# On the Existence of Three Positive Solutions for a Caputo Fractional Difference Equation

Authors: Lili Kong; Huiqin Chen; Luping Li; Shugui Kang
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we introduce the application of three fixed point theorem by discussing the existence of three positive solutions for a class of Caputo fractional difference equation boundary value problem. We establish the condition of the existence of three positive solutions for this problem.


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## Body

## 1. Introduction

After being proved to be a valuable tool in science and engineering fields, fractional difference equation has attracted attention of more and more scholars. And the existing results of positive solutions for boundary value problem of nonlinear fractional difference equations is the hot spot which has been discussed in recent years. So, a large number of scholars have devoted themselves to the study of fractional difference equations, such as [1–9].At the same time, the fixed point theory (see [10–12]) has also been widely applied to study the fractional difference equations. After that, many authors obtained the existence of positive solutions for the fractional difference equations by using the fixed point theorem (see [13–22]). For example, Jiraporn Reunsumrit and Thanin Sitthiwirattham [20] considered the nonlinear discrete fractional boundary value problem of the form(1)△αxt+ft+α−1,xt+α−1=0,t∈N0,T,△xα−2=0,xα+T=λΔ−βxη+β,where 1<α≤2,0<β≤1 and f:Nα−2,α+T×R⟶R is a continuous function. The authors employed some fixed-point theorems to obtain the existence, uniqueness of solutions, and the existence of positive solutions.Reunsumrit and Sitthiwirattham [21] studied the three-point fractional sum boundary value problem of the form(2)△Cαut+at+α−1fuθt+α−1=0,t∈N0,T,uα−3=Δ2uα−3=0,uT+α=λΔ−βuη+β,where 2<α≤3,0<β≤1 and f:0,+∞⟶0,+∞ is a continuous function. The authors employed Guo-Krasnoselskii’s fixed-point theorem to obtain the existence of at least one positive solution.Based on the above research results, this article considers the existence of three positive solutions for the nonlinear fractional difference equation boundary value problem(3)△Cνut=−ft+ν−1,ut+ν−1,uν−3=△ub+ν=△2uν−3=0,where t∈ℕ0b+1, b≥5 is an integer. f:ℕν−2b+ν×0,+∞⟶0,+∞ is continuous and f is not identically zero, 2<ν≤3, and △Cνut is the standard Caputo difference. Our analysis relies on Leggett-Williams fixed-point theorem to obtain sufficient conditions of the existence of three positive solutions for Caputo fractional boundary value problem (3). Chen et al. [22] considered the existence of positive solutions for (3). In this article, the authors obtained the existence of one or two positive solutions by means of the cone theoretic fixed-point theorems. Compared with [22], the application of Leggett-Williams fixed-point theorem makes our proving process simpler and the number of solutions increased. The research in this article shows that employing the Leggett-Williams fixed-point theorem to prove the existence of positive solutions for the fractional difference equation can get better results.In the remainder of this paper, we will present basic definitions and some lemmas in order to prove our main results in Section2. In Section 3, we establish some results for the existence of three positive solutions to problem (3). And some examples to corroborate our results are given in Section 4.


