# Multiple Positive Solutions of Nonlinear Boundary Value Problem for Finite Fractional Difference

Authors: Yansheng He; Mingzhe Sun; Chengmin Hou
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We consider a discrete fractional nonlinear boundary value problem in which nonlinear termf  is involved with the fractional order difference. And we transform the fractional boundary value problem into boundary value problem of integer order difference equation. By using a generalization of Leggett-Williams fixed-point theorem due to Avery and Peterson, we provide sufficient conditions for the existence of at least three positive solutions.


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## Body

## 1. Introduction

LetR and Z be the sets of real numbers and integers, respectively. For a,b∈R, N∈Z+ with b=a+N, define [a,b]Na={a,a+1,…,b}. Assume that T is a given positive integer with T>2. We consider the fractional difference boundary value problem (briefly FBVP) of the forms
(1)ΔφpΔν-1νxt+qtMM·ft+ν-1,xt+ν-1,Δt+ν-βν-1βxt=0,MMMMMMMMMMMMMMMMMM0t∈0,TN0,(2)Δν-1εxtt=T+1+ν-ε=0,x(ν-1)=ηΔν-1δxtt=ξ+1+ν-δ,
where η∈[0,1], ν,β,ε,δ∈(0,1) with β,δ≤ν≤ε, and φp(s)=|s|p-2s, p>1, ξ∈[0,T]N0, Δν-1ϱ(ϱ=ν,β,ε,δ) is fractional difference operator. We give the following assumptions:(H1)
f(t+ν-1,·,·):[ν-1,T+ν-1]Nν-1×[0,+∞)×R→[0,+∞) is continuous;(H2)
q(t) is nonnegative on [0,T]N0,q(t)≡0 does not hold on [0,T]N0, and ∑t=0T‍q(t)<∞.Fixed-point theorems and their applications in nonlinear problems have a long history, some of which are documented in Zeidler’s book [1]. There seems to be increasing interest in multiple fixed-point theorems and their applications to boundary value problems for ordinary differential equations or finite difference equations. The applications can be found in the papers [2–6]. An interest in triple solutions has evolved from the Leggett-Williams multiple fixed-point theorem [7]. And, lately, two triple fixed-point theorems by Avery and Peterson [4] and Avery [8] have been applied to obtain triple solutions of certain boundary value problems for ordinary differential equations as well as for their discrete analogues. On the other hand, fractional differential and difference “operators” are found themselves in concrete applications, and hence attention has to be paid to associated fractional difference and differential equations under various boundary or side conditions. For example, Atici and Eloe [9] explored some of the theories of a discrete conjugate FBVP in [9]. Similarly, in [10], a discrete right-focal FBVP was analyzed. Other recent advances in the theory of the discrete fractional calculus may be found in [11–26]. In particular, an interesting recent paper by Atici and Şengül [14] addressed the use of fractional difference equations in tumor growth modeling. Thus, it seems that there exists some promise in using fractional difference equations as mathematical models for describing physical problems in more accurate manners.In order to handle the existence problem for FBVP, various methods (among which are some standard fixed-point theorems) can be used. For example, in [10, 12, 27], authors investigated the existence to some boundary value problems by fixed-point theorems on a cone. In [28], we established the existence conditions for a boundary value problem by using the coincidence degree theory. In [29], authors pointed out the existence of multiple solutions for a FBVP with parameter by establishing the corresponding variational framework and using the mountain pass theorem, linking theorem, and Clark theorem in critical point theory. To the best of our knowledge, Leggett-Williams fixed-point theorem has not been used in discrete fractional boundary value problems. The aim of this paper is to establish the existence conditions for boundary value problem (1)-(2). The proof relies on the Leggett-Williams fixed-point theorem.Throughout this paper, we make the convention that∑i=jm‍x(i)=0 for m<j and denote φq=φp-1, 1/p+1/q=1.


## 2. Preliminaries

In this section, we collect some basic definitions and lemmas for manipulating discrete fractional operators. These and other related results can be found in [4, 12, 14, 16].For any integerβ, let Nβ=β,β+1,β+2,…. We define t(ν)∶=Γ(t+1)/Γt+1-ν, for any t and ν for which the right-hand side is defined. We also appeal to the convention that if t+1-ν is a pole of the Gamma function and t+1 is not a pole, then t(ν)=0.Definition 1.
