R E S E A R C H
Open Access
Positive solutions of a discrete nonlinear
third-order three-point eigenvalue problem
with sign-changing Green’s function
Chenghua Gao
1,2*and Tianmei Geng
1*Correspondence:
1Department of Mathematics,
Northwest Normal University, Lanzhou, Gansu 730070, People’s Republic of China
2School of Mathematics and
Statistics, Lanzhou University, Lanzhou, Gansu 730000, People’s Republic of China
Abstract
In this paper, by using the Krasnosel’skii fixed point theorem in a cone, we discuss the existence of positive solutions to the discrete third-order three-point boundary value problem
3u(t– 1) =
λ
a(t)f(t,u(t)), t∈[1,T– 2]Z,
u(0) =u(T) =
2u(η) = 0,
whereT> 4 is an integer, [1,T– 2]Z={1, 2,. . .,T– 2},
λ
> 0 is a parameter,η
∈ {T–12 ,. . .,T– 2}for oddT, andη
∈ {T–22 ,. . .,T– 2}for evenT. Despite the sign-changing Green’s function, we also give the explicit interval forλ
to guarantee the existence of positive solutions of the problem whenfsatisfies different growth assumptions.MSC: 39A10; 39A12
Keywords: third-order difference equation; three-point eigenvalue problem; sign-changing Green’s function; positive solutions; Krasnosel’skii’s fixed point theorem
1 Introduction
Let [a,b]Zdenote the integer set{a,a+ , . . . ,b}withb>a. In this paper, we consider the
existence of positive solutions for the discrete nonlinear third-order three-point BVP
u(t– ) =λa(t)f(t,u(t)), t∈[,T– ]Z,
u() =u(T) =u(η) = , (.)
whereT> is an integer,λ> is a parameter,f : [,T– ]Z×[, +∞)→[, +∞) is
con-tinuous,a: [,T– ]Z→(,∞), andηsatisfies the condition:
(H) η∈[T–,T– ]Z, ifTis an odd number, orη∈[T–,T– ]Z, ifTis an even number. Boundary value problems for third-order differential or difference equations arise in many problems of physics, control system and applied mathematics, such as the deflection of a curved beam having a constant or varying cross section, three-layer beam and the electromagnetic wave incident on a system of charges sets them into motion,etc.[]. In
recent years, the existence of positive solutions of third-order boundary value problems has been discussed by several authors. For example, in [–], by using different methods, such as the Krasnosel’skii’s fixed point theorem in a cone, the iterative technique, and the fixed point theory, the authors obtained the existence of positive solutions of the boundary value problems for third-order differential equations. For the discrete case, there are also several excellent results on the existence of positive solutions of the discrete third-order boundary value problems; see, for instance, [–] and the references therein. Specially, in [], Agarwal and Henderson considered the following discrete third-order nonlinear eigenvalue problems:
u(t) =λa(t)f(t,u(t)), t∈[,T]Z,
u() =u() =u(T+ ) = . (.)
By using the Krasnosel’skii fixed point theorem in a cone, they obtained the existence of positive solutions of (.) under both the case thatλ= and thatλ= . Later, by using the
Krasnosel’skii fixed point theorem, Anderson [] obtained the existence of positive solu-tions for a kind of discrete nonlinear third-order eigenvalue problems, Ji and Yang [] discussed the existence of positive solutions of a discrete third-order three-point right-focal boundary value problems, Karaca [] discussed the existence of positive solutions under more general boundary conditions, and Konget al.[] obtained the existence of positive solutions for a discrete boundary value problem of a third-order functional dif-ference equation. It can be seen that, in these papers, the key condition for obtaining the existence of positive solution is that the Green’s functions is positive. This condition guar-antees the positivity of corresponding summation operator. Now, the question is: when the Green’s function changes its sign, how can we guarantee the positivity of the correspond-ing summation operator? In , Wang and Gao [] discussed the existence of positive solutions of the following third-order difference equation boundary value problems:
u(t– ) =a(t)f(t,u(t)), t∈[,T– ]
Z,
u() =u(T) =u(η) = . (.)
