# Positive Solutions to Nonlinear Higher-Order Nonlocal Boundary Value Problems for Fractional Differential Equations

Authors: Mujeeb Ur Rehman; Rahmat Ali Khan
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We study existence of positive solutions to nonlinear higher-order nonlocal
boundary value problems corresponding to fractional differential equation of the type𝒟c0+δu(t)+f(t,u(t))=0, t∈(0,1), 0<t<1. u(1)=βu(η)+λ2, u′(0)=αu′(η)−λ1, u′′(0)=0, u′′′(0)=0⋯u(n−1)(0)=0, where, n−1<δ<n, n(≥3)∈ℕ, 0<η,α,β<1, the boundary parameters λ1,λ2∈ℝ+ and Dc0+δ is the Caputo fractional derivative. We use the classical tools from functional analysis to obtain
sufficient conditions for the existence and uniqueness of positive solutions to the boundary value
problems. We also obtain conditions for the nonexistence of positive solutions to the problem. We
include examples to show the applicability of our results.


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## Body

## 1. Introduction

Fractional calculus goes back to the beginning of the theory of differential calculus and is developing since the 17th century through the pioneering work of Leibniz, Euler, Abel, Liouville, Riemann, Letnikov, Weyl, and many others. Fractional calculus is the generalization of ordinary integration and differentiation to an arbitrary order. For almost 300 years, it was seen as interesting but abstract mathematical concept. Nevertheless the applications of fractional calculus just emerged in the last few decades in various areas of physics, chemistry, engineering, biosciences, electrochemistry, and diffusion processes. For details, we refer the readers to [1–5].The existence and uniqueness of solutions for fractional differential equations is well studied in [6–10] and references therein. It should be noted that most of the papers and books on fractional calculus are devoted to the solvability of initial value problems for fractional differential equations. In contrast, the theory of boundary value problems for nonlinear fractional differential equations has received attention quiet recently, and many aspects of the theory need to be further investigated.There are some recent development dealing with the existence and multiplicity of positive solutions to nonlinear boundary value problems for fractional differential equations, see, for example, [11–18] and the reference therein. However, few results can be found in the literature concerning the existence of positive solutions to nonlinear three-point boundary value problems for fractional differential equations. For example, Li and coauthors [19] obtained sufficient conditions for the existence and multiplicity results to the following three point fractional boundary value problem
(1.1)D0+αu(t)+f(t,u(t))=0,0<t<1,1<α≤2,u(0)=0,D0+βu(1)=aD0+βu(ξ),
where 𝒟0+α is standard Riemann-Liouville fractional order derivative.Bai [20] studied the existence and uniqueness of positive solutions to the following three-point boundary value problem for fractional differential equations
(1.2)D0+αu(t)+f(t,u(t))=0,0<t<1,1<α≤2,u(0)=0,βu(η)=u(1),
where 0<βηα-1<1, 0<η<1, 𝒟0+α is standard Riemann-Liouville fractional order derivative. The function f is assumed to be continuous on [0,1]×[0,∞).The purpose of the present work is to investigate sufficient conditions for the existence, uniqueness, and nonexistence of positive solutions to more general boundary value problems for higher-order nonlinear fractional differential equations(1.3)D0+δcu(t)+f(t,u(t))=0,t∈(0,1),n-1<δ<n,u(1)=βu(η)+λ2,u′(0)=αu′(η)-λ1,u′′(0)=0,u′′′(0)=0⋯u(n-1)(0)=0,
where, n-1<δ<n, n=3,4,5,…; 0<η,α,β<1, the boundary parameters λ1,λ2∈ℝ+ and 𝒟0+δc, is the Caputo fractional derivative. The function f is assumed to be continuous and nonnegative on [0,1]×[0,∞). To the best of our knowledge, existence and uniqueness of positive solution to the boundary value problem (1.3) have never been studied previously.This paper is organized as follows: in Section2, we recall some basic definitions and preliminary results which are needed for our main results. In Section 3, we study existence and uniqueness and nonexistence of positive solutions to the boundary value problem (1.3) under certain assumptions on the function f. Moreover, examples are provided to illustrate the applicability of main results.


## 2. Background Materials and Lemmas

For the convenience of the readers, in this section, we provide definitions of Riemann-Liouville fractional integral and fractional derivative and some of their basic properties which will be helpful in the forth coming investigations.Definition 2.1 (see [2, 5]).
