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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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A new method for converting boundary value problems for impulsive fractional differential equations to integral equations and its applications

Yuji Liu
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  • Department of Mathematics and Statics, Guangdong University of Finance and Economics, Guangzhou 510320, P. R. China
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Published Online: 2017-05-05 | DOI: https://doi.org/10.1515/anona-2016-0064

Abstract

In this paper, we present a new method for converting boundary value problems of impulsive fractional differential equations to integral equations. Applications of this method are given to solve some types of anti-periodic boundary value problems for impulsive fractional differential equations. Firstly by using iterative method, we prove existence and uniqueness of solutions of Cauchy problems of differential equations involving Caputo fractional derivative, Riemann–Liouville and Hadamard fractional derivatives with order q(0,1), see Theorem 2, Theorem 4, Theorem 6 and Theorem 8. Then we obtain exact expression of piecewise continuous solutions of these fractional differential equations see Theorem 1, Theorem 2, Theorem 3 and Theorem 4. Finally, four classes of integral type anti-periodic boundary value problems of singular fractional differential equations with impulse effects are proposed. Sufficient conditions are given for the existence of solutions of these problems. See Theorems 4.14.4. We allow the nonlinearity p(t)f(t,x) in fractional differential equations to be singular at t=0,1 and be involved a super-linear and sub-linear term. The analysis relies on Schaefer’s fixed point theorem. In order to avoid misleading readers, we correct the results in [28] and [65]. We establish sufficient conditions for the existence of solutions of an anti-periodic boundary value problem for impulsive fractional differential equation. The results in [68] are complemented. The results in [81] are corrected. See Lemma 5.1, Lemma 5.7, Lemma 5.10 and Lemma 5.13.

Keywords: Impulsive fractional differential equation; fractional Langevin equation; anti-periodic boundary value problem; Riemann–Liouville fractional derivative; Caputo fractional derivative; Hadamard fractional derivative; Caputo-type Hadamard fractional derivative; Picard iterative technique; fixed point theorem

MSC 2010: 34K37; 34K45; 34B37; 34B15; 34B10

1 Introduction

One knows that the fractional derivatives (Riemann–Liouville fractional derivative, Caputo fractional derivative and Hadamard fractional derivative and other types, see [33]) are actually nonlocal operators because integrals are nonlocal operators. Moreover, calculating time fractional derivatives of a function at some time requires all the past history and hence fractional derivatives can be used for modeling systems with memory.

Fractional order differential equations are generalizations of integer order differential equations. Using fractional order differential equations can help us to reduce the errors arising from the neglected parameters in modeling real life phenomena. Fractional differential equations have many applications, see [57, Chapter 10] and the books [34, 57, 60].

In recent years, there have been many results obtained on the existence and uniqueness of solutions of initial value problems or boundary value problems for nonlinear fractional differential equations, see [13, 15, 51, 55, 56, 59, 74, 79, 82].

Dynamics of many evolutionary processes from various fields such as population dynamics, control theory, physics, biology, and medicine, undergo abrupt changes at certain moments of time like earthquake, harvesting, shock, and so forth. These perturbations can be well approximated as instantaneous change of states or impulses. These processes are modeled by impulsive differential equations. In 1960, Milman and Myshkis introduced impulsive differential equations in their paper [52]. Based on their work, several monographs have been published by many authors like Samoilenko and Perestyuk [61], Lakshmikantham, Bainov and Simeonov [36], Bainov and Simeonov [10, 11], Bainov and Covachev [9], and Benchohra, Henderson and Ntonyas [16].

The fractional differential equation was extended to impulsive fractional differential equations, since Agarwal and Benchohra published the first paper on the topic [2] in 2008. Since then many authors [24, 25, 32, 46, 47, 42, 41, 54, 59, 75, 74, 78, 43, 48, 49] studied the existence or uniqueness of solutions of impulsive initial or boundary value problems for fractional differential equations. For examples, impulsive anti-periodic boundary value problems, see [3, 4, 2, 22, 44, 40, 63, 77], impulsive periodic boundary value problems, see [72, 67], impulsive initial value problems, see [18, 23, 53, 62], two-point, three-point or multi-point impulsive boundary value problems, see [8, 73, 80], and impulsive boundary value problems on infinite intervals, see [81].

