# Positive Solutions forp-Laplacian Fourth-Order Differential System with Integral Boundary Conditions

Authors: Jiqiang Jiang; Lishan Liu; Yonghong Wu
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

This paper investigates the existence of positive solutions for a class of singularp-Laplacian fourth-order
differential equations with integral boundary conditions. By using the fixed point theory in cones, explicit range for λ and
μ is derived such that for any λ and μ lie in their respective interval, the existence of at least one positive
solution to the boundary value system is guaranteed.


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## Body

## 1. Introduction

Boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics and so on. Fourth-order differential equations boundary value problems, including those with thep-Laplacian operator, have their origin in beam theory [1, 2], ice formation [3, 4], fluids on lungs [5], brain warping [6, 7], designing special curves on surfaces [8], and so forth. In beam theory, more specifically, a beam with a small deformation, a beam of a material that satisfies a nonlinear power-like stress and strain law, and a beam with two-sided links that satisfies a nonlinear powerlike elasticity law can be described by fourth order differential equations along with their boundary value conditions. For more background and applications, we refer the reader to the work by Timoshenko [9] on elasticity, the monograph by Soedel [10] on deformation of structure, and the work by Dulcska [11] on the effects of soil settlement. Due to their wide applications, the existence and multiplicity of positive solutions for fourth-order (including p-Laplacian operator) boundary value problems has also attracted increasing attention over the last decades; see [12–33] and references therein. In [28], Zhang and Liu studied the following singular fourth-order four-point boundary value problem
(1.1)(ϕp(u′′(t)))′′=f(t,u(t)),0<t<1,u(0)=u(1)-au(ξ)=u′′(0)=u′′(1)-bu′′(η)=0,
where ϕp(x)=|x|p-2x, p>1,  0<ξ, η<1,  0≤a, b<1, f∈C((0,1)×(0,∞),(0,∞)), f(t,x) may be singular at t=0 and/or t=1 and x=0. The authors gave sufficient conditions for the existence of one positive solution by using the upper and lower solution method, fixed point theorems, and the properties of the Green function.In [32], Zhang et al. discussed the existence and nonexistence of symmetric positive solutions of the following fourth-order boundary value problem with integral boundary conditions:
(1.2)(ϕp(u′′(t)))′′=w(t)f(t,u(t)),0<t<1,u(0)=u(1)=∫01g(s)u(s)ds,ϕp(u′′(0))=ϕp(u′′(1))=∫01h(s)ϕp(u′′(s))ds,
where ϕp(x)=|x|p-2x, p>1, w∈L1[0,1] is nonnegative, symmetric on the interval [0,1],f:[0,1]×[0,+∞)→[0,+∞) is continuous, f(1-t,x)=f(t,x) for all (t,x)∈[0,1]×[0,+∞), and g,h∈L1[0,1] are nonnegative, symmetric on [0,1].Motivated by the work of the above papers, in this paper, we study the existence of positive solutions of the following singular fourth-order boundary value system with integral boundary conditions:(1.3)(ϕp1(u′′(t)))′′=λp1-1a1(t)f1(t,u(t),v(t)),0<t<1,(ϕp2(v′′(t)))′′=μp2-1a2(t)f2(t,u(t),v(t)),u(0)=u(1)=∫01u(s)dξ1(s),ϕp1(u′′(0))=ϕp1(u′′(1))=∫01‍ϕp1(u′′(s))dη1(s),v(0)=v(1)=∫01‍v(s)dξ2(s),ϕp2(v′′(0))=ϕp2(v′′(1))=∫01‍ϕp2(v′′(s))dη2(s),
where λ and μ are positive parameters, ϕpi(x)=|x|pi-2x, pi>1, ϕqi=ϕpi-1, 1/pi+1/qi=1, ξi,ηi:[0,1]→ℝ+(i=1,2) are nondecreasing functions of bounded variation, and the integrals in (1.3) are Riemann-Stieltjes integrals, f1:[0,1]×ℝ0+×ℝ+→ℝ+ and f2:[0,1]×ℝ+×ℝ0+→ℝ+ are two continuous functions, and f1(t,x,y) may be singular at x=0 while f2(t,x,y) may be singular at y=0;     a1, a2:(0,1)→ℝ+ are continuous and may be singular at t=0 and/or t=1, in which ℝ+=[0,+∞), ℝ0+=(0,+∞).Compared to previous results, our work presented in this paper has the following new features. Firstly, our study is on singular nonlinear differential systems, that is,a1 and a2 in (1.3) are allowed to be singular at t=0 and/or t=1, meanwhile f1(t,x,y) is allowed to be singular at x=0 while f2(t,x,y) is allowed to be singular at y=0, which bring about many difficulties. Secondly, the main tools used in this paper is a fixed-point theorem in cones, and the results obtained are the conditions for the existence of solutions to the more general system (1.3). Thirdly, the techniques used in this paper are approximation methods, and a special cone has been developed to overcome the difficulties due to the singularity and to apply the fixed-point theorem. Finally, we discuss the boundary value problem with integral boundary conditions, that is, system (1.3) including fourth-order three-point, multipoint and nonlocal boundary value problems as special cases. To our knowledge, very few authors studied the existence of positive solutions for p-Laplacian fourth-order differential equation with boundary conditions involving Riemann-Stieltjes integrals. Hence we improve and generalize the results of previous papers to some degree, and so it is interesting and important to study the existence of positive solutions for system (1.3).The rest of this paper is organized as follows. In Section2, we present some lemmas that are used to prove our main results. In Section 3, the existence of positive solution for system (1.3) is established by using the fixed point theory in cone. Finally, in Section 4, one example is also included to illustrate the main results.Definition 1.1.
