# The Existence of Positive Solutions for Fractional Differential Equations with Sign Changing Nonlinearities

Authors: Weihua Jiang; Jiqing Qiu; Weiwei Guo
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the existence of at least two positive solutions to eigenvalue
problems of fractional differential equations with sign changing nonlinearities in more generalized
boundary conditions. Our analysis relies on the Avery-Peterson fixed point theorem
in a cone. Some examples are given for the illustration of main results.


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## Body

## 1. Introduction

The theory of fractional differential equations has become an important aspect of differential equations (see [1–8]). Boundary value problems of fractional differential equations have been investigated in many papers (see [9–46]). The existence of positive solutions to boundary value problems of fractional differential equations has been studied by many authors when nonlinearities are positive (see [9–24]). There are a few papers to study the existence of positive solutions of semipositone fractional differential equations. For example, using the Krasnoselskii fixed point theorem, Yuan et al. [9] discussed the existence of positive solutions for the singular positone and semipositone two-point boundary value problems
(1.1)D0+αu(t)=μa(t)f(t,u(t)),u(0)=u′(0)=u(1)=u′(1)=0,3<α≤4,
where μ>0,a and f are continuous. In [10], Wang et al. studied the existence of positive solutions for the singular semipositone two-point boundary value problems
(1.2)D0+αu(t)+λf(t,u(t))=0,(1.3)u(0)=u′(0)=u(1)=0,2<α≤3,
where λ>0, f is continuous.In [11], using Krasnoselskii fixed point theorem, Goodrich discussed the existence of at least one positive solutions for the system of fractional boundary value problems
(1.4)-D0+ν1y1(t)=λ1a1(t)f(y1(t),y2(t)),-D0+ν2y2(t)=λ2a2(t)g(y1(t),y2(t)),y1(i)(0)=y2(i)(0)=0,0≤i≤n-2,D0+αy1(t)∣t=1=ϕ1(y),D0+αy2(t)∣t=1=ϕ2(y),
where a1, a2, f, and g are nonnegative for t∈[0,1].Motivated by the excellent results mentioned above, in this paper, we investigate the existence of at least two positive solutions for the problem(1.5)D0+αy(t)+λf(t,y(t))=0,t∈[0,1],y(i)(0)=0,0≤i≤n-2,D0+βy(t)∣t=1=h(y),
where λ>0,α∈(n-1,n],n≥3,1≤β≤n-2<α-1, f∈C([0,1]×ℝ+,[-M,∞)),M>0, h∈C(C(ℝ+),ℝ+),ℝ+=[0,∞). The main tool is the Avery-Peterson theorem. To the best of our knowledge, this is the first paper dealing with eigenvalue problems of fractional differential equations with sign changing nonlinearities involving more general boundary conditions. Our results improve some of the earlier work presented in [10, 17, 46].


