# Multiple Positive Solutions to Multipoint Boundary Value Problem for a System of Second-Order Nonlinear Semipositone Differential Equations on Time Scales

Authors: Gang Wu; Longsuo Li; Xinrong Cong; Xiufeng Miao
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We study a system of second-order dynamic equations on time scales(p1u1∇)Δ(t)-q1(t)u1(t)+λf1(t,u1(t),u2(t))=0,t∈(t1,tn),(p2u2∇)Δ(t)-q2(t)u2(t)+λf2(t,u1(t), u2(t))=0, satisfying four kinds of different multipoint boundary value conditions, fi is continuous and semipositone. We derive an interval of λ such that any λ lying in this interval, the semipositone coupled boundary value problem has multiple positive solutions. The arguments are based upon fixed-point theorems in a cone.


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## Body

## 1. Introduction

In this paper, we consider the following dynamic equations on time scales:(1)(p1u1∇)Δ(t)-q1(t)u1(t)+λf1(t,u1(t),u2(t))=0,t∈(t1,tn),λ>0,(p2u2∇)Δ(t)-q2(t)u2(t)+λf2(t,u1(t),u2(t))=0,
satisfying one of the boundary value conditions
(2)α1u1(t1)-β1p1(t1)u1∇(t1)=0,γ1u1(tn)+δ1p1(tn)u1∇(tn)=∑i=2n-2‍biu1(ti),α2u2(t1)-β2p2(t1)u2∇(t1)=0,γ2u2(tn)+δ2p1(tn)u2∇(tn)=∑i=2n-2‍biu2(ti),(3)α1u1(t1)-β1p1(t1)u1∇(t1)=∑i=2n-2‍aiu1(ti),γ1u1(tn)+δ1p1(tn)u1∇(tn)=0,α2u2(t1)-β2p2(t1)u2∇(t1)=∑i=2n-2‍aiu2(ti),γ2u2(tn)+δ2p1(tn)u2∇(tn)=0,(4)α1u1(t1)-β1p1(t1)u1∇(t1)=0,γ1u1(tn)+δ1p1(tn)u1∇(tn)=∑i=2n-2‍biu1(ti),α2u2(t1)-β2p2(t1)u2∇(t1)=∑i=2n-2‍aiu2(ti),γ2u2(tn)+δ2p1(tn)u2∇(tn)=0,(5)α1u1(t1)-β1p1(t1)u1∇(t1)=∑i=2n-2‍aiu1(ti),γ1u1(tn)+δ1p1(tn)u1∇(tn)=0,α2u2(t1)-β2p2(t1)u2∇(t1)=0,γ2u2(tn)+δ2p1(tn)u2∇(tn)=∑i=2n-2‍biu2(ti),
where
(6)pi,qi:[t1,tn]→(0,+∞)withpi∈CΔ[t1,tn),qi∈C[t1,tn]fori=1,2;αi,βi,γi,δi∈[0,+∞)withαiγi+αiδi+βiγi>0fori=1,2,
and fi is continuously and nonegative functionsquad, ai,bi∈[0,+∞) for i∈{1,2,…,n}; the points ti∈𝕋κκ for i∈{1,2,…,n} with t1<t2<⋯<tn.In the past few years, the boundary value problems of dynamic equations on time scales have been studied by many authors (see [1–19] and references). Recently, multipoint boundary value problems on time scale have been studied, for instance, see [1–12].In 2006, Anderson and Ma [1] studied the second-order multiple time-scale eigenvalue problem:
(7)(py∇)Δ(t)-q(t)y(t)+λh(t)f(y)=0,t∈(t1,tn),λ>0,αy(t1)-βp(t1)y∇(t1)=∑i=2n-2‍aiy(ti),γy(tn)+δp(tn)y∇(tn)=∑i=2n-2‍biy(ti),
where the functions f:[0,+∞)→[0,+∞) and h:[t1,tn]→[0,+∞) are continuous. The authors discuss conditions for the existence of at least one positive solution to the second-order Sturm-Liouville-type multiple eigenvalue problem on time scales.In 2009, Feng et al. [2] studied
(8)(py∇)Δ(t)-q(t)y(t)=f(t,y),t∈(t1,tn),αy(t1)-βp(t1)y∇(t1)=∑i=2n-2‍aiy(ti),γy(tn)+δp(tn)y∇(tn)=∑i=2n-2‍biy(ti),
where the functions f(t,y)=∑j=1ncj(t)yvj, cj∈C([t1,tn],[0,∞)), vi∈[0,∞), j=1,2,…,n. This paper shows the existence of multiple positive solutions for the boundary value problem on time scales.In 2009, Topal and Yantir [3] studied the second-order nonlinear m-point boundary value problems
(9)uΔ∇(t)+a(t)uΔ(t)+b(t)u(t)+λq(t)f(t,u(t))=0,t∈(0,1)𝕋,u(ρ(0))=0,u(σ(1))=∑i=1m-2‍αiu(ηi),
where αi≥0,0<ηi<ηi+1<1; for  all i=1,2,…,m-2; a∈C([0,1],[0,+∞)), b∈C([0,1],(-∞,0]), f,q are continuously and nonegative functions. The authors deal with determining the value of λ, and the existences of multiple positive solutions of the equation are obtained. In 2010, Yuan and Liu [4] also study the second-order m-point boundary value problems; Yuan and Liu shows the existence of multiple positive solutions if f is semipositone and superlinear.Motivated by the above results mentioned, we study the second-order nonlinearm-point boundary value problem (1) with boundary condition (k), and nonlinear term may be singularity and semipositone.In this paper, the nonlinear termfi of (1) is suit to and semipositone and the superlinear case, we shall prove our two existence results for the problem (1) with (k) by using a nonlinear alternative of Leray-Schauder type and Krasnosel'skii fixed-point theorem. This paper is organized as follows. In Section 2, we start with some preliminary lemmas. In Section 3, we give the main result which state the sufficient conditions for (1) with m-point boundary value (k) to have existence of positive solutions (k=2,…,5).


