In this paper, we study the existence and uniqueness of solutions for the following boundary value problem of nonlinear fractional differential equation: D0+qCut=ft,ut, t∈0,1, u0=u′′0=0,D0+σ1Cu1=λI0+σ2u1, where 2<q<3, 0<σ1≤1, σ2>0, and λ≠Γ2+σ2/Γ2-σ1. The main tools used are nonlinear alternative of Leray-Schauder type and Banach contraction principle.

National Natural Science Foundation of China116610491. Introduction

Fractional calculus has wide applications in many fields of science and engineering, for example, fluid flow, biosciences, rheology, electrical networks, chemical physics, control theory of dynamical systems, and optics and signal processing [1].

Recently, nonlinear fractional differential equations have been discussed under the following boundary conditions (BCs for short):

Integer derivative BCs:

u0=u1=0,

u0+u′0=0, u1+u′1=0,

u0=u′1=u′′0=0,

u0=0, u′0+u′′0=0, u′1+u′′1=0,

u0=u0, u′0=u0∗, u′′T=uT,

u0=u′1=u′′0=⋯=un-10=0;

see papers [2–7], respectively.

Integer derivative and integral BCs:

αu0-βu′0=∫01gsusds, γu1+δu′1=∫01hsusds,

u0=u′0=u′′0=0, u1=λ∫0ηusds;

see papers [8, 9], respectively.

Integer and fractional derivative BCs:

u0=D0+σ1Cu1=0, u′′0=D0+σ2Cu1=0,

u0=u′′0=0, u′1=D0+σCu1,

u0=u′0=0, u′1=D0+σCu1,

u0=0, D0+βu1=∑i=1m-2ξiD0+βuηi,

u0=0, u1+D0+βu1=kuξ+lD0+βuη,

u0=0, D0+βu1=aD0+βuξ,

u0=u′0=⋯=un-20=0, D0+αu1=0;

see papers [10–16], respectively.

Integer derivative and fractional integral BCs:

u0=αI0+puη,

u0=0, u′1=I0+σu1;

see papers [17, 18], respectively.

Besides, there are some other BCs involved in fractional differential equations, such as nonlinear BCs; refer to [19, 20].

Motivated greatly by the above-mentioned works, in this paper, we study the following boundary value problem (BVP for short) of nonlinear fractional differential equation with fractional integral BCs as well as integer and fractional derivative(1)D0+qCut=ft,ut,t∈0,1,u0=u′′0=0,D0+σ1Cu1=λI0+σ2u1,where D0+qC and D0+σ1C denote the standard Caputo fractional derivatives and I0+σ2 denotes the standard Riemann-Liouville fractional integral. Throughout this paper, we always assume that 2<q<3, 0<σ1≤1, σ2>0, λ≠Γ(2+σ2)/Γ(2-σ1), and f:[0,1]×R→R is continuous.

In order to prove our main results, the following well-known fixed point theorems are needed.

Theorem 1 (nonlinear alternative of Leray-Schauder type [<xref ref-type="bibr" rid="B21">21</xref>]).

Let B be a Banach space with E⊆B closed and convex. Assume Ω is a relatively open subset of E with θ∈Ω and T:Ω¯→E is a continuous and compact map. Then either

Let (X,d) be a complete metric space and T:X→X be contractive. Then T has a unique fixed point in X.

2. Preliminaries

In this section, we always assume that N={1,2,3,…}, α,β>0, and [α] denotes the integer part of α. Now, for the convenience of the reader, we give the definitions of the Riemann-Liouville fractional integrals and fractional derivatives and the Caputo fractional derivatives on a finite interval of the real line, which may be found in [1].

Definition 3.

The Riemann-Liouville fractional integrals I0+αu and I1-αu of order α on [0,1] are defined by (2)I0+αut≔1Γα∫0tusdst-s1-α,I1-αut≔1Γα∫t1usdss-t1-α,respectively.

Definition 4.