It is worth to notice that the Green’s function changes its sign in this paper. Inspired by the work of the above papers, we try to establish some criteria for the existence of positive solutions of (.) in this paper. The Green’s function we construct in Section changes its sign and is more complicated than in the continuous case. This will takes lots of diffi-culties for us to obtain the existence of positive solutions. To overcome it, a new cone is introduced to overcome these obstacles. Meanwhile, we will point out that the condition (H) is optimal forηto obtain the existence of a positive solution of (.); see Remark . and Remark ..
The main tool we will use is the following fixed point theorem in a cone. See, for example [] and [].
Theorem . Let E be a Banach space and K a cone in E.Assume that andare
bounded open subsets of E such that ∈, ⊂,and A:K∩(\)→K is a
completely continuous operator such that either
(ii) Au ≥ uforu∈K∩∂andAu ≤ uforu∈K∩∂.
Then A has a fixed point in K∩(\).
2 Preliminaries
First, let us consider the following linear problem:
u(t– ) =y(t), t∈[,T– ]
Z,
u() =u(T) =u(η) = . (.)
Define the Green’s functionG(t,s) as follows. If (t,s)∈[,T]Z×[η+ ,T– ]Z, then
G(t,s) =
–(T–s)(T–s–)
, ≤t– <s≤T– , (t–T)(t+T––s)
, η<s≤t– ≤T– .
(.)
If (t,s)∈[,T]Z×[,η]Z, then
G(t,s) =
t–t–s–s+sT
, ≤t– <s≤η,
s(T–t), ≤s≤t– ≤T– . (.)
Meanwhile,
G(,s) =G(,s) =
–(T–s)(T–s–)
, η<s≤T– , sT–s–s
, ≤s≤η.
(.)
Ifη=T– , then the Green’s functionG(t,s) is defined only by (.) and (.).
In the rest of this paper, we always suppose that for two integersa,bwitha<band a functionf defined on [a,b]Z,at=bf(t) = .
Lemma . The problem(.)has a unique solution
u(t) =
T–
s=
G(t,s)y(s), (.)
where G(t,s)is defined in(.), (.),and(.).
Proof Summing froms= tos=t– at both sides of the equation in (.), then we get
u(t– ) =u() +
t–
s=
y(s).
Repeating the above process, we obtain
u(t– ) = (t– )u() +
t–
s=
(t–s– )y(s).
Summing froms= tos=tat both sides of the above equation, we have
u(t) =u() +t(t– )
u() +
t–
s=
By using the boundary conditionu() =u(T) =u(η) = , we get
u() +ηs=y(s) = ,
u() =ηs=T(T–)y(s) –Ts=–(T–s)(T–s–)y(s).
Therefore,
u(t) = η
s=
T(T– ) –t(t– ) y(s) –
T–
s=
(T–s)(T–s– ) y(s)
+
t–
s=
(t–s)(t–s– )
y(s). (.)
This implies that (.) holds.
Lemma . Suppose that(H)holds.Then the Green’s function G(t,s)has the following
properties:
(i) Ifs∈[,η]Z,thenG(t,s)is nonincreasing with respect tot∈[,T]Z. Ifs∈[η+ ,T– ]Z,thenG(t,s)is nondecreasing with respect tot∈[,T]Z.
(ii) G(t,s)changes its sign on[,T]Z×[,T– ]Z.In details,if(t,s)∈[,T]Z×[,η]Z,
thenG(t,s)≥.If(t,s)∈[,T]Z×[η+ ,T– ]Z,thenG(t,s)≤. (iii) Ifs>η,thenmaxt∈[,T]ZG(t,s) =G(T,s) = and
min
t∈[,T]ZG(t,s) =G(,s) =
–(T–s)(T–s– )
≥
–(T–η)(T–η– )
. (.)
Ifs≤η,thenmint∈[,T]ZG(t,s) =G(T,s) = and
max
t∈[,T]Z
G(t,s) =G(,s) = –s
–s+ sT
≤
–η–η+ ηT
. (.)
