# Uniqueness and Asymptotic Behavior of Positive Solutions for a Fractional-Order Integral Boundary Value Problem

Authors: Min Jia; Xin Liu; Xuemai Gu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study a model arising from porous media, electromagnetic, and signal processing of wireless communication system-𝒟tαx(t)=f(t,x(t),x'(t),x”(t),…,x(n-2)(t)),  0<t<1,
x(0)=x'(0)=⋯=x(n-2)(0)=0,
x(n-2)(1)=∫01x(n-2)(s)dA(s), where n-1<α≤n,
n∈ℕ and n≥2, 𝒟tα is the standard Riemann-Liouville derivative, ∫01x(s)dA(s) is linear functionals given by Riemann-Stieltjes integrals, A is a function of bounded variation, and dA can be a changing-sign measure. The existence, uniqueness, and asymptotic behavior of positive solutions to the singular nonlocal integral boundary value problem for fractional differential equation are obtained. Our analysis relies on Schauder's fixed-point theorem and upper and lower solution method.


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## Body

## 1. Introduction

Recently, fractional-order models have proved to be more accurate than integer order models, that is, there are more degrees of freedom in the fractional-order models.Indeed, we can find numerous applications in viscoelasticity, electrochemistry control, porous media, electromagnetic, and signal processing of wireless communication system. Especially, in application of the digital signal processing, the fractional digital signal processing can greatly improve the high frequency components of signal, enhance an intermediate frequency component of signal, and reserve nonlinear low frequency signal. According to this analysis, fractional differential equation applied to the edge information extraction will get higher signal to noise ratio than that of the traditional method based on one- or-two order differential equation.Many applications on digital signal processing were found recently, for example, Hartley et al. [1] derived a closed-loop control electromagnetic and signal processing system of equations with the Chua resistor
(1.1)𝒟tαx(t)=a𝒟tα-1(y(t)-x(t))-b7(4x(t)-x3(t)),dy(t)dt=x(t)-y(t).
Caputo [2] presented a fractional model system to describe the relationship between the electric field E and the current density D in the study of a electrochemical polarization medium,
(1.2)γ𝒟tνD(t)+αD(t)=σE(t)+δ𝒟tνE(t),
where γ, α, σ, δ are constants and ν is a real number. Anastasio [3] believed that signal processing of communication system of vestibule visual reflex effect is fractional, which can be described by the following model
(1.3)τ1r′(t)+r(t)=τ1τ2𝒟tα+1v(t)+τ1𝒟tαv(t),
where r(t) is the vestibular reflex nerve discharge rate, v(t) is the head of the rotational angular velocity, α is fraction.Motivated by systems (1.1)–(1.3) and their application background in electromagnetic and signal processing of wireless communication system, in this paper, we consider the existence, uniqueness, and asymptotic behavior of positive solutions for the higher nonlocal fractional differential equation
(1.4)-𝒟tαx(t)=f(t,x(t),x′(t),x′′(t),…,x(n-2)(t)),0<t<1,x(0)=x′(0)=⋯=x(n-2)(0)=0,x(n-2)(1)=∫01‍x(n-2)(s)dA(s),
where n-1<α≤n,  n∈ℕ and n≥2, 𝒟tα is the standard Riemann-Liouville derivative, ∫01‍x(s)dA(s) is linear functionals given by Riemann-Stieltjes integrals, A is a function of bounded variation, and dA can be a changing-sign measure, and f:(0,1)×(0,+∞)n-1→[0,+∞) is continuous, f may be singular at xi=0 and t=0,1.The nonlocal integral-boundary value problems represent a class of interesting and important problems arising in physical, biological, and chemical processes and have attracted the attention of Khan [4], Gallardo [5], Karakostas and Tsamatos [6], Ahmad et al. [7], Feng et al. [8], Corduneanu [9], and Agarwal and O’Regan [10].Whenα is an integer, a lot of work has been done dealing with nonlocal 3-point boundary value problems, see [11, 12]. In [12], Eloe and Ahmad studied the following nth-order nonlocal differential equation
