# Positive Solutions to a Generalized Second-Order Difference Equation with Summation Boundary Value Problem

Authors: Thanin Sitthiwirattham; Jessada Tariboon
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

By using Krasnoselskii's fixed point theorem, we study the existence of positive solutions to the three-point summation boundary value problemΔ2u(t-1)+a(t)f(u(t))=0, t∈{1,2,…,T}, u(0)=β∑s=1ηu(s), u(T+1)=α∑s=1ηu(s), where f is continuous, T≥3 is a fixed positive integer, η∈{1,2,...,T-1}, 0<α<(2T+2)/η(η+1), 0<β<(2T+2-αη(η+1))/η(2T-η+1), and Δu(t-1)=u(t)-u(t-1). We show the existence of at least one positive solution if f is either superlinear or sublinear.


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## Body

## 1. Introduction

The study of the existence of solutions of multipoint boundary value problems for linear second-order ordinary differential and difference equations was initiated by Ilin and Moiseev [1]. Then Gupta [2] studied three-point boundary value problems for nonlinear second-order ordinary differential equations. Since then, nonlinear second-order three-point boundary value problems have also been studied by many authors; one may see the text books [3, 4] and the papers [5–10]. However, all these papers are concerned with problems with three-point boundary condition restrictions on the difference of the solutions and the solutions themselves, for example,
(1.1)u(0)=0,u(T+1)=0,u(0)=0,au(s)=u(T+1),u(0)=0,u(T+1)-au(s)=b,u(0)-αΔu(0)=0,u(T+1)=βu(s),u(0)-αΔu(0)=0,Δu(T+1)=0,
and so forth.In [5], Leggett-Williams developed a fixed point theorem to prove the existence of three positive solutions for Hammerstein integral equations. Since then, this theorem has been reported to be a successful technique for dealing with the existence of three solutions for the two-point boundary value problems of differential and difference equations; see [6, 7]. In [8], X. Lin and W. Liu, using the properties of the associate Green’s function and Leggett-Williams fixed point theorem, studied the existence of positive solutions of the problem.In [9], Zhang and Medina studied the existence of positive solutions for second-order boundary value problems of difference equations by applying Krasnoselskii’s fixed point theorem. In [10], Henderson and Thompson used lower and upper solution methods to study the existence of multiple solutions for second-order discrete boundary value problems.We are interested in the existence of positive solutions of the following second-order difference equation with three-point summation boundary value problem (BVP):(1.2)Δ2u(t-1)+a(t)f(u(t))=0,t∈{1,2,…,T},u(0)=β∑s=1ηu(s),u(T+1)=α∑s=1ηu(s),
where f is continuous, T≥3 is a fixed positive integer, η∈{1,2,…,T-1}.The aim of this paper is to give some results for existence of positive solutions to (1.2), assuming that 0<α<(2T+2)/η(η+1), 0<β<(2T+2-αη(η+1))/η(2T-η+1), and f is either superlinear or sublinear. Set
(1.3)f0=lim⁡u→0+f(u)u,f∞=lim⁡u→∞f(u)u.
Then f0=0 and f∞=∞ correspond to the superlinear case, and f0=∞ and f∞=0 correspond to the sublinear case. Let ℕ be the nonnegative integer; we let ℕi,j={k∈ℕ|i≤k≤j} and ℕp=ℕ0,p. By the positive solution of (1.2), we mean that a function u(t):ℕT+1→[0,∞) and satisfies the problem (1.2).Recently, Sitthiwirattham [11] proved the existence of positive solutions for the boundary value problem with summation condition
(1.4)Δ2u(t-1)+a(t)f(u(t))=0,t∈{1,2,…,T},u(0)=0,u(T+1)=α∑s=1ηu(s),
where f is continuous, T≥3 is a fixed positive integer, η∈{1,2,…,T-1}, and 0<α<2T+2/η(η+1).Throughout this paper, we suppose the following conditions hold:(A1)
f∈C([0,∞),[0,∞));(A2)
a∈C(ℕT+1,[0,∞)) and there exists t0∈ℕη,T+1 such that a(t0)>0.The proof of the main theorem is based upon an application of the following Krasnoselskii’s fixed point theorem in a cone.Theorem 1.1 (see [12]).
LetE be a Banach space, and let K⊂E be a cone. Assume Ω1, Ω2 are open subsets of E with 0∈Ω1, Ω¯1⊂Ω2, and let
(1.5)A:K∩(Ω¯2∖Ω1)⟶K
be a completely continuous operator such that(i)
∥Au∥⩽∥u∥, u∈K∩∂Ω1, and ∥Au∥⩾∥u∥, u∈K∩∂Ω2, or(ii)
∥Au∥⩾∥u∥, u∈K∩∂Ω1, and ∥Au∥⩽∥u∥, u∈K∩∂Ω2.Then A has a fixed point in K∩(Ω¯2∖Ω1).


