# The Existence of Solutions to Integral Boundary Value Problems of Fractional Differential Equations at Resonance

Authors: Yumei Zou; Guoping He
Journal: Journal of Function Spaces
 (2017)

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## Abstract

This paper deals with the integral boundary value problems of fractional differential equations at resonance. By Mawhin’s coincidence degree theory, we present some new results on the existence of solutions for a class of differential equations of fractional order with integral boundary conditions at resonance. An example is also included to illustrate the main results.


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## Body

## 1. Introduction

In this paper, we are concerned with the following integral boundary value problem for nonlinear fractional differential equation:(1)-D0+pxt=ft,xt,D0+p-1xt,D0+p-2xt,t∈0,1,x0=x′0=0,x1=∫01xtdAt,where 2<p<3, D0+p is the standard Riemann-Liouville differentiation, f:[0,1]×R3→R, and f satisfies Carathéodory conditions; A(t) is right continuous on [0,1) and left continuous at t=1; ∫01x(t)dA(t) denotes the Riemann-Stieltjes integrals of x with respect to A. Our problem are at resonance, in the sense that, under the integral boundary conditions, we study the linear equation -D0+px(t)=0, t∈(0,1), which has nontrivial solutions.Recently, fractional differential equations have received considerable attentions not only because of a generalization of ordinary differential equations but also because they have played a significant role in science, engineering, economy, and other fields; see, for example, [1–3].WhenA(t)≡0, problem (1) is nonresonant. In [4], the authors studied the existence of positive solutions for the nonresonant case by Krasnosel’skii’s fixed point theorem. In [5], the author investigated the uniqueness of solutions for the nonresonant case by use of the u0-positive operator under a Lipschitz condition on f.In present, many papers are devoted to the integral boundary value problem for fractional differential equation under nonresonance conditions; see [4–20]. On the other hand, there are some papers studying integral boundary value problem for differential equation under resonant conditions; we refer the reader to [21–29].Motivated by the above results, in this paper, we consider the existence of solutions for the resonance integral boundary value problem (1) under nonlinear growth restriction of f. Our method is based upon the coincidence degree theorem of Mawhin.Now, we recall the essentials of the coincidence degree theory. LetX and Y be real Banach spaces, and let L:dom⁡L⊂X→Y be a Fredholm operator of index zero. If P:X→X and Q:Y→Y are continuous projectors such that Im⁡P=Ker⁡L, Ker⁡Q=Im⁡L, X=Ker⁡L⊕Ker⁡P, and Y=Im⁡L⊕Im⁡Q, then the inverse operator of L|dom⁡L∩Ker⁡P:dom⁡L∩Ker⁡P→Im⁡L exists and is denoted by KP (generalized inverse operator of L). If Ω is an open bounded subset of X such that dom⁡L∩Ω≠∅, the mapping N:X→Y will be called L-compact on Ω¯, if QN(Ω¯) is bounded and KP(I-Q)N:Ω¯→X is compact.The abstract equationLx=Nx is shown to be solvable in view of Theorem IV.13 [30].Theorem 1 (see [30]).
LetL be a Fredholm operator of index zero and let N be L-compact on Ω¯. Assume the following conditions are satisfied:(i)
Lx≠λNx for every (x,λ)∈[(dom⁡L\Ker⁡L)∩∂Ω]×(0,1).(ii)
Nx∉Im⁡L for every x∈Ker⁡L∩∂Ω.(iii)
deg⁡(QN|Ker⁡L,Ker⁡L∩Ω,0)≠0, where Q:Z→Z is a projector as above with Im⁡L=Ker⁡Q.
Then the equationLx=Nx has at least one solution in dom⁡L∩Ω¯.Throughout this paper, we always suppose that(H1)
∫01tp-1dA(t)=1, κ=∫01tpdA(t)-1≠0;(H2)
f:[0,1]×R×R×R→R satisfies the Carathéodory conditions; that is, f(·,u,v,w) is measurable for each fixed (u,v,w)∈R×R×R, f(t,·,·,·) is continuous for a.e. t∈[0,1], and for each r>0, there exists Φr∈L∞[0,1] such that |f(t,u,v,w)|≤Φr(t) for all |u|, |v|, and |w|≤r and for a.e. t∈[0,1].


