# Multiple Positive Solutions for Nonlinear Semipositone Fractional Differential Equations

Authors: Wen-Xue Zhou; Ji-Gen Peng; Yan-Dong Chu
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We present some new multiplicity of positive solutions results for nonlinear semipositone fractional boundary value problemD0+αu(t)=p(t)f(t,u(t))-q(t),0<t<1,u(0)=u(1)=u′(1)=0, where 2<α≤3 is a real number and D0+α is the standard Riemann-Liouville differentiation. One example is also given to illustrate the main result.


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## Body

## 1. Introduction

This paper is mainly concerned with the multiplicity of positive solutions of nonlinear fractional differential equation boundary value problem (BVP for short)(1.1)D0+αu(t)=p(t)f(t,u(t))-q(t),0<t<1,u(0)=u(1)=u′(1)=0,
where 2<α≤3 is a real number and D0+α is the standard Riemann-Liouville differentiation, and f,p,q is a given function satisfying some assumptions that will be specified later.In the last few years, fractional differential equations (in short FDEs) have been studied extensively the motivation for those works stems from both the development of the theory of fractional calculus itself and the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For an extensive collection of such results, we refer the readers to the monographs by Kilbas et al. [1], Miller and Ross [2], Oldham and Spanier [3], Podlubny [4], and Samko et al. [5].Some basic theory for the initial value problems of FDE involving the Riemann-Liouville differential operator has been discussed by Lakshmikantham and Vatsala [6–8], Babakhani and Daftardar-Gejji [9–11], and Bai [12], and others. Also, there are some papers that deal with the existence and multiplicity of solutions (or positive solution) for nonlinear FDE of BVPs by using techniques of nonlinear analysis (fixed point theorems, Leray-Schauders theory, topological degree theory, etc.), see [13–22] and the references therein.Bai and Lü [15] studied the following two-point boundary value problem of FDEs(1.2)D0+qu(t)+f(t,u(t))=0,u(0)=u(1)=0,0<t<1,1<q≤2,
where D0+q is the standard Riemann-Liouville fractional derivative. They obtained the existence of positive solutions by means of the Guo-Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem.Zhang [22] considered the existence and multiplicity of positive solutions for the nonlinear fractional boundary value problem(1.3)cD0+qu(t)=f(t,u(t)),0<t<1,u(0)+u′(0)=0,u(1)+u′(1)=0,
where 1<q≤2 is a real number, f:[0,1]×[0,+∞)→[0,+∞), and cD0+q is the standard Caputo’s fractional derivative. The author obtained the existence and multiplicity results of positive solutions by means of the Guo-Krasnosel’skii fixed point theorem.From the above works, we can see the fact that although the fractional boundary value problems have been investigated by some authors to the best of our knowledge, there have been few papers that deal with the boundary value problem (1.1) for nonlinear fractional differential equation. Motivated by all the works above, in this paper we discuss the boundary value problem (1.1), using the Guo-Krasnosel’skii fixed point theorem, and we give some new existence of multiple positive solutions criteria for boundary value problem (1.1).The paper is organized as follows. In Section2, we give some preliminary results that will be used in the proof of the main results. In Section 3, we establish the existence of multiple positive solutions for boundary value problem (1.1) by the Guo-Krasnosel’skii fixed point theorem. In the end, we illustrate a simple use of the main result.


