Existence of positive solutions of discrete third order three point BVP with sign changing Green’s function

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R E S E A R C H

Open Access

Existence of positive solutions of discrete

third-order three-point BVP with

sign-changing Green’s function

Youji Xu

1*

_{, Weiwei Tian}

1

_{and Chenghua Gao}

1

*_{Correspondence:}

[email protected]

1_{College of Mathematics and}

Statistics, Northwest Normal University, Lanzhou, P.R. China

Abstract

Consider positive solutions and multiple positive solutions for a discrete nonlinear third-order boundary value problem

3_u(t_{– 1) =}_a(t)f(t,_u(t)), _t_∈_[1,_T_{– 2]}

Z,

u(0) =

u(T) = 0,

2_{u(η) –}

_αu(T

_{– 1) = 0,}

which has the sign-changing Green’s function. HereT> 8 is a positive integer, [1,T– 1]Z={1, 2,. . .,T– 2},

α

∈[0,_T–11 ),a: [0,T– 2]Z→(0, +∞),

f: [1,T– 2]Z×[0,∞)→[0,∞) is continuous.

Keywords: Discrete third-order three-point boundary value problem; Positive solutions; Cone; Fixed point; Sign-changing Green’s function

1 Introduction

Multi-point boundary value problems for differential equations have a wide application in computational physics, economics, and modern biological fields [1]. In 1992, Gupta [2] studied solvability of differential equation three-point boundary value problem. Soon afterwards, there arose many results on multi-point nonlinear boundary value problems [3–6]. In 1999, Ma [7] studied the existence of positive solution for a second-order dif-ferential equation three-point boundary value problem. Thereafter, many results for the existence of positive solutions on multi-point boundary value problems have been stud-ied [7–19]. With the development of the computing science and the computer simulation, multi-point boundary value problems should be discretized, so we need to study corre-sponding difference equation [20–26]. In 1998, by using Krasnosel’skii’s fixed point theo-rem, Agarwal and Henderson [24] studied the discrete problem

⎧ ⎨ ⎩

3u(t– 1) =λa(t)f(t,u(t)), t∈[2,T]Z,

u(0) =u(1) =u(T+ 3) = 0.

They obtained the existence of positive solutions in two cases forλ= 1 andλ= 1. Later, there were many interesting results on the positive solutions to the discrete boundary

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value problems, see, for instance, [23–26] and the references therein. It is noted that Green’s functions are positive in most of these results. However, when the Green’s function is sign-changing, could we also obtain the existence of positive solutions to these kinds of problems?

In 2015, by using the Guo–Krasnosel’skii ﬁxed point theorem, Wang and Gao [25] stud-ied the existence of positive solutions to the discrete third-order three-point boundary value problem

⎧ ⎨ ⎩

3u(t– 1) +a(t)f(t,u(t)) = 0, t∈[1,T– 1]Z,

u(0) =u(T) =2u(η) = 0,

whereη∈[1, [3T₆2–3_T₊₃T–2]]Zand the Green’s function is sign-changing. In 2016, Geng and

Gao [26], by using the Guo–Krasnosel’skii ﬁxed point theorem, studied the discrete third-order three-point boundary value problem

⎧ ⎨ ⎩

3u(t– 1) +λa(t)f(t,u(t)) = 0, t∈[1,T– 1]Z,

u(0) =u(t) =2u(η) = 0,

whenfsatisﬁes some superlinear and sublinear condition on 0 and∞. For the continuous case, which has the sign-changing Green’s function, one can see [27–29] and the references therein. Inspired by the above works, in this paper, we consider the existence and multiple positive solutions to the following discrete nonlinear third-order three-point BVP:

⎧ ⎨ ⎩

3_u₍_t_{– 1) =}_a₍_t₎_f₍_t_,_u₍_t_)), _t_∈_[1,_T_{– 2]}

Z,

u(0) =u(T) = 0, 2u(η) –αu(T– 1) = 0, (1.1) whereT> 8 is a positive integer,α∈[0,_T1_–1),a: [1,T – 2]Z→(0, +∞),f : [1,T– 2]Z×

[0,∞)→[0,∞) is continuous,ηsatisﬁes (H0) η∈[T–2₂ + 1,T– 2]Z.

