# Positive Solutions to Periodic Boundary Value Problems of Nonlinear Fractional Differential Equations at Resonance

Authors: Lei Hu
Journal: International Journal of Differential Equations
 (2016)

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## Abstract

By Leggett-Williams norm-type theorem for coincidences due to O’Regan and Zima, we discuss the existence of positive solutions to fractional order with periodic boundary conditions at resonance. At last, an example is presented to demonstrate the main results.


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## Body

## 1. Introduction

Fractional differential equations are generalizations of ordinary differential equations to an arbitrary order. It has played a significant role in many fields, such as viscoelasticity, engineering, physics, and economics; see [1–5]. During the last ten years, there are a large number of papers dealing with the existence of solutions boundary value problem for fractional differential equations; see [6–10].Recently, there is an increasing tendency on discussion for the existence of positive solutions to boundary value problems of fractional differential equations which enriched many previous results.In [9], Yang and Wang considered the existence of solutions for the following two-point boundary value problems for fractional differential equations:(1)D0+αut=ft,ut,u′t,0<t<1,u0=0,u′0=u′1,where 1<α<2, D0+α denoting the Caputo fractional derivative. By using the Leggett-Williams norm-type theorem for coincidences due to O’Regan and Zima, the authors obtained the existence of positive solutions to the above problem.In [10], Hu discussed the existence of positive solutions for a boundary value problem of fractional differential inclusions with resonant boundary conditions:(2)D0+αut∈ft,ut,0≤t≤1,ui0=0,i=1,2,…,n-1,u0=u1,where n-1<α<n, n≥2, D0+α denotes the Caputo fractional derivative, f:[0,1]×R→F(R), and F(R) donates the family of nonempty compact and convex subsets of R. By using the Leggett-Williams theorem for coincidences of multivalued operators due to O’Regan and Zima, results on the existence of positive solutions were established.Periodic boundary value problems have profound practical background and wide range of applications, such as mechanics, biology, and engineering; see [11–14]. Recently, periodic boundary value conditions of fractional order have been studied by some authors, such as [15–17].In [16], Chen et al. studied the following periodic boundary value problem for fractional p-Laplacian equation:(3)D0+βϕPD0+αxt=ft,xt,D0+αxt,t∈0,T,x0=xT,D0+αx0=D0+αxT,where 0<α,β≤1, D0+α is a Caputo fractional derivative, and f:[0,T]×R2→R is continuous.In [17], Hu et al. considered the existence of solutions for the following periodic boundary value problem for fractional differential equation:(4)D0+αut=ft,ut,u′t,0<t<1,u0=u1,u′0=u′1,where 1<α<2, D0+α denotes the Caputo fractional derivative, and f:[0,1]×R2→R is continuous. By using the coincidence degree theory, the authors obtained the existence of solutions.From the above works, we notice that the study of positive solutions to periodic boundary value problems of fractional order at resonance is poor. Now, the question is as follows: though the existence of solutions to (4) is obtained, how can we get the positive solutions of it? The aim of this paper is to fill the gap in the relevant literature. Our main tool is the recent Leggett-Williams norm-type theorem for coincidences due to O’Regan and Zima [18].The rest of this paper is organized as follows. Section2, we give some necessary notations, definitions, and lemmas. In Section 3, we obtain the existence of positive solutions of (4) by Theorem 9. Finally, an example is given to illustrate our results in Section 4.