## 2. Preliminaries

In this section, we state the preliminary information that we need to prove the main results.In this paper, for our constructions, we shall consider the Banach spaceE=C[ρ(t1),tn] equipped with standard norm ∥x∥=maxρ(t1)≤t≤tn|x(t)|,x∈E; for each (x,y)∈E×E, we write ||(x,y)||1=||x||+||y||. Clearly, (E×E,||·||1) is a Banach space. Denote by ϕi1 and ϕi2 (i=1,2), the solutions of the equation
(10)(piui∇)Δ(t)-qi(t)ui(t)=0,t∈[t1,tn),
under the initial conditions
(11)ui(t1)=βi,pi(t1)ui∇(t1)=αi,ui(tn)=δi,pi(tn)ui∇(tn)=-γi,
respectively. So that ϕi1 and ϕi2 (i=1,2) satisfy
(12)(piϕi1∇)Δ(t)-qi(t)ϕi1(t)=0,t∈[t1,tn),ϕi1(t1)=βi,pi(t1)ϕi1∇(t1)=αi,(piϕi2∇)Δ(t)-qi(t)ϕi2(t)=0,t∈[t1,tn),ϕi2(tn)=δi,pi(tn)ϕi2∇(tn)=-γi,
respectively. For i=1,2, set di=αiϕi2(t1)-βipi(t1)ϕi2∇(t1)=γiϕi1(tn)+δipi(tn)ϕi1∇(tn), the Green's function of the corresponding homogeneous boundary value problem is defined by
(13)Gi(t,s)=1di{ϕi2(t)ϕi1(s),ρ(t1)≤s≤t≤tn,ϕi1(t)ϕi2(s),ρ(t1)≤t≤s≤tn,fori=1,2.
From Lemmas 3.1 and 3.3 in [1], we have the following lemma.Lemma 1.
Ifdi≠0, (u1,u2) is a solution of (1) with boundary value condition (k) if only and if
(14)u1(t)=λ∫t1tn‍H1k(t,s)f1(s,u1(s),u2(s))Δs,t∈[ρ(t1),tn],(15)u2(t)=λ∫t1tn‍H2k(t,s)f2(s,u1(s),u2(s))Δs,
where k=2,…,5, and
(16)Hi2(t,s)=Gi(t,s)+1di-∑j=2n-1bjϕi1(tj)∑j=2n-1‍bjGi(tj,s)ϕi1(t),(i=1,2),Hi3(t,s)=Gi(t,s)+1di-∑j=2n-1ajϕi2(tj)∑j=2n-1‍ajGi(tj,s)ϕi2(t),(i=1,2),H14(t,s)=G1(t,s)+1d1-∑j=2n-1biϕ11(ti)∑j=2n-1‍biG1(ti,s)ϕ11(t)=H12(t,s),H24(t,s)=G2(t,s)+1d2-∑j=2n-1aiϕ22(ti)∑j=2n-1‍aiG2(ti,s)ϕ22(t)=H23(t,s),H15(t,s)=G1(t,s)+1d1-∑j=2n-1aiϕ12(ti)∑j=2n-1‍aiG1(ti,s)ϕ12(t)=H13(t,s),H25(t,s)=G2(t,s)+1d2-∑j=2n-1biϕ21(ti)∑j=2n-1‍biG2(ti,s)ϕ21(t)=H22(t,s).For the rest of the paper, we need the following assumption:(C)0<∑j=2n-1‍bjϕi1(tj),∑j=2n-1‍ajϕi2(tj)<di,fori=1,2.Fromϕi1 is nondecreasing on [ρ(t1),tn], ϕi2 is nonincreasing on [ρ(t1),tn] (see [2, Proposition 2.3]), it is easy to verify the following inequalities:
(17)diGi(t,s)≤ϕi1(t)ϕi2(t),diGi(t,s)≤ϕi1(s)ϕi2(s),diGi(t,s)≥1∥ϕi1∥∥ϕi2∥ϕi1(t)ϕi2(t)ϕi1(s)ϕi2(s).Lemma 2.
The Green's functionGi(t,s) has properties
(18)Gi(t,s)≤Gi(t,t),di∥ϕi1∥∥ϕi2∥Gi(t,t)Gi(s,s)≤Gi(t,s)≤Gi(s,s).Lemma 3.