Under assumption (H0), the Green’s function of (1.1) changes its sign. The proof of our

main results is based upon the following well-known Guo–Krasnoselskii ﬁxed point the-orems [30].

Theorem 1.1 Let E be a Banach space and K⊂E be a cone.Assume thatΩ1,Ω2are open bounded subsets of E with0∈Ω1,Ω1⊂Ω2.If

A:K∩(Ω2\Ω1)→K

is a completely continuous operator such that

(i) Au ≤ u,u∈K∩∂Ω1,andAu ≥ u,u∈K∩∂Ω2, or

(ii) Au ≥ u,u∈K∩∂Ω1,andAu ≤ u,u∈K∩∂Ω2, then A has a ﬁxed point in K∩(Ω2\Ω1).

Theorem 1.2 Let E be a Banach space and K be a cone in E.For some p> 0,deﬁne Kp={x∈

K|x ≤p}.Assume that A:Kp→K is a compact operator;if x∈∂Kp={x∈k|x=p},

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(i) For∀x∈∂Kp,ifAx ≥ x,theni(A,Kp,K) = 0,

(ii) For∀x∈∂Kp,ifAx ≤ x,theni(A,Kp,K) = 1.

2 Preliminaries

Lemma 2.1 Suppose that(H0)holds.Then the linear problem

⎧ ⎨ ⎩

3u(t– 1) =y(t), t∈[1,T– 2]Z,

u(0) =u(T) = 0, 2u(η) –αu(T– 1) = 0 (2.1)

has a unique solution

u(t) =

T–2

s=1

G(t,s)y(s), (2.2)

where

G(t,s) =g(t,s) +k(t,s) +

⎧ ⎨ ⎩

T(T–1)–t(t–1)

2–2α(T–1) , s≤η,

0, η<s,

g(t,s) = –α(T–s– 1)[T(T– 1) –t(t– 1)] 2 – 2α(T– 1) –

(T–s)(T–s– 1) 2 ,

k(t,s) =

⎧ ⎨ ⎩

(t–s)(t–s–1)

2 , 0 <s≤t– 2≤T– 2,

0, 0≤t– 2 <s≤T– 2,

and

G(0,s) =G(1,s) =

⎧ ⎨ ⎩

–αT(T–1)(T–s–1)+T(T–1) 2–2α(T–1) –

(T–1)(T–s–1)

2 , s≤η,

–αT(T–1)(T–s–1) 2–2α(T–1) –

(T–1)(T–s–1)

2 , η<s.

Proof Summing both sides of equation (2.1) froms= 1 tos=t– 1, we get

2u(t– 1) =2u(0) +

t–1

s=1 y(s),

and

u(t– 1) = (t– 1)2u(0) +

t–2

s=1

(t–s– 1)y(s).

Then summing both sides fromτ= 1 toτ=t, we get

u(t) =u(0) +t(t– 1) 2

2_u_{(0) +}

t–2

s=1

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From boundary conditionsu(0) =u(T) = 0,2_u₍_η_{) –}_α_u₍_T_{– 1) = 0, we have}

⎧ ⎨ ⎩

u(0) +T(T₂–1)2u(0) +_sT₌₁–2(T–s)(T₂–s–1)y(s) = 0;

2_u_{(0) +}η

s=1y(s) –α(T– 1)2u(0) –α

T–2

s=1(T–s– 1)y(s) = 0.

Furthermore, we get

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

u(0) =T_s₌₁–2–αT_2–2(T–1)(_α₍_TT_–1)–s–1)y(s) +η_s₌₁_2–2T(_αT₍–1)_T_–1)y(s) –T_s₌₁–2(T–s)(T₂–s–1)y(s),

2u(0) =T_s₌₁–2_1–α(T_α–₍_Ts–1)_–1)y(s) –η_s=11–α(1T–1)y(s).