## 2. Background Materials and Preliminaries

For convenience, we first review some basic results about fractional sums and differences. For anyt∈ℕ0b+1 and ν>0, we define(4)tν¯=Γt+1Γt+1−ν,for which the right-hand side is defined. We appeal to the convention that if t+1−ν is a pole of the Gamma function and t+1 is not a pole, then tν¯=0.Definition 1.
Forν>0 and a function f defined on Na:=a,a+1,⋯, the ν-th fractional sum of f is defined by(5)△−νft=1Γν∑s=at−νt−s−1ν−1¯fs,where t∈a+ν,a+ν+1,…=ℕa+ν.Definition 2.
Forν>0 and a continuous function f:0,+∞⟶ℝ defined on Na, the ν-th Caputo fractional difference of f is given by(6)△Cνft=△−n−ν△nft=1Γn−ν∑s=at−n−νt−s−1n−ν−1¯△nfs,where n−1<ν<n. If ν=n, then ΔCνft=Δnft.Lemma 1 (see [13]).
Assume thatν>0 and f is defined on domains ℕa, then(7)△a+n−ν−ν△Cνft=ft−∑k=0n−1ckt−ak¯,where ck∈ℝ,k=0,1,…n−1 and n−1<ν≤n.Lemma 2 (see [22]).
Let2<ν≤3 and g:ℕν−2b+ν⟶ℝ be given. Then the solution of the FBVP(8)△Cνut=−gt+ν−1,uν−3=△ub+ν=△2uν−3=0,is given by(9)ut=∑s=0b+1Gt,sgs+ν−1,where function G:ℕν−2b+ν×ℕ0b+1⟶ℝ is defined by(10)Gt,s=ν−1t−ν+3b+ν−s−1ν−2¯−t−s−1ν−1¯Γν,0≤s<t−ν+1≤b+1,ν−1t−ν+3b+ν−s−1ν−2¯Γν,0≤t−ν+1≤s≤b+1.Here Gt,s is called the Green function of boundary value problem (8).Remark 1.
Notice thatGν−3,s=0, Gt,b+2=0. G could be extended to ℕν−3b+ν×ℕ0b+2, so we only discuss t,s∈ℕν−2b+ν×ℕ0b+1.Lemma 3 (see [22]).
The Green functionGt,s defined by (10) satisfies(i)
Gt,s>0, t,s∈ℕν−2b+ν×ℕ0b+1.(ii)
maxt∈ℕν−2b+νGt,s=Gb+ν,s, s∈ℕ0b+1.(iii)
minb+ν/4≤t≤3b+ν/4Gt,s≥1/4maxt∈ℕν−2b+νGt,s=1/4Gb+ν,s, s∈ℕ0b+1.Definition 3 (see [12]).
IfP is a cone of the real Banach space E, a mapping ψ:P⟶0,∞ is continuous and with(11)ψtx+1−ty≥tψx+1−tψy,x,y∈P,t∈0,1,is called a nonnegative concave continuous functional ψ on P.
Assume thatr, a, b are positive constants, we will employ the following notations(12)Pr=u∈P:u<r,P¯r=u∈P:u≤r,Pψ,a,b=u∈P:ψu≥a,u≤b.
Our existence criteria will be based on the following Leggett-Williams fixed-point theorem.Lemma 4 (see [12]).
LetE=E,⋅ be a Banach space, P⊂E be a cone of E, and c>0 be a constant. Suppose there exists a concave nonnegative continuous functional ψ on P with ψu≤u for u∈P¯c. Let A:P¯c⟶P¯c be a completely continuous operator. Assume there are numbers d, a, and b with 0<d<a<b≤c such that(i)
The setu∈Pψ,a,b:ψu>a is nonempty and ψAu>a for all u∈Pψ,a,b.(ii)
Au<d for u∈P¯d.(iii)
ψAu>a for all u∈Pψ,a,c with Au>b.Then A has at least three fixed points u1, u2, and u3∈P¯c. Furthermore, we have(13)maxt∈0,1u1t<d,a<mint∈0,1u2t<maxt∈0,1u2t<candd<maxt∈0,1u3t≤c,mint∈0,1u3t<a.