ForHik(t,s), k=2,…,5 and i=1,2, one has the conclusions  Hik(t,s)≤C*Gi(s,s) and
(19)c*ϕi1(t)Gi(s,s)≤Hi2(t,s)≤C*ϕi1(t),(i=1,2),c*ϕi2(t)Gi(s,s)≤Hi3(t,s)≤C*ϕi2(t),(i=1,2),c*ϕii(t)G1(s,s)≤Hi4(t,s)≤C*ϕii(t),(i=1,2),c*ϕ12(t)G1(s,s)≤H15(t,s)≤C*ϕ12(t),c*ϕ21(t)G1(s,s)≤H25(t,s)≤C*ϕ21(t),
where C*=C1+C2 and
(20)C1=maxi=1,2{1+∥ϕi1∥di-∑j=2n-1bjϕi1(tj)∑i=1n-1‍bj+∥ϕi2∥di-∑j=2n-1ajϕi2(tj)∑j=2n-1‍aj},C2=maxi=1,2{∥ϕi1∥+∥ϕi2∥di+1di-∑j=2n-1bjϕi1(tj)×∑j=2n-1‍bjGi(tj,tj)+1di-∑j=2n-1bjϕi2(tj)×∑j=2n-1‍ajG2(tj,tj)},c*=min{di∥ϕi1∥∥ϕi2∥1di-∑j=2n-1bjϕi1(tj)×∑j=2n-1‍bjGi(tj,tj),di∥ϕi1∥∥ϕi2∥1di-∑j=2n-1ajϕi2(tj)×∑j=2n-1‍ajGi(tj,tj);i=1,2}.Proof.
From Lemma2 and
(21)1di-∑j=2n-1bjϕi1(tj)∑j=2n-1‍bjGi(tj,s)ϕi1(t)≤1di-∑j=2n-1bjϕi1(tj)∑j=2n-1‍bjGi(s,s)∥ϕi1∥,1di-∑j=2n-1ajϕi2(tj)∑j=2n-1‍ajGi(tj,s)ϕi2(t)≤1di-∑j=2n-1ajϕi2(tj)∑j=2n-1‍ajGi(s,s)∥ϕi2∥,
we have
(22)Hik(t,s)≤C1Gi(s,s)≤C*Gi(s,s).
Fork=2 or 3, we have
(23)Hi2(t,s)≤∥ϕi2(t)∥diϕi1(t)+1di-∑j=2n-1biϕi1(ti)×∑j=2n-1‍bjGi(tj,tj)ϕi1(t)≤C*ϕi1(t),Hi3(t,s)≤∥ϕi1(t)∥diϕi2(t)+1di-∑j=2n-1aiϕi2(ti)×∑j=2n-1‍ajGi(tj,tj)ϕi2(t)≤C*ϕi2(t),Hi2(t,s)≥1di-∑j=2n-1bjϕi1(tj)×∑j=2n-1‍bjGi(tj,s)ϕi1(t)≥1di-∑j=2n-1bjϕi1(tj)×∑j=2n-1‍bjdi∥ϕ1∥∥ϕ2∥Gi(tj,tj)Gi(s,s)ϕi1(t)≥c*ϕi1(t)Gi(s,s),Hi3(t,s)≥1di-∑j=2n-1ajϕi2(tj)×∑j=2n-1‍ajGi(tj,s)ϕi2(t)≥1di-∑j=2n-1ajϕi1(tj)×∑j=2n-1‍ajdi∥ϕ1∥∥ϕ2∥Gi(tj,tj)Gi(s,s)ϕi2(t)≥c*ϕi2(t)Gi(s,s).
So, we have
(24)c*ϕi1(t)Gi(s,s)≤Hi2(t,s)≤C*ϕi1(t),(i=1,2),c*ϕi2(t)Gi(s,s)≤Hi3(t,s)≤C*ϕi2(t),(i=1,2).
SinceH14(t,s)=H12(t,s),H24(t,s)=H23(t,s),H15(t,s)=H13(t,s),H25(t,s)=H22(t,s), then we also have
(25)c*ϕ11(t)G1(s,s)≤H14(t,s)=H12(t,s)≤C*ϕ11(t),c*ϕ22(t)G1(s,s)≤H24(t,s)=H23(t,s)≤C*ϕ22(t),c*ϕ12(t)G1(s,s)≤H15(t,s)=H13(t,s)≤C*ϕ12(t),c*ϕ21(t)G1(s,s)≤H25(t,s)=H22(t,s)≤C*ϕ21(t).
The proof is complete.The following theorems will play a major role in our next analysis.Theorem 4 (see [20]).
LetX be a Banach space, and Ω⊂X closed and convex. Assume U is a relatively open subset of Ω with 0∈U, and let S:U¯→Ω be a compact, continuous map. Then either  (1)
Shas a fixed point in U¯, or(2)
there existsu∈∂U and ν∈(0,1), with u=νSu.Theorem 5 (see [21]).
LetX be a Banach space, and let P⊂X be a cone in X. Let  Ω1,Ω2 be bounded open subsets of X with 0∈Ω1⊂Ω¯1⊂Ω2, and let S:P→P be a completely continuous operator such that, either  (1)
∥Sw∥≤∥w∥, w∈P∩∂Ω1, ∥Sw∥≥∥w∥,w∈P∩∂Ω2, or(2)
∥Sw∥≥∥w∥, w∈P∩∂Ω1, ∥Sw∥≤∥w∥,w∈P∩∂Ω2. Then S has a fixed point in P∩Ω¯2∖Ω1.