Theνth fractional sum of f for ν>0 is defined by
(3)Δa-νft=1Γν∑s=at-ν‍t-s-1ν-1fs,
for t∈Na+ν. We also define the νth fractional difference for ν>0 by Δaνf(t)∶=ΔNΔaν-Nf(t), where t∈Na+N-ν, and N∈N is chosen so that 0≤N-1<ν≤N.Definition 2.
LetX be a real Banach space. A nonempty closed convex set P⊂X is called a cone of X if it satisfies the following conditions:(1)
x∈P, λ≥0 implies λx∈P;(2)
x∈P, -x∈P implies x=0.
Every coneP⊂X induces a partial ordering “≤” on X defined by x≤y if and only if y-x∈P.Definition 3.
Given a coneP in a real Banach space X, a functional ψ:P→R is said to be increasing on P, provided that ψ(x)≤ψ(y) for all x,y∈P with x≤y.Letγ and θ be nonnegative continuous convex functionals on P,α a nonnegative continuous concave functional on P, and ψ a nonnegative continuous functional on P. Then, for positive real numbers a,b,c, and d, we define the following convex sets:P(γ,d)={x∈P∣γ(x)<d},P(γ,α,b,d)={x∈P∣b≤α(x),γ(x)≤d},P(γ,θ,α,b,c,d)={x∈P∣b≤α(x),θ(x)≤c,γ(x)≤d} and a closed set R(γ,ψ,a,d)={x∈P∣a≤ψ(x),γ(x)≤d}.The following fixed-point theorem due to Avery and Peterson is fundamental in the proof of our main results.Lemma 4 (see [4]).
LetX be a Banach space and let P⊂X be a cone. Let γ and θ be nonnegative continuous convex functionals on P, α a nonnegative continuous concave functional on P, and ψ a nonnegative continuous functional on P satisfying ψ(λx)≤λψ(x) for 0≤λ≤1, such that, for some positive numbers M and d,α(x)≤ψ(x) and x≤Mγ(x) for all x∈P(γ,d)¯. Suppose Q:P(γ,d)¯→P(γ,d)¯ is completely continuous and there exist positive numbers a,b,c with a<b such that(S1)
{x∈P(γ,θ,α,b,c,d)∣α(x)>b}≠∅ and α(Qx)>b for x∈P(γ,θ,α,b,c,d);(S2)
α(Qx)>b for x∈P(γ,α,b,d) with θ(Qx)>c;(S3)
0∉R(γ,ψ,a,d) and ψ(Qx)<a for x∈R(γ,ψ,a,d) with ψ(x)=a.Then Q has at least three fixed points x1,x2,x3∈P(γ,d)¯ such thatγ(xi)≤d for i=1,2,3;b<α(x1), ψ(x3)<a, a<ψ(x2) with α(x2)<b.Lemma 5 (see [16]).
Lety:Na→R and ν>0 with N-1<ν≤N, and then
(4)Δaνyt=∑k=0t-a+ν‍-1kνkyt+ν-k,
for t∈Na+N-ν, where νk=Γ(ν+1)/Γ(k+1)Γ(ν-k+1).Lemma 6 (see [16]).
Letf:Na→R be given and suppose k∈N0 and ν>0. Then, for t∈Na+ν,
(5)Δa-νΔkf(t)=Δak-νft-∑j=0k-1‍Δjf(a)Γν-k+j+1t-a(ν-k+j).
Moreover, ifμ>0 with M-1≤μ≤M, then, for t∈Na+M-μ+ν,
(6)Δa+M-μ-νΔaμftM=Δaμ-νft-∑j=0M-1‍Δaj-M+μfa+M-μΓν-M+j+1MMMMMMMMMM·t-a-M+μ(ν-M+j).Lemma 7 (see [16]).
Letf:Na→R be given and suppose μ,ν>0 with N-1<μ≤N. Then, for t∈Na+ν+N-μ,
(7)Δa+νμΔa-νft=Δaμ-νft.Lemma 8 (see [16]).
Letf:Na→R be given and suppose ν>0 with N-1<ν<N. Then, for t∈Na-ν+N,
(8)Δaνft=1Γ-ν∑s=at+ν‍t-s-1-ν-1fs,N-1<ν<N;ΔNft,ν=N.Lemma 9 (see [16]).
