MPEMathematical Problems in Engineering1563-51471024-123XHindawi10.1155/2020/50935355093535Research ArticleNontrivial Solutions for a System of Second-Order Discrete Boundary Value Problemshttps://orcid.org/0000-0002-4578-8714SuHua1https://orcid.org/0000-0001-8777-9624WangYongqing2https://orcid.org/0000-0001-6537-4167XuJiafa2ChenChuanjun1School of Mathematics and Quantitative EconomicsShandong University of Finance and EconomicsJinanShandong 250014Chinasdufe.edu.cn2School of Mathematical SciencesQufu Normal UniversityQufuShandong 273165Chinaqfnu.edu.cn202024820202020170720201008202024820202020Copyright © 2020 Hua Su et al.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

National Social Science Fund of China (NSSFC)18BTY015Government of Shandong ProvinceJ16LI01
1. Introduction

Nonlinear discrete problems appear in many mathematical models, such as computer science, mechanical engineering, control systems, economics, and fluid mechanics (see ). 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  and the references therein). For example, in , 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)Δ2xi1+fxi=0,i1,n,x0=0=xn+1,and the following discrete second-order system:(2)Δ2xi1+fxi,yi=0,i1,n,Δ2yi1+gxi,yj=0,i1,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., Δxi1=xixi1, and Δ2xi1=ΔΔxi1.

In , 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)Δν1νϕpΔν1νyt=fyt+ν1,t0,T,yν1=yν+T,Δν1νyν1=Δν1νyν+T,where ν0,1 is a real number, Δν1ν is a discrete fractional operator, and ϕps=sp2s is the p-Laplacian with s,p>1.

Coupled systems of discrete problems have also been investigated by many authors; some results can be found in a series of papers  and the references cited therein (also see some results on differential systems ). For example, in , the authors used the Guo–Krasnosel’skii fixed point theorem to study the following systems of three-point discrete boundary value problems:(4)Δ2un1+λanfun,vn=0,n1,2,,N1,Δ2vn1+μbngun,vn=0,u0=βuη,uN=αuη,v0=βvη,vN=αvη,where N4, η1,2,,N1,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 , 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)Δv1vxt=f1t+v1,xt+v1,yt+v1,t0,T,Δv1vyt=f2t+v1,xt+v1,yt+v1,t0,T,xv1=xv+T,yv1=yv+T.

By discrete Jensen’s inequality, the authors adopted some appropriate nonnegative concave and convex functions to characterize the coupling behavior of the nonlinearities fii=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)Δ2uk1+fk,vk=0,k1,2,,T,Δ2vk1+gk,uk=0,u0=uT+1=v0=vT+1=0,where T>2 is a fixed positive integer number, uk=uk+1uk,Δ2uk=uk,  and  f,g:1,2,,T×,+ are continuous and satisfy the following conditions:

(H1) There exist three nonnegative functions aik,bikbik0,kT1 and βii=1,2 on + such that(7)fk,va1kb1kβ1v,gk,ua2kb2kβ2u,u,v,tT1,

where T1:=1,2,,T.

(H2) limv+β1v/v=0, limu+β2u/u=0.

(H3) lim infv+fk,v/v>λ1, lim infu+gk,u/u>λ1 uniformly on kT1, where λ1=4sin2π/2T+2.

(H4) lim supv0fk,v/v<λ1, limsupu0gk,u/u<λ1 uniformly on kT1.

Now, we state our main result here.

Theorem 1.

Suppose that (H1)–(H4) hold. Then, (6) has at least one nontrivial solution.

2. Preliminaries

Let E be the Banach space of real valued functions defined on the discrete interval T2 with the norm u=maxkT2uk, where T2:=0,1,2,,T+1. Define the following sets:(8)P=uE:uk0,kT2,P0=uE:minkT1uk1Tu,and Br=xE:x<r for r>0. Then, P,P0 are cones on E, and Br is an open ball in E.

Lemma 1 (see [<xref ref-type="bibr" rid="B11">11</xref>, <xref ref-type="bibr" rid="B15">15</xref>]).

Let hkCT1. Then, the discrete boundary value problem(9)Δ2uk1+hk=0,kT1,u0=uT+1=0,has a solution with the form(10)uk=l=1TGk,lhl,kT2,where(11)Gk,l=1T+1lT+1k,1lk1T,kT+1k,0klT.

Furthermore, Gk,l has the following properties (see [13, 15]):

Gk,l>0 and Gk,l=Gl,k, for k,lT1×T1.

Gl,l/TGk,lGl,l, for k,lT1×T1.

By Lemma 1, system (6) is equivalent to(12)uk=l=1TGk,lfl,vl,kT2,vk=l=1TGk,lgl,ul,kT2.

