# Existence and Nonexistence of Positive Solutions for High-Order Fractional Differential Equation Boundary Value Problems at Resonance

Authors: Yongqing Wang; Huiqing Wang
Journal: Journal of Function Spaces
 (2020)

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## 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.


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## Body

## 1. Introduction

We study the high-order fractional boundary value problem (FBVP) at resonance:(1)−D0+αxt=ft,xt,n−1<α<n,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi,where n > 3, 0 < ξ1 < ⋯ < ξm < 1, and ηi > 0 (i = 1, …, m) satisfying ∑i=1mηiξiα−1=1 and D0+α denotes the Riemann–Liouville derivative. It is clear that λ = 0 and ctα−1,c∈ℝ, solve the linear problem:(2)−D0+αxt=λxt,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi,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 [1–13] 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 [14–23]). The nonlocal FBVPs have especially drawn much attention (see [24–31]). For instance, in [14], the authors investigated the Dirichlet-type FBVP:(3)−D0+αxt=ft,xt,1<α≤2,0<t<1,x0=x1=0.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:(4)−D0+αxt=ft,xt,n−1<α<n,0<t<1,x0=x′0=⋯=xn−20=0,D0+βx1=∫0ηatD0+γxtdt.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 ([32–37]). 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(5)gz=∑i=0+∞i+1α+1−ni+1α−nziΓi+1α.Clearly,gz has a unique positive root written as σ, that is,(6)gσ=0.For convenience, we use the notations:(7)f0=liminfz⟶0+mint∈0,1ft,zz,f∞=liminfz⟶+∞mint∈0,1ft,zz,f0=limsupz⟶0+maxt∈0,1ft,zz,f∞=limsupz⟶+∞maxt∈0,1ft,zz,ht=tα−1Eα,ασtα,h1t=tα+1−nEα,ασtα,Gt,s=1h1hth1−s,0≤t≤s≤1,hth1−s−h1−sh1,0≤s≤t≤1,Kt,s=Gt,s+∑i=1mηiGξi,shth1−∑i=1mηihξi,where(8)Eα,αz=∑i=0+∞ziΓi+1α.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.
Letϕ ∈ C(0, 1) ∩ L1[0, 1]. Then, the FBVP(9)−D0+αxt+σxt=ϕt,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi,has a unique solution:(10)xt=∫01Kt,sϕsds.Proof.
First, it follows fromηi > 0, 0 < ξi < 1, and ∑i=1m−2ηiξiα−1=1 that(11)h1−∑i=1mηihξi=∑j=1+∞σj1−∑i=1m−2ηiξij+1α−1Γj+1α>0.
By [38], we can write the solution of (9) as(12)xt=−∫0tht−sϕsds+c1ht+c2h′t+⋯+cnhn−1t.
The conditions(13)x0=x′0=x″0=⋯=xn−20=0yields that(14)c2=c3=⋯=cn=0.
Therefore, we have(15)x1=c1h1−∫01h1−sϕsds,xξi=c1hξi−∫0ξihξi−sϕsds.
It follows from(16)x1=∑i=1mηixξithat(17)c1=∫01h1−sϕsds−∑i=1mηi∫0ξihξi−sϕsdsh1−∑i=1mηihξi.
Then, the solution of (9) is(18)xt=−∫0tht−sϕsds+∫01h1−sϕsds−∑i=1mηi∫0ξihξi−sϕsdsh1−∑i=1mηihξiht=∫01hth1−sϕsds−∫0th1ht−sϕsdsh1−∫01h1−sϕsdsh1ht+∫01h1−sϕsdsh1−∑i=1mηihξiht−∑i=1mηi∫0ξihξi−sϕsdsh1−∑i=1mηihξiht=∫01Gt,sϕsds+∑i=1mηi∫01hξih1−sϕsds−∫0ξih1hξi−sϕsdsh1h1−∑i=1mηihξi=∫01Kt,sϕsds.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], where(19)Λ1=1h1Γα2,Λ2=h′12h1s0,in which, s0 ∈ (0, 1) satisfies s0=1−s0α−2.Proof.