Recently, in [28, 71, 83], the authors studied the existence and uniqueness of solutions of the following boundary value problem of impulsive fractional differential equation:

{D0+qCx(t)=f(t,x(t)),t(ti,ti+1],i0,Δx|t=ti=Ii(x(ti-)),i,ax(0)+bx(T)=x0,(1.1)

where q(0,1], D0+qC is the standard Caputo fractional derivative of order q, 0={0,1,,m} and ={1,2,,m}, f:[0,T]× is a jointly continuous function, Ik: (k) are continuous functions, and 0=t0<t1<<tm<tm+1=T,

Δx|t=tk=limttk+x(t)-limttk-x(t)=x(tk+)-x(tk-)

and x(tk+),x(tk-) represent the right and left limits of x(t) at t=tk, respectively, a,b,x0 are constants with a+b0. It was proved in [28] that if f is a jointly continuous function and there is a constant λ¯[0,1-1p] for some p(1,11-q] and L>0 such that |f(t,x)|L(1+|x|λ¯] for each t[0,T] and x. Then BVP (1.1) has at least one solution. We find that this result is wrong. For example, consider the following special problem of (1.1):

{D0+αCx(t)=t,t(ti,ti+1],i=0,1,Δx(t1)=-2x(t1),x(0)+x(1)=0,(1.2)

where α(0,1), 0=t0<t1<t2=1 with t1(12)1α+1. It is easy to see that f(t,x)=t, I1(x)=-2x, a=b=1 and x0=0. One can see that f is a jointly continuous function and there is a constant λ¯[0,1-1p] for some p(1,11-q] and L>0 such that |f(t,x)|L(1+|x|λ¯] for each t[0,T] and x, (λ¯=0,L=1). But BVP (1.2) has no solution. In fact, if (1.2) has a solution x, then

x(t)={c0+tα+1Γ(α+1),t(0,t1],c0+c1+tα+1Γ(α+1),t(t1,1].

Thus we get

2c0+c1+1Γ(α+1)=0,c1=-2(c0+t1α+1Γ(α+1)).

So 2t1α+1=1, which contradicts to t1(12)1α+1. Hence BVP (1.2) (which is called a anti-periodic boundary value problem) has no solution. It is needed to make a correction for (1.1), see Lemma 5.1.

In [71], it was proved that if f is a jointly continuous function and there exist Cf,Mf>0 and q1[0,1] such that |f(t,x)|Cf|x|q1+Mf for all t[0,T] and x, there exist CI,MI>0 and q2[0,1] such that |Ii(x)|CI|x|q2+MI for all i and x, there exist KIi>0 such that |Ii(u)-Ii(v)|KIi|u-v| for all u,v and i with i=1mKIi<1, then BVP (1.1) has at least one solution when

K¯=max{|a|,|b|}|a+b|i=1mKIi<1.

In [68], the authors actually studied the existence of solutions of the following boundary value problem for the impulsive fractional Langevin equations:

{D0+βC[D0+αCx(t)+λx(t)]=f(t,x(t)),t[0,1]{t1,t2,,tm},Δx(ti)=Ii,i,x(0)=x(ηi)=x(1)=0,i,(1.3)

where α,β(0,1) with α+β<1, λ>0, D0+*C is the standard Caputo fractional derivative of order *, see [68, Definitions 2 and 3], ={1,,m}, f:[0,1]× is a given function, 0=t0<t1<t2<<tm<tm+1=1, Ii (i) are constants, ηi(ti,ti+1] (i=0,1,2,,m-1),

ΔIx(ti)=x(ti+)-x(ti-)=limttk+x(t)-limttk-x(t).

It is claimed in [68] that the presentation of solutions of

DtβC[D0+αCx(t)+λx(t)]=f(t),t(ti,ti+1],i0,

is given by

x(t)=𝐄α(-λtα)bi-1-𝐄α(-λtα)λai+0t(t-s)α+β-1𝐄α,α+β(-λ(t-s)α)f(s)𝑑s,

where ai,bi (i0) are constants. We find that this claim is wrong, see Lemma 5.7.

The concept of “fractional relaxation differential equations” was introduced by Hilfer (see [30, (118), p. 115]) and [84]. That is, Hilfer considered the following generalized Hilfer fractional relaxation Cauchy problem

Da+α,βHx(t)=-Kx(t),t(a,b),Ia+1-γRLx(a)=x0,

where α(0,1),β[0,1],αγ=α+β-αβ<1, Da+α,βH denotes the left-sided Hilfer fractional derivative of order α (see [30, Definition 3.3, p. 113] and type β, Ia+1-γRL denotes the left-sided Riemann–Liouville fractional integral (see [30, Definition 3.2, p. 112]), and K is called “fractional relaxation factor”. The solution of this problem can be written as (see [30, (124), p. 115]) x(t)=x0tγ-1𝐄α,α+β(1-γ)(-Ktα). Obviously, the fractional relaxation factor K will generate the famous generalized Mittag-Leffler function 𝐄α,α+β(1-γ)(-Ktα), which plays an important role in the study of various fractional relaxation differential equation [84].

