# Resonant Integral Boundary Value Problems for Caputo Fractional Differential Equations

Authors: Wenjie Ma; Shuman Meng; Yujun Cui
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

This paper deals with the following Caputo fractional differential equations with Riemann-Stieltjes integral boundary conditionsDc0+αut=ft,ut,u′t,u′′t,t∈0,1,u0=u′′0=0,u1=∫01‍utdAt, where Dc0+α denotes the standard Caputo derivative, α∈(2,3]; ∫01x(t)dA(t) denotes the Riemann-Stieltjes integrals of x with respect to A. By mean of coincidence degree theory, we obtain the existence of solutions for the above fractional BVP at resonance. In the end, according to the main results, we give a typical example.


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## Body

## 1. Introduction

Fractional differential equations have various applications in chemical engineering, blood flow phenomena, physics, signal and image processing, fitting of experimental data, control theory, etc. Fractional differential equations also are a fundamental tool for the description of process of natural phenomena. In past years, many authors studied the existence of solutions or positive solutions of various fractional differential equations such as fractional boundary value problem at resonance [1–9], fractional boundary value problem at nonresonance [10–23], Caputo fractional derivative problem [24, 25], impulsive problem [26], conjugate boundary value problems [27, 28], integral boundary value problem [29], variational structure problem [30], fractional p-Laplace problem [31], fractional lower and upper solution problem [32, 33], solitons [34, 35], etc.For example, Alberto et al. [11] considered the following fractional differential equations: (1)Dcαut+ft,ut=0,t∈0,1,u0=u′′0=0,u1=λ∫01usds,where λ∈(0,2), α∈(2,3), Dc0+α denotes the Caputo fractional derivative, and f:[0,1]×[0,∞)→[0,∞) is a continuous function. The existence of solutions is obtained by using the Guo-Krasnoselskii fixed point theorem. Similar arguments for related issues are treated in [12].In [7], by mean of coincidence degree theory, the existence of solution of integral BVP for nonlinear differential equation is proved:(2)-D0+pxt=ft,xt,D0+p-1xt,D0+p-2xt,t∈0,1,x0=x′0=0,x1=∫01xtdAt,where p∈(2,3), D0+p is the Riemann-Liouville derivative, f:[0,1]×R3→R, and f satisfies Caratheodory conditions. Similar method for different nonlinear differential equations, with either Caputo or Riemann-Liouville derivatives, and using suitable coincidence degree theory in each case, has allowed proving the existence of solutions in many papers; see also [1–5, 24, 25, 33].In [5], the authors studied nonlocal fractional differential equation: (3)D0+αut+ft,ut=0,t∈0,1,u0=0,u1=ηuξ,where ηξα-1=1, ξ∈(0,1), α∈(1,2), D0+α is the Riemann-Liouville derivative, and f:[0,1]×[0,∞)→R is continuous function. They solved the above differential equation by using iterative technique and the fixed point index theory. We can also refer to [16, 17].In this paper, we are concerned with the existence of solution for the following differential equation with boundary value conditions:(4)Dc0+αut=ft,ut,u′t,u′′t,t∈0,1,u0=u′′0=0,u1=∫01utdAt,where Dc0+α denotes the Caputo fractional derivative, α∈(2,3], and f:[0,1]×R3→R is a continuous function. The BVP (4) is a resonance problem, because ∫01‍tdA(t)=1, if and only if its associated linear homogeneous BVP (5)Dc0+αxt=0,t∈0,1,x0=x′′0=0,x1=∫01xtdAthave nontrivial solution. Here, we assume that the following two conditions hold:(A1)∫01‍tdA(t)=1, ∫01‍tαdA(t)≠1.(A2)f:[0,1]×R3→R is a continuous function.As is known to all, there are few studies on such boundary value problems. Therefore, by mean of coincidence degree theory, we consider the existence of solutions for fractional differential equations (4) with Riemann-Stieltjes integral boundary conditions at resonance.