## 2. Preliminaries

We now state and prove several lemmas before stating our main results.Lemma 2.1.
Letβ≠(2T+2-αη(η+1))/η(2T-η+1). Then, for y∈C(ℕT+1,[0,∞)), the problem
(2.1)Δ2u(t-1)+y(t)=0,t∈N1,T,(2.2)u(0)=β∑s=1ηu(s),u(T+1)=α∑s=1ηu(s),
has a unique solution
(2.3)u(t)=βη(η+1)+2t(1-βη)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)y(s)-β(T+1)-(β-α)t(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s-1)y(s)-∑s=1t-1(t-s)y(s),t∈NT+1.Proof.
From (2.1), we get
(2.4)Δu(t)-Δu(t-1)=-y(t),Δu(t-1)-Δu(t-2)=-y(t-1),⋮Δu(1)-Δu(0)=-y(1).
We sum the above equations to obtain(2.5)Δu(t)=Δu(0)-∑s=1ty(s),t∈NT,
we denote ∑s=pqy(s)=0, if p>q. Similarly, summing the above equation from t=0 to t=h, we get
(2.6)u(h+1)=u(0)+(h+1)Δu(0)-∑s=1h(h+1-s)y(s),h∈NT,
changing the variable from h+1 to t, we have
(2.7)u(t)=u(0)+tΔu(0)-∑s=1t-1(t-s)y(s):A+Bt-∑s=1t-1(t-s)y(s),t∈NT+1.
We sum (2.7) from s=1,2,…,η,
(2.8)∑s=1ηu(s)=ηA+η(η+1)2B-∑s=1η-1∑l=1η-s(l-s)y(s)=ηA+η(η+1)2B-12∑s=1η-1(η-s)(η-s+1)y(s).
By (2.2) from u(0)=β∑s=1ηu(s), we get
(2.9)(1-βη)A-βη(η+1)2B=-β2∑s=1η-1(η-s)(η-s+1)y(s),
and from u(T+1)=α∑s=1ηu(s), we obtain
(2.10)(1-αη)A+(T+1-αη(η+1)2)B=∑s=1T(T-s+1)y(s)-α2∑s=1η-1(η-s)(η-s+1)y(s).
Therefore,
(2.11)A=βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)y(s)-β(T+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)y(s),B=2(1-βη)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)y(s)+β-α(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)y(s).
Hence, (2.1)-(2.2) has a unique solution
(2.12)u(t)=βη(η+1)+2t(1-βη)(2t+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)y(s)-β(T+1)-(β-α)t(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)y(s)-∑s=1t-1(t-s)y(s),t∈NT+1.Lemma 2.2.
Let0<α<(2T+2)/η(η+1), 0<β<(2T+2-αη(η+1))/η(2T-η+1). If y∈C(ℕT+1,[0,∞)) and y(t)⩾0 for t∈ℕ1,T, then the unique solution u of (2.1)-(2.2) satisfies u(t)⩾0 for t∈ℕT+1.Proof.
From the fact thatΔ2u(t-1)=u(t+1)-2u(t)+u(t-1)=-y(t)≤0, we know that u(t)≥(u(t+1)+u(t-1))/2, so u(t+1)/(t+1)<u(t)/t.
Hence(2.13)u(T+1)-u(0)T+1<u(η)-u(0)η+1,η∈N1,T,
since u(T)⩾0 and u(0)≥0 imply that u(t)⩾0 for t∈ℕT+1.
Moreover, fromu(i)>(i/η)u(η)+((η-i)/η)u(0), we get(2.14)∑s=1ηu(s)>[1ηu(η)+η-1ηu(0)]+[2ηu(η)+η-2ηu(0)]+⋯+[ηηu(η)+η-ηηu(0)]=1ηu(η)[1+2+⋯+η]+1ηu(0)[(η-1)+(η-2)+⋯+0]=1ηu(η)[12η(η+1)]+1ηu(0)[η2-12η(η+1)]=12(η+1)u(η)+12(η-1)u(0).
Combining (2.14) with (2.2), we can get
(2.15)u(0)>β(η+1)2-β(η-1)u(η),
again combining (2.2), (2.14), and (2.15), we obtain
(2.16)u(T+1)>α(η+1)2-β(η-1)u(η),
such that
(2.17)2-β(η-1)>2-βη>2-2T+2-αη(η+1)2T-η+1=2(T-η)+αη(η+1)2T-η+1>0.
By using (2.13), (2.15), and (2.16), we obtain
(2.18)2-2βηηu(η)⩾(α-β)(η+1)T+1u(η).
If u(0)<0, then u(η)<0. It implies that (2T+2-αη(η+1))/η(2T-η+1)⩽β, a contradiction to β<(2T+2-αη(η+1))/η(2T-η+1). If u(T)<0, then u(η)<0, and the same contradiction emerges. Thus, it is true that u(0)⩾0, u(T)⩾0, together with (2.13), we have
(2.19)u(t)⩾0,t∈NT+1.
This proof is complete.Lemma 2.3.
Letαη(η+1)>2T+2, β>max{(2T+2-αη(η+1))/η(2T-η+1),0}. If y∈C(ℕT+1,[0,∞)) and y(t)⩾0 for t∈ℕ1,T, then problem (2.1)-(2.2) has no positive solutions.Proof.
