By employing known Guo-Krasnoselskii fixed point theorem, we investigate the eigenvalue interval for the existence and nonexistence of at least one positive solution of nonlinear fractional differential equation with integral boundary conditions.

1. Introduction

Fractional calculus has been receiving more and more attention in view of its extensive applications in the mathematical modelling coming from physical and other applied sciences; see books [1–5]. Recently, the existence of solutions (or positive solutions) of nonlinear fractional differential equation has been investigated in many papers (see [6–28] and references cited therein). However, in terms of the eigenvalue problem of fractional differential equation, there are only a few results [29–33].

To the best of author’s knowledge, no paper has considered the eigenvalue problem of the following nonlinear fractional differential equation with integral boundary conditions:
(1)CDαu(t)+λf(t,u(t))=0,0<t<1,n<α≤n+1,n≥2,n∈N.u(0)=u′′(0)=u′′′(0)=⋯=u(n)(0)=0,u(1)=ξ∫01u(s)ds,
where 0<ξ<2, CDα is the Caputo fractional derivative, and f:[0,1]×[0,∞)→[0,∞) is a continuous function.

Our proof is based upon the properties of the Green function and Guo-Krasnoselskii’s fixed point theorem given in [34]. Our purpose here is to give the eigenvalue interval for nonlinear fractional differential equation with integral boundary conditions. Moreover, according to the range of the eigenvalue λ, we establish some sufficient conditions for the existence and nonexistence of at least one positive solution of the problem (1).

2. Preliminaries

For the convenience of the readers, we first present some background materials.

Definition 1.

For a function f:[0,∞)→ℝ, the Caputo derivative of fractional order α is defined as
(2)CDαf(t)=1Γ(n-α)∫0t(t-s)n-α-1f(n)(s)ds,111111111111111111111n=[α]+1,
where [α] denotes the integer part of the real number α.

Definition 2.

The Riemann-Liouville fractional integral of order α for a function f is defined as
(3)Iαf(t)=1Γ(α)∫0t(t-s)α-1f(s)ds,α>0,
provided that such integral exists.

Lemma 3.

Let α>0; then
(4)IαCDαu(t)=u(t)+C0+C1t+C2t2+⋯+Cn-1tn-1,
for some Ci∈ℝ, i=1,2,…,n, n=[α]+1.

Lemma 4 (see [<xref ref-type="bibr" rid="B34">34</xref>]).

Let E be a Banach space, and let P⊂E be a cone. Assume that Ω1, Ω2 are open subsets of E with 0∈Ω1, Ω¯1⊂Ω2, and let T:P∩(Ω¯2∖Ω1)→P be a completely continuous operator such that

∥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).

Lemma 5.

Let n<α≤n+1, n≥2, n∈N, and ξ≠2. Assume y∈C[0,1]; then the unique solution of the problem
(5)CDαu(t)+y(t)=0,0<t<1,u(0)=u′′(0)=u′′′(0)=⋯=u(n)(0)=0,u(1)=ξ∫01u(s)ds,
is given by the expression
(6)u(t)=∫01G(t,s)y(s)ds,
where
(7)G(t,s)={2t(1-s)α-1(α-ξ+ξs)-(2-ξ)α(t-s)α-1(2-ξ)Γ(α+1),111111111111111111111111110≤s≤t≤1,2t(1-s)α-1(α-ξ+ξs)(2-ξ)Γ(α+1),0≤t≤s≤1.

Proof.

It is well known that the equation CDαu(t)+y(t)=0 can be reduced to an equivalent integral equation:
(8)u(t)=-Iαy(t)-∑i=0nbiti=-∫0t(t-s)α-1Γ(α)y(s)ds-∑i=0nbiti,
for some bi∈ℝ(i=0,1,2,…,n).

By the conditions u(0)=u′′(0)=u′′′(0)=⋯=u(n)(0)=0 and u(1)=ξ∫01u(s)ds, we can get that b0=b2=b3=⋯=bn=0 and
(9)b1=-∫01(1-s)α-1Γ(α)y(s)ds-ξ∫01u(s)ds.
Hence, we have
(10)u(t)=-∫0t(t-s)α-1Γ(α)y(s)ds+t∫01(1-s)α-1Γ(α)y(s)ds+ξt∫01u(s)ds.

