Abstract

We investigate positive solution for high-order fractional differential equations with resonant boundary value conditions. By using the spectral theory and the fixed point index theory, both the nonexistence and existence results of the resonant boundary value problems are given.

1. Introduction

We study the high-order fractional boundary value problem (FBVP) at resonance:where n > 3, 0 < ξ1 < ⋯ < ξm < 1, and ηi > 0 (i = 1, …, m) satisfying and denotes the Riemann–Liouville derivative. It is clear that λ = 0 and , solve the linear problem:which implies that FBVP (1) happens to be at resonance.

By virtue of the widespread applications, various differential equations have been studied by many researchers (see [113] and the references therein). Fractional-order models can describe many processes more accurately than integer-order models, and a great deal of papers focusing on FBVPs appeared in recent years (see [1423]). The nonlocal FBVPs have especially drawn much attention (see [2431]). For instance, in [14], the authors investigated the Dirichlet-type FBVP:

By using the contraction mapping principle, Cui et al. established the uniqueness results of solution to FBVP (3).

Zhang and Zhong [24, 25] considered the following FBVP:

In [24], the authors established multiplicity results of positive solutions to (4) by using fixed point theorems. In [25], Zhang and Zhong considered uniqueness of solution for FBVP (4) by using the contraction mapping principle.

While there are many papers focusing on nonlocal problems, papers on resonant boundary value problems (RBVPs) are relatively scarce ([3237]). It should be noted that nothing but positive solution is meaningful in some practical problems. The main tool used to consider positive solution for RBVPs is the Leggett–Williams theorem for coincidence (see [35, 37]).

Motivated by the above work, we focus on deducing some necessary conditions and sufficient conditions of the existence of positive solutions to the RBVP (1). This article has some new features compared with some existing papers. Firstly, the main tools used in this article are the spectral theory and fixed point index theory which is different from [35, 37]. Secondly, some necessary conditions to RBVP (1) are established by virtue of spectral theory. Finally, we deduce some sufficient conditions concerning the behavior of the nonlinearity f at 0 and at ∞.

2. Basic Definitions and Preliminaries

Let

Clearly, has a unique positive root written as σ, that is,

For convenience, we use the notations:where

We assume that the following assumption holds throughout this paper:(H0) f is continuous on [0, 1] × [0, +∞)We also list here some hypotheses to be used later:(H1) f(t, ctα−1)  0, ∀c > 0(H2) f(t, z) ≥ −σz, ∀ (t, z) ∈ [0, 1] × [0, +∞)(H3) f < 0 < f0(H4) f0 < 0 < f

Lemma 1. Let ϕ ∈ C(0, 1) ∩ L1[0, 1]. Then, the FBVPhas a unique solution:

Proof. First, it follows from ηi > 0, 0 < ξi < 1, and thatBy [38], we can write the solution of (9) asThe conditionsyields thatTherefore, we haveIt follows fromthatThen, the solution of (9) is

Lemma 2. G(t, s) satisfies the following:(1)G(1 − s, 1 − t) = G(t, s), ∀s, t ∈ [0, 1].(2)G(t, s) ≥ Λ1s(1 − s)α−1(1 − t)tα−1, ∀s, t ∈ [0, 1].(3)G(t, s) ≤ Λ2s(1 − s)α−1, ∀s, t ∈ [0, 1], wherein which, s0 ∈ (0, 1) satisfies .

Proof. Since (1) is trivially true, we only need to verify (2) and (3). When 1 ≥ t > s ≥ 0, it is easy to getTherefore,which implies thatThen, we haveWhen 1 ≥ s ≥ t ≥ 0, we can getIt follows from (23) and (24) that (2) holds.
In the sequel, we prove that (3) holds.
When 1 ≥ s ≥ t ≥ 0, we can getWhen 1 ≥ t > s ≥ 0, we haveNotice that h′(t) is nondecreasing on [0, 1], from which we can getTherefore,Let s0 be the unique root of the equation s = (1 − s)α−2 on [0, 1]. Then,By direct calculation, we getTherefore, (26)–(30) yield thatThen, it follows from (25) and (31) that (3) holds.

Corollary 1. It follows from (1) and (3) that

Lemma 3. Green’s function K(t, s) satisfies the following:(1)K(t, s) ≥ Λ3s(1 − s)α−1tα−1, ∀s, t ∈ [0, 1].(2)K(t, s) ≤ Λ4s(1 − s)α−1, ∀s, t ∈ [0, 1].(3)K(t, s) ≤ Λ4tα−1, ∀s, t ∈ [0, 1], where

Proof. First, it follows from (2) of Lemma 2 thatthat is, (1) holds.
By (3) of Lemma 2, we haveSo, (2) holds.
It follows from Corollary 1 thatTherefore,So, (3) holds.

Let E = C[0, 1] with , θ is the zero element, and Br = {x ∈ E: ‖x‖ < r}. Define two cones P and Q by

Clearly, RBVP (1) is equivalent to the auxiliary FBVP:

Denote

Then, the fixed point x of A is a solution of RBVP (1). It is clear that L: P ⟶ Q is continuous and the first eigenvalue of L is λ1 = σ. Moreover, ψ(t) = tα−1 is a eigenfunction corresponding to λ1, that is,

Set

By virtue of [13], one has the sequence {r(Ln)} that is nondecreasing and

Lemma 4 (see [39]). Let E be a Banach space, QE be a cone, Ω ⊂ E be a bounded open set, and A: be a completely continuous operator. Assume that ∃x0 ∈ Q\{θ} such thatThen, i(A, Ω ∩ Q, Q) = 0.