## 2. Preliminaries and Lemmas

In this section, first we provide recall some necessary basic definitions and lemmas of the fractional calculus theory, which will be used in this paper. For more details, we refer to books [1–3] for details.Definition 2 (see [1]).
The Riemann-Liouville fractional integral of orderp>0 of a function f:0,∞→R is given by (2)I0+pft=1Γp∫0tt-sp-1fsds, provided that the right-hand side is pointwise defined on (0,∞).Definition 3 (see [1]).
The Riemann-Liouville fractional derivative of orderp>0 of a continuous function f:(0,∞)→R is given by (3)D0+pft=1Γn-pddtn∫0tfst-sp-n+1ds,where n-1≤p<n, provided that the right-hand side is pointwise defined on (0,∞).Lemma 4 (see [1]).
Assume thatu∈L1[0,1]. If α, β, p>0, then (4)I0+αI0+βut=I0+α+βut,D0+pI0+put=ut.Lemma 5 (see [1]).
Assume thatI0+N-pu∈ACN-1[0,1], p>0. Then one has (5)I0+pD0+put=ut-∑k=0N-1tp-k-1Γp-kdN-k-1dtN-k-1I0+n-pu0,t∈0,1, where N=[p]+1.Lemma 6 (see [4]).
Letσ∈C(0,1)∩L(0,1) and 2<p≤3, then the unique solution of(6)D0+pxt+σt=0,t∈0,1,x0=x′0=0,x1=0is given by (7)xt=∫01Gt,sσsds, where G(t,s) is Green’s function given by (8)Gt,s=1-sp-1tp-1-t-sp-1Γp,0≤s≤t≤1,1-sp-1tp-1Γp,0≤t≤s≤1.Lemma 7 (see [4]).
The functionG(t,s) defined by (8) satisfies G(t,s)>0 and t,s∈(0,1).We use the classical Banach spaceC[0,1] with the norm x∞=maxt∈[0,1]xt and L1[0,1] with the norm x1=∫01|x(t)|dt. We also use the Banach space X(9)X=x:0,1⟶R∣x,D0+p-1x,D0+p-2x∈C0,1 with the norm xX=maxx∞,D0+p-1x∞,D0+p-2x∞.DefineL:domL⊂X→Y=L1[0,1] and N:X→Y as follows: (10)Lxt=-D0+pxt,Nxt=ft,xt,D0+p-1xt,D0+p-2xt,t∈0,1,where dom⁡L=x∈X∣D0+px∈Y,x0=x′0=0,x1=∫01xtdAt.Then integral boundary value problems (1) can be rewritten as follows:(11)Lxt=Nxt,x∈dom⁡L.Lemma 8.
The operatorL is a Fredholm operator of index zero.Proof.
Firstly, we show that(12)Ker⁡L=ctp-1:c∈R. By Lemma 5, u∈Ker⁡L means that u(t)=c1tp-1+c2tp-2+c3tp-3. It follows from u(0)=u′(0)=0 that c2=c3=0. That is, Ker⁡L=ctp-1:c∈R.
Now we prove(13)Im⁡L=y∈Y:∫011-sp-1ysds-∫01∫0tt-sp-1ysdsdAt=0. In fact, if u∈dom⁡L and Lu=y, then by Lemma 5,(14)ut=-I0+pyt+c1tp-1+c2tp-2+c3tp-3. By the boundary condition, we obtain c2=c3=0, (15)∫01utdAt=-1Γp∫01∫0tt-sp-1ysdsdAt+c1∫01tp-1dAt=-1Γp∫01∫0tt-sp-1ysdsdAt+c1,u1=-1Γp∫011-sp-1ysds+c1. The above equalities imply that(16)∫011-sp-1ysds=∫01∫0tt-sp-1ysdsdAt. On the other hand, if y∈Y satisfies ∫01(1-s)p-1y(s)ds=∫01∫0t(t-s)p-1y(s)dsdA(t), let (17)ut=-1Γp∫0tt-sp-1ysds+tp-1Γp∫011-sp-1ysds. By a simple computation, we can obtain that u∈dom⁡L and Lu=y; that is, y∈Im⁡L.
Clearly,dimKer⁡L=1 and Im⁡L is closed. It follows from Y1=Y∖ImL that (18)Y1=y1∈Y:y1=pκ∫011-sp-1ysds-∫01∫0tt-sp-1ysdsdAt,y∈Y,where κ=∫01tpdA(t)-1≠0. In fact, for each y∈Y, we have(19)∫011-sp-1ys-y1sds-∫01∫0tt-sp-1ys-y1sdsdAt=1-pκ∫011-sp-1ds-∫01∫0tt-sp-1dsdAt·∫011-sp-1ysds-∫01∫0tt-sp-1ysdsdAt=0 which shows that y-y1∈Im⁡L. This together with Y1∩ImL=θ implies that Y=Y1⊕Im⁡L. Note that dim⁡Y1=1 and thus codimIm⁡L=1. Therefore, L is a Fredholm operator of index zero. The proof is completed.Next, define the projectionsP:X→X by (20)Pyt=y1tp-1and Q:Y→Y by(21)Qyt=pκ∫011-sp-1ysds-∫01∫0tt-sp-1ysdsdAt.Clearly, Im⁡P=KerL and Ker⁡Q=Im⁡L.The generalized inverse operator ofL, KP:Im⁡L→dom⁡L∩Ker⁡P can be defined by(22)Kpyt=∫01Gt,sysds=tp-1Γp∫011-sp-1ysds-1Γp∫0tt-sp-1ysds.In fact, ify∈ImL, then (23)LKpy=-D0+p-1Γp∫0tt-sp-1ysds+tp-1Γp∫011-sp-1ysds=y.Forx∈domL∩KerP, Lx=y, we have(24)-D0+pxt=yt,t∈0,1,x0=x′0=0,x1=0. Then by Lemma 6, we obtain KpLx=x whenever x∈domL∩KerP.Using (21) and (22), we write (25)KpI-QNyt=∫01Gt,sNysds-tp-11-tκΓp·∫011-sp-1Nysds-∫01∫0tt-sp-1NysdsdAt.By a standard method, we obtain the following lemma.Lemma 9.
K
p
(
I
-
Q
)
N
:
Y
→
Y is completely continuous.Lemma 10.
Fory∈Y, (26)Kpy∞≤1Γpy1,D0+p-1Kpy∞≤2y1,D0+p-2Kpy∞≤2y1.Moreover, (27)KpyX≤2y1.Proof.
It is easy to see that(28)Kpyt=tp-1Γp∫011-sp-1ysds-1Γp∫0tt-sp-1ysds,D0+p-1Kpyt=∫011-sp-1ysds-∫0tysds,D0+p-2Kpyt=t∫011-sp-1ysds-∫0tt-sysds. Then by Lemma 7, we obtain (29)Kpy∞≤1Γpmaxt,s∈0,1⁡Gt,sy1=1Γpy1D0+p-1Kpy∞≤2y1,D0+p-2Kpy∞≤2y1.It follows that (30)KpyX≤max1Γp,2y1=2y1. The proof is completed.


