# Separated Boundary Value Problems of Sequential Caputo and Hadamard Fractional Differential Equations

Authors: Jessada Tariboon; Asawathep Cuntavepanit; Sotiris K. Ntouyas; Woraphak Nithiarayaphaks
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, we discuss the existence and uniqueness of solutions for new classes of separated boundary value problems of Caputo-Hadamard and Hadamard-Caputo sequential fractional differential equations by using standard fixed point theorems. We demonstrate the application of the obtained results with the aid of examples.


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## Body

## 1. Introduction

Fractional differential equations have been of increasing importance for the past decades due to their diverse applications in science and engineering such as biophysics, bioengineering, virology, control theory, signal and image processing, blood flow phenomena, etc.; see [1–6]. Many interesting results of the existence of solutions of various classes of fractional differential equations have been obtained; see [7–15] and the references therein.Sequential fractional differential equations are also found to be of much interest [16, 17]. In fact, the concept of sequential fractional derivative is closely related to the nonsequential Riemann-Liouville derivatives, for details, see [3]. For some recent results on boundary value problems for sequential fractional differential equations; see [18–22] and references cited therein.In this paper, we discuss existence and uniqueness of solutions for two sequential Caputo-Hadamard and Hadamard-Caputo fractional differential equations subject to separated boundary conditions as(1)DpCDqHxt=ft,xt,t∈a,b,α1xa+α2DqHxa=0,β1xb+β2DqHxb=0,and(2)DqHDpCxt=ft,xt,t∈a,b,α1xa+α2DpCxa=0,β1xb+β2DpCxb=0,where DpC and DqH are the Caputo and Hadamard fractional derivatives of orders p and q, respectively, 0<p,q≤1, f:[a,b]×R→R is a continuous function, a>0 and αi,βi∈R, i=1,2.It can be observed that the sequential Caputo-Hadamard and Hadamard-Caputo fractional differential equations in (1) and (2) are different type when p=1 and q=1, since (3)ddttddtxt=td2xtdt2+dxtdt=ft,xt,and (4)tddtddtxt=td2xtdt2=ft,xt,for t∈(a,b), respectively.The rest of the paper is arranged as follows. In Section2, we establish basic results that lay the foundation for defining a fixed point problem equivalent to the given problems (1) and (2). The main results, based on Banach’s contraction mapping principle, Krasnoselskii’s fixed point theorem, and nonlinear alternative of Leray-Schauder type, are obtained in Section 3. Illustrating examples are discussed in Section 4.