Then we have

u(t) =

T–2

s=1

–α(T–s– 1)[T(T– 1) –t(t– 1)] 2 – 2α(T– 1) –

(T–s)(T–s– 1) 2

y(s)

+

t–2

s=1

(t–s)(t–s– 1) 2 y(s)

+

η

s=1

T(T– 1) –t(t– 1)

2 – 2α(T– 1) y(s).

Lemma 2.2 Suppose that(H0)holds.Then the Green’s function G(t,s)changes its sign on

[0,T]Z×[0,T]Z.More precisely,

(i) Ifs∈[1,η]Z,thentG(t,s)≤0,G(t,s)≥0for∀t∈[0,T]Z.Furthermore,

min

t∈[0,T]ZG(t,s) =G(T,s) = 0,

max

t∈[0,T]_ZG(t,s) =G(0,s)≤

T(T– 1)(1 +αη) 1 –α(T– 1) .

(ii) Ifs∈[η+ 1,T– 2]Z,thentG(t,s)≥0,G(t,s)≤0for∀t∈[0,T]Z.Furthermore,

max

t∈[0,T]ZG(t,s) =G(T,s) = 0,

min

t∈[0,T]ZG(t,s) =G(0,s)≥

–T(T– 1)(1 +αη) 1 –α(T– 1) .

Proof (i) Ifs∈[1,η]Z, we have

tG(t,s) =

⎧ ⎨ ⎩

αt(T–s–1)–2t

1–α(T–1) , t– 2≤s, αt(T–s–1)–2t

1–α(T–1) +t–s, s≤t– 2.

Ifs≤t– 2, sinceα(T– 1) – 1 < 0, we get

tG(t,s) =

2αt(T–s– 1) – 2t

2 – 2α(T– 1) +t–s

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=–2sα+ 2s[α(T– 1) – 1] 2 – 2α(T– 1) < 0.

Ift– 2≤s, sinceα<_T1_–1, so 2 – 2α(T– 1) > 0,α(T–s– 1) – 1 < 0, we have

tG(t,s) =

2αt(T–s– 1) – 2t

2 – 2α(T– 1) < 0.

For∀t∈[0,T– 1]Z,tG(t,s)≤0. Ifs∈[1,η]Z,G(t,s)≥0. So min

t∈[0,T]ZG(t,s) =G(T,s) = 0,

max

t∈[0,T]ZG(t,s) =G(0,s)

=T(T– 1)[1 –α(T–s– 1)] 2 – 2α(T– 1) –

(T–s)(T–s– 1) 2

≤T(T– 1)[1 –α(T–η– 1)]

2 – 2α(T– 1) –

(T–η)(T–η– 1) 2

≤T(T– 1)(1 +αη)

1 –α(T– 1) . (ii) Ifs∈[η+ 1,T]Z, we have

tG(t,s) =

⎧ ⎨ ⎩

2αt(T–s–1)

2–2α(T–1), s≥t– 2, 2αt(T–s–1)

2–2α(T–1)+t–s, s≤t– 2.

Ift– 2≤s, sinceα<_T1_–1, 2 – 2α(T– 1) > 0, we have

tG(t,s) =

2αt(T–s– 1) 2 – 2α(T– 1) > 0. Ifs≤t– 2, thent–s> 0, so

tG(t,s) =

2αt(T–s– 1)

2 – 2α(T– 1) +t–s> 0. Ifs∈[η+ 1,T]Z, for∀t∈[0,T]Z, we have

max

t∈[0,T]ZG(t,s) =G(T,s) = 0,

min

t∈[0,T]_ZG(t,s) =G(0,s)

=–αT(T– 1)(T–s– 1) 2 – 2α(T– 1) –

(T–s)(T–s– 1) 2

≥–αT(T– 1)(T–η– 1)

2 – 2α(T– 1) –

(T–η)(T–η– 1) 2

≥–T(T– 1)(1 +αη)

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In conclusion, ifs∈[1,η]Z,G(t,s) is decreasing, thenG(t,s)≥0, sincemint∈[0,T]ZG(t,s) =

0; ifs∈[η+ 1,T]Z,G(t,s) is increasing, thenG(t,s)≤0 sincemaxt∈[0,T]ZG(t,s) = 0. So, the

Green’s functionG(t,s) is a sign-changing function on [0,T]Z×[0,T]Z.