For a functionϕ:(a,∞)→ℝ, the Riemann-Liouville fractional integral of order α>0 is defined by
(2.1)Ia+αϕ(t)=1Γ(α)∫at(t-s)α-1ϕ(s)ds,
The provided that the integral on the right hand side exists. Forα,β≥0, the fractional integral satisfies the semigroup property
(2.2)I0+αI0+βϕ(t)=I0+α+βϕ(t)=I0+βI0+αϕ(t)almosteverywhereon[0,1].
In addition, if ϕ∈C[0,1] or if α+β≥1, then the identity is true for every t∈[0,1].Definition 2.2 (see [2, 5]).
The standard Riemann-Liouville fractional derivative of orderα>0 of a function ϕ:(a,∞)→ℝ is given by 𝒟a+αϕ(t)=(d/dt)nℐa+n-αϕ(t), where a∈ℝ, n=[s]+1, provided that the right hand side is pointwise defined on (a,∞).Definition 2.3.
For a given functionϕ:(a,∞)→ℝ, the Caputo fractional derivative of order α>0 is defined by 𝒟a+αcϕ(t)=ℐa+n-αϕ(n)(t), where a∈ℝ, n=[s]+1.Lemma 2.4 (see [2]).
Ifα>β>0, then 𝒟0+βcℐ0+αϕ(t)=ℐ0+α-βϕ(t). In particular, if m is positive integer and δ>m, then (dm/dtm)(ℐ0+δϕ(t))=ℐ0+δ-mϕ(t).The following two lemmas play a fundamental role to obtain an equivalent integral representation to the boundary value problem (1.3).Lemma 2.5 (see [2]).
Letα>0, then
(2.3)I0+αD0+αcϕ(t)=ϕ(t)-∑k=0n-1ϕk(0)k!tk=ϕ(t)+c1+c2t+c3t2+⋯+cntn-1,
where ci=ϕk-1(0)/(k-1)!, i=1,2,…,n and n-1<α≤n.Lemma 2.6 (see [2]).
Forα>0, the fractional differential equation 𝒟0+αcϕ(t)=0 has a general solution ϕ(t)=c1+c2t+c3t2+⋯+cntn-1, where ci∈ℝ, i=1,2,…,n and n-1<α≤n.For the existence of positive solutions to the boundary value problem (1.3), we use the following fixed point theorem due to Krasnosel’skii.Theorem 2.7 (see [21]).
Letℰ be a Banach space, and let 𝒫⊂ℰ be a cone. Assume that Ω1, Ω2 are open subsets of ℰ such that 0∈Ω1⊂Ω¯1⊂Ω2. Let 𝒜:𝒫∩(Ω¯2∖Ω1)→ℰ be completely continuous such that either  (i)
∥𝒜u∥≤∥u∥, for u∈𝒫∩∂Ω1 and ∥𝒜u∥≥∥u∥, for u∈𝒫∩∂Ω2, or(ii)
∥𝒜u∥≥∥u∥, for u∈𝒫∩∂Ω1 and ∥𝒜u∥≤∥u∥, for u∈𝒫∩∂Ω2.Then, 𝒜 has a fixed point in 𝒫∩(Ω¯2∖Ω1).


## 3. Main Results

Lemma 3.1.
Leth∈C[0,1], then the unique solution of the linear problem
(3.1)D0+δcu(t)+h(t)=0,t∈(0,1),n-1<δ<n,n(≥3)∈N,(3.2)u(1)=βu(η)+λ2,u′(0)=αu′(η)-λ1,u′′(0)=0,u′′′(0)=0⋯u(n-1)(0)=0
is given by
(3.3)u(t)=∫01G(t,s)h(s)ds+∫01H(t;η,s)h(s)ds+ψ(t),
where,
(3.4)G(t,s)={(1-s)δ-1-(t-s)δ-1Γ(δ),s≤t,(1-s)δ-1Γ(δ),t≤s,H(t;η,s)={β[(1-s)δ-1-(η-s)δ-1](1-β)Γ(δ)+α[1-βη-(1-β)t](η-s)δ-2(1-α)(1-β)Γ(δ-1),s≤η,β(1-s)δ(1-β)Γ(δ-1),η≤s,
and ψ(t)=((1-βη-(1-β)t)/(1-α)(1-β))λ1+λ2/(1-β).Proof.
