# On the Nonlinear Fractional Differential Equations with Caputo Sequential Fractional Derivative

Authors: Hailong Ye; Rui Huang
Journal: Advances in Mathematical Physics
 (2015)

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## Abstract

The purpose of this paper is to investigate the existence of solutions to the following initial value problem for nonlinear fractional differential equation involving Caputo sequential fractional derivativeDc0α2Dc0α1yxp-2Dc0α1yx=fx,yx, x>0, y(0)=b0, Dc0α1y(0)=b1, where Dc0α1, Dc0α2 are Caputo fractional derivatives, 0<α1, α2≤1, p>1, and b0,b1∈R. Local existence of solutions is established by employing Schauder fixed point theorem. Then a growth condition imposed to f guarantees not only the global existence of solutions on the interval [0,+∞), but also the fact that the intervals of existence of solutions with any fixed initial value can be extended to [0,+∞). Three illustrative examples are also presented. Existence results for initial value problems of ordinary differential equations with p-Laplacian on the half-axis follow as a special case of our results.


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## Body

## 1. Introduction

This paper deals with the following initial value problem for nonlinear fractional differential equation involving Caputo sequential fractional derivative:(1)D0α2cDc0α1yxp-2Dc0α1yx=fx,yx,x>0,y0=b0,Dc0α1y0=b1,where Dc0α1, D0α2c are Caputo fractional derivatives, 0<α1, α2≤1, p>1, and b0,b1∈R, and xσf(x,y) is continuous on [0,+∞)×R, 0≤σ<α2. When p=2, the equation in (1) becomes a sequential fractional differential equation. Here, we follow the definition of sequential fractional derivative presented by Podlubny [1]:(2)Dνyx=Dν1Dν2⋯Dνmyx,m∈N+,where the symbol Dνi (i=1,2,…,m) means the Riemann-Liouville derivative or the Caputo derivative. It is easy to see that (2) is a generalized expression presented by Miller and Ross in [2].Fractional differential equations have been of great interest for the past three decades; see the monographs [1–4] and the papers of [5–8]. This is due to the intensive development of the theory of fractional calculus itself as well as its applications. Apart from diverse areas of pure mathematics, fractional differential equations can be used in modeling of various fields of science and engineering such as rheology, self-similar dynamical processes, porous media, fluid flows, viscoelasticity, electrochemistry, control, electromagnetic, and many other branches of science. For details, see [9–12] and the references therein.Recently, we note that the investigation for the existence of solutions of sequential fractional differential equations associated with a variety of initial and boundary value conditions has attracted the considerable attention of researchers. Here, we mention some works on them. In [13, 14], the authors investigated a class of Riemann-Liouville sequential fractional differential equation (3)D0αy′+axy=0,x≥0,and obtained existence, nonexistence, and asymptotic property of the solutions by some nonlinear analysis methods. In [15, 16], the authors considered the existence and uniqueness of the solutions for initial value problems involving Riemann-Liouville sequential fractional derivative by using monotone iterative method and fixed point method. As for sequential fractional differential equations associated with boundary value conditions, we refer the reader to [17–22]. For example, Chen and Tang [21] considered the existence of the solutions for the following Caputo sequential fractional differential equation: (4)atDc0+αut′=ft,ut,u′t,Dc0+αut,t∈0,1,with m-point boundary value conditions by the coincidence degree continuation theorem. After that, in 2015, Jiang [22] investigated the following Riemann-Liouville sequential fractional differential equations with p-Laplacian at resonance: (5)D0+βφpD0+αut+ft,ut,D0+α-1ut,D0+αut=0,t∈0,1,where φp(s)=|s|p-2s, p>1. By the extension of the continuous theorem in [23] and constructing suitable operators, they obtained the existence of solutions satisfied integral boundary value conditions.In view of the facts that the Laplace transform of the Caputo derivative allows utilization of initial values of classical integer-order derivatives and that the Caputo derivative of a constant is 0, sequential fractional differential equations involving the Caputo fractional derivative have more clear physical interpretations than those involving the Riemann-Liouville fractional derivative (see [3, 4]).To the best of our knowledge, there is no paper dealing with the existence of solutions of Caputo sequential fractional differential equations with initial value conditions. In our latest paper [24], by virtue of uniform Lipschitz continuity of xσf(x,·) on R for every x>0, we proved the existence and uniqueness of solutions of problem (1). Now, in this paper, we are concerned with the initial value problem (1) without uniform Lipschitz continuity of xσf(x,·). By fractional Taylor expansion theorem, we first obtain an integral equation equivalent to the initial value problem (1), to which local existence of solutions is established utilizing Schauder fixed point theorem. Then a growth condition imposed to f guarantees not only the global existence of solutions on the interval [0,+∞), but also the fact that the intervals of existence of solutions with any fixed initial value can be extended to [0,+∞). In addition, existence results for initial value problems of ordinary differential equations with p-Laplacian on the half-axis follow as a special case of our results.The paper is organized as follows. In Section2, we present some necessary definitions and preliminary results that will be used in our discussions. The main results and their proofs are given in Section 3. In Section 4, we will give three examples to illustrate our results.


