now∀u∈P, leth=Fu, then
h∈Y+. Using Lemma ,i.e., (Qλ,μu)(t) = (HFu)(t)≥(TFu)(t),t∈[, ] and by Lemma ,
for allu∈P, we have
(TFu)(t)≤C
G(v,v)G(v,s)G(s,τ)(Fu)(τ)dτds dv, ∀t∈[, ].
Thus,
G(v,v)G(v,s)G(s,τ)(Fu)(τ)dτds dv≥
CTFu. ()
On the other hand, by Lemma and (), we have
(TFu)(t)≥δG(t,t)
G(v,v)G(v,s)G(s,τ)(Fu)(τ)dτds dv
≥δG(t,t)
CTFu≥δG(t,t)
C( –L)Qu, ∀t∈[, ].
Thus,Qλ,μ(P)⊂P.
3 Main results
Lemma Let f(t,u)be nondecreasing in u for t∈[, ]and f(t, ) >c> for all t∈[, ], wherec is a constant and L< .Then there existsλ> andμ> such that the operator Qλ,μhas a fixed point uat(λ,μ)with u∈P\{θ}.
Proof Setu(t) = (Qλ,μu)(t), where
u(t) =
–L
G(t,v)G(v,τ)G(τ,s)ds dτdv.
It is easy to see that u(t)∈P. Let λ =M–fu and μ =min(Nfu–;μ∗∗), where Mfu=
maxt∈[,]f(t,u(t)) andNfu=maxt∈[,]u(t)
G(t,s)u(s)ds, respectively. ThenMfu>
andNfu> , and from Lemma , we obtain
u(t) =u(t) =
–L
G(t,v)G(v,τ)G(τ,s)ds dτdv
≥ –L
G(t,v)G(v,τ)G(τ,s)
×
λfs,u(s)
+μu(t)
G(t,s)u(s)ds ds dτdv
=
–L(Tλ,μu)(t)≥(Qλ,μu)(t) =u(t). It is easy to see that
un(t) = (Qλ,μun–)(t)
+ (TG)nTλ,μun–+· · ·
≥Tλ,μun–+ (TG)Tλ,μun–+ (TG)Tλ,μun–+· · · + (TG)nTλ,μun–+· · ·
=un–(t).
Indeed, forh,h∈Y+, leth(t)≥h(t), then from (), we haveu(t) =Th≥Th=u(t). Using equation () and (), we obtain
–u+λu=
G(t,v)G(v,τ)h(τ)dτdv, t∈[, ] ()
and
u()– (λ+λ)u+λλu=
G(t,v)h(v)dv, t∈[, ]. ()
Then by (), we have fort∈[, ]
u(t) –u(t) =λ
u(t) –u(t)–
G(t,v)G(v,τ)h(t) –h(t)dτdv≤,
becauseλ< , and finally, from (), we have
u() (t) –u() (t) = (λ+λ)
u(t) –u(t)–λλ
u(t) –u(t)
+
G(t,v)
h(t) –h(t)
dτdv≥, t∈[, ]
becauseλ+λ≤ andλλ≤. From the equation
(Gu)(t) = –A(t) –au()–B(t) –bu–C(t) –cu+μD(t)
G(t,s)u(s)ds≥,
t∈[, ]
we have
(Gu)(t) – (Gu)(t)
= –A(t) –au() (t) –u() (t)–B(t) –bu(t) –u(t)
–C(t) –cu(t) –u(t)+μD(t)
G(t,s)u(t) –u(t)ds≥,
t∈[, ] ()
i.e., (Gu)(t)≥(Gu)(t) for allt∈[, ]. Finally, ifh=Fuandh=Fu, then
(Hh)(t) =T(h) + (TG)(Th) + (TG)(Th) +· · ·+ (TG)n(Th) +· · ·
i.e.,
Qλ,μu
= (HFu)(t) =T(Fu) + (TG)(TFu) + (TG)(TFu) +· · ·+ (TG)n(TFu) +· · ·
≥(HFu)(t) =T(Fu) + (TG)(TFu) + (TG)(TFu) +· · ·+ (TG)n(TFu) +· · ·
=Qλ,μu, ()
and from (), it follows that foru,u∈Y+, ifu(t)≥u(t) then, we have
Qλ,μu≥Qλ,μu. ()
Setu(t) =u(t) andun(t) = (Qλ,μun–)(t),n= , , . . . ,t∈[, ]. Then u(t) =u(t)≥u(t)≥ · · · ≥un(t)≥ · · · ≥LG(t,t),
where
L=λδδδcCCC.
