# Positive Solutions of a Fractional Boundary Value Problem with Changing Sign Nonlinearity

Authors: Yongqing Wang; Lishan Liu; Yonghong Wu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We discuss the existence of positive solutions of a boundary value problem of nonlinear fractional differential equation with changing sign nonlinearity. We first derive some properties of the associated Green function and then obtain some results on the existence of positive solutions by means of the Krasnoselskii's fixed point theorem in a cone.


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## Body

## 1.  Introduction

Recently, much attention has been paid to the existence of solutions for fractional differential equations due to its wide range of applications in engineering, economics, and many other fields, and for more details see, for instance, [1–17] and the references therein. In most of the works in literature, the nonlinearity needs to be nonnegative to get positive solutions [10–17]. In particular, by using the Krasnosel’skii fixed-point theorem and the Leray-Schauder nonlinear alternative, Bai and Qiu [14] consider the positive solution for the following boundary value problem:(P)cD0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
where 2<α≤3 is a real number, cD0+α is the Caputo fractional derivative, f:(0,1]×[0,∞)→[0,∞) is continuous and singular at t=0.To the best of our knowledge, there are only very few papers dealing with the existence of positive solutions of semipositone fractional boundary value problems due to the difficulties in finding and analyzing the corresponding Green function. The purpose of this paper is to establish the existence of positive solutions to the following nonlinear fractional differential equation boundary value problem:(1.1)cD0+αu(t)+λf(t,u(t))=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
where 2<α≤3 is a real number, cD0+α is the Caputo fractional derivative, λ is a positive parameter, and  f may change sign and may be singular at t=0,1. In this paper, by a positive solution to (1.1), we mean a function u∈C[0,1], which is positive on (0,1] and satisfies (1.1).The rest of the paper is organized as follows. In Section2, we present some preliminaries and lemmas that will be used to prove our main results. We also develop some properties of the associated Green function. In Section 3, we discuss the existence of positive solutions of the semipositone BVP (1.1). In Section 4, we give two examples to illustrate the application of our main results.