A vector(u,v)∈(C2[0,1]∩C4(0,1))×(C2[0,1]∩C4(0,1)) is said to be a positive solution of system (1.3) if and only if (u,v) satisfies (1.3) and u(t)>0, v(t)≥0 or u(t)≥0, v(t)>0 for any t∈(0,1).LetK be a cone in a Banach space E. For 0<r<R<+∞, let Kr={x∈K:∥x∥<r}, ∂Kr={x∈K:∥x∥=r}, and K¯r,R={x∈K:r≤∥x∥≤R}. The proof of the main theorem of this paper is based on the fixed point theory in cone. We list one lemma [34, 35] which is needed in our following argument.Lemma 1.2.
LetK be a positive cone in real Banach space E and T:K¯r,R→K a completely continuous operator. If the following conditions hold(i)
∥Tx∥≤∥x∥ for x∈∂KR;(ii)
there existse∈∂K1 such that x≠Tx+me for any x∈∂Kr and m>0. Then T has a fixed point in K¯r,R.Remark 1.3.
If (i) and (ii) are satisfied forx∈∂Kr and x∈∂KR, respectively. Then Lemma 1.2 is still true.


## 2.  Preliminaries and Lemmas

The basic space used in this paper isE=C[0,1]×C[0,1]. Obviously, the space E is a Banach space if it is endowed with the norm as follows:
(2.1)‖(u,v)‖:=‖u‖+‖v‖,‖u‖=max⁡0≤t≤1|u(t)|,‖v‖=max⁡0≤t≤1|v(t)|for any(u,v)∈E. Denote C+[0,1]={u∈C[0,1]:u(t)≥0,0≤t≤1}. For convenience, we list the following assumptions:(H1)
a1,a2:(0,1)→ℝ+ are continuous and
(2.2)0<Li:=(∫01‍e(s)ai(s)ds)qi-1<+∞,i=1,2,
where e(s)=s(1-s), s∈[0,1].(H2)
ξi,ηi:[0,1]→ℝ+(i=1,2) are nondecreasing functions of bounded variation, and αi∈[0,1), βi∈[0,1), where
(2.3)αi=∫01‍dξi(s),βi=∫01‍dηi(s),i=1,2.(H3)
f1:[0,1]×ℝ0+×ℝ+→ℝ+,f2:[0,1]×ℝ+×ℝ0+→ℝ+ are continuous and satisfy
(2.4)f1(t,x,y)≤g1(t,x)+h1(t,y),∀(t,x,y)∈[0,1]×R0+×R+,f2(t,x,y)≤g2(t,x)+h2(t,y),∀(t,x,y)∈[0,1]×R+×R0+,where g1,h2:[0,1]×R0+→R+ are continuous and nonincreasing in the second variable, and g2,h1:[0,1]×R+→R+ are continuous and for any constant r>0,
(2.5)0<∫01e(s)a1(s)g1(s,r)ds<+∞,0<∫01‍e(s)a2(s)h2(s,r)ds<+∞.Similar to the proof of Lemmas 2.1 and  2.2 in [32], the following two lemmas are valid.Lemma 2.1.
If(H2) holds, then for any y∈L(0,1), the boundary value problem
(2.6)-x′′(t)=ϕqi(y(t)),0<t<1,x(0)=x(1)=∫01x(s)dξi(s)
has a unique solution
(2.7)x(t)=∫01Hi(t,s)ϕqi(y(s))ds,
where
(2.8)Hi(t,s)=G(t,s)+11-αi∫01G(τ,s)dξi(τ),i=1,2,G(t,s)={s(1-t),0≤s≤t≤1,t(1-s),0≤t≤s≤1.Lemma 2.2.