In [65], the authors studied the following anti-periodic BVP for impulsive fractional differential equations with constant coefficients of the form:

{D0+γCx(t)+λx(t)=f(t,x(t)),t[0,1]{t1,t2,,tm},Δx(ti)=x(ti+)-x(ti-)=xi,i,x(0)+x(1)=0,(1.5)

where λ>0, xi (i), D0+γC is the Caputo fractional derivative, see [65, Definition 2.2 and Definition 2.3], f:[0,1]×, 0=t0<t1<<tm<tm+1=1. It is claimed (see [65, Lemma 3.1]) that

{D0+γCx(t)+λx(t)=h(t),t[0,1]{t1,t2,,tm},Δx(ti)=x(ti+)-x(ti-)=xi,i,x(0)+x(1)=0,

has a unique solution

x(t)=𝐄γ(-λtγ)1+𝐄γ(-λ)[i=1mxi𝐄γ(-λtiγ)-01(1-s)γ-1𝐄γ,γ(-λ(1-s)γ)h(s)𝑑s]-𝐄γ(-λtγ)i=k+1mxi𝐄γ(-λtiγ)+0t(t-s)γ-1𝐄γ,γ(-λ(t-s)γ)h(s)𝑑s,t(tk,tk+1],k0.

This lemma is also wrong, see Lemma 5.10.

In [84], Zhang and Wang studied the following nonlocal Cauchy problems for implicit impulsive fractional relaxation differential systems:

{D0,tγCx(t)=-ax(t)+f(t,x(t),y(t)),t[0,1]{t1,t2,,tm},D0,tγCy(t)=-by(t)+g(t,x(t),y(t)),t[0,1]{t1,t2,,tm},x(0)=i=1mαix(τi),y(0)=i=1mβiy(τi),Δx(ti)=Ii(x(ti)),Δy(ti)=Ji(y(ti)),i=1,2,,m,(1.6)

where a,b>0 are fractional relaxation factors, D0,tγC is the generalized Caputo fractional derivative of order γ(0,1) with the lower limit zero, f,g:[0,1]×× and αi,βi are real numbers with 1+i=1mαi0, 1+i=1mβi0. Next, Ii,Ji: and ti,τi are given points satisfying 0t1t2tmtm+1=1 and τi(ti,ti+1), i=1,2,,m, Δx(ti)=limtti+x(t)-x(ti), Δy(ti)=limtti+y(t)-y(ti).

It is claimed [84] that one can make straightforward fractional calculations to show that the above system can be written as the following integral system:

x(t)=-αi=1mαi[0<ti<τi𝐄γ(-atiγ)Ii(x(ti))+0<ti<t𝐄γ(-atγ)Ii(x(ti))+0τi(τi-s)γ-1𝐄γ,γ(-a(τi-s)γ)f(s,x(s),y(s))ds]+0t(t-s)γ-1𝐄γ,γ(-a(t-s)γ)f(s,x(s),y(s))𝑑s,y(t)=-βi=1mβi[0<ti<τi𝐄γ(-btiγ)Ji(y(ti))+0<ti<t𝐄γ(-btγ)Ji(y(ti))+0τi(τi-s)γ-1𝐄γ,γ(-b(τi-s)γ)g(s,x(s),y(s))ds]+0t(t-s)γ-1𝐄γ,γ(-b(t-s)γ)g(s,x(s),y(s))𝑑s,

where

𝐄γ(z)=χ=0zχΓ(γχ+1),𝐄γ,γ(z)=χ=0zχΓ(γ(χ+1)),α=1+i=1mαi𝐄γ(-aτiγ),β=1+i=1mβi𝐄γ(-bτiγ).

However, this claim is also wrong, see Lemma 5.13.

In a fractional differential equation, there exist two cases involving its fractional derivatives: the first case is Dα=D0+α in which the fractional derivative has a unique start point. The second case is Dα=Dti+α in which Dα has multiple start points, i.e.,

Dα=Dti+α.

Concerning the second case, readers may find some discussions in [14, 72, 45, 77], and there is no confusion problem for this case.

In [26], Feckan and Zhou pointed out that the formula of solutions for impulsive fractional differential equations in [1, 12, 17] is incorrect and gave their correct formula. In [14, 70, 69], the authors established a general framework to find the solutions for impulsive fractional boundary value problems and obtained some sufficient conditions for the existence of the solutions to a kind of impulsive fractional differential equations, respectively. In [64], the authors illustrated their comprehensions for the counterexample in [26] and criticized the viewpoint in [26, 70, 69]. Next, in [27], Feckan, Zhou and Wang expounded for the counterexample in [26] and provided further five explanations in the paper. The concept of solution of an impulsive fractional differential equation is still an open question. This is the third motivation of this paper for studying the solvability of boundary value problems of impulsive fractional differential equations.