## 2. Preliminaries

In this section, we introduce some basic definitions and notations (see the monographs [1, 2] for further details) and give several useful preliminary results which are used throughout this paper.Definition 1.
Letα>0. The Riemann-Liouville fractional integral of a function y:(0,+∞)→R of order α is given by (6)J0αyx≔1Γα∫0xx-tα-1ytdtprovided that the right-hand side is pointwise defined on (0,+∞). Here and in what follows Γ is the Gamma function.The fractional integration operatorJaα has that the following semigroup property (7)J0αJ0βyx=J0βJ0αyx=J0α+βyxholds for all α>0, β>0.Definition 2.
Letα>0 and let n be the smallest integer that exceeds α. The Riemann-Liouville fractional derivative of a continuous function y:(0,+∞)→R of order α is given by (8)D0αyx≔1Γn-αddxn∫0xx-tn-α-1ytdtprovided that the right-hand side is pointwise defined on (0,+∞).Obviously, the Riemann-Liouville fractional differentiation is the left inverse of the Riemann-Liouville fractional integration for continuous functiony in the following sense: (9)D0αJ0αyx=yxfor α>0. However, it is not the right inverse. More precisely, we have the following fractional Taylor expansion theorem.Theorem 3.
Let0<α<1. Assume that y is such that J01-αy is absolutely continuous. Then (10)J0αD0αyx=yx-xα-1Γαlimz→0+⁡J01-αyz,x≥0.Definition 4.
Letα>0 and let n be the smallest integer that exceeds α. The Caputo fractional derivative of a continuous function y:(0,+∞)→R of order α is given by (11)Dc0αyx≔D0αy-∑k=0n-1yk0k!·kxprovided that the right-hand side is pointwise defined on (0,+∞).The definition of solutions of the initial value problem (1) is given as follows.Definition 5.
LetT>0; a function y is called a solution of (1) on [0,T], if (i)
y,  Dcα1y∈C([0,T]), Dc0α2(Dc0α1yp-2Dc0α1y)∈C((0,T]);(ii)
y satisfies problem (1) on (0,T].Definition 6.
A functiony is called a solution of (1) on [0,+∞), if for any T>0, y is a solution of problem (1) on [0,T].In order to study the existence of solutions of (1), we should transform problem (1) into an equivalent integral equation. We need the following two lemmas.Lemma 7.
Letα,μ>0; then one has (12)∫abb-tα-1t-aμ-1dt=ΓαΓμΓα+μb-aα+μ-1.Lemma 8.
Let0<μ<α≤1. If g is a continuous function defined on [0,+∞), then ∫0x‍(x-t)α-1t-μg(t)dt is continuous with respect to x in [0,+∞).Proof.
Letx0>0 and take δ0∈(0,x0). For δ<δ0, we have(13)∫0x0+δx0+δ-tα-1t-μgtdt-∫0x0x0-tα-1t-μgtdt≤∫0x0+δx0+δ-tα-1t-μgtdt-∫0x0x0+δ-tα-1t-μgtdt+∫0x0x0+δ-tα-1t-μgtdt-∫0x0x0-tα-1t-μgtdt=∫x0x0+δx0+δ-tα-1t-μgtdt+∫0x0x0+δ-tα-1-x0-tα-1t-μgtdt.Since g is continuous and bounded in the neighborhood of x0, we conclude that(14)∫x0x0+δx0+δ-tα-1t-μgtdt≤Cx0+δα-1∫x0x0+δ1-tx0+δα-1t-μdt≤Cx0+δα-μ∫x0/x0+δ11-zα-1z-μdz≤Cδααx0μ,(15)∫0x0x0+δ-tα-1-x0-tα-1t-μgtdt≤Cx0α-μ∫01x0+δx0-zα-1-1-zα-1z-μdz=Cx0α-μ∫011-zα-1z-μdz-∫01+δ/x01+δx0-zα-1z-μdz+∫11+δ/x01+δx0-zα-1z-μdz≤Cx0α-μ∫011-zα-1z-μdz-∫01+δ/x01+δx0-zα-1z-μdz+Cx0α-μ∫11+δ/x01+δx0-zα-1z-1-μdz≤Cx0α-μΓαΓ1-μΓα+1-μ1-1+δx0α-μ+δx0α-μ,where C=sup0≤x≤x0+δ0gx. Combining (13), (14) with (15), we arrive at (16)limδ→0⁡∫0x0+δx0+δ-tα-1t-μgtdt-∫0x0x0-tα-1t-μgtdt=0.In addition, it is easy to see that (17)∫0xxα-1t-μgtdt≤Cxα-μ.Therefore, ∫0x‍(x-t)α-1t-μg(t)dt is continuous with respect to x in [0,+∞).Remark 9.