(1.5)u(n)(t)+a(t)f(u)=0,0<t<1,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=ηu(ξ),
where 0<ξ<1, 0<ηξn-1<1. By applying the fixed point theorem in cones, the authors prove the existence of at least one positive solution when a and f is continuous and f is either sublinear or superlinear. Recently, Hao et al. [13] studied the existence of positive solutions for the BVP (1.5) with integer order n, and nonlinear term is replaced by a(t)f(t,x(t)), where a can be singular at t=0,1, f can be singular at x=0 and no singularity at t=0,1.Ifα is fractional, Yuan [14] studied the (n-1,1)-type conjugate boundary value problem
(1.6)𝒟tαu(t)+f(t,u(t))=0,0<t<1,n-1<α≤n,n≥3,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(1)=0,
where f is continuous and semipositone, 𝒟tα is the standard Riemann-Liouville derivative. By giving properties of Green’s function and using the Guo-Krasnosel’skii fixed-point theorem on cones, the existence of multiple positive solutions were obtained. More recently, Zhang [15] considered the following BVP whose nonlinear term and boundary condition contain derivatives of unknown functions
(1.7)𝒟tαx(t)+q(t)f(x,x′,…,x(n-2))=0,0<t<1,n-1<α≤n,x(0)=x′(0)=⋯=x(n-2)(0)=x(n-2)(1)=0,
where 𝒟tα is the standard Riemann-Liouville fractional derivative of order α, q may be singular at t=0 and f may be singular at x=0,  x'=0,…,x(n-2)=0, by using fixed-point theorem of the mixed monotone operator, the unique existence result of positive solution to problem (1.7) was established. Other some recent results, see [16–20]Motivated by the results mentioned above, in this paper, we study the existence, uniqueness, and asymptotic behavior of positive solutions for the BVP (1.4) where the nonlinear terms and boundary conditions all involve derivatives of unknown functions and with Riemann-Stieltjes integral boundary condition, moreover f may be singular at xi=0 and t=0,1. Our main tool relies on Schauder’s fixed-point theorem and upper and lower solution method.


## 2. Preliminaries and Lemmas

In this section, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literatures.Definition 2.1 (see [21, 22]).
The Riemann-Liouville fractional integral of orderα>0 of a function x:(0,+∞)→ℝ is given by the following:
(2.1)Iαx(t)=1Γ(α)∫0t(t-s)α-1x(s)ds
provided that the right-hand side is pointwise defined on (0,+∞).Definition 2.2 (see [21, 22]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x:(0,+∞)→ℝ is given by the following:
(2.2)𝒟tαx(t)=1Γ(n-α)(ddt)n∫0t‍(t-s)n-α-1x(s)ds,
where n=[α]+1, [α] denotes the integer part of number α, provided that the right-hand side is pointwise defined on (0,+∞).Proposition 2.3 (see [21, 22]).
(1) Ifx∈L1(0,1),ν>σ>0, then
(2.3)IνIσx(t)=Iν+σx(t),𝒟tσIνx(t)=Iν-σx(t),𝒟tσIσx(t)=x(t).
(2) If α>0,σ>0, then
(2.4)𝒟tαtσ-1=Γ(σ)Γ(σ-α)tσ-α-1.Proposition 2.4 (see [21, 22]).
Letα>0, and f(x) is integrable, then
(2.5)Iα𝒟tαf(x)=f(x)+c1xα-1+c2xα-2+⋯+cnxα-n,
where ci∈ℝ(i=1,2,…,n), n is the smallest integer greater than or equal to α.Our discussion is based on the assumption-1<α-n≤0 in this paper.Letx(t)=In-2y(t),y(t)∈C[0,1], by standard discuss, we easily reduce order the BVP (1.4) to the following equivalent BVP
(2.6)-𝒟tα-n+2y(t)=f(t,In-2y(t),In-3y(t),…,I1y(t),y(t)),y(0)=0,y(1)=∫01‍y(s)dA(s).In fact, letx(t)=In-2y(t),y(t)∈C[0,1], then
(2.7)𝒟tαx(t)=dndtnIn-αx(t)=dndtnIn-αIn-2y(t)=dndtnI2n-α-2y(t)=𝒟tα-n+2y(t),x′(t)=𝒟t1In-2y(t)=In-3y(t),x′′(t)=𝒟t2In-2y(t)=In-4y(t),⋮x(n-3)(t)=𝒟tn-3In-2y(t)=I1y(t),x(n-2)(t)=𝒟tn-2In-2y(t)=y(t).
By (2.7), we have x(n-2)(0)=y(0)=0 and
(2.8)y(1)=∫01‍y(s)dA(s).