## 3. Main Results

In this section, we will use Theorem1 to prove the existence of solutions to IBVP (1). To obtain our main theorem, we need the following conditions:(H3)
There exist functionsρ,α,β,γ∈L1[0,1] such that, for all (u,v,w)∈R3,t∈[0,1], (31)ft,u,v,w≤ρt+αtu+βtv+γtw.H4
There exists a constantB>0 such that either for each c∈R:|c|>B(32)cQNctp-1>0or for eachc∈R:|c|>B, (33)cQNctp-1<0.(H5)
There exists a constantM>0 such that if |D0+p-1x(t)|>M for all t∈[0,1], then QNx≠0.(H6)
There exists a constantM>0 such that if |D0+p-1x(t)|+|D0+p-2x(t)|>M for all t∈[0,1], then QNx≠0.Theorem 11.
Suppose (H1)–(H5) hold. Then IBVP (1) has at least one solution in Y, provided (34)α1+β1+γ1<15.Proof.
Set(35)Ω1=x∈domL∖KerL:Lx=λNxfor  someλ∈0,1. Take x∈Ω1. Since Lx=λNx, so λ≠0, Nx∈ImL=KerQ, and hence (36)QNx=0. Thus, from (H5), there exists t0∈[0,1] such that (37)D0+p-1xt0<M. Noticing that(38)D0+p-1xt=D0+p-1xt0+∫t0tD0+pxsds,we obtain(39)D0+p-1xt≤M+∫t0tNxsds≤M+Nx1.Observe that (I-P)x∈domL∩KerP for all x∈Ω1. Then by Lemma 10, (40)I-PxX=KpLI-PxX=KpLxX≤2Lx1≤2Nx1,(41)D0+p-1I-Px∞≤D0+p-1KpLx∞≤2Lx1≤2Nx1.Using (39) and (41), we have (42)Γpx1=D0+p-1x1tp-1=D0+p-1Pxt=D0+p-1xt-I-Pxt≤D0+p-1xt+D0+p-1I-Pxt≤M+3Nx1, so that (43)xX≤I-PxX+PxX≤1ΓpNx1+x1·tp-1X≤2Nx1+1ΓpM+3Nx1Γp=M+5Nx1; that is, for all x∈Ω1, (44)xX≤M+5Nx1. Applying (H3), we have (45)xX≤M+5ρ1+α1x∞+β1D0+p-1x∞+γ1D0+p-2x∞≤M+5ρ1+5α1+β1+γ1xX. Therefore, Ω1 is bounded.
Let(46)Ω2=x∈KerL:Nx∈ImL. Then for x∈Ω2, x=ctp-1 for some c∈R. So, QNx(t)=0. By (H4), we have |c|≤B. Therefore, Ω2 is bounded.
We define the isomorphismJ:KerL→ImQ by (47)Jctp-1=c. If (32) holds, then let (48)Ω3=x∈KerL:λJx+1-λQNx=0,λ∈0,1. For x=ctp-1∈Ω3, (49)λc=-1-λQNctp-1. If λ=1, then c=0. Otherwise, if |c|>B, in view of (H4), one has (50)c1-λQNctp-1>0 which contradicts λc2≥0. Thus Ω3 is bounded.
If (33) holds, then define the set (51)Ω3=x∈KerL:-λJx+1-λQNx=0,λ∈0,1, where J is as above. Similar to the above argument, we can show that Ω3 is bounded too.
Next, we will prove that all the assumptions of Theorem1 are satisfied. Let Ω be given any bounded open subset of Y such that ⋃i=13Ωi¯⊂Ω. By Lemma 9, KP(I-Q)N:Ω→Y is compact; thus N is L-compact on Ω¯. Clearly, assumptions (i) and (ii) of Theorem 1 are fulfilled.
