# Positive Solutions for a System of Semipositone Fractional Difference Boundary Value Problems

Authors: Cheng Chen; Jiafa Xu; Donal O’Regan; Zhengqing Fu
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Using the fixed point index, we establish two existence theorems for positive solutions to a system of semipositone fractional difference boundary value problems. We adopt nonnegative concave functions and nonnegative matrices to characterize the coupling behavior of our nonlinear terms.


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## Body

## 1. Introduction

In this paper we study the existence of positive solutions for the system of fractional difference boundary value problems involving semipositone nonlinearities:(1)-Δν-3νxt=ft+ν-1,xt+ν-1,yt+ν-1,t∈0,b+2N0,-Δν-3νyt=gt+ν-1,xt+ν-1,yt+ν-1,t∈0,b+2N0,xν-3=Δν-3αxtt=ν-α-2=Δν-3βxtt=ν+b+2-β=0,yν-3=Δν-3αytt=ν-α-2=Δν-3βytt=ν+b+2-β=0,where 2<ν≤3, 1<β<2, ν-β>1, 0<α<1, b>3(b∈N), and Δν-3ν is a discrete fractional operator. For the nonlinear terms f,g, we assume the following.(H0)f,g:[ν-1,b+ν+1]Nν-1×R+×R+→R are two continuous functions; moreover, there exists a positive constant M>0 such that (2)f,gt,x,y≥-M,for  allt,x,y∈ν-1,b+ν+1Nν-1×R+×R+.Note that, in this paper, we use[a,b]Na to stand for {a,a+1,a+2,…,b} with b-a∈N1, where Na≔{a,a+1,a+2,⋯}.Fractional calculus has been applied in physics, chemistry, aerodynamics, biophysics, and blood flow phenomena. For example,CD4+T cells’ infections can be depicted by a fractional order model(3)Dα1T=s-KVT-dT+bI,Dα2I=KVT-b+δI,Dα3V=NδI-cV,where Dαi(i=1,2,3) are fractional derivatives (see [1, 2]); we also refer the reader to [1–45] and the references therein. In [3], the authors considered the existence of positive solutions for the semipositone discrete fractional system(4)-Δν1y1t=λ1f1t+ν1-1,y1t+ν1-1,y2t+ν2-1,t∈1,b+1N,-Δν2y2t=λ2f2t+ν2-1,y1t+ν1-1,y2t+ν2-1,t∈1,b+1N,y1ν1-2=y1ν1+b+1=0,y2ν2-2=y2ν2+b+1=0,where ν1,ν2∈1,2. Using the Guo-Krasnosel’skiĭ fixed point theorem, the authors showed that the problem has positive solutions for sufficiently small values of λ1,λ2>0. The growth conditions on fi(i=1,2) are superlinear; i.e.,(5)limy1+y2→+∞⁡fit,x,yy1+y2=+∞,limy1+y2→0+⁡fit,x,yy1+y2=0,uniformly for t∈νi,νi+bNνi-2. Using conditions of (5) type the existence of solutions for various fractional boundary value problems was considered in [1, 4–6, 9, 11–13].In this paper, we use the fixed point index to obtain two existence theorems for positive solutions to (1) with semipositone nonlinearities. We adopt some appropriate nonnegative concave functions and nonnegative matrices to characterize the coupling behavior of our nonlinear terms. Moreover, the growth conditions on +∞ of our nonlinearities f,g are an improvement of (5); see conditions (H1) and (H3) in Section 3.


## 2. Preliminaries

We first recall some background materials from discrete fractional calculus; for more details we refer the reader to [10].Definition 1.
We definetν_≔Γ(t+1)/Γ(t+1-ν) for any t,ν∈R for which the right-hand side is well-defined. We use 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 2.
Forν>0, the ν-th fractional sum of a function f is(6)Δa-νft=1Γν∑s=at-νt-s-1ν-1_fs,fort∈Na+ν.We also define the ν-th fractional difference for ν>0 by(7)Δaνft=ΔNΔaν-Nft,fort∈Na+N-ν,where N∈N with 0≤N-1<ν≤N.