## 2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [4, 5] and present preliminary results needed in our proofs later.Definition 1 (see [5]).
For an at leastn-times differentiable function g:[a,∞)→R, the Caputo derivative of fractional order q is defined as (5)DqCgt=1Γn-q∫att-sn-q-1gnsds,n-1<q<n,n=q+1,where [q] denotes the integer part of the real number q.Definition 2 (see [5]).
The Riemann-Liouville fractional integral of orderq of a function g:[a,∞)→R is defined as (6)IqgRLt=1Γq∫atgst-s1-qds,q>0,provided the integral exists.Definition 3 (see [5]).
For an at leastn-times differentiable function g:[a,∞)→R, the Caputo-type Hadamard derivative of fractional order q is defined as (7)DqHgt=1Γn-q∫atlogtsn-q-1δngsdss,n-1<q<n,n=q+1,where δ=td/dt, log(·)=loge(·).Definition 4 (see [5]).
The Hadamard fractional integral of orderq is defined as (8)IqHgt=1Γq∫atlogtsq-1gsdss,q>0,provided the integral exists.Lemma 5 (see [5]).
Forq>0, the general solution of the fractional differential equation CDqu(t)=0 is given by (9)ut=c0+c1t-a+⋯+cn-1t-an-1,where ci∈R, i=0,1,2,…,n-1(n=[q]+1).In view of Lemma5, it follows that (10)IqRLDqCut=ut+c0+c1t-a+⋯+cn-1t-an-1, for some ci∈R,  i=0,1,2,…,n-1 (n=[q]+1).Lemma 6 (see [23]).
Letu∈ACδn[a,b] or  Cδn[a,b] and q∈C, where Xδn[a,b]={g:[a,b]→C:δn-1g(t)∈X[a,b]}. Then, we have (11)IqHDqHut=ut-∑k=0n-1cklogtak,where ci∈R, i=0,1,2,…,n-1(n=[q]+1).In order to define the solution of the boundary value problem (1), we consider the linear variant(12)DpCDqHxt=yt,t∈a,b,α1xa+α2DqHxa=0,β1xb+β2DqHxb=0,where y∈C([a,b],R).Lemma 7.
Let(13)Ω≔β1α2-α1β1logb/aqΓq+1+β2≠0.Then, the unique solution of the separated boundary value problem of sequential Caputo and Hadamard fractional differential equation (12) is given by the integral equation(14)xt=β1Ωα1logt/aqΓq+1-α2IqHIpRLyb+β2Ωα1logt/aqΓq+1-α2IpRLyb+IqHIpRLyt,t∈a,b.Proof.
Taking the Riemann-Liouville fractional integral of orderp to the first equation of (12), we get(15)DqHxt=c1+IpRLyt,c1∈R.Again taking the Hadamard fractional integral of order q to the above equation, we obtain(16)xt=c2+c1logt/aqΓq+1+IqHIpRLyt,c2∈R.Substituting t=a in (15)-(16) and applying the first boundary condition of (12), it follows that(17)α2c1+α1c2=0.For t=b in equations (15)-(16) and using the second boundary condition of (12), it yields(18)c1β1logb/aqΓq+1+β2+β1c2=-β1IqHIpRLyb-β2IpRLyb.Solving the linear system of (17) and (18) for finding two constants c1,c2, we get(19)c1=α1β1ΩIqHIpRLyb+α1β2ΩIpRLyband(20)c2=-β1α2ΩIqHIpRLyb-α2β2ΩIpRLyb.Substituting constants c1 and c2 in (16), we get the integral equation (14). The converse follows by direct computation. The proof is completed.In the same way, we can prove the following lemma, which concerns a linear variant of problem (2):(21)DqHDpCxt=zt,t∈a,b,α1xa+α2DpCxa=0,β1xb+β2DpCxb=0,where z∈C([a,b],R).Lemma 8.
Let(22)Ω∗≔β1α2-α1β1b-apΓp+1+β2≠0.Then, the unique solution of the separated boundary value problem of sequential Caputo and Hadamard fractional differential equation (21) is given by the integral equation(23)xt=β1Ω∗α1t-apΓp+1-α2IpRLIqHzb+β2Ω∗α1t-apΓp+1-α2IqHzb+IpRLIqHzt,t∈a,b.


