We discuss the existence of positive solutions to a class of fractional boundary value problem with changing sign nonlinearity and advanced arguments Dαx(t)+μh(t)f(x(a(t)))=0,t∈(0,1),2<α≤3,μ>0,x(0)=x′(0)=0,x(1)=βx(η)+λ[x],β>0, and η∈(0,1), where Dα is the standard Riemann-Liouville derivative, f:[0,∞)→[0,∞) is continuous, f(0)>0, h :[0,1]→(−∞,+∞), and a(t) is the advanced argument. Our analysis relies on a nonlinear alternative of Leray-Schauder type. An example is given to illustrate our results.

1. Introduction

Fractional differential equations (FDEs) have been of great interest for the past three decades. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, and so forth (see [1, 2]). Therefore, the theory of FDEs has been developed very quickly. There has been a significant development in fractional differential equations in recent years; see [1–30].

In [5], the author studied existence of positive solutions in case of the nonlinear fractional differential equation as follows:
(1)Dsu=λa(t)f(u),0<t<1,u(0)=0,
where 0<s<1,Ds is the standard Riemann-Liouville fractional derivative, f:[0,∞)→[0,∞) is continuous, and a:[0,1]→R. In [10], the author applied the Avery-Peterson fixed point theorem to obtain sufficient conditions of the existence of multiple solutions to the following problem:
(2)x′′′(t)+h(t)f(x(α(t)))=0,t∈(0,1),x(0)=x′′(0)=0,x(1)=βx(η)+λ[x],β>0,η∈(0,1),
where f:[0,∞)→[0,∞) is continuous and h(t) is a nonnegative continuous function defined on [0,1].

Motivated by [5, 10], in this paper, we consider the existence of positive solution of the following boundary value problem for nonlinear fractional differential equation with changing sign nonlinearity and advanced arguments:
(3)Dαx(t)+μh(t)f(x(a(t)))=0,t∈(0,1),2<α≤3,μ>0,x(0)=x′(0)=0,x(1)=βx(η)+λ[x],β>0,η∈(0,1),
where λ denotes a linear functional on C[0,1] given by λ[x]=∫01x(t)dΛ(t) involving a Stieltjes integral with a suitable function Λ of bounded variation. It is important to indicate that we did not assume that λ[x] is positive to all positive x. The measure dΛ can be a signed measure.

Put J=[0,1]; let us introduce the following assumptions:

f:[0,∞)→[0,∞) is continuous, and f(0)>0;

a∈C(J,J), and t≤a(t) on J;

h:[0,1]→(-∞,+∞) may change sign; h is not identically zero on any subinterval on J;

0<βηα-1+λ[p]<1, where p(t)=tα-1.

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 convenience, we also present the necessary definitions from fractional calculus theory here. These definitions can be found in the recent literature.

Definition 1.

The fractional integral of order α>0 of a function x:(0,+∞)→R is given by
(4)I0+αx(t)=1Γ(α)∫0t(t-s)α-1x(t)ds,
provided that the right-hand side is pointwise defined on (0,+∞).

Definition 2.

The fractional derivative of order α>0 of a continuous function x:(0,+∞)→R is given by
(5)D0+αx(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α-1x(t)ds,
where n=[α]+1 and [α] denotes the integral part of number α, provided that the right-hand side is pointwise defined on (0,+∞).

Lemma 3.

Let α>0,x∈C(0,1)∩L(0,1); then
(6)I0+αD0+αx(t)=x(t)+c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈R(i=1,2,…,n),n being the smallest integer greater than or equal to α.

Consider the following boundary value problem:
(7)Dαx(t)+y(t)=0,t∈(0,1),2<α≤3,x(0)=x′(0)=0,x(1)=βx(η)+λ[x],β>0,η∈(0,1).

Lemma 4.

Assume that βηα-1≠1 and y∈C(J,R); then problem (7) has the unique solution given by the following formula:
(8)x(t)=tα-11-βηα-1λ[x]+βtα-11-βηα-1∫01k(η,s)y(s)ds+∫01k(t,s)y(s)ds,
where
(9)k(t,s)=1Γ(α){[t(1-s)]α-1,0≤t≤s≤1,[t(1-s)]α-1-(t-s)α-1,0≤s≤t≤1.