The Riemann-Liouville fractional derivatives D0+αu and D1-αu of order α on [0,1] are defined by (3)D0+αut≔ddtnI0+n-αut=1Γn-αddtn∫0tusdst-sα-n+1,D1-αut≔-ddtnI1-n-αut=1Γn-α-ddtn∫t1usdss-tα-n+1,respectively, where n=[α]+1.

Definition 5.

Let D0+α[u(s)](t)≡(D0+αu)(t) and D1-α[u(s)](t)≡(D1-αu)(t) be the Riemann-Liouville fractional derivatives of order α. Then the Caputo fractional derivatives CD0+αu and CD1-αu of order α on [0,1] are defined by (4)D0+αCut≔D0+αus-∑k=0n-1uk0k!skt,D1-αCut≔D1-αus-∑k=0n-1uk1k!1-skt, respectively, where(5)n=α+1,α∉N,α,α∈N.

Lemma 6 (see [<xref ref-type="bibr" rid="B23">23</xref>]).

If α+β>1, then the equation (I0+αI0+βu)(t)=(I0+α+βu)(t), t∈[0,1], is satisfied for u∈L10,1.

Lemma 7 (see [<xref ref-type="bibr" rid="B23">23</xref>]).

Let β>α. Then the equation D0+αCI0+βut=I0+β-αut, t∈[0,1], is satisfied for u∈C0,1.

Lemma 8 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let n be given by (5). Then the following relations hold:

For k∈{0,1,2,…,n-1}, D0+αCtk=0.

If β>n, then D0+αCtβ-1=Γβ/Γβ-αtβ-α-1.

Lemma 9 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let n be given by (5) and u∈Cn0,1. Then (6)I0+αD0+αCut=ut+c0+c1t+c2t2+⋯+cn-1tn-1,where ci∈R, i=0,1,…,n-1.

For any x∈L10,1, we define (7)xL1=∫01xtdt.

Lemma 10.

Let u∈L1[0,1] be nonnegative. Then I0+α+1ut≤I0+αuL1, t∈0,1.

Proof.

For any t∈0,1, we have (8)I0+α+1ut=1Γα+1∫0tust-s-αds=1αΓα∫0tust-sαds=1Γα∫0tus∫str-sα-1drds=1Γα∫0t∫0rusr-s1-αdsdr≤∫011Γα∫0rusr-s1-αdsdr=∫01I0+αurdr=I0+αuL1.

3. Main Results

In the remainder of this paper, for any nonnegative function g∈L1[0,1], we denote (9)Mg=I0+q-1gL1+Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2λI0+q+σ2g1+I0+q-σ1g1and for any y∈C[0,1], we use the norm(10)y∞=maxt∈0,1yt.

Lemma 11.

Let y∈C[0,1] be a given function. Then the BVP(11)D0+qCut=yt,t∈0,1,u0=u′′0=0,D0+σ1Cu1=λI0+σ2u1has a unique solution (12)ut=∫01Gt,sysds,t∈0,1,where(13)Gt,s=-Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλ1-sq+σ2-1Γq+σ2-1-sq-σ1-1Γq-σ1+t-sq-1Γq,0≤s≤t≤1,0,0≤t≤s≤1.

Proof.

It follows from the equation in (11) and Lemma 9 that(14)ut=I0+qyt-c0-c1t-c2t2,t∈0,1.So, (15)u′t=I0+q-1yt-c1-2c2t,t∈0,1,(16)u′′t=I0+q-2yt-2c2,t∈0,1.In view of (14), (16), and the BCs u(0)=u′′0=0, we get (17)c0=c2=0,and so, (18)ut=I0+qyt-c1t,t∈0,1. Then, by using Lemmas 6, 7, and 8, we may obtain(19)D0+σ1Cut=I0+q-σ1yt-c1Γ2Γ2-σ1t1-σ1,t∈0,1,I0+σ2ut=I0+q+σ2yt-c1Γ2Γ2+σ2t1+σ2,t∈0,1,which together with the BC (CD0+σ1u)1=λI0+σ2u1 implies that (20)c1=Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2λI0+q+σ2y1-I0+q-σ1y1.Therefore, the BVP (11) has a unique solution (21)ut=I0+qyt-Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2λI0+q+σ2y1-I0+q-σ1y1t=∫0t-Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλ1-sq+σ2-1Γq+σ2-1-sq-σ1-1Γq-σ1+t-sq-1Γqysds+∫t1-Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλ1-sq+σ2-1Γq+σ2-1-sq-σ1-1Γq-σ1ysds=∫01Gt,sysds,t∈0,1.