## 3. Main Results

We make the following assumptions:(H1)
fi(t,u1,u2)∈C([t1,tn]×[0,+∞)2,(-∞,+∞)), moreover there exists a function g(t)∈L1([t1,tn],(0,+∞)) such that fi(t,u1,u2)≥-g(t), for any t∈(t1,tn), ui∈[0,+∞), i=1,2.(H1*)
fi(t,u1,u2)∈C((t1,tn)×[0,+∞)2,(-∞,+∞)) may be singular at t=t1,tn; moreover, there exists a function g(t)∈L1((t1,tn),(0,+∞)) such that fi(t,u1,u2)≥-g(t), for any t∈(t1,tn), ui∈[0,+∞).(H2)
fi(t,0,0)>0, for t∈[t1,tn]  (i=1,2).(H3)
There exists[θ1,θ2]⊂(t1,tn) such that limu1+u2↑+∞mint∈[θ1,θ2](fi(t,u1,u2)/(u1+u2))=+∞(i=1,2).(H4)
∫t1tn‍Gi(s,s)g(s)∇s<+∞ and ∫t1tn‍Gi(s,s)fi(s,z1,z2)∇s<+∞ for any zi∈[0,m], m>0 is any constant (i=1,2).In fact, we only consider the system
(26)(p1x1∇)Δ(t)-q1(t)x1(t)+λ(f1(t,[x1(t)-v1k(t)]*,[x2(t)-v2k(t)]*)+g(t)(t,[x1(t)-v1k(t)]*)=0,λ>0,(p2x2∇)Δ(t)-q2(t)x2(t)+λ(f2(t,[x1(t)-v1k(t)]*,[x2(t)-v2k(t)]*)+g(t)[x1(t)-v1k(t)]*)=0,λ>0,
with one of the boundary value conditions
(27)α1x1(t1)-β1p1(t1)x1∇(t1)=0,γ1x1(tn)+δ1p1(tn)x1∇(tn)=∑i=2n-2‍bix1(ηi),α2x2(t1)-β2p2(t1)x2∇(t1)=0,γ2x2(tn)+δ2p1(tn)x2∇(tn)=∑i=2n-2‍bix2(ηi),α1x1(t1)-β1p1(t1)x1∇(t1)=∑i=2n-2‍aix1(ηi),γ1x1(tn)+δ1p1(tn)x1∇(tn)=0,α2x2(t1)-β2p2(t1)x2∇(t1)=∑i=2n-2‍aix2(ηi),γ2x2(tn)+δ2p1(tn)x2∇(tn)=0,α1x1(t1)-β1p1(t1)x1∇(t1)=0,γ1x1(tn)+δ1p1(tn)x1∇(tn)=∑i=2n-2‍bix1(ηi),α2x2(t1)-β2p2(t1)x2∇(t1)=∑i=2n-2‍aix2(ηi),γ2x2(tn)+δ2p1(tn)x2∇(tn)=0,α1x1(t1)-β1p1(t1)x1∇(t1)=∑i=2n-2‍aix1(ηi),γ1x1(tn)+δ1p1(tn)x1∇(tn)=0,α2x2(t1)-β2p2(t1)x2∇(t1)=0,γ2x2(tn)+δ2p1(tn)x2∇(tn)=∑i=2n-2‍bix2(ηi),
where
(28)y(t)*={y(t),y(t)≥0,0,y(t)<0,
and vik(t)=λ∫t1tn‍Hik(t,s)g(s)Δs. For k=2,…,5, from Lemma 1, (v1k(t),v2k(t)) is the solution of the equation
(29)(p1v1∇)Δ(t)-q1(t)v1(t)+λg(t)=0,λ>0,t1<t<tn,(p2v2∇)Δ(t)-q2(t)v2(t)+λg(t)=0,λ>0,
respectively, satisfying the following boundary value conditions:
(30)α1v1(t1)-β1p1(t1)v1∇(t1)=0,γ1v1(tn)+δ1p1(tn)v1∇(tn)=∑i=2n-2‍biv1(ηi),α2v2(t1)-β2p2(t1)v2∇(t1)=0,γ2v2(tn)+δ2p1(tn)v2∇(tn)=∑i=2n-2‍biv2(ηi),α1v1(t1)-β1p1(t1)v1∇(t1)=∑i=2n-2‍aiv1(ηi),γ1v1(tn)+δ1p1(tn)v1∇(tn)=0,α2v2(t1)-β2p2(t1)v2∇(t1)=∑i=2n-2‍aiv2(ηi),γ2v2(tn)+δ2p1(tn)v2∇(tn)=0,α1v1(t1)-β1p1(t1)v1∇(t1)=0,γ1v1(tn)+δ1p1(tn)v1∇(tn)=∑i=2n-2‍biv1(ηi),α2v2(t1)-β2p2(t1)v2∇(t1)=∑i=2n-2‍aiv2(ηi),γ2v2(tn)+δ2p1(tn)v2∇(tn)=0,α1v1(t1)-β1p1(t1)x1∇(t1)=∑i=2n-2‍aiv1(ηi),γ1v1(tn)+δ1p1(tn)x1∇(tn)=0,α2v2(t1)-β2p2(t1)x2∇(t1)=0,γ2v2(tn)+δ2p1(tn)x2∇(tn)=∑i=2n-2‍biv2(ηi).We will show that there exists a solution(x1k,x2k) to the boundary value problem (k-) of the system (26) with xik(t)≥vik(t),t∈[ρ(t1),tn]. If this is true, then uik(t)=xik(t)-vik(t) is a nonnegative solution (positive on (ρ(t1),tn)) of the system (1) with the boundary value problem (k), (wherei=1,2;k=2,…,5,k-=k+11). Since for any t∈(t1,tn), from
(31)(pixik∇)Δ(t)-qi(t)xik∇(t)=(pi(uik+vik)∇)Δ(t)-qi(t)(uik+vik)∇(t)=-λ(fi(t,[x1k(t)-v1k(t)]*,[x2k(t)-v2k(t)]*)+g(t))=-λ(fi(t,u1k(t),u2k(t))+g(t)),
we have
(32)(piuik∇)Δ(t)-qi(t)uik(t)=-λfi(t,u1k(t),u2k(t)).