From the above discussion, it is of most importance and is interesting to search a new method for converting an impulsive boundary value problem for fractional differential equation to an equivalent integral equation. This is the motivation of this paper.

In this paper, we will study the existence of solutions of four classes of impulsive integral type boundary value problems of singular fractional differential systems. The first one is as follows:

{D0+αRLx(t)-λx(t)=p(t)f(t,x(t)),t(ti,ti+1],i0,x(1)+limt0t1-αx(t)=01ϕ(s)G(s,x(s))𝑑s,limtti+(t-ti)1-αx(t)=I(ti,x(ti)),i,(1.7)

where

  • (a)

    0<α<1, λ, D0+αRL is the Riemann–Liouville fractional derivative of order α,

  • (b)

    0=t0<t1<t2<<tm<tm+1=1, 0={0,1,2,,m} and ={1,2,,m},

  • (c)

    ϕ:(0,1) satisfies ϕ|(ti,ti+1), i0,

  • (d)

    p:(0,1) is measurable and satisfies the growth conditions: there exist constants k,l with k>-1 and max{-α,-k-1}<l0 such that |p(t)|tk(1-t)l, t(0,1),

  • (e)

    f,G defined on (0,1]× are impulsive II-Carathéodory functions, I:{ti:i}× is a discrete II-Carathéodory function.

The second one is the following:

{D0+αCx(t)-λx(t)=p(t)f(t,x(t)),t(ti,ti+1],i0,x(1)+limt0+x(t)=01ϕ(s)G(s,x(s))𝑑s,limtti+x(t)-x(ti)=I(ti,x(ti)),i,(1.8)

where

  • (f)

    0<α<1, λ, D0+αC is the Caputo fractional derivative of order α, ti satisfies (b), ϕ:(0,1) satisfies (c), p:(0,1) is measurable and satisfies that there exist constants k,l with k>-1, l0,l0 with α+l>0, α+k+l>0 such that |p(t)|tk(1-t)l, t(0,1),

  • (g)

    f,G defined on (0,1]× are impulsive I-Carathéodory functions, I:{ti:i}× is a discrete I-Carathéodory function.

We emphasize that much work on fractional boundary value problems involves either Riemann–Liouville or Caputo type fractional differential equations, see [5, 6, 7, 4]. Another kind of fractional derivatives that appears side by side to Riemann–Liouville and Caputo derivatives in the literature is the fractional derivative due to Hadamard introduced in 1892 [29], which differs from the preceding ones in the sense that the kernel of the integral (in the definition of Hadamard derivative) contains logarithmic function of arbitrary exponent. Recent studies can be seen in [19, 20, 21].

Thirdly we study the following impulsive integral type boundary value problems of singular fractional differential systems:

{D1+αRLHx(t)-λx(t)=p(t)f(t,x(t)),t(ti,ti+1],i0,x(e)+limt1+(logt)1-αx(t)=1eϕ(s)G(s,x(s))𝑑s,limtti+(logtti)1-αx(t)=I(ti,x(ti)),i,(1.9)

where

  • (h)

    0<α<1, λ, D1+αRLH is the Hadamard fractional derivative of order α,

  • (i)

    1=t0<t1<t2<<tm<tm+1=e, ϕL1(1,e), 0={0,1,2,,m}, ={1,2,,m}, the function p:(1,e) is measurable and satisfies the growth conditions: there exist constants k and l with k>-1 and max{-α,-k-1}<l0 such that |p(t)|(logt)k(1-logt)l, t(1,e),

  • (j)

    f,G defined on (1,e]× are impulsive III-Carathéodory functions, I:{ti:i}× is a discrete III-Carathéodory function.

Finally, we study the following impulsive integral type boundary value problems of singular fractional differential systems:

{D1+αCHx(t)-λx(t)=p(t)f(t,x(t)),t(ti,ti+1],i0,x(e)+limt1+x(t)=1eϕ(s)G(s,x(s))𝑑s,limtti+x(t)-x(ti)=I(ti,x(ti)),i,(1.10)

where

  • (k)

    0<α<1, λ, D1+αCH is the Caputo type Hadamard fractional derivative of order α,

  • (l)

    1=t0<t1<t2<<tm<tm+1=e, 0={0,1,2,,m} and ={1,2,,m}, ϕL1(1,e), and the function p:(1,e) is measurable and satisfies that there exist constants k and l with k>-1 and max{-α,-k-α}<l0 such that |p(t)|(logt)k(1-logt)l, t(1,e),

  • (m)

    f,G defined on (1,e]× are impulsive I-Carathéodory functions, I:{ti:i}× is a discrete I-Carathéodory function.