Theorem 5.

Let X be a Banach space with C⊂X closed and convex. Assume that U is a relatively open subset of C with 0∈C and A:U¯→C is a continuous, compact map. Then either

A has a fixed point in U¯ or

there exist u∈∂U and τ∈(0,1) with u=τAu.

3. Existence of Positive Solutions

Let us denote by X=C[0,1] the Banach space of all continuous real functions on [0,1] endowed with the sup norm and let K be the cone:
(10)K={x∈X,x(t)≥0,t∈J}.

Lemma 6.

Let assumptions (H_{
1})–(H_{
4}) hold. Moreover, we assume that assumptions (H_{5})-(H_{6}) hold with

h:[0,1]→(-∞,+∞) is continuous, h(0)≠0, and there is σ>1 such that
(11)tα-1Δ-ρ(βρΔ∫01k(η,s)h+(s)ds+∫01κ(s)h+(s)ds)+∫01k(t,s)h+(s)ds+βtα-1Δ∫01k(η,s)h+(s)ds≥σ[tα-1Δ-ρ(βρΔ∫01k(η,s)h-(s)ds+∫01κ(s)h-(s)ds)+∫01k(t,s)h-(s)ds+βtα-1Δ∫01k(η,s)h-(s)ds],

where Δ=1-βηα-1,ρ=λ[p],h+(t)=max{0,h(t)}andh-(t)=max{0,-h(t)}. Then, for every 0<δ<1, there exists a positive number μ¯ such that, for 0<μ<μ¯, the nonlinear fractional differential equation,
(12)Dαx(t)+μh+(t)f(x(a(t)))=0,f,t∈(0,1),2<α≤3,μ>0,x(0)=x′(0)=0,x(1)=βx(η)+λ[x],β>0,η∈(0,1),
has a positive solution x¯μ with ∥x¯μ∥→0 as μ→0 and
(13)x¯μ(t)≥μδf(0)m(t),
where
(14)m(t)=tα-1Δ-ρ(βρΔ∫01k(η,s)h+(s)ds+∫01κ(s)h+(s)ds)+βtα-1Δ∫01k(η,s)h+(s)ds+∫01k(t,s)h+(s)ds.Proof.

It is easy to know from (9), (H_{
5}), and (H_{
6}) that m(t)>0,t∈(0,1]. By Lemma 4, (12) has a unique solution in X:
(15)x(t)=tα-11-βηα-1λ[x]+βtα-11-βηα-1μ∫01k(η,s)h+(s)f(x(a(s)))ds+μ∫01k(t,s)h+(s)f(x(a(s)))ds.
For x∈C(J,R+), we define two operators T and S by
(16)Tx(t)=tα-1Δλ[x]+μFx(t),Sx(t)=tα-1Δ-ρμλ[Fx]+μFx(t),
where
(17)Fx(t)=βtα-1Δ∫01k(η,s)h+(s)f(x(a(s)))ds+∫01k(t,s)h+(s)f(x(a(s)))ds,λ[Fx]=βρΔ∫01k(η,s)h+(s)f(x(a(s)))ds+∫01κ(s)h+(s)f(x(a(s)))ds.
It is easy to show that T:K→K and S:K→K are completely continuous. We claim that operators T and S have the same fixed points in K. In fact, let x=Sx; then
(18)λ[x]=ρΔ-ρμλ[Fx]+μλ[Fx]=ΔΔ-ρμλ[Fx].
So
(19)x(t)=Sx(t)=tα-1Δ-ρμλ[Fx]+μFx(t)=tα-1Δλ[x]+μFx(t)=Tx(t).
Let x=Tx; then λ[x]=(ρ/Δ)λ[x]+μλ[Fx]. So λ[x]=(Δ/(Δ-ρ))μλ[Fx], and hence
(20)x(t)=Tx(t)=tα-1Δλ[x]+μFx(t)=tα-1Δ-ρμλ[Fx]+μFx(t)=Sx(t).
This shows that fixed points of S are solutions of (12). We will apply the nonlinear alternative of Leray-Schauder type to prove that S has at least one fixed point for small μ.