## 2. Preliminaries

For the convenience of the readers, we present here some necessary definitions and lemmas from fractional calculus theory. For more details see [1, 2].Definition 2.1.
The Riemann-Liouville fractional integral of orderα>0 of a function y:(0,∞)→ℝ is given by
(2.1)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
provided that the right-hand side is pointwise defined on (0,∞).Definition 2.2.
The Riemann-Liouville fractiontal derivative of orderα>0 of a function y:(0,∞)→ℝ is given by
(2.2)D0+αy(t)=1Γ(n-α)dndtn∫0t‍(t-s)n-α-1y(s)ds,
provided that the right-hand side is pointwise defined on (0,∞), where n=[α]+1.Lemma 2.3.
Assumef∈C[0,1], q≥p≥0, then
(2.3)D0+pI0+qf(t)=I0+q-pf(t).Lemma 2.4.
Assumeα>0, then(1)
Ifλ>-1, λ≠α-i, i=1,2,…,[α]+1, t>0, then
(2.4)D0+αtλ=Γ(λ+1)Γ(λ-α+1)tλ-α.(2)
D0+αtα-i=0, i=1,2,…,n.(3)
D0+αI0+αu(t)=u(t),  for a.e. t∈[0,1], where u∈L1[0,1].(4)
D0+αu(t)=0 if and only if
(2.5)u(t)=c1tα-1+c2tα-2+⋯+cntα-n,for some ci∈ℝ, i=1,2,…,n, where n is the least integer greater than or equal to α.Lemma 2.5 (see[11]).
Giveng∈C[0,1], y is a solution of the problem
(2.6)D0+αy(t)+g(t)=0,t∈[0,1],y(i)(0)=0,0≤i≤n-2,D0+βy(t)∣t=1=h(y),
if and only if it satisfies
(2.7)y(t)=Γ(α-β)Γ(α)tα-1h(y)+∫01‍G(t,s)g(s)ds,
where
(2.8)G(t,s)={tα-1(1-s)α-β-1-(t-s)α-1Γ(α),0≤s≤t≤1,tα-1(1-s)α-β-1Γ(α),0≤t≤s≤1.Lemma 2.6 (see[11]).
G
(
t
,
s
) is continuous on [0,1]×[0,1] and
(2.9)0≤G(t,s)≤G(1,s),t,s∈[0,1].Lemma 2.7.
G
(
t
,
s
)
≥
t
α
-
1
G
(
1
,
s
)
,
t
,
s
∈
[
0,1
].Proof.
Fors≤t,
(2.10)G(t,s)=tα-1(1-s)α-β-1-(t-s)α-1Γ(α)=tα-1(1-s)α-β-1-(1-s/t)α-1Γ(α)≥tα-1(1-s)α-β-1-(1-s)α-1Γ(α)=tα-1G(1,s).
For s>t,
(2.11)G(t,s)=tα-1(1-s)α-β-1Γ(α)≥tα-1(1-s)α-β-1-(t-s)α-1Γ(α)=tα-1G(1,s).By simple calculation, we can get(2.12)∫01‍G(1,s)ds=β(α-β)Γ(α+1),∫1/21‍G(1,s)ds=2βα-α+β2α(α-β)Γ(α+1).By Lemma2.5, we can easily get the following lemma.Lemma 2.8.
The boundary value problem(2.13)D0+αu(t)+1=0,0<t<1,u(i)(0)=D0+βu(t)∣t=1=0,0≤i≤n-2,
has a unique solution
(2.14)u(t)=tα-1Γ(α)(1α-β-tα).
Obviously,u satisfies
(2.15)βtα-1(α-β)Γ(α+1)≤u(t)≤tα-1(α-β)Γ(α),t∈[0,1].Lemma 2.9.
y
~
≥
λ
M
u is a solution of the following problem:
(2.16)D0+αy~+λ[f(t,y~-λMu)+M]=0,t∈[0,1],y~(i)(0)=0,0≤i≤n-2,D0+βy~(t)∣t=1=h(y~-λMu),
if and only if y=y~-λMu is a positive solution of (1.5).Proof.
In fact, ify is a positive solution of the problem (1.5), by Lemma 2.8, we get that y satisfies
(2.17)D0+α(y+λMu)+λ[f(t,y)+M]=0,t∈[0,1],(y+λMu)(i)(0)=0,0≤i≤n-2,D0+β(y+λMu)∣t=1=h(y).
Take y~=y+λMu. Then y~ satisfies (2.16) and y~≥λMu.
On the other hand, ify~ is a solution of (2.16) and y~≥λMu. Take y=y~-λMu. By Lemma 2.8, we can easily get that y satisfies (1.5). Clearly, y≥0.Define functionsh~, f~ and an operator T:C[0,1]→C[0,1] by
