# New Sequential Fractional Differential Equations with Mixed-Type Boundary Conditions

Authors: Haiyan Zhang; Yaohong Li; Jingbao Yang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we introduce new sequential fractional differential equations with mixed-type boundary conditionsCDq+kCDq−1ut=ft,ut,CDq−1ut,t∈0,1,α1u0+β1u1+γ1Iruη=ε1,η∈0,1,α2u′0+β2u′1+γ2Iru′η=ε2, where q∈1,2 is a real number, k,r>0,αi,βi,γi,εi∈ℝ,i=1,2,CDq is the Caputo fractional derivative, and the boundary conditions include antiperiodic and Riemann-Liouville fractional integral boundary value cases. Our approach to treat the above problem is based upon standard tools of fixed point theory and some new inequalities of norm form. Some existence results are obtained and well illustrated through the aid of examples.


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## Body

## 1. Introduction

In this paper, we focus on sequential fractional differential equations with mixed-type boundary conditions.(1)CDq+kCDq−1ut=ft,ut,CDq−1ut,t∈0,1,α1u0+β1u1+γ1Iruη=ε1,η∈0,1,α2u′0+β2u′1+γ2Iru′η=ε2,where q∈1,2 is a real number and k,r>0,αi,βi,γi,εi∈ℝ, i=1,2,CDq is the Caputo fractional derivative of order q. The nonlinearity term f contains the unknown function and its lower order fractional derivatives. The new boundary conditions include antiperiodic and Riemann-Liouville fractional integral boundary value cases which can be regarded as the linear combination of the values of the unknown function and its first derivatives at the end points of interval, and the Riemann-Liouville fractional integral value of the unknown function and its first derivatives at an interior point of interval.Fractional differential equations have attracted significant attention for their wide application in many fields of engineering and applied sciences (see [1–10]). Sequential fractional differential equations as an importance branch have also received wide attention; for instance, see [11–16]. Motivated by the HIV infection model and its application background in [12], the existence and uniqueness of solutions for the following sequential fractional differential system are obtained by means of Leray-Schauder’s alternative and Banach’s contraction principle
(2)CDp+λ1CDp−1ut=f1t,ut,vt,t∈0,1,CDp+λ2CDp−1vt=f2t,ut,vt,t∈0,1,u0=u′0=0,u1=avξ,v0=v′0=0,v1=buη,where λi>0 (i=1,2) is a parameter; 2<p, q≤3, CDp, and CDq are the Caputo fractional derivatives; and the nonlinearity terms f1,f2: 0,1×ℝ×ℝ⟶ℝ are the given continuous function.Antiperiodic boundary conditions arise in the mathematical problems of certain physical phenomena and processes. Recently, many scholars paid attention to solvability for fractional differential equations involving antiperiodic boundary conditions (see [17–21]). For example, in [21], the authors considered the nonlinear antiperiodic boundary value problems
(3)CDq+kCDq−1ut=ft,ut,q∈2,3,t∈0,T,α1u0+γ1uT=a,α2u′0+γ2u′T=b,α3u″0+γ3u″T=c,where CDq is the Caputo fractional derivatives of order q, αi, γi (i=1,2,3), a,b,c∈ℝ, k>0, f is a continuous function.Integral boundary conditions are believed to be more reasonable than the local boundary conditions, which can describe modeling of blood flow, cellular systems, population dynamics, heat transmission, etc. There are a number of results about fractional differential equations and partial differential equations with integral boundary condition; we refer the reader to see [17, 20, 22–43] and the references cited therein. In [20], the authors discussed the following fractional differential equation with integral boundary conditions given by
(4)CDα+β+kCDβxt=ft,xt,t∈0,1,x0=0,x1=μ∫01xsds,CDαx0+CDαx1=0,where CDα and CDβ are the Caputo fractional derivatives; 0<α<1, 1<β≤2, k>0, and μ>0 are real numbers; and f is a given continuous function.Observing the results of the above literature, an interesting and important question is whether antiperiodic and integral boundary conditions can be unified in a system. If we have unified the conditions, how can we obtain the existence of the solutions? Through a literature search, the sequential fractional differential equation (1) has not been given up to now.Now in this paper, we shall discuss the problem (1) by using the standard tools of fixed point theory and some new inequalities of norm from.