As a result, we will concentrate our study on (26) with the boundary value problem (k-).Employing Lemma1, we note that (x1k(t),x2k(t)) is a solution of the system (26) with boundary value (k-) if and only if
(33)x1k(t)=λ∫t1tn‍H1k(t,s)(f1(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s)[x1k(s)-v1k(s)]*)Δs,t∈[ρ(t1),tn],x2k(t)=λ∫t1tn‍H2k(t,s)(f2(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s)(s,[x1k(s)-v1k(s)]*)Δs.We define a conePij  (i,j=1,2) by
(34)Pij={x∈E∣x(t)≥c*C*ϕij(t)∥x∥,t∈[ρ(t1),tn]}.
It is clearly that Pij×Pmn is a cone of E×E, (i,j,m,n=1,2). Define the integral operator T2:P11×P21→E×E, T3:P12×P22→E×E, T4:P12×P21→E×E, T5:P21×P12→E×E, by
(35)Tk(x1k,x2k)=(T1k(x1k,x2k),T2k(x1k,x2k)),(k=2,…,5),
where operators Tik are defined by
(36)Tik(x1k,x2k)(t)=λ∫t1tn‍Hik(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s)(s,[x1k(s)-v1k(s)]*)Δs,t∈[ρ(t1),tn],
where i=1,2. Clearly, if (x1k,x2k) is a fixed point of Tk, then (x1k,x2k) is a solution of system (26) with (k-)  (k=2,…,5,k-=k+11).Fork=2,…,5, from (35) and Lemma 3, we have Tk(x1k,x2k)(t)≥0 on [0,1], for (x1k,x2k)∈Pij×Pmn, we have
(37)Tik(x1k,x2k)(t)=λ∫t1tn‍Hik(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≤C*λ×∫ρ(t1)σ(tn)‍Gi(s,s)(fi(s,[x(s)-v(s)]*)+g(s))Δs,
then ∥Tik(x1k,x2k)∥≤C*λ∫ρ(t1)σ(tn)‍Gi(s,s)(f(t,[x(t)-v(t)]*)+g(t))Δs.On the other hand, whenk=2, we have
(38)Ti2(x1k,x2k)(t)=λ∫t1tn‍Hi2(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≥λ∫t1tn‍c*ϕi1(t)Gi(s,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≥c*C*ϕi1(t)λ×∫ρ(t1)σ(tn)‍C*Gi(s,s)(fi(s,[x(s)-v(s)]*)+g(s))Δs≥c*C*ϕi1(t)∥Ti2(x1k,x2k)∥.
Thus, Ti2(P11×P21)⊂Pi1. Hence T2(P11×P21)⊂P11×P21.Whenk=3, we have
(39)Ti3(x1k,x2k)(t)=λ∫t1tn‍Hi3(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s)[x1k(s)-v1k(s)]*)Δs≥λ∫t1tn‍c*ϕi2(t)Gi(s,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≥c*C*ϕi2(t)λ×∫ρ(t1)σ(tn)‍C*Gi(s,s)(fi(s,[x(s)-v(s)]*)+g(s))Δs≥c*C*ϕi2(t)∥Ti3(x1k,x2k)∥.
Thus, Ti3(P12×P22)⊂Pi2. Hence T3(P12×P22)⊂P12×P22.Similarly discussion, we also haveT4(P12×P21)⊂P12×P21, T5(P21×P12)⊂P21×P12. In addition, standard arguments show that Tk is a completely continuous operator.For simplicity, we adopt the notation:P14=P25:=P12 and P24=P15:=P21, then, we can write Tk(P1(k-1)×P2(k-1))⊂P1(k-1)×P2(k-1), that is, Tik(P1(k-1)×P2(k-1))⊂Pi(k-1), (i=1,2,k=2,…,5).Theorem 6.
Suppose that(H1)-(H2) hold. Then there exists a constant λ¯>0 such that, for any 0<λ≤λ¯, (1) with boundary value condition (k) has at least one positive solution (k=2,…,5).Proof.
Fixδ∈(0,1) and k (k=2,…,5). From (H2) let 0<ε<1 be such that
(40)fi(t,z1,z2)≥δfi(t,0,0),fort1≤t≤tn,0≤zi≤ε,i=1,2.
Letf¯(ε)=maxt1≤t≤tn,0≤z1,z2≤ε{maxi=1,2{fi(t,z1,z2)}+g(t)}, and c=∫t1tn‍C*Gi(s,s)Δs. We have
(41)limz↓0f¯(z)z=+∞.
