# Positive Solutions for Systems of Second-Order Difference Equations

Authors: Johnny Henderson; Rodica Luca
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

We study the existence and nonexistence of positive solutions of some systems of nonlinear second-order difference equations subject to multipoint boundary conditions which contain some positive constants.


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## Body

## 1. Introduction

The mathematical modeling of many nonlinear problems from computer science, economics, mechanical engineering, control systems, biological neural networks, and others leads to the consideration of nonlinear difference equations (see [1, 2]). In the last decades, many authors have investigated such problems by using various methods, such as fixed point theorems, the critical point theory, upper and lower solutions, the fixed point index theory, and the topological degree theory (see, e.g., [3–15]).In this paper, we consider the system of nonlinear second-order difference equations:SΔ2un-1+snfvn=0,n=1,N-1¯,Δ2vn-1+tngun=0,n=1,N-1¯, with the multipoint boundary conditions BCu0=∑i=1paiuξi+a0,uN=∑i=1qbiuηi,v0=∑i=1rcivζi,vN=∑i=1ldivρi+b0, where N,p,q,r,l∈N, N≥2, Δ is the forward difference operator with stepsize 1, Δun=un+1-un, Δ2un-1=un+1-2un+un-1, n=k,m¯ means that n=k,k+1,…,m for k,m∈N, ai∈R for all i=1,p¯, bi∈R for all i=1,q¯, ci∈R for all i=1,r¯, di∈R for all i=1,l¯, ξi∈N for all i=1,p¯, ηi∈N for all i=1,q¯, ζi∈N for all i=1,r¯, ρi∈N for all i=1,l¯, 1≤ξ1<⋯<ξp≤N-1, 1≤η1<⋯<ηq≤N-1, 1≤ζ1<⋯<ζr≤N-1, 1≤ρ1<⋯<ρl≤N-1, and a0 and b0 are positive constants.Under some assumptions on the functionsf and g, we will prove the existence of positive solutions of problem S-BC. By a positive solution of S-BC we mean a pair of sequences ((un)n=0,N¯,(vn)n=0,N¯) satisfying S and BC with un,vn≥0 for all n=0,N¯ and un>0 for all n=0,N-1¯, vn>0 for all n=1,N¯. We will also give sufficient conditions for the nonexistence of positive solutions for this problem. System S with the multipoint boundary conditions αu0-βΔu0=0, uN=∑i=1m-2aiuξi+a0, γv0-δΔv0=0, and vN=∑i=1p-2bivηi+b0 (a0,b0>0) has been investigated in [16]. Some systems of difference equations with parameters, subject to multipoint boundary conditions, were studied in [17, 18], by using the Guo-Krasnosel’skii fixed point theorem. We also mention the paper [19], where the authors investigated the existence and multiplicity of positive solutions for the system Δ2un-1+f(n,vn)=0, Δ2vn-1+g(n,un)=0, n=1,N-1¯, with the multipoint boundary conditions BC with a0=b0=0, by using some theorems from the fixed point index theory.In Section2, we present some auxiliary results which investigate a second-order difference equation subject to multipoint boundary conditions. In Section 3, we will prove our main results, and in Section 4, we will present an example which illustrates the obtained theorems. Our main existence result is based on the Schauder fixed point theorem which we present now.Theorem 1.
LetX be a Banach space and Y⊂X a nonempty, bounded, convex, and closed subset. If the operator A:Y→Y is completely continuous (continuous and compact, i.e., mapping bounded sets into relatively compact sets), then A has at least one fixed point.