Proof (i) As we know, iftG(t,s)≥ (≤), thenG(t,s) is nondecreasing
(nonincreas-ing) with respect tot. Now, we discuss the sign oftG(t,s). From (.), we know that
tG(,s) = whenevers≤ηors>η. Moreover, by (.) and (.),
G(,s) =
sT–s–s–
, s≤η, –(T–s)(T–s–)
, s>η.
ThentG(,s) = – ifs≤ηandtG(,s) = ifs>η. Ift≥, then the proof will be divided
into two cases.
Case . (t,s)∈[,T– ]Z×[η+ ,T– ]Z. Ifs>t– , then by (.),tG(t,s) = . If
s≤t– , thentG(t,s) =t–s> . Consequently,G(t,s) is nondecreasing with respect to
t∈[,T]Z.
Case. (t,s)∈[,T– ]Z×[,η]Z. Ifs>t– , thentG(t,s) = –t< . Ifs≤t– , then
tG(t,s) = –s< . Therefore,G(t,s) is nonincreasing with respect tot∈[,T]Z.
(iii) Ifs>η, then, from (i), we knowG(t,s) is nondecreasing with respect tot, then we havemaxt∈[,T]ZG(t,s) =G(T,s) = and
min
t∈[,T]ZG(t,s) =G(,s) =
–(T–s)(T–s– )
.
Furthermore, the functionz(s) = –(T–s)(T–s– ) is increasing fors< (T– )/. Com-bining this withη≤T– and (T– )/ >s>T– , we get the inequality in (.).
Meanwhile, ifs≤η,G(t,s) is nonincreasing with respect tot. Thenmint∈[,T]G(t,s) =
G(T,s) = and
max
t∈[,T]Z
G(t,s) =G(,s) = –s
–s+ sT
.
Moreover, sinceηsatisfies (H) and (T– )/ <s<η, we obtain thatG(,s)≤–η
–η+ηT
.
Remark . Ifη=T– , then we will find thatG(t,s)≥ andG(t,s)≡. This case has been discussed by several authors; see, for instance, [–]. So, in the rest of this paper, we could suppose thatη<T– .
Remark . Before we consider the existence of positive solutions of (.), we may discuss
the existence of positive solution of a more special problem
u(t– ) = , t∈[,T– ]
Z,
u() =u(T) =u(η) = . (.)
We will see that (H) is a necessary and sufficient condition for the existence of positive solutions to (.). To some extent, this explains why we chooseηwhich satisfies (H).
From Lemma ., we know that (.) has a solutionu(t) as follows:
u(t) =t
– ( +η)t+ (η+ )t–T+ ( +η)T– (η+ )T
.
For the sake of convenience, let
φ(t) =t– ( +η)t+ (η+ )t–T+ ( +η)T– (η+ )T.
Obviously,u(t)≥⇔φ(t)≥. By direct computation, we getφ(t) =t(t– – η),
φ(t)≥ fort> + ηandφ(t)≤ for <t≤ + η. Furthermore, if + ηis an integer, thenφ( + η) = . Now, we prove that (H) is a necessary and sufficient condi-tion ofφ(t)≥,t∈[,T]Z. In fact, ifφ(t)≥, thenφ(T) = implies thatφ(T– )≥. If
φ(T–) > , thenφ(T–) < . This implies thatη>T–
. Ifφ(T–) = , thenφ(T– ) > . Otherwise,φ(t)≡,t∈[,T]Z. This contradictsφ(t)≡. Therefore,φ(T– ) < and
LetE={u: [,T]Z→R|u() =u(T) =u(η) = }. ThenE is a Banach space under (.),satisfies the following inequality:
Proof By Lemma .,u(t) is concave fort∈[,η+ ]Z. Then
u(t) –u()
t ≥
u(η+ ) –u()
η+ , t∈[,η+ ]Z. (.)
From Lemma ., we see thatu(t) is nonincreasing fort∈[,T]Z, which implies thatu() =
u. Combining this with (.), we obtain
u(t)≥η+ –t
η+ u() =
η+ –t
η+ u, t∈[,η+ ]Z. This implies
min
t∈[T–θ,θ]Zu(t) =u(
θ)≥η+ –θ
η+ u=θ
∗u.