## 2. Preliminaries

For the convenience of the reading, we introduce the theorem and lemma of fractional calculus theory.Lemma 1 (see [36]).
Putα∈(0,+∞), f∈L(0,1)∩C(0,1); the fractional differential equation(6)Dc0+αfx=0has the general solution f(x)=A0+A1x+⋯+An-1xn-1,Ak∈R,k=0,1…,n-1, where n is the smallest integer greater than or equal to α.Lemma 2 (see [36]).
Given thatf∈Cn[0,1], then (7)I0+αDc0+αfx=fx-f0-f′0x-⋯-fn-10n-1!xn-1,where n is the smallest integer greater than or equal to α.LetY,Z be two Banach spaces and L:D(L)⊂Y→Z be a linear operator. Assume that L is a Fredholm operator of index zero; P:Y→Y, Q:Z→Z are two continuous projectors such that (8)ImP=kerL,kerQ=ImL,Y=kerL⊕kerP,Z=ImL⊕ImQ.Thus (9)LKerP∩DL:KerP∩DL→ImLis invertible. We denote the inverse of the mapping by KP (generalized inverse operator of L). If Ω is an open bounded subset of Y such that D(L)∩Ω≠∅, N:Y→Z will be called L-compact on Ω¯ if QN(Ω¯) is bounded and KP(I-Q)N:Ω¯→Y is compact.Theorem 3 (see [37]).
L : D ( L ) ⊂ Y → Z is a linear operator, and L is a Fredholm operator of index zero, and N is L-compact on Ω¯. The equation Lψ=Nψ has at least one solution in D(L)∩Ω¯ if and only if the following three conditions are satisfied:(1)
Lψ≠λNψ for every (ψ,λ)∈[(D(L)∖kerL)∩∂Ω]×(0,1).(2)
Nψ∉ImL for every ψ∈KerL∩∂Ω.(3)
deg⁡QNKerL,Ω∩KerL,0≠0, where Q:Y→Y is a projection such that ImL=KerQ.In this paper, letY=C2[0,1] with the norm x=maxx∞,x′∞,x′′∞ in which x∞=maxt∈[0,1]|x(t)|. Let Z=C[0,1] with the norm x∞=maxt∈[0,1]|x(t)|. Operator L is defined as L:Lu(t)=Dc0+αu(t) with (10)DL=u∈Y:Dc0+αut∈C0,1,u0=u′′0=0,u1=∫01utdAt.Define operatorN:Y→Z as follows: (11)Nux=ft,ut,u′t,u′′t.And so problem (4) becomes Lu=Nu,u∈D(L).