Moreover -𝒟tα-n+2y(t)=f(t,In-2y(t),In-3y(t),…,I1y(t),y(t)). Thus, (1.4) is transformed into (2.6).On the other hand, ify∈C([0,1],[0,+∞)) is a solution for problem (2.6). Then, from Proposition 2.3 and (2.7), one has
(2.9)-𝒟tαx(t)=-𝒟tα-n+2y(t)=f(t,In-2y(t),In-3y(t),…,I1y(t),y(t))=f(t,x(t),x′(t),x′′(t),…,x(n-2)(t)),0<t<1.
Notice for any α>0,
(2.10)Iαy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
which implies that Iαy(0)=0. So x(0)=0, and from (2.7), for i=1,2,…,n-2, we have x(i)(0)=0, and
(2.11)x(n-2)(1)=∫01‍x(n-2)(s)dA(s).
Consequently, the BVP (2.6) is transformed into the BVP (1.4).Applying Propositions2.3 and 2.4, by standard discuss, we have the following Lemma.Lemma 2.5.
Giveny∈L1(0,1), then the problem
(2.12)𝒟tα-n+2y(t)+h(t)=0,0<t<1,y(0)=0,y(1)=0,
has the unique solution
(2.13)y(t)=∫01‍G(t,s)h(s)ds,
where G(t,s) is the Green function of the BVP (2.12) and is given by the following:
(2.14)G(t,s)={tα-n+1(1-s)α-n+1-(t-s)α-n+1Γ(α-n+2),0≤s≤t≤1,tα-n+1(1-s)α-n+1Γ(α-n+2),0≤t≤s≤1.By Proposition2.4, the unique solution of the problem
(2.15)𝒟tα-n+2y(t)=0,0<t<1,y(0)=0,y(1)=1,
is tα-n+1. Let
(2.16)𝒞=∫01‍tα-n+1dA(t),
and define
(2.17)𝒢A(s)=∫01‍G(t,s)dA(t).
Then the Green function for the nonlocal BVP (2.6) is (the detail see [23] or [16])
(2.18)K(t,s)=tα-n+11-𝒞𝒢A(s)+G(t,s).Throughout paper, we always assume the following holds.(
H
0
)
A is a function of bounded variation such that 𝒢A(s)≥0 for s∈[0,1] and 0≤𝒞<1, where 𝒞 is defined by (2.16).Lemma 2.6.
Suppose(H0) holds, then the Green function defined by (2.18) satisfies (1)
K(t,s)>0, for all t,s∈(0,1).(2)
(2.19)tα-n+11-𝒞𝒢A(s)≤K(t,s)≤ℋ(s)tα-n+1,where
(2.20)ℋ(s)=(1-s)α-n+1Γ(α-n+2)+𝒢A(s)1-𝒞.Proof.
(1) is obvious. For (2), by (2.14) and (2.18), we have
(2.21)H(t,s)=tα-n+11-𝒞𝒢A(s)+G(t,s)≤tα-n+1(1-s)α-n+1Γ(α-n+2)+tα-n+11-𝒞𝒢A(s)=ℋ(s)tα-n+1,H(t,s)=tα-n+11-𝒞𝒢A(s)+G(t,s)≥tα-n+11-𝒞𝒢A(s).Definition 2.7.
A continuous functionψ(t) is called a lower solution of the BVP (2.6), if it satisfies
(2.22)-𝒟tα-n+2ψ(t)(t)≤f(t,In-2ψ(t),In-3ψ(t),…,I1ψ(t),ψ(t)),ψ(0)≥0,ψ(1)≥∫01‍ψ(s)dA(s).Definition 2.8.
A continuous functionϕ(t) is called a upper solution of the BVP (2.6), if it satisfies
(2.23)-𝒟tα-n+2ϕ(t)(t)≥f(t,In-2ϕ(t),In-3ϕ(t),…,I1ϕ(t),ϕ(t)),ϕ(0)≤0,ϕ(1)≤∫01‍ϕ(s)dA(s).
It follows from Lemma2.5, we have the following maximum principle.Lemma 2.9 (maximum principle).
Ify∈C([0,1],ℝ) satisfies
(2.24)y(0)=0,y(1)=∫01‍y(s)dA(s),
and 𝒟tα-n+2y(t)≤0 for any t∈(0,1). Then
(2.25)y(t)≥0,t∈[0,1].