We give an exact expression of continuous solutions of linear fractional differential equations by using the Picard iterative method. By using the mathematical induction method, we obtain an exact expression of piecewise continuous solutions of linear fractional differential equations. By using these results, we convert boundary value problems for impulsive fractional differential equations to integral equations technically.

In order to get solutions of a boundary value problem for impulsive fractional differential equations, we firstly define a Banach space X. Then we transform the boundary value problem into an integral equation and define a nonlinear operator T on X. Finally, we prove that T has a fixed point in X. This fixed point is just a solution of the boundary value problem. Three difficulties occur in known papers: one is how to transform the boundary value problem into a integral equation; the other one is how to define and prove a Banach space and the completely continuous property of the nonlinear operator defined; the third one is to choose a suitable fixed point theorem and impose suitable growth conditions on functions to get the fixed points of the operator.

To the best of the authors knowledge, no one has studied the existence of solutions of BVPs (1.7), (1.8), (1.9), (1.10). This paper fills this gap. Another purpose of this paper is to illustrate the similarity and difference of these three kinds of fractional differential equations. We obtain results on the existence of at least one solution for BVPs (1.7), (1.8), (1.9), (1.10), respectively. Some examples are given to illustrate the efficiency of the main theorems. For simplicity we only consider the left-sided operators here. The right-sided operators can be treated similarly.

The remainder of this paper is as follows: in Section 2, we present related definitions, especially the concept of solution of an impulsive fractional differential equation is proposed. In Section 3 some preliminary results are given. (By iterative method, we prove firstly the existence and uniqueness of solutions of the initial value problems of linear fractional differential equations, the results obtained generalize known ones. Then the exact piecewise continuous solutions of the impulsive linear fractional differential equations are obtained. Finally, we prove preliminary results for establishing existence of solutions of BVPs (1.7), (1.8), (1.9), (1.10) in the next section.) In Section 4, the main theorems (Theorems 4.14.4) and their proofs are given (we establish sufficient conditions for the existence of solutions (see Definitions 2.142.17) of BVPs (1.7), (1.8), (1.9), (1.10), respectively). In Section 5, we apply the methods proposed in this paper to solve five classes of boundary value problems more general than BVP (1.1), BVP (1.3), BVP (1.5) and BVP (1.6), respectively, see Theorems 5.35.15. The results in [68] are complemented. The results in [81] are corrected. See Lemmas 5.1, 5.7, 5.10 and 5.13.

2 Definitions

For the convenience of the readers, we firstly present the necessary definitions from the fractional calculus theory. These definitions and results can be found in the literature [34, 57, 60].

Let the Gamma function, Beta function and the classical Mittag-Leffler special function be defined by

Γ(α)=0+xα-1e-x𝑑x,α>0,𝐁(α,β)=01xα-1(1-x)β-1𝑑x,α>0,β>0,𝐄α,β(x)=v=0+xvΓ(αv+β),α>0,β>0,x.

It is known that the Gamma function satisfies Γ(α+1)=αΓ(α), the Beta function satisfies 𝐁(α,β)=Γ(α)Γ(β)Γ(α+β), and

|𝐄α,β(x)|𝐄α,β(|x|)𝐄α,β(|y|)

for all x,y with |x||y|.

Definition 2.1 ([34]).

Let c and g:(c,+). The left Riemann–Liouville fractional integral of order α>0 of g is given by

Ic+αg(t)=1Γ(α)ct(t-s)α-1g(s)𝑑s,t>c,

provided that the right-hand side exists.

Definition 2.2 ([34]).

Let c and g:(c,+). The left Riemann–Liouville fractional derivative of order α>0 of g is given by

Dc+αRLg(t)=1Γ(n-α)dndtnctg(s)(t-s)α-n+1𝑑s,t>c

where α<n<α+1, i.e., n=α, provided that the right-hand side exists.

Definition 2.3 ([34]).

Let c and g:(c,+). The left Caputo fractional derivative of order α>0 of g is given by

Dc+αCg(t)=1Γ(n-α)ctg(n)(s)(t-s)α-n+1𝑑s,t>c,

where α<n<α+1, i.e., n=α, provided that the right-hand side exists.

Definition 2.4 ([34]).

Let c>0 and g:(c,+). The left Hadamard fractional integral of order α>0 of g is given by

Ic+αHg(t)=1Γ(α)ct(logts)α-1g(s)dss,t>c,

provided that the right-hand side exists.

Definition 2.5 ([34]).

Let c>0 and g:(c,+). The left Hadamard fractional derivative of order α>0 of g is given by

Dc+αRLHg(t)=1Γ(n-α)(tddt)nct(logts)n-α-1g(s)dss,t>c,

where α<n<α+1, i.e., n=α, provided that the right-hand side exists.

Definition 2.6 ([31]).