## 2. Preliminary and Lemmas

In this paper, we provide some necessary definitions and lemmas of the Caputo fractional calculus; for more information, see the books [1–3].Definition 1.
The Caputo derivative of fractional orderα>0 for a n-times continuously differentiable function f: 0,∞⟶ℝ is defined as
(5)CDαfx=1Γn−α∫0xx−tn−α−1fnxdt,n−1<α<n,where n=α+1 and α denotes the integer part of number α.Definition 2.
The Riemann-Liouville fractional integral of orderα for a function f is defined as
(6)Iαfx=1Γα∫0xx−tα−1ftdt,α>0,where Γα=∫0∞tα−1e−tdt.Lemma 3.
The Caputo fractional differential equationCDαut=0 has the general solution for α>0
(7)ut=c1+c2t+c3t2+⋯+cntn−1,where ci∈ℝ, i=1,2,⋯,n, and n is given as in Definition 1.Lemma 4.
Letht∈C0,1 and u∈C20,1. Then, the following sequential fractional differential equations
(8)CDq+kCDq−1ut=ht,0<t<1,1<q≤2,α1u0+β1u1+γ1Iruη=∈1,r>0,0<η<1,α2u′0+β2u′1+γ2Iru′η=∈2,have a unique solution
(9)ut=Δ1−Δ2e−ktΔ2Δ3Φh1,hη+1Δ2ψh1,hη+∫0te−kt‐sIq−1hsds,where
(10)Δ1=α1+β1e−k+γ1∫0ηη−sr−1Γre−ksds,Δ2=α1+β1+γ1ηrΓr+1,Δ3=−kα2−ke−kβ2−kγ2∫0ηη−sr−1Γre−ksds,Φh1,hη=−β2Iq−1h1−kβ2∫01e−k1−sIq−1hsds+γ2∫0ηη−sr−1ΓrIq−1hsds‐k∫0ηη−sr−1Γr∫0se−ks−mIq−1hmdmds−∈2,Ψh1,hη=−γ1∫0ηη−sr−1Γr∫0se−ks‐mIq−1hmdmds−β1∫01e−k1−sIq−1hsds+∈1.Proof.
Using Lemma3, the general solution of the fractional differential equations CDq+kCDq−1ut=ht can be written as
(11)ut=A0e−kt+A1+∫0te−kt−sIq−1hsds,where A0,A1∈ℝ. Differentiating (11) with respect to t, we get
(12)u′t=−kA0e−kt+Iq−1ht−k∫0te−kt−sIq−1hsds.
Applying the boundary condition (8) in (11) and (12), we obtain
(13)Δ1A0+Δ2A1=Ψh1,hη,Δ3A0+Φh1,hη=0.
A simultaneous solution of equation (13) leads to
(14)A0=−1Δ3Φh1,hη,A1=Δ1Δ2Δ3Φh1,hη+1Δ2Ψh1,hη.
SubstitutingA0 and A1 to (11), we obtain the desired solution in (9). The converse of the lemma follows by direct computation. The proof is completed.Remark 5.
Caputo fractional differentiating (11) with respect to t, we obtain