Then, we can define operators T,S:EE by(13)Tvk=l=1TGk,lfl,vl,Suk=l=1TGk,lgl,ul,and operator A:E×EE×E by(14)Au,vk=Tv,Suk.

Note that T,S,A are completely continuous operators (see ), and u,v solves (6) if and only if u,v is a fixed point of the operator A.

Lemma 2 (see [<xref ref-type="bibr" rid="B7">7</xref>, <xref ref-type="bibr" rid="B15">15</xref>]).

Let ϕk=sinkπ/T+1,kT2. Then, λ1l=1TGk,lϕl=ϕk,kT1.

Define a linear operator as follows:(15)Lxk=l=1TGk,lxl,kT2.

Then, we have(16)Lϕk=1λ1ϕk,and we have the following lemma.

Lemma 3.

If xP, then LxP0.

This is a direct result by Lemma 1 (ii), so we omit the proof.

Remark 1.

ϕP0 in Lemma 2.

Lemma 4 (see [<xref ref-type="bibr" rid="B25">25</xref>, Theorem A.3.3]).

Let Ω be a bounded open set in a Banach space E and T:ΩE be a continuous compact operator. If there exists x0E\0 such that(17)xTxμx0,xΩ,μ0,then the topological degree degIT,Ω,0=0.

Lemma 5 (see [<xref ref-type="bibr" rid="B25">25</xref>, Lemma 2.5.1]).

Let Ω be a bounded open set in a Banach space E with 0Ω and T:ΩE be a continuous compact operator. If(18)Txμx,xΩ,μ1,then the topological degree degIT,Ω,0=1.

3. Main Results

In order to obtain the Proof of Theorem 1, we first provide a lemma.

Lemma 6.

There exists a sufficiently large R>0 such that(19)degIA,BR,0=0.

Proof.

By (H3), there exist ε1>0 and X1>0 such that(20)fk,vλ1+ε1v,gk,uλ1+ε1u,kT1,u,v>X1.

Note that when kT1,u,vX1, the functions fk,v and gk,u are bounded, so we can choose some appropriate positive numbers M1,M2 such that(21)fk,vλ1+ε1vM1,gk,uλ1+ε1uM2,kT1,u,v,where(22)M1=maxkT1,u,vX1fk,v+λ1+ε1X1,M2=maxkT1,u,vX1gk,u+λ1+ε1X1.

From (H2), for any given ε,ε˜>0 with ε1εb1>0,ε1ε˜b2>0, there is X2>X1 such that(23)β1vεv,β2uε˜u,u,v>X2.

Let β1=maxvX2β1v and β2=maxuX2β2u. Then,(24)β1vεv+β1,β2uε˜u+β2,u,v.

Thus, we have(25)fk,vλ1+ε1va1kb1kβ1vM1λ1+ε1va1kb1kεv+β1M1λ1+ε1εb1|v|a1kβ1b1kM1,kT1,v,(26)gk,uλ1+ε1ε˜b2ua2kβ2b2kM2,kT1,u.

Note that ε,ε˜ can be chosen arbitrarily small, so we can let(27)R>maxN1,N2,N3,N4,where(28)N1=2l=1TGl,la1l+β1b1l+M112εl=1TGl,lb1l,N2=2l=1TGl,la2l+β2b2l+M212ε˜l=1TGl,lb2l,N3=λ1T+1+Tε1εb1l=1TGl,la1l+a2l+β1b1l+β2b2l+M1+M2ε1εb1λ1T+1+Tε1εb1εl=1TGl,lb1l+ε˜l=1TGl,lb2l,N4=λ1T+1+Tε1εb2l=1TGl,la1l+a2l+β1b1l+β2b2l+M1+M2ε1εb2λ1T+1+Tε1εb2εl=1TGl,lb1l+ε˜l=1TGl,lb2l.

Now, we prove(29)u,vAu,vμϕ,ϕ,u,vBR,μ0,where ϕk=sinkπ/T+1,kT2. We argue this claim by indirection. Suppose that there exist u,vBR,μ0 such that(30)u,vAu,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 Lemma 3, 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+M1P,gl,ul+a2l+b2lβ2ul+M2P.

So, from Lemma 3 and Remark 1, we have(36)v+u˜,u+v˜P0.

Note that u,vBR, and using (24), R>N1,andR>N2, we have(37)v˜l=1TGl,la1l+b1lβ1vl+M1l=1TGl,la1l+b1lεv+β1+M1<R2,u˜l=1TGl,la2l+b2lε˜u+β2+M2<R2.

It is noted that u=v=R,u+u˜+v˜P0, and v+u˜+v˜P0. Therefore, we get(38)uk+u˜k+v˜k1Tu+u˜+v˜1Tuu˜+v˜1Tuu˜+v˜,vk+u˜k+v˜k1Tv+u˜+v˜1Tvu˜+v˜1Tvu˜+v˜.