Ify∈C([0,+∞)), then xσf(x,y(x)) is continuous on [0,+∞). According to Lemma 8, the function (18)J0α2fx,yx=1Γα2∫0xx-tα2-1ft,ytdtis also continuous in [0,+∞) and J0α2f(0,y(0))=0.Now we are ready to transform problem (1) into an equivalent integral equation. For the reader’s convenience, we list two special notations that will be used in the following paper: q=p/(p-1) and ϕq(s)=|s|q-2s for s∈R.Proposition 10.
A functiony defined in C([0,T]) is a solution of problem (1) if and only if it satisfies the following integral equation on [0,T]:(19)yx=b0+1Γα1∫0xx-tα1-1Φytdt,where (20)Φyx=ϕqϕpb1+1Γα2∫0xx-tα2-1ft,ytdt.Proof.
First we prove the necessity. Lety∈C([0,T]) be a solution of problem (1) and define (21)gx=Dc0α1yxp-2Dc0α1yx,and then g∈C([0,T]) and g(0)=ϕp(b1). According to Definition 4 and Theorem 3, the differential equation of problem (1) can be transformed into the following form: (22)fx,yx=DJ01-α2g-g0x.Obviously, J01-α2(g-g(0)) is absolutely continuous on [0,T]. Combining with Theorem 3, we have (23)gx=g0+J0α2Dc0α2gx=ϕpb1+J0α2f·,y·x.That is, (24)Dc0α1yx=ϕqϕpb1+J0α2f·,y·x.Applying Definition 4 and Theorem 3 again, we have (25)Dc0α1yx=DJ01-α1y-y0x.Obviously, J01-α1(y-y(0))∈C1([0,T]). Combining with Theorem 3, we have (26)yx=y0+J0α1D0α1y-y0x=y0+J0α1Dc0α1yx=b0+J0α1Φyx.Therefore, y satisfies the integral equation (19).
Next, we prove the sufficiency. Lety∈C([0,T]) be a solution of the integral equation (19). Combining with Definition 1, (19) reduces to(27)yx=b0+J0α1Φyx.From Remark 9, we see that J0α1Φy∈C([0,T]) and J0α1Φy(0)=0. That is, y∈C([0,T]) and y(0)=b0. Applying the operator Dcα1 to both sides of (27), we obtain that (28)Dc0αyx=Dc0αb0+J0α1Φyx=D0αJ0α1Φyx=Φyx.Then we have Dc0αy∈C([0,T]) and Dc0αy(0)=Φy(0)=b1. By virtue of g, we transform the above equation into the following form:(29)gx=ϕpb1+J0α2f·,y·x.Similarly, applying the operator Dcα1 to both sides of (29), we arrive at (30)Dc0α2Dc0α1yxp-2Dc0α1yx=fx,yx.Therefore, y is a solution of (1) on [0,T]. Summing up, we complete the proof of Proposition 10.SinceT can be chosen arbitrarily large in Proposition 10, according to Definition 6, we have the following result.Corollary 11.
Lety∈C([0,+∞)); then y is a solution of problem (1) if and only if it satisfies the integral equation (19) on [0,+∞).By Corollary11, we obtain the following two existence results.Remark 12.
Ifb0=b1=0, and f(x,0)=0 for x∈(0,+∞), then the constant function y=0 is a solution of problem (1).Remark 13.
Iff(x,y)=f(x) for (x,y)∈(0,+∞)×R, then problem (1) has a solution on [0,+∞).The following fixed point lemma is the main tool in the proofs of our results.Lemma 14 (Schauder fixed point theorem).
LetΩ be a closed, convex, and nonempty subset of a Banach space Y, and let S:Ω→Ω be a mapping such that Sy:y∈Ω is a relatively compact subset in Y. Then S has at least one fixed point in Ω.