(15)CDq−1ut=kA0Γ2−q∫0tt−s1−qe−ksds+1Γ2−q∫0tt−s1−qIq−1hsds−kΓ2−q∫0tt−s1−q∫0se−ks−mIq−1hmdmds,where A0 is defined as (14).
SetC0,1 is all the continuous functions on 0,1, Cq−10,1=u∈C0,1:CDq−1u∈C0,1. Let E=Cq−10,1,·q−1 denotes the Banach space endowed with the norm defined by uq−1=u+CDq−1u=sup0≤t≤1ut+sup0≤t≤1CDq−1ut. For the convenience of the proofs in the next main results, we first give the bounds for integrals arising from the sequel, which are very important for us to establish the existence of solutions for problem (1).Lemma 6.
Suppose thath∈C0,1,ℝ. Then, we have
(i)
Φh1,hη≤β22−e−k/Γq+γ2ηq+r2−e−kη/ΓqΓr+1h+∈2
(ii)
Ψh1,hη≤β11−e−k/kΓq+γ1ηq+r1−e−kη/kΓqΓr+1h+∈1Proof.
(i)
Obviously, we have(16)Iq−1hs≤1Γq‐1∫0ss−ωq−2dωh=sqΓqh,Iq−1h1≤1Γqh,∫0se−ks−mIq−1hmdm≤sqΓq∫0se−ks−mdmh=sqΓq1−e−kskh,∫01e−k1−sIq−1hsds≤1Γq∫01e−k1−sdsh=1−e−kkΓqh,∫0ηη−sr‐1ΓrIq−1hsds≤ηqΓq∫0ηη−sΓrdsh≤ηq+rΓqΓr+1h.
Furthermore,(17)∫0ηη−sr−1Γr∫0se−ks−mIq−1hmdm≤ηq+r1−e−kηkΓqΓr+1h.Hence,
(18)Φh1,hη≤β2Iq−1h1+kβ2∫01e−k1−sIq−1hsds+γ2∫0ηη−sr−1ΓrIq−1hsds+k∫0ηη−sr−1Γr∫0se−ks−mIq−1hmdmds+∈2≤β2Γq+β21−e−kΓq+γ2ηq+rΓqΓr+1+γ2ηq+r1−e−kηΓqΓr+1h+∈2.
When the proof of (ii) is similar to (i), we omit it.
In view of Lemma4 and Remark 5, replacing ht by ft,ut,CDq−1ut in (9) and (15), we transform the solution of problem (1) into the fixed point of operator equation u=Fu , where operator F:E⟶E is defined as
(19)Fut=Δ1−Δ2e−ktΔ2Δ3Φf1,u1,CDq−1u1,fη,uη,CDq−1uη+1Δ2Ψf1,u1,CDq−1u1,fη,uη,CDq−1uη+∫0te−kt−sIq−1fs,usCDq−1usds,(20)DCq−1Fut=−kΓ2−qΦf1,u1,CDq−1u1,fη,uη,CDq−1uη∫0tt−s1−qe−ksds+1Γ2−q∫0tt−s1−qIq−1fs,us,CDq−1usds−kΓ2−q∫0tt−s1−q∫0se−ks−mIq−1fm,um,CDq−1umdmds.Lemma 7 (see [44]).
LetT:E→E be completely continuous (i.e., a map restricted to any bounded set in E is compact). Let
(21)F=x∈E:x=λTx,forsome0<λ<1.
Then, either setF is unbounded or F has at least one fixed point.Lemma 8 (see [45]).
LetE be a Banach space, D⊂E be closed and F:D⟶D a strict contraction, i.e., ∣Fx−Fy∣≤k∣x−y∣ for some k∈0,1 and all x,y∈D. Then, F has a unique fixed point.