Indeed, by Lemma , we have
un(t) =Qλ,μun–= (HF)(un–)≥(TF)(un–)
≥λ
G(t,v)G(v,τ)G(τ,s)f
s,un–(s)
ds dτdv
≥λc
G(t,v)G(v,τ)G(τ,s)ds dτdv
≥λcδδδCCCG(t,t).
Now,f(t,u) nondecreasing inufort∈[, ], Lemma , and the Lebesgue convergence theorem guarantee that{un}∞n=={Qλ,μu}∞n=decreases to a fixed pointu∈P\{θ}of the operatorQλ,μ.
Lemma Suppose that(H)-(H)hold,and L< .Set
Sλ,μ=
u∈P:Qλ,μu=u, (λ,μ)∈A
,
where A⊂[a,∞)×[b,∞)for some constants a> ,b> .Then there exists a constant CA
such thatu<CAfor all u∈Sλ,μ.
Proof Suppose, to the contrary, that there exists a sequence {un}∞n= such that limn→∞un= +∞, whereun∈P is a fixed point of the operatorQλ,μ at (λn,μn)∈A
(n= , , . . .). Then
un(t)≥kun fort∈
,
,
wherek=δ
ChooseJ> , so that
JaδδδCCmmk> ,
andl> such that
f(t,u)≥Ju foru>landt∈
, ,
andN, so thatuN>
l
k. Now,
(QλN,μNuN)
≥(TFuN)
≥λN
G
,v G(v,τ)G(τ,s)f
s,uN(s)
ds dτdv
≥λNδδδCCG
,
G(s,s)fs,uN(s)
ds
≥λNδδδCCG
,
G(s,s)fs,uN(s)
ds
≥
aδδδCCmmJuN(t)
≥
aδδδCCmmJkuN>uN,
and so,
uN=QλN,μNuN≥(TF)uN≥(TFuN)
>uN,
which is a contradiction.
Lemma Suppose that L< , (H)and(H)hold and that the operator Qλ,μhas a
posi-tive fixed point in P atλ> andμ> .Then for every(λ,μ)∈(,λ)×(,μ)there exists
a function u∈P\{θ}such that Qλ,μu=u.
Proof Letu(t) be a fixed point of the operatorQλ,μat (λ,μ). Then
u(t) =Qλ,μu(t)≥Qλ,μu(t),
where <λ<λ, <μ<μ. Hence
G(t,v)G(v,τ)G(τ,s)
λfs,u(s)+μu(s)
G(s,p)u(p)dp ds dτdv
≥
G(t,v)G(v,τ)G(τ,s)
×
λf
s,u(s)+μu(s)
Set
(Tλ,μu)(t) =
G(t,v)G(v,τ)G(τ,s)
×
λf
s,u(s)+μu(s)
G(s,p)u(p)dp ds dτdv
and
(Qλ,μu)(t) =Tλ,μu+ (TG)Tλ,μu+ (TG)
T
λ,μu+· · ·+ (TG) nT
λ,μu+· · ·
u(t) =u(t) andun(t) =Qλ,μun–. Then
u(t) =u(t) =Tλ,μu+ (TG)Tλ,μu+ (TG)Tλ,μu+· · ·+ (TG)nTλ,μu+· · ·
≥Tλ,μu+ (TG)Tλ,μu+ (TG)
T
λ,μu+· · ·+ (TG) nT
λ,μu+· · ·=u(t)
and
un(t) =Qλ,μun–=Tλ,μun–+ (TG)Tλ,μun–+ (TG)
T
λ,μun–+· · ·
+ (TG)nTλ,μun–+· · ·
≥Tλ,μun–+ (TG)Tλ,μun–+ (TG)
T
λ,μun–+· · ·
+ (TG)nT
λ,μun–+· · ·=un–(t)
becausef(t,u) is nondecreasing inufort∈[, ] andTλ,μuis also nondecreasing inu.
Thus
u(t)≥u(t)≥ · · · ≥un(t)≥un+(t)≥ · · · ≥LG(t,t), ()
where
L=λcδδδCCC.