If(H2) holds, then for any z∈L(0,1), the boundary value problem
(2.9)-y′′(t)=z(t),0<t<1,y(0)=y(1)=∫01y(s)dηi(s)
has a unique solution
(2.10)y(t)=∫01Ki(t,s)z(s)ds,
where
(2.11)Ki(t,s)=G(t,s)+11-βi∫01G(τ,s)dηi(τ),i=1,2.Remark 2.3.
Fort, s∈[0,1], we have
(2.12)e(t)e(s)≤G(t,s)≤e(s)ore(t)≤max⁡t∈[0,1]e(t)=14.Remark 2.4.
If(H2) holds, it is easy to testify Hi(t,s) defined by (2.8) that:
(2.13)ρie(s)≤Hi(t,s)≤γie(s)≤14γi<γi,t,s∈[0,1],i=1,2,
where
(2.14)γi=11-αi,ρi=∫01‍e(τ)dξi(τ)1-αi,i=1,2.Remark 2.5.
From (2.11), we can prove that the properties of Ki(t,s)(i=1,2) are similar to those of Hi(t,s)(i=1,2).Lemma 2.6.
Forx>0, y>0, we have
(2.15)ϕqi(x+y)≤{2qi-1[ϕqi(x)+ϕqi(y)],qi≥2ϕqi(x)+ϕqi(y),1<qi<2,≤2qi-1[ϕqi(x)+ϕqi(y)],qi>1,i=1,2,(2.16)ϕqi(x)>ϕqi(y)>ϕqi(0)=0,x>y>0,qi>1,i=1,2.Proof.
The proof of this lemma is easy, and we omit it.Let(2.17)K={min⁡t∈[0,1](u,v)∈C+[0,1]×C+[0,1]:u,vareconcaveon⁡[0,1],min⁡t∈[0,1]u(t)≥Λ‖u‖,min⁡t∈[0,1]v(t)≥Λ‖v‖},
where
(2.18)Λ=min⁡{ρ1σ1q1-1γ1ν1q1-1,ρ2σ2q2-1γ2ν2q2-1},σi=∫01‍e(s)dηi(s)1-βi,νi=11-βi,i=1,2.It is easy to see thatK is a cone of E. For any 0<r<R, let Kr,R={(u,v)∈K:r<∥u∥<R,r<∥v∥<R}.Remark 2.7.
By the definition ofρi, σi, γi, νi(i=1,2), we have 0<Λ<1.
To overcome singularity, we consider the following approximate problem of (1.3):
(2.19)(ϕp1(u′′(t)))′′=λp1-1a1(t)f1n(t,u(t),v(t)),0<t<1,(ϕp2(v′′(t)))′′=μp2-1a2(t)f2n(t,u(t),v(t)),u(0)=u(1)=∫01‍u(s)dξ1(s),ϕp1(u′′(0))=ϕp1(u′′(1))=∫01‍ϕp1(u′′(s))dη1(s),v(0)=v(1)=∫01‍v(s)dξ2(s),ϕp2(v′′(0))=ϕp2(v′′(1))=∫01‍ϕp2(v′′(s))dη2(s),
where n is a positive integer and
(2.20)f1n(t,u,v)=f1(t,max⁡{u,n-1},v),f2n(t,u,v)=f2(t,u,max⁡{v,n-1}).
Clearly,fin∈C([0,1]×ℝ+×ℝ+,ℝ+)(i=1,2).
By Lemmas2.1 and 2.2, for each n∈ℕ, λ>0, μ>0, let us define operators Anλ:K→C[0,1],  Bnμ:K→C[0,1], and Tn:K→E by
(2.21)Anλ(u,v)(t)=λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds,(2.22)Bnμ(u,v)(t)=μ∫01‍H2(t,s)ϕq2(∫01‍K2(s,τ)a2(τ)f2n(τ,u(τ),v(τ))dτ)ds,
and Tn(u,v)=(Anλ(u,v),Bnμ(u,v)), respectively.Lemma 2.8.
Assume that(H1)–(H3) hold, then for each λ>0, μ>0, n∈ℕ, Tn:K¯r,R→K is a completely continuous operator.Proof.
Letλ>0, μ>0, and n∈ℕ be fixed. For any (u,v)∈K, by (2.21), we have
(2.23)(Anλ(u,v))′′(t)=-λϕq1(∫01‍K1(t,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)≤0,Anλ(u,v)(0)=Anλ(u,v)(1)=λ∫01‍H1(0,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≥0,
which implies that Anλ is nonnegative and concave on [0,1]. Similarly, by (2.22) we can obtain that Bnμ is nonnegative and concave on [0,1]. For any (u,v)∈K and t∈[0,1], it follows from (2.13) that
(2.24)Anλ(u,v)(t)=λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≤λγ1ν1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds.