## 2. Preliminaries

First of all, we present the necessary definitions and lemmas from fractional calculus theory. For more details, see [1].Definition 1 (see [1]).
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,∞)→R is given by(5)I0+αft=1Γα∫0tt-sα-1fsds,provided that the right-hand side is pointwise defined on (0,∞).Definition 2 (see [1]).
The Caputo fractional derivative of orderα>0 of a continuous function f:(0,∞)→R is given by(6)D0+αft=1Γn-α∫0tt-sn-α-1fnsds,where n-1<α≤n, provided that the right-hand side is pointwise defined on (0,∞).Lemma 3 (see [1]).
The fractional differential equation(7)D0+αyt=0has solution y(t)=c0+c1t+⋯+cn-1tn-1, ci∈R, i=0,1,…,n-1, and n=[α]+1.
Furthermore, fory∈ACn[0,1],(8)I0+αD0+αyt=yt-∑k=0n-1yk0k!tk,D0+αI0+αyt=yt.Lemma 4 (see [1]).
The relation(9)Ia+αIa+βfx=Ia+α+βfxis valid in the following case: β>0, α+β>0, and f(x)∈L1(a,b).In the following, let us recall some definitions on Fredholm operators and cones in Banach space (see [19]).LetX, Y be real Banach spaces. Consider a linear mapping L:dom⁡L⊂X→Y and a nonlinear operator N:X→Y. Assume that(A1)
L is a Fredholm operator of index zero; that is, Im⁡L is closed and dim⁡ker⁡L=codimIm⁡L<∞.This assumption implies that there exist continuous projections P:X→X and Q:Y→Y such that Im⁡P=kerL and kerQ=Im⁡L. Moreover, since dim⁡Im⁡Q=codim⁡Im⁡L, there exists an isomorphism J:Im⁡Q→kerL. Denote by LP the restriction of L to kerP∩dom⁡L. Clearly, LP is an isomorphism from kerP∩dom⁡L to Im⁡L; we denote its inverse by KP:Im⁡L→kerP∩dom⁡L. It is known that the coincidence equation Lx=Nx is equivalent to(10)x=P+JQNx+KPI-QNx.Let C be a cone in X such that(i)
μx∈C for all x∈C and μ≥0,(ii)
x,-x∈C implies x=θ.It is well known that C induces a partial order in X by(11)x⪯yiffy-x∈C.The following property is valid for every cone in a Banach spaceX.Lemma 5 (see [18]).
LetC be a cone in X. Then for every u∈C∖{0} there exists a positive number σ(u) such that(12)x+u≥σuu∀x∈C.Letγ:X→C be a retraction, that is, a continuous mapping such that γ(x)=x for all x∈C. Set(13)Ψ≔P+JQN+KPI-QN,Ψγ≔Ψ∘γ.We use the following result due to O’Regan and Zima.Theorem 6 (see [18]).
LetC be a cone in X and let Ω1, Ω2 be open bounded subsets of X with Ω¯1⊂Ω2 and C∩(Ω¯2∖Ω1)≠∅. Assume (A1) and the following assumptions hold:(A2)
QN:X→Y is continuous and bounded and KP(I-Q)N:X→X is compact on every bounded subset of X.(A3)
Lx≠λNx for all x∈C∩∂Ω2∩Im⁡L and λ∈(0,1).(A4)
γ maps subsets of Ω¯2 into bounded subsets of C.(A5)
deg{[I-(P+JQN)γ]|kerL,kerL∩Ω2,0}≠0.(A6)
There existsu0∈C∖{0} such that x≤σ(u0)Ψx for x∈C(u0)∩∂Ω1, where C(u0)={x∈C:μu0⪯xforsomeμ>0} and σ(u0) such that x+u0≥σ(u0)x for every x∈C.(A7)
(P+JQN)γ(∂Ω2)⊂C.(A8)
Ψγ(Ω¯2∖Ω1)⊂C.
Then the equationLx=Nx has a solution in the set C∩(Ω¯2∖Ω1).