Lemma 12.

Let g∈L1[0,1] be nonnegative. Then (22)∫01Gt,sgsds≤Mg,t∈0,1.

Proof.

In view of Lemma 10, we have (23)∫01Gt,sgsds=∫0tGt,sgsds+∫t1Gt,sgsds≤∫0tΓ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλ1-sq+σ2-1Γq+σ2+1-sq-σ1-1Γq-σ1+t-sq-1Γqgsds+∫t1Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλ1-sq+σ2-1Γq+σ2+1-sq-σ1-1Γq-σ1gsds=1Γq∫0tgst-s1-qds+Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλΓq+σ2∫01gs1-s1-q-σ2ds+1Γq-σ1∫01gs1-s1-q+σ1ds=I0+qgt+Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2tλI0+q+σ2g1+I0+q-σ1g1≤I0+q-1gL1+Γ2+σ2Γ2-σ1λΓ2-σ1-Γ2+σ2λI0+q+σ2g1+I0+q-σ1g1=Mg,t∈0,1.

Now, we define an operator T:C[0,1]→C[0,1] by (24)Tut=∫01Gt,sfs,usds,t∈0,1.Obviously, u is a solution of the BVP (1) if and only if u is a fixed point of T.

Theorem 13.

Assume that f(t,0)≢0, t∈(0,1), and there exist nonnegative functions g1,g2∈L1[0,1], nonnegative increasing continuous function ϕ defined on [0,+∞), and r>0 such that(25)ft,x≤g1t+g2tϕx,t,x∈0,1×R,(26)Mg1+ϕrMg2<r.Then the BVP (1) has one nontrivial solution.

Proof.

Let Ω=u∈C[0,1]:u∞<r. Since G(t,s) and f(t,x) are continuous on 0,1×0,1 and 0,1×R, respectively, we may denote(27)L=maxt,s∈0,1×0,1Gt,s,(28)H=maxt,x∈0,1×-r,rft,x.

First, we prove that T:Ω¯→C[0,1] is continuous. Suppose that un (n=1,2,…), u0∈Ω¯, and un-u0∞→0 (n→∞). Then for any n and s∈[0,1], we have uns≤r. This together with (27) and (28) implies that, for any n and t∈[0,1], (29)Gt,sfs,uns≤LH,s∈0,1.By applying Lebesgue dominated convergence theorem, we get (30)limn→∞Tunt=limn→∞∫01Gt,sfs,unsds=∫01Gt,sfs,u0sds=Tu0t,t∈0,1,which indicates that T:Ω¯→C[0,1] is continuous.

Next, we show that T:Ω¯→C[0,1] is compact. Assume that K is a subset of Ω¯. Then for any u∈K, we have(31)us≤r,s∈0,1.In what follows, we will prove that T(K) is relatively compact. On the one hand, for any y∈T(K), there exists u∈K such that y=Tu, and so, it follows from (27), (28), and (31) that (32)yt=Tut=∫01Gt,sfs,usds≤∫01Gt,sfs,usds≤LH,t∈0,1,which shows that T(K) is uniformly bounded. On the other hand, for any ε>0, since G(t,s) is uniformly continuous on 0,1×0,1, there exists δ>0 such that, for any t1,t2∈0,1 with t1-t2<δ,(33)Gt1,s-Gt2,s<εH,s∈0,1.For any y∈T(K), there exists u∈K such that y=Tu, and so, for any t1,t2∈0,1 with t1-t2<δ, it follows from (28), (31), and (33) that (34)yt1-yt2=Tut1-Tut2=∫01Gt1,s-Gt2,sfs,usds≤∫01Gt1,s-Gt2,sfs,usds≤H∫01Gt1,s-Gt2,sds<ε,which indicates that T(K) is equicontinuous. By Arzela-Ascoli theorem, we know that T(K) is relatively compact. Therefore, T:Ω¯→C[0,1] is compact.