Leta∈R and μ>0 be given. Then,
(9)Δt-a(μ)=μt-a(μ-1),
for any t for which both sides are well-defined. Furthermore, for ν>0,
(10)Δa+μ-νt-aμ=μ-νt-aμ+ν,fort∈Na+μ+ν,Δa+μνt-a(μ)=μ(ν)t-a(μ-ν),fort∈Na+μ+N-ν.


## 3. Triple Positive Solutions

In this section, we impose growth conditions onf to obtain the triple positive solutions for FBVP (1)-(2).For the sake of convenience, for0<δ≤ν≤ε, ξ∈[0,T]N0, η∈[0,1], p>1, we set the following notations:
(11)κt,ρ=t+1+ν-ρν-ρ-1Γν-ρ,fort,ρ∈-2,TN-2×0,1,(12)τ=-∑s=0T‍κT-s-1,ε-ηκT,ε1-ηκξ,δ·∑s=0ξ+1‍κξ-s-1,δ,(13)M1=∑i=0T‍qi1-τp-1,M2=η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφq∑j=sT‍qj,M3=∑j=0T‍φq∑i=jT‍qi,N=max⁡⁡M2,M3,m=ηξ+2ξ+1+ν-δ(ν-δ-1)1-ηκξ,δΓν-δ+1.Then, from (11), we immediately obtain some properties of κ(t,ϱ) as follows:
(14)κt,ϱ∈1,t,ϱ∈-2×(0,1),0,1,t,ϱ∈-1,TN-1×(0,ν),0,t,ϱ∈-1,TN-1×ν,-1,0,t,ϱ∈-1,TN-1×ν,1.We will also need the following elementary facts:(15)τ∈0,δ≤ν=ε,0,1,δ≤ν<ε.Indeed, ifδ≤ν=ε, then, by (12) and (14), clearly, τ=0.Ifδ=ν<ε, then
(16)τ=-∑s=0T‍κT-s-1,ε-ηκT,ε=1-∑s=0T+1‍kT-s-1,ε-ηκT,ε=1-1Γν-ε∑s=0T+1‍T+ν-ε-sν-ε-1-ηκT,ε=1-T+ν-ε⋯ν-εT+1!1+(ν-ε)+1+ν-εν-ε2!+⋯+T+ν-ε⋯ν-εT+1!-ηκT,ε=1-T+ν-ε⋯ν-εT+1!1+ν-ε1+ν-ε2+2+ν-ε⋯ν-ε3!+⋯+T+ν-ε⋯ν-εT+1!-ηκT,ε=1-T+ν-ε⋯ν-εT+1!1+ν-ε2+ν-ε2!1+ν-ε3+⋯+T+ν-ε⋯ν-εT+1!-ηκT,ε=⋯=1-T+1+ν-εT+ν-ε⋯1+ν-εT+1!-ηT+1+ν-εT+ν-ε⋯ν-εT+2!=1-T+1+ν-εT+ν-ε⋯1+ν-εT+1!·1+ην-εT+2=1-T+2κT,εν-ε1-ηε-νT+2.
It follows that 0<τ<1.Ifδ<ν<ε, then, since
(17)1+ε-νξ+2ν-δT+2κξ,δM≤1+ε-νν-δκξ,δM=(ε-δ)ξ+1+ν-δ(ν-δ-1)Γ(ν-δ+1)M=(ε-δ)Γ(ξ+2+ν-δ)Γ(ξ+3)Γ(ν-δ+1)M=ε-δξ+1+ν-δ⋯ν-δ+1ξ+2⋯2M<1,τ=-∑s=0T‍κ(T-s-1,ε)-ηκT,ε1-ηκξ,δM·∑s=0ξ+1‍κ(ξ-s-1,δ)=1-∑s=0T+1‍κT-s-1,ε-ηκT,ε1-ηκξ,δM·∑s=0ξ+1‍κξ-s-1,δ=1-(T+2)κ(T,ε)ν-ε-ηκ(T,ε)κ(ξ,δ)(ξ+2)1-ηκξ,δν-δ=1-T+2κT,εν-ε·1-ε-νν-δηξ+2κξ,δ1-ηκξ,δT+2=1-T+2κT,εν-ε·1-ηκξ,δ-ε-νξ+2ν-δT+2ηκξ,δ·1-ηκξ,δ-1ε-νξ+2ν-δT+2=1-T+2κT,εν-ε·1-1+ε-νξ+2/ν-δT+2ηκξ,δ1-ηκξ,δ.