Let ϵ>0 be such that
(21)f(x(a(t)))≥δf(0),0≤x(a(t))≤ϵ,∀t∈[0,1].
Suppose that 0<μ<ϵ/2∥m∥f¯(ϵ)∶=μ¯, where f¯(t)=max0≤s≤tf(s); then
(22)f¯(∥x∥)=max0≤|x(a(t))|≤∥x∥f(x(a(t))),∀t∈[0,1].
Since limt→0+(f¯(t)/t)=+∞,f¯(ϵ)/ϵ<1/2μ∥m∥, there exists a unique Rμ∈(0,ϵ) such that
(23)f¯(Rμ)Rμ=12μ∥m∥.

Let x∈K and τ∈(0,1) be such that x=τSx. We claim that ∥x∥≠Rμ. In fact,
(24)x(t)=τtα-1Δ-ρμ(βρΔ∫01k(η,s)h+(s)f(x(a(s)))dsmmmmmmmm+∫01κ(s)h+(s)f(x(a(s)))dsβρΔ)+τμ(βtα-1Δ∫01k(η,s)h+(s)f(x(a(s)))dsmmmmmm+∫01k(t,s)h+(s)f(x(a(s)))dsβtα-1Δ)≤tα-1Δ-ρμf¯(∥x∥)×(∫01k(η,s)h+(s)ds+∫01κ(s)h+(s)ds)+μf¯(∥x∥)×(βtα-1Δ∫01k(η,s)h+(s)ds+∫01k(t,s)h+(s)ds)=μf¯(∥x∥)m(t)≤μf¯(∥x∥)∥m∥.
That is, f¯(∥x∥)/∥x∥≥1/μ∥m∥, which implies that ∥x∥≠Rμ. Let U={x∈K:∥x∥<Rμ}. By Theorem 5, S has a fixed point x¯μ∈U¯. Moreover, combining (21) with the expression of operator S, we obtain that
(25)x¯μ(t)≥μδf(0)m(t),∀t∈(0,1].
Hence (12) has a positive solution x¯μ(t). Note that Rμ→0 as μ→0; we get that ∥x¯μ∥→0 as μ→0.

Theorem 7.

Suppose that (H_{
1})–(H_{
6}) hold. Then there exists a positive number μ*>0 such that (3) has at least one positive solution for μ∈(0,μ*).

Proof.

Let
(26)ω(t)=tα-1Δ-ρ(βρΔ∫01k(η,s)h-(s)ds+∫01κ(s)h-(s)ds)+βtα-1Δ∫01k(η,s)h-(s)ds+∫01k(t,s)h-(s)ds.
Then ω(t)≥0 for each t∈(0,1]. We have m(t)≥σω(t),σ>1. Choose c∈(0,1) such that σc>1. There is b>0 such that f(x(a(t)))≤σcf(0) for x∈[0,b]; then
(27)ω(t)f(x(a(t)))≤cm(t)f(0)fort∈(0,1],x∈[0,b].
Fix δ∈(c,1), and let μ*>0 be such that
(28)∥x¯μ∥+μδf(0)∥m∥≤b,μ∈(0,μ*),
where x¯μ is given by Lemma 6, and
(29)|f(x1(a(t)))-f(x2(a(t)))|≤f(0)δ-c2,
for x1,x2∈[0,b] with |x1-x2|≤μ*δf(0)∥m∥.

Let μ∈(0,μ*). We look for a solution xμ of the form x¯μ+vμ, where x¯μ is the solution of (12), given by Lemma 6. Thus vμ solves the following equation:
(30)Dαvμ=μh+(t)(f1′-f2′)-μh-(t)f1′,vμ(0)=vμ′(0)=0,vμ(1)=βvμ(η)+λ[vμ],
where f1′=f(x¯μ(a(t))+vμ(a(t))),f2′=f(x¯μ(a(t))).