## 3. Main Results

LetE=C[0,1],  and
(3.1)P={y∈E:thereexistpositivenumbers0<ly<1,Ly>1suchthatlytα-n+1≤y(t)≤Lytα-n+1,t∈[0,1]}
Clearly, tα-n+1∈P, so P is nonempty. For any y∈P, define an operator T by the following:
(3.2)(Ty)(t)=∫01‍K(t,s)f(s,In-2y(s),In-3y(s),…,I1y(s),y(s))ds.Let(3.3)κi(t)=Γ(α-n+2)Γ(α-i)tα-i-1,i=0,1,2,…,n-2,
then
(3.4)κ0(t)=In-2sα-n+1,κ1(t)=In-3sα-n+1,…,κn-3(t)=I1sα-n+1,κn-2(t)=tα-n+1.The conditions imposed onf are the following.(H1)  f∈C((0,1)×(0,∞)n-1,[0,+∞)), and f(t,x0,x1,x2,…,xn-2) is decreasing in xi>0 for i=0,1,2,…,n-2;(H2)for any λi>0, f(t,λ0tn-2,λ1tn-3,…,λn-3t,λn-2)≢0,  t∈(0,1), and
(3.5)0<∫01‍ℋ(s)f(s,λ0κ0(s),λ1κ1(s),λ2κ2(s),…,λn-2κn-2(s))ds<+∞.Lemma 3.1.
Suppose(H0)–(H2) hold, then T is well defined and T(P)⊂P.Proof.
For anyy∈P, by the definition of P, there exists two positive numbers 0<ly<1,Ly>1 such that lytα-n+1≤y(t)≤Lytα-n+1 for any t∈[0,1]. It follows from (2.19) and (H1)-(H2) that
(3.6)(Ty)(t)=∫01‍K(t,s)f(s,In-2y(s),In-3y(s),…,I1y(s),y(s))ds≤∫01‍ℋ(s)f(s,In-2lysα-n+1,…,I1lysα-n+1,lysα-n+1)ds=∫01‍ℋ(s)f(s,lyκ0(s),lyκ1(s),…,lyκn-2(s))ds<+∞.
Now takeC=maxt∈[0,1]y(t), then for any s∈(0,1),
(3.7)𝒢A(s)f(s,C(n-2)!sn-2,C(n-3)!sn-3,…,C1!s,C)≢0,
by (H2). Thus by the continuity of f(t,x0,x1,…,xn-2), we have
(3.8)∫01‍𝒢A(s)f(s,C(n-2)!sn-2,C(n-3)!sn-3,…,C1!s,C)ds>0.
This yields
(3.9)∫01‍𝒢A(s)f(s,In-2C,In-3C,…,I1C,C)ds=∫01‍𝒢A(s)f(s,C(n-2)!sn-2,C(n-3)!sn-3,…,C1!s,C)ds>0.
By (2.19) and (3.6)–(3.9), we have
(3.10)(Ty)(t)=∫01‍K(t,s)f(s,In-2y(s),In-3y(s),…,I1y(s),y(s))ds≥tα-n+11-𝒞∫01‍𝒢A(s)f(s,In-2C,In-3C,…,I1C,C)ds≥ly′tα-n+1,
where
(3.11)ly′=min{1,11-𝒞∫01‍𝒢A(s)f(s,In-2C,In-3C,…,I1C,C)ds}.
On the other hand, it follows from (2.19) that
(3.12)(Ty)(t)=∫01‍K(t,s)f(s,In-2y(s),In-3y(s),…,I1y(s),y(s))ds≤tα-n+1∫01‍ℋ(s)f(s,lyκ0(s),lyκ1(s),…,lyκn-2(s))ds,≤Ly′tα-n+1,
where
(3.13)Ly′=max{1,∫01‍ℋ(s)f(s,lyκ0(s),lyκ1(s),…,lyκn-2(s))ds}.
It follows from (3.6)–(3.12) that T is well defined and T(P)⊂P.Take(3.14)L=max{1,∫01‍ℋ(s)f(s,κ0(s),κ1(s),…,κn-2(s))ds},l=min{1,11-𝒞∫01‍𝒢A(s)f(s,κ0(s),κ1(s),…,κn-2(s))ds}.
By (H2) and Weierstrass distinguishing method, we know
(3.15)a(t)=∫01‍K(t,s)f(s,κ0(s),κ1(s),…,κn-2(s))ds
is a continuous function on [0,1], that is, a(t)∈E.Theorem 3.2 (existence).
Suppose(H0)–(H2) hold, and if
(3.16)∫01‍𝒢A(s)f(s,Llκ0(s),Llκ1(s),…,Llκn-2(s),Llκn-1(s))ds≥1-𝒞.
Then the BVP (1.4) has at least a positive solution w(t).Proof.
