# Nontrivial Solutions for a System of Second-Order Discrete Boundary Value Problems

Authors: Hua Su; Yongqing Wang; Jiafa Xu
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this work, we shall study the existence of nontrivial solutions for a system of second-order discrete boundary value problems. Under some conditions concerning the eigenvalues of relevant linear operator, we use the topological degree theory to obtain our main results.


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## Body

## 1. Introduction

Nonlinear discrete problems appear in many mathematical models, such as computer science, mechanical engineering, control systems, economics, and fluid mechanics (see [1–4]). Owing to the wide applications, in recent years, there are a large number of researchers paying special attention in this direction (we refer to some results [5–15] and the references therein). For example, in [5], the authors used the Guo–Krasnosel’skii fixed point theorem to study the existence of positive solutions for the following second-order discrete boundary value problem:(1)Δ2xi−1+fxi=0,i∈1,n,x0=0=xn+1,and the following discrete second-order system:(2)Δ2xi−1+fxi,yi=0,i∈1,n,Δ2yi−1+gxi,yj=0,i∈1,n,x0=xn+1=y0=yn+1=0,where n is a positive integer, 1,n=1,2,…,n,Δ is the forward difference operator, i.e., Δxi−1=xi−xi−1, and Δ2xi−1=ΔΔxi−1.In [6], the authors used the monotone iterative technique to investigate the existence and uniqueness of positive solutions for the following discrete p-Laplacian fractional boundary value problem:(3)Δν−1νϕpΔν−1νyt=fyt+ν−1,t∈0,Tℤ,yν−1=yν+T,Δν−1νyν−1=Δν−1νyν+T,where ν∈0,1 is a real number, Δν−1ν is a discrete fractional operator, and ϕps=sp−2s is the p-Laplacian with s∈ℝ,p>1.Coupled systems of discrete problems have also been investigated by many authors; some results can be found in a series of papers [11–15] and the references cited therein (also see some results on differential systems [16–24]). For example, in [11], the authors used the Guo–Krasnosel’skii fixed point theorem to study the following systems of three-point discrete boundary value problems:(4)Δ2un−1+λanfun,vn=0,n∈1,2,…,N−1,Δ2vn−1+μbngun,vn=0,u0=βuη,uN=αuη,v0=βvη,vN=αvη,where N≥4, η∈1,2,…,N−1,α>0,β>0,λ,μ>0. They offered some values for the parameters λ,μ to yield a positive solution for the above system.In [12], the authors used the fixed point index to study the positive solutions for the following system of first-order discrete fractional boundary value problems:(5)Δv−1vxt=f1t+v−1,xt+v−1,yt+v−1,t∈0,Tℤ,Δv−1vyt=f2t+v−1,xt+v−1,yt+v−1,t∈0,Tℤ,xv−1=xv+T,yv−1=yv+T.By discrete Jensen’s inequality, the authors adopted some appropriate nonnegative concave and convex functions to characterize the coupling behavior of the nonlinearitiesfii=1,2.Motivated by the aforementioned works, in this paper, by means of the topological degree theory, we study the existence of nontrivial solutions for the following system of second-order discrete boundary value problems:(6)Δ2uk−1+fk,vk=0,k∈1,2,…,T,Δ2vk−1+gk,uk=0,u0=uT+1=v0=vT+1=0,where T>2 is a fixed positive integer number, △uk=uk+1−uk,Δ2uk=△△uk,andf,g:1,2,…,T×ℝ⟶ℝℝ≔−∞,+∞ are continuous and satisfy the following conditions:(H1) There exist three nonnegative functionsaik,bikbik≢0,k∈T1 and βii=1,2 on ℝ+ such that(7)fk,v≥−a1k−b1kβ1v,gk,u≥−a2k−b2kβ2u,∀u,v∈ℝ,t∈T1,whereT1:=1,2,…,T.(H2)limv⟶+∞β1v/v=0, limu⟶+∞β2u/u=0.(H3)liminfv⟶+∞fk,v/v>λ1, liminfu⟶+∞gk,u/u>λ1 uniformly on k∈T1, where λ1=4sin2π/2T+2.(H4)lim supv⟶0fk,v/v<λ1, limsupu⟶0gk,u/u<λ1 uniformly on k∈T1.Now, we state our main result here.Theorem 1.
Suppose that (H1)–(H4) hold. Then, (6) has at least one nontrivial solution.