Suppose that problem (2.1)-(2.2) has a positive solution u satisfying u(t)⩾0, t∈ℕT+1, and there is a τ0∈ℕ1,T such that u(τ0)>0.
Ifu(T+1)>0, then ∑s=1ηu(s)>0. It implies(2.20)u(0)=β∑s=1ηu(s)>2T+2-αη(η+1)η(2T-η+1)∑s=1ηu(s)⩾η(T+1)(u(0)+u(η))-η(η+1)u(T+1)η(2T-η+1),
that is,
(2.21)u(T+1)-u(0)T+1>u(η)-u(0)η+1,
which is a contradiction to (2.13).
Ifu(T+1)=0, then ∑s=1ηu(s)ds=0. When τ0∈ℕ1,η-1, we get u(τ0)>u(T)=0>u(η), which contradicts to (2.13). When τ0∈ℕη+1,T, we get u(η)⩽0=u(0)<u(τ0), which contradicts to (2.13) again. Therefore, no positive solutions exist.LetE=C(ℕT+1,[0,∞)), then E is a Banach space with respect to the norm
(2.22)‖u‖=sup⁡t∈NT+1|u(t)|.Lemma 2.4.
Let0<α<(2T+2)/η(η+1), 0<β<(2T+2-αη(η+1))/η(2T-η+1). If y∈C(ℕT+1,[0,∞)) and y(t)⩾0, then the unique solution to problem (2.1)-(2.2) satisfies
(2.23)inf⁡t∈NT+1u(t)⩾γ‖u‖,
where
(2.24)γ:=min⁡{α(η+1)(T+1-η)(T+1)(2-β(η-1))-αη(η+1),αη(η+1)(2-β(η-1))(T+1),β(η+1)(T+1-η)(2-β(η-1))(T+1),βη(η+1)(2-β(η-1))(T+1)}.Proof.
Letu(t) be maximal at t=τ1, when τ1∈ℕ1,T and ∥u∥=u(τ1). We divide the proof into two cases.
Case i.
Ifu(0)⩾u(T+1) and inft∈ℕT+1u(t)=u(T+1), then either 0⩽τ1⩽η<T+1 or 0<η<τ1<T+1, if 0⩽τ1⩽η<T+1, then
(2.25)u(τ1)⩽u(T+1)+u(T+1)-u(η)T+1-η(τ1-(T+1))⩽u(T+1)+u(T+1)-u(η)T+1-η(0-(T+1))⩽u(T+1)[1-((T+1)-(T+1)(2-β(η-1))/α(η+1)T+1-η)]⩽u(T+1)[(T+1)(2-β(η-1))-αη(η+1)α(T+1)(T+1-η)].
This implies
(2.26)inf⁡t∈NT+1u(t)⩾α(T+1)(T+1-η)(T+1)(2-β(η-1))-αη(η+1)‖u‖.
Similarly, if0<η<τ1<T+1, from(2.27)u(η)η⩾u(τ1)τ1⩾u(τ1)T+1,
together with (2.16), we have
(2.28)u(T+1)⩾αη(η+1)(2-β(η-1))(T+1)u(τ1).
This implies
(2.29)inf⁡t∈NT+1u(t)⩾αη(η+1)(2-β(η-1))(T+1)‖u‖.
Case ii.
Ifu(0)⩽u(T+1) and inft∈ℕT+1u(t)=u(0), then either 0<τ1<η<T+1 or 0<η⩽τ1⩽T+1, by (2.13). If 0<τ1<η<T+1, from
(2.30)u(η)T+1-η⩾u(τ1)T+1-τ1⩾u(τ1)T+1,
together with (2.15), we have
(2.31)u(0)>β(η+1)(T+1-η)(2-β(η-1))(T+1)u(τ1).
Hence,
(2.32)inf⁡t∈NT+1u(t)>β(η+1)(T+1-η)(2-β(η-1))(T+1)‖u‖.
If 0<η⩽τ1⩽T+1, from
(2.33)u(τ1)T+1⩽u(τ1)τ1⩽u(η)η,
together with (2.15), we have
(2.34)u(0)<βη(η+1)(2-β(η-1))(T+1)u(τ1).
This implies
(2.35)inf⁡t∈NT+1u(t)<βη(η+1)(2-β(η-1))(T+1)‖u‖.
This completes the proof.In the rest of the paper, we assume that0<α<(2T+2)/η(η+1), T∈ℕ1,T; 0<β<(2T+2-αη(η+1))/η(2T-η+1). It is easy to see that the BVP (1.2) has a solution u=u(t) if and only if u is a solution of the operator equation
(2.36)Au(t)≜βη(η+1)+2t(1-βη)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)u(s)f(u(s))-β(T+1)-(β-α)t(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)u(s)f(u(s))-∑s=1t-1(t-s)u(s)f(u(s)).Denote
(2.37)K={u∈E:u⩾0,min⁡t∈NT+1u(t)⩾γ‖u‖},whereγ is defined in (2.24).It is obvious thatK is a cone in E. Since Au=u and from Lemmas 2.2 and 2.4, then A(K)⊂K. It is also easy to check that A:K→K is completely continuous. In the following, for the sake of convenience, set(2.38)Λ1=(2T+2)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s),Λ2=γ(2-βη+β)(T-η)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1Tsa(s).