In view of Lemma2.5, (3.1) is equivalent to the integral equation
(3.5)u(t)=-I0+δh(t)+c1+c2t+c3t2+⋯+cntn-1.
Using Lemma 2.4, we obtain
(3.6)u′(t)=-I0+δ-1h(t)+c2+2c3t+⋯+(n-1)cntn-2,u′′(t)=-I0+δ-2h(t)+2c3+⋯+(n-1)(n-2)cntn-3,⋮u(n-1)(t)=-I0+δ-(n-1)h(t)+(n-1)!cn,
where c1,c2,…,cn∈ℝ. The boundary conditions, u′′(0)=0, u′′′(0)=0,…,u(n-1)(0)=0, lead to c3=0, c4=0,…,cn=0. Using the boundary conditions u′(0)=αu′(η)-λ1, u(1)=βu(η)+λ2 and the fact that I0+δ-1h(0)=0, we obtain
(3.7)(1-α)c2=-αI0+δ-1h(η)-λ1,(1-β)c1+(1-βη)c2=I0+δh(1)-βI0+δh(η)+λ2,
which implies that c2=-(α/(1-α))I0+δ-1h(η)-(λ1/(1-α)) and
(3.8)c1=11-βI0+δh(1)-β1-βI0+δh(η)+α(1-βη)(1-α)(1-β)I0+δ-1h(η)+(1-βη)λ1(1-α)(1-β)+λ2(1-β).
Hence, the unique solution of the linear fractional boundary value problem (3.1), (3.2) is given by
(3.9)u(t)=-I0+δh(t)+11-βI0+δh(1)-β1-βI0+δh(η)+α(1-βη)-α(1-β)t(1-α)(1-β)I0+δ-1h(η)+(1-βη(1-α)(1-β)-t1-α)λ1+λ21-β=-I0+δh(t)+I0+δh(1)+β1-βI0+δh(1)-β1-βI0+δh(η)+α(1-βη)-α(1-β)t(1-α)(1-β)I0+δ-1h(η)+(1-βη(1-α)(1-β)-t1-α)λ1+λ21-β=∫0t(1-s)δ-1-(t-s)δ-1Γ(δ)h(s)ds+∫t1(1-s)δ-1Γ(δ)h(s)ds+∫0η(β(1-s)δ-1-β(η-s)δ-1(1-β)Γ(δ)+(α(1-βη)-α(1-β)t)(η-s)δ-2(1-α)(1-β)Γ(δ-1))h(s)ds+∫η1β(1-s)δ-1(1-β)Γ(δ)h(s)ds+(1-βη(1-α)(1-β)-t1-α)λ1+λ21-β=∫01G(t,s)h(s)ds+∫01H(t;η,s)h(s)ds+ψ(t).Lemma 3.2.
The functionsG(t,s) and H(t;η,s) satisfy the following properties:  (i)
G(t,s)≥0, H(t;η,s)≥0 and G(t,s)≤G(s,s) for all 0≤s, t≤1,(ii)
Fors∈(0,1), 0<ξ<τ<1, minξ≤t≤τG(t,s)≥(1-τδ-1)max0≤t≤1G(t,s)=(1-τδ-1)G(s,s),(iii)
β(1-ηδ-1)(1-s)δ-1≤(1-β)Γ(δ)H(t;η,s)<2(δ-1)(1-s)δ-2,(iv)
fors∈(0,1),0<ξ<τ<1, minξ≤t≤τH(t;η,s)≥(1-τ)max0≤t≤1H(t;η,s).Proof.
(i) Forδ>1, in view of the expression for G(t,s), it follows that G(t,s)≥0 and G(t,s)≤G(s,s) for all 0≤s, t≤1.
Fors≤η and 0<β<1, we have 1-βη-(1-β)t>(1-βη)(1-t). Hence, H(t,η,s)>0. Also, we note that H(t,η,s)>0 for s≥η.
(ii) Now,(3.10)max⁡t∈[0,1]G(t,s)=(1-s)δ-1Γ(δ).
Since
(3.11)G(t,s)=1Γ(δ)[(1-s)δ-1-(t-s)δ-1]=max⁡t∈[0,1]G(t,s)-(t-s)δ-1Γ(δ)≥max⁡t∈[0,1]G(t,s)-τδ-1(1-s/τ)δ-1Γ(δ),t∈[ξ,τ],
which implies that
(3.12)min⁡t∈[ξ,τ]G(t,s)≥(1-τδ-1)max⁡t∈[0,1]G(t,s),fors∈(0,1).