## 2. Auxiliary Results

In this section, we present some auxiliary results from [17] related to the following second-order difference equation with the multipoint boundary conditions:(1)Δ2un-1+yn=0,n=1,N-1¯,u0=∑i=1paiuξi,uN=∑i=1qbiuηi,where N,p,q∈N, N≥2, ai∈R for all i=1,p¯, bi∈R for all i=1,q¯, ξi∈N for all i=1,p¯, ηi∈N for all i=1,q¯, 1≤ξ1<⋯<ξp≤N-1, and 1≤η1<⋯<ηq≤N-1.Lemma 2 (see [17]).
Ifai∈R for all i=1,p¯, bi∈R for all i=1,q¯, ξi∈N for all i=1,p¯, ηi∈N for all i=1,q¯, 1≤ξ1<⋯<ξp≤N-1, 1≤η1<⋯<ηq≤N-1, Δ1=1-∑i=1qbi∑i=1paiξi+1-∑i=1paiN-∑i=1qbiηi≠0, and yn∈R for all n=1,N-1¯, then the solution of (1) is given by un=∑j=1N-1G1(n,j)yj for all n=0,N¯, where Green’s function G1 is defined by (2)G1n,j=g0n,j+1Δ1N-n1-∑k=1qbk+∑i=1qbiN-ηi∑i=1paig0ξi,j+1Δ1n1-∑k=1pak+∑i=1paiξi∑i=1qbig0ηi,j,n=0,N¯,j=1,N-1¯,g0n,j=1NjN-n,1≤j≤n≤N,nN-j,0≤n≤j≤N-1.Lemma 3 (see [17]).
Ifai≥0 for all i=1,p¯, ∑i=1pai<1, bi≥0 for all i=1,q¯, ∑i=1qbi<1, ξi∈N for all i=1,p¯, 1≤ξ1<ξ2<⋯<ξp≤N-1, ηi∈N for all i=1,q¯, and 1≤η1<η2<⋯<ηq≤N-1, then Green’s function G1 of problem (1) satisfies G1(n,j)≥0 for all n=0,N¯, j=1,N-1¯. Moreover, if yn≥0 for all n=1,N-1¯, then the unique solution un, n=0,N¯, of problem (1) satisfies un≥0 for all n=0,N¯.Lemma 4 (see [17]).
Assume thatai≥0 for all i=1,p¯, ∑i=1pai<1, bi≥0 for all i=1,q¯, ∑i=1qbi<1, ξi∈N for all i=1,p¯, 1≤ξ1<ξ2<⋯<ξp≤N-1, ηi∈N for all i=1,q¯, and 1≤η1<η2<⋯<ηq≤N-1. Then Green’s function G1 of problem (1) satisfies the following inequalities:(a)
G1(n,j)≤I1(j), ∀n=0,N¯,j=1,N-1¯, where (3)I1j=g0j,j+1Δ1N-∑i=1qbiηi∑i=1paig0ξi,j+1Δ1N-∑i=1paiN-ξi∑i=1qbig0ηi,j.(b)
For everyc∈{1,…,N/2}, one has (4)minn=c,N-c¯⁡G1n,j≥γ1I1j≥γ1G1n′,j,∀n′=0,N¯,j=1,N-1¯,where(5)γ1=min⁡cN-1,c1-∑k=1qbk+∑i=1qbiN-ηiN-∑i=1qbiηi,c1-∑k=1pak+∑i=1paiξiN-∑i=1paiN-ξi>0,andN/2 is the largest integer not greater than N/2.Lemma 5 (see [17]).
Assume thatai≥0 for all i=1,p¯, ∑i=1pai<1, bi≥0 for all i=1,q¯, ∑i=1qbi<1, ξi∈N for all i=1,p¯, 1≤ξ1<ξ2<⋯<ξp≤N-1, ηi∈N for all i=1,q¯, 1≤η1<η2<⋯<ηq≤N-1, c∈{1,…,N/2}, and yn≥0 for all n=1,N-1¯. Then the solution un,n=0,N¯, of problem (1) satisfies the inequality minn=c,N-c¯un≥γ1maxm=0,N¯um.We can also formulate similar results as Lemmas2–5 above for the discrete boundary value problem(6)Δ2vn-1+hn=0,n=1,N-1¯,v0=∑i=1rcivζi,vN=∑i=1ldivρi,where N,r,l∈N, N≥2, ci∈R for all i=1,r¯, di∈R for all i=1,l¯, ζi∈N for all i=1,r¯, ρi∈N for all i=1,l¯, 1≤ζ1<⋯<ζr≤N-1, 1≤ρ1<⋯<ρl≤N-1, and hn∈R for all n=1,N-1¯. We denote by Δ2,γ2,G2, and I2 the corresponding constants and functions for problem (6) defined in a similar manner as Δ1,γ1,G1, and I1, respectively.