We have 0<τ<1.Now, suppose thatx(t) is a solution of the problem (1)-(2), and let y(t)=Δν-1-(1-ν)x(t). Then, from Lemmas 6, 7, 8, and 9, we have x(t)=Δ01-νy(t)=Δ0-νΔy(t)+(y(0)/Γ(ν))t(ν-1), and
(18)Δt+ν-βν-1βx(t)M=Δt+ν-βν-1βΔ0-νΔyt+y(0)Γ(ν)Δt+ν-βν-1βt(ν-1)M=1Γ(-β)∑s=νt+ν‍t+ν-β-s-1(-β-1)Δ0-νΔy(s)MMMMMMM+t+ν-β-s-1-β-1Δ0-νΔyss=ν-1∑s=νt+ν‍MM+y(0)Γ(ν)ν-1(β)t+ν-β(ν-β-1)M=Δt+ν-βνβΔ0-νΔy(t)+y(0)Γ(ν-β)t+ν-β(ν-β-1)M=Δt+ν-β0-(ν-β)Δy(t)+y(0)Γ(ν-β)t+ν-β(ν-β-1)M=∑s=0t‍κt-s-2,βΔys+y0κt-1,β.
Similarly, we have
(19)Δν-1εxtt=T+1+ν-εM=∑s=0T+1‍κT-s-1,εΔy(s)+y0κT,ε,Δν-1δxtt=ξ+1+ν-δM=∑s=0ξ+1‍κ(ξ-s-1,δ)Δy(s)+y(0)κ(ξ,δ).
Thus, by the transformation y(t)=Δν-1-(1-ν)x(t), the problem (1)-(2) is equivalent to the following problem (20)–(22):
(20)ΔφpΔyt+qt·ft+ν-1,xt+ν-1,Δt+ν-βν-1βxt=0,0000000000000000000000000000000t∈0,TN0,(21)ΔyT+1M=-∑s=0T‍κT-s-1,εΔys-y0κT,ε,(22)y(0)M=η∑s=0ξ+1‍κξ-s-1,δΔys+ηy0κξ,δ,
where
(23)xt+ν-1=Δ01-νyt+ν-1,Δt+ν-βν-1βx(t)M=∑s=0t‍κt-s-2,βΔys+y0kt-1,β.Now, suppose thaty(t) is a solution of (20)–(22). We will show that
(24)yt≥0,Δyt≥0,Δ2y(t)≤0.
Firstly, it is easy to see that Δ2y(t)≤0 from Δφp(Δy(t))≤0.Secondly, ifδ≤ν=ε, then Δy(T+1)=0 by (21) and (22). If δ≤ν<ε, we have
(25)ΔyT+1M=-∑s=0T‍κT-s-1,εΔy(s)MM-ηκ(T,ε)1-ηκ(ξ,δ)∑s=0ξ+1‍κ(ξ-s-1,δ)Δy(s)M≥∑s=0ξ+1‍-∑s=0T‍κT-s-1,εMMMM-ηκT,ε1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δΔyT+1M=τΔyT+1.So we obtain thatΔy(T+1)≥0 and hence Δy(t)≥0 for t∈[0,T+1]N0. Note that y(0)=(η∑s=0ξ+1‍(κ(ξ-s-1,δ)Δy(s)))/(1-ηκ(ξ,δ))≥0. Thus (24) holds.Next, clearlyy(t)=Δν-1-(1-ν)x(t)=1/Γ(1-ν)∑s=ν-1t+ν-1‍(t-s-1)(-ν)x(s)≥0 if x(t)≥0.On the other hand, ify(t) is a solution of (20)–(22), then, from x(t)=Δ01-νy(t)=∑k=0t+1-ν‍-1k1-νky(t+1-ν-k), we have
(26)x(ν-1)=y(0)≥0,y(1)≥x(ν)=y(1)-1-νy(0)≥νy1,y2≥xν+1=y2-1-νy1+1-ν-ν2!y(0)≥ν1+ν2!y(2),⋮yt≥xt+ν-1=yt-1-νyt-1+⋯+1-ν-ν⋯-ν-t+2t!y0≥ν(1+ν)⋯t+ν-1t!y(t)=1Γ(ν)t+ν-1(ν-1)y(t)≥1ΓνT+ν-1ν-1yt,t∈0,TN0;that is,(27)T+ν-1ν-1Γνy(t)≤xt+ν-1≤yt,t∈0,TN0.This implies that the problem (1)-(2) has positive solutions if and only if the problem (20)–(22) has positive solutions. In the sequel, we will concretely consider the boundary problem (20)–(22).Summing (20) from t to T, we find that
(28)φp(Δy(T+1))-φpΔytM=-∑i=tT‍qifi+ν-1,xi+ν-1,Δi+ν-βν-1βxi.
