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.
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: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: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 Section 2, 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 order α>0 of a function u:(0,+∞)→R is given by
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 order α>0 of a function u:(0,+∞)→R is given by
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]).
Given y(t)∈C(0,1)∩L(0,1), the unique solution of the problem
cD0+αu(t)+y(t)=0,0<t<1,u(0)=u′(1)=u′′(0)=0
is
u(t)=∫01G(t,s)y(s)ds,
where
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 function G(t,s) has the following properties:
G(t,s)≤(1/Γ(α-1))t(1-s)α-2, fort,s∈[0,1],
G(t,s)≤(1/Γ(α-1))(α-2+s)(1-s)α-2, for t,s∈[0,1],
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) When 0≤s≤t≤1, as 2<α≤3, we have
∂G(t,s)∂t=(1-s)α-2-(t-s)α-2Γ(α-1)≥0,
therefore
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, since 0<α-2≤1, we have
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) When 0≤t≤s≤1, we have
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, as α-1≥α-2+s for 0≤s≤1, we have
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 Lemma 2.4, there exists K>0 such that the positive solution u in [14] satisfies
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
Au(t)=∫01G(t,s)f(s,u(s))ds.
For any u∈Q, by (1) of Lemma 2.4, we have
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
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)∥.
If u is a positive solution of (P), then u is a fixed point of A in Q, therefore
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
-r(t)≤f(t,x)≤z(t)g(x),
where r,z∈C((0,1),[0,+∞)), g∈C([0,+∞),[0,+∞)).
(H3) There exists [a,b]⊂(0,1) such that
liminfu→+∞mint∈[a,b]f(t,u)u=+∞.
(H4) There exists [c,d]⊂(0,1) such that
liminfu→+∞mint∈[c,d]f(t,u)>2(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds,limu→+∞g(u)u=0.
Lemma 2.6.
Assume that (H1) and (H2) hold, then the boundary value problem
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
ω(t)≤t∫011Γ(α-1)(1-s)α-2r(s)ds,t∈[0,1].
Proof.
By Lemma 2.3, 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
ω(t)=∫01G(t,s)r(s)ds≤t∫011Γ(α-1)(1-s)α-2r(s)ds.
The proof is completed.
Let E=C[0,1] be endowed with the maximum norm ∥u∥=max0≤t≤1|u(t)|. Define a cone P byP={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: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}.
LetTu(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 any u∈P, Lemma 2.4 implies that
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
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 that T:P→P is a completely continuous operator. The proof is completed.
Lemma 2.8 (see [18]).
Let E 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
∥Tu∥≤∥u∥,u∈P∩∂Ω1 and ∥Tu∥≥∥u∥,u∈P∩∂Ω2, or
∥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 SolutionsTheorem 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.
Choose r1>((α-1)/Γ(α-1))∫01(1-s)α-2r(s)ds. Let
λ*=min{1,r1Γ(α-1)[g*(r1)+1]∫01(α-2+s)(1-s)α-2(z(s)+r(s))ds},
where
g*(r)=maxx∈[0,r]g(x).
In the rest of the proof, we suppose λ∈(0,λ*).
For any u∈∂Pr1, noting that
u(t)≥tα-1r1,t∈[0,1]
and using (2.20), we have
0≤t[r1α-1-∫011Γ(α-1)(1-s)α-2r(s)ds]≤u(t)-λω(t)≤r1.
Therefore,
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,
‖Tu‖≤‖u‖,∀u∈∂Pr1.
Now choose a real number
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
f(t,x)>Lx.
Select
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
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
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
u(t)-λω(t)>ar24>N,t∈[a,b].
Hence we get
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,
‖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
f(t,u)≥2(α-1)2∫01(1-s)α-2r(s)ds∫cd(α-2+s)(1-s)α-2ds.
Select
λ*=NΓ(α-1)c∫01(1-s)α-2r(s)ds.
In the rest of the proof, we suppose λ>λ*.
Let
R1=2λ(α-1)∫01(1-s)α-2r(s)dsΓ(α-1).
Then, for any u∈∂PR1, we have
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,
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
‖Tu‖≥‖u‖,∀u∈∂PR1.
On the other hand, as g(t) is continuous on [0,+∞), from the second limit of (H4), we have
limu→+∞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
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
g(u)≤g*(x)≤εx.
Select
R2≥max{X0,R1,2λΓ(α-1)∫01(α-2+s)(1-s)α-2r(s)ds}.
Then, for any u∈∂PR2, we get
‖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,
‖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 Theorem 3.2 is valid if (H4) is replaced by (H4*): there exists [c,d]⊂(0,1) such that
liminfu→+∞mint∈[c,d]f(t,u)=+∞;limu→+∞g(u)u=0.
4. ExamplesExample 4.1.
Consider the following problem
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
liminfu→+∞mint∈[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
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
liminfu→+∞mint∈[1/4,3/4]f(t,u)=+∞;limu→+∞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.
Acknowledgments
The first and second authors were supported financially by the National Natural Science Foundation of China (11071141, 11101237) and the Natural Science Foundation of Shandong Province of China (ZR2011AQ008, ZR2011AL018). The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant.
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