(2.18)h~(y)=h(max⁡{y-λMu,0}),f~(t,y)=f(t,max⁡{y-λMu,0})+M.Ty(t)=Γ(α-β)Γ(α)tα-1h~(y)+λ∫01‍G(t,s)f~(s,y(s))ds.Obviously,y≥λMu is a fixed point of the operator T if and only if y-λMu is a positive solution of the problem (1.5).TakeX=C[0,1] with norm ∥x∥=maxt∈[0,1]|x(t)|. Define a cone P by
(2.19)P={y∈C[0,1]∣y(t)≥tα-1‖y‖,t∈[0,1]}.Lemma 2.10.
T
:
P
→
P is a completely continuous operator.Proof..
Takey∈P. By Lemmas 2.6 and 2.7, we get
(2.20)Ty(t)≥tα-1[Γ(α-β)Γ(α)h~(y)+λ∫01‍G(1,s)f~(s,y(s))ds]≥tα-1‖Ty‖.
So, T:P→P. Let Ω⊂P be bounded. It follows from the continuity of h, f that there exist constants M1 and M2 such that h~(y)≤M1 and f~(t,y)≤M2 for t∈[0,1],y∈Ω. Thus,
(2.21)‖Ty‖≤Γ(α-β)Γ(α)M1+λM2∫01‍G(1,s)ds.
That is T(Ω) is bounded. For y∈Ω,t1,t2∈[0,1],
(2.22)|Ty(t1)-Ty(t2)|≤M1Γ(α-β)Γ(α)|t1α-1-t2α-1|+λM2∫01‍|G(t1,s)-G(t2,s)|ds.
By the uniform continuity of tα-1 and G(t,s), we get that T(Ω) is equicontinuous. Obviously, T:P→P is continuous. By the Arzela-Ascoli theorem, we get that T:P→P is completely continuous.Definition 2.11.
A mapϕ is said to be a nonnegative, continuous, and concave functional on a cone P of a real Banach space E if and only if ϕ:P→ℝ+ is continuous and
(2.23)ϕ(tx+(1-t)y)≥tϕ(x)+(1-t)ϕ(y),
for all x,y∈P and t∈[0,1].Definition 2.12.
A mapΦ is said to be a nonnegative, continuous, and convex functional on a cone P of a real Banach space E iff Φ:P→ℝ+ is continuous and
(2.24)Φ(tx+(1-t)y)≤tΦ(x)+(1-t)Φ(y),
for all x,y∈P and t∈[0,1].Letφ and Θ be nonnegative, continuous, and convex functional on P, Φ a nonnegative, continuous, and concave functional on P, and Ψ a nonnegative continuous functional on P. Then, for positive numbers a, b, c, and d, we define the following sets:
(2.25)P(φ,d)={x∈P:φ(x)<d},P(φ,Φ,b,d)={x∈P:b≤Φ(x),φ(x)≤d},P(φ,Θ,Φ,b,c,d)={x∈P:b≤Φ(x),Θ(x)≤c,φ(x)≤d},R(φ,Ψ,a,d)={x∈P:a≤Ψ(x),φ(x)≤d}.We will use the following fixed point theorem of Avery and Peterson to study the problem (1.5).Theorem 2.13 (see [47]).
LetP be a cone in a real Banach space E. Let φ and Θ be nonnegative, continuous, and convex functionals on P, Φ a nonnegative, continuous, and concave functional on P, and Ψ a nonnegative continuous functional on P satisfying Ψ(kx)≤kΨ(x) for 0≤k≤1, such that for some positive numbers M and d,
(2.26)Φ(x)≤Ψ(x),‖x‖≤Mφ(x)
for all x∈P(φ,d)¯. Suppose that
(2.27)T:P(φ,d)¯⟶P(φ,d)¯
is completely continuous and there exist positive numbers a, b, c with a<b, such that the following conditions are satisfied:(S1)
{x∈P(φ,Θ,Φ,b,c,d):Φ(x)>b}≠∅ and Φ(Tx)>b for x∈P(φ,Θ,Φ,b,c,d);(S2)
Φ(Tx)>b for x∈P(φ,Φ,b,d) with Θ(Tx)>c;(S3)
0∉R(φ,Ψ,a,d) and Ψ(Tx)<a for x∈R(φ,Ψ,a,d) with Ψ(x)=a.
ThenT has at least three fixed points x1,x2,x3∈P(φ,d)¯, such that
(2.28)φ(xi)≤d,fori=1,2,3,b<Φ(x1),a<Ψ(x2),Φ(x2)<b,Ψ(x3)<a.