Thus(2.25)‖Anλ(u,v)‖≤λγ1ν1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds.
On the other hand, by (2.13) and (2.18), we have
(2.26)Anλ(u,v)(t)=λ∫01H1(t,s)ϕq1(∫01K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≥λρ1σ1q1-1∫01e(s)ϕq1(∫01‍e(τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≥ρ1σ1q1-1γ1ν1q1-1‖Anλ(u,v)‖≥Λ‖Anλ(u,v)‖.
This implies that(2.27)min⁡t∈[0,1]Anλ(u,v)(t)≥Λ‖Anλ(u,v)‖.
Similar to (2.27), we also have
(2.28)min⁡t∈[0,1]Bnμ(u,v)(t)≥Λ‖Bnμ(u,v)‖.
Therefore,Tn(K)⊂K.
Next, we prove thatTn:K¯r,R→K is completely continuous. Suppose (um,vm)∈K¯r,R and (u0,v0)∈K¯r,R with ∥(um,vm)-(u0,v0)∥→0(m→∞). We notice that t∈[0,1]fin(t,um(t),vm(t))-fin(t,u0(t),v0(t))→0(m→∞). Using the Lebesgue dominated convergence theorem, we have
(2.29)|ϕq1p1-1(∫01‍K1(s,τ)a1(τ)f1n(τ,um(τ),vm(τ))dτ)-ϕq1p1-1(∫01‍K1(s,τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)|≤ν1∫01e(τ)a1(τ)|f1n(τ,um(τ),vm(τ))-f1n(τ,u0(τ),v0(τ))|dτ⟶0,m⟶∞.
Therefore,(2.30)‖Anλ(um,vm)-Anλ(u0,v0)‖≤λγ1∫01‍e(s)|ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,um(τ),vm(τ)dτ))-ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)|ds⟶0,m⟶∞.
Similarly, we also have(2.31)‖Bnμ(um,vm)-Bnμ(u0,v0)‖⟶0,m⟶∞.
SoAnλ:K¯r,R→C[0,1] and Bnβ:K¯r,R→C[0,1] are continuous. Therefore, Tn:K¯r,R→K is also continuous.
LetD⊂K¯r,R be any bounded set, then for any (u,v)∈D, we have (u,v)∈K, r≤∥u∥≤R, r≤∥v∥≤R, and then 0<Λr≤u(τ)≤R, 0<Λr≤v(τ)≤R for any τ∈[0,1]. By (H3), we have
(2.32)Lr:=(∫01‍e(τ)a1(τ)g1(s,rΛ)dτ)q1-1<+∞.
It is easy to show thatAnλ(D) is uniformly bounded. In order to show that Tn is a compact operator, we only need to show that Anλ(D) is equicontinuous. By the uniformly continuity of H1(t,s) on [0,1]×[0,1], for all ε>0, there is δ>0 such that for any t1, t2, s∈[0,1] and |t1-t2|<δ, we have
(2.33)|H1(t1,s)-H1(t2,s)|<ε.
This together with (2.15) and (2.32) implies
(2.34)|Anλ(u,v)(t1)-Anλ(u,v)(t2)|≤λ∫01|H1(t1,s)-H1(t2,s)|ϕq1(∫01K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds<ελν1q1-1ϕq1(∫01e(τ)a1(τ)[g1(τ,max⁡⁡{u(τ),n-1})+h1(τ,v(τ))]dτ)≤ελν1q1-1ϕq1(∫01e(τ)a1(τ)[g1(τ,rΛ)+h1(τ,v(τ))]dτ)≤ελν1q1-12q1-1[ϕq1(∫01e(τ)a1(τ)g1(τ,rΛ)dτ)+ϕq1(∫01e(τ)a1(τ)h1(τ,v(τ))dτ)]≤ελν1q1-12q1-1[Lr+L1(max⁡y∈[rΛ,R]τ∈[0,1]h1(τ,y))q1-1],|t1-t2|<δ,(u,v)∈D.
This means thatAnλ(D) is equicontinuous. By the Arzela-Ascoli theorem, Anλ(D) is a relatively compact set and that Anλ:K¯r,R→C[0,1] is a completely continuous operator.
In the same way, we can show thatBnμ:K¯r,R→C[0,1] is also completely continuous, and so Tn:K¯r,R→K is completely continuous. Now since λ, μ, and n are given arbitrarily, the conclusion of this lemma is valid.