Indeed, by Lemma , we have
un(t) =Qλ,μun–= (HF)(un–)≥Tλ,μ(un–)
≥λ
G(t,v)G(v,τ)G(τ,s)fs,un–(s)
ds dτdv
≥λc
G(t,v)G(v,τ)G(τ,s)ds dτdv≥λcδδδCCCG(t,t).
Lemma implies that{Qnλu}∞n=decreases to a fixed pointu∈P\{θ}.
Lemma Suppose that L< , (H)-(H)hold.Let
=λ> ,μ> :Qλ,μhave at least one fixed point at(λ,μ)in P
.
Proof Suppose, to the contrary, that there exists a fixed point sequence{un}∞n=⊂P of Qλ,μat (λn,μn) such thatlimn→∞λn=∞and <μn<μ∗∗. Then there are two cases to
be considered: (i) there exists a subsequence{uni}∞n=such thatlimi→∞uni=∞, which is impossible by Lemma , so we only consider the next case: (ii) there exists a constant H> such thatun≤H,n= , , , , . . . . In view of (H) and (H), we can choosel> such thatf(t, ) >lH, and further,f(t,un) >lHfort∈[, ]. We know that
un=Qλn,μnun≥Tλn,μnun.
Letvn(t) =Tλn,μnun,i.e.,un(t)≥vn(t). Then it follows that
–v()n +av()n +bvn+cvn=λnf(t,un) +μnun(t)
G(t,p)un(p)dp, <t< . ()
Multiplying () bysinπtand integrating over [, ], and then using integration by parts on the left side of (), we have
vn(t)sinπt dt=λn
f(t,un)sinπt dt+μn
un(t)sinπt
G(t,p)un(p)dp dt.
Next, assume that (ii) holds. Then
un(t)sinπt dt≥
vn(t)sinπt dt
=λn
f(t,un)sinπt dt+μn
un(t)sinπt
G(t,p)un(p)dp dt
and
H
sinπt dt≥un
sinπt dt≥
un(t)sinπt dt≥
vn(t)sinπt dt
=λn
f(t,un)sinπt dt+μn
un(t)sinπt
G(t,p)un(p)dp dt
≥λnlH
sinπt dt
lead to≥λnl, which is a contradiction. The proof is complete.
Lemma Suppose that L< , (H)-(H)hold.Let
μ=
λ> : (λ,μ)∈andμis fixed,
and letλμ=supμ.Thenμ= (,λμ],whereis defined in Lemma.
Proof By Lemma , it follows that (,λ)×(,μ)⊂. We only need to prove (λμ,μ)∈.
We may choose a distinct nondecreasing sequence{λn}∞n=⊂such thatlimn→∞λn=λμ.
{un}∞n=,is uniformly bounded, so it has a subsequence, denoted by{unk} ∞
k=, converging to
u∈P. Note that
un=Tλn,μun+ (TG)Tλn,μun+ (TG)Tλn,μun+· · ·+ (TG)nTλn,μun+· · ·
=Qλn,μun. ()
Taking the limit asn→ ∞on both sides of (), and using the Lebesgue convergence theorem, we have
u=Tλ,μu+ (TG)Tλ,μun+ (TG)Tλ,μu+· · ·+ (TG)nTλ,μu+· · ·
which shows thatQλ,μhas a positive fixed pointuat (λ,μ).
Theorem Suppose that(H)-(H)hold,and L< .For fixedμ∈(,μ),then there exists atλ> such that()has at least two,one and has no positive solutions for <λ<λ,
λ=λforλ>λ,respectively.
Proof Suppose that (H) and (H) hold. Then there existsλ> andμ> such that
Qλ,μhas a fixed pointuλ,μ∈P\{θ}atλ=λandμ=μ. In view of Lemma ,Qλ,μalso
has a fixed pointuλ,μ<uλ,μ,uλ,μ∈P\{θ}, and <λ<λ, <μ<μ,μ∈(,μ). For
<λ<λ\, there existsδ
> such that
f(t,uλ,μ+δ) –f(t,uλ,μ)≤f(t, )
λ
λ –
fort∈[, ], <δ≤δ. In this case, it is easy to see that
Tλ,μ(uλ,μ,+δ) =λ
G(t,v)G(v,τ)G(τ,s)fs,uλ,μ(s) +δds dτdv
+μ
G(t,v)G(v,τ)G(τ,s)uλ,μ(s) +δ
×
G(s,p)uλ,μ(p) +δdp ds dτdv
≤λ
G(t,v)G(v,τ)G(τ,s)f
s,uλ(s)ds dτdv
+μ
G(t,v)G(v,τ)G(τ,s)uλ,μ(s)
×
G(s,p)uλ,μ(p)dp ds dτdv=Tλ,μuλ,μ.