By (2.18) and (3.2), we have
(3.17)-𝒟tα-n+2(Ty)(t)=f(t,In-2y(t),In-3y(t),…,I1y(t),y(t)),(Ty)(0)=0,(Ty)(1)=∫01‍(Ty)(s)dA(s).
It follows from (2.19) that
(3.18)ltα-n+1≤a(t)≤Ltα-n+1,
which implies that a(t)∈P.  Let
(3.19)b(t)=a(t)l,
then we also have
(3.20)tα-n+1≤b(t)≤Lltα-n+1.
Thus according to the fact that the operatorT is nonincreasing relative to y, Lemma 3.1 and (3.20), we have
(3.21)Tb(t)≤Ttα-n+1=a(t)≤a(t)l=b(t).
and Tb(t)∈P. By (2.19), (3.12), and (3.20), one has
(3.22)Tb(t)≥11-𝒞tα-n+1∫01‍𝒢A(s)f(s,In-2b(s),In-3b(s),…,I1b(s),b(s))ds≥11-𝒞tα-n+1∫01‍𝒢A(s)f(s,Llκ0(s),Llκ1(s),…,Llκn-2(s))ds≥tα-n+1.
Thus, by (3.20), (3.21) and the operator T is nonincreasing relative to y, we have
(3.23)𝒟tα-n+2b(t)+f(t,In-2b(t),In-3b(t),…,I1b(t),b(t))=-1lf(s,In-2tα-n+1,…,I1tα-n+1,tα-n+1)+f(t,In-2b(t),…,I1b(t),b(t))≤-1lf(s,In-2tα-n+1,…,I1tα-n+1,tα-n+1)+f(s,In-2tα-n+1,…,I1tα-n+1,tα-n+1)≤0,𝒟tα-n+2Tb(t)+f(t,In-2Tb(t),In-3Tb(t),…,I1Tb(t),Tb(t))=-f(t,In-2b(t),In-3b(t),…,I1b(t),b(t))+f(t,In-2Tb(t),In-3Tb(t),…,I1Tb(t),Tb(t))≥0,t∈[0,1].
However (3.17) implies that b(t), Tb(t) satisfy boundary conditions of BVP (2.6). Let
(3.24)ψ(t)=Tb(t),ϕ(t)=1lTtα-n+1=1la(t)=b(t),
then (3.21), (3.23) imply that ψ(t),ϕ(t) are lower and upper solution of BVP (2.6), respectively, and ψ(t),ϕ(t)∈P.
Define the functionF and the operator A in E by the following:
(3.25)F(t,y)={f(t,In-2ψ(t),In-3ψ(t),…,I1ψ(t),ψ(t)),y<ψ(t),f(t,In-2y(t),In-3y(t),…,I1y(t),y(t)),ψ(t)≤y≤ϕ(t),f(t,In-2ϕ(t),In-3ϕ(t),…,I1ϕ(t),ϕ(t)),y>ϕ(t),(3.26)(By)(t)=∫01‍K(t,s)F(s,y(s))ds,∀y∈E.
Clearly, F:[0,1]×[0,+∞)→[0,+∞) is continuous by (3.25). Consider the following boundary value problem
(3.27)-𝒟tα-n+2y(t)=F(t,y),0<t<1,y(0)=0,y(1)=∫01‍y(s)dA(s).
Obviously, a fixed point of the operator B is a solution of the BVP (3.27).
For ally∈E, it follows from Lemma 2.6 and (3.22) that
(3.28)(By)(t)≤∫01‍ℋ(s)F(s,y(s))ds≤∫01‍ℋ(s)f(s,In-2ψ(s),In-3ψ(s),…,I1ψ(s),ψ(s))ds≤∫01‍ℋ(s)f(s,In-2sα-n+1,…,I1sα-n+1,sα-n+1)ds=∫01‍ℋ(s)f(s,κ0(s),κ1(s),…,κn-2(s))ds<+∞.
So B is bounded. It is easy to see B:E→E is continuous from the continuity of F and K.
LetΩ⊂E be bounded, that is, there exists a positive constant N>0 such that ||y||≤N, for all y∈Ω. Let L=max0≤t≤1,0≤y≤N|F(t,y)|+1, and since K(t,s) is uniformly continuous on [0,1]×[0,1], then for any ε>0 and s∈[0,1], there exists δ>0 such that
(3.29)|K(t1,s)-K(t2,s)|<εL,
for |t1-t2|<δ. Then
(3.30)|By(t1)-By(t2)|≤∫01‍|F(s,y(s))K(t1,s)-K(t2,s)||F(s,y(s))|ds<ε.