Remark Now, we give a brief explanation for the reason why we choose

η∈

T– 2 2

+ 1,T– 2

Z

. (2.3)

Consider the problem

⎧ ⎨ ⎩

3u(t– 1) = 1, t∈[1,T– 2]Z,

u(0) =u(T) = 0, 2u(η) –αu(T– 1) = 0.

(2.4)

From Lemma2.1, we get

u(t) = 1

12 – 12α(T– 1)

3αt(t– 1) –T(T– 1)(T– 1)(T– 2) + 6ηT(T– 1) –t(t– 1)– 2T(T– 1)(T– 2)1 –α(T– 1) + 2t(t– 1)(t– 2)1 –α(T– 1)

= φ(t) 12 – 12α(T– 1), where

φ(t) = 3αt(t– 1) –T(T– 1)(T– 1)(T– 2) + 6ηT(T– 1) –t(t– 1) – 2T(T– 1)(T– 2)1 –α(T– 1)+ 2t(t– 1)(t– 2)1 –α(T– 1).

Clearly,u(t)≥0 is equivalent toφ(t)≥0, and

φ(t) = 6t(t– 1)1 –α(T– 1)– 2η+α(T– 1)(T– 2).

Letφ(t) = 0. Thent= 0 ort= 1 +2η–_1–α(T_α₍–1)(_T_–1)T–2). Therefore,

φ(t) > 0 ⇔ t>2η+ 1 –α(T– 1)

2

1 –α(T– 1) .

Now, we claim that if (2.3) holds, thenφ(t) is a positive solution of (2.4). In fact, if (2.3) holds, then 2η+1–_1–_αα₍_T(T_–1)–1)2 ≥T– 1 in this case, this implies that

φ(t)≤0, t∈[0,T– 1]Z.

More precisely,φ(0) = 0,φ(t) < 0 fort∈[1,T– 2]Zandφ(T– 1) < 0 for2η+1–α(T–1)

2

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(8)

≤y(η)

Combining this with the monotonicity ofy, we get that

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= y(η)

Combining this with the fact thaty∈K0and the monotonicity ofy, we get that

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Proof From Lemma2.3,u(t) is concave on [0,η+ 2]Z. So,u(t) satisﬁes

u(t) –u(0)

t ≥

u(η+ 2) –u(0)

η+ 2 , t∈[0,η+ 2]Z.

Meanwhile, from Lemma2.3,u(t) is non-increasing on [0,T]Z, which implies thatu(0) =

u. Therefore,

u(t)≥η+ 2 –t

η+ 2 u(0) =

η+ 2 –t

η+ 2 u,

min

t∈[T–θ,θ]Zu(t) =u(θ)≥

η+ 2 –θ η+ 2 u=θ

∗_u_.

3 Main results

(H1) f : [1,T– 2]Z×[0,∞)→[0,∞)is continuous. Foru∈[0,∞),f(t,u)is a decreasing function with respect tot, and fort∈[1,T– 2]Z,f(t,u)is an increasing function with respect tou.

(H2) a: [1,T– 2]Z→[0,∞)is a decreasing function.

Let

K=

u∈K0| min

t∈[T–θ,θ]_Zu(t)≥θ

∗_u_.

ThenKis a cone inE. Deﬁne the operatorS:K→Eas

Su(t) =

T–2

s=1

G(t,s)a(s)fs,u(s). (3.1)

Lemma 3.1 S:K→K is completely continuous.

Proof It is obvious thatS:K →E is completely continuous since the Banach spaceE

is ﬁnite dimensional. Now, let us prove that S:K →K, that is to say, for anyu∈K,

Su∈K.