## 3. Main Results

We define a concave functionΦ(x)=mint∈[1/2,1]|x(t)| and convex functions Ψ(x)=Θ(x)=φ(x)=∥x∥.Theorem 3.1.
Assume that there exists a constant0<l<Γ(α)/Γ(α-β), such that h(y)≤l∥y∥ for y∈P. In addition, suppose that there exist constants k, a, b, c, d with k>22α-1βΓ(α)/(Γ(α)-Γ(α-β)l), [Γ(α)-Γ(α-β)l]αM/βΓ(α)<a<b-[Γ(α)-Γ(α-β)l]αM/βΓ(α)<2α-1b<c<d, such that the following conditions hold:(C1)
f(t,y)≤d-M,  for (t,y)∈[0,1]×[0,d];(C2)
f(t,y)≥kb-M,  for (t,y)∈[1/2,1]×[b-[Γ(α)-Γ(α-β)l]αM/βΓ(α),c];(C3)
f(t,y)≤a-M,  for (t,y)∈[0,1]×[0,a].
Then the problem (1.5) has at least two positive solutions for
(3.1)22α-1(α-β)Γ(α+1)k[α(2β-1)+β]<λ<α(α-β)[Γ(α)-Γ(α-β)l]β.Proof.
Takey∈P(φ,d)¯. By ∥max{y-λMu,0}∥≤d, (C1), Lemma 2.6, (2.12), and (3.1), we have
(3.2)‖Ty‖≤Γ(α-β)Γ(α)ld+λd∫01‍G(1,s)ds=[Γ(α-β)Γ(α)l+λ∫01‍G(1,s)ds]d≤d.
This means that T:P(φ,d)¯→P(φ,d)¯.
It is easy to see that{y∈P(φ,Θ,Φ,b,c,d):Φ(y)>b}≠∅. y∈P(φ,Θ,Φ,b,c,d) implies mint∈[1/2,1]y(t)≥b, ∥y∥≤c. It follows from (2.15) and (3.1) that mint∈[1/2,1](y-λMu)≥b-α[Γ(α)-Γ(α-β)l]M/βΓ(α). By (C2), (2.12), (3.1), and Lemma 2.7, we get
(3.3)Φ(Ty)=min⁡t∈[1/2,1]Ty(t)≥λmin⁡t∈[1/2,1]∫01‍G(t,s)f~(s,y(s))ds≥(12)α-1λkb∫1/21‍G(1,s)ds>b.
So, the condition (S1) of Theorem 2.13 holds.
Takey∈P(φ,Φ,b,d) with Θ(Ty)>c. By Ty∈P, we get
(3.4)min⁡t∈[1/2,1]Ty(t)≥min⁡t∈[1/2,1]tα-1‖Ty‖≥12α-1‖Ty‖>12α-1c>b.
Thus, (S2) holds.
Bya>0, we have 0∉R(φ,Ψ,a,d). Take y∈R(φ,Ψ,a,d) with Ψ(y)=a. By (C3), we get
(3.5)Ψ(Ty)=‖Ty‖≤Γ(α-β)Γ(α)la+λa∫01‍G(1,s)ds=[Γ(α-β)Γ(α)l+λ∫01‍G(1,s)ds]a≤a.
By Theorem 2.13, we get that T has at least three fixed points y1,y2,y3∈P(φ,d)¯ such that ∥yi∥≤d,i=1,2,3, and
(3.6)b<Φ(y1),a<Ψ(y2),Φ(y2)<b,Ψ(y3)<a.
Ify∈P and ∥y∥>a, by (2.15) and (3.1), we have
(3.7)y(t)≥tα-1‖y‖>atα-1>λMu(t).
Obviously,∥y1∥>b>a and ∥y2∥>a. So, y1-λMu,y2-λMu are two positive solutions of (1.5). The proof is completed.