This implies that B(Ω) is equicontinuous.
By the means of the Arzela-Ascoli theorem, we haveB:E→E is completely continuous. Thus, by using Schauder’s fixed-point theorem, B has at least a fixed-point w such that w=Bw.
Now we prove(3.31)ψ(t)≤w(t)≤ϕ(t),t∈[0,1].
Letz(t)=ϕ(t)-w(t),  t∈[0,1]. By ϕ(t) is upper solution of BVP (2.6) and w is fixed point of B, we know
(3.32)w(0)=0,w(1)=∫01‍w(s)dA(s).
From the definition of F and (3.22), we obtain
(3.33)f(t,In-2ϕ(t),In-3ϕ(t),…,I1ϕ(t),ϕ(t))≤F(t,y(t))≤f(t,In-2ψ(t),In-3ψ(t),…,I1ψ(t),ψ(t))≤f(t,In-2tα-n+1,…,I1tα-n+1,tα-n+1),t∈[0,1].
Thus (3.16) and (3.33) imply
(3.34)𝒟tα-n+2z(t)=𝒟tα-n+2ϕ(t)-𝒟tα-n+2w(t)=-1lf(t,In-2tα-n+1,…,I1tα-n+1,tα-n+1)+F(t,w(t))≤-1lf(t,In-2tα-n+1,…,I1tα-n+1,tα-n+1)+f(t,In-2tα-n+1,…,I1tα-n+1,tα-n+1)≤0,∀t∈[0,1].
By (3.32),  (3.34), and Lemma 2.9, we know z(t)≥0 which implies w(t)≤ϕ(t) on [0,1]. By the same way, it is easy to prove w(t)≥ψ(t) on [0,1]. So we obtain
(3.35)ψ(t)≤w(t)≤ϕ(t),t∈[0,1].
Consequently, F(t,w(t))=f(t,In-2w(t),In-3w(t),…,I1w(t),w(t)),  t∈[0,1]. Then w(t) is a positive solution of the BVP (2.6), which implies that x(t)=In-2w(t) is a positive solution of the BVP (1.4).Theorem 3.3 (asymptotic behavior).
Suppose the conditions of Theorem3.2 are satisfied. Then there exist two constants ℬ1,ℬ2 such that the positive solution x(t) of the BVP (1.4) satisfies
(3.36)ℬ1tα-1≤x(t)≤ℬ2tα-1,limt→0+x(t)tα-1=0,
where
(3.37)ℬ1=Γ(α-n+2)Γ(α),ℬ2=Lℬ1.Proof.
By (3.22) and (3.35), we know
(3.38)w(t)≥ψ(t)≥tα-n+1.
On the other hand, it follows from (3.38) and (2.19) that
(3.39)w(t)=∫01‍K(t,s)f(s,In-2w(s),In-3w(s),…,I1w(s),w(s))ds≤tα-n+1∫01‍ℋ(s)f(s,In-2sα-n+1,…,I1sα-n+1,sα-n+1)ds=tα-n+1∫01‍ℋ(s)f(s,κ0(s),κ1(s),…,κn-2(s),κn-1(s))ds≤Ltα-n+1.
Then
(3.40)tα-n+1≤w(t)≤Ltα-n+1.
Since(3.41)In-2tα-n+1=1Γ(n-2)∫0t‍(t-s)n-3sα-n+1ds=Γ(α-n+2)Γ(α)tα-1,
so it follows from (3.40) that
(3.42)Γ(α-n+2)Γ(α)tα-1≤In-2w(t)=x(t)≤LΓ(α-n+2)Γ(α)tα-1,
That is,
(3.43)ℬ1tα-1≤x(t)≤ℬ2tα-1.
In the end, fromn-1<α≤n and L’Hospital Rule, we have
(3.44)Limt→0+x(t)tα-1=limt→0+x(n-2)(t)(α-1)(α-2)⋯(α-n+2)tα-n+1=limt→0+x(n-1)(t)(α-1)(α-2)⋯(α-n+1)tα-n=0.Theorem 3.4 (uniqueness).
Suppose the conditions of Theorem3.2 are satisfied and α=n. Then the positive solution x(t) of the BVP (1.4) is unique.Proof.
We only need the BVP (2.6) has unique solution. Suppose that w1,w2 are C[0,1] positive solutions to the BVP (2.6). We may assume, without loss of generality, that there exists t*∈(0,1) such that w2(t*)-w1(t*)=max{w2(t)-w1(t)}>0. Let
(3.45)a=inf{t1∣0≤t1<t*,w2(t)≥w1(t),t∈(t1,t*]};b=inf{t2∣t*<t2≤1,w2(t)≥w1(t),t∈(t*,t2]};z(t)=w2(t)-w1(t),t∈[0,1].