Setλ¯=ε/4cf¯(ε), since for any 0<λ≤λ¯, fix the λ∈(0,λ¯], we always have
(42)limz↓0f¯(z)z=+∞,f¯(ε)ε<14cλ,
Then there exists a R0∈(0,ε] such that
(43)f¯(R0)R0=14cλ.
LetUk={(x1k,x2k)∈P1(k-1)×P2(k-1):∥(x1k,x2k)∥1<R0}, (x1k,x2k)∈∂Uk and ν∈(0,1) be such that (x1k,x2k)=νTk(x1k,x2k), that is, xik=νTik(x1k,x2k)  (i=1,2). We claim that ∥(x1k,x2k)∥1≠R0. In fact for (x1k,x2k)∈∂Uk and ∥(x1k,x2k)∥1=R0, we have
(44)xik=νTik(x1k,x2k)≤λ∫t1tn‍Hik(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≤λ∫t1tn‍C*Gi(s,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≤λ∫t1tn‍C*Gi(s,s)f¯(R0)Δs≤λcf¯(R0).
It follows that
(45)R0=∥(x1k,x2k)∥1≤2λcf¯(R0),
that is,
(46)f¯(R0)R0≥12cλ>14cλ=f¯(R0)R0,
which implies that ∥(x1k,x2k)∥1≠R0. By the nonlinear alternative of Leray-Schauder type, Tk has a fixed point (xi1,xi2)∈Uk¯. Moreover combining (40) and the fact that R0<ε, we obtain
(47)xik=λ∫t1tn‍Hik(t,s)(fi(t,[x1k(t)-v1k(t)]*,[x2k(t)-v2k(t)]*)+g(t))Δs≥λ∫t1tn‍Hik(t,s)(δf(s,0,0)+g(t))Δs≥λ∫t1tn‍Hik(t,s)g(t)Δs=vik(t)fort∈(ρ(t1),tn).
ThenTk has a positive fixed point (xi1,xi2) and ∥(xi1,xi2)∥1≤R0<1; that is, (xi1,xi2) is a positive solution of the boundary value problem (26) with xik>vik(t) for t∈(t1,tn).
Letuik(t)=xik(t)-vik(t)≥0  (i=1,2), then (u1k,u2k) is a nonnegative solution (positive on (ρ(t1),tn)) of the boundary value problem (1).Theorem 7.
Suppose that(H1*) and (H3)-(H4) hold. Then there exists a constant λ*>0 such that, for any 0<λ≤λ*, (1) with boundary value condition (k) has at least one positive solution (k=2,…,5).Proof.
We fixk (k=2,…,5). Let Ω1={(x1k,x2k)∈E×E:∥xik∥<R1,i=1,2}, where R1=max{1,r} and r=(C*2/c*)∫t1tn‍g(s)Δs. Choose
(48)λ*=min{1,R12(R+1)-1,R12r},
where R=∫t1tn‍C*Gi(s,s)(max0≤z1,z2≤R1fi(s,z1,z2)+g(s))Δs and R≥0.
Then for any(x1k,x2k)∈(P1(k-1)×P2(k-1))∩∂Ω1, xik(s)-vik(s)≤xik(s)≤∥xik∥≤R1  (i=1,2) and for 0<λ≤λ*, we have
(49)∥Tik(x1k,x2k)(t)∥≤λ∫t1tn‍C*Gi(s,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≤λ∫t1tn‍C*Gi(s,s)(max0≤z1,z2≤R1fi(s,z1,z2)+g(s))Δs≤λR≤R12.
This implies
(50)∥Tk(x1k,x2k)∥1≤R1≤∥(x1k,x2k)∥1,(x1k,x2k)∈(P1(k-1)×P2(k-1))∩∂Ω1.
Choose a constantN>1 such that
(51)λNγc*2(∥ϕi1∥+∥ϕi2∥)∫θ1θ2‍Gi(s,s)ϕi1(s)ϕi2(s)Δs≥1,
where γ=minkminθ1≤t≤θ2{ϕ1k(t),ϕ2k(t)}.
By assumption(H3) and (H4), there exists a constant B>R1 such that
(52)fi(t,z1,z2)z1+z2>N,thatis,fi(t,z1,z2)>N(z1+z2),fort∈[θ1,θ2],z1+z2>B(i=1,2).
ChooseR2=max{R1+1,2λr,2C*(B+1)/c*γ} and let Ω2={(x1k,x2k)∈E×E:∥xik∥<R2,i=1,2}. We note that x(t)≥(c*/C*)ϕij(t)∥x∥  forallx∈Pij, by Lemma 3, we have Hik(t,s)≤(C*2/c*)(x(t)/∥x∥). Then for any (x1k,x2k)∈(P1(k-1)×P2(k-1))∩∂Ω2, we have ∥x1k∥=R2 or ∥x2k∥=R2. Without loss of generality let ∥x1k∥=R2, so we have
(53)x1k(t)-v1k(t)=x1k(t)-λ∫t1tn‍H1k(t,s)g(s)Δs≥x1k(t)-λ∫t1tn‍C*2c*x1k(t)∥x1k∥g(s)Δs=x1k(t)-λx1k(t)∥x1k∥∫t1tn‍C*2c*g(s)Δs≥x1k(t)-x1k(t)∥x1k∥λr≥x1k(t)-x1k(t)R2λr≥(1-λrR2)x1k(t)≥12x1k(t)≥0,t∈[ρ(t1),tn].