## 2. Preliminaries

LetE be the Banach space of real valued functions defined on the discrete interval T2 with the norm u=maxk∈T2uk, where T2:=0,1,2,…,T+1. Define the following sets:(8)P=u∈E:uk≥0,∀k∈T2,P0=u∈E:mink∈T1uk≥1Tu,and Br=x∈E:x<r for r>0. Then, P,P0 are cones on E, and Br is an open ball in E.Lemma 1 (see [11, 15]).
Lethk∈CT1. Then, the discrete boundary value problem(9)Δ2uk−1+hk=0,k∈T1,u0=uT+1=0,has a solution with the form(10)uk=∑l=1TGk,lhl,k∈T2,where(11)Gk,l=1T+1lT+1−k,1≤l≤k−1≤T,kT+1−k,0≤k≤l≤T.Furthermore,Gk,l has the following properties (see [13, 15]):(i)
Gk,l>0 and Gk,l=Gl,k, for k,l∈T1×T1.(ii)
Gl,l/T≤Gk,l≤Gl,l, for k,l∈T1×T1.By Lemma1, system (6) is equivalent to(12)uk=∑l=1TGk,lfl,vl,k∈T2,vk=∑l=1TGk,lgl,ul,k∈T2.Then, we can define operatorsT,S:E⟶E by(13)Tvk=∑l=1TGk,lfl,vl,Suk=∑l=1TGk,lgl,ul,and operator A:E×E⟶E×E by(14)Au,vk=Tv,Suk.Note thatT,S,A are completely continuous operators (see [11]), and u,v solves (6) if and only if u,v is a fixed point of the operator A.Lemma 2 (see [7, 15]).
Letϕk=sinkπ/T+1,k∈T2. Then, λ1∑l=1TGk,lϕl=ϕk,∀k∈T1.
Define a linear operator as follows:(15)Lxk=∑l=1TGk,lxl,∀k∈T2.Then, we have(16)Lϕk=1λ1ϕk,and we have the following lemma.Lemma 3.
Ifx∈P, then Lx∈P0.This is a direct result by Lemma1 (ii), so we omit the proof.Remark 1.
ϕ∈P0 in Lemma 2.Lemma 4 (see [25, Theorem A.3.3]).
LetΩ be a bounded open set in a Banach space E and T:Ω⟶E be a continuous compact operator. If there exists x0∈E\0 such that(17)x−Tx≠μx0,∀x∈∂Ω,μ≥0,then the topological degree degI−T,Ω,0=0.Lemma 5 (see [25, Lemma 2.5.1]).
LetΩ be a bounded open set in a Banach space E with 0∈Ω and T:Ω⟶E be a continuous compact operator. If(18)Tx≠μx,∀x∈∂Ω,μ≥1,then the topological degree degI−T,Ω,0=1.