## 3. Main Results

Set(14)ℬ=u:ℕν−3b+ν⟶ℝ,uν−3=△ub+ν=△2uν−3=0.Thenℬ is a Banach space with respect to the norm u=maxt∈ℕν−3b+νut. We define a cone in ℬ by(15)P=u∈ℬ:ut≥0,minb+ν/4≤t≤3b+ν/4ut≥14u.Now consider the operatorT defined by(16)Tut=∑s=0b+1Gt,sfs+ν−1,us+ν−1.It is easy to see thatu=ut is a solution of the FBVP (3) if and only if u=ut is a fixed point of T. We shall obtain conditions for the existence of three fixed points of T. First, we notice that T is a summation operator on a discrete finite set. Hence, T is trivially completely continuous. From (16),(17)minb+ν/4≤t≤3b+ν/4Tut≥14∑s=0b+1Gb+ν,sfs+ν−1,us+ν−1≥14maxt∈ℕν−2b+ν∑s=0b+1Gt,sfs+ν−1,us+ν−1=14Tu,hence, TP⊂P.We will discuss the existence of three fixed points ofT by using Lemma 4. Thus, the conditions for the existence of the three positive solutions of (3) are obtained. For this purpose, let the nonnegative concave continuous function ψ on P be defined by(18)ψu=minb+ν/4≤t≤3b+ν/4ut.Denote(19)f0=limsupu⟶0maxt∈ℕν−2b+νft,uu,f∞=limsupu⟶+∞maxt∈ℕν−2b+νft,uu,l=∑s=0b+1Gb+ν,s,m=∑s=b+ν/4−ν+13b+ν/4−ν+114Gb+ν,s.Suppose that the functionft,u satisfies the following condition(C) ft,u is a nonnegative continuous function on 0,1×0,+∞ and there exists tn⟶0 such that ftn,utn>0,n=1,2,⋯.Theorem 1.
Assume condition (C) holds and there exist constants 0<d<a such that(C1)
ft,u<1/ld for t,u∈ℕν−2b+ν×0,d(C2)
ft,u≥1/ma for t,u∈b+ν/4,3b+ν/4×a,c, where c>4a(C3)
ft,u≤κu+β for t,u∈ℕν−2b+ν×0,+∞, where κ,β are positive numbersThen the boundary value problem (3) has at least three positive solutions u1,u2, and u3.Proof.
Setc>maxβl/1−κl,4a, then for u∈P¯c, from (C3), we have(20)Tu=maxt∈ℕν−2b+ν∑s=0b+1Gt,sfs+ν−1,us+ν−1≤∑s=0b+1Gb+ν,sκus+ν−1+β≤κu+β∑s=0b+1Gb+ν,s=κu+βl<c,namely, Tu∈Pc. Therefore T:P¯c⟶P¯c is a completely continuous operator. From (C1), we can get(21)Tu=maxt∈ℕν−2b+ν∑s=0b+1Gt,sfs+ν−1,us+ν−1≤∑s=0b+1Gb+ν,sfs+ν−1,us+ν−1<1ld∑s=0b+1Gb+ν,s=d.
Therefore, assumption (ii) of Lemma4 is satisfied.
We chooseφ0=5a/2 for t∈b+ν/4,3b+ν/4, then φ0∈u∈Pψ,a,4a:ψu>a which implies u∈Pψ,a,4a:ψu>a≠∅. Hence, if u∈Pψ,a,4a, then a≤ut≤4a for b+ν/4≤t≤3b+ν/4. Thus,(22)ψTu=minb+ν/4≤t≤3b+ν/4Tut=minb+ν/4≤t≤3b+ν/4∑s=0b+1Gt,sfs+ν−1,us+ν−1>∑s=b+ν/4−ν+13b+ν/4−ν+114Gb+ν,sfs+ν−1,us+ν−1≥1ma∑s=b+ν/4−ν+13b+ν/4−ν+114Gb+ν,s=a,from which we see that ψTu>a for all u∈Pψ,a,c. This shows that condition (i) of Lemma 4 is satisfied.
Finally, foru∈Pψ,a,c with Tu>4a, we get(23)ψTu=minb+ν/4≤t≤3b+ν/4Tut≥14∑s=0b+1Gb+ν,sfs+ν−1,us+ν−1=14∑s=0b+1maxt∈ℕν−2b+νGt,sfs+ν−1,us+ν−1≥14maxt∈ℕν−2b+ν∑s=0b+1Gt,sfs+ν−1,us+ν−1=14Tu>a,this shows that condition (iii) of Lemma 4 is satisfied. By the use of Lemma 4, the boundary value problem (3) has at least three solutions u1,u2, and u3. Take into account that condition (C) holds, we have uit>0,0<t<1,i=1,2,3. The proof is completed.Theorem 2.
Assume condition (C) holds. There exist constants 0<d<a<4a<c such that (C1), (C2), and (C4) are satisfied, where(C4)
ft,u≤1/lc for t,u∈ℕν−2b+ν×0,c.Then the boundary value problem (3) has at least three positive solutions u1,u2, and u3 such that(24)maxt∈ℕν−2b+νu1t<d,a<mint∈b+ν/4,3b+ν/4u2t<maxt∈ℕν−2b+νu2t<c,d<maxt∈ℕν−2b+νu3t⩽c,mint∈b+ν/4,3b+ν/4u3t<a.Proof.
From (C4), we get(25)Tu=maxt∈ℕν−2b+ν∑s=0b+1Gt,sfs+ν−1,us+ν−1≤∑s=0b+1Gb+ν,sfs+ν−1,us+ν−1<1lc∑s=0b+1Gb+ν,s=c.Therefore, T:P¯c⟶P¯c. The remainder of proof is essentially the same as that of Theorem 1 and is therefore omitted. By Lemma 4, the boundary value problem (3) has at least three positive solutions u1,u2, and u3 satisfying(26)maxt∈ℕν−2b+νu1t<d,a<mint∈b+ν/4,3b+ν/4u2t<maxt∈ℕν−2b+νu2t<c,d<maxt∈ℕν−2b+νu3t≤c,mint∈b+ν/4,3b+ν/4u3t<a.The proof is complete.Theorem 3.
Assume condition (C) holds. There exist constants0<d<a such that (C1), (C2) are satisfied, and(C5)
f∞<1/l.Then the boundary value problem (3) has at least three positive solutions.Proof.
By (C5), there exist0<σ<1/l and R>0, when u≥R, we have(27)ft,u≤σu.Set M=maxt,u∈ℕν−2b+ν×0,Rft,u, consequently(28)0≤ft,u≤σu+M,0≤u<+∞.This shows that condition (C3) of Theorem 1 is satisfied. By Theorem 1, the boundary value problem (3) has at least three positive solutions. The proof is completed.Theorem 4.
Assume there exist two positive constantsa,c (0<4a<c) such that conditions (C), (C2), and (C4) hold. And function ft,u satisfies(C6)
f0<1/l.Then the boundary vale problem (3) has at least three positive solutions.Proof.
In line with (C6), it is easy to see that there exists a positive constantd<a such that for u<d, we have(29)ft,ut<1lu.
Namely,(30)ft,ut<1ld,u<d.
This implies that conditions of Theorem2 are satisfied. By Theorem 2, the boundary vale problem (3) has at least three positive solutions. The proof is completed.
In the light of the proof of Theorem3 and Theorem 4, we obtain one theorem as follows.Theorem 5.
Assume conditions (C), (C5), and (C6) hold. Suppose that there exists a positive constantsa such that ft,u≥1/ma for t,u∈ℕν−2b+ν×a,4a. Then the boundary vale problem (3) has at least three positive solutions.