## 4. Example

For convenience, we define the following notations:(4.1)[a,b]:={x:x∈R,a≤x≤b},(a,b]:={x:x∈R,a<x≤b},fora,b∈R,a<b.Example 4.1.
Consider the following boundary value problem:(4.2)D0+5/2y(t)+λf(t,y(t))=0,t∈[0,1],y(0)=y′(0)=0,y′(1)=h(y),
where h(y)=∫01‍y(s)dg(s),g(s) is a bounded variation function on [0,1] with 0<⋁01(g)≤1<3/2,
(4.3)f(t,y)={sin⁡(t-12)π-1-y,(t,y)∈[0,1]×[0,6],sin⁡(t-12)π-1+601(y-6)-y,(t,y)∈[0,1]×(6,7],sin⁡(t-12)π+600-y,(t,y)∈[0,1]×(7,900],sin⁡(t-12)π+570,(t,y)∈[0,1]×(900,+∞).
Corresponding to the problem (1.5), we get that α=5/2,β=1,n=3,h(y)≤∥y∥⋁01(g). Take l=1,k=50,M=a=6,b=12,c=36,d=620.
By simple calculation, we can get that the conditions of Theorem3.1 are satisfied. So, when (9/35)π<λ<(15/16)π, the problem (4.2) has at least two positive solutions.Example 4.2.
Consider the following boundary value problem:(4.4)D0+7/2y(t)+λf(t,y(t))=0,t∈[0,1],y(0)=y′(0)=y′′(0)=0,D0+3/2y(t)∣t=1=h(y),
where h(y)=∫01‍y(s)dg(s),g(s) is a bounded variation function on [0,1] with 0<⋁01(g)≤(15/8)π-1,
(4.5)f(t,y)={-34cos⁡π2t-14y,(t,y)∈[0,1]×[0,1],-34cos⁡π2t+3205(y-1)-14y,(t,y)∈[0,1]×(1,1.2],-34cos⁡π2t+641-14y,(t,y)∈[0,1]×(1.2,12],-34cos⁡π2t+641-123,(t,y)∈[0,1]×(12,+∞).
Corresponding to the problem (1.5), we get that α=7/2,β=3/2,n=4,h(y)≤∥y∥⋁01(g). Take l=(15/8)π-1,k=320,M=a=1,b=2,c=12,d=642.
Obviously,f∈C([0,1]×ℝ+,[-1,∞)),h(y)≤l∥y∥. By simple calculation, we can get that k,l,a,b,c,d,α,β,M satisfy 0<l<Γ(α)/Γ(α-β),k>22α-1βΓ(α)/(Γ(α)-Γ(α-β)l), and
(4.6)[Γ(α)-Γ(α-β)l]αMβΓ(α)<a<b-[Γ(α)-Γ(α-β)l]αMβΓ(α)<2α-1b<c<d.
It is easy to see thatf(t,y)+1≤642, for (t,y)∈[0,1]×[0,642], and f(t,y)+1≤1, for (t,y)∈[0,1]×[0,1]. So, conditions (C1) and (C3) of Theorem 3.1 hold.
For(t,y)∈[1/2,1]×[2-56/45√π,12],f(t,y)+1≥kb=640. Therefore, condition (C2) of Theorem 3.1 holds. So, for (21/8(7√2-2))π<λ<14/3, the problem (4.4) has at least two positive solutions.
Specially, in Example4.2, we take
(4.7)g(t)={0,t<ξ,1,t≥ξ,
where 0<ξ<1 and all other conditions remain unchanged. Then h(y)=y(ξ). Clearly, ∨01(g)=1<(15/8)π-1. The problem (4.4) has at least two positive solutions for (21/8(7√2-2))π<λ<14/3.



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*Source: 180672-2012-07-09.xml*