Leth:[ν-1,b+ν+1]Nν-1→R be a continuous function. Then we consider the fractional difference boundary value problems(8)-Δν-3νyt=ht+ν-1,t∈0,b+2N0,yν-3=Δν-3αytt=ν-α-2=Δν-3βytt=ν+b+2-β=0,where ν,α,β,b are as in (1). The following two lemmas are in [9], so we omit their proofs.Lemma 3 (see [9], Lemma 4).
Problem (8) has a unique solution(9)yt=∑s=0b+2Gt,shs+ν-1,t∈ν-1,b+ν+1Nν-1,where(10)Gt,s=1Γνtν-1_ν+b-β-s+1ν-β-1_ν+b-β+2ν-β-1_-t-s-1ν-1_,0≤s<t-ν+1≤b+2,tν-1_ν+b-β-s+1ν-β-1_ν+b-β+2ν-β-1_,0≤t-ν+1≤s≤b+2.Lemma 4 (see [9], Lemma 5).
Green’s function (10) has the following properties.
(i)G(t,s)>0, (t,s)∈[ν-1,b+ν+1]Nν-1 × [0,b+2]N0,
(ii)q∗(t)G(b+ν+1,s) ≤ G(t,s) ≤ G(b+ν+1,s), (t,s) ∈ [ν-1,b+ν+1]Nν-1 × [0,b+2]N0, where q∗(t) = tν-1_/b+ν+1ν-1_.
Letφ(s+ν-1)=G(b+ν+1,s) for s∈[0,b+2]N0. Then φ(t)=G(b+ν+1,t-ν+1) for t∈[ν-1,b+ν+1]Nν-1. From Lemma 4, the following inequalities are satisfied:(11)∑t=ν-1b+ν+1q∗tφt·φs+ν-1≤∑t=ν-1b+ν+1Gt,sφt≤∑t=ν-1b+ν+1φt·φs+ν-1,fors∈0,b+2N0.For convenience, we let(12)κ1=∑t=ν-1b+ν+1q∗tφtandκ2=∑t=ν-1b+ν+1φt.
LetE be the collection of all maps from [ν-3,b+ν+1]Nν-3 to R equipped with the max norm, ·. Then E is a Banach space. Define a set P⊂E by P=y∈E:y(t)≥0,t∈[ν-1,b+ν+1]Nν-1. Then P is a cone in E. Note that E×E is a Banach space with the norm (x,y)≔max{x,y}, and P×P is a cone in E×E.
From Lemma3, for all t∈[ν-1,b+ν+1]Nν-1, we have that (1) is equivalent to(13)xt=∑s=0b+2Gt,sfs+ν-1,xs+ν-1,ys+ν-1,yt=∑s=0b+2Gt,sgs+ν-1,xs+ν-1,ys+ν-1,where G is defined in (10).Lemma 5 (see [46]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set and that A:Ω¯∩P→P is a continuous compact operator. If there exists ω0∈P∖{0} such that (14)ω-Aω≠λω0,∀λ≥0,ω∈∂Ω∩P,then i(A,Ω∩P,P)=0, where i denotes the fixed point index on P.Lemma 6 (see [46]).
LetE be a real Banach space and P a cone on E. Suppose that Ω⊂E is a bounded open set with 0∈Ω and that A:Ω¯∩P→P is a continuous compact operator. If(15)ω-λAω≠0,∀λ∈0,1,ω∈∂Ω∩P,then i(A,Ω∩P,P)=1.