Now, for s≤η, we have
(3.13)H(t;η,s)=β[(1-s)δ-1-(η-s)δ-1](1-β)Γ(δ)+α[1-βη-(1-β)t](η-s)δ-2(1-α)(1-β)Γ(δ-1)≤β(1-s)δ-1(1-β)Γ(δ)+α(η-s)δ-2(1-α)(1-β)Γ(δ)<2(δ-1)(1-s)δ-2(1-α)(1-β)Γ(δ).
For s≥η, obviously, H(η,s)<2(δ-1)(1-s)δ-2/(1-α)(1-β)Γ(δ). From the expression of H(η,s), it clearly follows that
(3.14)H(t;η,s)≥β(1-ηδ-1)(1-s)δ-1(1-β)Γ(δ).
(iii) From the definition ofH(t;η,s), we have
(3.15)∂∂t(H(t;η,s))=-(η-s)δ-2(1-α)Γ(δ-1)≤0.
Therefore, H(t;η,s) is nonincreasing in t, so its minimum value occurs at t=τ for t∈[ξ,τ], and its maximum value occurs at t=0 for t∈[0,1]. That is,
(3.16)min⁡ξ≤t≤τH(t;η,s)=β[(1-s)δ-1-(η-s)δ-1](1-β)Γ(δ)+α[1-βη-(1-β)τ](η-s)δ-2(1-β)Γ(δ),(3.17)max⁡0≤t≤1H(t;η,s)=β[(1-s)δ-1-(η-s)δ-1](1-β)Γ(δ)+α(1-βη)(η-s)δ-2(1-β)Γ(δ).
Since (1-s)δ-1-(η-s)δ-1≥0 and 1-τ<1, therefore
(3.18)(1-s)δ-1-(η-s)δ-1≥(1-τ)((1-s)δ-1-(η-s)δ-1).
Also, as 1-β≤1-βη, therefore
(3.19)1-βη-(1-β)τ≥1-βη-(1-βη)τ=(1-τ)(1-βη).
Subletting (3.18) and (3.19) in (3.16), we have
(3.20)min⁡ξ≤t≤τH(t;η,s)≥(1-τ){β[(1-s)δ-1-(η-s)δ-1](1-β)Γ(δ)+α(1-βη)(η-s)δ-2(1-β)Γ(δ)}=(1-τ)max⁡0≤t≤1H(t;η,s).Remark 3.3.
Forλ1,λ2>0, minξ≤t≤τψ(t)≥(1-τ)max0≤t≤1ψ(t).Proof.
As(d/dt)ψ(t)=-λ1/(1-α)<0, therefore ψ(t) is a decreasing function. Hence,
(3.21)max⁡0≤t≤1ψ(t)=ψ(0)=(1-βη)λ1(1-α)(1-β)+λ2(1-β),min⁡ξ≤t≤τψ(t)=ψ(τ)=[1-βη-(1-β)τ]λ1(1-α)(1-β)+λ2(1-β)≥1-τ1-β(1-βη1-αλ1+λ2)=(1-τ)max⁡0≤t≤1ψ(t).
Thus, we have minξ≤t≤τψ(t)≥(1-τ)max0≤t≤1ψ(t).
Letℬ=C[0,1] be endowed with the norm ∥u∥=max0≤t≤1|u(t)|. Define a cone 𝒫⊂ℬ by
(3.22)P={u∈B:min⁡ξ≤t≤τ⁡u(t)≥(1-τ)‖u‖},
and an operator 𝒜:ℬ→ℬ by
(3.23)Au(t)=∫01G(t,s)f(s,u(s))ds+∫01H(t;η,s)f(s,u(s))ds+ψ(t).
By Lemma 3.1, the boundary value problem (1.3) has a solution if and only if 𝒜 has a fixed point.Lemma 3.4.
The operator𝒜:𝒫→𝒫 is completely continuous.Proof.
Firstly, we show that𝒜(P)⊂P. From (3.23), Lemma 3.2, and Remark 3.3, we have
(3.24)min⁡ξ≤t≤τ(Au(t))≥(1-τδ-1)∫01G(s,s)f(s,u(s))ds+(1-τ)∫01max⁡t∈[0,1]H(t;η,s)f(s,u(s))ds+(1-τ)max⁡t∈[0,1]ψ(t)≥(1-τ)∫01G(s,s)f(s,u(s))ds+∫01max⁡t∈[0,1]H(t;η,s)f(s,u(s))ds+max⁡t∈[0,1]ψ(t)≥(1-τ)‖Au‖.