## 2. Preliminaries and Lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature such as [1, 4, 15].Definition 2.1 (see [1, 4]).
The Riemann-Liouville fractional integral of orderα(α>0) of a function f:(0,+∞)→ℝ is given by
(2.1)I0+αf(t)=∫0t(t-s)α-1Γ(α)f(s)ds,
provided that the right side is pointwise defined on (0,+∞), where Γ is the gamma function.Definition 2.2 (see [1, 4]).
The Riemann-Liouville fractional derivative of orderα(α>0) of a continuous function f:(0,+∞)→ℝ is given by
(2.2)(D0+αf)(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α+1f(s)ds,
provided that the right side is pointwise defined on (0,+∞), where n=[α]+1 and [α] denotes the integer part of α.Lemma 2.3 (see [15]).
Letα>0. If one assumes u∈C(0,1)∩L(0,1), then fractional differential equation
(2.3)Dαu(t)=0
has
(2.4)u(t)=C1tα-1+C2tα-2+⋯+CNtα-N,Ci∈R,i=1,2,…,N,
as unique solutions, where N is the smallest integer greater than or equal to α.Lemma 2.4 (see [15]).
Assume thath∈C(0,1)∩L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L(0,1). Then
(2.5)IαDαh(t)=h(t)+C1tα-1+C2tα-2+⋯+CNtα-N,
for some Ci∈ℝ, i=1,2,…,N, where N is the smallest integer greater than or equal to α.In the following, we present Green’s function of the fractional differential equation boundary value problem.Lemma 2.5.
Leth∈C[0,1] and 2<α≤3, then the unique solution of
(2.6)Dαu(t)+h(t)=0,0<t<1,u(0)=u(1)=u′(1)=0
is given by
(2.7)u(t)=∫01G(t,s)h(s)ds,
where G(t,s) is Green’s function given by
(2.8)G(t,s)=1Γ(α){(1-s)α-1tα-1-(t-s)α-1,if0≤s≤t≤1,(1-s)α-1tα-1,if0≤t≤s≤1.The following properties of Green’s function form the basis of our main work in this paper.Lemma 2.6.
The functionG(t,s) defined by (2.8) possesses the following properties:(i)
G(t,s)=G(1-s,1-t) for t,s∈(0,1);(ii)
tα-1(1-t)s(1-s)α-1≤G(t,s)Γ(α)≤(α-1)s(1-s)α-1 for t, s∈(0,1);(iii)
tα-1(1-t)s(1-s)α-1≤G(t,s)Γ(α)≤(α-1)tα-1(1-t) for t, s∈(0,1);(iv)
G(t,s)>0 for t,s∈(0,1).The following Krasnosel’skii’s fixed point theorem will play a major role in our next analysis.Lemma 2.7 (see [23]).
LetX be a Banach space, and let P⊂X be a cone in X. Assume Ω1,Ω2 are open subsets of X with 0∈Ω1⊂Ω¯1⊂Ω2, and let 𝒜:P→P be a completely continuous operator such that either(i)
∥𝒜u∥≤∥u∥,u∈P∩∂Ω1, ∥𝒜u∥≥∥u∥,u∈P∩∂Ω2, or(ii)
∥𝒜u∥≥∥u∥,u∈P∩∂Ω1, ∥𝒜u∥≤∥u∥,u∈P∩∂Ω2.Then 𝒜 has a fixed point in P∩(Ω¯2∖Ω1).