## 3. Main Results

Letω be a solution of(16)-Δν-3νyt=1,t∈0,b+2N0,yν-3=Δν-3αytt=ν-α-2=Δν-3βytt=ν+b+2-β=0,where ν,α,β,b are as in (1). Define z=Mω, and then, from Lemmas 3 and 4, we have (17)zt=Mωt=M∑s=0b+2Gt,s≤M∑s=0b+2φs+ν-1=M∑s=ν-1b+ν+1φs=Mκ2.We note that (1) has a positive solution (x,y)∈(P×P)∖{0} if and only if (x~,y~)=(x+z,y+z) is a solution of the fractional difference boundary value problems(18)-Δν-3νxt=f~t+ν-1,xt+ν-1-zt+ν-1,yt+ν-1-zt+ν-1,t∈0,b+2N0,-Δν-3νyt=g~t+ν-1,xt+ν-1-zt+ν-1,yt+ν-1-zt+ν-1,t∈0,b+2N0,xν-3=Δν-3αxtt=ν-α-2=Δν-3βxtt=ν+b+2-β=0,yν-3=Δν-3αytt=ν-α-2=Δν-3βytt=ν+b+2-β=0,and (x~,y~)(t)≥(z,z)(t) for t∈[ν-1,b+ν+1]Nν-1, where ν,α,β,b are as in (1) and (19)f~t,x,y=ft,x,y+M,t∈ν-1,b+ν+1Nν-1,x,y≥0,ft,x,0+M,t∈ν-1,b+ν+1Nν-1,x>0,y<0,ft,0,y+M,t∈ν-1,b+ν+1Nν-1,x<0,y>0,ft,0,0+M,t∈ν-1,b+ν+1Nν-1,x,y<0,and(20)g~t,x,y=gt,x,y+M,t∈ν-1,b+ν+1Nν-1,x,y≥0,gt,x,0+M,t∈ν-1,b+ν+1Nν-1,x>0,y<0,gt,0,y+M,t∈ν-1,b+ν+1Nν-1,x<0,y>0,gt,0,0+M,t∈ν-1,b+ν+1Nν-1,x,y<0.Note that for (x1,y1)(t)≥(x2,y2)(t), we mean x1(t)≥x2(t), y1(t)≥y2(t) for all t∈[ν-1,b+ν+1]Nν-1.For(x,y)∈P×P, and t∈[ν-1,b+ν+1]Nν-1, we define the operators(21)B1x,yt=∑s=0b+2Gt,sf~s+ν-1,xs+ν-1-zs+ν-1,ys+ν-1-zs+ν-1,B2x,yt=∑s=0b+2Gt,sg~s+ν-1,xs+ν-1-zs+ν-1,ys+ν-1-zs+ν-1,and(22)Bx,yt=B1,B2x,yt.Then (H0) and using the Arzelà-Ascoli theorem in a standard way establish that B:P×P→P×P is a completely continuous operator. It is clear that (x,y)∈(P×P)∖{0} is a positive solution for (18) if and only if (x,y)∈(P×P)∖{0} is a fixed point of B.LetP0={y∈P:y(t)≥q∗(t)y,∀t∈[ν-1,b+ν+1]Nν-1}. Then from Lemma 4 we have(23)BiP×P⊂P0,i=1,2.If we seek a fixed point (x~,y~) of B then x~,y~∈P0 and(24)wt-zt≥q∗tw-Mκ2≥q0w-Mκ2,fort∈ν-1,b+ν+1Nν-1,where w=x~,y~, and q0=mint∈[ν-1,b+ν+1]Nν-1q∗(t)>0, so as a result if x~,y~≥q0-1Mκ2 then (x~,y~)(t)≥(z,z)(t) for t∈[ν-1,b+ν+1]Nν-1 (i.e., (x~-z,y~-z)(t) is a positive solution for (1)).For convenience, we usec1,c2,… to stand for different positive constants. Let Bϱ≔{x∈E:x<ϱ} for ϱ>0. Now, we list our assumptions on f~,g~ (the first two are needed for Theorem 10 and the last two are needed for Theorem 11):(H1) There existp,q∈C(R+,R+) such that(i)p is concave and strictly increasing on R+,(ii) there existsc1>0 such that(25)f~t,x,y≥x+py-c1,g~t,x,y≥qx-c1,∀t,x,y∈ν-1,b+ν+1Nν-1×R+×R+,(iii) there is aγ1>0 such that κ1(1+γ1κ1)>1 and(26)pκ2qxt+ν-1-zt+ν-1≥κ2γ1xt+ν-1-zt+ν-1-c1for x∈R+ and t∈[0,b+2]N0.(H2) For any(t,x,y) ∈ [ν-1,b+ν+1]Nν-1 × 0,q0-1Mκ2 × 0,q0-1Mκ2, assume(27)f~,g~t,x,y<q0-1M.(H3) There existei≥0(i=1,2,3,4) with e12+e22≠0,e32+e42≠0 such that(i)κ=(1-e1κ2)(1-e4κ2)-e2e3κ22>0, e1,e4<κ2-1,(ii) there existc2>0 such that (28)f~t,x,yg~t,x,y≤e1x+e2y+c2e3x+e4y+c2,∀t,x,y∈ν-1,b+ν+1Nν-1×R+×R+.(H4) For any(t,x,y) ∈ [ν-1,b+ν+1]Nν-1 × [0,q0-1Mκ2] × [0,q0-1Mκ2], assume (29)f~,g~t,x,y>q0-2M.Example 7.