## 3. Main Results

In this section, we continue to give the following assumptions.( A 3 ) There exist functions a(t),b(t),c(t),d(t)∈C[0,1] with b∞+c∞+d∞<Γ(α-1) such that (12)ft,u,v,w≤at+btu+ctv+dtw,t∈0,1.( A 4 ) There exists positive constant F such that, for y∈DomL, if y′t>F for t∈[0,1], then (13)∫011-sα-1Nysds-∫01∫0tt-sα-1NysdsdAt≠0.( A 5 ) There exists constant a0>0 such that either(14)c∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt<0or(15)c∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt>0holds if |c|>a0.Based on the above, we will prove the following important theorem.Theorem 4.
Suppose that(A1)-(A5) is satisfied; there exists at least one solution of the fractional differential equation (4) in Y.For convenience, we first give some important lemmas as follows.Lemma 5.
Assume that(A1) is satisfied; then L:D(L)⊂Y→Z is Fredholm operator with index zero, and a linear continuous projector Q:Z→Z is defined by (16)Qyt=α1-∫01tαdAt∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt.Furthermore, define a linear operator Kp:ImL→D(L)∩kerP as follows: (17)Kpyt=1Γα∫0tt-sα-1ysds-tΓα∫011-sα-1ysdssuch that Kp=(LD(L)∩kerP)-1.Proof.
It follows from Lemma1 and Lu=0 that (18)ut=c1+c2t+c3t2,c1,c2,c3∈R.According to u(0)=u′′0=0,u(1)=∫01‍u(t)dA(t), we have (19)kerL=ct:c∈R.Moreover, we can obtain that (20)ImL=y∈Z:∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt=0.On one hand, suppose y∈ImL; then there exist u∈D(L):(21)y=Lu∈Z.Then we have (22)ut=I0+αyt+c1+c2t+c3t2.This together with (A1), u(0)=u′′0=0,u(1)=∫01‍u(t)dA(t), we can get c1=c3=0 and (23)∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt=0.And so we obtain that (24)ImL⊂y∈Z:∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt=0.On the other hand, if y∈Z satisfies (25)∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt=0,we let (26)ut=I0+αyt-tI0+αy1=1Γα∫0t1-sα-1ysds-tΓα∫011-sα-1ysds.Then we conclude that (27)u0=u′′0=0,u1=I0+αy1-1Γα∫011-sα-1ysds=0and (28)∫01utdAt=∫011Γα∫0t1-sα-1ysdsdAt-∫01tdAt1Γα∫011-sα-1ysds=∫011Γα∫0t1-sα-1ysdsdAt-1Γα∫011-sα-1ysds=0.That is, u∈D(L), then y∈ImL. In conclusion, (29)ImL=y∈Z:∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt=0.
We give projection operatorP:Y→Y as (30)Put=u1t.We get KerP=u∈Y∣u1=0, P2u=Pu, and ImP=kerL. For any u∈Y, together with u=(u-Pu)+Pu, we have Y=kerP+kerL. It is easy to obtain that kerL∩kerP=0 which implies (31)Y=kerP⊕kerL.
Next operatorQ:Z→Z is defined as follows: (32)Qyt=α1-∫01tαdAt∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt.Noting that (33)Q2yt=QQyt=α1-∫01tαdAt∫011-sα-1Qysds-∫01∫0tt-sα-1QysdsdAt=Qytα1-∫01tαdAt∫011-sα-1ds-∫01∫0tt-sα-1dsdAt=Qytmeans Q is a projection operator. And obviously, kerQ=ImL. For any y∈Z, let y1=y-Qy, (34)Qy1t=Qy-Qyt=α1-∫01tαdAt∫011-sα-1ys-Qysds-∫01∫0tt-sα-1ys-QysdsdAt=α1-∫01tαdAt∫011-sα-1ysds-∫01∫0tt-sα-1ysdsdAt+α1-∫01tαdAt∫011-sα-1Qysds-∫01∫0tt-sα-1QysdsdAt=Qyt-Q2yt=0,and we have y1∈ImL, so Z=ImL+ImQ. Moreover, by simple calculation, we can get ImQ∩ImL=0. Above all, Z=ImL⊕ImQ.
To summarize, we can know thatImL is closed subspace of Z; dimkerL=codimImL=1; and so L is a Fredholm operator of index zero.
Next we will prove thatKp is the inverse of LdomL∩kerP. It is clear that(35)LpKpyt=Dc0+αKpyt=yt,y∈ImL.For each u∈domL∩kerP, we have u(0)=u′′0=0,u(1)=0 and (36)KpLput=I0+αDc0+αut-tI0+αDc0+α1=ut+u′0t-tu1+u′0=ut.This implies that KpLpu=u. So Kp=(LdomL∩kerP)-1.Lemma 6.
Assume thatΩ is a bounded open subset of Y, then N is L-compact on Ω¯ if domL∩Ω¯≠∅.Proof.