Since (1) is trivially true, we only need to verify (2) and (3). When 1 ≥t > s ≥ 0, it is easy to get(20)h1″t=tα−1−ngσtα≤gσ=0.
Therefore,(21)∂∂sh1th11−s−h1t−sh11=h1′t−sh11−h1th1′1−s≥0,which implies that(22)h1th11−s≥h1t−sh11.
Then, we have(23)h1Gt,s=tn−21−sn−2h1th11−s−t−sn−2h1t−sh11≥h1th11−stn−21−sn−2−t−sn−2≥h1th11−stn−31−sn−3t1−s−t−s≥tα+1−n1−sα+1−nΓα2tn−31−sn−3s1−t=s1−ttα−21−sα−2Γα2≥s1−ttα−11−sα−1Γα2.
When 1 ≥s ≥ t ≥ 0, we can get(24)Gt,s=hth1−sh1≥tα−11−sα−1h1Γα2≥s1−ttα−11−sα−1h1Γα2.
It follows from (23) and (24) that (2) holds.
In the sequel, we prove that (3) holds.
When 1 ≥s ≥ t ≥ 0, we can get(25)Gt,s=hth1−sh1≤h1tα−11−sα−1≤h1s1−sα−1.
When 1 ≥t > s ≥ 0, we have(26)Gt,s≤hth1−sh1≤h1tα−11−sα−1≤h11−sα−1.
Notice thath′(t) is nondecreasing on [0, 1], from which we can get(27)∂∂sGt,s≤h′1−sh1−hth1.
Therefore,(28)Gt,s=∫0s∂∂τGt,τdτ≤∫0sh′−τh1−hth1dτ=h1−h1−sh1−hth1≤h′12s1−sh1.
Lets0 be the unique root of the equation s = (1 − s)α−2 on [0, 1]. Then,(29)mins,1−sα−2≤s1−sα−2s0,s∈0,1.
By direct calculation, we get(30)h1<h′1.
Therefore, (26)–(30) yield that(31)Gt,s≤minh11−sα−1,h′12s1−sh1≤min1−sα−2,s×h′121−sh1≤h′12s1−sα−1h1s0.
Then, it follows from (25) and (31) that (3) holds.Corollary 1.
It follows from (1) and (3) that(32)Gt,s≤Λ21−ttα−1,∀s,t∈0,1.Lemma 3.
Green’s functionK(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(33)Λ3=Λ11−∑i=1mηiξiαΓαh1−∑i=1mηihξi,Λ4=Λ21+h1∑i=1mηih1−∑i=1mηihξi.Proof.
First, it follows from (2) of Lemma2 that(34)Kt,s≥∑i=1mηiGξi,shth1−∑i=1mηihξi≥Λ1s1−sα−1∑i=1mηi1−ξiξiα−1h1−∑i=1mηihξi×ht≥Λ11−∑i=1mηiξiαh1−∑i=1mηihξi×s1−sα−1×tα−1Γα=Λ3s1−sα−1tα−1,that is, (1) holds.
By (3) of Lemma2, we have(35)Kt,s=Gt,s+∑i=1mηiGξi,shth1−∑i=1mηihξi≤Λ2s1−sα−11+∑i=1mηih1h1−∑i=1mηihξi=Λ4s1−sα−1.
So, (2) holds.
It follows from Corollary1 that(36)Gt,s≤Λ21−ttα−1≤Λ2tα−1.
Therefore,(37)Kt,s≤Gt,s+∑i=1mηiGξi,sh1tα−1h1−∑i=1mηihξi≤Λ2tα−11+∑i=1mηiξiα−1h1h1−∑i=1mηihξi≤Λ2tα−11+h1∑i=1mηih1−∑i=1mηihξi=Λ4tα−1.
So, (3) holds.LetE = C[0, 1] with x=max0≤t≤1xt, θ is the zero element, and Br = {x ∈ E: ‖x‖ < r}. Define two cones P and Q by(38)P=x∈E:xt≥0,0<t<1,Q=x∈E:∃lx>0,s.t.,lxtα−1≥xt≥Λ3xΛ4tα−1,0<t<1.Clearly, RBVP (1) is equivalent to the auxiliary FBVP:(39)−D0+αxt+σxt=ft,xt+σxt,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi.Denote(40)Axt=∫01Kt,sfs,xs+σxsds,Lxt=∫01Kt,sxsds.Then, the fixed pointx 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,(41)ψ=σLψ.Set(42)Lnxt=∫1/n1Kt,sxsds,n=1,2,3,….By virtue of [13], one has the sequence {r(Ln)} that is nondecreasing and(43)limn⟶+∞rLn=rL=σ−1.Lemma 4 (see [39]).
LetE be a Banach space, Q ⊂ E be a cone, Ω ⊂ E be a bounded open set, and A: Ω¯∩Q⟶Q be a completely continuous operator. Assume that ∃x0 ∈ Q\{θ} such that(44)x−Ax≠λx0,∀λ≥0,x∈∂Ω∩Q.Then, i(A, Ω ∩ Q, Q) = 0.Lemma 5 (see [39]).
LetE be a Banach space, Q ⊂ E be a cone, Ω ⊂ E be a bounded open set, θ ∈ Ω, and A: Ω¯∩Q⟶Q be a completely continuous operator. Assume that(45)Ax≠λx,∀λ≥1,x∈∂Ω∩Q, and then, i(A, Ω ∩ Q, Q) = 1.