Letp(y)=y19/20 and q(x)=x2 for x,y∈R+. Then, for any ω>0, we have(30)liminfx→+∞⁡pωqxx=liminfx→+∞⁡ω19/20x19/10x=+∞.Let f(t,x,y)=(1/2κ2)x+(1/2κ2eβ1+cos⁡tx)y-M and g(t,x,y)=(1/eβ2+sin⁡tx)(q0-1M)1-β3κ2-β3xβ3-M, where β1,β2>1, β3>2, for (t,x,y)∈[ν-1,b+ν+1]Nν-1×R+×R+. Then, for any (t,x,y)∈[ν-1,b+ν+1]Nν-1×[0,q0-1Mκ2]×[0,q0-1Mκ2], we have(31)ft,x,y+M=12κ2x+12κ2eβ1+costxy<12κ2q0-1Mκ2+12κ2q0-1Mκ2=q0-1M,and(32)gt,x,y+M=1eβ2+sintxq0-1M1-β3κ2-β3xβ3<q0-1M1-β3κ2-β3q0-1Mκ2β3=q0-1M.Also,(33)liminfx→+∞,y→+∞⁡ft,x,y+Mx+py=liminfx→+∞,y→+∞⁡1/2κ2x+1/2κ2eβ1+cos⁡txyx+y19/20=+∞,fort∈ν-1,b+ν+1Nν-1,and(34)liminfx→+∞⁡gt,x,y+Mqx=liminfx→+∞⁡1/eβ2+sin⁡txq0-1M1-β3κ2-β3xβ3x2=+∞,uniformly  ont,y∈ν-1,b+ν+1Nν-1×R+.Thus, (H1)-(H2) are satisfied.Example 8.
Letf(t,x,y)=q0-2Meq0-1Mκ2+esintxye-(x+y)/2-M and g(t,x,y)=q0-2Meq0-1Mκ2+ecostxye-(x+y)/2-M, for (t,x,y)∈[ν-1,b+ν+1]Nν-1×R+×R+. Then, for any (t,x,y)∈[ν-1,b+ν+1]Nν-1×[0,q0-1Mκ2]×[0,q0-1Mκ2], we have(35)ft,x,y+M,gt,x,y+M>q0-2Meq0-1Mκ2e-q0-1Mκ2=q0-2M.Also,(36)limsupx→+∞,y→+∞⁡ft,x,y+Me1x+e2y=limsupx→+∞,y→+∞⁡q0-2Meq0-1Mκ2+esintxye-x+y/2e1x+e2y=0,and(37)limsupx→+∞,y→+∞⁡gt,x,y+Me3x+e4y=limsupx→+∞,y→+∞⁡q0-2Meq0-1Mκ2+ecostxye-x+y/2e3x+e4y=0,for t∈[ν-1,b+ν+1]Nν-1. Thus, (H3)-(H4) hold.Remark 9.
(i) In (H1), the growth condition for nonlinear termf depends on two variables x,y; however, in [7], this corresponding condition only involves one variable.
(ii) When nonlinear termsf,g grow sublinearly at +∞, nonnegative matrices are used to depict the coupling behavior of our nonlinearities. This is different from condition (H4) in [7].Theorem 10.
Suppose that (H0)-(H2) hold. Then (1) has at least one positive solution.Proof.