## 3. Main Results

Before starting and introducing the main results, we list our assumption forf:
(i)
H0:f:0,1×ℝ×ℝ⟶ℝ is continuous
(ii)
H1: there exist positive constantsa11,a12,a13∈ℝ+ such that
(22)ft,u,v≤a11uσ1+a12vσ2+a13,∀t∈0,1,u,v∈ℝ,0<σ1,σ2<1
(iii)
H2: there exist positive constantsa21,a22,a23∈ℝ+ such that
(23)ft,u,v≤a21uτ1+a22vτ2+a23,∀t∈0,1,u,v∈ℝ,0<τ1,τ2>1
(iv)
H3: there exist positive constantsa31,a32,a33∈ℝ+ such that
(24)ft,u,v≤a31u+a32v+a33,∀t∈0,1,u,v∈ℝ
(v)
H4: there exist positive constantsa41,a42,a43∈ℝ+ such that
(25)ft,u,v≤a41uρ1+a42tvρ2+a43,∀t∈0,1,u,v∈ℝ,0<ρ1<1<ρ2
(vi)
H5: there exist positive constantsa51,a52∈ℝ+ such that
(26)ft,u,v−ft,u¯,v¯≤a51u−u¯+a52v−v¯,∀t∈0,1,u,v,u¯,v¯∈ℝFor convenience, we introduce the following symbols:(27)L1=β22−e−kΓq+γ2ηq+r2−e−kηΓqΓr+1,L2=β11−e−kkΓq+γ1ηq+r1−e−kηkΓqΓr+1,L3=Δ1+Δ2L1+Δ3L2Δ2Δ3+1kΓq+kΓqL1+2Γ3−qΓq,L4=Δ1+Δ2ε2+Δ3ε1Δ2Δ3+kε2Γ3−q.Theorem 9.
Suppose that (H0) and (H1) hold. Then, problem (1) has at least one solution.Proof.
We first define a ball inE as BR=uu∈E,uq−1≤R, where
(28)R≥max3L3a111/1−σ1,3L3a21/1−σ1,33L3a13+L4.
Then, we show thatT:BR⟶BR. For ∀u∈BR, using Lemma 6 and the condition (H1), we have
(29)Fts≤∆1+∆2e−kt∆2∆3Φf1,u1,CDq−1u1,fη,uη,CDq−1uη+1∆2Ψf1,u1,CDq−1u1,fη,uη,CDq−1uη+∫0te−kt−sIq−1fs,us,CDq−1usds≤∆1+∆2∆2∆3L1a11Rσ1+a12Rσ2+a13+ε2+1∆2L2a11Rσ1+a12Rσ2+a13+ε1+1kΓqa11Rσ1+a12Rσ2+a13≤∆1+∆2L1+∆3L2∆2∆3+1kΓqa11Rσ1+a12Rσ2+a13+∆1+∆2ϵ2+∆3ϵ1∆2∆3,(30)CDq−1Fut≤kΓ3−qΦf1,u1,CDq−1u1,fη,uη,CDq−1uη+1Γ3−qΓqfs,us,CDq−1us+1Γ3−qΓqfs,us,CDq−1us≤kL1Γq+2Γ3−qΓqa11Rσ1+a12Rσ2+a13+kε2Γ3−q.
From (29) and (30), we obtain
(31)∥Fu∥q−1≤L3a11Rσ1+L3a12Rσ2+L4≤13R+13R=R.
This meansF:BR⟶BR.
From the formula (19), it is easy to know that operators Fut,CDq−1Fut are continuous on 0,1. Now, we show that operator F is equicontinuous. Set
(32)M=maxt∈0,1ft,ut,CDq−1ut,∀u∈BR.
Lett1,t2∈0,1t1<t2, we have
(33)Fut2−Fut1≤e−kt2−e−kt1Δ3Φf1,u1,CDq−1u1,fη,uη,CDq−1uη+∫0t1e−kt2−s−e−kt1−sIq−1fs,us,CDq−1usds+∫t1t2e−kt2−sIq−1fs,us,CDq−1usds≤L1M+ε2Δ3e−kt1−e−kt2+MΓq∫0t1e−kt2−s−e−kt1−sds+MΓq∫t1t2e−kt2−sds≤L1M+ε2Δ3e−kt1−e−kt2+MΓq21−e−kt2−t1+e−kt1−e−kt2,(34)CDq−1Fut2−CDq−1Fut1≤kΓ2−qΦf1,u1,CDq−1u1,fη,uη,CDq−1uη×∫0t2t2−s1−qe−ksds−∫0t1t1−s1−qe−ksds+1Γ2−q∫0t2t2−s1−qIq−1fs,us,CDq−1usds−∫0t1t1−s1−qIq−1fs,us,CDq−1usds+kΓ2−q∫0t2t2−s1−q∫0se−ks−mIq−1fm,um,CDq−1umdmds−∫0t1t1−s1−q∫0se−ks−mIq−1fm,um,CDα−1umdmds≤kL1M+kε2Γ2−q∫0t1t2−s1−q−t1−s1−qe−ksds+∫t1t2t2−s1−qe−ksds+2MΓqΓ2−q∫0t1t2−s1−q−t1−s1−qds+∫t1t2t2−s1−qds.