3 Existence results
In this section, we are concerned with the existence of at least one positive solution of the problem (.). Assume that
(H) f : [,T– ]Z×[,∞)→[,∞)is continuous, the mappingt→f(t,u)is decreasing for eachu∈[,∞)and the mappingu→f(t,u)is increasing for eacht∈[,T– ]Z;
(H) a: [,T– ]Z→[,∞)is decreasing. Define the coneKby
K= u∈P min
t∈[T–θ,θ]Zu(t)≥θ
∗u.
Define the operatorTλ:K→Eby
Tλu(t) =λ
T–
s=
G(t,s)a(s)fs,u(s).
From Lemma . to Lemma ., we know thatTλ:K→K. Meanwhile, sinceEis finite dimensional,Tλ:K→Kis completely continuous. Therefore, ifuis a fixed point ofTλin
K, thenuis positive solution of (.).
Since –η–η+ ηT> ,G(θ,s)≥, andG(θ,s)≡ fors∈[T–θ,θ]Z, we could define two positive constants as follows:
A=
T–
s=
–η–η+ ηT
a(s), B= θ
s=T–θ
G(θ,s)a(s).
Lemma . Suppose that(H), (H),and(H)hold.If there exist two positive constants r
and R with r=R such that
(A) f(t,u)≤λrA,(t,u)∈[,T– ]Z×[,r];
(A) f(t,u)≥λRB,(t,u)∈[,T– ]Z×[θ∗R,R].
This combines with the fact thatTλu(t)≥ fort∈[,T]Z, we have
Tλu ≥ u, foru∈K∩∂. (.)
Applying Theorem .(i) to (.) and (.), we getTλhas a fixed pointu∈K∩(\), and thenuis a positive solution of (.) withr≤ u ≤R.
Theorem . Suppose that(H), (H),and(H)hold.Assume that
(A) f:= lim
u→+t∈[,maxT–] Z
f(t,u)
u = ,
f∞:= lim
u→∞t∈[,minT–]Z f(t,u)
u =∞ (super linear case)
or
(A) f:= lim
u→+t∈[,minT–]Z f(t,u)
u =∞,
f∞:= lim
u→∞t∈[,maxT–]Z f(t,u)
u = (sublinear case).
Then for anyλ∈(,∞), (.)has at least one positive solution.
Proof Super linear case. From (A),f= , then there exists a constantR> such that
f(t,u)≤ R
λA, (t,u)∈[,T– ]Z×[,R].
Furthermore, sincef∞=∞, there existR>Rsuch that
f(t,u)≥ u
θ∗λB≥ θ∗R
θ∗λB= R
λB, (t,u)∈[,T– ]Z×
θ∗R,R
.
Now, by Lemma ., (.) has a positive solutionu∈K.
Sublinear case. Sincef=∞, there existsR> such that
f(t,u)≥ u
θ∗λB, (t,u)∈[,T– ]Z×
,R. (.)
Set={u∈E:u<R}. Then, ifu∈K∩∂,
min
s∈[T–θ,θ]Zu(s)≥θ
∗u=θ∗R
. (.)
Then, by (.)-(.) and the fact thatG(θ,s)≥ fors∈[T–θ,θ]Z, we get
Tλu(θ) =λ
T–
s=
G(θ,s)a(s)fs,u(s)≥λ
θ
s=T–θ
G(θ,s)a(s) u(s)
θ∗λB=R
.
This implies that
On the other hand, sincef∞= , there existsR> such that
f(t,u)≤ u
λA, (t,u)∈[,T– ]Z×[R,∞). (.)
We consider two cases: f is bounded andf is unbounded. Iff is bounded,i.e., there exists a constantM> , such thatf≤M, then we takeR=max{R,λMA}and={u∈
K:u<R}. Iffis unbounded, then we takeR>max{R,R}such thatf(t,u)≤f(t,R), for (t,u)∈[,T – ]Z×[,R] and={u∈K:u<R}. Now, similar to the proof of (.), we will get
Tλu ≤ u, foru∈K∩∂. (.)
Therefore, by Theorem ., we obtain a positive solutionuof problem (.).
Theorem . Suppose that(H), (H),and(H)hold.If <Af<θ∗Bf∞<∞,then for
eachλ∈( θ∗Bf∞,
Af), (.)has at least one positive solution.