## 3. Main Results

In this section, we will give and prove our main results in this paper.Theorem 15.
For any fixed initial valuesb0 and b1, there exists a sufficiently small constant T∗>0 such that problem (1) has a solution on [0,T∗].Proof.
For any given positive constantK, choose T∗>0 sufficiently small which will be determined later. Let (31)Ω1=y∈C0,T∗;sup0≤x≤T∗⁡yx-b0≤K.Obviously, Ω1 is a closed, convex, and nonempty subset of C([0,T∗]). On this set Ω1 we define the operator S: (32)Syx≔b0+1Γα1∫0xx-tα1-1Φytdt,where (33)Φyx=ϕqϕpb1+1Γα2∫0xx-tα2-1ft,ytdt.According to Proposition 10, in what follows, it suffices to show that the operator S has a fixed point in Ω1.
Firstly, we will show thatSy∈Ω1 for any y∈Ω1. To this end we begin by noting that for any y∈Ω1, Φy∈C([0,T∗]) by Remark 9. Then we obtain that Sy∈C([0,T∗]). Furthermore, for x∈[0,T∗] we have (34)Ayx-b0=1Γα1∫0xx-tα1-1Φytdt≤1Γα1∫0xx-tα1-1ϕqb1p-1+MT∗,KΓ1-σΓα2-σ+1tα2-σdt≤T∗α1Γ1+α1b1p-1+MT∗,KΓ1-σΓα2-σ+1T∗α2-σq-1,where (35)MT∗,K=supx∈0,T∗,y∈b0-K,b0+K⁡xσfx,y.Now we can choose T∗>0 so small that(36)T∗α1Γ1+α1b1p-1+MT∗,KΓ1-σΓα2-σ+1T∗α2-σq-1≤K,which means that Sy∈Ω1; that is, S maps the set Ω1 to itself.
Secondly, we will also show that the family of functionsSΩ1=Sy:y∈Ω1 is a relatively compact set. That is to say, we need to show that SΩ1 is uniformly bounded and equicontinuous on [0,T∗]. The uniform boundedness follows from the definition of Ω1. As for the equicontinuity, for 0≤x1≤x2≤T∗ and y∈Ω1, we see that (37)Ayx1-Ayx2=1Γα1∫0x1x1-tα1-1Φytdt-∫0x2x2-tα1-1Φytdt≤1Γα1∫0x1x1-tα1-1-x2-tα1-1Φytdt+∫x1x2x2-tα1-1Φytdt≤CΓα1∫0x1x1-tα1-1-x2-tα1-1dt+∫x1x2x2-tα1-1dt≤2CΓα1+1x2-x1α1,where C>0 is independent of x1, x2, and y. Therefore SΩ1 is uniformly bounded and equicontinuous on [0,T∗], and thus SΩ1 is a relatively compact subset of C([0,T∗]). By Lemma 14 there exists y∈Ω1 such that Sy=y and y is a solution of problem (1) on [0,T∗]. The proof is completed.We have proved the local existence of solutions of problem (1) in Theorem 15. However, maximal intervals of existence of those solutions are not necessarily [0,+∞), which change according to initial values b0 and b1. We give an example to illustrate it.Example 16.
Consider the following initial value problem for second-order ordinary differential equation(38)y′′x=2y3,x>0,y′0=1c2,y0=1c,where c≠0 is a constant. When c<0, it is easy to see that the function (39)yx=1c-xis a solution of (38) on [0,+∞) and [0,+∞) is the maximum interval of existence. When c>0, however, the function y is a solution of (38) on [0,c) and [0,c) is the maximum interval of existence.Under certain growth condition on the nonlinearityf in (1), we will show that any solutions obtained in Theorem 15 can be extended to the interval [0,+∞) as solutions of problem (1).Theorem 17.