We first claim that there exists a sufficiently large positive numberR>q0-1Mκ2 such that(38)x,y≠Bx,y+λx0,y0,∀x,y∈∂BR×BR∩P×P,λ≥0,where x0,y0∈P0 are two given functions. Suppose not. Then there exist (x,y)∈∂(BR×BR)∩(P×P) and λ≥0 such that (x,y)=B(x,y)+λ(x0,y0), and so(39)xt=B1x,yt+λx0t,yt=B2x,yt+λy0t,fort∈ν-1,b+ν+1Nν-1.This implies x(t)≥B1(x,y)(t), and y(t)≥B2(x,y)(t) for t∈[ν-1,b+ν+1]Nν-1. From (H1) we have(40)xt≥B1x,yt≥∑s=0b+2Gt,sxs+ν-1-zs+ν-1+pys+ν-1-zs+ν-1-c1≥∑s=0b+2Gt,sxs+ν-1+∑s=0b+2Gt,spys+ν-1-zs+ν-1-c3≥∑s=0b+2Gt,sxs+ν-1+∑s=0b+2Gt,spys+ν-1-pzs+ν-1-c3≥∑s=0b+2Gt,sxs+ν-1+∑s=0b+2Gt,spys+ν-1-c4,fort∈ν-1,b+ν+1Nν-1,and(41)yt≥B2x,yt≥∑s=0b+2Gt,sqxs+ν-1-zs+ν-1-c1≥∑s=0b+2Gt,sqxs+ν-1-zs+ν-1-c3,fort∈ν-1,b+ν+1Nν-1.As a result, for t∈[0,b+2]N0, we have(42)pyt+ν-1+pc3≥pyt+ν-1+c3≥p∑s=0b+2Gt+ν-1,sqxs+ν-1-zs+ν-1=p∑s=0b+2Gt+ν-1,sκ2-1κ2qxs+ν-1-zs+ν-1≥∑s=0b+2Gt+ν-1,sκ2-1pκ2qxs+ν-1-zs+ν-1≥∑s=0b+2Gt+ν-1,sκ2-1κ2γ1xs+ν-1-zs+ν-1-c1≥γ1∑s=0b+2Gt+ν-1,sxs+ν-1-zs+ν-1-c5≥γ1∑s=0b+2Gt+ν-1,sxs+ν-1-c6.Thus(43)pyt+ν-1≥γ1∑s=0b+2Gt+ν-1,sxs+ν-1-c7,and, therefore,(44)xt≥∑s=0b+2Gt,sxs+ν-1+∑s=0b+2Gt,spys+ν-1-c4≥∑s=0b+2Gt,sxs+ν-1+∑s=0b+2Gt,sγ1∑τ=0b+2Gs+ν-1,τxτ+ν-1-c7-c4≥∑s=0b+2Gt,sxs+ν-1+γ1∑s=0b+2Gt,s∑τ=0b+2Gs+ν-1,τxτ+ν-1-c8,fort∈ν-1,b+ν+1Nν-1.Multiply both sides of the above inequality by φ(t) and sum from ν-1 to b+ν+1 and together with (11) we obtain(45)∑t=ν-1b+ν+1xtφt=∑t=0b+2xt+ν-1φt+ν-1≥∑t=0b+2φt+ν-1∑s=0b+2Gt+ν-1,sxs+ν-1+γ1∑s=0b+2Gt+ν-1,s∑τ=0b+2Gs+ν-1,τxτ+ν-1-c8≥κ1+γ1κ12∑t=0b+2xt+ν-1φt+ν-1-c9=κ1+γ1κ12∑t=ν-1b+ν+1xtφt-c9.From (23), (39), and x0∈P0 we have x∈P0. This implies(46)κ1x=x∑t=ν-1b+ν+1q∗tφt≤∑t=ν-1b+ν+1xtφt≤c9κ11+γ1κ1-1,and(47)x≤κ1-1c9κ11+γ1κ1-1.Note that, from (23), (39), and y0∈P0, we find y∈P0. Moreover, we may assume y(t)≢0, for t∈[ν-1,b+ν+1]Nν-1. Then y>0 and p(y)>0. Thus, from the concavity of p, we have(48)κ1y=y∑t=ν-1b+ν+1q∗tφt≤∑t=ν-1b+ν+1ytφt=∑t=0b+2yt+ν-1φt+ν-1=ypy∑t=0b+2yt+ν-1ypyφt+ν-1≤ypy∑t=0b+2pyt+ν-1φt+ν-1.This implies that(49)py≤κ1-1∑t=0b+2pyt+ν-1φt+ν-1.