## 2.  Basic Definitions and Preliminaries

In this section, we present some preliminaries and lemmas that are useful to the proof of our main results. For the convenience of the reader, we also present here some necessary definitions from fractional calculus theory. These definitions can be found in the recent literature.Definition 2.1.
The Riemann-Liouville fractional integral of orderα>0 of a function   u:(0,+∞)→R is given by
(2.1)I0+αu(t)=1Γ(α)∫0t(t-s)α-1u(s)ds
provided that the right-hand side is pointwise defined on (0,+∞).Definition 2.2.
The Caputo’s fractional derivative of orderα>0   of a function u:(0,+∞)→R is given by
(2.2)cD0+αu(t)=1Γ(n-α)∫0t(t-s)n-α-1u(n)(s)ds,
where n-1<α≤n, provided that the right-hand side is pointwise defined on (0,+∞).Lemma 2.3 (see [14]).
Giveny(t)∈C(0,1)∩L(0,1),  the unique solution of the problem
(2.3)cD0+αu(t)+y(t)=0,0<t<1,u(0)=u′(1)=u′′(0)=0
is
(2.4)u(t)=∫01G(t,s)y(s)ds,
where
(2.5)G(t,s)=1Γ(α){(α-1)t(1-s)α-2,0≤t≤s≤1,(α-1)t(1-s)α-2-(t-s)α-1,0≤s≤t≤1.Lemma 2.4.
The functionG(t,s) has the following properties:(1)
G(t,s)≤(1/Γ(α-1))t(1-s)α-2, fort,s∈[0,1],(2)
G(t,s)≤(1/Γ(α-1))(α-2+s)(1-s)α-2, for  t,s∈[0,1],(3)
G(t,s)≥(1/Γ(α))(α-2+s)t(1-s)α-2, for t,s∈[0,1].Proof.
It is obvious that (1) holds. In the following, we will prove (2) and (3).
(i) When0≤s≤t≤1, as 2<α≤3, we have
(2.6)∂G(t,s)∂t=(1-s)α-2-(t-s)α-2Γ(α-1)≥0,
therefore
(2.7)G(t,s)≤G(1,s)=(α-2+s)Γ(α)(1-s)α-2≤1Γ(α-1)(α-2+s)(1-s)α-2.
On the other hand, since0<α-2≤1, we have
(2.8)G(t,s)=(α-1)t(1-s)α-2-(t-s)(t-s)α-2Γ(α)≥(α-1)t(1-s)α-2-(t-s)(1-s)α-2Γ(α)=(α-2)t+sΓ(α)(1-s)α-2≥(α-2)t+stΓ(α)(1-s)α-2=1Γ(α)(α-2+s)t(1-s)α-2.
(ii) When0≤t≤s≤1, we have
(2.9)G(t,s)=(α-1)t(1-s)α-2Γ(α)≤(α-1)s(1-s)α-2Γ(α)≤1Γ(α-1)(α-2+s)(1-s)α-2.
On the other hand, asα-1≥α-2+s for 0≤s≤1, we have
(2.10)G(t,s)=(α-1)t(1-s)α-2Γ(α)≥1Γ(α)(α-2+s)t(1-s)α-2.
The proof is completed.Remark 2.5.
By Lemma2.4, there exists K>0 such that the positive solution u in [14] satisfies
(2.11)u(t)≥tα-1∥u∥,u(t)≤Kt,
where ∥u∥=max0≤t≤1|u(t)|.Proof.
In [14], the positive solution of (P) is equivalent to the fixed point of A in Q, where Q={u(t)∈C[0,1]:u(t)≥0} and
(2.12)Au(t)=∫01G(t,s)f(s,u(s))ds.
For anyu∈Q, by (1) of Lemma 2.4, we have
(2.13)Au(t)≤tΓ(α-1)∫01(1-s)α-2f(s,u(s))ds.
On the other hand, by (2),(3) of Lemma 2.4, we get
(2.14)Au(t)≥tΓ(α)∫01(α-2+s)(1-s)α-2f(s,u(s))ds,Au(t)≤1Γ(α-1)∫01(α-2+s)(1-s)α-2f(s,u(s))ds,
which implies Au(t)≥(t/(α-1))∥Au(t)∥.
Ifu is a positive solution of (P), then u is a fixed point of A in Q, therefore
(2.15)u(t)≥tα-1‖u‖,u(t)≤Kt,
where K=(1/(Γ(α-1)))∫01(1-s)α-2f(s,u(s))ds. The proof is completed.For the convenience of presentation, we list here the hypotheses to be used later.(H1)f∈C((0,1)×[0,+∞),(-∞,+∞)) and satisfies
(2.16)-r(t)≤f(t,x)≤z(t)g(x),
where r,z∈C((0,1),[0,+∞)), g∈C([0,+∞),[0,+∞)).(H2)0<∫01r(s)ds<+∞, 0<∫01(α-2+s)(1-s)α-2(z(s)+r(s))ds<+∞.(H3)  There exists [a,b]⊂(0,1) such that
(2.17)liminf⁡u→+∞min⁡t∈[a,b]f(t,u)u=+∞.(H4)  There exists [c,d]⊂(0,1) such that
(2.18)liminf⁡u→+∞min⁡t∈[c,d]f(t,u)>2(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds,lim⁡u→+∞g(u)u=0.Lemma 2.6.
Assume that(H1) and (H2) hold, then the boundary value problem
(2.19)cD0+αu(t)+r(t)=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
has a unique solution ω(t)=∫01G(t,s)r(s)ds with
(2.20)ω(t)≤t∫011Γ(α-1)(1-s)α-2r(s)ds,t∈[0,1].Proof.
By Lemma2.3, we have that ω(t)=∫01G(t,s)r(s)ds is the unique solution of (2.19). By (1) of Lemma 2.4, we have
(2.21)ω(t)=∫01G(t,s)r(s)ds≤t∫011Γ(α-1)(1-s)α-2r(s)ds.
The proof is completed.LetE=C[0,1] be endowed with the maximum norm ∥u∥=max0≤t≤1|u(t)|. Define a cone P by(2.22)P={u(t)∈E:u(t)≥tα-1‖u‖}.
Set Br={u(t)∈E:∥u∥<r}, Pr=P∩Br, ∂Pr=P∩∂Br.Next we consider the following boundary value problem:(2.23)cD0+αu(t)+λ[f(t,[u(t)-λω(t)]+)+r(t)]=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
where λ>0, ω(t) is defined in Lemma 2.6, [u(t)-λω(t)]+=max{u(t)-λω(t),0}.Let(2.24)Tu(t)=λ∫01G(t,s)[f(s,[u(s)-λω(s)]+)+r(s)]ds.
It is easy to check that u is a solution of (2.23) if and only if u is a fixed point of  T.Lemma 2.7.
T:P→P is a completely continuous operator.Proof.
For anyu∈P, Lemma 2.4 implies that
(2.25)Tu(t)≥λtΓ(α)∫01(α-2+s)(1-s)α-2[f(s,[u(s)-λω(s)]+)+r(s)]ds.
On the other hand
(2.26)Tu(t)≤λΓ(α-1)∫01(α-2+s)(1-s)α-2[f(s,[u(s)-λω(s)]+)+r(s)]ds.
Then Tu(t)≥(t/(α-1))∥Tu(t)∥, which implies T:P→P.
According to the Ascoli-Arzela theorem, we can easily get thatT:P→P is a completely continuous operator. The proof is completed.Lemma 2.8 (see [18]).
LetE be a real Banach space, and let   P⊂E  be a cone. Assume that  Ω1 and Ω2 are two bounded open subsets of E  with  θ∈Ω1,Ω¯1⊂Ω2,   and T:P∩(Ω¯2∖Ω1)→P is a  completely continuous operator such that either(1)
∥Tu∥≤∥u∥,u∈P∩∂Ω1 and ∥Tu∥≥∥u∥,u∈P∩∂Ω2, or(2)
∥Tu∥≥∥u∥,u∈P∩∂Ω1 and ∥Tu∥≤∥u∥,u∈P∩∂Ω2.Then T has a fixed point in P∩(Ω¯2∖Ω1).