Indeed, we have
λ
G(t,v)G(v,τ)G(τ,s)fs,uλ(s) +δds dτdv
–λ
=λ
G(t,v)G(v,τ)G(τ,s)fs,uλ(s) +δ–fs,uλ(s)ds dτdv
–λ–λ
G(t,v)G(v,τ)G(τ,s)fs,uλ(s)ds dτdv
≤λ–λ
G(t,v)G(v,τ)G(τ,s)f(s, )ds dτdv
–λ–λ
G(t,v)G(v,τ)G(τ,s)f
s,uλ(s)ds dτdv
=λ–λ
G(t,v)G(v,τ)G(τ,s)f(s, ) –fs,uλ(s)ds dτdv≤.
Similarly, it is easy to see that
μ
G(t,v)G(v,τ)G(τ,s)uλ,μ(s) +δ
G(s,p)uλ,μ(p) +δdp ds dτdv
–μ
G(t,v)G(v,τ)G(τ,s)uλ,μ(s)
G(s,p)uλ,μ(p)dp ds dτdv≤.
Moreover, from (), it follows that forTλ,μ(uλ,μ+δ)≤Tλ,μuλ,μwe have
GTλ,μ(uλ,μ+δ)≤G(Tλ,μuλ,μ).
Finally, we have
(TG)Tλ,μ(uλ,μ+δ)≤(TG)Tλ,μuλ,μ.
By induction, it is easy to see that
(TG)nTλ,μ(uλ,μ+δ)≤(TG)nTλ,μuλ,μ, n= , , . . . . ()
Hence, using (), we have
Qλ,μ(uλ,μ+δ) =Tλ,μ(uλ,μ+δ) + (TG)Tλ,μ(uλ,μ+δ)
+ (TG)Tλ,μ(uλ,μ+δ) +· · ·+ (TG)nTλ,μ(uλ,μ+δ) +· · ·
≤Tλ,μuλ,μ+ (TG)Tλ,μuλ,μ+ (TG)Tλ,μuλ,μ+· · ·
+ (TG)nTλ,μuλ,μ+· · ·
=Qλ,μ(uλ,μ)
i.e.,
Qλ,μ(uλ,μ+δ) –Qλ,μ(uλ,μ)≤,
so that
SetDuλ,μ ={u∈C[, ] : –δ<u(t) <uλ,μ +δ}. Then Qλ,μ:P∩Duλ,μ →P is
com-pletely continuous. Furthermore,Qλ,μu=υuforυ≥ andu∈P∩∂Duλ,μ. Indeed set
u∈P∩∂Duλ,μ. Then there existst∈[, ] such thatu(t) =u=uλ,μ+δand
(Qλ,μu)(t) =
Tλ,μ(u) + (TG)Tλ,μ(u) + (TG)Tλ,μ(u) +· · ·+ (TG)nTλ,μ(u) +· · ·
(t)
≤Tλ,μ(uλ,μ+δ) + (TG)Tλ,μ(uλ,μ+δ) + (TG)Tλ,μ(uλ,μ+δ) +· · ·
+ (TG)nTλ,μ(uλ,μ+δ) +· · ·(t) =Qλ,μ(uλ,μ+δ)(t)
<uλ,μ(t) +δ=u(t)≤υu(t), υ≥.
By Lemma ,i(Qλ,μ,P∩∂Duλ,μ,P) = .
Letkbe such that
u(t)≥ku fort∈
,
.
We know thatlimu→∞f(t,u)
u =∞uniformly fort∈[, ], so we may chooseJ> , so that
λJδδδCCmCk> ,
l>uλ,μ+δ> , so that
f(t,u)≥Ju foru>landt∈
,
.
SetR=l
k andPR={u∈P:u<R}. ThenQλ,μ:PR→Pis completely continuous. It
is easy to obtain
(Qλ,μu)(t)≥(Tλ,μu)(t)≥λ
G(t,v)G(v,τ)G(τ,s)f
s,u(s)ds dτdv
≥λδδδCCG(t,t)
G(s,s)fs,u(s)ds
≥λδδδCCG(t,t)
G(s,s)fs,u(s)ds
≥
λδδδCCmCJu(t)≥
λδδδCCmCJku>u
fort∈[, ] andu∈∂PR. Nowu(t)≥ku=kR=l, and so
Qλ,μu>u.