## 3. Main Results

In this section, we give the existence theorems for problem (4). We write Banach space X=Y=C[0,1] with the norm x=maxt∈[0,1]|x(t)|.Define the operatorL:dom⁡L→X by(14)Lu=D0+αu,where(15)dom⁡L=x∈X:D0+αut∈Y,u0=u1,u′0=u′1.Define the operator(16)N:X⟶Yby(17)Nut=ft,ut,u′t.Then problem (4) can be written by Lu=Nu, u∈domL.Let(18)
G
t
,
s
=
1
+
t
-
s
α
-
1
1
-
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2
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1
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1
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1
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+
1
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+
1
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α
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2
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-
1
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+
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+
t
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+
1
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0
≤
s
<
t
≤
1
,
1
-
Γ
α
1
-
s
α
α
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1
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2
α
-
1
+
1
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s
1
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α
t
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-
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α
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+
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+
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α
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1
α
Γ
α
+
1
+
t
Γ
α
+
1
,
0
≤
t
<
s
≤
1
.Let κ be a constant, which is in (0,1) and satisfies(19)κGt,s<1.Lemma 7.
The mappingL:dom⁡L⊂Z is a Fredholm operator of index zero. Furthermore, the operator KP:Im⁡L→dom⁡L∩kerP can be written as(20)KPyt=∫01kt,sysds,t∈0,1,where(21)
k
t
,
s
≔
t
-
s
α
-
1
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α
-
Γ
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≤
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-
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α
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1
-
1
Γ
α
1
-
s
α
-
1
,
0
≤
t
<
s
≤
1
.Proof.
By Lemma3, D0+αu(t)=0 has solution(22)ut=c0+c1t,c0,c1∈R.According to the boundary value conditions of (4), we have(23)kerL=c,c∈R≅R1.Let y∈Im⁡L, so there exists a function u(t)∈domL which satisfies Lu(t)=y(t). By Lemma 3, we have(24)ut=I0+αyt+c0+c1t.By u′0=u′(1), we can obtain ∫01‍(1-s)α-2y(s)ds=0. On the other hand, suppose y∈Y satisfies ∫01‍(1-s)α-2y(s)ds=0. Let u(t)=I0+αy(t)-I0+αy(t)|t=1·t. We can easily prove u(t)∈domL. Thus, we conclude that(25)Im⁡L=y∈Y:∫011-sα-2ysds=0.
Consider the linear operatorP:X→X defined by(26)Pxt=α-1∫011-sα-2xsds,t∈0,1.Define the operator Q:Y→Y by(27)Qyt=α-1∫011-sα-2ysds,t∈0,1.For u(t)∈X, we get(28)PPu=Pα-1∫011-sα-2usds=α-1∫011-sα-2usds=Pu.So we have P2=P. In the same way, Q2=Q. We notice that Im⁡P=kerL and kerQ=Im⁡L. It follows from IndL=dim⁡kerL-codim⁡Im⁡L=0 that L is a Fredholm mapping of index zero.
Next, we will prove that the operatorKP is the inverse of L|domL∩kerP.
In fact, foru(t)∈domL∩kerP, we have D0+αu(t)=y(t). By Lemma 3, we have u(t)=I0+αy(t)+c0+c1t. According to u(0)=u(1), we get(29)c1=-I0+αy1.By ut∈kerP, that is, (α-1)∫01‍(1-s)α-2u(s)ds=0, we have(30)c0=-ΓαI0+2α-1y1-c1α.Then, we have(31)KPyt=ut=I0+αyt+c0+c1t=I0+αyt-ΓαI0+2α-1y1+1αI0+αy1-I0+αy1t=1Γα∫0tt-sα-1ysds-ΓαΓ2α-1∫011-s2α-2ysds+1αΓα∫011-sα-1ysds+1Γα∫011-sα-1tysds=∫01kt,sysds.
It is easy to see thatLKPy(t)=y(t). Hence, KP=(L|domL∩kerP)-1. This completes the proof of Lemma 7.Lemma 8.
AssumeΩ⊂X is an open bounded set such that dom⁡(L)∩Ω¯≠∅; then N is L-compact on Ω¯.Proof.
By the continuity off, we can obtain that QN(Ω¯) and KP(I-Q)N(Ω¯) are bounded. Hence, for u(t)∈Ω¯, t∈[0,1], there exists a positive constant T such that(32)I-QNut≤T,Nut≤T.Thus, in the view of Arzela-Ascoli theorem, we need only to prove that KP(I-Q)N(Ω¯) is equicontinuous.
For0≤t1<t2≤1, u∈Ω¯, we have(33)KPI-QNut2-KPI-QNut1=I0+αI-QNutt=t2-ΓαI0+2α-1I-QNu1+1αI0+αI-QNu1-I0+αI-QNu1t2-I0+αI-QNutt=t1+ΓαI0+2α-1I-QNu1-1αI0+αI-QNu1+I0+αI-QNu1t1≤1Γα∫0t2t2-sα-1I-QNusds-∫0t1t1-sα-1I-QNusds+1Γα∫011-sα-1I-QNusds·t2-t1≤1Γα∫0t1t2-sα-1-t1-sα-1I-QNusds+1Γα∫t1t2t2-sα-1I-QNusds+TΓα+1t2-t1≤TΓα+1t2α-t1α+t2-t1α+t2-t1.Notice that tα, t are uniformly continuous on [0,1]. Thus, we have that KP(I-Q)N(Ω¯) is equicontinuous on [0,1]. The proof is completed.Theorem 9.
Assume that(H1)
there exist positive constantsb1, b2, b3, c1, c2, and B with(34)B>c2c1+3b2c2α-1b1c1+3b3α-1b1.For allt∈[0,1] and u(t)∈[0,B], one has(35)-κut≤ft,ut,u′t≤-c1ut+c2,ft,ut,u′t≤-b1ft,ut,u′t+b2ut+b3;(H2)
there existb∈(0,B), t0∈[0,1], ρ∈(0,1], δ∈(0,1), and q(t)∈L1[0,1], q(t)≥0 on [0,1], h(x)∈C((0,b],R+) such that G(t0,s)>0 and f(t,u,u′)≥q(t)h(u) for (t,u,u′)∈[0,1]×(0,b]×R. Moreover, h(u)/uρ is nonincreasing on (0,b] and(36)α-1hbb∫01Gt0,s1-sα-2qsds≥1-δδρ.