From (40) and Lemma 4 we obtain(50)xt+c4≥∑s=0b+2Gt,spys+ν-1≥∑s=0b+2q∗tφs+ν-1pys+ν-1≥q0∑t=0b+2pyt+ν-1φt+ν-1.Combining the above two inequalities, we get(51)py≤κ1q0-1xt+c4≤κ1q0-1κ1-1c9κ11+γ1κ1-1+c4.From (H1), limz→+∞p(z)=+∞, and thus there exists M1>0 such that y≤M1.
Hence, we havex≤κ1-1c9/(κ1(1+γ1κ1)-1) and y≤M1. As a result, choosing R>max{q0-1Mκ2,κ1-1c9/(κ1(1+γ1κ1)-1),M1} we have a contradiction (recall in general ∂(A×B)=(∂A×B¯)∪(A¯×∂B)). Thus (38) is true. Consequently Lemma 5 (with R chosen above) implies(52)iB,BR×BR∩P×P,P×P=0.
Now we show that(53)x,y≠λBx,y,∀x,y∈∂Bq0-1Mκ2×Bq0-1Mκ2∩P×P,λ∈0,1.Suppose not. Then there exist (x,y)∈∂(Bq0-1Mκ2×Bq0-1Mκ2)∩(P×P),λ0∈[0,1] such that (x,y)=λ0B(x,y). This implies that(54)xt≤B1x,yt,yt≤B2x,yt,fort∈ν-1,b+ν+1Nν-1.Hence, x≤B1(x,y) and y≤B2(x,y). However, from (H2) we have(55)B1x,yt=∑s=0b+2Gt,sf~s+ν-1,xs+ν-1-zs+ν-1,ys+ν-1-zs+ν-1<∑s=0b+2φs+ν-1q0-1M=q0-1Mκ2,for all t∈[ν-1,b+ν+1]Nν-1. This implies B1(x,y)<q0-1Mκ2. Similarly, B2(x,y)<q0-1Mκ2. Thus, note that (x,y)∈∂(Bq0-1Mκ2×Bq0-1Mκ2)∩(P×P), and we have(56)x,y=max⁡x,y≤max⁡B1x,y,B2x,y<q0-1Mκ2=x,y.Clearly, this is a contradiction. Thus (53) is true. It follows from Lemma 6 that(57)iB,Bq0-1Mκ2×Bq0-1Mκ2∩P×P,P×P=1.
From (52) and (57) we have(58)iB,BR×BR∖B¯q0-1Mκ2×B¯q0-1Mκ2∩P×P,P×P=0-1=-1.Therefore the operator B has at least one fixed point (x,y) in ((BR×BR)∖(B¯q0-1Mκ2×B¯q0-1Mκ2))∩(P×P) with x,y≥q0-1Mκ2, and then (x-z,y-z)(t) is a positive solution for (1). This completes the proof.Theorem 11.
Suppose that (H0), (H3), and (H4) hold. Then (1) has at least one positive solution.Proof.