Letu∈K. Thenu∈K0, which implies thatu(t)≤0 anduis decreasing ont. Therefore, by (H1),f(t,u(t)) is a decreasing function oft. Lety(t) :=a(t)f(t,u(t)). Then, from (H1)

and (H2), we obtain thaty(t)≥0 andyis also a decreasing function oft. Thus,y∈K0. Furthermore, by (3.1), we know that

3(Su)(t– 1) =y(t), t∈[1,T– 2]Z, (3.2)

and

(Su)(0) = (Su)(T) = 0, 2(Su)(η) –α(Su)(T– 1) = 0. (3.3)

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Lemma2.4and the factSu∈K0, we know that

min

t∈[T–θ,θ]_Z(Su)(t)≥θ

∗_Su_.

Therefore,Su∈KandS:K→Kis completely continuous. From (3.1) and Lemma3.1, we know that ifuis a ﬁxed point ofSinK, thenuis a positive solution of (1.1). In the rest of this paper, we try to proveShas at least one or two ﬁxed point(s) inKby using Theorem1.1and Theorem1.2.

Let

A=

T–2

s=1

T(T– 1)(1 +αη)

1 –α(T– 1) a(s), B=

θ

s=T–θ

θ(T–θ)[2 –α(θ– 1)] 2 – 2α(T– 1) a(s).

Theorem 3.1 Suppose that(H0), (H1),and(H2)hold.If there exist two constants r and R

(r=R)such that

(A1) f(t,u)≤_Ar,(t,u)∈[1,T– 2]Z×[0,r];

(A2) f(t,u)≥R_B,(t,u)∈[1,T– 2]Z×[θ∗R,R],

then problem (1.1) has at least one positive solution u ∈ K with min{r,R} ≤ u ≤

max{r,R}.

Proof Without loss of generality, suppose thatr<R, the other case could be treated sim-ilarly. LetΩ1={u∈E:u<r}. From Lemma 2.2, G(t,s)≤0 for s∈[η+ 1,T – 2]Z;

G(t,s)≥0 fors∈[1,η]Z. Since (A1), we get, for∀u∈K∩∂Ω1,

Su= max

t∈[0,T]Z

T–2

s=1

G(t,s)a(s)fs,u(s)

≤ max

t∈[0,T]Z T–2

s=1

G(t,s)a(s)fs,u(s)

≤

T–2

s=1

T(T– 1)(1 +αη) 1 –α(T– 1) a(s)f

s,u(s)

≤

T–2

s=1

T(T– 1)(1 +αη) 1 –α(T– 1) a(s)

r A

=r.

So, foru∈K∩∂Ω1,

Su ≤ u. (3.4)

LetΩ2={u∈E:u<R}. Then, foru∈K∩∂Ω2,

Su(T–θ) =

T–2

s=1

G(T–θ,s)a(s)fs,u(s)≥

T–θ

s=θ

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Then

Therefore, (3.5) holds. This implies that

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Then, by Theorem1.1,Shas at least one ﬁxed pointu∈Kanduwill be a positive solution

of problem (1.1).

Theorem 3.2 Suppose that(H0), (H1),and(H2)hold.If one of the following conditions holds:

(B1) f0:= lim

u→0+_t_∈_[1,max_T_–2]

Z

f(t,u)

u = 0, f∞:=ulim→∞t∈[1,maxT–2]_Z f(t,u)

u =∞, or

(B2) f0:= lim

u→0+_t_∈_[1,max_T_–2]

Z

f(t,u)

u =∞, f

∞_:= _lim

u→∞t∈[1,maxT–2]Z

f(t,u)

u = 0, then(1.1)has at least one positive solution.

Proof Firstly, we prove the case that (B1) holds. Sincef0_{= 0, there exists a constant}_R1_{> 0}

such that

f(t,u)

u ≤

R1

A, (t,u)∈[1,T– 2]Z×[0,R1].

Sincef∞=∞, there exists a constantR2>R1such that

f(t,u)≥ u

θ∗B≥

θ∗R2

θ∗B = R2

B, (t,u)∈[1,T– 2]Z×

θ∗R2,R2. From Theorem3.1, problem (1.1) has at least one positive solutionu∈K.

Secondly, suppose that (B2) holds. Sincef0=∞, there exists a constantr1> 0 such that

f(t,u)≥ u

θ∗B, (t,u)∈[1,T– 2]Z×[0,r1].