## 3. Main Results

For notational convenience, we denote by(3.1)Mi=6(ρiσqi-1ΛLi)-1,Ni=(γiνiqi-1Li)-1,i=1,2,f1α=(limsup⁡x→αsup⁡y∈R+t∈[0,1]f1(t,x,y)ϕp1(x))q1-1,f2α=(limsup⁡y→αsup⁡x∈R+t∈[0,1]f2(t,x,y)ϕp2(y))q2-1,f1α=(liminf⁡x→αinf⁡y∈R+t∈[0,1]f1(t,x,y)ϕp1(x))q1-1,f2α=(liminf⁡y→αinf⁡x∈R+t∈[0,1]f2(t,x,y)ϕp2(y))q2-1,
where α denotes 0 or ∞. The main results of this paper are the following.Theorem 3.1.
Assume that(H1)–(H3) hold. Then we have:(C1)
Iff10, f1∞, f20∈(0,∞) and M1/f1∞<N1/f10, then for each λ∈(M1/f1∞,N1/f10), μ∈(0,N2/f20), the system (1.3) has at least one positive solution.(C2)
Iff10, f20, f2∞∈(0,∞) and M2/f2∞<N2/f20, then for each λ∈(0,N1/f10), μ∈(M2/f2∞,N2/f20), the system (1.3) has at least one positive solution.(C3)
Iff10=0, f1∞=∞, 0<f20<∞, then for each λ∈(0,∞), μ∈(0,N2/f20), the system (1.3) has at least one positive solution.(C4)
If0<f10<∞, f20=0, f2∞=∞, then for each λ∈(0,N1/f10), μ∈(0,∞), the system (1.3) has at least one positive solution.(C5)
Iffi0=0, fi∞=∞(i=1,2), then for each λ∈(0,∞), μ∈(0,∞), the system (1.3) has at least one positive solution.(C6)
If0<f10<∞, f1∞=∞ or f2∞=∞, 0<f20<∞, then for each λ∈(0,N1/f10), μ∈(0,N2/f20), the system (1.3) has at least one positive solution.(C7)
Iff10=0, 0<f1∞<∞, and f10=0, 0<f2∞<∞, then for each λ∈(M1/f1∞,∞), μ∈(0,∞) or λ∈(0,∞), μ∈(M2/f2∞,∞), the system (1.3) has at least one positive solution.Proof.
We only prove the condition in which(C1) holds. The other cases can be proved similarly.
Letλ∈(M1/f1∞,N1/f10), μ∈(0,N2/(f20)), choose ε1>0 such that f1∞-ε1>0 and
(3.2)M1f1∞-ε1≤λ≤N1f10+ε1,0<μ≤N2f20+ε1.
It follows fromfi0∈(0,∞) of (C1) that there exists r1>0 such that for any t∈[0,1],
(3.3)f1(t,x,y)≤(f10+ε1)p1-1ϕp1(x)≤[r1(f10+ε1)]p1-1,0<x≤r1,y≥0,(3.4)f2(t,x,y)≤(f20+ε1)p2-1ϕp2(y)≤[r1(f20+ε1)]p2-1,x≥0,0<y≤r1.
LetKr1={(u,v)∈K:∥u∥<r1,∥v∥<r1}. For any (u,v)∈∂Kr1, n>1/r1, by (2.13), (3.3), we have
(3.5)‖Anλ(u,v)‖=max⁡t∈[0,1]λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≤λγ1ν1q1-1ϕq1(∫01‍e(τ)a1(τ)(f10+ε1)p1-1ϕp1(u(τ))dτ)≤λγ1ν1q1-1(f10+ε1)L1r1=λN1-1(f10+ε1)r1.
Similarly, we also have(3.6)‖Bnμ(u,v)‖≤μN2-1(f20+ε1)r1.
Therefore, we have(3.7)‖Tn(u,v)‖=‖Anλ(u,v)‖+‖Bnμ(u,v)‖≤[λN1-1(f10+ε1)+μN2-1(f20+ε1)]r1≤2r1=‖(u,v)‖.
On the other hand, byf1∞>f1∞-ε1>0, there exists R0>0 such that
(3.8)f1(t,x,y)≥(f1∞-ε1)p1-1ϕp1(x),t∈[0,1],x≥R0,y≥0.
LetR1>max{2r1,Λ-1R0}, KR1={(u,v)∈K:∥u∥<R1,∥v∥<R1}. Next, we take (φ1,φ2)=(1,1)∈∂K1, and for any (u,v)∈∂KR1, m>0, n∈ℕ, we will show
(3.9)(u,v)≠Anλ(u,v)+m(φ1,φ2).