Hence, we have 𝒜(P)⊂P.
Next, we show that𝒜 is uniformly bounded. For fixed ℓ>0, consider a bounded subset ℳ of 𝒫 defined by
(3.25)M={u∈P:‖u‖≤l,l∈R+},
and define 𝒦=max0≤u≤ℓf(t,u(t))+1, then for u∈ℳ, we have
(3.26)|Au(t)|≤∫01G(t,s)|f(s,u(s))|ds+∫01H(t;η,s)|f(s,u(s))|ds+max⁡t∈[0,1]ψ(t)≤KΓ(δ)∫01(1-s)δ-1ds+2K(δ-1)(1-α)(1-β)Γ(δ)∫01(1-s)δ-2ds+(1-βη)λ1+(1-α)λ2(1-α)(1-β)≤K(1-α)(1-β)(I0+δ(1)+2I0+δ(1))+(1-βη)λ1+(1-α)λ2(1-α)(1-β)=(3δ-1)Kδ(1-α)(1-β)(δ-1)+(1-βη)λ1+(1-α)λ2(1-α)(1-β).
Hence, 𝒜(ℳ) is bounded.
Finally, we show that𝒜 is equicontinuous. Define σ=δ(1-α)Γ(δ)ε/𝒦[δ(1-α)+αηα-1+λ1] and choose t>τ such that t-τ<σ. Then, for all ε>0 and u∈ℳ, we have
(3.27)|Au(t)-Au(τ)|=∫01(G(t,s)-G(τ,s))f(s,u(s))ds+∫01(H(t;η,s)-H(t;η,s))f(s,u(s))ds-λ11-α(t-τ)≤K∫01G(t,s)-G(τ,s)ds+∫01H(t;η,s)-H(t;η,s)ds+λ11-α(t-τ)=K1Γ(δ)∫0t((t-s)δ-1-(τ-s)δ-1)ds+α(t-τ)(1-α)Γ(δ-1)×∫0η(η-s)δ-2ds+λ11-α(t-τ)=KδΓ(δ)[tδ-τδ+αηδ-1+λ11-α(t-τ)].
Using the mean value theorem, we obtain tδ-τδ≤δ(t-τ)<δσ. Hence, it follows that
(3.28)|Au(t)-Au(τ)|<Kσ[δ(1-α)+αηα-1+λ1]δ(1-α)Γ(δ)<ε.
By means of Arzela-Ascoli theorem 𝒜:𝒫→𝒫 is completely continuous operator.For convenience, we introduce following notations:(3.29)Kα,β,δ1=(3I0+δψ1(1)+6I0+δ-1ψ1(1)(1-α)(1-β))-1,Kξ,β,τ,δ2=((1-τδ-1)1-βIξ+δψ2(τ))-1,cα=13(1-α),cβ=13(1-β),γ=1-τδ-1.Theorem 3.5.
If there exist constantsρ1,ρ2∈ℝ+ such that ρ2>ρ1 and functions ψ1,ψ2∈L[0,1] such that  (i)
f(t,u)≥ρ2𝒦ξ,β,τ,δ2ψ2(t), for(t,u)∈[0,1]×[γρ2,ρ2],(ii)
f(t,u)≤ρ1𝒦α,β,η1ψ1(t), for(t,u)∈[0,1]×[0,ρ1], then the boundary value problem (1.3) has at least one positive solution for λ1, λ2 small enough and has no positive solution for λ1, λ2 large enough.Proof.
By Lemma3.4, the operator 𝒜 is completely continuous. The proof is divided into two steps.Step 1.
We prove that the boundary value problem (1.3) has at least one positive solution.