Then problem (4) has at least one positive solution on [0,1].Proof.
According to Lemmas7 and 8, we have that conditions (A1) and (A2) of Theorem 6 are satisfied.
Consider the cone(37)C=x∈X:xt≥0,t∈0,1.
Let(38)Ω1=x∈X:δx<xt<b,t∈0,1,Ω2=x∈X:xt<B,t∈0,1.Obviously, Ω1 and Ω2 are bounded and(39)Ω¯1=x∈X:δx≤xt≤b,t∈0,1⊂Ω2.Moreover, C∩(Ω¯2∖Ω1)≠∅. Let J=I and (γx)(t)=|x(t)| for x∈X; then γ is a retraction and maps subsets of Ω¯2 into bounded subsets of C, which means that (A4) holds.
Next, we will show that (A3) holds. Suppose that there existu0∈∂Ω2∩C∩dom⁡L and λ0∈(0,1) such that Lu0=λ0Nu0; that is, D0+αu0(t)=λ0f(t,u0(t),u0′(t)), t∈[0,1]. In view of (H1), we get(40)D0+αu0t=λ0ft,u0t,u0′t≤-λ0b1ft,u0t,u0′t+λ0b2u0t+λ0b3=-b1λ0ft,u0t,u0′t+λ0b2u0t+λ0b3=-b1D0+αu0t+λ0b2u0t+λ0b3≤-b1D0+αu0t+b2u0t+b3,(41)D0+αu0t=λ0ft,u0t,u0′t≤-λ0c1u0t+λ0c2.In view of D0+αu0(t)=λ0f(t,u0(t))∈Im⁡L, from the definition of Im⁡L and (41), we obtain(42)0=∫011-sα-2D0+αu0sds≤∫011-sα-2-λ0c1u0s+λ0c2dswhich gives(43)∫011-sα-2u0sds≤c2α-1c1.From (40), we have(44)0=∫011-sα-2D0+αu0sds≤∫011-sα-2-b1D0+αu0t+b2u0t+b3ds=-b1∫011-sα-2D0+αu0sds+b2∫011-sα-2u0sds+b3α-1.By (43), we have(45)∫011-sα-2D0+αu0sds≤b2b1∫011-sα-2u0sds+b3α-1b1≤b2c2α-1b1c1+b3α-1b1.According to the function expression of k(t,s), it is easy to see that |k(t,s)|≤3(1-s)α-2, s,t∈[0,1]. From (43) and the equation u0=(I-P)u0+Pu0=KPL(I-P)u0+Pu0=Pu0+KPLu0, we can get(46)u0=α-1∫011-sα-2u0sds+∫01kt,sD0+αu0sds≤α-1·c2α-1c1+∫01kt,s·D0+αu0sds=c2c1+∫01kt,s1-sα-2·1-sα-2D0+αu0sds≤c2c1+3∫011-sα-2D0+αu0sds≤c2c1+3b2c2α-1b1c1+3b3α-1b1.Then, we have(47)B=u0≤c2c1+3b2c2α-1b1c1+3b3α-1b1,which contradicts (H1). Hence (A3) holds.
To prove (A5), consideru(t)∈kerL∩Ω¯2; then u(t)≡c. For c∈[-B,B] and λ∈[0,1], we have(48)Hc,λ=I-λP+JQNγc=c-λα-1∫011-sα-2cds-λα-1∫011-sα-2fs,c,0ds=c-λc-λα-1∫011-sα-2fs,c,0ds=c-λα-1∫011-sα-2fs,c,0+cds.By use of proof by contradiction, it is easy to show that H(c,λ)=0 implies c≥0. Suppose H(B,λ)=0 for some λ∈(0,1]; then we have(49)0=B-λB-λα-1∫011-sα-2fs,Bds.According to (H1), we have(50)0≤B1-λ=λα-1∫011-sα-2fs,B,0ds≤λ-c1B+c2<0,which is a contradiction. In addition, if λ=0, then B=0, which is impossible. As a result, for x∈kerL∩∂Ω2 and λ∈[0,1], we have H(x,λ)≠0. Thus,(51)degI-P+JQNγkerL,kerL∩Ω2,0=degH·,1,kerL∩Ω2,0=degH·,0,kerL∩Ω2,0=degI,kerL∩Ω2,0=1≠0.So (A5) holds.
Next, we prove (A6). Letu0(t)≡1, t∈[0,1]; then u0∈C∖{0}, C(u0)={x∈C:x(t)>0,t∈[0,1]}. We take σ(u0)=1. Let x∈C(u0)∩∂Ω1; then 0<x≤b and x(t)≥δx on [0,1].
By (H2), for everyx∈C(u0)∩∂Ω1, we have(52)Ψxt0=P+JQN+KPI-QNxtt=t0=α-1∫011-sα-2xsds+α-1∫01Gt0,s1-sα-2fs,xs,x′sds≥α-1∫011-sα-2δxds+α-1∫01Gt0,s1-sα-2fs,xs,x′sds≥δx+α-1∫01Gt0,s1-sα-2qshxsds=δx+α-1∫01Gt0,s1-sα-2qs·hxsxρsxρsds≥δx+α-1∫01Gt0,s1-sα-2qs·hxsxρs·δρxρds≥δx+δρxρα-1∫01Gt0,s1-sα-2qs·hbbρds=δx+δρx·b1-ρx1-ρα-1∫01Gt0,s1-sα-2qshbbds≥δx+δρx·α-1∫01Gt0,s1-sα-2qshbbds≥δx+δρx·1-δδρ=x.
Thus, for allx∈C(u0)∩∂Ω1, we have x≤σ(u0)Ψx; that is, (A6) holds.
Forx∈∂Ω2, by (H2) and (19), we have(53)P+JQN∘γxt=Pxt+JQNxt=α-1∫011-sα-2xsds+α-1∫011-sα-2fs,xs,xs′ds≥α-1∫011-sα-21-κxsds≥0.Thus, [(P+JQN)∘γ]x(t)⊂C for x∈∂Ω2. Then (A7) holds.
Next, we prove (A8). Forx(t)∈Ω¯2∖Ω1, by (H2), we have(54)
Ψ
γ
x
t
=
P
+
J
Q
N
+
K
P
I
-
Q
N
∘
γ
x
t
=
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+
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P
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t
=
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x
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=
α
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1
∫
0
1
1
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α
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2
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d
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+
α
-
1
∫
0
1
1
-
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t
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f
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x
s
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x
s
′
d
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>
α
-
1
∫
0
1
1
-
s
α
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2
x
s
d
s
+
α
-
1
∫
0
1
1
-
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α
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2
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t
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s
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κ
x
s
d
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>
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0
1
1
-
s
α
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2
x
s
1
-
κ
G
t
,
s
d
s
≥
0
.Hence, Ψγ(Ω¯2∖Ω1)⊂C; that is, (A8) holds.
Hence, applying Theorem6, BVP (4) has a positive solution u∗(t) on [0,1] with b≤u∗(t)≤B. This completes the proof.