Suppose that there exist constantμ∈[0,p-1) and two nonnegative continuous functions c1(x), c2(x) defined on [0,+∞) such that(40)fx,y≤1xσc1x+c2xyμ,x,y∈0,+∞×R.Then maximum intervals of existence of any solutions of (1) obtained in Theorem 15 are [0,+∞).Proof.
LetyT be a solution of (1) on [0,T] where T>0. By Definition 6, it suffices to prove that for any fixed ρ>0, there exists a function y~ as a solution of (1) on [0,T+ρ] such that y~(x)=yT(x) for 0≤x≤T. The proof will be completed by applying Lemma 14. Choose K>0 which will be determined later and denote (41)Ω2=y∈CT,T+ρ;supT≤x≤T+ρ⁡yx≤K.Obviously, Ω2 is a closed, convex, and nonempty subset of C([T,T+ρ]). On this set Ω2 we define the operator S2: (42)S2yx≔zx+1Γα1∫Txx-tα1-1Φ~ytdt,T≤x≤T+ρ,where (43)zx=b0+1Γα1∫0Tx-tα1-1ΦyTtdt,ΦyTt=ϕqϕpb1+1Γα2∫0tt-sα2-1fs,yTsds,Φ~yt=ϕqϕpb1+1Γα2∫0Tt-sα2-1fs,yTsds+1Γα2∫Ttt-sα2-1fs,ysds.It is easy to see that S2y∈C([T,T+ρ]) for any y∈Ω2. Further, by virtue of (40) we have (44)S2yx≤zx+1Γα1∫0xx-tα1-1Φ~ytdt≤zx+1Γα1∫0xx-tα1-1ϕqM0+1Γα2∫Ttt-sα2-1fs,ysdsdt≤zx+1Γα1∫0xx-tα1-1ϕqM0+1Γα2∫Ttt-sα2-1s-σc1s+Kμc2sdsdt≤zx+T+ρα1Γα1+1ϕqM0+1Γα2∫TT+ρx-tα2-1t-σc1t+Kμc2tdt≤supT≤x≤T+ρ⁡zx+T+ρα1Γα1+1M0+M1+M2Kμq-1,where (45)M0=b1p-1+1Γα2∫0TT-sα2-1fs,yTds,M1=1Γα2∫TT+ρx-tα2-1t-σc1tdt,M2=1Γα2∫TT+ρx-tα2-1t-σc2tdt.Since 0≤μ<p-1, that is, 0≤μ(q-1)<1, we can choose K so large that (46)S2yx≤K,x∈T,T+ρ.Then S2 maps the set Ω2 to itself. To apply Lemma 14, we also need to show that the family of functions S2Ω2={S2y:y∈Ω2} is a relatively compact set. That is to say, we need to show that S2Ω2 is uniformly bounded and equicontinuous on [T,T+ρ]. The uniform boundedness follows from the definition of Ω2. As for the equicontinuity, we see that, for T≤x1≤x2≤T+ρ and y∈Ω2, (47)S2yx1-S2yx2≤zx1-zx2+1Γα1∫Tx1x1-tα1-1Φ~ytdt-∫Tx2x2-tα1-1Φ~ytdt≤1Γα1∫0Tx1-tα1-1-x2-tα1-1ΦyTtdt+∫Tx1x1-tα1-1-x2-tα1-1Φ~ytdt+∫x1x2x2-tα1-1Φ~ytdt≤CΓα1∫0Tx1-tα1-1-x2-tα1-1dt+∫Tx1x1-tα1-1-x2-tα1-1dt+∫x1x2x2-tα1-1dt≤3CΓα1+1x1-x2α1,where C>0 is independent of x1, x2, and y. Therefore S2Ω2 is uniformly bounded and equicontinuous on [T,T+ρ], and thus S2Ω2 is a relatively compact subset in C([T,T+ρ]). By Lemma 14 there exists yT+ρ∈Ω2 such that S2yT+ρ=yT+ρ. Noting that yT(T)=yT+ρ(T), we define (48)y~x=yTx,x∈0,T,yT+ρx,x∈T,T+ρ.From the definition of S2, we see that y~∈C([0,T+ρ]) is a solution of (1) on [0,T+ρ], which completes the proof.In a special case, an estimate of increasing rate for solutions asx→+∞ is made.Corollary 18.
Suppose thatxσf(x,y) is bounded on [0,+∞)×R. Let y be a solution of (1) on [0,+∞); then y(x)=O(xα1+(α2-σ)/(p-1))  (x→+∞).Proof.
According to Corollary11, we have (49)yx=b0+1Γα1∫0xx-tα1-1Φytdt.Note that (50)Φyx=ϕqϕpb1+1Γα2∫0xx-tα2-1ft,ytdt≤b1p-1+1Γα2∫0xx-tα2-1ft,ytdtq-1≤b1p-1+MΓα2∫0xx-tα2-1t-σdtq-1=b1p-1+MΓ1-σΓα2-σ+1xα2-σq-1≤qb1+qMΓ1-σΓα2-σ+1q-1xα2-σ/p-1,and we obtain (51)yx≤b0+1Γα1∫0xx-tα1-1Φytdt=c1+c2xα1+c3xα1+α2-σ/p-1,where c1,c2,c3>0 is independent of x and y. Since 0≤σ<α2, we conclude that (52)yx=Oxα1+α2-σ/p-1x⟶+∞.