## 3. Main Results

Theorem 1.
Assume thatf ≥ 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, then(46)x=Ax=∫01Kt,sfs,xs+σxsds.
First, we will show that(47)∫01s1−sα−1fs,xsds>0.
If otherwise, we have(48)∫01s1−sα−1fs,xsds=0,which implies f(t, x(t)) ≡ 0 on [0, 1]. Then, (1) yields that x satisfies(49)D0+αxt=0,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi.
It is clear that the solution of the above equation isctα−1. Then, f(t, ctα−1) ≡ 0, which contradicts with (H1). So, (47) holds.
By Lemma3, we can get that ∃ ϱ1>ϱ2 > 0, s.t.(50)ϱ1tα−1≥xt≥ϱ2tα−1,that is,(51)ϱ1ψ≥x≥ϱ2ψ.
Therefore,(52)x=∫01Kt,sfs,xsds+σLx≥Λ3tα−1∫01s1−sα−1fs,xsds+σLx=σΛ3∫01s1−sα−1fs,xsdsLψ+σLx≥σΛ3∫01s1−sα−1fs,xsdsLxϱ1+σLx=σΛ3∫01s1−sα−1fs,xsdsϱ1+σLx.
By induction, we have(53)x≥σΛ3∫01s1−sα−1fs,xsdsϱ1+σnLnx,n=1,2,3,…,which implies(54)x≥σΛ3∫01s1−sα−1fs,xsdsϱ1+σnLnϱ2ψ=Λ3∫01s1−sα−1fs,xsdsϱ1+1nϱ2ψ.
This is impossible since(55)Λ3∫01s1−sα−1fs,xsdsϱ1+1n⟶+∞,n⟶∞.Theorem 2.
Assume thatf ≤ 0 and (H1) holds. RBVP (1) cannot have any positive solution.Proof.
Suppose that RBVP (1) has a positive solution x, then we have(56)D0+αxt=−ft,xt,0<t<1,x0=x′0=x″0=⋯=xn−20=0,x1=∑i=1mηixξi.
It follows from(57)D0+αxt=−ft,xt,x0=x′0=x″0=⋯=xn−20=0,such that(58)xt=−1Γα∫0tt−sα−1fs,xsds+c1tα−1.
Noticingf ≤ 0, we can get(59)tα−1∫0t1−sα−1fs,xsds≤∫0tt−sα−1fs,xsds≤0.
This and (58) yield(60)limt⟶0+xttα−1=c1,which implies c1 ≥ 0.
Next, we will prove thatc1 > 0. If otherwise, it follows from (58) that(61)xt=−1Γα∫0tt−sα−1fs,xsds.
By consideringx1=∑i=1mηixξi and ∑i=1mηiξiα−1=1, we have(62)∑i=1mηi∫01ξi−ξisα−1fs,xsds=∑i=1mηi∫0ξiξi−sα−1fs,xsds,which implies f(t, x(t)) ≡ 0. Thus, (61) yields x = θ, which contradicts with the hypothesis.
From (58), there exists c2 > c1 such that(63)c1tα−1≤xt≤c2tα−1.
Then,(64)x=∫01Kt,sfs,xsds+σLx≤Λ4tα−1∫01fs,xsds+σLx=σΛ4∫01fs,xsdsLψ+σLx≤σΛ4∫01fs,xsdsLxc1+σLx=σΛ4∫01fs,xsdsc1+σLx.
By induction, we have(65)x≤σΛ4∫01fs,xsdsc1+σnLnx,n=1,2,3,…,which implies(66)x≤σΛ4∫01fs,xsdsc1+σnLnc2ψ=Λ4∫01fs,xsdsc1+1nc2ψ.