## 3. Main Results

We present first the assumptions that we will use in the sequel:(H1)
ξi∈N for all i=1,p¯, 1≤ξ1<⋯<ξp≤N-1, ai≥0 for all i=1,p¯, ∑i=1pai<1, ηi∈N for all i=1,q¯, 1≤η1<⋯<ηq≤N-1, bi≥0 for all i=1,q¯, ∑i=1qbi<1, ζi∈N for all i=1,r¯, 1≤ζ1<⋯<ζr≤N-1, ci≥0 for all i=1,r¯, ∑i=1rci<1, ρi∈N for all i=1,l¯, 1≤ρ1<⋯<ρl≤N-1, and di≥0 for all i=1,l¯, ∑i=1ldi<1.(H2)
The constantssn,tn≥0 for all n=1,N-1¯, and there exist i0,j0∈{1,…,N-1} such that si0>0, tj0>0.(H3)
f,g:[0,∞)→[0,∞) are continuous functions and there exists c0>0 such that f(u)<c0/L, g(u)<c0/L for all u∈[0,c0], where L=max{∑i=1N-1siI1(i),∑i=1N-1tiI2(i)} and I1,I2 are defined in Section 2.(H4)
f,g:[0,∞)→[0,∞) are continuous functions and satisfy the conditions limu→∞f(u)/u=∞,limu→∞g(u)/u=∞.Our first theorem is the following existence result for problemS-BC.Theorem 6.
Assume that assumptions(H1)–(H3) hold. Then problem S-BC has at least one positive solution for a0>0 and b0>0 sufficiently small.Proof.
We consider the problems(7)Δ2hn-1=0,n=1,N-1¯,h0=∑i=1paihξi+1,hN=∑i=1qbihηi,(8)Δ2kn-1=0,n=1,N-1¯,k0=∑i=1rcikζi,kN=∑i=1ldikρi+1.
Problems (7) and (8) have the solutions(9)hn=1Δ1-n1-∑i=1qbi+N-∑i=1qbiηi,n=0,N¯,kn=1Δ2n1-∑i=1rci+∑i=1rciζi,n=0,N¯,respectively, where Δ1 and Δ2 are defined in Section 2. By assumption (H1) we obtain hn>0 for all n=0,N-1¯ and kn>0 for all n=1,N¯.
We define the sequences(xn)n=0,N¯, (yn)n=0,N¯ by (10)xn=un-a0hn,yn=vn-b0kn,n=0,N¯, where ((un)n=0,N¯,(vn)n=0,N¯) is a solution of S-BC. Then S-BC can be equivalently written as(11)Δ2xn-1+snfyn+b0kn=0,n=1,N-1¯,Δ2yn-1+tngxn+a0hn=0,n=1,N-1¯,with the boundary conditions(12)x0=∑i=1raixξi,xN=∑i=1qbixηi,y0=∑i=1rciyζi,yN=∑i=1ldiyρi.
Using Green’s functionsG1 and G2 from Section 2, a pair ((xn)n=0,N¯,(yn)n=0,N¯) is a solution of problem (11)-(12) if and only if it is a solution for the problem(13)xn=∑i=1N-1G1n,isif∑j=1N-1G2i,jtjgxj+a0hj+b0ki,n=0,N¯,yn=∑i=1N-1G2n,itigxi+a0hi,n=0,N¯,where (hn)n=0,N¯, (kn)n=0,N¯ are given in (9).
We consider the Banach spaceX=RN+1 with the norm u=maxn=0,N¯|un|, u=(un)n=0,N¯, and we define the set M={(xn)n=0,N¯,0≤xn≤c0,∀n=0,N¯}⊂X.
We also define the operatorE:M→X by (14)Ex=∑i=1N-1G1n,isif∑j=1N-1G2i,jtjgxj+a0hj+b0kin=0,N¯,x=xnn=0,N¯∈M.
For sufficiently smalla0>0 and b0>0, by (H3), we deduce(15)fyn+b0kn≤c0L,gxn+a0hn≤c0L,∀n=0,N¯,∀xnn,ynn∈M.
Then, by using Lemma3, we obtain E(x)n≥0 for all n=0,N¯ and x=(xn)n=0,N¯∈M. By Lemma 4, for all x∈M, we have (16)∑j=1N-1G2i,jtjgxj+a0hj≤∑j=1N-1I2jtjgxj+a0hj≤c0L∑j=1N-1tjI2j≤c0,∀i=1,N-1¯,Exn≤∑i=1N-1I1isif∑j=1N-1G2i,jtjgxj+a0hj+b0ki≤c0L∑i=1N-1siI1i≤c0,∀n=0,N¯. Therefore E(M)⊂M.