At last, we will prove that (iii) of Theorem1 is satisfied. Let H(x,λ)=±λJx+(1-λ)QNx. According to above argument, we know (52)Hx,λ≠0forx∈KerL∩∂Ω. Thus, by the homotopy property of degree (53)deg⁡QNKerL,Ker⁡L∩Ω,0=deg⁡H·,0,Ker⁡L∩Ω,0=deg⁡H·,1,Ker⁡L∩Ω,0=deg⁡±J,Ker⁡L∩Ω,0≠0.
Then by Theorem1, Lx=Nx has at least one solution in domL∩Ω¯, so that IBVP (1) has a solution. The proof is completed.Theorem 12.
Suppose (H1)–(H4) and (H6) hold. Then IBVP (1) has at least one solution in Y, provided (54)α1+β1+γ1<Γp-1.Proof.
As in the proof of Theorem11, x∈Ω1, implies from (H6), there exist t0∈[0,1] such that(55)D0+p-1xt0<M,D0+p-2xt0<M.Noticing that(56)D0+p-1xt=D0+p-1xt0+∫t0tD0+pxsds,D0+p-2xt=D0+p-2xt0+∫t0tD0+p-1xsds, we obtain(57)D0+p-1xt≤M+∫t0tNxsds≤M+Nx1.Hence, we have (58)D0+p-2xt≤D0+p-2xt0+∫t0tD0+p-1xsds≤2M+Nx1.
In view ofx(0)=0, I3-px(0)=0. From this together with Lemma 5, for x∈domL, we have (59)xt=I0+p-2D0+p-2xt+I3-px0Γp-2tp-3=I0+p-2D0+p-2xt=1Γp-2∫0tt-sp-3D0+p-2xsds. Thus, by (58), (60)xt≤1Γp-2∫0tt-sp-3D0+p-2xsds≤1Γp-12M+Nx1.Considering (57), (58), and (60), and applying (H3), we get (61)xX=max⁡x∞,D0+p-1x∞,D0+p-2x∞≤max⁡M+Nx1,2M+Nx1,1Γp-12M+Nx1≤1Γp-1Nx1+2MΓp-1≤ρ1+α1x∞+β1D0+p-1x∞+γ1D0+p-2x∞Γp-1+2MΓp-1≤1Γp-1α1+β1+γ1xX+ρ1+2MΓp-1,xX≤ρ1+2Mα1+β1+γ1-Γp-1. Therefore, Ω1 is bounded. The rest of the proof repeats that of Theorem 11.Example 13.
Consider the IBVP(62)-D0+5/2xt=t+124sin⁡xt+112D0+p-1xt+124arctan⁡D0+p-2xt,x0=x′0=0,x1=52∫01xtdt. Let p=5/2 and A(t)=5/2t, then 5/2∫01x3/2dt=1 and 5/2∫01x5/2dt=5/7; thus (H1) is satisfied. Let (63)ft,u,v,w=t+124sin⁡u+112v+124arctan⁡w, so that α1=γ1=1/24, β1=1/12, and α1+β1+γ1=1/6<1/5, which verifies (H2) and (34).
TakingM=15, we have |D0+p-1x(t)|>15, (64)ft,xt,D0+p-1xt,D0+p-2xt≥112D0+p-1xt-2+π48>0,ifD0+p-1xt>15;ft,xt,D0+p-1xt,D0+p-2xt≤112D0+p-1xt+50+π48<0,ifD0+p-1xt<-15,(65)QNxt=pκ∫011-sp-1Nxsds-∫01∫0tt-sp-1NxsdsdAt=pκ∫011-sp-1sNxsds=pκ∫011-sp-1sfs,xs,D0+p-1xs,D0+p-2xsds≠0.
Hence(H5) holds. Finally, taking B=10, when |c|>10, (66)cft,ctp-1,cD0+p-1tp-1,cD0+p-2tp-1≥c212Γp-c50+π48>0, then we obtain cQN(ctp-1)>0; that is, condition (H4) is satisfied. It follows from Theorem 11 that IBVP (62) has at least one solution.



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