Put ∫01u(s)ds=A; then, from (10), we deduce that
(11)A=∫01u(t)dt=-∫01∫0t(t-s)α-1Γ(α)y(s)dsdt+∬01t(1-s)α-1Γ(α)y(s)dsdt+∫01ξAtdt=-∫01(1-s)ααΓ(α)y(s)ds+12∫01(1-s)α-1Γ(α)y(s)ds+12ξA,
which implies that
(12)A=-22-ξ∫01(1-s)ααΓ(α)y(s)ds+12-ξ∫01(1-s)α-1Γ(α)y(s)ds.

Replacing this value in (10), we obtain the following expression for function u(t):
(13)u(t)=-∫0t(t-s)α-1Γ(α)y(s)ds+t∫01(1-s)α-1Γ(α)y(s)ds-2ξ2-ξ∫01t(1-s)ααΓ(α)y(s)ds+ξ2-ξ∫01t(1-s)α-1Γ(α)y(s)ds=-∫0t(t-s)α-1Γ(α)y(s)ds+∫012t(1-s)α-1(α-ξ+ξs)(2-ξ)αΓ(α)y(s)ds=∫0t(((2-ξ)Γ(α+1))-1(2t(1-s)α-1(α-ξ+ξs)-(2-ξ)α(t-s)α-1)×((2-ξ)Γ(α+1))-1y(s)ds)+∫t12t(1-s)α-1(α-ξ+ξs)(2-ξ)Γ(α+1)y(s)ds=∫01G(t,s)y(s)ds.
This completes the proof.

Lemma 6.

Let G be the Green function, which is given by the expression (7). For 0<λ<2, the following property holds:
(14)tG(1,s)≤G(t,s)≤2αξ(α-2)G(1,s),∀t,s∈(0,1).

The proof is similar to that of Lemma 2.4 in [7], so we omit it.

Consider the Banach space X=C[0,1] with general norm
(15)∥u∥=supt∈[0,1]|u(t)|.
Define the cone P={u∈X:u(t)≥(ξ(α-1)/2α)t∥u∥}.

Suppose u is a solution of (1). It is clear from Lemma 5 that
(16)u(t)=λ∫01G(t,s)f(s,u(s))ds,∀t∈[0,1].

Define the operator Sλ:P→X as follows:
(17)(Sλu)(t)=λ∫01G(t,s)f(s,u(s))ds,∀t∈[0,1].

Lemma 7.

Sλ:P→P is completely continuous.

Proof.

Since 0<ξ<2, it is obvious that G(t,s)≥0. So we have
(18)∥Sλu∥=supt∈[0,1]λ∫01G(t,s)f(s,u(s))ds≤λ∫012αξ(α-2)G(1,s)f(s,u(s))ds,(Sλu)(t)=λ∫01G(t,s)f(s,u(s))ds≥ξ(α-2)2αtλ∫012αξ(α-2)G(1,s)f(s,u(s))ds≥ξ(α-2)2αt∥Sλu∥.
Therefore, Sλ(P)⊂P. The other proof is similar to that in [7], so we omit it.

3. Main Result

For convenience, we list the denotation:
(19)F0=limu→0+supt∈[0,1]f(t,u(t))u,F∞=limu→+∞supt∈[0,1]f(t,u(t))u,f0=limu→0+inft∈[0,1]f(t,u(t))u,f∞=limu→+∞inft∈[0,1]f(t,u(t))u.

Next, we will establish some sufficient conditions for the existence and nonexistence of positive solution for problem (1).

Theorem 8.

Let l∈(0,1) be a constant. Then for each
(20)λ∈((ξ(α-2)lf∞2α∫01sG(1,s)ds)-1,(2αF0ξ(α-2)∫01G(1,s)ds)-1),
problem (1) has at least one positive solution.

Proof.

First, for any ε>0, from (20) we have
(21)(ξ(α-2)l(f∞-ε)2α∫01sG(1,s)ds)-1≤λ≤(2α(F0+ε)ξ(α-2)∫01G(1,s)ds)-1.