## 4. Examples

This section, we present two examples to illustrate our results. Setν=33/16,b=19, by estimating, we then have l≈304.4632,m≈33.5505.Example 1.
We take(31)ft,u=t40000+125u4,t,u∈116,33716×0,1,t40000+3100+1100u,t,u∈116,33716×1,+∞.
There exist constantsd=1/3 and a=3/2 such that(32)ft,u=t40000+125u4≤0.00102039<1ld,t,u∈116,33716×0,13,ft,u=t40000+3100+1100u≥0.045>1ma,t,u∈33764,101164×32,152,ft,u≤3372560000+125u,t,u∈116,33716×0,+∞.
All the conditions of Theorem1 hold. Thus, this moment, by virtue of Theorem 1, we know that the boundary value problem (3) has three positive solutions.Example 2.
We take(33)ft,u=2000.002⋅2−tu2,t,u∈116,33716×0,1,2−t2000u1/2+0.002u,t,u∈116,33716×1,+∞.There exist constants d=0.0000016 and a=1.008 such that(34)f∞≤0.002<1l,t,u∈116,33716×0,+∞,ft,u≥0.030639>1ma,t,u∈33764,101164×126125,504125,ft,u≤0.003064d≤1ld,t,u∈116,33716×0,d.
All the conditions of Theorem3 hold. Thus, in this case, by Theorem 3, we know that the boundary value problem (3) has three positive solutions.



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