## 3. Main Results

In this section, we establish some new existence results for the fractional differential equation (1.1). Given a∈L1(0,1), we write a a≻0, if a≥0 for t∈[0,1], and it is positive in a set of positive measure.Let us list the following assumptions:(H1)f:[0,1]×[0,∞)→[0,∞) is continuous, p,q≻0;(H2) there existsθ∈(0,1/2), such that(3.1)∫θ1-θp(s)s(1-s)α-1ds>0.In view of Lemmas2.5 and 2.6, we obtain the following.Lemma 3.1.
Letq∈L1[0,1] with q>0 on (0,1), and γ(t) is the unique solution of
(3.2)D0+αu(t)=q(t),0<t<1,u(0)=u(1)=u′(1)=0,
Then
(3.3)0≤γ(t)≤α-1Γ(α)∥q∥1tα-1(1-t):=Ctα-1(1-t),fort∈[0,1],
where  𝒞=(α-1/Γ(α))∥q∥1, ∥q∥1=∫01|q(t)|dt.Next, we consider(3.4)D0+αu(t)=p(t)g(t,u(t)-γ(t)),0<t<1,u(0)=u(1)=u′(1)=0,
where(3.5)g(t,u)={f(t,u),ifu≥0f(t,0),ifu<0,
Then (3.4) is equivalent to the following integral equation:(3.6)u(t)=∫01G(t,s)p(s)g(s,u(s)-γ(s))ds.Lemma 3.2.
Letu(t)≥γ(t) for t∈[0,1], and u(t) is positive solution of the problem (3.4). Then u(t)-γ(t) is positive solution of the problem (1.1).Proof.
In fact, letx(t)=u(t)-γ(t). Then x(t)≥0 and u(t)=x(t)+γ(t). Since u(t) is positive solution of the problem (3.4), we have
(3.7)D0+α[x(t)+γ(t)]=p(t)g(t,x(t)),0<t<1,(x+γ)(0)=(x+γ)(1)=(x+γ)′(1)=0.
So
(3.8)D0+αx(t)=p(t)g(t,x(t))-q(t),0<t<1,x(0)=x(1)=x′(1)=0.
For our constructions, we will consider the Banach spaceE=C[0,1] equipped with standard norm ∥u∥=max0≤t≤1|u(t)|,u∈E.
Define a coneK by
(3.9)K={u∈E:u(t)≥tα-1(1-t)α-1∥u∥,∀t∈[0,1],α∈(2,3]}.
Let the operator𝒜:K→E be defined by the formula
(3.10)(Ax)(t):=∫01G(t,s)p(s)g(s,x(s)-γ(s))ds,0≤t≤1,u∈K.Lemma 3.3.
Assume that (H1) holds. Then𝒜(K)⊂K.Proof.
Notice from (3.10) and Lemma 2.6 that, for x∈K, 𝒜x(t)≥0 on [0,1] and
(3.11)‖Ax‖≤1Γ(α)∫01(α-1)s(1-s)α-1p(s)g(s,x(s)-γ(s))ds.
On the other hand, we have
(3.12)Ax(t)≥1Γ(α)∫01tα-1(1-t)s(1-s)α-1p(s)g(s,x(s)-γ(s))ds≥tα-1(1-t)(α-1)Γ(α)∥Ax∥.
Thus we have 𝒜(K)⊂K. The proof is finished.It is standard that𝒜:K→K is continuous and completely continuous.For convenience, we introduce the following notations:Ñ=((α-1)∥p∥1/Γ(α))max0≤s≤1s(1-s)α-1, N¯=σ∫θ1-θ(s(1-s)α-1/Γ(α))p(s)ds, σ=minθ≤t≤1-θ(1-t)tα-1.Theorem 3.4.
Assume that (H1) and (H2) are satisfied. Also suppose the following conditions are satisfied:(A1)
there exists a constantR1>(α-1)𝒞 such that Ñf(t,u)≤R1 for all (t,u)∈[0,1]×[0,R1];(A2)
there exists a constantR2>2R1 such that N¯f(t,u)>R2 for all (t,u)∈[0,1]×[σR2,R2];(A3)
limu→+∞max0≤t≤1(f(t,u)/u)=0.Then the problem (1.1) has at least two positive solutions.Proof.
To show that (1.1) has at least two positive solutions, we will assume the problem (3.4) has at least two positive solutions x1 and x2 with R1≤∥x1∥<R2<∥x2∥≤R3.
We now show(3.13)‖Ax‖≤‖x‖,forx∈K∩∂Ω1,
To see this, letΩ1={x∈K∣∥x∥<R1}, then for x∈K∩∂Ω1, t∈[0,1], by Lemma 3.1 and (A1), we have
(3.14)x(t)-γ(t)≤x(t)≤∥x∥=R1,x(t)-γ(t)≥tα-1(1-t)α-1R1-Ctα-1(1-t)≥(R1α-1-C)tα-1(1-t)≥0.
Thus, we see, from Lemma 2.6 and (A1), that
(3.15)∥Ax∥=max⁡0≤t≤1(Au)(t)=max⁡0≤t≤1∫01G(t,s)p(s)g(s,x(s)-γ(s))ds≤(α-1)Γ(α)∫01s(1-s)α-1p(s)f(s,x(s)-γ(s))ds≤(α-1)Γ(α)∫01R1Ñs(1-s)α-1p(s)ds≤R1,
from which we see that ∥𝒜x∥≤∥x∥, for x∈K∩∂Ω1.
Next we now show(3.16)∥Ax∥≥∥x∥,forx∈K∩∂Ω2.
To see this, let Ω2={x∈K|∥x∥<R2}; then, for x∈K∩∂Ω2, t∈[0,1], by R2>2R1, we have
(3.17)x(t)-γ(t)≥tα-1(1-t)α-1R2-Ctα-1(1-t)≥tα-1(1-t)2(α-1)R2.
Forx∈∂Ω2; t∈[θ,1-θ]σ, then, it follows from (3.17) that
(3.18)R2≤tα-1(1-t)2(α-1)R2≤x(t)-γ(t)≤R2.
In view of (A2), (3.17) and Lemma 2.6, we have that for all x∈∂Ω2, t∈[θ,1-θ]σ(3.19)∥Au∥≥∫01G(t,s)p(s)g(s,x(s)-γ(s))ds≥tα-1(1-t)∫01s(1-s)α-1Γ(α)p(s)f(s,x(s)-γ(s))ds>tα-1(1-t)∫θ1-θs(1-s)α-1R2Γ(α)N¯p(s)ds≥σ∫θ1-θs(1-s)α-1R2Γ(α)N¯p(s)ds=R2,
from which we see that ∥𝒜x∥>∥x∥, for x∈K∩∂Ω2.
On the other hand, letε>0, where
(3.20)ε(α-1)Γ(α)max⁡0≤t≤1⁡t(1-t)α-1‖p‖1≤1.
Supposing that (A3) holds, one can findN>R2>0, so that
(3.21)f(t,u)≤εu,∀t∈[0,1],u≥N.
Setting
(3.22)R3=(α-1)max⁡0≤t≤1⁡t(1-t)α-1∥p∥1max⁡(t,u)∈[0,1]×[0,N]f(t,u)(Γ(α)-ε(α-1)max⁡0≤t≤1⁡t(1-t)α-1∥p∥1)+N,
then R3>N>R2, and so
(3.23)∥Au∥=max⁡0≤t≤1∫01G(t,s)p(s)g(s,x(s)-γ(s))ds≤∫01(α-1)s(1-s)α-1Γ(α)p(s)max⁡(s,u)∈[0,1]×[0,N]f(s,u)ds+∫01(α-1)s(1-s)α-1Γ(α)p(s)ε(x(s)-γ(s))ds≤R3,
from which we see that ∥𝒜x∥≤∥x∥, for x∈K∩∂Ω3.
In view of Lemma2.7, the problem (3.4) has at least two positive solutions x1 and x2 with R1≤∥x1∥<R2<∥x2∥≤R3. Since R2>R1>(α-1)𝒞, we have
(3.24)x1(t)-γ(t)≥tα-1(1-t)α-1R1-Ctα-1(1-t)≥(R1α-1-C)tα-1(1-t)≥0,x2(t)-γ(t)≥tα-1(1-t)α-1R2-Ctα-1(1-t)≥(R2α-1-C)tα-1(1-t)≥0.
Therefore x1,x2 are solutions of the problem (1.1). This completes the proof.Theorem 3.5.
Suppose that (H1), (H2) are satisfied. Furthermore assume that(A4)
there exists a constantR1>2(α-1)𝒞 such that N¯f(t,u)≥R1 for all (t,u)∈[0,1]×[σR1,R1];(A5)
there exists a constantR2>max{R1,(R1/N̅)Ñ} such that Ñf(t,u)<R2 for all (t,u)∈[0,1]×[0,R2];(A6)
limu→+∞minθ≤t≤1-θ(f(t,u)/u)=+∞. Then the problem (1.1) has at least two positive solutions.