It is easy to get thatf(t, x(t)) ≢ 0 (If otherwise, (58) yields x(t) = c1tα−1. Then, D0+αxt=0, which implies f(t, c1tα−1) ≡ 0, which contradicts with (H1)). Therefore, we have(67)0<Λ4∫01fs,xsdsc1+1<1.
Letn ⟶ ∞, 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 Lemma3, we have that A: Q ⟶ Q is completely continuous. In the sequel, we will prove that A has a positive fixed point.
Byf0 > 0, there exists r1 > 0, s.t.(68)ft,z≥0,∀t,z∈0,1×0,r1.
Suppose the operatorA has no fixed point on ∂Br1∩Q. Next, we will show(69)x−Ax≠λψ,∀λ>0,x∈∂Br1∩Q.
If otherwise, there existλ0 > 0 and x1∈∂Br1∩Q, s.t.(70)x1−Ax1=λ0ψ.
Thus,(71)x1≥λ0ψ.
Denote(72)λ∗=supμ:x1≥λψ.
Then,λ∗ ≥ λ0 and x1 ≥ λ∗ψ. Noticing Ax1 ≥ σLx1, we have(73)x1=Ax1+λ0ψ≥σLx1+λ0ψ≥λ∗σLψ+λ0ψ=λ∗+λ0ψ.This contradicts with the notation of λ∗. Thus, (69) holds. Then, it follows from Lemma 4 that(74)iA,Br1∩Q,Q=0.
Byf∞ < 0, there exist ɛ ∈ (0, 1) and r2 > r1, s.t.(75)ft,z≤−σεz,∀t,z∈0,1×r2,+∞.
Denote(76)S=x∈Qx=λAx,0≤λ≤1.
Next, we will showS is bounded.
For anyx ∈ S, let x¯t=minxt,r2. Then, we have(77)ft,xt≤−σεxt+ft,x¯t+σx¯t,t∈0,1.
Clearly, we have(78)Ax¯=∫01Kt,sfs,x¯s+σx¯sds≤max0,1×0,r2Λ4×ft,z+σz.
SetT1 = σ(1 −ɛ)L. Then,(79)x=λAx≤Ax≤σ1−εLxt+Ax¯≤T1x+M.Here,(80)M=max0,1×0,r2Λ4×ft,z+σz.
Therefore,(81)I−T1x≤M.
Clearly, we haver(T1) = 1 − ɛ. So, the inverse operator of I − T1 exists. Thus, (81) yields(82)x≤MI−T1−1,which implies S is bounded.
Letr3=r2+MI−T1−1. Then, it follows from Lemma 5 that(83)iA,Br3∩Q,Q=1.
By (74) and (83), we have(84)iA,Br3\B¯r1∩Q,Q=1.
This meansA has a fixed point on Br3\B¯r1∩Q. So, RBVP (1) has a positive solution.Theorem 4.
Assume that (H2) and (H4) hold. RBVP (1) has a positive solution.Proof.
It follows fromf0 < 0 that ∃R1 > 0, s.t.(85)ft,z≤0,∀t,z∈0,1×0,R1.
SetT2 = σL. Clearly, we have r(T2) = 1.
Suppose the operatorA has no fixed point on ∂BR1∩Q. Next, we will show(86)Ax≠λx,∀λ>1,x∈∂BR1∩Q.