In (33) and (34), letting t1 ⟶ t2, then,
(35)Fut2−Fut1⟶0,CDq−1Fut2−CDq−1Fut1⟶0,
That is, ast1 ⟶ t2,
(36)Fut2−Fut2q−1⟶0.
Therefore,FBR is an equicontinuous set. Furthermore, it is uniformly bounded because of FBR⊂BR. Applying the Arzelà-Ascoli theorem, we can infer that F is a completely continuous operator.
ConsiderV=u∈BR∣u=μFu,0<μ<1 and show that V is bounded. For u∈V, we know uq−1<Fuq−1≤R. By Lemma 7, problem (1) has at least one solution in BR.Theorem 10.
Suppose that (H0) and (H2) hold. If3L3a23+L4≤min3L3a211/1−τ1,3L3a221/1−τ2, then problem (1) has at least one solution.Proof.
The proof is similar to Theorem9. We just need to make sure that R satisfies 3L3a23+L4≤R≤min3L3a211/1−τ1,3L3a221/1−τ2 in BR.Theorem 11.
Suppose that (H0) and (H3) hold. IfL3a31+a32<1, then problem (1) has at least one solution.Proof.
The proof is similar to Theorem9, we omit it.Theorem 12.
Suppose that (H0) and (H4) hold. Ifmax3L3a411/1−ρ1,3L3a43+L4≤3L3a421/1−ρ2, then problem (1) has at least one solution.Proof.
The proof is similar to Theorem10, we omit it.Theorem 13.
Suppose that (H0) and (H5) hold. IfL3a51+a52<1, then problem (1) has a unique solution.Proof.
Definesupt∈0,1ft,0,0=M<∞, such that r≥L3M+L4/1−L3a51+a52.
First, we show thatTBr⊂Br, where Br=uu∈E:uα−1≤r. For ∀u∈Br, by direct calculation, we have
(37)Fut≤Δ1+Δ2Δ2Δ3Φf1,u1,CDq−1u1−f1,0,0+f1,0,0,fη,uη,CDq−1uη−fη,0,0+fη,0,0+1Δ2Ψf1,u1,CDq−1u1−f1,0,0+f1,0,0,fη,uη,CDq−1uη−fη,0,0+fη,0,0+∫0te−kt−sIq−1fs,us,CDq−1us−fs,0,0+fs,0,0ds≤Δ1+Δ2Δ2Δ3L1a51u+a52CDq−1u+M+ε2+1Δ2L2a51u+a52CDq−1u+M+ε1+1kΓqa51u+a52CDq−1u+M≤Δ1+Δ2L1+Δ3L2Δ2Δ3+1kΓqa51+a52uq−1+M+Δ1+Δ2ε2+Δ3ε1Δ2Δ3,(38)CDq−1Fut≤kΓ3−qΦf1,u1,CDq−1u1−f1,0,0+f1,0,0,fη,uη,CDq−1uη−fη,0,0+fη,0,0+2Γ3−qΓqfs,us,CDq−1us−fs,0,0+fs,0,0≤kΓ3−qL1a51u+a52CDq−1u+M+ε2+2Γ3−qΓqa51u+a52CDq−1u+M≤kΓqL1+2Γ3−qΓqa51+a52uq−1+M+kε2Γ3−q.
Combining (37) and (38), we obtain
(39)Fuα−1≤L3a51+a52uq−1+L3M+L4≤r.
Now, for anyu,v∈Br, we have
(40)Fut−Fvt≤Δ1+Δ2Δ2Δ3Φf1,u1,CDq−1u1,fη,uη,CDα−1uη−Φf1,v1,CDα−1v1,fη,vη,CDα−1vη+1Δ2Ψf1,u1,CDα−1u1,fη,uη,CDα−1uη−Ψf1,v1,CDα−1v1,fη,vη,CDα−1vη+∫0te−kt−sIq−1fs,us,CDα−1us−fs,vs,CDα−1vsds≤Δ1+Δ2L1+Δ3L2Δ2Δ3+1kΓqa51u−v+a52CDα−1u−CDα−1v≤Δ1+Δ2L1+Δ3L2Δ2Δ3+1kΓqa51+a52u−vα−1.
Similar to (40), one has
(41)CDα−1Fut−CDα−1Fvt≤kΓqL1+2Γ3−qΓqa51+a52u−vα−1.
Therefore,(42)Fu−Fvα−1≤L3a51+a52u−vα−1.