Using standard arguments, we deduce thatE is completely continuous. By Theorem 1, we conclude that E has a fixed point x=(xn)n=0,N¯∈M. This element together with y=(yn)n=0,N¯ given by (13) represents a solution for (11)-(12). This shows that our problem S-BC has a positive solution ((un)n=0,N¯,(vn)n=0,N¯) with un=xn+a0hn,vn=yn+b0kn,n=0,N¯, (un>0 for all n=0,N-1¯ and vn>0 for all n=1,N¯) for sufficiently small a0>0 and b0>0.In what follows, we present sufficient conditions for the nonexistence of positive solutions ofS-BC.Theorem 7.
Assume that assumptions(H1), (H2), and (H4) hold. Then problem S-BC has no positive solution for a0 and b0 sufficiently large.Proof.
We suppose that((un)n=0,N¯,(vn)n=0,N¯) is a positive solution of S-BC. Then ((xn)n=0,N¯,(yn)n=0,N¯) with xn=un-a0hn,yn=vn-b0kn,n=0,N¯, is a solution for (11)-(12), where (hn)n=0,N¯ and (kn)n=0,N¯ are the solutions of problems (7) and (8), respectively (given by (9)). By (H2) there exists c∈{1,2,…,N/2} such that i0,j0∈{c,…,N-c} and then ∑i=cN-csiI1(i)>0 and ∑i=cN-ctiI2(i)>0. By using Lemma 3 we have xn≥0,yn≥0 for all n=0,N¯, and by Lemma 5, we obtain minn=c,N-c¯xn≥γ1x and minn=c,N-c¯yn≥γ2y, where γ1 and γ2 are defined in Section 2.
Using now (9), we deduce that (17)minn=c,N-c¯⁡hn=hN-c=hN-ch0h,minn=c,N-c¯⁡kn=kc=kckNk.
Therefore, we obtain(18)minn=c,N-c¯⁡xn+a0hn≥γ1x+a0hN-ch0h≥r1x+a0h≥r1x+a0h,minn=c,N-c¯⁡yn+b0kn≥γ2y+b0kckNk≥r2y+b0k≥r2y+b0k, where r1=min{γ1,hN-c/h0} and r2=min{γ2,kc/kN}.
We now consider(19)R=minγ2r1∑i=cN-ctiI2i,γ1r2∑i=cN-csiI1i-1>0.
By using(H4), for R defined above, we conclude that there exists M0>0 such that f(u)>2Ru, g(u)>2Ru for all u≥M0. We consider a0>0 and b0>0 sufficiently large such that(20)minn=c,N-c¯⁡xn+a0hn≥M0,minn=c,N-c¯⁡yn+b0kn≥M0.By (H2), (11), (12), and the above inequalities, we deduce that x>0 and y>0.
Now, by using Lemma4 and the above considerations, we have (21)yc=∑i=1N-1G2c,itigxi+a0hi≥γ2∑i=1N-1I2itigxi+a0hi≥γ2∑i=cN-cI2itigxi+a0hi≥2Rγ2∑i=cN-cI2itixi+a0hi≥2Rγ2∑i=cN-cI2itiminj=c,N-c¯⁡xj+a0hj≥2Rγ2r1∑i=cN-cI2itix+a0h≥2x+a0h≥2x.Therefore, we obtain(22)x≤yc2≤y2.
In a similar manner, we deduce(23)xc=∑i=1N-1G1c,isifyi+b0ki≥γ1∑i=1N-1I1isifyi+b0ki≥γ1∑i=cN-cI1isifyi+b0ki≥2Rγ1∑i=cN-cI1isiyi+b0ki≥2Rγ1∑i=cN-cI1isiminj=c,N-c¯⁡yj+b0kj≥2Rγ1r2∑i=cN-cI1isiy+b0k≥2y+b0k≥2y. So, we obtain(24)y≤xc2≤x2.