Using R>N3, we have(39)ε1εb1l=1TGk,lvl+u˜l+v˜lλ1+ε1εb1l=1TGk,lu˜l+v˜lε1εb1Tl=1TGk,lRu˜+v˜λ1+ε1εb1l=1TGk,lu˜+v˜0,and R>N4 implies that(40)ε1ε˜b2l=1TGk,lul+u˜l+v˜lλ1+ε1ε˜b2l=1TGk,lu˜l+v˜l0.

Consequently, we obtain(41)Tvk+v˜k=l=1TGk,lfl,vl+a1l+b1lβ1vl+M1l=1TGk,lλ1+ε1εb1vla1lβ1b1lM1+a1l+b1lβ1vl+M1l=1TGk,lλ1+ε1εb1vlβ1b1l+b1lεvl+β1λ1+ε1εb1l=1TGk,lvlλ1+ε1εb1l=1TGk,lvl+u˜l+v˜lλ1+ε1εb1l=1TGk,lu˜l+v˜lλ1l=1TGk,lvl+u˜l+v˜lλ1l=1TGk,lvl+u˜l,Suk+u˜k=l=1Tgk,lgl,vl+a2l+b2lβ2ul+M2l=1TGk,lλ1+ε1ε˜b2ula2lβ2b2lM2+a2l+b2lβ2ul+M2λ1+ε1ε˜b2l=1TGk,lulλ1+ε1ε˜b2l=1TGk,lul+u˜l+v˜lλ1+ε1ε˜b2l=1TGk,lu˜l+v˜lλ1l=1TGk,lul+u˜l+v˜lλ1l=1TGk,lul+v˜l.

As a result, we get(42)Tvk+Suk+u˜k+v˜kλ1Lu+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+2μϕkλ1Lu+v+u˜+v˜k+2μϕk2μϕk.

Define μ=supSμsupμ>0:u+v+u˜+v˜2μϕ. Then, Sμ, μμ and u+v+u˜+v˜2μϕ. From ϕ=λ1Lϕ, we obtain(44)λ1Lu+v+u˜+v˜λ1L2μϕ=2μλ1Lϕ=2μϕ.

Hence,(45)u+v+u˜+v˜λ1Lu+v+u˜+v˜+2μϕ2μ+μϕ,which contradicts the definition of μ. Therefore, (29) holds, and from Lemma 4, we obtain(46)degIA,BR,0=0.

This completes the proof.

Proof of Theorem <xref ref-type="statement" rid="thm1">1</xref>.

From (H4), there exist ε20,λ1 and r0,R such that(47)fk,vλ1ε2v,gk,uλ1ε2u,kT1,u,vwithu,vr.

This implies that(48)Tvk=l=1TGk,lfl,vll=1TGk,lfl,vlλ1ε2l=1TGk,lvl,Suk=l=1TGk,lgl,ull=1TGk,lgl,ulλ1ε2l=1TGk,lul.

Consequently, we have(49)Tvk+Sukλ1ε2l=1TGk,lul+vl.

Now, we prove that(50)u,vμAu,v,for all u,vBr and μ0,1. We argue by contradiction. Suppose that there exist u,vBr and μ0,1 such that(51)u,v=μAu,v.

Therefore,(52)uk=μTvk,andvk=μSuk,kT1.

Hence, we have(53)uk+vkTvk+Sukλ1ε2l=1TGk,lul+vl.

From Lemma 1 (i) and Lemma 2, we have(54)λ1k=1TGk,lϕk=ϕl,lT1.

Multiplying both sides of (53) by sinkπ/T+1, then summing from 1 to T, and using (54), we obtain(55)k=1Tuk+vksinkπT+1λ1ε2k=1TsinkπT+1l=1TGk,lul+vl=λ1ε2λ1l=1Tul+vlsinlπT+1.

This implies that(56)k=1Tuk+vksinkπT+1=0.

Because sinkπ/T+100 for kT1, we have uk+vk0,kT1. This contradicts u,vBr. Therefore, (50) holds, and Lemma 5 implies that(57)degIA,Br,0=1.

Combining this with Lemma 6, we have(58)degIA,BRB¯r,0=degIA,BR,0degIA,Br,0=1.

Therefore, the operator A has at least one fixed point in BR/B¯r, and (6) has at least one nontrivial solution. This completes the proof.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Authors’ Contributions

This study was carried out in collaboration among all authors. All authors read and approved the final manuscript.

Acknowledgments

This study was supported by the project of National Social Science Fund of China (NSSFC) (18BTY015) and Shandong Province Higher Educational Science and Technology Program (J16LI01).

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