Now, we need to prove the existence of vμ. Consider the following equation:
(31)Dαv=μh+(t)(f1-f2)-μh-(t)f1,v(0)=v′(0)=0,v(1)=βv(η)+λ[v],
where
(32)f1=f(x¯μ(a(t))+v(a(t))),f2=f(x¯μ(a(t))).
Obviously, (31) is equivalent to the operator equation:
(33)Sv(t)=tα-1Δ-ρμ(βρΔ∫01k(η,s)h+(s)(f1-f2)dsmmmmmmmm+∫01κ(s)h+(s)(f1-f2)dsβρΔ)+μ(βtα-1Δ∫01k(η,s)h+(s)(f1-f2)dsmmmmmm+∫01k(t,s)h+(s)(f1-f2)dsβtα-1Δ)-tα-1Δ-ρμ(βρΔ∫01k(η,s)h-(s)f1dsmmmmmmmmm+∫01κ(s)h-(s)f1dsβρΔ)-μ(βtα-1Δ∫01k(η,s)h-(s)f1dsmmmmmm+∫01k(t,s)h-(s)f1dsβtα-1Δ).
It is easy to show that operator S:X→X is completely continuous. Let v∈X and τ∈(0,1) such that v=τSv. That is,
(34)v(t)=τtα-1Δ-ρμ(βρΔ∫01k(η,s)h+(s)(f1-f2)dsmmmmmmmm+∫01κ(s)h+(s)(f1-f2)dsβρΔ)+τμ(βtα-1Δ∫01k(η,s)h+(s)(f1-f2)dsmmmmmm+∫01k(t,s)h+(s)(f1-f2)dsβtα-1Δ)-τtα-1Δ-ρμ(βρΔ∫01k(η,s)h-(s)f1dsmmmmmmmmm+∫01κ(s)h-(s)f1dsβρΔ)-τμ(βtα-1Δ∫01k(η,s)h-(s)f1dsmmmmmm+∫01k(t,s)h-(s)f1dsβtα-1Δ).
We claim that ∥v∥≠μδf(0)∥m∥. Suppose on the contrary that ∥v∥=μδf(0)∥m∥. Then, by (28) and (29), we get
(35)∥x¯μ+v∥≤∥x¯μ∥+∥v∥≤b,|f1-f2|≤f(0)δ-c2.
From (27), we get
(36)ω(t)f(x(a(t)))≤cm(t)f(0),t∈(0,1].
Using (34)–(36), for each t∈(0,1], we obtain that
(37)|v(t)|≤tα-1Δ-ρμ(βρΔ∫01k(η,s)h+(s)|f1-f2|dsmmmmmim+∫01κ(s)h+(s)|f1-f2|dsβρΔ)+μ(βtα-1Δ∫01k(η,s)h+(s)|f1-f2|dsmmmm+∫01k(t,s)h+(s)|f1-f2|dsβtα-1Δ)+tα-1Δ-ρμ(βρΔ∫01k(η,s)h-(s)f1dsmmmmimmm+∫01κ(s)h-(s)f1dsβρΔ)+μ(βtα-1Δ∫01k(η,s)h-(s)f1dsmmmim+∫01k(t,s)h-(s)f1dsβtα-1Δ)≤tα-1Δ-ρμf(0)δ-c2×(βρΔ∫01k(η,s)h+(s)ds+∫01κ(s)h+(s)ds)+μf(0)δ-c2(βtα-1Δ∫01k(η,s)h+(s)dsmmmmmmmmmm+∫01k(t,s)h+(s)dsβtα-1Δ)+tα-1Δ-ρμf¯(b)(βρΔ∫01k(η,s)h-(s)dsmmmmmmmmmm+∫01κ(s)h-(s)dsβρΔ)+μf¯(b)(βtα-1Δ∫01k(η,s)h-(s)dsmmmmmmm+∫01k(t,s)h-(s)dsβtα-1Δ)=μf(0)δ-c2m(t)+μf¯(b)ω(t)≤μf(0)δ-c2m(t)+μcf(0)m(t)=μf(0)δ+c2m(t).
In particular,
(38)∥v∥≤μf(0)δ+c2∥m∥<μf(0)δ∥m∥,
which is a contradiction. And so the claim is proved. Let U={x∈X:∥x∥<μδf(0)∥m∥}. By Theorem 5, S has a fixed point vμ∈U¯. Consequently, ∥vμ∥≤μδf(0)∥m∥. This proves that there exists vμ this is the solution of (30). Hence vμ satisfies (37) and Lemma 6; then we get
(39)xμ(t)≥x¯μ(t)-|vμ(t)|≥μδf(0)m(t)-μf(0)δ+c2m(t)=μf(0)δ-c2m(t)>0;
that is, xμ is a positive solution of (3). So the proof of Theorem 7 is complete.