## 3. Main Results

Now we are in the position to establish the main result.Theorem 3.1.
The BVP (1.2) has at least one positive solution in the case(H1)
f0=0 and f∞=∞ (superlinear) or(H2)
f0=∞ and f∞=0 (sublinear).Proof.
Superlinear Case
Let(H1) hold. Since f0=limu→0+(f(u)/u)=0 for any ɛ∈(0,Λ1-1], there exists ρ* such that
(3.1)f(u)⩽ɛuforu∈[0,ρ*].
Let Ωρ*={u∈E:∥u∥<ρ*} for any u∈K∩∂Ωρ*. From (3.1), we get
(3.2)Au(t)=2t(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))-β(T+1)-(β-α)t(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)a(s)f(u(s))-∑s=1t-1(t-s)a(s)f(u(s))⩽2t(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))⩽2(T+1)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))⩽ερ*(2T+2)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)=ɛΛ1ρ*⩽ρ*=‖u‖,
which yields
(3.3)‖Au‖⩽‖u‖foru∈K∩∂Ωρ*.
Further, since f∞=limu→∞(f(u)/u)=∞, then, for any M*∈[Λ2-1,∞), there exists ρ*>ρ* such that
(3.4)f(u)⩾M*uforu⩾γρ*.
Set Ωρ*={u∈E:∥u∥<ρ*} for u∈K∩∂Ωρ*.
Sinceu∈K, mint∈NTu(t)⩾γ∥u∥=γρ*. Hence, for any u∈K∩Ωρ*, from (3.4) and (2.23), we get
(3.5)Au(η)=(2-βη+β)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))-β(T+1)-(β-α)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η(η-s)(η-s+1)a(s)f(u(s))-∑s=1η-1(η-s)a(s)f(u(s))=(2-βη+β)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))+1(2T+2-αη(η+1))-βη(2T-η+1)×∑s=1η-1(η-s)[-(2-βη+β)T+(β(T-η)+αη+1)s+(η-1)β]a(s)f(u(s))=(2-βη+β)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))-T(2-βη+β)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)a(s)f(u(s))+β(t-η)+αη+1(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(ηs-s2)a(s)f(u(s))+(η-1)β(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)a(s)f(u(s))⩾(2-βη+β)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))-T(2-βη+β)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(η-s)a(s)f(u(s))=(2-βη+β)(T-η)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1Tsa(s)f(u(s))⩾γρ*M*(2-βη+β)(T-η)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1Tsa(s)=M*Λ2ρ*⩾ρ*=‖u‖,
which implies
(3.6)‖Au‖⩾‖u‖foru∈K∩∂Ωρ*.
Therefore, from (3.3), (3.6), and Theorem 1.1, it follows that A has a fixed point in K∩(Ω¯ρ*∖Ωρ*) such that ρ*⩽∥u∥⩽ρ*.
Sublinear Case
Let(H2) hold. In view of f0=limu→0+(f(u)/u)=∞ for any M*∈[Λ2-1,∞), there exists r*>0 such that
(3.7)f(u)⩾M*ufor0⩽u⩽r*.
Set Ωr*={u∈E:∥u∥<r*} for u∈K∩∂Ωr*. Since u∈K, then mint∈ℕT+1u(t)⩾γ∥u∥=γr*. Thus, from (3.7) for any u∈K∩∂Ωr*, we can get