Lemma 5 (see [39]). Let E be a Banach space, QE be a cone, Ω ⊂ E be a bounded open set, θ ∈ Ω, and A: be a completely continuous operator. Assume that and then, i(A, Ω ∩ Q, Q) = 1.

3. Main Results

Theorem 1. Assume that f ≥ 0 and (H1) holds. RBVP (1) cannot have any positive solution.

Proof. We will use the method of contradiction to prove this theorem. Suppose that RBVP (1) has a positive solution x, thenFirst, we will show thatIf otherwise, we havewhich implies f(t, x(t)) ≡ 0 on [0, 1]. Then, (1) yields that x satisfiesIt is clear that the solution of the above equation is ctα−1. Then, f(t, ctα−1) ≡ 0, which contradicts with (H1). So, (47) holds.
By Lemma 3, we can get that ∃ ϱ1 >ϱ2 > 0, s.t.that is,Therefore,By induction, we havewhich impliesThis is impossible since

Theorem 2. Assume that f ≤ 0 and (H1) holds. RBVP (1) cannot have any positive solution.

Proof. Suppose that RBVP (1) has a positive solution x, then we haveIt follows fromsuch thatNoticing f ≤ 0, we can getThis and (58) yieldwhich implies c1 ≥ 0.
Next, we will prove that c1 > 0. If otherwise, it follows from (58) thatBy considering and , we havewhich implies f(t, x(t)) ≡ 0. Thus, (61) yields x = θ, which contradicts with the hypothesis.
From (58), there exists c2 > c1 such thatThen,By induction, we havewhich impliesIt is easy to get that f(t, x(t))  0 (If otherwise, (58) yields x(t) = c1tα−1. Then, , which implies f(t, c1tα−1) ≡ 0, which contradicts with (H1)). Therefore, we haveLet n ⟶ ∞, and (66) yields x = θ, which contradicts with the hypothesis.

Corollary 2. Assume that (H1) holds. The necessary condition for RBVP (1) to have positive solutions is that f admits changing sign.

Theorem 3. Assume that (H2) and (H3) hold. RBVP (1) has a positive solution.

Proof. First, by virtue of the Arzela–Ascoli theorem and Lemma 3, we have that A: Q ⟶ Q is completely continuous. In the sequel, we will prove that A has a positive fixed point.
By f0 > 0, there exists r1 > 0, s.t.Suppose the operator A has no fixed point on . Next, we will showIf otherwise, there exist λ0 > 0 and , s.t.Thus,DenoteThen,  ≥ λ0 and x1 ≥ ψ. Noticing Ax1 ≥ σLx1, we haveThis contradicts with the notation of . Thus, (69) holds. Then, it follows from Lemma 4 thatBy f < 0, there exist ɛ ∈ (0, 1) and r2 > r1, s.t.DenoteNext, we will show S is bounded.
For any x ∈ S, let . Then, we haveClearly, we haveSet T1 = σ(1 −ɛ)L. Then,Here,Therefore,Clearly, we have r(T1) = 1 − ɛ. So, the inverse operator of I − T1 exists. Thus, (81) yieldswhich implies S is bounded.
Let . Then, it follows from Lemma 5 thatBy (74) and (83), we haveThis means A has a fixed point on . So, RBVP (1) has a positive solution.

Theorem 4. Assume that (H2) and (H4) hold. RBVP (1) has a positive solution.

Proof. It follows from f0 < 0 that ∃R1 > 0, s.t.Set T2 = σL. Clearly, we have r(T2) = 1.
Suppose the operator A has no fixed point on . Next, we will showIf otherwise, there exist λ0 > 1 and , s.t. Ax1 = λ0x1. Noticing thatwe haveBy induction, we havewhich impliesThis contradicts with r(T2) = 1. So, (86) holds. By Lemma 5, we getFrom f > 0, we know that there exist δ > 0 and R2 > R1, s.t.Since {r(Ln)} is nondecreasing and converges to σ−1, one can choose m satisfiesIt follows from Lemma 3 and the Krein–Rutman theorem that there exists ψm satisfiesLetFor any , we haveThen,Suppose the operator A has no fixed point on . In the following, we will showIf otherwise, then there exist λ0 > 0 and such thatDenoteThen,  ≥ λ0 and x1 ≥ ψm. Noticing (97), we havewhich contradicts with the notation of . Hence, (98) holds. By virtue of Lemma 4, one hasFrom (91) and (102), one hasTherefore, RBVP (1) has a positive solution on .

4. Examples

Example 1. Let us consider the existence of positive solutions for the RBVP:withIt is clear that (H0) and (H1) hold. Then, Theorem 1 yields that RBVP (104) has no positive solution.

Example 2. Consider the following RBVP:withBy direct calculation, we have , f = +∞. Thus, (H4) holds. Moreover,which implies . Therefore, (H2) holds. It follows from Theorem 4 that RBVP (106) has a positive solution.

5. Conclusions

In this article, we focus on the existence results of positive solutions to high-order FBVPs at resonance. The main tool used in this article is different from most of articles dealing with positive solution for resonant problems. First, we rewrite the RBVPs as equivalent nonresonant FBVPs. By means of the spectral theory, we obtain the necessary conditions for RBVP (1). By means of the fixed point index theory, we establish some sufficient conditions for RBVP (1) concerning the behavior of the nonlinearity f at 0 and at ∞.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

This work was supported by the Natural Science Foundation of Shandong Province of China (ZR2017MA036), the National Natural Science Foundation of China (11871302), the Project of Shandong Province Higher Educational Science and Technology Program (J18KA217), and the International Cooperation Program of Key Professors by Qufu Normal University.