If otherwise, there existλ0 > 1 and x1∈∂BR1∩Q, s.t. Ax1 = λ0x1. Noticing that(87)Ax1=∫01Kt,sfs,x1s+σx1sds≤T2x1,we have(88)λ0x1≤T2x1.
By induction, we have(89)λ0nx1≤T2nx1,n=1,2,…,which implies(90)rT2=limi⟶+∞T2ii≥λ0.This contradicts with r(T2) = 1. So, (86) holds. By Lemma 5, we get(91)iA,BR1∩Q,Q=1.
Fromf∞ > 0, we know that there exist δ > 0 and R2 > R1, s.t.(92)ft,z≥δz,∀t,z∈0,1×R2,+∞.
Since {r(Ln)} is nondecreasing and converges to σ−1, one can choose m satisfies(93)rLm>1σ+δ.
It follows from Lemma3 and the Krein–Rutman theorem that there exists ψm satisfies(94)Lmψm=rLmψm.
Let(95)R3=Λ4R2mα−1Λ3+R2.
For anyx∈∂BR3∩Q, we have(96)xt≥Λ3xΛ4tα−1>R2,∀t∈1m,1.
Then,(97)Axt≥∫1/m1Kt,sfs,xs+σxsds≥∫1/m1Kt,sδxs+σxsds=δ+σLmxt.
Suppose the operatorA has no fixed point on ∂BR3∩Q. In the following, we will show(98)x−Ax≠λψm,∀x∈∂BR3∩Q,λ>0.
If otherwise, then there existλ0 > 0 and x1∈∂BR3∩Q such that(99)x1−Ax1=λ0ψm.
Denote(100)λ∗=supμ:x1≥λψm.
Then,λ∗ ≥ λ0 and x1 ≥ λ∗ψm. Noticing (97), we have(101)x1=Ax1+λ0ψm≥δ+σLmx1+λ0ψm≥δ+σLmλ∗ψm+λ0ψm=δ+σλ∗rLmψm+λ0ψm≥λ∗+λ0ψm,which contradicts with the notation of λ∗. Hence, (98) holds. By virtue of Lemma 4, one has(102)iA,BR3∩Q,Q=0.
From (91) and (102), one has(103)iA∗,BR\B¯r3∩Q,Q=−1.
Therefore, RBVP (1) has a positive solution on BR3\B¯R1∩Q.


## 4. Examples

Example 1.
Let us consider the existence of positive solutions for the RBVP:(104)−D0+7/2xt=ft,xt,0<t<1,x0=x′0=x″0=0,x1=19x19+7x49,with(105)ft,z=z2cost.
It is clear that (H0) and (H1) hold. Then, Theorem 1 yields that RBVP (104) has no positive solution.Example 2.
Consider the following RBVP:(106)−D0+5/2xt=ft,xt,0<t<1,x0=x′0=0,x1=2x19+83x14+2x49,with(107)ft,z=z2cost−z4.
By direct calculation, we havef0=−1/4, f∞ = +∞. Thus, (H4) holds. Moreover,(108)g14=−13π+∑k=1+∞4−k5k/2+3/2Γ5k/2−1/2≤−13π+13∑k=1+∞4−kΓ2k=−13π+e1/2−e−1/212≈−0.1012136,which implies σ>1/4. 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 ∞.



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