## 4. Example

To illustrate how our main result can be used in practice, we present here an example.Let us consider the following fractional differential equation at resonance:(55)D0+1.5ut=ft,ut,u′t,0<t<1,u0=u1,u′0=u′1,where(56)ft,ut,u′t=14001+t-t2u2t-8ut+12ut+130e-u′t.Corresponding to BVP (4), we have that α=1.5 and(57)
G
t
,
s
=
1
+
t
-
s
0.5
1
-
s
0.5
0.5
Γ
1.5
-
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1.5
1
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1.5
0.5
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2
+
1
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s
1
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1.5
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0.5
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2.5
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1.5
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2.5
+
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1.5
Γ
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+
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Γ
2.5
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0
≤
s
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t
≤
1
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1
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Γ
1.5
1
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1.5
0.5
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2
+
1
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1
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1.5
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0.5
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2.5
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2.5
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2.5
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0
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s
≤
1
.By simple calculation, we can get that if t∈[0,1] and u∈[0,1], one has(58)-115ut≤ft,ut,u′t≤120ut+110,ft,ut,u′t≤-39ft,ut,u′t+12ut+85.So, we can choose κ=1/15, B=1, c1=1/20, c2=1/10, b1=39, b2=1/2, and b3=8/5. Furthermore, we can verify(59)B>c2c1+3b2c2α-1b1c1+3b3α-1b1.So, (H1) is satisfied.Takingρ=1, h(u)=(1/20)u, and q(t)=(33/80)(1+t-t2), we have(60)ft,ut,u′t≥qthu=120ut·33851+t-t2,ut∈0,1,t∈0,1.Let t0=0; then(61)G0,s=1-2Γ1.51-s1.5+2Γ2.51-s+Γ1.52-23Γ2.5>0.Moreover, takeb=1/2∈[0,1], and we have that(62)α-1hbb∫01Gt0,s1-sα-2qsds≈0.98>1-δδρ=0.97,huuρ=1/20utuρt=120is nonincreasing on (0,1/2]δ=0.9995. So, conditions (H1)-(H2) of Theorem 9 are satisfied; then BVP (55) has a positive solution on [0,1].



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*Source: 9260726-2016-11-09.xml*