LetΩ1={u∈E:u<r1}. Ifu∈K∩∂Ω1, we have

min

s∈[T–θ,θ]Zu(s)≥θ

∗_u₌_θ∗_r

1.

Therefore, similar to the proof of (3.6), we have

Su ≥ u foru∈K∩∂Ω1.

On the other hand, sincef∞= 0, there exists a constantr2> 0 such that

f(t,u)≤ u

A, (t,u)∈[1,T– 2]Z×[r2,∞).

If f is bounded, then there exists a constant N > 0 such that f ≤N. So, we choose

R=max{2r1,NA}. Iff is unbounded, then letR>max{2r1,r2}such thatf(t,u)≤f(t,R2).

LetΩ2={u∈K:u<R}for∀(t,u)∈[1,T– 2]Z×[0,R2]. Similar to the proof of

Theo-rem3.1, we get, for∀u∈k∩∂Ω2,

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Thus, by Theorem1.1,Shas at least one ﬁxed pointu∈K∩Ω2\Ω1, which is a positive

solution of problem (1.1).

Theorem 3.3 Assume that(H0), (H1),and(H2)hold.If

(C1) f0:=limu→0+min_t_∈_[1,_T_–2]_Z=f(t,u)

u = +∞,f∞:=limu→∞mint∈[1,T–2]Z= f(t,u)

u = +∞,

and

(C2) There exists a constantp> 0such thatf(t,u) <γpfor0≤u≤pandt∈[1,T– 2]Z,

where

γ =

T(T– 1)(1 +αη) 1 –α(T– 1)

T–2

s=1 a(s)

–1

,

then problem(1.1)has at least two positive solutions u1and u2with0≤ u1 ≤p≤ u2. Proof ChooseM> 0 such that

Mθ∗θ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1)

θ

s=T–θ

a(s)≥1.

Sincef0= +∞, there exists a constantrwith 0 <r<psuch thatf(t,u)≥Mufor 0≤u≤r. Then, for∀u∈∂Kr, we have

Su(T–θ) =

T–2

s=1

G(T–θ,s)a(s)fs,u(s)

≥ θ

s=T–θ

≥Mθ∗

θ

s=T–θ

G(T–θ,s)a(s)u

≥Mθ∗θ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1)

θ

s=T–θ a(s)u

≥ u.

From Theorem1.2, we get

i(S,Kr,K) = 0.

Sincef∞= +∞, there exists a constantR1> 0 such thatf(t,u)≥Mufor∀u≥R1. Choose R>max{p,R1

θ∗}, then for∀u∈∂KR, mint∈[T–θ,θ]Zu(t)≥θ∗u>R1. Similar to the above

proof, we have

Su ≥ u foru∈∂KR.

Therefore,

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From (C2), for∀u∈∂Kp,

Su= max

t∈[0,T]Z

T–2

s=1

G(t,s)a(s)fs,u(s)

≤ max

t∈[0,T]Z T–2

s=1

_G(t,s)a(s)fs,u(s)

≤

T–2

s=1

T(T– 1)(1 +αη) 1 –α(T– 1) a(s)f

s,u(s) =u.

Therefore, for∀u∈∂Kp,Tu ≤ u. By Theorem1.2,

i(S,Kp,K) = 1.

Thus,

i(S,KR\K˚p,K) = –1, i(S,Kp\K˚r,K) = 1.

So,Shas a ﬁxed pointu1inKp\K˚rand another ﬁxed pointu2 inKR\K˚p. So, problem

(1.1) has at least two positive solutionsu1andu2with 0≤ u1 ≤p≤ u2.

Theorem 3.4 Assume that(H0), (H1),and(H2)hold.If

(D1) f0:=limu→0+max_t_∈_[1,_T_–2]_Zf(t,u)

u = 0,f∞:=limu→∞maxt∈[1,T–2]Zf(tu,u)= 0,

and

(D2) there exists a constantp> 0such thatf(t,u) >βpforθ∗p≤u≤pandt∈[1,T– 2]Z,

where

β=

θ∗θ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1)

θ

s=T–θ a(s)

–1

,

then(1.1)has at least two positive solutions u1and u2with0≤ u1 ≤p≤ u2.