Thus
(54)minθ1≤t≤θ2{[x1k(t)-v1k(t)]*+[x2k(t)-v2k(t)]*}≥minθ1≤t≤θ2{x1k(t)-v1k(t)}≥minθ1≤t≤θ2{12x1k(t)}≥minθ1≤t≤θ2{c*2C*ϕ1k(t)∥x1k∥,c*2C*ϕ2k(t)∥x1k∥}=c*2C*R2minθ1≤t≤θ2{ϕ1k(t),ϕ2k(t)}≥B+1>B.
Now sinceB>R1, it follows that(55)Tik(x1k,x2k)(t)=λ∫t1tn‍Hik(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≥λ∫θ1θ2‍Hik(t,s)(fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)+g(s))Δs≥λ∫θ1θ2‍Hik(t,s)fi(s,[x1k(s)-v1k(s)]*,[x2k(s)-v2k(s)]*)Δs≥λ∫θ1θ2‍Hik(t,s)N([x1k(s)-v1k(s)]*+[x2k(s)-v2k(s)]*)Δs≥λ∫θ1θ2‍Hik(t,s)N(x1k(s)-v1k(s))Δs≥λ∫θ1θ2‍c*minθ1≤t≤θ2{ϕ1k(t),ϕ2k(t)}Gi(s,s)N(x1k(s)-v1k(s))Δs≥λ∫θ1θ2‍c*minθ1≤t≤θ2{ϕ1k(t),ϕ2k(t)}Gi(s,s)N2x1k(s)ds≥λminθ1≤t≤θ2{ϕ1k(t),ϕ2k(t)}×∫θ1θ2‍c*Gi(s,s)c*N2(∥ϕi1∥+∥ϕi2∥)ϕi1(s)ϕi2(s)∥x1k∥Δs≥λNγc*2(∥ϕi1∥+∥ϕi2∥)∫θ1θ2‍Gi(s,s)ϕi1(s)ϕi2(s)ΔsR2≥R2,t∈[θ1,θ2].This implies
(56)∥Tk(x1k,x2k)∥1≥∥(x1k,x2k)∥1,(x1k,x2k)∈(P1(k-1)×P2(k-1))∩∂Ω2.
For the Krasnosel'skii's fixed point theorem, one deduces that Tk has a fixed point (x1k,x2k) with R1<∥(x1k,x2k)∥<R2⇔R1<∥x1k∥+∥x2k∥<R2.
Sincer≤R1<∥xik∥<R2  (i=1,2), then
(57)xik(t)-vik(t)=xik(t)-λ∫t1tn‍Hik(t,s)g(s)Δs≥xik(t)-λ∫t1tn‍C*2c*xik(t)∥xik∥g(s)Δs=xik(t)-λxik(t)∥xik∥r≥xik(t)-λxik(t)=(1-λ)xik(t)≥(1-λ)c*C*ϕi1(t)ϕi2(t)∥ϕi1∥+∥ϕi2∥∥xik∥>0,t∈(ρ(t1),tn).
Thus(x1k,x2k) is a positive solution of the boundary value problem (26) with xik(t)>vik(t)  (i=1,2) for t∈(ρ(t1),tn).
Letuik(t)=xik(t)-vik(t)≥0  (i=1,2), then (u1k,u2k) is a nonnegative solution (positive on (ρ(t1),tn)) of the boundary value problem (1).Since condition(H1) implies conditions (H1*) and (H4) then from the proof of Theorems 6 and 7, we immediately have the following theorem.Theorem 8.
Suppose that(H1)–(H3) hold. Then (1) with boundary value condition (k) has at least two positive solutions for λ>0 sufficiently small (k=2,…,5).In fact with0<λ<min{λ¯,λ*} then (1) with boundary value condition (k) has at least two positive solutions.Remark 9.
In Theorems6–8, we use the assumption condition 16. If we have not the condition 16, that is, ai=bi=0, then the system (1) and boundary condition (k) are
(58)(p1u1∇)Δ(t)-q1(t)u1(t)+λf1(t,u1(t),u2(t))=0,t∈(t1,tn),λ>0,(p2u2∇)Δ(t)-q2(t)u2(t)+λf2(t,u1(t),u2(t))=0,α1u1(t1)-β1p1(t1)u1∇(t1)=0,γ1u1(tn)+δ1p1(tn)u1∇(tn)=0,α2u2(t1)-β2p2(t1)u2∇(t1)=0,γ2u2(tn)+δ2p1(tn)u2∇(tn)=0.
From Lemma 2, an argument similar to those in Theorems 6–8 yields the following theorems.Theorem 10.
Suppose that(H1) and (H2) hold. Then there exists a constant λ¯>0 such that, for any 0<λ≤λ¯, the boundary value problem (58) has at least one positive solution.Theorem 11.
Suppose that(H1*) and (H3)-(H4) hold. Then there exists a constant λ*>0 such that, for any 0<λ≤λ*, the boundary value problem (58) has at least one positive solution.Theorem 12.
Suppose that(H1)–(H4) hold. Then the boundary value problem (58) has at least two positive solutions for λ>0 sufficiently small.


## 4. Example

To illustrate the usefulness of the results, we give some examples.Example 13.