Thus,
(29)Δy(t)=φq∑i=tT‍qifi+ν-1,xi+ν-1,Δi+ν-βν-1βxiMMMM+∑i=tT‍φpΔyT+1.Summing (29) from 0 to t-1, it follows that
(30)y(t)-y(0)=∑j=0t-1‍φq∑i=jT‍qifi+ν-1,xi+ν-1,Δi+ν-βν-1βxiMMMMM+∑i=jT‍φpΔyT+1,
which, together with (22), implies that
(31)y(t)=η1-ηκξ,δM·∑s=0ξ+1‍κξ-s-1,δφqMMM·∑j=sT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxjMMMMM+φpΔyT+1∑j=sT‍M+∑j=0t-1‍φq∑i=jT‍qifΔi+ν-βν-1βi+ν-1,xi+ν-1,Δi+ν-βν-1βxiMMMMM+∑i=jT‍φpΔyT+1.Next, letB={y:[0,T+2]N0→R} be endowed with the norm y=max⁡{y∞,Δy∞}, where y∞=max⁡t∈[0,T+2]N0|y(t)|. Choose the cone P⊂B defined by
(32)P=Δ2y∈B:yt≥0fort∈0,T+2N0,MMΔy(t)≥0fort∈0,T+1N0,MMΔ2y(t)≤0,fort∈0,TN0and2122hold.Let the nonnegative continuous concave functionalα, the nonnegative continuous convex functional θ,γ, and the nonnegative continuous functional ψ be defined on the cone P by
(33)γy=max⁡t∈0,T+1N0Δyt,αy=min⁡t∈[h,T+2]N0yt,ψ(y)=θ(y)=max⁡t∈0,T+2N0yt,
where h=[T+2/2] and [x] is the greatest integer not greater than x. Clearly, γy=Δy0, ψ(y)=θ(y)=y(T+2), and α(y)=y(h).Fort∈[0,T+2]N0, define an operator Q:P→B by
(34)Qyt=η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφqM·∑j=sT‍qjfj+ν-1,xj+ν-1,Δj+ν-βν-1βxjMMM+∑j=sT‍φpΔyT+1M+∑i=0t-1‍φq∑j=iT‍qj·fj+ν-1,xj+ν-1,j+ν-βΔν-1βxjMMMMMM+∑j=sT‍φpΔyT+1.We next require a preliminary lemma.Lemma 10.
LetQ be defined by the above equation. If y∈P, then(i)
Δ(Qy)(t)≥0 for t∈[0,T+1]N0, (Qy)(t)≥0 for t∈[0,T+2]N0;(ii)
Δ(Qy)(T+1)=-∑s=0T‍κ(T-s-1,ε)Δ(Qy)(s)-(Qy)(0)κ(T,ε), Qy0=η/1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δΔQys;(iii)
Q:P→P is completely continuous;(iv)
finding positive solutions of FBVP (20)–(22) is equivalent to finding fixed points of the operator Q on P;(v)
Ify∈P,y(t)≥t/T+2y∞=t/T+2y(T+2), for t∈[0,T+2]N0.The proof is simple and omitted.By Lemma10 and (33), for all y∈P(γ,d)¯⊂P, the functionals defined above satisfy
(35)hT+2θ(y)≤α(y)≤θ(y)=ψy.Furthermore, since(36)y(T+2)-y(0)T+2≤Δy0,y(T+2)M≤y(0)+T+2Δy0M≤η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ+T+2Δy0M=η(ξ+2)ξ+1+ν-δν-δ-11-ηκξ,δΓν-δ+1+T+2Δy0,
we have
(37)y=max⁡yT+2,Δy0≤T+2+mΔy0.Therefore,α(y)≤ψ(y) and y≤Mγ(y) are satisfied, where M=T+2+m.We now put growth condition onf such that the boundary value problem (20)–(22) has at least three positive solutions belonging to the cone P. Then (1)-(2) has at least three positive solutions.Theorem 11.