In view of Lemma ,i(Qλ,μ,PR,P) = . By the additivity of the fixed point index,
i(Qλ,μ,PR\P∩Duλ,μ,P) =i(Qλ,μ,PR,P) –i(Qλ,μ,P∩Duλ,μ,P) = –.
ThusQλ,μhas a fixed point in{P∩Duλ,μ}\{θ}and has another fixed point inPR\P∩
Duλ,μ by choosingλ
Let us introduce the notationμ= in the equation of (), then we have
–u()+A(t)u()+B(t)u+C(t)u=λf(t,u),
u() =u() =u() =u() =u()() =u()() = .
()
In this case, we can prove the following theorem, which is similar to Theorem .
Theorem Suppose that(H)-(H)hold,and L< .Then there exists atλ> such that
()has at least two,one and has no positive solutions for <λ<λ,λ=λ forλ>λ,
respectively.
We follow exactly the same procedure, described in detail in the proof of Theorem for
μ= .
Let us introduce the following notations forμ= andλ=
TFu(t) =
G(t,v)G(v,s)G(s,τ)fτ,u(τ)dτds dv,
Qu:=HFu=TFu+ (TG)TFu+ (TG)TFu+· · ·+ (TG)nTFu+· · ·,
()
i.e.,Qu=Q,u=HFu.
Lemma Suppose that (H), (H) and (H) hold, and L < . Then for any u ∈ C+[, ]\{θ},there exist real numbers S
u≥su> such that
sug(t)≤(Qu)(t)≤Sug(t), for t∈[, ],
where g(t) =G(t,τ)G(τ,v)G(v,v)dv dτ.
Proof For anyu∈C+[, ]\{θ}from Lemma , we have
(Qu)(t) = (HFu)(t)≤ –L
G(t,v)G(v,τ)G(τ,s)f
s,u(s)ds dτdv
≤ C –Lsmax∈[,]f
s,u(s)
G(t,τ)G(τ,v)G(v,v)dv dτ
= C –Lsmax∈[,]f
s,u(s)g(t) =Sug(t) fort∈[, ].
Note that for anyu∈C+[, ]\{θ}, there exists an interval [a
,b]⊂(, ) and a number p> such thatu(t)≥pfort∈[a,b]. In addition, by (H), there exists s> andu∈ (,∞) such thatf(t,u)≥sfort∈[a,b]. Ifp≥u, thenf(t,u)≥f(t,p)≥f(t,u)≥s; ifp<u, thenf(t,u)≥f(t,p)≥f(t, p
up)≥(
p
u)αs. Hence
(Qu)(t)≥(TFu)(t)
=
G(t,v)G(v,τ)G(τ,s)fs,u(s)ds dτdv
≥δ
≥δg(t)
b
a
G(s,s)fs,u(s)ds dτdv
≥(b–a)δg(t)mG
p u
α
=sug(t),
where mG=mins∈[a,b]G(s,s),g(t) =
G(t,v)G(v,τ)G(τ,τ)dτdv,su= (b–a)×
δmG(up)α.
Theorem Suppose that(H), (H)and(H)hold,L< andλ= .Then
(i) ()has a unique positive solutionu∈C+[, ]\{θ}satisfying mug(t)≤u(t)≤Mug(t) fort∈[, ],
where <mu<Muare constants. (ii) For anyu(t)∈C+[, ]\{θ},the sequence
un(t) = (Qun–)(t) = (HFun–)(t)
=TFun–+ (TG)TFun–+ (TG)TFun–+· · ·+ (TG)nTFun–+· · ·
(n= , , . . .)converges uniformly to the unique solutionu,and the rate of
convergence is determined by
un(t) –u(t)=O
–dαn,
where <d< is a positive number.
Proof In view of (H), (H) and (H),Q:C+[, ]→C+[, ] is a nondecreasing opera-tor and satisfiesQ(ρu)≥ραQ(u) fort∈[, ] andu∈C+[, ]. Indeed, letu
(t)≤u(t),
u,u∈C+[, ], sincef(s,u) is nondecreasing inu, then by usingf(s,u(s))≤f(s,u(s)),
fort∈[, ], it follows that
TFu(t) =
G(t,v)G(v,τ)G(τ,s)fs,u(s)
ds dτdv
≤
G(t,v)G(v,τ)G(τ,s)f
s,u(s)
ds dτdv=TFu(t).