## 3. Main Results

In order to obtain the Proof of Theorem1, we first provide a lemma.Lemma 6.
There exists a sufficiently largeR>0 such that(19)degI−A,BR,0=0.Proof.
By (H3), there existε1>0 and X1>0 such that(20)fk,v≥λ1+ε1v,gk,u≥λ1+ε1u,∀k∈T1,u,v>X1.
Note that whenk∈T1,u,v≤X1, the functions fk,v and gk,u are bounded, so we can choose some appropriate positive numbers M1,M2 such that(21)fk,v≥λ1+ε1v−M1,gk,u≥λ1+ε1u−M2,∀k∈T1,u,v∈ℝ,where(22)M1=maxk∈T1,u,v≤X1fk,v+λ1+ε1X1,M2=maxk∈T1,u,v≤X1gk,u+λ1+ε1X1.
From (H2), for any givenε,ε˜>0 with ε1−εb1>0,ε1−ε˜b2>0, there is X2>X1 such that(23)β1v≤εv,β2u≤ε˜u,∀u,v>X2.
Letβ1∗=maxv≤X2β1v and β2∗=maxu≤X2β2u. Then,(24)β1v≤εv+β1∗,β2u≤ε˜u+β2∗,u,v∈ℝ.
Thus, we have(25)fk,v≥λ1+ε1v−a1k−b1kβ1v−M1≥λ1+ε1v−a1k−b1kεv+β1∗−M1≥λ1+ε1−εb1|v|−a1k−β1∗b1k−M1,∀k∈T1,v∈ℝ,(26)gk,u≥λ1+ε1−ε˜b2u−a2k−β2∗b2k−M2,∀k∈T1,u∈ℝ.
Note thatε,ε˜ can be chosen arbitrarily small, so we can let(27)R>maxN1,N2,N3,N4,where(28)N1=2∑l=1TGl,la1l+β1∗b1l+M11−2ε∑l=1TGl,lb1l,N2=2∑l=1TGl,la2l+β2∗b2l+M21−2ε˜∑l=1TGl,lb2l,N3=λ1T+1+Tε1−εb1∑l=1TGl,la1l+a2l+β1∗b1l+β2∗b2l+M1+M2ε1−εb1−λ1T+1+Tε1−εb1ε∑l=1TGl,lb1l+ε˜∑l=1TGl,lb2l,N4=λ1T+1+Tε1−εb2∑l=1TGl,la1l+a2l+β1∗b1l+β2∗b2l+M1+M2ε1−εb2−λ1T+1+Tε1−εb2ε∑l=1TGl,lb1l+ε˜∑l=1TGl,lb2l.
Now, we prove(29)u,v−Au,v≠μϕ,ϕ,∀u,v∈∂BR,μ≥0,where ϕk=sinkπ/T+1,k∈T2. We argue this claim by indirection. Suppose that there exist u,v∈∂BR,μ≥0 such that(30)u,v−Au,v=μϕ,ϕ.(31)uk=Tvk+μϕk=∑l=1TGk,lfl,vl+μϕk,(32)vk=Suk+μϕk=∑l=1TGk,lgl,ul+μϕk.(33)v˜k=∑l=1TGk,la1l+b1lβ1vl+M1,u˜k=∑l=1TGk,la2l+b2lβ2ul+M2.
Then by Lemma3, u˜,v˜∈P0, and we also have(34)uk+v˜k=∑l=1TGk,lfl,vl+a1l+b1lβ1vl+M1+μϕk,vk+u˜k=∑l=1TGk,lgl,ul+a2l+b2lβ2ul+M2+μϕk.
Using (24) and (25), we have(35)fl,vl+a1l+b1lβ1vl+M1∈P,gl,ul+a2l+b2lβ2ul+M2∈P.
So, from Lemma3 and Remark 1, we have(36)v+u˜,u+v˜∈P0.
Note thatu,v∈∂BR, and using (24), R>N1,andR>N2, we have(37)v˜≤∑l=1TGl,la1l+b1lβ1vl+M1≤∑l=1TGl,la1l+b1lεv+β1∗+M1<R2,u˜≤∑l=1TGl,la2l+b2lε˜u+β2∗+M2<R2.
It is noted thatu=v=R,u+u˜+v˜∈P0, and v+u˜+v˜∈P0. Therefore, we get(38)uk+u˜k+v˜k≥1Tu+u˜+v˜≥1Tu−u˜+v˜≥1Tu−u˜+v˜,vk+u˜k+v˜k≥1Tv+u˜+v˜≥1Tv−u˜+v˜≥1Tv−u˜+v˜.
UsingR>N3, we have(39)ε1−εb1∑l=1TGk,lvl+u˜l+v˜l−λ1+ε1−εb1∑l=1TGk,lu˜l+v˜l≥ε1−εb1T∑l=1TGk,lR−u˜+v˜−λ1+ε1−εb1∑l=1TGk,lu˜+v˜≥0,and R>N4 implies that(40)ε1−ε˜b2∑l=1TGk,lul+u˜l+v˜l−λ1+ε1−ε˜b2∑l=1TGk,lu˜l+v˜l≥0.