Let0<β, δ≤ν<ε<1, and suppose that H1-(H2) hold. In addition, assume that there exist positive numbers k,a,b,d with a<b≤h/T+2d such that the following conditions are satisfied:(H3)
f(t+ν-1,u,v)≤1/M1φp(d) for (t,u,v)∈[0,T]N0×[0,(T+2+m)d]×[0,(T+2+m)d],(H4)
f(t+ν-1,u,v)>φp((T+2)b/hM3) for (t,u,v)∈[h,T]Nh×[(T+ν-1)(ν-1)/Γ(ν)b,T+2/hb]×[0,(T+2+m)d],(H5)
f(t+ν-1,u,v)≤min⁡{(p-1)(((ε-ν)a/(∑i=0T‍q(i))1/p-1+a/(Nk))p-2a/Nk,φp(a/Nk)} for (t,u,v)∈[0,T]N0×[0,a]×[0,(T+2+m)d],(H6)
L=M2+M3/Nk+η(ε-ν)∑s=0ξ+1‍κ(ξ-s-1,δ)+(ε-ν)(T+2)≤1.Then FBVP (1)-(2) has at least three positive solutions x1,x2, and x3 satisfying
(38)x1(t)=Δ01-νy1(t),x2(t)=Δ01-νy2(t),x3(t)=Δ01-νy3(t),max⁡t∈0,T+1N0Δyit≤d,fori=1,2,3,b<min⁡t∈h,T+2Nhy1t,max⁡t∈0,T+2N0y3t<a,a<max⁡t∈0,T+2N0y2t,withmin⁡t∈h,T+2Nhy2t<b.Proof.
By the definition of operatorQ and its properties (i)–(v), it suffices to show that the conditions of Lemma 4 hold with respect to Q.
Firstly, ify∈P(γ,d)¯, we know γ(y)=max⁡t∈[0,T+1]N0|Δy(t)|≤d. From (37), we have
(39)max⁡t∈0,T+2N0yt≤T+2+md.
Since κ(t,β) is monotonic decreasing in variable t, then
(40)0≤Δt+ν-βν-1βxt=∑s=0t‍κt-s-2,βΔy(s)+y0κt-1,β=∑s=1t‍κt-s-2,βΔys+y1κt-2,βM-y0κt-2,β-κt-1,β≤∑s=2t‍κ(t-s-2,β)Δy(s)+y2κt-3,βM-y1κt-3,β-κt-2,β≤⋯≤yt+1≤T+2+md,Δy(T+1)M=-∑s=0T‍κ(T-s-1,ε)Δy(s)MM-ηκT,ε1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δΔysM≤-∑s=0T‍κT-s-1,εMMM-ηκT,ε1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δd=τd.
Thus, assumption(H3) and Lemma 10 imply that
(41)γQyM=max⁡t∈[0,T+1]N0ΔQyt=ΔQy0M=φq∑i=0T‍qifi+ν-1,xi+ν-1,Δi+ν-βν-1βxiMMMMM+∑i=0T‍φpΔyT+1M≤φq1M1φpd∑i=0T‍qi+τp-1φpdM=φq1M1∑i=0T‍qi+τp-1φpd=d.
Hence, Q:P(γ,d)¯→P(γ,d)¯.
To check condition(S1) of Lemma 4, we choose y0(t) as follows for t∈[0,T+2]N0:
(42)y0(t)M=-2bT+22t-T+22+T+2hb,tiseven;-2bT+32t-T+22+T+2hb,tisodd.
It is easy to verify thaty0∈P(γ,θ,α,b,T+2/hb,d) and α(y0)=min⁡t∈[h,T+2]Nh|y0(t)|=y0(h)>b. So {y∈P(γ,θ,α,b,T+2/hb,d)|α(y)>b}≠∅.