## 3. Main Results

We set some abbreviate notations for sequential Riemann-Liouville and Hadamard fractional integrals of a function with two variables as(24)IqHIpRLfxϕ=1ΓqΓp∫aϕ∫aslogϕsq-1s-rp-1fr,xrdrdss, and (25)IpRLIqHfxϕ=1ΓpΓq∫aϕ∫asϕ-sp-1logsrq-1fr,xrdrrds, where ϕ∈{t,b}. Also we use this one for a single Riemann-Liouville and Hadamard fractional integrals of orders p and q, respectively.In this section, we will use fixed point theorems to prove the existence and uniqueness of solution for problems (1) and (2). To accomplish our purpose, we define the Banach space C=C([a,b],R), of all continuous functions on [a,b] to R endowed with the norm x=sup{|x(t)|,t∈[a,b]}. In addition, we define the operator K:C→C by(26)Kxt=β1Ωα1log⁡t/aqΓq+1-α2IqHIpRLfxb+β2Ωα1log⁡t/aqΓq+1-α2IpRLfxb+IqHIpRLfxt,where Ω≠0 is defined by (13) and fx(t)=f(t,x(t)). Note that the separated boundary value problem (1) has solutions if and only if x=Kx has fixed points.For computational convenience we put(27)Ω1=β1Ωα1logb/aqΓq+1+α2IqHIpRL1b+β2Ωα1logb/aqΓq+1+α2IpRL1b+IqHIpRL1b.To prove the existence theorems of problem (2), we define the operator A:C→C by(28)Axt=β1Ω∗α1t-apΓp+1-α2IpRLIqHfxb+β2Ω∗α1t-apΓp+1-α2IqHfxb+IpRLIqHfxt.Now, we prove the existence and uniqueness result for problem (1). For problem (2) the proof is similar and omitted.Theorem 9.
Suppose that
( H 1 ) there exists a function ψ(t)>0,t∈[a,b], such that(29)ft,x-ft,y≤ψtx-yfor  allt∈a,bandx,y∈R.If ψ∗Ω1<1, where ψ∗=sup{ψ(t):t∈[a,b]}, then the separated boundary value problem (1) has a unique solution on [a,b].Proof.
Firstly, we define a ballBr as Br={x∈C:x≤r}, where the constant r satisfies (30)r≥MΩ11-ψ∗Ω1,where M=sup{f(t,0):t∈[a,b]}. Next, we will show that KBr⊂Br. For any x∈Br and using the triangle inequality fx≤fx-f0+f0, we have(31)Kxt≤β1Ωα1logb/aqΓq+1+α2IqHIpRLfxb+β2Ωα1logb/aqΓq+1+α2IpRLfxb+IqHIpRLfxt≤β1Ωα1logb/aqΓq+1+α2IqHIpRLfx-f0+f0b+β2Ωα1logb/aqΓq+1+α2IpRLfx-f0+f0b+IqHIpRLfx-f0+f0b≤β1Ωα1logb/aqΓq+1+α2IqHIpRLψ∗r+Mb+β2Ωα1logb/aqΓq+1+α2IpRLψ∗r+Mb+IqHIpRLψ∗r+Mb=ψ∗Ω1r+MΩ1≤r, which implies that KBr⊂Br. Let x,y∈Br, then (32)Kxt-Kyt≤β1Ωα1logb/aqΓq+1+α2IqHIpRLfx-fyb+β2Ωα1logb/aqΓq+1+α2IpRLfx-fyb+IqHIpRLfx-fyt≤β1Ωα1logb/aqΓq+1+α2ψ∗x-yIqHIpRL1b+β2Ωα1logb/aqΓq+1+α2ψ∗x-yIpRL1b+ψ∗x-yIqHIpRL1t=ψ∗Ω1x-y,which yields that Kx-Ky≤ψ∗Ω1x-y. Since ψ∗Ω1<1, we deduce that the operator K is a contraction. By Banach contraction mapping principle the operator K has a unique fixed point, which leads that problem (1) has a unique solution on [a,b].Theorem 10.
Let(H1) in Theorem 9 holds. If ψ∗Ω1∗<1, where(33)Ω1∗=β1Ω∗α1b-apΓp+1+α2IpRLIqH1b+β2Ω∗α1b-apΓp+1+α2IqH1b+IpRLIqH1b, then the separated boundary value problem (2) has a unique solution on [a,b].Our second existence result is based on Krasnoselskii’s fixed point theorem.Theorem 11 ((Krasnoselskii’s fixed point theorem) [24]).
LetQ be a closed, bounded, convex, and nonempty subset of a Banach space X. Let A,B be operators such that (a)
Ax+By∈Q where x,y∈Q;(b)
A is compact and continuous;(c)
B is a contraction mapping.
Then there existsz∈Q such that z=Az+Bz.Theorem 12.
Letf:[a,b]×R→R be a continuous function satisfying (H1) in Theorem 9. In addition, assume that
(H2) ft,x≤φ(t), ∀(t,x)∈[a,b]×R and φ∈C([a,b],R+).
If(34)ψ∗IqHIpRL1b<1,then the separated boundary value problem (1) has at least one solution on [a,b].Proof.
LetBρ={x∈C:x≤ρ}, where a constant ρ satisfying ρ≥φ∗Ω1 and φ∗=sup{φ(t):t∈[a,b]}. We decompose the operator K into two operators K1 and K2 on Bρ with(35)K1xt=β1Ωα1logt/aqΓq+1-α2IqHIpRLfxb+β2Ωα1logt/aqΓq+1-α2IpRLfxb,t∈a,b,K2xt=IqHIpRLfxt,t∈a,b.Note that the ball Bρ is a closed, bounded, and convex subset of the Banach space C.
Now, we will show thatK1x+K2y∈Bρ for satisfying condition (a) of Theorem 11. Setting x,y∈Bρ, then we have(36)K1xt+K2yt≤β1Ωα1logb/aqΓq+1+α2IqHIpRLfxb+β2Ωα1logb/aqΓq+1+α2IpRLfxb+IqHIpRLfyt≤φ∗α1β1Ωlogb/aqΓq+1+β1α2ΩIqHIpRL1b+φ∗α1β2Ωlogb/aqΓq+1+α2β2ΩIpRL1b+φ∗IqHIpRL1b=φ∗Ω1≤ρ.This means that K1x+K2y∈Bρ. To prove that K2 is a contraction mapping, for x,y∈Bρ, we have(37)K2x-K2y≤IqHIpRLfx-fyb≤ψ∗IqHIpRL1bx-yby condition (H1), which is a contraction, by (34). Therefore, the condition (c) of Theorem 11 is satisfied. Next we will show that the operator K1 is compact and continuous. By using the continuity of the function f on [a,b]×R, we can conclude that the operator K1 is continuous. For x∈Bρ, it follows that (38)K1x≤φ∗Ω2, where(39)Ω2=β1Ωα1logb/aqΓq+1+α2IqHIpRL1b+β2Ωα1logb/aqΓq+1+α2IpRL1b,which implies that the set K1Bρ is uniformly bounded. Now we are going to prove that K1Bρ is equicontinuous. For τ1, τ2∈[a,b] such that τ1<τ2 and for x∈Bρ, we have (40)K1xτ2-K1xτ1≤α1β1ΩΓq+1logτ2aq-logτ1aqIqHIpRLfxb+α1β2ΩΓq+1logτ2aq-logτ1aqIpRLfxb≤φ∗Ω2logτ2aq-logτ1aq,which is independent of x and also tends to zero as τ1→τ2. Hence the set K1Bρ is equicontinuous. Therefore the set K1Bρ is relatively compact. By applying the Arzelá-Ascoli theorem, the operator K1 is compact on Bρ. Therefore the operators K1 and K2 satisfy the assumptions of Theorem 11. By the conclusion of Theorem 11, we get that the separated boundary value problem (1) has at least one solution on [a,b]. This completes the proof.Theorem 13.