Otherwise, there exist(u0,v0)∈∂KR1 and m0>0 such that
(3.10)(u0,v0)=Anλ(u0,v0)+m0(φ1,φ2).
From(u0,v0)∈∂KR1, we know that ∥u0∥=R1 or ∥v0∥=R1. Without loss of generality, we may suppose that ∥u0∥=R1, then u0(τ)≥Λ∥u0∥=ΛR1>R0 for any τ∈[0,1]. So, by (2.13), (3.8), we have
(3.11)u0(t)=λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)(f1∞-ε1)p1-1ϕp1(u0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)(f1∞-ε1)p1-1(ΛR1)p1-1dτ)ds+m0=16λρ1σ1q1-1(f1∞-ε1)ΛR1L1+m0=λM1-1(f1∞-ε1)R1+m0>R1.
This implies thatR1>R1, which is a contradiction. This yields that (3.9) holds. By (3.7), (3.9), and Lemma 1.2, for any n>1/r1and λ∈(M1/f1∞,N1/f10), μ∈(0,N2/f20), we obtain that Tn has a fixed point (un,vn) in K¯r1,R1 satisfying r1<∥un∥<R1,r1<∥vn∥<R1.
Let{(un,vn)}n≥n1 be the sequence of solutions of boundary value problems (2.19), where n1>1/r1is a fixed integer. It is easy to see that they are uniformly bounded. Next we show that {un}n≥n1 are equicontinuous on [0,1]. From (un,vn)∈K¯r1,R1, we know that R1≥un(τ)≥Λ∥un∥≥Λr1, R1≥vn(τ)≥Λ∥vn∥≥Λr1, τ∈[0,1]. For any ε>0, by the continuous of H1(t,s) in [0,1]×[0,1], there exists δ1>0 such that for any t1, t2, s∈[0,1] and |t1-t2|<δ1, we have
(3.12)|H1(t1,s)-H1(t2,s)|<ε.
This combining with (2.15), (2.32) implies that for any t1, t2∈[0,1] and |t1-t2|<δ1, we have
(3.13)|un(t1)-un(t2)|≤λ∫01‍|H1(t1,s)-H1(t2,s)|ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,un(τ),vn(τ))dτ)ds<ελν1q1-1ϕq1(∫01‍e(τ)a1(τ)[g1(τ,max⁡⁡{un(τ),n-1})+h1(τ,vn(τ))]dτ)≤ελν1q1-12q1-1[ϕq1(∫01‍e(τ)a1(τ)g1(τ,rΛ)dτ)+ϕq1(∫01‍e(τ)a1(τ)h1(τ,v(τ))dτ)]≤ελν1q1-12q1-1[Lr+L1(max⁡y∈[r1Λ,R1]τ∈[0,1]h1(τ,y))q1-1].
Similarly,{vn}n≥n1 are also equicontinuous on [0,1]. By the Ascoli-Arzela theorem, the sequence {(un,vn)}n≥n1 has a subsequence being uniformly convergent on [0,1]. From Lemma 2.2, we know that
(3.14)un′′(s)=λp1-1∫01‍K1(s,τ)a1(τ)f1n(τ,un(τ),vn(τ))dτ,vn′′(s)=μp2-1∫01‍K2(s,τ)a2(τ)f2n(τ,un(τ),vn(τ))dτ.
Since the properties ofKi(t,s)(i=1,2) are similar to those of Hi(t,s)(i=1,2), so (un′′,vn′′) have the similar properties of (un,vn), that is, (un′′,vn′′) also has a subsequence being uniformly convergent on [0,1]. Without loss of generality, we still assume that {(un,vn)}n≥n1 itself uniformly converges to (u,v) on [0,1] and {(un′′,vn′′)}n≥n1 itself uniformly converges to (u′′,v′′) on [0,1], respectively. Since {(un,vn)}n≥n1∈K¯r1,R1⊂K, so we have un≥0, vn≥0. By (2.19), we have
(3.15)un(t)=un(12)+(t-12)un′(12)-∫1/2tds∫1/2sϕq1(un′′p1-1(12)+(s2-12)un′′′p1-1(12)-∫1/2s2ds1∫1/2s1λp1-1a1(τ)f1n(τ,un(τ),vn(τ))dτ)ds2,t∈(0,1),(3.16)vn(t)=vn(12)+(t-12)vn′(12)-∫1/2tds∫1/2sϕq2(vn′′p2-1(12)+(s2-12)vn′′′p2-1(12)-∫1/2s2ds1∫1/2s1μp2-1a2(τ)f2n(τ,un(τ),vn(τ))dτ)ds2,t∈(0,1).