DefineΩρ1={u∈ℬ:∥u∥<ρ1} an open subset of ℬ and choose λ1, λ2 such that λ1≤cαρ1 and λ2≤cβρ1. Then, for any u∈𝒫∩∂Ωρ1, we have ∥u∥=ρ1 and in view of Lemma 3.2 and (3.23), it follows that
(3.30)|Au(t)|=|∫01G(t,s)f(s,u(s))ds+∫01H(t;η,s)f(s,u(s))ds+(1-βη-(1-β)t(1-α)(1-β))λ1+λ21-β|≤I0+δf(1,u(1))+2(1-α)(1-β)I0+δ-1f(1,u(1))+(1-βη)λ1+(1-α)λ2(1-α)(1-β)≤Kα,β,δ1ρ1[I0+δψ1(1)+2I0+δ-1ψ1(1)(1-α)(1-β)]+λ11-α+λ21-β≤ρ13+ρ13+ρ13=‖u‖,
which implies that ∥𝒜u∥≤∥u∥, for u∈𝒫∩∂Ωρ1.
DefineΩρ2={u∈𝒫:∥u∥<ρ2}. For any t∈[ξ,τ] and u∈𝒫∩∂Ωρ2, using Lemma 3.2, we have
(3.31)min⁡ξ≤t<τu(t)≥(1-τδ-1)‖u‖.
Therefore, by (3.3) and Remark 3.3, we have the following estimate:
(3.32)|Au(t)|=|∫01G(t,s)f(s,u(s))ds+∫01H(t;η,s)f(s,u(s))ds+(1-βη-(1-β)t(1-α)(1-β))λ1+λ21-β|≥[(1-τδ-1)+β(1-ηδ-1)1-β]Iξ+δf(τ,u(τ))+(1-βη-(1-β)t(1-α)(1-β))λ1+λ21-β≥ρ2Kξ,β,τ,δ2[(1-τδ-1)+β(1-τδ-1)1-β]Iξ+δψ2(τ)=ρ2Kξ,β,τ,δ2((1-τδ-1)1-β)Iξ+δψ2(τ)=ρ2=‖u‖,
which implies that ∥𝒜u∥≥∥u∥ for u∈𝒫∩∂Ωρ2
Hence, by Theorem2.7, it follows that 𝒜 has a fixed point u in 𝒫∩(Ω̅ρ2∖Ωρ1).Step 2.
Now, we prove that for large values ofλ1, λ2, the boundary value problem (1.3) has no positive solution. Otherwise, for i=1,2, there exists 0<λi1<λi2⋯<λin<⋯, with limn→∞λin=+∞, such that for any positive integer n, the boundary value problem
(3.33)D0+δu(t)+f(t,u(t))=0,t∈(0,1),0<t<1c,u(1)=βu(η)+λ2n,u′(0)=αu′(η)-λ1n,u′′(0)=0,u′′′(0)=0⋯u(n-1)(0)=0
has a positive solution given by
(3.34)un(t)=∫01G(t,s)f(s,un(s))ds+∫01H(t,s)f(s,un(s))ds+(1-βη-(1-β)t(1-α)(1-β))λ1n+λ2n1-β.
If t>n, then un(t)≥(1-t)λ1n/(1-α)(1-β)+λ2n/(1-β) and if t≤n, then un(t)≥(1-η)λ1n/(1-α)(1-β)+λ2n/(1-β). Therefore, un→∞ as n→∞.
Letρ̂1=min((1/2)ρ2,λ1/cα,λ2/cβ) and define Ωρ̂1={u∈𝒫:ρ1<∥un∥<ρ̂1}. For u∈Ωρ̂1, using (3.3) and Remark 3.3, we have
(3.35)‖un(t)‖≥∫01G(t,s)f(s,un(s))ds+∫01H(t;η,s)f(s,un(s))ds+(1-βη-(1-β)t(1-α)(1-β))λ1n+λ2n1-β≥12[(1-τδ-1)+β(1-ηδ-1)1-β]Iξ+δf(1,un(1))≥ρ2Kα,β,η21-β[(1-β)(1-τδ-1)+β(1-ηδ-1)]Iξ+δψ2(1)>2ρ̂1>2‖un‖,
which is a contradiction. Hence, the boundary value problem (1.3) have no positive solution for λ1, λ2 large enough.Theorem 3.6.
Assume that(A) is satisfied and there exists real-valued function ϕ(t)∈L[0,1] such that
(3.36)|f(t,u)-f(t,v)|≤ϕ(t)|u-v|,fort∈[0,1],u,v∈[0,∞),I0+δϕ(1)+2(1-α)(1-β)I0+δ-1ϕ(1)<1,
then the boundary value problem (1.3) has a unique positive solution.Proof.