On the one hand, by the definition of F0, there exists r1>0 such that, for any u∈[0,r1], we have
(22)f(t,u)≤(F0+ε)u.
Choose Ω1={u∈X:∥u∥≤r1}. For u∈P∩∂Ω1, we have
(23)∥Sλu∥=supt∈[0,1]λ∫01G(t,s)f(s,u(s))ds≤λ∫012αξ(α-2)G(1,s)(F0+ε)u(s)ds≤λ2α(F0+ε)ξ(α-2)∫01G(1,s)ds∥u∥≤∥u∥.

On the other hand, by the definition of F∞, there exists r2>r1 such that, for any u∈[r2,+∞), we have
(24)f(t,u)≥(f∞-ε)u.
Take Ω2={u∈X:∥u∥≤r2}. For u∈P∩∂Ω2, we have
(25)∥Sλu∥≥(Sλu)(l)≥λ∫01lG(1,s)(f∞-ε)u(s)ds≥λlξ(α-2)f∞2α∫01sG(1,s)ds∥u∥≥∥u∥.
According to (23), (25), and Lemma 4, Sλ has at least one fixed point u∈P∩(Ω¯2∖Ω1) with r1≤∥u∥≤r2, which is a positive solution of (1).

Remark 9.

If F0=0 and f∞=∞, then we can get
(26)2αF0ξ(α-2)∫01G(1,s)ds=0,ξ(α-2)lf∞2α∫01sG(1,s)ds=+∞.
Theorem 8 implies that, for λ∈(0,+∞), problem (1) has at least one positive solution.

Theorem 10.

Let l∈(0,1) be a constant. Then for each
(27)λ∈((ξ(α-2)lf02α∫01sG(1,s)ds)-1,(2αF∞ξ(α-2)∫01G(1,s)ds)-1),
problem (1) has at least one positive solution.

Proof.

First, it follows from (27) that, for any ε>0,
(28)(ξ(α-2)l(f0-ε)2α∫01sG(1,s)ds)-1≤λ≤(2α(F∞+ε)ξ(α-2)∫01G(1,s)ds)-1.

By the definition of f0, there exists r1>0 such that, for any u∈[0,r1], we have
(29)f(t,u)≥(f0+ε)u.
Choose Ω1={u∈X:∥u∥≤r1}. For u∈P∩∂Ω1, we have ∥u∥=r1. Similar to the proof in Theorem 8, it holds from (28) and (29) that
(30)∥Sλu∥≥(Sλu)(l)≥λlξ(α-2)f02α∫01sG(1,s)ds∥u∥≥∥u∥.

Note F∞=limu→+∞supt∈[0,1]f(t,u(t))/u. There exists r3>r1, such that
(31)f(t,u)≤(F∞+ε)u,u∈(r3,+∞).
We consider the problem on two cases. (I) Suppose f is bounded. There exists M>0, such that f(t,u(t))≤M, ∀u∈(r3,+∞). Choose r4=max{r3,Mλ(2α/ξ(α-2))∫01G(1,s)ds}. Let Ω2′={u∈X:∥u∥≤r4}. For u∈P∩∂Ω2′, we have
(32)∥Sλu∥=supt∈[0,1]λ∫01G(t,s)f(s,u(s))ds≤λ∫012αξ(α-2)G(1,s)f(s,u(s))ds≤λM2αξ(α-2)∫01G(1,s)ds∥u∥≤r4=∥u∥.

(II) Suppose f is unbounded. There exists r5>r3 such that
(33)f(t,u(t))≤u,u∈(r5,+∞).

Let Ω2′′={u∈X:∥u∥≤r5}. For u∈P∩∂Ω2′′, we have
(34)∥Sλu∥≤λ∫012αξ(α-2)G(1,s)f(s,u(s))ds≤λ2α(F∞+ε)ξ(α-2)∫01G(1,s)ds∥u∥≤∥u∥.
Combining (I) and (II), take Ω2={u∈X:∥u∥≤r2}; here, r2≥max{r4,r5}. Then for u∈P∩∂Ω2, we have
(35)∥Sλu∥≤∥u∥.

Hence, (30) and (42) together with Lemma 4 imply that Sλ has at least one fixed point u∈P∩(Ω¯2∖Ω1) with r1≤∥u∥≤r2, which is a positive solution of (1).