## 3.  Existence of Positive Solutions

Theorem 3.1.
Suppose that(H1)–(H3) hold. Then there exists λ*>0 such that the boundary value problem (1.1) has at least one positive solution for any λ∈(0,λ*).Proof.
Chooser1>((α-1)/Γ(α-1))∫01(1-s)α-2r(s)ds. Let
(3.1)λ*=min⁡{1,r1Γ(α-1)[g*(r1)+1]∫01(α-2+s)(1-s)α-2(z(s)+r(s))ds},
where
(3.2)g*(r)=max⁡x∈[0,r]g(x).
In the rest of the proof, we suppose λ∈(0,λ*).
For anyu∈∂Pr1, noting that
(3.3)u(t)≥tα-1r1,t∈[0,1]
and using (2.20), we have
(3.4)0≤t[r1α-1-∫011Γ(α-1)(1-s)α-2r(s)ds]≤u(t)-λω(t)≤r1.
Therefore,
(3.5)Tu(t)=λ∫01G(t,s)[f(s,[u(s)-λω(s)]+)+r(s)]ds≤λΓ(α-1)∫01(α-2+s)(1-s)α-2[z(s)g([u(s)-λω(s)]+)+r(s)]ds≤λΓ(α-1)[g*(r1)+1]∫01(α-2+s)(1-s)α-2[z(s)+r(s)]ds<λ*Γ(α-1)[g*(r1)+1]∫01(α-2+s)(1-s)α-2[z(s)+r(s)]ds≤r1.
Thus,
(3.6)‖Tu‖≤‖u‖,∀u∈∂Pr1.
Now choose a real number(3.7)L>4Γ(α)λa∫ab(α-2+s)(1-s)α-2ds.
By (H3), there exists a constant N>0 such that for any t∈[a,b], x≥N, we have
(3.8)f(t,x)>Lx.
Select
(3.9)r2>max⁡⁡{r1,2(α-1)Γ(α-1)∫01(1-s)α-2r(s)ds,4Na}.
Then for any u∈∂Pr2, we have u(t)-λω(t)≥0,t∈[0,1]. Moreover, by the selection of r2 we have
(3.10)1Γ(α-1)∫01(1-s)α-2r(s)ds<r22(α-1).
Thus for any t∈[a,b], as 1<α-1≤2, we get
(3.11)u(t)-λω(t)≥t[r2α-1-1Γ(α-1)∫01(1-s)α-2r(s)ds]>ar22(α-1)≥ar24.
Noting that r2>4N/a, we have
(3.12)u(t)-λω(t)>ar24>N,t∈[a,b].
Hence we get
(3.13)Tu(1)=λΓ(α)∫01(α-2+s)(1-s)α-2[f(s,[u(s)-λω(s)]+)+r(s)]ds≥λΓ(α)∫ab(α-2+s)(1-s)α-2f(s,[u(s)-λω(s)]+)ds≥λLΓ(α)∫ab(α-2+s)(1-s)α-2[u(s)-λω(s)]ds≥λLar24Γ(α)∫ab(α-2+s)(1-s)α-2ds>r2.
Thus,
(3.14)‖Tu‖≥‖u‖,∀u∈∂Pr2.
By Lemma 2.8, T has a fixed point u such that r1≤∥u∥≤r2. Since ∥u∥≥r1, by (3.4) we have u(t)-λω(t)>0,t∈(0,1]. Let u̅(t)=u(t)-λω(t). As ω(t) is the solution of (2.19) and u(t) is the solution of (2.23), u̅(t) is a positive solution of the singular semipositone boundary value problem (1.1). The proof is completed.Theorem 3.2.
Suppose that(H1),  (H2), and (H4) hold. Then there exists λ*>0 such that the boundary value problem (1.1) has at least one positive solution for any λ∈(λ*,+∞).Proof.
By the first limit of(H4), we have that there exists N>0 such that, for any t∈[c,d] and u≥N, we have
(3.15)f(t,u)≥2(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds.
Select
(3.16)λ*=NΓ(α-1)c∫01(1-s)α-2r(s)ds.
In the rest of the proof, we suppose λ>λ*.
Let(3.17)R1=2λ(α-1)∫01(1-s)α-2r(s)dsΓ(α-1).
Then, for any u∈∂PR1, we have
(3.18)u(t)-λω(t)≥t[R1α-1-λ∫011Γ(α-1)(1-s)α-2r(s)ds]=λtΓ(α-1)∫01(1-s)α-2r(s)ds≥λ*tΓ(α-1)∫01(1-s)α-2r(s)ds=Ntc,
and therefore u(t)-λω(t)≥N on t∈[c,d], u∈∂PR1. Then,
(3.19)Tu(t)=λ∫01G(t,s)[f(s,[u(s)-λω(s)]+)+r(s)]ds≥λ∫cdG(t,s)f(s,[u(s)-λω(s)]+)ds≥2λ(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds∫cdG(t,s)ds≥2tλ(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds∫cd(α-2+s)Γ(α)(1-s)α-2ds=2λt(α-1)Γ(α-1)∫01(1-s)α-2r(s)ds=tR1,
which implies
(3.20)‖Tu‖≥‖u‖,∀u∈∂PR1.
On the other hand, asg(t) is continuous on [0,+∞), from the second limit of (H4), we have
(3.21)lim⁡u→+∞g*(u)u=0,
where g*(u) is defined by (3.2). In fact, by limu→+∞g(u)/u=0, for any ϵ>0, there exists N1>0 such that for any u>N1 we have 0≤g(u)<ϵu. Let N=max{N1,g*(N1)/ϵ}, for any u>N we have 0≤g*(u)<ϵu+g*(N1)<2ϵu. Therefore, limu→+∞g*(u)/u=0. For
(3.22)ϵ=Γ(α-1)[2λ∫01(α-2+s)(1-s)α-2z(s)ds]-1,
there exists X0>0 such that when x≥X0, for any 0≤u≤x, we have
(3.23)g(u)≤g*(x)≤εx.
Select(3.24)R2≥max⁡{X0,R1,2λΓ(α-1)∫01(α-2+s)(1-s)α-2r(s)ds}.
Then, for any u∈∂PR2, we get
(3.25)‖Tu‖≤λΓ(α-1)∫01(α-2+s)(1-s)α-2[z(s)g([u(s)-λω(s)]+)+r(s)]ds≤λεR2Γ(α-1)∫01(α-2+s)(1-s)α-2z(s)ds+λΓ(α-1)∫01(α-2+s)(1-s)α-2r(s)ds≤R22+R22=R2.
Thus,
(3.26)‖Tu‖≤‖u‖,∀u∈∂PR2.
By Lemma 2.8, T has a fixed point u such that R1≤∥u∥≤R2. Since R1≤∥u∥, by (3.18), we have  u(t)-λω(t)>0,t∈(0,1]. Let  u̅(t)=u(t)-λω(t). As ω(t) is a solution of (2.19) and u(t) is a solution of (2.23), u̅(t) is a positive solution of the singular semipositone boundary value problem (1.1). The proof is completed.Corollary 3.3.
The conclusion of Theorem3.2 is valid if   (H4)  is replaced by  (H4*):  there exists [c,d]⊂(0,1)  such that
(3.27)liminf⁡u→+∞min⁡t∈[c,d]f(t,u)=+∞;lim⁡u→+∞g(u)u=0.