Evidently,
(3.46)t*∈(a,b),w2(t)≥w1(t),f(t,In-2w2(t),In-3w2(t),…,I1w2(t),w2(t))≤f(t,In-2w1(t),In-3w1(t),…,I1w1(t),w2(t)),t∈[a,b].
Thus
(3.47)𝒟tα-n+2z(t)=z′′(t)≥0,t∈[a,b].
By the boundary conditions of the BVP (2.6), it is easy to check that there exist the following two possible cases:
(3.48)(1)z(a)=0,z(b)=0;(2)z(a)=0,z(b)>0;Case 1.
Fromz′′(t)≥0 and z(a)=0,z(b)=0, we get that z(t)≤0,t∈[a,b], which implies a contradiction with w2(t*)>w1(t*).Case 2.
In this case we haveb=1 and z′(t*)=0. Since z′(t) is increasing on [a,b], we also have z′(t) is increasing on t∈[a,t*], that is, z′(t)≤z′(t*)=0 for any t∈[a,t*]. Thus z(t) is nonincreasing on t∈[a,t*], which implies that z(t*)≤z(a)=0, this is a contradiction with w2(t*)>w1(t*).
Therefore the BVP (2.6) has unique solution, and the BVP (1.4) also has unique solution.Remark 3.5.
We only get the uniqueness of positive solution for the BVP (1.4) as α is integer, but it remains unknown as to whether an analogous unique result holds for the case n-1<α<n. We believe the result should also hold and try to study this problem, but it is a pity that we failed to find the effective method to solve this problem.Remark 3.6.
The BVP (1.4) allows the nonlinearity f(t,x0,x2,…,xn-2) has singularity at t=0,1 and xi=0,  i=0,1,…,n-2. If f(t,x0,x2,…,xn-2) is continuous at xi=0,  i=0,1,…,n-2, that is
f
∈
C
(
(
0,1
)
×
[
0
,
∞
)
n
-
1
,
[
0
,
+
∞
)
), and f(t,x0,x1,x2,…,xn-2) is decreasing in xi>0 for i=0,1,2,…,n-2;(3.49)0<∫01‍ℋ(s)f(s,0,0,…,0)ds<+∞.
then the BVP (1.4) has at least a positive solution x(t), and there exists a constant ℬ3>0 such that
(3.50)0≤x(t)≤ℬ3tα-1.
If α=n, the positive solution of the BVP (1.4) also is unique.Proof.
In fact, in Theorem3.2 the set P is replaced by the following:
(3.51)P1={x∈E:x(t)≥0,t∈[0,1]}
and (3.21) can be replaced by the following:
(3.52)0≤ϕ(t)=T(0),0≤ψ(t)=(Tϕ)(t)≤T(0)=ϕ(t).
Clearly ϕ(t),ψ(t)∈P1, and
(3.53)𝒟tα-n+2ϕ(t)+f(t,In-2ϕ(t),In-3ϕ(t),…,I1ϕ(t),ϕ(t))=-f(t,0,0,…,0)+f(t,In-2T(0),In-3T(0),…,I1T(0),T(0))≤0,𝒟tα-n+2ψ(t)+f(t,In-2ψ(t),In-3ψ(t),…,I1ψ(t),ψ(t))=-f(t,In-2ϕ(t),In-3ϕ(t),…,I1ϕ(t),ϕ(t))+f(t,In-2ψ(t),In-3ψ(t),…,I1ψ(t),ψ(t))≥0,t∈[0,1].
Thus the rest of proof is similar to those of Theorem 3.2.
On the other hand, by Lemma2.6 and (G2), the positive solution of the BVP (2.6) satisfies
(3.54)0≤w(t)≤ϕ(t)=T0=∫01‍K(t,s)f(s,0,0,…,0)ds≤∫01‍ℋ(s)f(s,0,0,…,0)dstα-n+1,
and then
(3.55)0≤In-2w(t)=x(t)≤ℬ3tα-1,
where
(3.56)ℬ3=Γ(α-n+2)Γ(α)∫01‍ℋ(s)f(s,0,0,…,0)ds.
Moreover if α=n, the positive solution of the BVP (1.4) also is unique.Example 3.7.