Assume that(H1) and (H2) are fulfilled. If ψ∗[IpRL(IqH(1))(b)]<1, then the separated boundary value problem (2) has at least one solution on [a,b].The above theorem can be proved by applying Krasnoselskii’s fixed point theorem to the operatorA defined in (28).Remark 14.
If the operatorsK1 and K2 are interchanged, then we have the existence results as follows: (i)
Ifψ∗Ω2<1, then problem (1) has at least one solution on [a,b].(ii)
Ifψ∗Ω2∗<1, then problem (2) has at least one solution on [a,b], where (41)Ω2∗=β1Ω∗α1b-apΓp+1+α2IpRLIqH1b+β2Ω∗α1b-apΓp+1+β2IqH1b.However, in application to existence theory, the computation of values IqH(IpRL(1))(b) and IpRL(IqH(1))(b) is easier than Ω2 and Ω2∗, respectively.The third existence result will be proved by applying Leray-Schauder nonlinear alternative.Theorem 15 ((nonlinear alternative for single valued maps) [25]).
LetE be a Banach space, C a closed, convex subset of E,U an open subset of C, and 0∈U. Suppose that D:U¯→C is a continuous; compact (that is, D(U¯) is a relatively compact subset of C) map. Then either (i)
D has a fixed point in U¯ or(ii)
there is ax∈∂U (the boundary of U in C) and ν∈(0,1) with x=νD(x).Let us state and prove the existence theorem.Theorem 16.
Suppose that
(H3) there exist a continuous nondecreasing function ξ:[0,∞)→(0,∞) and a function η∈C([a,b],R+) such that(42)ft,x≤ηtξxfor  eacht,x∈a,b×R;
(H4) there exists a constant K>0 such that(43)KηξKΩ1>1. Then the separated boundary value problem (1) has at least one solution on [a,b].Proof.
Let the operatorK be defined in (26). Let us prove that the operator K maps bounded sets (balls) into bounded sets in C. For a constant λ>0, we define a bounded ball Bλ={x∈C:x≤λ}. Then for t∈[a,b], one has (44)Kxt≤β1Ωα1logb/aqΓq+1+α2IqHIpRLfxb+β2Ωα1logb/aqΓq+1+α2IpRLfxb+IqHIpRLfxt≤ηξxβ1Ωα1logb/aqΓq+1+α2IqHIpRL1b+ηξxβ2Ωα1logb/aqΓq+1+α2IpRL1b+ηξxIqHIpRL1b≤ηξλΩ1, which implies that(45)Kx≤ηξλΩ1.After that we will show that the operator K maps bounded sets into equicontinuous sets of C. Let τ1,τ2 be any two points in [a,b] such that τ1<τ2. Then for x∈Bλ, we have (46)Kxτ2-Kxτ1≤α1β1ΩΓq+1logτ2aq-logτ1aqIqHIpRLfxb+α1β2ΩΓq+1logτ2aq-logτ1aqIpRLfxb+IqHIpRLfxτ2-IqHIpRLfxτ1≤ηξλα1β1ΩΓq+1logτ2aq-logτ1aqIqHIpRL1b+ηξλα1β2ΩΓq+1logτ2aq-logτ1aqIpRL1b+ηξλIqHIpRL1τ2-IqHIpRL1τ1. As τ1→τ2, the right-hand side of the above inequality tends to zero independently of x∈Bρ. Hence, by applying the Arzelá-Ascoli theorem, the operator K:C→C is completely continuous.
The result will be followed from the Leray-Schauder nonlinear alternative if we prove the boundedness of the set of the solutions to equationx=νKx for ν∈(0,1). Let x be a solution of the operator equation x=Kx. Then, for t∈[a,b], by directly computation, we have(47)xt≤ηξxΩ1, which leads to (48)xηξxΩ1≤1. From the assumption (H4), there exists a positive constant K such that x≠K. Let us set (49)U=x∈C:x<K. It is easy to see that the operator K:U¯→C is continuous and completely continuous. From the choice of U, there is no x∈∂U such that x=νKx for some ν∈(0,1). Therefore, by the nonlinear alternative of Leray-Schauder type (Theorem 15), we deduce that the operator K has a fixed point x∈U¯ which is a solution of problem (1). The proof is completed.Theorem 17.
Assume that the condition(H3) in Theorem 16 is satisfied. If a positive constant K1 satisfying(50)K1ηξK1Ω1∗>1,then the separated boundary value problem (2) has at least one solution on [a,b].The next two special cases can be obtained by settingη(t)=1, t∈[a,b] and ξ(y)=Ey+G, y∈[0,∞) with two constants E≥0, G>0.Corollary 18.
Letf:[a,b]×R→R be a continuous function satisfying ft,x≤Ex+G, for all x∈R. Then (i)
ifEΩ1<1, then the separated boundary value problem (1) has at least one solution on [a,b];(ii)
ifEΩ1∗<1, then the separated boundary value problem (2) has at least one solution on [a,b].