We show there exists a positive constantR>q0-1Mκ2 such that(59)x,y≠λBx,y,∀x,y∈∂BR×BR∩P×P,λ∈0,1.Suppose not. Then there exist (x,y)∈∂(BR×BR)∩(P×P),λ0∈[0,1] such that (x,y)=λ0B(x,y). This implies that(60)xt≤B1x,yt,yt≤B2x,yt,fort∈ν-1,b+ν+1Nν-1.From (H3) we have(61)xt≤∑s=0b+2Gt,se1xs+ν-1-zs+ν-1+e2ys+ν-1-zs+ν-1+c2≤∑s=0b+2Gt,se1xs+ν-1+e2ys+ν-1+c10,∀t,x,y∈ν-1,b+ν+1Nν-1×R+×R+.Similarly, we have(62)yt≤∑s=0b+2Gt,se3xs+ν-1+e4ys+ν-1+c10,∀t,x,y∈ν-1,b+ν+1Nν-1×R+×R+.Consequently, for all t∈0,b+2N0, multiply both sides of the above two inequalities by φ(t) and sum from ν-1 to b+ν+1 and together with (11) we obtain(63)∑t=ν-1b+ν+1xtφt=∑t=0b+2xt+ν-1φt+ν-1≤∑t=0b+2φt+ν-1∑s=0b+2Gt,se1xs+ν-1+e2ys+ν-1+c10≤e1κ2∑t=ν-1b+ν+1xtφt+e2κ2∑t=ν-1b+ν+1ytφt+c10∑t=ν-1b+ν+1φt.Similarly, we have(64)∑t=ν-1b+ν+1ytφt=∑t=0b+2yt+ν-1φt+ν-1≤∑t=0b+2φt+ν-1∑s=0b+2Gt,se3xs+ν-1+e4ys+ν-1+c10≤e3κ2∑t=ν-1b+ν+1xtφt+e4κ2∑t=ν-1b+ν+1ytφt+c10∑t=ν-1b+ν+1φt.Consequently, we obtain(65)1-e1κ2-e2κ2-e3κ21-e4κ2∑t=ν-1b+ν+1xtφt∑t=ν-1b+ν+1ytφt≤κ2c10κ2c10.From (H3)(i) we have(66)∑t=ν-1b+ν+1xtφt∑t=ν-1b+ν+1ytφt≤1κ1-e4κ2e2κ2e3κ21-e1κ2κ2c10κ2c10=κ2c101+e2κ2-e4κ2κκ2c101+e3κ2-e1κ2κ.Note that x,y∈P0 from the fact that Bi(P×P)⊂P0(i=1,2). This implies(67)x∑t=ν-1b+ν+1q∗tφt≤∑t=ν-1b+ν+1xtφt≤κ2c101+e2κ2-e4κ2κ,y∑t=ν-1b+ν+1q∗tφt≤∑t=ν-1b+ν+1ytφt≤κ2c101+e3κ2-e1κ2κ.Hence(68)x≤κ2c101+e2κ2-e4κ2κ1κ,y≤κ2c101+e3κ2-e1κ2κ1κ.Thus if we choose R>max⁡​q0-1Mκ2, κ2c10(1+e2κ2-e4κ2)/κ1κ, and κ2c10(1+e3κ2-e1κ2)/κ1κ​ we have a contradiction. Thus (59) is true. Lemma 6 (with R chosen above) implies(69)iB,BR×BR∩P×P,P×P=1.
We next prove that(70)x,y≠Bx,y+λx0,y0,∀x,y∈∂Bq0-1Mκ2×Bq0-1Mκ2∩P×P,λ≥0,where x0,y0∈P are two fixed functions. Indeed, if not, there exist (x,y)∈∂(Bq0-1Mκ2×Bq0-1Mκ2)∩(P×P),λ0≥0 such that (x,y)=B(x,y)+λ0(x0,y0). This implies that(71)xt≥B1x,yt,yt≥B2x,yt,fort∈ν-1,b+ν+1Nν-1.Hence, x≥B1(x,y) and y≥B2(x,y). However, from (H4) we have(72)B1x,yt=∑s=0b+2Gt,sf~s+ν-1,xs+ν-1-zs+ν-1,ys+ν-1-zs+ν-1>∑s=0b+2q∗tφs+ν-1q0-2M≥q0-1Mκ2,for all t∈[ν-1,b+ν+1]Nν-1. This implies B1(x,y)>q0-1Mκ2. Similarly, B2(x,y)>q0-1Mκ2. Thus, note that (x,y)∈∂(Bq0-1Mκ2×Bq0-1Mκ2)∩(P×P), and we have(73)x,y=max⁡x,y≥max⁡B1x,y,B2x,y>q0-1Mκ2=x,y.This is a contradiction. So (70) is true. It follows from Lemma 5 that(74)iB,Bq0-1Mκ2×Bq0-1Mκ2∩P×P,P×P=0.
From (69) and (74) we have(75)iB,BR×BR∖B¯q0-1Mκ2×B¯q0-1Mκ2∩P×P,P×P=1-0=1.Therefore the operator B has at least one fixed point (x,y) in ((BR×BR)∖(B¯q0-1Mκ2×B¯q0-1Mκ2))∩(P×P) with x,y≥q0-1Mκ2, and then (x-z,y-z)(t) is a positive solution for (1). This completes the proof.



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*Source: 6835028-2018-07-18.xml*