Now, we will prove that (a) of Theorem 1 is fulfilled. Suppose on the contrary that (b) of Theorem 1 is satisfied; that is, there exists u∈∂Ω and η∈(0,1) such that u=ηTu. Then, in view of (25), (26), and Lemma 12, we have (35)ut=ηTut≤Tut=∫01Gt,sfs,usds≤∫01Gt,sfs,usds≤∫01Gt,sg1s+g2sϕusds≤∫01Gt,sg1sds+ϕr∫01Gt,sg2sds≤Mg1+ϕrMg2<r,t∈0,1,which shows that (36)u∞<r.This contradicts the fact u∈∂Ω.

So, it follows from Theorem 1 that T has a fixed point u∗, which is a desired solution of the BVP (1). At the same time, since f(t,0)≢0,t∈(0,1), we know that the zero function is not a solution of the BVP (1). Therefore, u∗ is a nontrivial solution of the BVP (1).

Theorem 14.

Assume that there exists a nonnegative function g3∈L1[0,1] such that(37)ft,x-ft,y≤g3tx-y,t∈0,1,x,y∈R,(38)Mg3<1.Then the BVP (1) has a unique solution.

Proof.

For any u,v∈C[0,1], in view of (37) and Lemma 12, we have (39)Tut-Tvt=∫01Gt,sfs,us-fs,vsds≤∫01Gt,sfs,us-fs,vsds≤∫01Gt,sg3sus-vsds≤u-v∞∫01Gt,sg3sds≤Mg3u-v∞,t∈0,1.This indicates that (40)Tu-Tv∞≤Mg3u-v∞,which together with (38) implies that T is contractive. So, it follows from Theorem 2 that T has a unique fixed point, and so, the BVP (1) has a unique solution.

Example 15.

We consider the BVP(41)D0+5/2Cut=t-t2ut,t∈0,1,u0=u′′0=0,D0+1/2Cu1=12I0+3/2u1.

Let f(t,x)=t-t/2x, t,x∈[0,1]×R. Then f:[0,1]×R→R is continuous and ft,0≠0, t∈0,1.

If we choose g1(t)=t, g2(t)=t/2, t∈[0,1], and ϕ(y)=y, y∈[0,+∞), then it is easy to verify that (25) is satisfied.

Since q=5/2, σ1=1/2, σ2=3/2, and λ=1/2, a direct calculation shows that (42)Γ2+σ2Γ2-σ1=154,Mg1=6656+4305π43680π,Mg2=6656+4305π87360π.If we choose r=1, then (26) is fulfilled.

Therefore, it follows from Theorem 13 that the BVP (41) has one nontrivial solution.

Example 16.

We consider the BVP(43)D0+5/2Cut=tπutarctanut-12ln1+u2t,t∈0,1,u0=u′′0=0,D0+1/2Cu1=12I0+3/2u1.

Let f(t,x)=t/πxarctanx-1/2ln1+x2, t,x∈0,1×R. Then f:[0,1]×R→R is continuous.

If we choose g3t=t/2,t∈[0,1], then we may assert that (37) is satisfied. In fact, for any t∈[0,1], if x=y, then (37) is obvious. When x≠y, we may suppose that x<y. In this case, by Lagrange mean value theorem, there exists ξ∈(x,y) such that, for any t∈[0,1], (44)ft,x-ft,y=tπxarctanx-12ln1+x2-yarctany+12ln1+y2=tπarctanξx-y≤g3tx-y;that is, (37) is satisfied.

On the other hand, in view of Mg3=Mg2=6656+4305π/87360π, we know that (38) is fulfilled.

Therefore, it follows from Theorem 14 that the BVP (43) has a unique solution.

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This paper is supported by the National Natural Science Foundation of China (11661049).

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