## 4. An Example

As an application of the main results, we consider(4.1)D5/2y(t)=f(y)-1t,0<t<1,y(0)=y′(0)=y(1)=0,
Set(4.2)f(y)={-2(y-7)2+1100,if0≤y≤7,-2(y-7)+1100,if7≤y≤450,(y-450)2+214,ify≥450,
Then we have 𝒞=(α-1)/Γ(α)∥q∥1≈2.25676,Ñ=((α-1)∥q∥1/Γ(α))max0≤s≤1s(1-s)α-1≈0.4, letting, θ=1/4, then σ=minθ≤t≤1-θ(1-t)tα-1≈0.09375,N¯=σ∫1/43/4(s(1-s)α-1/Γ(α))ds≈0.008, choosing R1=7, R2=450, then R1>2(α-1)𝒞=6.77, R2>max{R1,(R1/N̅)Ñ}=max{7,350}=350; therefore, we have N¯f(u)=0.008[-2(y-7)2+1100]≥8.156>R1,(t,u)→[1/4,3/4]×[0.65625,7], Ñf(u)=0.4[-2(y-7)2+1100]≤440<R2,(t,u)→[0,1]×[0,7], Ñf(u)=0.4[-2(y-7)+1100]≤440<R2,(t,u)→[0,1]×[7,450], and limy→+∞(f(y)/y)=limy→+∞(((y-450)2+214)/y)=+∞.It is clear thatf:[0,1]×[0,∞)→[0,∞) is continuous. Since all the conditions of Theorem 3.5 are satisfied, the problem (4.1) has at least two positive solutions.



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*Source: 850871-2012-02-22.xml*