By (22) and (24), we obtain x≤y/2≤x/4, which is a contradiction, because x>0. Then, for a0 and b0 sufficiently large, problem S-BC has no positive solution.Similar results as Theorems6 and 7 can be obtained if instead of boundary conditions BC we have BC1u0=∑i=1paiuξi,uN=∑i=1qbiuηi+a0,v0=∑i=1rcivζi+b0,vN=∑i=1ldivρi or BC2u0=∑i=1paiuξi+a0,uN=∑i=1qbiuηi,v0=∑i=1rcivζi+b0,vN=∑i=1ldivρi or BC3u0=∑i=1paiuξi,uN=∑i=1qbiuηi+a0,v0=∑i=1rcivζi,vN=∑i=1ldivρi+b0, where a0 and b0 are positive constants.For problemS–BC1, instead of sequences (hn)n=0,N¯ and (kn)n=0,N¯ from the proof of Theorem 6, the solutions of problems(25)Δ2h~n-1=0,n=1,N-1¯,h~0=∑i=1paih~ξi,h~N=∑i=1qbih~ηi+1,(26)Δ2k~n-1=0,n=1,N-1¯,k~0=∑i=1rcik~ζi+1,k~N=∑i=1ldik~ρiare (27)h~n=1Δ1n1-∑i=1pai+∑i=1paiξi,n=0,N¯,k~n=1Δ2-n1-∑i=1ldi+N-∑i=1ldiρi,n=0,N¯, respectively. By assumption (H1) we obtain h~n>0, for all n=1,N¯, and k~n>0 for all n=0,N-1¯.For problemS–BC2, instead of sequences (hn)n=0,N¯ and (kn)k=0,N¯ from Theorem 6, the solutions of problems (7) and (26) are (hn)n=0,N¯ and (k~n)n=0,N¯, respectively, which satisfy hn>0, for all n=0,N-1¯, and k~n>0 for all n=0,N-1¯. For problem S–BC3, instead of sequences (hn)n=0,N¯ and (kn)k=0,N¯ from Theorem 6, the solutions of problems (25) and (8) are (h~n)n=0,N¯ and (kn)n=0,N¯, respectively, which satisfy h~n>0, for all n=1,N¯, and kn>0 for all n=1,N¯.Therefore we also obtain the following results.Theorem 8.
Assume that assumptions(H1)–(H3) hold. Then problem S–BC1 has at least one positive solution (un>0, for all n=1,N¯, and vn>0 for all n=0,N-1¯) for a0>0 and b0>0 sufficiently small.Theorem 9.
Assume that assumptions(H1), (H2), and (H4) hold. Then problem S–BC1 has no positive solution (un>0, for all n=1,N¯, and vn>0 for all n=0,N-1¯) for a0 and b0 sufficiently large.Theorem 10.
Assume that assumptions(H1)–(H3) hold. Then problem S–BC2 has at least one positive solution (un>0, for all n=0,N-1¯, and vn>0 for all n=0,N-1¯) for a0>0 and b0>0 sufficiently small.Theorem 11.
Assume that assumptions(H1), (H2), and (H4) hold. Then problem S–BC2 has no positive solution (un>0, for all n=0,N-1¯, and vn>0 for all n=0,N-1¯) for a0 and b0 sufficiently large.Theorem 12.
Assume that assumptions(H1)–(H3) hold. Then problem S–BC3 has at least one positive solution (un>0, for all n=1,N¯, and vn>0 for all n=1,N¯) for a0>0 and b0>0 sufficiently small.Theorem 13.
Assume that assumptions(H1), (H2), and (H4) hold. Then problem S–BC3 has no positive solution (un>0, for all n=1,N¯, and vn>0 for all n=1,N¯) for a0 and b0 sufficiently large.