Let c>0 and g:(c,+). The left Caputo type Hadamard fractional derivative of order α>0 of g is given by

Dc+αCHg(t)=1Γ(n-α)ct(logts)n-α-1(sdds)ng(s)dss,t>c,

where α<n<α+1, i.e., n=α, provided that the right-hand side exists.

Definition 2.7.

We call F:i=0m(ti,ti+1)× an impulsive I-Carathéodory function if it satisfies

  • (i)

    tF(t,u) is measurable on (ti,ti+1) (i0) for any u,

  • (ii)

    uF(t,u) is continuous on for almost all t(ti,ti+1) (i0),

  • (iii)

    for each r>0 there exists Mr>0 such that

    |F(t,u)|Mr,t(ti,ti+1),|u|r,i0.

Definition 2.8.

We call F:i=0m(ti,ti+1)× an impulsive II-Carathéodory function if it satisfies

  • (i)

    tF(t,(t-ti)α-1u) is measurable on (ti,ti+1) (i=0,1) for any u,

  • (ii)

    uF(t,(t-ti)α-1u) is continuous on for almost all t(ti,ti+1) (i0),

  • (iii)

    for each r>0 there exists Mr>0 such that

    |F(t,(t-ti)α-1u)|Mr,t(ti,ti+1),|u|r,i0.

Definition 2.9.

We call F:i=0m(ti,ti+1)× an impulsive III-Carathéodory function if it satisfies

  • (i)

    tF(t,(logtti)α-1u) is measurable on (ti,ti+1) (i0) for any u,

  • (ii)

    uF(t,(logtti)α-1u) is continuous on for almost all t(ti,ti+1) (i0),

  • (iii)

    for each r>0 there exists Mr>0 such that

    |F(t,(logtti)α-1u)|Mr,t(ti,ti+1),|u|r,i0.

Definition 2.10.

We call I:{ti:i}× an discrete I-Carathéodory function if it satisfies

  • (i)

    (u,v)I(ti,u) is continuous on 2,

  • (ii)

    for each r>0 there exists Mr>0 such that |I(ti,u)|Mr,|u|r.

Definition 2.11.

We call I:{ti:i0}× an discrete II-Carathéodory function if it satisfies

  • (i)

    uI(ti,(ti-ti-1)α-1u) is continuous on ,

  • (ii)

    for each r>0 there exists Mr>0 such that |I(ti,(ti-ti-1)α-1u)|Mr,|u|r.

Definition 2.12.

We call I:{ti:i0}× an discrete III-Carathéodory function if it satisfies

  • (i)

    uI(ti,(logtiti-1)α-1u) is continuous on ,

  • (ii)

    for each r>0 there exists Mr>0 such that |I(ti,(logtiti-1)α-1u)|Mr,|u|r.

Definition 2.13 ([50]).

Let E and F be Banach spaces. A operator T:EF is called a completely continuous operator if T is continuous and maps any bounded set into relatively compact set.

Let a<b be constants. The following Banach spaces are used:

  • (i)

    C(a,b] denotes the set of all continuous functions on (a,b] with the limit limta+x(t) existing, and the norm x=supt(a,b]|x(t)|.

  • (ii)

    C1-α(a,b] the set of all continuous functions on (a,b] with the limit limta+(t-a)1-αx(t) existing, the norm x1-α=supt(a,b](t-a)1-α|x(t)|.

  • (iii)

    LC1-α(a,b] denotes the set of all continuous functions on (a,b] with the limit limta+(logta)1-αx(t) existing, and the norm x=supt(a,b](logta)1-α|x(t)|.

Let m be a positive integer and 0={0,1,2,,m}, 0=t0<t1<<tm<tm+1=1. The following Banach spaces are also used in this paper:

PmC1-α(0,1]={x:(0,1]:x|(ti,ti+1]C1-α(ti,ti+1],i0}

with the norm

x=xPmC1-α=max{supt(ti,ti+1](t-ti)1-α|x(t)|:i0},

and

PmC(0,1]={x:(0,1]:x|(ti,ti+1]C(ti,ti+1],i0}

with the norm

x=xPmC(0,1]=max{supt(ti,ti+1]|x(t)|:i0}.

Let 1=t0<t1<<tm<tm+1=e. We also use the Banach spaces

LPmC1-α(1,e]={x:(1,e]:x|(ti,ti+1]C(ti,ti+1],i0,limtti+(logtti)1-αx(t),i0,exist}

with the norm

x=xLPmC1-α=max{supt(ti,ti+1](logtti)1-α|x(t)|:i0},

and

PmC(1,e]={x:(1,e]:x|(ti,ti+1]C(ti,ti+1],i0}

with the norm

x=xPmC=max{supt(ti,ti+1]|x(t)|:i0}.