Foru,v∈𝒫, using (3.23) and Lemma 3.2, we obtain
(3.37)|Au(t)-Av(t)|≤∫01G(t,s)|f(s,u(s))-f(s,v(s))|ds+∫01H(t;η,s)|f(s,u(s))-f(s,v(s))|ds<‖u-v‖[∫01(1-s)δ-1Γ(δ)ϕ(s)ds+∫012(δ-1)(1-s)δ-2(1-α)(1-β)Γ(δ)ϕ(s)ds]=‖u-v‖[I0+δϕ(1)+2(1-α)(1-β)I0+δ-1ϕ(1)]<‖u-v‖,
where we have taken into account (1-s)δ-1≤(1-s)δ-2, we obtain the relation ℐ0+δϕ(1)≤(1/(δ-1))ℐ0+δ-1ϕ(1). Hence, it follows by Banach contraction principle that the boundary value problem (1.3) has a unique positive solution.Example 3.7.
Consider the boundary value problem(3.38)D0+δcu(t)=t2e(3/2)t+18(1+t2)+t2u2(1+sin⁡⁡t)16π,t∈(0,1),u(1)=βu(η)+λ2,u′(0)=αu′(η)-λ1,u′′(0)=0,u′′′(0)=0,
where, δ=7/2, η=1/2, α=2/3, and β=1/3. Let f(t,u)=(t2e(3/2)t+1)/8(1+t2)+t2u2(1+sint)/16π, (t,u)∈[0,1]×[0,∞]. For ρ1=1,ρ2=550, we observe that
(3.39)f(t,u)≤18(e+1π)t2,for(t,u)∈[0,1]×[0,1],f(t,u)≥(1e)t21+t2,for(t,u)∈[0,1]×[283,550].
Taking ψ1(t)=t2 and ψ2(t)=t2/(1+t2). For ξ=1/4, τ=3/4, by computations, we find that cα=1/9, cβ=2/9, Iξ+δψ2(τ)=(1/60π)(497tan-1(2/7)-tan-1(2)-11322/21)≈0.001291, also ℐ0+δψ1(1)=128/10395π. Therefore, 𝒦α,β,δ1≈0.583566 and 𝒦ξ,β,τ,δ2≈0.000993874. Hence,
(3.40)f(t,u)≤ρ1Kα,β,δ1ψ1(t)≈0.583566t2,for(t,u)∈[0,1]×[0,1],f(t,u)≥ρ2Kξ,β,τ,δ2ψ2(t)≈0.546631(t21+t2),for(t,u)∈[0,1]×[283,550].
Assumptions (i) and (ii) of the Theorem 3.5 are satisfied. Therefore, the boundary value problem (3.38) has a positive solution for λ1∈[0,1/9], λ2∈[0,2/9] and no positive solution for λ1>1/9, λ2>2/9.Example 3.8.
Consider the boundary value problem(3.41)D0+δcu(t)=e-33t(11πe33tcos⁡⁡(25t)+500sin⁡2t)u2(14π+425e(1/8)t)(1+u),(3.42)u(1)=βu(η)+λ2,u′(0)=αu′(η)-λ1,u′′(0)=0,u′′′(0)=0,
where δ=7/2, α=2/3, β=3/5, η=1/2, and λ1,λ2∈ℝ. Let f(t,u)=e-33t(11πe33tcos(25t)+500sin2t)u2/(14π+425e(1/8)t)(1+u), (t,u)∈[0,1]×[0,∞]. For u,v∈𝒫, t∈[0,1], we have the following estimate:
(3.43)|f(t,u)-f(t,v)|≤e-33t(11πe33t+500)(14π+425e(1/8)t)(|u21+u+v21+v|)≤e-33t(11πe33t+500)(14π+425e(1/8)t)(|u-v|(u+v+uv)(1+u)(1+v))≤e-33t(11πe33t+500)(14π+425e(1/8)t)(|u-v|).
Let ϕ(t)=e-33t(11πe33t+500)/(14π+425e(1/8)t), then by computations, we have ℐ0+δ-1ϕ(1)≈0.013469, (2δ-α-1)ℐ0+δ-1ϕ(1)/(δ-1)(1-α)(1-β)≈0.215504<1. All the conditions of Theorem 3.6 are satisfied. Therefore, by Theorem 3.6, the boundary value problem (3.41), (3.41) has a unique positive solution.



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*Source: 501230-2010-12-06.xml*