By the continuity off, we obtain QN(Ω¯) and (I-Q)N(Ω¯) are bounded, so there exists constant M>0, such that I-QNut≤M for u∈Ω¯ and t∈[0,1]. ∀t1,t2∈[0,1],t1<t2,u∈Ω¯, let (Ou)(s)=(I-Q)Nu(s), and we have (37)KPI-QNut2-KPI-QNut1≤1Γα∫0t2t2-sα-1Ousds-∫0t1t1-sα-1Ousds+t2-t1Γα∫011-sα-1Ousds≤MΓα∫0t1t2-sα-1-t1-sα-1ds+∫t1t2t2-sα-1ds+Mt2-t1Γα∫011-sα-1ds=MΓα+1t2α-t1α+t2-t1,KPI-QNu′t2-KPI-QNu′t1=1Γα-1∫0t2t2-sα-2Ousds-∫0t1t1-sα-2Ousds≤MΓα-1∫0t1t2-sα-2-t1-sα-2ds+∫t1t2t2-sα-2ds=MΓαt2α-1-t1α-1,and(38)KPI-QNu′′t2-KPI-QNu′′t1=1Γα-2∫0t2t2-sα-3Ousds-∫0t1t1-sα-3Ousds≤MΓα-2∫0t1t2-sα-3-t1-sα-3ds+∫t1t2t2-sα-3ds=MΓα-1t2α-2-t1α-2.As a result of tα,tα-1,tα-2 are uniformly continuous on [0,1]. So we have that KP(I-Q)N(Ω¯)⊂C[0,1],KPI-QN′(Ω¯)⊂C[0,1],KPI-QN′′(Ω¯)⊂C[0,1] are equicontinuous, and, by Ascoli-Arzela theorem, we have that KP(I-Q)N:Ω¯→Y is compact. Thus, N is L-compact.Lemma 7.
SetΩ1=u∈kerL:λu+β1-λQNu=0,λ∈0,1 is bounded if conditions (A1), (A2), and (A5) are satisfied, and (39)β=-1,if14holds;1,if15holds.Proof.
Suppose thatu∈Ω1 we have u(t)=ct,c∈R and (40)λct+β1-λα1-∫01tαdAt·∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt=0If λ=0, by condition (A5), we have |c|≤a0. If λ=1, then c=0. If λ∈(0,1), we suppose |c|≥a0, then (41)λc2t=-β1-λcα1-∫01tαdAt·∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt<0which contradicts with λc2>0.Lemma 8.
SetΩ2=u∈DL∖kerL:Lu=λNu,λ∈0,1 is bounded if (A1)-(A4) are satisfied.Proof.
Takeu∈Ω2, then Nu∈ImL. Thus we have(42)∫011-sα-1Nysds-∫01∫0tt-sα-1NysdsdAt=0.According to (A4), there exists constant ξ∈(0,1) such that (43)u′ξ≤F.Because u′t≤u′ξ+|∫ξt‍u′′(s)ds|≤F+u′′∞,(44)u′∞≤F+u′′∞.If u∈D(L), we have u(0)=0, then (45)ut=u0+∫0tu′sds≤u′∞,and therefore u∞≤u′∞.
ByLu=λNu, we have (46)ut=λΓα∫0tt-sα-1fs,us,u′s,u′′sds-u0-u′0t-u′′02t2.Then we obtain that(47)u′′t=λΓα-2∫0tt-sα-3fs,us,u′s,u′′sds-u′′0.According to the boundary condition u′′(0)=0, one has(48)u′′t=λΓα-2∫0tt-sα-3fs,us,u′s,u′′sds.So we get (49)u′′t≤1Γα-2∫0tt-sα-3fs,us,u′s,u′′sds≤1Γα-2∫011-sα-3fs,us,u′s,u′′sds≤1Γα-2∫011-sα-3as+bsus+csu′s+dsu′′sds≤1Γα-1a∞+b∞u∞+c∞u′∞+d∞u′′∞≤1Γα-1a∞+b∞F+u′′∞+c∞F+u′′∞+d∞u′′∞≤1Γα-1a∞+b∞F+c∞F+b∞+c∞+d∞u′′∞.Thus, from b∞+c∞+d∞<Γ(α-1), we obtain that (50)u′′∞≤a∞+Fb∞+c∞Γα-1-b∞-c∞-d∞=M1.So we have (51)u=max⁡u∞,u′∞,u′′∞≤M1+F.Therefore, Ω2 is bounded.Lemma 9.
SetΩ3=u:u∈kerL,Nu∈ImL is bounded if (A1), (A2), and (A5) hold.Proof.
Letu∈Ω3, then u(t)=ct,c∈R,Nu∈ImL, and we can get (52)∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt=0.According to (A5) we have |c|≤a0, that is to say, Ω3 is bounded.In summary, we can prove Theorem4.Proof of Theorem4.
Assume thatΩ⊃⋃i=13Ωi¯∪0 is bounded open subset of X, and, according to Lemma 6, we get that N is L-compact on Ω¯. Form Lemmas 8 and 9, we have (1)
Lu≠λNu,for  all(u,λ)∈[(DL∖kerL)∩∂Ω]×(0,1);(2)
Nu∉ImL,for  allu∈KerL∩∂Ω.
SetH(u,λ)=λu+β(1-λ)QNu. It follows from Lemma 7 that we get H(u,λ)≠0, u∈∂Ω∩KerL, so we have (53)deg⁡QNKerL,Ω∩KerL,0=deg⁡βJ,Ω∩KerL,0≠0.According to the above proof and Theorem 3, Theorem 4 holds.Example 10.
Considering the following fractional differential equation,(54)Dc0+5/2ut=t2+tu2t1+u2t+14πu′tarctan⁡u′t+14sin⁡u′′t,t∈0,1,u0=u′′0=0,u1=2∫01utdt.Obviously, α=5/2, A(t)=2t, then ∫01‍tdA(t)=1, and thus it is at resonance. Let (55)ft,u,v,w=t2+tu21+u2+14πvarctan⁡v+14sin⁡w,then (56)ft,u,v,w≤1+t2+12v+14w,where (57)c∞+d∞=12+14=34<π2=Γ32.
TakingF=16, then we have u′t≥16, (58)ft,ut,u′t,u′′t≥-12+4πarctan⁡1-14>0.and (59)∫011-sα-1Nusds-∫01∫0tt-sα-1NusdsdAt=∫011-sα-1Nusds-2∫01∫s1t-sα-1dtNusds=∫011-sα-1-21-sααNusds=∫011-sα-1α-2+2sαfs,us,u′s,u′′sds>0.That is, (A4) holds. Finally, taking a0=16, when c>16, we conclude that (60)∫011-sα-1fs,cs,c,0ds-∫01∫0tt-sα-1fs,cs,c,0dsdAt>0then (A5) is satisfied. It follows from Theorem 4 that (54) has at least one solution.



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*Source: 5438592-2018-08-07.xml*