Proof From (D1), for∀> 0, there existsM1> 0, if u> 0,t∈[0,T]Z, we have f(t,u)≤

M1+u. Then, for∀u∈K,

Su= max

t∈[0,T]Z

T–2

s=1

G(t,s)a(s)fs,u(s)

≤T(T– 1)(1 +αη)

1 –α(T– 1)

T–2

s=1

a(s)(M1+u).

Choose> 0 suﬃciently small andR>psuﬃciently large, then for∀u∈∂KR,Su ≤ u,

from Theorem1.2, we have

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In a similar way, if 0 <r<p,

i(S,KR,K) = 1.

From (D2), for∀u∈∂Kp,

Su(T–θ) =

T–2

s=1

≥ θ

s=T–θ

> βθ∗p θ

s=T–θ

G(T–θ,s)a(s)

≥βθ∗pθ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1)

θ

s=T–θ a(s) =p.

Then, for∀u∈∂Kp,Su ≥ u. From Theorem1.2, we have

i(S,Kp,K) = 0.

Then, problem (1.1) has at least two solutionsu1andu2with 0≤ u1 ≤p≤ u2. 4 Example

Example4.1 Consider the discrete three-point boundary problem

⎧ ⎨ ⎩

3u(t– 1) –a(t)f(t,u(t)) = 0, t∈[1, 7]Z, u(0) =u(9) = 0, 2_u_{(7) –}1

9u(8) = 0,

(4.1)

wherea(t) =9–₁₀t, and

f(t,u) =

⎧ ⎨ ⎩

15 –t+₁₀₀₀u , (t,u)∈[1, 7]Z×[0, 1000],

3

√

u+ 6 –t, (t,u)∈[1, 7]Z×[1000,∞].

SinceT= 9 andα=1₉,η∈[4, 7]Z. Without loss of generality, letη= 4. Then, by the direct calculation, we getθ∈[5, 6]Z. Chooseθ= 5, thenθ∗=η+2–η+2θ=

1 6. So,

A=

T–2

s=1

T(T– 1)(1 +αη)

1 –α(T– 1) a(s) = 936

7

s=1

9 –s

10 = 3276,

B=

θ

s=T–θ

θ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1) a(s) = 140

5

s=4

9 –s

10 = 136.

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Example4.2 In this example, we continue to consider problem (4.1) with

f(t,u) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

u2(10–t)

10 , (t,u)∈[1, 7]Z×[0, 1],

3

√ u+9–t

10 , (t,u)∈[1, 7]Z×[1, 3],

3

√ u+1890–t

1000 , (t,u)∈[1, 7]Z×[3,∞).

Continue to takeα=1₉,η= 4,θ= 5, andθ∗=1₆. Then

β=

θ∗θ(T–θ)[2 –α(θ– 1)]

2 – 2α(T– 1)

θ

s=T–θ a(s)

–1

= 3 68.

Furthermore, if we choose p= 1, then forθ∗p≤u≤p, f(t,u)≥βp= 3

68. From

Theo-rem3.4, problem (4.1) has at least two positive solutionsu1andu2with 0≤ u1 ≤p≤ u2.

Acknowledgements

The authors express their gratitude to the referee for reading the paper carefully and making several corrections and remarks.

Funding

This work was supported by the National Natural Science Foundation of China (No. 71561024). The key projects of Natural Science Foundation of Gansu province of China (No. 18JR3RA084).

Availability of data and materials

The datasets used and analysed during the current study are available from the corresponding author on reasonable request.

Competing interests

The authors declare that they do not have any competing interests in this manuscript.

Consent for publication

The authors conﬁrm that the work described has not been published before (except in the form of an abstract or as part of a published lecture, review, or thesis), that its publication has been approved by all co-authors.

Authors’ contributions

All authors contributed equally to this manuscript. All authors read and approved the ﬁnal manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 30 January 2019 Accepted: 17 May 2019

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