Therefore, ify∈P(γ,θ,α,b,T+2/hb,d), then b≤y(t)≤T+2/hb, |Δy(t)|≤d for t∈[h,T+2]Nh.
By (27), we have (T+ν-1)(ν-1)b/Γ(ν)≤x(t+ν-1)≤T+2/hb. According to assumption (H4) and Lemma 10, we get
(43)α(Qy)=min⁡t∈[h,T+2]NhQyt≥hT+2max⁡t∈[0,T+1]N0Qyt=hT+2η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφq·∑j=sT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxj+φpΔyT+1∑j=sT‍+∑i=0T+1‍φq∑j=iT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxj+φpΔyT+1∑j=iT‍≥hT+2∑i=0T‍φq∑j=iT‍qj·fΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxj∑j=iT‍≥hT+2∑i=0T‍φq∑j=iT‍qjφpT+2bhM3=b.
This shows that condition(S1) of Lemma 4 is satisfied.
Secondly, from (35), we have
(44)αQy≥hT+2θQy>b,
for all y∈P(γ,α,b,d) with θ(Qy)>T+2/hb. Thus, condition (S2) of Lemma 4 is satisfied.
We finally exhibit that(S3) of Lemma 4 is also satisfied. Clearly, as ψ(0)=0<a, we have 0∉R(γ,ψ,a,d). Suppose that y∈R(γ,ψ,a,d) with ψ(y)=a. Since κ(t,ε) is monotonic increasing in variable t, we have
(45)ΔyT+1=-∑s=0T‍κT-s-1,εΔys-y0κT,ε=-∑s=1T‍κT-s-1,εΔys-y1κT-1,εM+κT-1,ε-κT,εy0≤-∑s=1T‍κT-s-1,εΔys-y1κT-1,ε≤-∑s=2T‍κT-s-1,εΔys-y2κT-2,ε≤⋯≤-κ-1,εyT+1≤ε-νa.
Next, we consider two cases.
Case i. Consider p≥2. By (H5) and the inequality xp-1+yp-1≤(x+y)p-1 for x,y∈R+, we can obtain that
(46)ψQy=max⁡t∈[0,T+2]N0Qyt≤η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφq·∑j=sT‍qjfj+ν-1,xj+ν-1,Δj+ν-βν-1βxj+φpε-νa∑j=sT‍+∑i=0T+1‍φq∑j=iT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxj+φpε-νa∑j=iT‍≤η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφq·∑j=sT‍qjφpaNk+φpε-νa+∑i=0T+1‍φq∑j=iT‍qjφpaNk+φpε-νa=η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ·∑j=sT‍qj1/p-1aNk+ε-νa+∑i=0T+1‍∑j=sT‍qj1/p-1aNk+ε-νa≤aM2Nk+ηε-ν1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δa+aM3Nk+T+2ε-νa≤∑s=0ξ+1‍M2+M3Nk+ηε-ν1-ηκξ,δ·∑s=0ξ+1‍κξ-s-1,δ+T+2ε-νa≤a.Case ii. Consider 1<p<2. By (H5)(H6) and the inequality (x+y)p-1-yp-1≥(p-1)(x+y)p-2x, for x,y∈R+, we can have
(47)ψQy=max⁡t∈[0,T+2]N0Qyt≤η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ·∑j=sT‍qj1/p-1φq1∑j=sT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,0Δj+ν-βν-1βxj+φpε-νa1∑j=sT‍qj+a∑i=0T+1‍∑j=iT‍qj1/p-1φq1∑j=iT‍qjfΔj+ν-βν-1βj+ν-1,xj+ν-1,Δj+ν-βν-1βxj+φpε-νa1∑j=iT‍qj≤aη1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ∑j=sT‍qj1/p-1·φqp-1ε-ν∑j=sT‍qj1/p-1+1Nkp-2·1Nk+φpε-ν∑j=sT‍qj1/p-1+a∑i=0T+1‍∑j=iT‍qj1/p-1·φqp-1ε-ν∑j=iT‍qj1/p-1+1Nkp-2·1Nk+φpε-ν∑j=iT‍qj1/p-1≤aη1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ·∑j=sT‍qj1/p-1ε-ν∑j=sT‍qj1/p-1+1Nk+a∑i=0T+1‍∑j=iT‍qj1/p-1ε-ν∑j=iT‍qj1/p-1+1Nk≤aη1-ηκξ,δNk∑s=0ξ+1‍κξ-s-1,δ∑j=sT‍qj1/p-1+aηε-ν1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δ+ε-νT+2a+aNk∑i=0T+1‍∑j=iT‍qj1/p-1≤aNkM2+aηε-ν∑s=0ξ+1‍κξ-s-1,δ+ε-νT+2a+aNkM3=La≤a.