Moreover, from (), it follows that forTFu(t)≤TFu(t)
G(TFu)(t)≤G(TFu)(t) fort∈[, ]. ()
Finally, sincef(s,u) is nondecreasing inu, then by using form (),f(s,G(TFu)(t))≤
f(s,G(TFu)(t)), fort∈[, ], we have
(TG)TF(u) =
G(t,v)G(v,τ)G(τ,s)fs,G(TFu)(s)
ds dτdv
≤
G(t,v)G(v,τ)G(τ,s)fs,G(TFu)(s)
ds dτdv
i.e.,
(TG)TF(u)≤(TG)TFu.
By induction, it is easy to see that
(TG)nTF(u)≤(TG)nTFu, n= , , . . . . ()
Hence, using (), we have
Q(u) =TF(u) + (TG)TF(u) + (TG)TF(u) +· · ·+ (TG)nTF(u) +· · ·
≤TF(u) + (TG)TF(u) + (TG)TF(u) +· · ·+ (TG)nTF(u) +· · ·
=Q(u). ()
Now, we show thatQ:C+[, ]→C+[, ] satisfiesQ(ρu)≥ραQ(u) for t∈[, ] and
u∈C+[, ]. Note that
TF(ρu) =
G(t,v)G(v,τ)G(τ,s)f
s,ρu(s)ds dτdv
≥ρα
G(t,v)G(v,τ)G(τ,s)fs,u(s)ds dτdv
=ραTF(u).
Moreover, from (), it follows that forTF(ρu)≥ραTF(u), G(TFρu)(t)≥GραTF(u)(t)
=ραGTF(u)(t) fort∈[, ]. Finally, we have
(TG)TF(ρu)(t) =
G(t,v)G(v,τ)G(τ,s)fs,G(TFρu)(s)ds dτdv
≥
G(t,v)G(v,τ)G(τ,s)fs,ραGTF(u)(s)ds dτdv ≥ρα
G(t,v)G(v,τ)G(τ,s)fs,GTF(u)(s)ds dτdv
=ρα(TG)TF(u)(t), i.e.,
(TG)(TFρu)(t)≥ρα(TG)TF(u)(t). By induction, it is easy to see that
Hence, using () andρ∈(, ),α∈(, ), we have
Q(ρu) =TF(ρu) + (TG)TF(ρu) + (TG)TF(ρu) +· · ·+ (TG)nTF(ρu) +· · ·
≥ραTF(u) +ρα(TG)TF(u) +ρα(TG)TF(u) +· · ·+ραn+(TG)nTF(u) +· · · ≥ραTF(u) +ρα(TG)TF(u) +ρα(TG)TF(u) +· · ·+ρα(TG)nTF(u) +· · · =ραTF(u) + (TG)TF(u) + (TG)TF(u) +· · ·+ (TG)nTF(u) +· · ·
=ραQ(u). ()
By Lemma , there exists <sg≤Sgsuch that
sug(t)≤Qg(t)≤Sug(t).
Let
s=supsg:sug(t)≤Qg(t)
, S=infSg:Qg(t)≤Sug(t)
.
PickmsandMssuch that
<ms<min
,s–α ()
and
max,S–α=Ms<∞. ()
Setu(t) =msg(t),v(t) =Msg(t),un=Qun–, andvn=Qvn–,n= , , . . . . From () and
(), we have
msg(t) =u(t)≤u(t)≤ · · · ≤un(t)≤ · · · ≤vn(t)≤ · · · ≤v(t)≤v(t) =Msg(t). ()
Indeed, from ()ms< , andmαs–s> , we have
u(t) =Q(u) =Qmsg(t)
≥mαsQg(t)≥mαssg(t) =mαs–smsg(t) =mαs–su(t)≥u(t),
and by induction
un+(t) =Q(un)≥Q(un–) =un(t).
From (),Ms> , andMαs–S< , we have
v(t) =Q(v)≤MαsQg(t)=MαsQ
Ms
v =MsαQ(g) ≤Mα
and by induction
vn+(t) =Q(vn)≤Q(vn–) =vn(t).