SinceL3a51+a52<1, F is a contraction operator. Using Lemma 8, operator F has a unique fixed point which is the unique solution of problem (1).
In order to illustrate our main results, we consider the following sequential fractional differential equations:(43)CD1.5+CD0.5ut=ft,ut,CD0.5ut,t∈0,1,0.1u0+0.2u1−I2u0.5=0.01,0.2u′0+0.1u′1+2I2u′0.5=0.02,where q=1.5,k=1,r=2,η=0.5,α1=0.1,β1=0.2,γ1=−1,ε1=0.01,α2=0.2,β2=0.1,γ1=2, and ε2=0.02.Example 1.
Letft,ut,CD0.5ut=0.01ut0.2+0.02CD0.5ut0.1+0.151−t2. So we have
(44)ft,ut,CD0.5ut≤0.01ut0.2+0.02CD0.5ut0.1+0.15,where a11=0.01, a12=0.02, a13=0.15, σ1=0.2, and σ2=0.1. Theorem 9 implies that problem (43) has at least one solution.Example 2.
Letft,ut,CD0.5ut=0.01ut1.2+0.02CD0.5ut1.1+0.011−t2. By direct calculation, we obtain that
(45)Δ1=0.093438,Δ2=0.246808,Δ3=−0.407603,L1=0.323142,L2=0.161275,L3=5.785852,L4=0.130729.
In the meantime, we have(46)ft,ut,CD0.5ut≤0.01ut1.2+0.02CD0.5ut1.1+0.01,where a21=0.01,a22=0.22,a23=0.01,τ1=1.2, and τ2=1.1. So we obtain 3L3a23+L4=0.565763<min3L3a211/1−τ1,3L3a221/1−τ2=1.111599. Theorem 10 implies that problem (43) has at least one solution.Example 3.
Letft,ut,CD0.5ut=0.01ut+0.02CD0.5ut+0.011−t2. So we have
(47)ft,ut,CD0.5ut−ft,vt,CD0.5vt≤0.01ut−vt+0.02CD0.5ut−CD0.5vt,where a51=0.01 and a52=0.02. Combining with the calculation result of Example 2, we can obtain that L3a51+a52=0.173576<1. Hence, Theorem 13 implies that problem (43) has a unique solution.


## 4. Conclusions

We have obtained some existence results for new sequential fractional differential equations by using some nonlinear growth conditions(48)ft,uv≤b11uθ1+b12vθ1+b13,θ1,θ2∈0,+∞,which is different from the existing linear condition. Obviously, these results are easy to verify and apply (see Example 1 and Example 3).On the other hand, we note that our results contain some special types of results by fixing the parameters in the given problem (1). For instance, let α1=1, β1=1, γ1=0, α2=1, β2=1, and γ2=0, then the results of this paper are the following sequential fractional differential equations with the boundary value conditions of the form u0+u1=ϵ1 and u′0+u′1=ϵ2. Further, letting α1 = 0, β1 = 1, γ1 = − 1, ϵ1 = 0, α2 = 0, β2 = 1, γ2 = − 1, and ϵ2 = 0, we obtain the results for the boundary conditions u1=Iruη and u′1=Iru′η.



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