So, condition (S3) of Lemma 4 is satisfied. Therefore, Lemma 4 implies that the FBVP (20)–(22) has at least three positive solutions y1,y2, and y3 satisfying (38); that is, the FBVP (1)-(2) has at least three positive solutions x1,x2, and x3 satisfying (38). The proof is complete.Theorem 12.
Letδ≤ν=ε and (H1)–(H4) hold. In addition, assume that the following condition is satisfied:(H7)
f≤φpa/M2+M3.Then the FBVP (1)-(2) has at least three positive solutions x1,x2, and x3 satisfying (38).Proof.
Ifε=ν, (21) is equivalent to Δy(T+1)=0 by (14). Similar to discussion in Theorem 11, we know that (24) holds.
Fort∈[0,T+2]N0, define an operator Q:P→B by(48)Qy(t)=η1-ηκξ,δ∑s=0ξ+1‍κξ-s-1,δφqM·∑j=sT‍qjMMM·fj+ν-1,xj+ν-1,Δj+ν-βν-1βxj∑j=sT‍M+∑i=0t-1φq∑j=iT‍qj‍MMMMMMM·fj+ν-1,xj+ν-1,Δj+ν-βν-1βxj∑j=iT‍.The rest of the proof is similar to Theorem 11, so it is omitted. The proof is complete.Example 13.
Consider the boundary value problem:(49)ΔφpΔ-0.050.95xtMM+ft+ν-1,xt+ν-1,Δt+0.05-0.050.9xtM=0,t∈0,10N0,Δ-0.050.96xtt=10.99=0,x-0.05=12Δ-0.050.94xtt=2.01.
Compared with (1), we have q(t)=1, p=3, q=3/2, ξ=1, T=10, ν=0.95, β=0.9, ε=0.96, δ=0.94, η=1/2, h=6, and
(50)ft+ν-1,u,v=sint+ν-1+49+v1500,0≤u≤500;sint+ν-1+49+111.5u-5002+v1500,500≤u≤1000;sint+ν-1+49+111.55002+v1500,u≥1000.
Then, FBVP (49) has at least three positive solutions.Proof.
Choosea=500, b=1800, d=5000, p=3, q=3/2, and k=2.
By computation, we knowτ≈0.0021, M1≈11, M2≈1.56, M3=∑j=010‍(11-j)1/2≈25.78, N=M3, and m≈0.52.
It is easy to see that0<a<b<2d. And f satisfies that
(51)f(t+ν-1,u,v)M≤50+25000011.5+62590.211500≈21830.21M<2270663.03≈1M1φpd,
for (t,u,v)∈[0,10]N0×[0,62590.21]×[0,62590.21];
(52)ft+ν-1,u,vM≥49+500211.5≈21788>19488M≈360025.782=φpT+2bhM3,
for (t,u,v)∈[6,10]N0×[1800Γ(10+ν)/Γ(11)Γ(ν),3600]×[0,62590.21];
(53)ft+ν-1,u,v<50+62590.211500≈91.73≤5002×25.782<94.04≈φpa2N,
for (t,u,v)∈[0,10]N0×[0,500]×[0,65000]. Consider
(54)L=1.56+25.782×25.78+12×2.01×1.01×0.012+12×0.01≈0.65<1.
Thus, the conditions of Theorem 11 are satisfied. Therefore, the FBVP (49) has at least three positive solutions satisfying
(55)x1t=Δ00.05y1t,x2t=Δ00.05y2t,x3(t)=Δ00.05y3t,max⁡t∈[0,11]N0Δyit≤5000,fori=1,2,3,1800<min⁡t∈[6,12]N6y1t,max⁡t∈[0,12]N0y3t<500,500<max⁡t∈[0,12]N0y2t,withmin⁡t∈[6,12]N6y2t<1800.



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*Source: 147975-2014-11-12.xml*