Consequently, we obtain(41)Tvk+v˜k=∑l=1TGk,lfl,vl+a1l+b1lβ1vl+M1≥∑l=1TGk,lλ1+ε1−εb1vl−a1l−β1∗b1l−M1+a1l+b1lβ1vl+M1≥∑l=1TGk,lλ1+ε1−εb1vl−β1∗b1l+b1lεvl+β1∗≥λ1+ε1−εb1∑l=1TGk,lvl≥λ1+ε1−εb1∑l=1TGk,lvl+u˜l+v˜l−λ1+ε1−εb1∑l=1TGk,lu˜l+v˜l≥λ1∑l=1TGk,lvl+u˜l+v˜l≥λ1∑l=1TGk,lvl+u˜l,Suk+u˜k=∑l=1Tgk,lgl,vl+a2l+b2lβ2ul+M2≥∑l=1TGk,lλ1+ε1−ε˜b2ul−a2l−β2∗b2l−M2+a2l+b2lβ2ul+M2≥λ1+ε1−ε˜b2∑l=1TGk,lul≥λ1+ε1−ε˜b2∑l=1TGk,lul+u˜l+v˜l−λ1+ε1−ε˜b2∑l=1TGk,lu˜l+v˜l≥λ1∑l=1TGk,lul+u˜l+v˜l≥λ1∑l=1TGk,lul+v˜l.
As a result, we get(42)Tvk+Suk+u˜k+v˜k≥λ1Lu+v+u˜+v˜k.
In view of (31) and (32), we see(43)uk+vk+u˜k+v˜k=Tvk+Suk+u˜k+v˜k+2μϕk≥λ1Lu+v+u˜+v˜k+2μϕk≥2μϕk.
Defineμ∗=supSμ≔supμ>0:u+v+u˜+v˜≥2μϕ. Then, Sμ≠∅, μ∗≥μ and u+v+u˜+v˜≥2μ∗ϕ. From ϕ=λ1Lϕ, we obtain(44)λ1Lu+v+u˜+v˜≥λ1L2μ∗ϕ=2μ∗λ1Lϕ=2μ∗ϕ.
Hence,(45)u+v+u˜+v˜≥λ1Lu+v+u˜+v˜+2μϕ≥2μ∗+μϕ,which contradicts the definition of μ∗. Therefore, (29) holds, and from Lemma 4, we obtain(46)degI−A,BR,0=0.
This completes the proof.Proof of Theorem1.
From (H4), there existε2∈0,λ1 and r∈0,R such that(47)fk,v≤λ1−ε2v,gk,u≤λ1−ε2u,∀k∈T1,u,v∈ℝwithu,v≤r.
This implies that(48)Tvk=∑l=1TGk,lfl,vl≤∑l=1TGk,lfl,vl≤λ1−ε2∑l=1TGk,lvl,Suk=∑l=1TGk,lgl,ul≤∑l=1TGk,lgl,ul≤λ1−ε2∑l=1TGk,lul.
Consequently, we have(49)Tvk+Suk≤λ1−ε2∑l=1TGk,lul+vl.
Now, we prove that(50)u,v≠μAu,v,for all u,v∈∂Br and μ∈0,1. We argue by contradiction. Suppose that there exist u,v∈∂Br and μ∈0,1 such that(51)u,v=μAu,v.
Therefore,(52)uk=μTvk,andvk=μSuk,k∈T1.
Hence, we have(53)uk+vk≤Tvk+Suk≤λ1−ε2∑l=1TGk,lul+vl.
From Lemma1 (i) and Lemma 2, we have(54)λ1∑k=1TGk,lϕk=ϕl,∀l∈T1.
Multiplying both sides of (53) by sinkπ/T+1, then summing from 1 to T, and using (54), we obtain(55)∑k=1Tuk+vksinkπT+1≤λ1−ε2∑k=1TsinkπT+1∑l=1TGk,lul+vl=λ1−ε2λ1∑l=1Tul+vlsinlπT+1.
This implies that(56)∑k=1Tuk+vksinkπT+1=0.
Becausesinkπ/T+1≥0≢0 for k∈T1, we have uk+vk≡0,k∈T1. This contradicts u,v∈∂Br. Therefore, (50) holds, and Lemma 5 implies that(57)degI−A,Br,0=1.
Combining this with Lemma6, we have(58)degI−A,BRB¯r,0=degI−A,BR,0−degI−A,Br,0=−1.
Therefore, the operatorA has at least one fixed point in BR/B¯r, and (6) has at least one nontrivial solution. This completes the proof.



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