Consider the existence of positive solutions for the nonlinear fractional differential equation(3.57)-𝒟t7/2x(t)=10t-1/4[(x′)-1/3+x-1/8],0<t<1,x(0)=x′(0)=0,x′(1)=∫01‍x′(s)dA(s),
where
(3.58)A(t)={0,t∈[0,12),32,t∈[12,34),1,t∈[34,1].
Then the BVP (3.58) is equivalent to the following 4-point BVP with coefficients of both signs
(3.59)-𝒟t7/2x(t)=10t-1/4[(x′)-1/3+x-1/8],0<t<1,x(0)=x′(0)=0,x′(1)=32x′(12)-12x′(34),
Conclusion: The BVP (3.59) has at least a positive solution x(t) such that
(3.60)25t5/2≤x(t)≤14.886t5/2.Proof.
Clearly,(3.61)0≤𝒞=∫01‍t5/2dA(t)=1-[∫1/23/4‍32dt5/2+∫3/41‍dt5/2]≈0.0217<1,G(t,s)={G1(t,s)=t3/2(1-s)3/2Γ(5/2),0≤t≤s≤1,G2(t,s)=t3/2(1-s)3/2-(t-s)3/2Γ(5/2),0≤s≤t≤1.
Thus,
(3.62)𝒢A(s)={32G2(12,s)-12G2(34,s),0≤s<12,32G1(12,s)-12G2(34,s),12≤s<34,32G1(12,s)-12G1(34,s),34≤s≤1,={((3/2)2-(3/4)3)(1-s)3/2+2((3/4)-s)3/2-6((1/2)-s)3/23π,0≤s<12,((3/2)2-(3/4)3)(1-s)3/2+2((3/4)-s)3/23π,12≤s<34,((3/2)2-(3/4)3)(1-s)3/23π,34≤s≤1,ℋ(s)=(1-s)3/2Γ(5/2)+𝒢A(s)0.9783.
Clearly, (H0) and (H1) hold.
On the other hand, for anyλi>0, i=0,1, f(t,λ0t,λ1)=λ0-1/8t-3/8+λ1-1/3t-1/4≢0, t∈(0,1), and
(3.63)κ0(s)=25s5/2,κ1(s)=s3/2,0<∫01‍ℋ(s)f(s,λ0κ0(s),λ1κ1(s))ds=10∫01‍[(1-s)3/2Γ(5/2)+𝒢A(s)0.9783][(25λ0)-1/8s-9/16+λ1-1/3s-3/4]ds<+∞.
Thus (H2) is satisfied.
Now we computeL,l. Since
(3.64)∫01‍𝒢A(s)f(s,κ0(s),κ1(s))ds=10∫01‍𝒢A(s)[(25)-1/8s-9/16+s-3/4]ds=10∫01/2‍[((3/2)2-(3/4)3)(1-s)3/2+2((3/4)-s)3/2-6((1/2)-s)3/23π]×[(25)-1/8s-9/16+s-3/4]ds+10∫1/23/4‍((3/2)2-(3/4)3)(1-s)3/2+2((3/4)-s)3/23π[(25)-1/8s-9/16+s-3/4]+10∫3/41‍((3/2)2-(3/4)3)(1-s)3/23π[(25)-1/8s-9/16+s-3/4]ds=2.600,∫01‍ℋ(s)f(s,κ0(s),κ1(s))ds=10∫01‍ℋ(s)[(25)-1/8s-9/16+s-3/4]ds=10∫01‍[(1-s)3/2Γ(5/2)+𝒢A(s)0.9783][(25)-1/8s-9/16+s-3/4]ds=37.215.
Thus
(3.65)L=max{1,∫01‍ℋ(s)f(s,κ0(s),κ1(s))ds}=max{1,37.215}=37.215,l=min{1,11-𝒞∫01‍𝒢A(s)f(s,κ0(s),κ1(s))ds}=min{1,2.658}=2.658.
So(3.66)∫01‍𝒢A(s)f(s,Llκ0(s),Llκ1(s))ds=10∫01‍𝒢A(s)[(14×25)-1/8s-9/16+14-1/3s-3/4]ds=4.540≥1-𝒞=0.9783,
which implies that (3.16) holds. Then the BVP (1.4) has at least a positive solution x(t), and there exist two constants
(3.67)ℬ1=Γ(α-n+2)Γ(α)=25,ℬ2=Lℬ1=14.886,
such that
(3.68)25t5/2≤x(t)≤14.886t5/2.



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*Source: 294694-2012-10-02.xml*