(3.8)Au(η)=(2-βη+β)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))-β(T+1)-(β-α)η(2T+2-αη(η+1))-βη(2T-η+1)∑s=1η-1(η-s)(η-s+1)a(s)f(u(s))-∑s=1η-1(η-s)a(s)f(u(s))⩾γr*M*(2-βη+β)(T-η)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1Tsa(s)=M*Λ2r*⩾r*=‖u‖,
which yields
(3.9)‖Au‖⩾‖u‖foru∈K∩∂Ωr*.
Sincef∞=limu→∞(f(u)/u)=0, then, for any ɛ1∈(0,Λ1-1], there exists r0>r* such that(3.10)f(u)⩽ɛ1uforu∈[r0,∞).
We have the following two cases.
Case i.
Suppose thatf(u) is unbounded, then, from f∈C([0,∞),[0,∞)), we know that there is r*>r0 such that
(3.11)f(u)⩽f(r*)foru∈[0,r*].
Since r*>r0, then, from (3.10) and (3.11), one has
(3.12)f(u)⩽f(r*)⩽ɛ1r*foru∈[0,r*].
For u∈K, ∥u∥=r*, from (3.12), we obtain
(3.13)Au(t)⩽(2T+2)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))⩽ɛ1r*(2T+2)(1-βη)+βη(η+1)(2T+2-α(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)=ɛ1Λ1r*⩽r*=‖u‖.
Case ii.
Suppose thatf(u) is bounded, say f(u)⩽N for all u∈[0,∞). Taking r*⩾max{N/ɛ1,r*}, for u∈K, ∥u∥=r*, we have
(3.14)Au(t)⩽(2T+2)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)f(u(s))⩽N(2T+2)(1-βη)+βη(η+1)(2T+2-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)⩽ɛ1r*(2T+2)(1-βη)+βη(η+1)(2T-αη(η+1))-βη(2T-η+1)∑s=1T(T-s+1)a(s)=ɛ1Λ1r*⩽r*=‖u‖.
Hence, in either case, we may always set Ωr*={u∈E:∥u∥<r*} such that
(3.15)‖Au‖⩽‖u‖foru∈K∩∂Ωr*.
Hence, from (3.9), (3.15), and Theorem 1.1, it follows that A has a fixed point in K∩(Ω¯ρ*∖Ωρ*) such that r*⩽∥u∥⩽r*. The proof is complete.


## 4. Some Examples

In this section, in order to illustrate our result, we consider some examples.Example 4.1.
Consider the BVP(4.1)Δ2u(t-1)+t2uk=0,t∈N1,4,u(0)=13∑s=12u(s),u(5)=23∑s=12u(s).
Set α=2/3, β=1/3, η=2, T=4, a(t)=t2, and f(u)=uk.
We can show that(4.2)0<α=23<53=2T+2η(η+1),0<β=13<37=2T+2-αη(η+1)η(2T-η+1).
Case I.
k∈(1,∞). In this case, f0=0,f∞=∞, and H1 of Theorem 3.1 holds. Then BVP (4.1) has at least one positive solution.
Case II.
k∈(0,1). In this case, f0=∞,f∞=0, and H2 of Theorem 3.1 holds. Then BVP (4.1) has at least one positive solution.Example 4.2.
Consider the BVP(4.3)Δ2u(t-1)+ette(πsin⁡u+2cos⁡uu2)=0,t∈N1,4,u(0)=14∑s=13u(s),u(5)=13∑s=13u(s).
Set α=1/3, β=1/4, η=3, T=4, a(t)=ette, f(u)=(πsinu+2cosu)/u2.
We can show that(4.4)0<α=13<56=2T+2η(η+1),0<β=14<13=2T+2-αη(η+1)η(2T-η+1).
Through a simple calculation we can getf0=∞,f∞=0. Thus, by H2 of Theorem 3.1, we can get BVP (4.3) has at least one positive solution.



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*Source: 569313-2012-05-09.xml*