Letd=ms
Ms. Then
un≥dα
n
vn. ()
In factu= dvis clear. Assume that () holds withn=k(kis a positive integer),i.e., uk≥dα
k
vk. Then
uk+=Q(uk)≥Q
dαkv
k
≥dαkαQ(v
k) =dα
k+
Q(vk) =dα
k+
vk+.
By induction, it is easy to see that () holds. Furthermore, in view of (), () and (), we have
≤un+z–un≤vn–un≤
–dαnv= –dαnM
sg(t)
and
un+z–un ≤ vn–un ≤
–dαnMsg,
wherezis a nonnegative integer. Thus, there existsu∈C+[, ] such that
lim
n→∞un(t) =nlim→∞vn(t) =u
(t) fort∈[, ]
andu(t) is a fixed point ofQand satisfies
mgg(t)≤u(t)≤Mgg(t).
This means thatu∈C+
[, ], whereC+[, ] ={u∈C+[, ],u(t) > fort∈(, )}.
Next we show thatuis the unique fixed point ofQinC+
[, ]. Suppose, to the contrary,
that there exists anotheru∈C+[, ] such thatQu=u. We can suppose that
u(t)≤u(t), u(t)=u(t) fort∈[, ].
Letτ=sup{ <τ < :τu≤u≤τ–u}. Then <τ ≤ andτu≤u≤τ–u. We assert
τ = . Otherwise, <τ< , and then
u=Qu≥Qτu≥ταQu=ταu, u=Qu≥Q(τu)≥ταQ(u) =ταu.
This means thatταu≤u≤(τα)–u, which is a contradiction of the definition ofτ,
Let us introduce the following notations forμ=
Tλu(t) :=TFu(t) =λ
G(t,v)G(v,s)G(s,τ)fτ,u(τ)dτds dv,
Qλu:=HFu=TFu+ (TG)TFu+ (TG)TFu+· · ·+ (TG)nTFu+· · ·
=Tλu+ (TG)Tλu+ (TG)Tλu+· · ·+ (TG)nTλu+· · ·,
i.e.,Qλu=λQu, whereQis given by ().
Theorem Suppose that(H), (H), (H)and L< hold.Then()has a unique positive solution uλ(t)for any <λ≤.
Proof Theorem implies that forλ= , the operatorQλhas a unique fixed pointu∈
C+[, ], that isQu=u. Then from Lemma , for everyλ
∈(, ), there exists a
func-tionu∈P\{θ}such thatQλu=u.
Thus,uλis a unique positive solution of () for every <λ≤.
4 Application
As an application of Theorem , consider the sixth-order boundary value problem
–u()+ – .tu()+ (. – .sinπt)u+C(– +cos.πt)u
=.t( –t) +uϕ+λ +sinπt+u, <t< ,
–ϕ+ ϕ=μu, <t< ,
u() =u() =u() =u() =u()() =u()() = ,
ϕ() =ϕ() = ,
()
for a fixedλ= ,λ= –,λ= andκ= . In this case,a=λ+λ+λ= ,b= –λλ–
λλ–λλ = , andc=λλλ= –. We haveA(t) = – .t,B(t) = . – .sinπt, C(t) = – +cos.πt,D(t) = .t( –t) andf(t,u) = +sinπt+u. It is easy to see thatπ+ aπ–bπ+c= ,. > ,a=supt∈[,]A(t),b=inft∈[,]B(t) andc=supt∈[,]C(t). Note also thatK=max≤t≤[–A(t) +B(t) –C(t) – (–a+b–c)] = ,D=maxt∈[,]
G(t,v)dv= ., C=maxt∈[,]D(t) = .,d=maxt∈[,]G(t,s)ds= .,μ∗∗=
–DK
DCd =
. and DK= . < . Thus, if <μ< ., then the conditions of Theorem (noteL=D(K+μCd) < ) are fulfilled (in particular, (H)-(H) are satisfied). As a result, Theorem can be applied to ().
Competing interests
The authors did not provide this information.
Authors’ contributions
The authors did not provide this information.
Author details
1Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, 78363-8202, USA. 2Department of Analysis, University of Miskolc, Egyetemvaros, 3515, Hungary.3School of Mathematics, Statistics and
Applied Mathematics, National University of Ireland, Galway, Ireland.
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Cite this article as:Agarwal et al.:Positive solutions for a sixth-order boundary value problem with four parameters.