Proof For anyλ∈(θ∗Bf∞,Af), there existsε> such that
θ∗B(f∞–ε)≤λ≤
A(f+ε). (.)
By the definition off, there existsR
> such thatf(t,u)≤(f+ε)u, for (t,u)∈[,T– ]Z×[,R]. Let={u∈K:u<R}, then similar to the proof of (.), ifu∈K∩∂, then
Tλu ≤λ
–η–η+ ηT
T–
s=
a(s)f+εu.
Furthermore, by (.), we get
Tλu ≤ u, foru∈K∩∂. (.)
On the other hand, by the definition off∞, there existsRsuch thatf(t,u)≥(f∞–ε)u, for
(t,u)∈[,T– ]Z×[R,∞). LetR=max{R,Rθ∗}and={u∈K:u<R}. Ifu∈K withu=R, thenmins∈[T–θ,θ]Zu(s)≥θ∗u ≥R. Therefore, similar to the discussion from (.)-(.), we get
Tλu(θ)≥λ θ
s=T–θ
G(θ,s)(f∞–ε)θ∗ua(s) =λθ∗B(f∞–ε).
Combining this with (.), we get
Tλu ≥ u, foru∈K∩∂. (.)
By Theorem .(i), (.) has at least one positive solutionu∈K.
Theorem . Suppose that(H), (H),and(H)hold.If <Af∞<θ∗Bf<∞,then for
eachλ∈(θ∗Bf,Af∞), (.)has at least one positive solution.
Theorem . Suppose that(H), (H),and(H)hold.Then the following results hold. () Iff∞=∞, <f<∞then for eachλ∈(,Af), (.)has at least one positive
solution.
() Iff=∞, <f∞<∞then for eachλ∈(,Af∞), (.)has at least one positive
solution. () Iff=∞, <f
∞<∞then for eachλ∈(θ∗Bf∞,∞), (.)has at least one positive
solution.
() Iff∞= , <f<∞then for eachλ∈(θ∗Bf,∞), (.)has at least one positive
solution.
Remark . Theηwe choose in this paper is optimal for obtaining the existence of pos-itive solution. Otherwise, suppose thatη∈[,T–– ]Zfor evenT andη∈[, [T–]]Zfor oddT. We will see that even the linear problem (.) will not have a positive solution. In fact, by the analysis in Remark ., we know that ifη<T–, thenφ(T– ) > . Com-bining this with the boundary conditionφ(T) = , we getφ(T– ) < . Therefore,φ(t) is not a positive solution of (.). Therefore, (H) is an optimal condition for obtaining the existence of a positive solution.
Example . Consider the following discrete nonlinear third-order three-point eigen-value problem:
u(t– ) =λa(t)f(t,u(t)), t∈[, ]Z,
u() =u() =u() = , (.)
where a(t) =t– t+ ,f satisfies the assumption (H
), according to the hypotheses in Theorem .-Theorem ., we will givef(t,u(t)) the specific forms. Then finding the positive solutions of the problem (.) is equivalent to finding the fixed point the operator equation
Tu(t) :=λ
s=
G(t,s)a(s)fs,u(s).
Sinces∈[, ]Z, ifs> thens= . Furthermore, we get the expression ofG(t,s) as follows.
(i) (t,s)∈[, ]Z× {}. Sinces= , we know that ifs≤t– , thent= and
G(t,s) =G(, ) = in the case thats≤t– . Therefore, the expression ofG(t,s)is
G(t,s) =
, s≤t– , –, s>t– .
(ii) (t,s)∈[, ]Z×[, ]Z. Then
G(t,s) =
s( –t), s≤t– ,
t–t–s+s
(iii) (.) has at least one positive solution.
Iff(t,s) = ( –t)h(s), where satisfies the condition in Theorem .. So, we omit this here.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgements
The authors are very appreciated to the reviewer for his/her valuable suggestions. Supported by NSFC (11401479, 11301059), Natural Science Foundation of Gansu Province (145RJYA237), China Postdoctoral Science Foundation (2014M562472) and Postdoctoral Science Foundation of Gansu Province.
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