Now, we propose new concepts of solution of an impulsive fractional differential equation with Caputo fractional derivative, Riemann–Liouville fractional derivative, Caputo type Hadamard fractional derivative and Riemann–Liouville type Hadamard fractional derivative respectively.

Definition 2.14.

A function x:(0,1] is called a solution of BVP (1.7) if x|(ti,ti+1] is continuous on (ti,ti+1] (i0), the limits limtti+(t-ti)1-αx(t) (i0) exist, D0+αRLx:(0,1] is measurable on (0,1], and x satisfies all equations in (1.7).

Definition 2.15.

A function x:(0,1] is called a solution of BVP (1.8) if x|(ti,ti+1] is continuous on (ti,ti+1] (i0), the limits limtti+x(t) (i0) exist, D0+αCx:(0,1] is measurable on (0,1] and x satisfies all equations in (1.8).

Definition 2.16.

A function x:(1,e] is called a solution of BVP (1.9) if x|(ti,ti+1] is continuous on (ti,ti+1] (i0), the limits limtti+(logt-logti)1-αx(t) (i0) exist, D0+αRLHx:(1,e] is measurable on (1,e], and x satisfies all equations in (1.9).

Definition 2.17.

A function x:(1,e] is called a solution of BVP (1.10) if x|(ti,ti+1] is continuous on (ti,ti+1] (i0), the limits limtti+x(t) (i0) exist, D0+αCHx:(1,e] is measurable on (1,e], and x satisfies all equations in (1.10).

3 Preliminary results

This section is divided into three subsections. By Picard iterative method, in Section 3.1, we prove the existence and uniqueness of solutions of the initial value problems for linear fractional differential equations, the results obtained generalize known ones. In Section 3.2, the exact piecewise continuous solutions of the impulsive linear fractional differential equations are obtained. In Sections 3.33.6, we prove preliminary results for establishing existence of solutions of BVPs (1.7), (1.8), (1.9), (1.10).

3.1 Picard iterative method for linear fractional differential equations

In [35, 37, 38, 39]. the basic theory of initial value problems for fractional differential equations involving Riemann–Liouville differential operators of order q(0,1) were investigated. The existence and uniqueness of solutions of the following initial value problems (3.1.1)–(3.1.4) of fractional differential equations were discussed under the assumption that fCr[0,1]. We will establish existence and uniqueness results for these problems under more weaker assumptions, see Assumptions A1A4 in this subsection.

Let η, and let F,A:(0,1) and B,G:(1,e) be continuous functions. We will consider the following four classes of initial value problems of nonhomogeneous linear fractional differential equations:

{D0+𝜶Cx(t)=A(t)x(t)+F(t),t(0,1),limt0+x(t)=η,(3.1.1){D0+𝜶RLx(t)=A(t)x(t)+F(t),(0,1),limt0+t1-αx(t)=η,(3.1.2){D0+𝜶RLHx(t)=B(t)x(t)+G(t),t(1,e),limt1+(logt)1-αx(t)=η,(3.1.3){D0+𝜶CHx(t)=B(t)x(t)+G(t),t(1,e),limt1+x(t)=η,(3.1.4)

To get solutions of (3.1.1), we need the following assumptions:

Assumption A1.

There exist constants ki>-1 and li0 with li>max{-α,-α-ki} (i=1,2), MA0 and MF0 such that |A(t)|MAtk1(1-t)l1 and |F(t)|MFtk2(1-t)l2 for all t(0,1).

Choose Picard function sequence as

ϕ0(t)=η+0t(t-s)α-1Γ(α)F(s)𝑑s,t[0,1],ϕn(t)=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s,t[0,1],n=1,2,.

Claim 3.1.

One has ϕnC[0,1].

Proof.

One sees ϕ0C[0,1] by the assumption imposed on F in Assumption A1. Then ϕ1 is continuous on [0,1] by

|0t(t-s)α-1Γ(α)A(s)ϕ0(s)𝑑s|ϕ00t(t-s)α-1Γ(α)|A(s)|𝑑sϕ00t(t-s)α-1Γ(α)MA|η|sk1(1-s)l1𝑑sMAϕ0|η|0t(t-s)α+l1-1Γ(α)sk1𝑑s=MA|η|ϕ0tα+k1+l101(1-w)α+l1-1Γ(α)wk1𝑑w=MA|η|ϕ0tα+k1+l1𝐁(α+l1,k1+1)Γ(α)0as t0+.

By the mathematical induction method, we can prove that ϕnC[0,1]. ∎

Claim 3.2.

The sequence {ϕn} is convergent uniformly on [0,1].

Proof.