## 4. Example

To illustrate our main results, we will present three examples.Example 19.
Consider the initial value problem for sequential fractional differential equation(53)Dc00.1Dc00.5yx=x+yλ1+x1+y,x>0,y0=1,Dc00.5y0=1,where λ∈(0,2). Here p=2, α1=0.5, α2=0.1, and f(x,y)=(x+|y|λ)/(1+x)(1+|y|). It is easy to see that (54)
f
x
,
y
≤
x
1
+
x
+
y
λ
1
+
x
,
λ
∈
0,1
,
x
1
+
x
+
y
λ
-
1
1
+
x
,
λ
∈
1,2
,
x
1
+
x
+
y
0.5
1
+
x
,
λ
=
1
,for (x,y)∈[0,∞)×R. According to Theorem 17, problem (53) has at least one positive solution on [0,+∞) for any fixed λ∈(0,2). Let yλ be a positive solution of (53) on [0,+∞). Since fx,y is bounded on [0,+∞)×R when λ∈(0,1], we further have an estimate of increasing rate for yλ by Corollary 18: (55)yλx=Ox0.6x⟶+∞.Example 20.
Consider the initial value problem for nonlinear fractional differential equation(56)Dc00.75Dc00.5yxp-2Dc00.5yx=x-0.5+x-0.25sin⁡y,x>0,y0=1,Dc00.5y0=0,where p>1. Here α1=0.5, α2=0.75, and f(x,y)=x-0.5+x-0.25siny. Choosing σ=0.5∈[0,0.75), we see that (57)xσfx,y=x0.5x-0.5+x-0.25sin⁡y≤1+x0.25yμfor (x,y)∈(0,∞)×R, where μ∈(0,p-1). According to Theorem 17, problem (56) has at least one solution on [0,+∞). Let y be a solution of (56) on [0,+∞). By simple computation similar to that in the proof of Corollary 18, we further have an estimate of increasing rate for y: (58)yx=Oxp/2p-1x⟶+∞.Example 21.
Consider the initial value problems for ordinary differential equations withp-Laplacian(59)y′xp-2y′x′=ln⁡x+x-0.5arctan⁡y,x>0,y0=1,Dc00.5y0=0,where p>2. Here α1=α2=1 and f(x,y)=lnx+x-0.5arctany. Choosing σ=0.5∈[0,0.75), we see that (60)xσfx,y=x0.5ln⁡x+x-0.5arctan⁡y≤x0.5ln⁡x+yfor (x,y)∈(0,∞)×R. According to Theorem 17, problem (59) has at least one solution on [0,+∞). Let y be a solution of (59) on [0,+∞). By simple computation similar to that in the proof of Corollary 18, we further have an estimate of increasing rate for y: (61)yx=Oxp+1/p-1x⟶+∞.



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