## 4. An Example

We considerN=20, sn=c^/(n+1), tn=d^/n for all n=1,19¯, c^>0, d^>0, p=2, q=3, r=1, l=2, a1=1/2, a2=1/3, ξ1=4, ξ2=16, b1=1/3, b2=1/4, b3=1/5, η1=5, η2=10, η3=15, c1=3/4, ζ1=10, d1=1/3, d2=1/5, ρ1=3, and ρ2=18. We also consider the functions f,g:[0,∞)→[0,∞), f(x)=a~xα1/2x+1, and g(x)=b~xα2/3x+2, for all x∈[0,∞), with a~,b~>0 and α1,α2>2. We have limx→∞f(x)/x=limx→∞g(x)/x=∞.Therefore, we consider the system of second-order difference equationsS0Δ2un-1+c^a~vnα1n+12vn+1=0,n=1,19¯,Δ2vn-1+d^b~unα2n3un+2=0,n=1,19¯, with the multipoint boundary conditions BC0u0=12u4+13u16+a0,u20=13u5+14u10+15u15,v0=34v10,v20=13v3+15v18+b0,where a0 and b0 are positive constants.We have∑i=12ai=5/6<1, ∑i=13bi=47/60<1, ∑i=11ci=3/4<1, and ∑i=12di=8/15<1. The functions I1 and I2 are given by (28)I1j=2565j671-j220,1≤j≤4,7820671+70j61-j220,5≤j≤9,12620671+290j671-j220,10≤j≤15,30700671-864j671-j220,16≤j≤19.I2j=145j63-j220,1≤j≤2,250147+85j49-j220,3≤j≤9,2560147+8j49-j220,10≤j≤17,3460147-26j147-j220,18≤j≤19.Hence, we deduce that assumptions(H1), (H2), and (H4) are satisfied. In addition, by using the above functions I1 and I2, we obtain A~:=∑i=119I1(i)/(i+1)≈30.1784002, B~:=∑i=119I2(i)/i≈23.63831254, and then L=max{c^A~,d^B~}. We choose c0=1 and if we select a~, b~ satisfying the conditions a~<3/L=3min1/(c^A~),1/(d^B~), b~<5/L=5min1/(c^A~),1/(d^B~), then we conclude that f(x)≤a~/3<1/L, g(x)≤b~/5<1/L for all x∈[0,1]. For example, if c^=1, d^=2, then for a~≤0.063 and b~≤0.105 the above conditions for f and g are satisfied. So, assumption (H3) is also satisfied. By Theorems 6 and 7 we deduce that problem S0-BC0 has at least one positive solution (here un>0 and vn>0 for all n=0,20¯) for sufficiently small a0>0 and b0>0 and no positive solution for sufficiently large a0 and b0.By the proofs of Theorems6 and 7 we can find some intervals for a0 and b0 such that problem S0-BC0 has at least one positive solution, or it has no positive solution. We consider a~=0.063, b~=0.105, c^=1, d^=2, c0=1, L=max{A~,2B~}=2B~ (as above), α1=3, and α2=4. Then Δ1=671/180, Δ2=147/20, and the sequences (hn)n=0,20¯ and (kn)n=0,20¯ from (9) are hn=-39n+2310/671 and kn=5n+150/147 for all n=0,20¯. We also obtain hmax=h0=210/61 and kmax=k20=250/147. If we choose b0≤(f-1(1/L)-1)/kmax and a0≤(g-1(1/L)-1)/hmax, then inequalities (15) are satisfied. Because f-1(1/L)≈1.0031 and g-1(1/L)≈1.00213, for a0≤6.17·10-4 and b0≤18.21·10-4 problem S0-BC0 has at least one positive solution.Now we choosec=5 (the constant from the beginning of the proof of Theorem 7), and then we obtain γ1=γ2=5/19, h15=1725/671, k5=175/147, r1=min{γ1,h15/h0}=5/19, r2=min{γ2,k5/k20}=5/19, ∑i=515tiI2(i)≈27.88321391, ∑i=515siI1(i)≈18.41083231, and R≈0.78432087 (given by (19)). For R~:=2R+0.1, the inequalities f(x)/x≥R~ and g(y)/y≥R~ are satisfied for x≥M0′≈53.468121231 and y≥M0′′≈7.2166269, respectively. We consider M0=max{M0′,M0′′}=M0′, and then for a0≥M0/h15 and b0≥M0/k5, inequalities (20) are satisfied. Therefore, if a0≥20.7984 and b0≥44.9133, problem S0-BC0 has no positive solution.



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