Theorem 11.

Assume F0<+∞ and F∞<+∞. Problem (1) has no positive solution provided
(36)λ<(2αkξ(α-2)∫01G(1,s)ds)-1,
where k is a constant defined in (38).

Proof.

Since F0<+∞ and F∞<+∞, together with the definitions of F0 and F∞, there exist positive constants k1, k2, r1, and r2 satisfying r1<r2 such that
(37)f(t,u)≤k1u,u∈[0,r1],f(t,u)≤k2u,u∈[r2,+∞].
Take
(38)k=max{k1,k2,sup(t,u)∈(0,1)×(k1,k2)f(t,u)u}.

It follows that f(t,u)≤ku for any u∈(0,+∞). Suppose that v(t) is a positive solution of (1). That is,
(39)(Sλv)(t)=v(t),∀t∈J.
In sequence,
(40)∥v∥=∥Sλv∥=supt∈[0,1]λ∫01G(t,s)f(s,v(s))ds≤λ∫012αξ(α-2)G(1,s)f(s,v(s))ds≤λk2αξ(α-2)∫01G(1,s)ds∥v∥<∥v∥,
which is a contradiction. Hence, (1) has no positive solution.

Theorem 12.

Assume f0>0 and f∞>0. Problem (1) has no positive solution provided
(41)λ>(ξk(α-2)2α∫01s2G(1,s)ds)-1,
where k is a constant defined in (43).

Proof.

Since f0>0 and f∞>0, together with the definitions of f0 and f∞, there exist positive constants k1, k2, r1, and r2 satisfying r1<r2 such that
(42)f(t,u)≥k1u,u∈[0,r1],f(t,u)≥k2u,u∈[r2,+∞].
Take
(43)k=min{k1,k2,inf(t,u)∈(0,1)×(k1,k2)f(t,u)u}.
It follows that f(t,u)≥ku for any u∈(0,+∞). Suppose that v(t) is a positive solution of (1). That is,
(44)(Sλv)(t)=v(t),∀t∈J.
In sequence,
(45)∥v∥≥λ∫01sG(1,s)f(s,v(s))ds≥λkξ(α-2)2α∫01s2G(1,s)ds∥v∥>∥v∥,
which is a contradiction. Hence, (1) has no positive solution.

Example 13.

Consider the fractional differential equation
(46)CD5/2u(t)+λf(t,u(t))=0,0<t<1,u(0)=u′′(0)=0,u(1)=∫01u(s)ds.
In this example, take
(47)f(t,u(t))=(500u2+u)(7-t2)u+7.
Obviously, we have
(48)F0=limu→0+supt∈[0,1](500u2+u)(7-t2)u(u+7)=1,f∞=limu→+∞inft∈[0,1](500u2+u)(7-t2)u(u+7)=3000.

Since α=5/2 and ξ=1, through a computation, we can get
(49)∫01G(1,s)ds=∫012t(1-s)α-1(α-ξ+ξs)-(2-ξ)α(t-s)α-1(2-ξ)Γ(α+1)ds=∫012(1-s)3/2(3/2+s)-(5/2)(1-s)3/2Γ(7/2)ds≤1Γ(7/2),∫01sG(1,s)ds=∫012s(1-s)3/2(3/2+s)-(5/2)s(1-s)3/2Γ(7/2)ds=∫01s(1-s)3/22Γ(7/2)ds≥235Γ(7/2).

Choose l=2/3; we have
(50)(ξ(α-2)lf∞2α∫01sG(1,s)ds)-1≤7Γ(7/2)80<Γ(7/2)10≤(2αF0ξ(α-2)∫01G(1,s)ds)-1.
Theorem 8 implies that, for λ∈(7Γ(7/2)/80, Γ(7/2)/10), the problem (46) has at least one positive solution.

Remark 14.

In particular, if we take f(t,u(t))=u2(1+t) in Example 13, then F0=0 and f∞=∞. Remark 9 implies that problem (46) has at least one positive solution for λ∈(0,+∞).

Conflict of Interests

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

Acknowledgments

This work is supported by the NNSF of China (no. 61373174) and the Natural Science Foundation for Young Scientists of Shanxi Province, China (no. 2012021002-3).

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