## 4. Examples

In this section, we present some examples to illustrate our results.Example 1.
Consider the following sequential Caputo-Hadamard fractional differential equations with separated boundary conditions(51)D1/2CD1/3Hxt=ft,xt,t∈12,52,14x12+35D1/3Hx12=0,58x52+79D1/3Hx52=0.Herep=1/2, q=1/3, a=1/2, b=5/2, α1=1/4, α2=3/5, β1=5/8, and β2=7/9. From given information, we find that Ω=-0.0244992447, I1/3HI1/2RL(1)(5/2)=1.622871815, and I1/2RL(1)(5/2)=1.595769121 which yield Ω1=87.06444876.(i) Letf:[1/2,5/2]×R→R with(52)ft,x=cos2⁡πt2t-1/2+90x2+xx+1+12.It follows that (53)ft,x-ft,y≤cos2⁡πtt-1/2+90x-y≔ψtx-y.Then condition (H1) is satisfied with ψ∗=1/90. Thus ψ∗Ω1=0.9673827640<1. Hence, by Theorem 9, problem (51) with (52) has a unique solution on [1/2,5/2].(ii) Givenf:[1/2,5/2]×R→R by(54)ft,x=cos2⁡πt2t-1/2+2x2+xx+1+12.Observe that the function f defined in (54) satisfies (H1) with ψ∗=1/2. But the Theorem 9 can not be applied to this case because the value of ψ∗Ω1=43.53222438>1. However, by the benefit of Theorem 12, we have ψ∗I1/3HI1/2RL(1)(5/2)=0.8114359075<1. By the conclusion of Theorem 12, problem (51) with (54) has at least one solution on [1/2,5/2].Example 2.
Consider the following sequential Hadamard-Caputo fractional differential equations with separated boundary conditions(55)D3/4HD2/5Cxt=gt,xt,t∈18,78,311x18+π4D2/5Cx18=0,29x78+213D2/5Cx78=0.Hereq=3/4, p=2/5, a=1/8, b=7/8, α1=3/11, α2=π/4, β1=2/9, and β2=2/13. From above information, we can find that Ω∗=0.03840540910, I2/5RLI3/4H(1)(7/8)=1.526044488, and I3/4H(1)(7/8)=1.792656288 which can be computed the value of Ω1∗=15.74791264.(i) The functiong:[1/8,7/8]×R→R is defined by(56)gt,x=8sin4⁡πt263+8tx6x4+1+1.Setting η(t)=(8sin4⁡πt/(263+8t)) and ξ(y)=y2+1, we see that the condition (H3) of Theorem 16 is satisfied with the above function g(t,x). In addition, we can find that η=1/33. Then there exists a constant K such that K∈(0.7350333746,1.360482441) satisfying inequality (50). Therefore, applying Theorem 17, problem (55) with (56) has at least one solution on [1/8,7/8].(ii) Letg:[1/8,7/8]×R→R by(57)gt,x=e-t-1/8216x8x7+1+341+t2.It is easy to see that the function g(t,x) defined in (57) can be expressed as gt,x≤1/16x+(3/4). Then (1/16)Ω1∗=0.9842445400<1. Using (ii) of the Corollary 18, the problem (55) with (57) has at least one solution on [1/8,7/8].



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