Consider the boundary value problem(59)u′′-u=-λ((u+v)a+1(t-t2)1/2cos(2π(u+v))),-1<t<1,λ>0,v′′-v=-λ((u-1)2+v2+1(t-t2)1/2sin(2πu)),u(-1)=v(1)=0,u(1)=au(0),v(-1)=bv(0),
where a>1. Then if λ>0 is sufficiently small, (59) has a positive solution u with u(t)>0 for t∈(0,1).To see this, we will apply Theorem7 with
(60)f1(t,u,v)=(u+v)a+1(t2-t4)1/4cos(2π(u+v)),f2(t,u,v)=(u-1)2+v2+1(t2-t4)1/4sin(2πu),g1(t)=g2(t)=g(t)=1(t2-t4)1/4.
Clearly for t∈(0,1),
(61)fi(t,u,v)+g(t)>0,fort∈(0,1)i=1,2,limu+v↑+∞inffi(t,u,v)u+v=+∞forallt∈[θ1,θ2]⊂(0,1).
Now (H1*), (H3), and  (H4) hold. We note that the boundary condition of (59) is in accord with (4), and from [1], we have
(62)ϕ11=ϕ21=et+1-e-t-12,ϕ12=ϕ22=e-t+1-et-12,d1=d2=sinh(2).
Then
(63)G1(t,s)=G2(t,s)=1d1{ϕ12(t)ϕ11(s),ρ(t1)≤s≤t≤tn,ϕ11(t)ϕ12(s),ρ(t1)≤t≤s≤tn,H14(t,s)=G1(t,s)+1d1-aϕ11(0)aG1(0,s)ϕ11(t),H24(t,s)=G2(t,s)+1d2-bϕ22(0)bG2(0,s)ϕ22(t).
Note r=∫t1tn‍(C*2/c*)g(s)Δs. Let R1=r+1 and we have
(64)R=∫t1tn‍C*Gi(s,s)(max0≤z1,z2≤R1fi(s,z1,z2)+g(s))Δs≤∫t1tn‍C*Gi(s,s)[2a+2R1a+2+2(s2-s4)1/4]Δs≤∫-11‍C*e44[2a+2R1a+2+2(s2-s4)1/4]Δs=∫01‍C*e42[2a+2R1a+2+2(s2-s4)1/4]Δs≤∫01‍C*e42[2a+2R1a+2+2(s-s2)1/2]Δs≤2a+1C*e4(R1a+2+π).Also let(65)λ*=min{1,R12a+2C*e4(R1a+2+π)-1,R12r}.
Now, if λ<λ*, Theorem 7 guarantees that (59) has a positive solutions (u,v) with ∥u∥≥1 and ∥v∥≥1.Example 14.
Consider the boundary value problem:(66)(p1u1∇)Δ(t)-q1(t)u1(t)=-λ(eu1+u22+7cos(2πtu1)),t1<t<tn,λ>0,(p2u2∇)Δ(t)-q2(t)u2(t)=-λ((u1-1)2+u22+5sin(2πtu2))
satisfying one of the boundary value conditions (k), (k=2,…,5).
Then ifλ>0 is sufficiently small, (66) has two solutions (u11,u12), (u21,u22) with uij(t)>0 for t∈(0,1),i,j=1,2.
To see this, we will apply Theorem8 with
(67)f1(t,u1,u2)=eu1+u22+7cos(2πtu1),f2(t,u1,u2)=(u1-1)2+u22+5sin(2πtu2),g1(t)=g2(t)=g(t)=8.
Clearly, for t∈(0,1),
(68)fi(t,u1,u2)+g(t)≥1>0,f1(t,0,0)=8>0,f2(t,0,0)=3>0,limu+v↑+∞inffi(t,u1,u2)u1+u2=+∞,i=1,2.
Now (H1)–(H4) hold. Let δ=1/100, ε=1/8, and we have
(69)fi(t,u1,u2)≥δfi(t,0,0),for0≤t≤1,0≤ui≤ε,i=1,2.
Furthermore let f¯(ε)=max0≤t≤1,0≤u1,u2≤ε{maxi=1,2fi(t,u1,u2)+g(t)}, and c=∫t1tn‍C*Gi(s,s)Δs. Note
(70)ε4cf¯(ε)≥132c(e+8)>1352c.
Letλ¯=1/352c. Now, if 0<λ<λ¯ then 0<λ<ε/4cf¯(ε) and Theorem 6 guarantees that (66) has positive solutions (u11,u12) with ∥u1j∥≤(1/8)(j=1,2).
Next noter=8C*2(tn-t1)/c* and let R1=r+2 so we have
(71)R=∫01‍C*Gi(s,s)(max0≤z1,z2≤R1fi(s,z1,z2)+g(s))Δs≤∫01‍C*Gi(s,s)(eR1+2R12+7+8)Δs≤∫01‍C*Gi(s,s)Δs(eR1+2R12+15)≤(eR1+2R12+15)c.
Also let(72)λ*=min{1,R12(eR1+2R12+15)c,R12r}.
Now, if λ<λ*, Theorem 7 guarantees that (59) has a positive solutions (u21,u22) with ∥u2j∥≥2, j=1,2.
Thus, ifλ<min{λ¯,λ*}, Theorem 8 guarantees that (66) has two solutions (u11,u12) and (u21,u22) with uij>0 for t∈(0,1),i,j=1,2.



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*Source: 679316-2013-03-27.xml*