From (3.15) and (3.16), we know that {un′(1/2)}n≥n1, {vn′(1/2)}n≥n1, {un′′(1/2)}n≥n1, {vn′′(1/2)}n≥n1, {un′′′(1/2)}n≥n1, {vn′′′(1/2)}n≥n1 are bounded sets. Without loss of generality, we may assume (un′(1/2),vn′(1/2))→(c1,d1),(un′′(1/2),vn′′(1/2))→(c2,d2),(un′′′(1/2),vn′′′(1/2))→(c3,d3) as n→∞. Then by (3.15), (3.16), and the Lebesgue dominated convergence theorem, we have
(3.17)u(t)=u(12)+c1(t-12)-∫1/2t‍ds∫1/2sϕq1(c2p1-1+c3p1-1(s2-12)-∫1/2s2‍ds1∫1/2s1λp1-1a1(τ)f1(τ,u(τ),v(τ))dτ)ds2,t∈(0,1),(3.18)v(t)=v(12)+d1(t-12)-∫1/2t‍ds∫1/2s‍ϕq2(d2p2-1+d3p2-1(s2-12)-∫1/2s2ds1∫1/2s1μp2-1a2(τ)f2(τ,u(τ),v(τ))dτ)ds2t∈(0,1).
By (3.17) and (3.18), direct computation shows that
(3.19)(ϕp1(u′′(t)))′′=λp1-1a1(t)f1(t,u(t),v(t)),(ϕp2(v′′(t)))′′=μp2-1a2(t)f2(t,u(t),v(t)),0<t<1.
On the other hand,(u,v) satisfies the boundary condition of (1.3). In fact, un(0)=un(1)=∫01‍un(s)dξ1(s), vn(0)=vn(1)=∫01‍vn(s)dξ2(s),  ϕp1(un′′(0))=ϕp1(un′′(1))=∫01‍ϕp1(un′′(s))dη1(s),  ϕp2(vn′′(0))=ϕp2(vn′′(1))=∫01‍ϕp2(vn′′(s))dη2(s), and so the conclusion holds by letting n→∞.Theorem 3.2.
Assume that(H1)–(H3) hold. Then we have:(D1)
Iff10, f1∞, f2∞∈(0,∞) and M1/f10<N1/f1∞, then for each λ∈(M1/f10,N1/f1∞), μ∈(0,N2/f2∞), the system (1.3) has at least one positive solution.(D2)
Iff1∞, f20, f2∞∈(0,∞) and M2/f20<N2/f2∞, then for each λ∈(0,N1/f1∞), μ∈(M2/f20,N2/f2∞), the system (1.3) has at least one positive solution.(D3)
Iff10=∞, f1∞=0, 0<f2∞<∞, then for each λ∈(0,∞), μ∈(0,N2/(f2∞)), the system (1.3) has at least one positive solution.(D4)
If0<f1∞<∞, f20=∞, f2∞=0, then for each λ∈(0,N1/f1∞), μ∈(0,∞), the system (1.3) has at least one positive solution.(D5)
Iffi0=∞, fi∞=0(i=1,2), then for each λ∈(0,∞), μ∈(0,∞), the system (1.3) has at least one positive solution.(D6)
If0<f1∞<∞, f10=∞ or f20=∞, 0<f2∞<∞, then for each λ∈(0,N1/f1∞), μ∈(0,N2/f2∞), the system (1.3) has at least one positive solution.(D7)
Iff1∞=0, 0<f10<∞, and f2∞=0, 0<f20<∞, then for each λ∈(M1/f10,∞), μ∈(0,∞) or λ∈(0,∞), μ∈(M2/f20,∞), the system (1.3) has at least one positive solution.Proof.
We may suppose that condition(D1) holds. Similarly, we can prove the other cases.
Letλ∈(M1/f10,N1/f1∞), μ∈(0,N2/f2∞). We can choose ε2>0 such that N1-ε2>0, N2-ε2>0 and
(3.20)λf1∞<N1-ε2,μf2∞<N2-ε2.
It follows from(D1) and (2.16) that there exists R2*>0 such that for any t∈[0,1](3.21)f1(t,x,y)≤(1λ(N1-ε2))p1-1ϕp1(x),x≥R2*,y≥0,(3.22)f2(t,x,y)≤(1λ(N2-ε2))p2-1ϕp2(y),x≥0,y≥R2*.