4. An Example

In this section, we give an example to illustrate the result of this paper. Consider the following nonlinear fractional differential equation:
(40)D5/2x(t)-μ(45-t)(x(4)(t)+sin2x(t)+110)=0,x(0)=x′(0)=0,x(1)=12x(12)+∫01x(t)(3t-1)dt.

Let f(x(a(t)))=x(4)(t)+sin2x(t)+1/10,a(t)=t and h(t)=4/5-t. Obviously, all assumptions (H_{
1})–(H_{
3}) hold. In the following, we will verify that assumptions (H_{
4})–(H_{
6}) hold also.

(i) It is obvious that
(41)βηα-1+λ[p]=βηα-1+∫01tα-1(3t-1)dt=βηα-1+2α-1α(α+1)=142+1635
implies (H_{
4}).

(ii) By direct calculation, we have
(42)∫01(3t-1)dt=12,∫01tα-1(3t-1)dt=1635,κ(s)=∫01k(t,s)(3t-1)dt=1α(α+1)(1-s)α-1s(3s+2α-4)=435(1-s)3/2s(3s+1)≥0,
so assumption (H_{
5}) holds.

(iii) Finally, we check assumption (H_{
6}). It means that there exists ϵ>0 such that m(t)≥(1+ϵ)ω(t),t∈(0,1]. Note that
(43)h+(t)=max{0,h(t)}={45-t,0≤t≤45,0,45<t≤1,h-(t)=max{0,-h(t)}={0,0≤t≤45,t-45,45<t≤1.
We now verify that there exists ϵ1>0 such that
(44)tα-1Δ-ρβρΔ∫01k(η,s)h+(s)ds≥(1+ϵ1)tα-1Δ-ρβρΔ∫01k(η,s)h-(s)ds,t∈(0,1];
that is,
(45)∫01k(12,s)(45-s)ds≥ϵ1∫4/51k(12,s)(s-45)ds,mmmmmimmmt∈(0,1].
By simple calculation, we get
(46)∫01k(12,s)(45-s)ds=132700,∫4/51k(12,s)(s-45)ds=27×54×10.
Setting ϵ1∈(0,(13×52×5)/4), then inequality (44) holds. Similarly, there exists ϵ2,ϵ3,ϵ4>0 such that
(47)tα-1Δ-ρ∫01κ(s)h+(s)ds≥(1+ϵ2)tα-1Δ-ρ∫01κ(s)h-(s)ds,t∈(0,1],βtα-1Δ∫01k(η,s)h+(s)ds≥(1+ϵ3)βtα-1Δ∫01k(η,s)h-(s)ds,t∈(0,1],∫01k(t,s)h+(s)ds≥(1+ϵ4)∫01k(t,s)h-(s)ds,t∈(0,1].
Let ϵ=min{ϵ1,ϵ2,ϵ3,ϵ4}. By (44)–(47), we obtain that there exists ϵ>0 such that
(48)m(t)≥(1+ϵ)ω(t),t∈(0,1].
Thus assumption (H_{
6}) holds. By applying Theorem 7, we know that there exists a number μ*>0 such that (40) has at least one positive solution for μ∈(0,μ*).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

Zhaocai Hao acknowledges the support from NSFC (11371221) and the Education Department of Shandong Province Science and Technology Plan Project (J13LI01). The authors are grateful to the anonymous referees for their helpful suggestions and comments.

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