## 4.  Examples

Example 4.1.
Consider the following problem(4.1)cD0+5/2u(t)+λf(t,u)=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
where f(t,u)=u2/t(1-t)+lnt. Let  z(t)=1/t(1-t),r(t)=-lnt, g(u)=u2. By direct calculation, we have ∫01r(t)dt=1,∫01z(t)dt=π, and
(4.2)liminf⁡u→+∞min⁡t∈[1/4,3/4]f(t,u)u=+∞.
So all conditions of Theorem 3.1 are satisfied. By Theorem 3.1, BVP (4.1) has at least one positive solution provided λ is sufficiently small.Example 4.2.
Consider the following problem(4.3)cD0+9/4u(t)+λf(t,u)=0,0<t<1,u(0)=u′(1)=u′′(0)=0,
where   f(t,u)=ln(1+u)/t(1-t)+lnt. Let z(t)=1/t(1-t),r(t)=-lnt,g(u)=ln(1+u). By direct calculation, we have ∫01r(t)dt=1, ∫01z(t)dt=π, and
(4.4)liminf⁡u→+∞min⁡t∈[1/4,3/4]f(t,u)=+∞;lim⁡u→+∞g(u)u=0.
So all conditions of Theorem 3.2 are satisfied. By Theorem 3.2, BVP (4.3) has at least one positive solution provided λ is sufficiently large.



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