LetR2=Λ-1R2*, KR2={(u,v)∈K:∥u∥<R2,∥v∥<R2}. For any (u,v)∈∂KR2, n∈ℕ, by (2.13), (3.21), we have
(3.23)‖Anλ(u,v)‖=max⁡t∈[0,1]λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u(τ),v(τ))dτ)ds≤λγ1ν1q1-1ϕq1(∫01‍e(τ)a1(τ)(1λ(N1-ε2))p1-1ϕp1(u(τ))dτ)≤λγ1ν1q1-11λ(N1-ε2)L1R2<R2.
Similarly, by (3.22) we have ∥Bnμ(u,v)∥<R2. Therefore,
(3.24)‖Tn(u,v)‖=‖Anλ(u,v)‖+‖Bnμ(u,v)‖≤2R2=‖(u,v)‖,(u,v)∈∂KR2,n∈N.
On the other hand, chooseε3>0 such that M1+ε3<λf10. By the condition f10∈(0,∞) of (D1) and (2.16), there exists r2*>0 such that
(3.25)f1(t,x,y)≥(1λ(M1+ε3))p1-1ϕp1(x),t∈[0,1],0<x≤r2*,y≥0.
Let0<r2<min{R2,r2*},  Kr2={(u,v)∈K:∥u∥<r2,∥v∥<r2}. Next, we take (φ1,φ2)=(1,1)∈∂K1, n>1/r2, and for any (u,v)∈∂Kr2, m>0, we will show
(3.26)(u,v)≠Anλ(u,v)+m(φ1,φ2).
Otherwise, there exist(u0,v0)∈∂Kr2 and m0>0 such that
(3.27)(u0,v0)=Anλ(u0,v0)+m0(φ1,φ2).
From(u0,v0)∈∂Kr2, we know that ∥u0∥=r2 or ∥v0∥=r2. Without loss of generality, we may suppose that ∥u0∥=r2, then u0(τ)≥Λ∥u0∥≥Λr2 for any τ∈[0,1]. So, we have
(3.28)u0(t)=λ∫01‍H1(t,s)ϕq1(∫01‍K1(s,τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)f1n(τ,u0(τ),v0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)(1λ(M1+ε3))p1-1ϕp1(u0(τ))dτ)ds+m0≥λρ1σ1q1-1∫01‍e(s)ϕq1(∫01‍e(τ)a1(τ)(M1+ε3)p1-1(Λr2)p1-1dτ)ds+m0=16λρ1σ1q1-11λ(M1+ε3)Λr2L1+m0>r2.
This implies thatr2>r2, which is a contradiction. This yields that (3.26) holds. By (3.24), (3.26), and Lemma 1.2, for any n>1/r2and λ∈(M1/f10,N1/f1∞), μ∈(0,N2/f2∞), we obtain that Tn has a fixed point (un,vn) in K¯r2,R2 and r2<∥un∥<R2,  r2<∥vn∥<R2. The rest of proof is similar to Theorem 3.1.


## 4. An Example

Example 4.1.
We consider system (1.3) with p1=3/2,   p2=7/3,  a1(t)=1/(t(1-t),  a2(t)=1/((1-t)t),
(4.1)f1(t,u,v)=t2+1u+1+sin⁡(v2+v+t),(t,u,v)∈[0,1]×R0+×R+,f2(t,u,v)=2+sin⁡(u+ln⁡(t+1))+t4+t+3v,(t,u,v)∈[0,1]×R+×R0+.Obviously,a1,  a2 are singular at t=0 and t=1, f1(t,u,v) is singular at u=0 and f2(t,u,v) is singular at v=0. Choose g1(t,u)=(t2+1)/u, h1(t,v)=1+sin(v2+v+t), g2(t,u)=2+sin(u+ln(t+1)), and h2(t,v)=(t4+t+3)/√v. Let
(4.2)ξ1(s)={0,s∈[0,13),15,s∈[13,23),14,s∈[23,1],ξ2(s)={0,s∈[0,12),17,s∈[12,34),13,s∈[34,1],η1(s)={0,s∈[0,12),35,s∈[12,1],η2(s)={0,s∈[0,12),47,s∈[12,1].By direct calculation, we haveα1=1/4,  α2=1/3,  β1=3/5,  β2=4/7, ∫01‍e(s)ai(s)ds=(2/3)(i=1,2). It is easy to check that f10=f20=∞, f1∞=f2∞=0, and the conditions (H1)–(H3) and (D5) are satisfied. By Theorem 3.2, system (1.3) has at least one positive solution provided λ, μ∈(0,+∞).Remark 4.2.
Example4.1 not only implies that f1(t,u,v), f2(t,u,v) can be singular at u=0 and v=0, respectively, but also indicates that there is a large number of functions that satisfy the conditions of Theorem 3.2. In addition, the condition (D5) is also easy to check.



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*Source: 293734-2012-08-08.xml*