In fact, we have for t[0,1] that

|ϕ1(t)-ϕ0(t)|=|0t(t-s)α-1Γ(α)A(s)ϕ0(s)𝑑s|MA|η|ϕ00t(t-s)α-1Γ(α)sk1(1-s)l1𝑑sMA|η|ϕ00t(t-s)α+l1-1Γ(α)sk1𝑑s=MA|η|ϕ0tα+k1+l1𝐁(α+l1,k1+1)Γ(α).

So

|ϕ2(t)-ϕ1(t)|=|0t(t-s)α-1Γ(α)A(s)[ϕ1(s)-ϕ0(s)]𝑑s|0t(t-s)α-1Γ(α)MAsk1(1-s)l1(MA|η|ϕ0sα+k1+l1𝐁(α+l1,k1+1)Γ(α))𝑑s|η|ϕ0MA20t(t-s)α+l1-1Γ(α)sα+2k1+l1𝐁(α+l1,k1+1)Γ(α)𝑑s=|η|ϕ0MA2t2α+2k1+2l1𝐁(α+l1,k1+1)Γ(α)𝐁(α+l1,α+2k1+l1+1)Γ(α).

Now suppose that

|ϕj(t)-ϕj-1(t)||η|ϕ0MAjtjα+jk1+jl1i=0j-1𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α).

We get

|ϕj+1(t)-ϕj(t)|=|0t(t-s)α-1Γ(α)A(s)[ϕj(s)-ϕj-1(s)]𝑑s|0t(t-s)α-1Γ(α)MA(|η|ϕ0MAjsjα+jk1+jl1i=0j-1𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α))sk1(1-s)l1𝑑s0t(t-s)α+l1-1Γ(α)MA(|η|ϕ0MAjsjα+jk1+jl1i=0j-1𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α))sk1𝑑s|η|ϕ0MAj+1t(j+1)α+(j+1)k1+(j+1)l1i=0j𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α).

From the mathematical induction method, we get for every n=1,2, that

|ϕn+1(t)-ϕn(t)||η|ϕ0MAn+1t(n+1)α+(n+1)k1+(n+1)l1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α)|η|ϕ0MAn+1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α),t[0,1].

Consider

n=1+un=n=1+|η|ϕ0MAn+1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α).

One sees for sufficiently large n that

un+1un=MA𝐁(α+l1,(n+1)α+(n+1)k1+(n+1)l1)Γ(α)=MA01(1-x)α+l1-1x(n+1)α+(n+1)k1+(n+1)l1𝑑xMA0δ(1-x)α+l1-1x(n+1)α+(n+1)k1+(n+1)l1𝑑x+MAδ1(1-x)α+l1-1𝑑xwith δ(0,1)MA0δ(1-x)α+l1-1𝑑xδ(n+1)α+(n+1)k1+(n+1)l1+MAα+l1δα+l1MAα+l1δ(n+1)α+(n+1)k1+(n+1)l1+MAα+l1δα+l1.

For any ϵ>0, it is easy to see that there exists δ(0,1) such that

Maα+l1δα+l1<ϵ2.

For this δ, there exists an integer N>0 sufficiently large such that

Maα+l1δ(n+1)α+(n+1)k1+(n+1)l1<ϵ2

for all n>N. So 0<un+1un<ϵ2+ϵ2=ϵ for all n>N. It follows that limn+un+1un=0. Then n=1+un is convergent. Hence

ϕ0(t)+[ϕ1(t)-ϕ0(t)]+[ϕ2(t)-ϕ1(t)]++[ϕn(t)-ϕn-1(t)]+,t[0,1]

is uniformly convergent. Then {ϕn(t)} is convergent uniformly on [0,1]. ∎

Claim 3.3.

The function ϕ defined on [0,1] by ϕ(t)=limn+ϕn(t) is a unique continuous solution of the integral equation

x(t)=ϕ0(t)+1Γ(α)0t(t-s)α-1A(s)x(s)𝑑s,t[0,1].

Proof.

By ϕ(t)=limn+ϕn(t) and the uniformly convergence, we see ϕ(t) is continuous on [0,1] by defining x(t)|t=0=limt0+x(t). It follows that

ϕ(t)=limn+ϕn(t)=limn+[ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s]=ϕ0(t)+limn+0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)limn+ϕn-1(s)ds=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕ(s)𝑑s.

Then ϕ is a continuous solution of (3.1.5) defined on [0,1].

Suppose that ψ defined on [0,1] is also a solution of (3.1.5). Then

ψ(t)=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ψ(s)𝑑s,t(0,1].

We need to prove that ϕ(t)ψ(t) on [0,1]. Then

|ψ(t)-ϕ0(t)|=|0t(t-s)α-1Γ(α)|A(s)ψ(s)|𝑑s||η|ψMAtα+k1+l1𝐁(α+l1,k1+1)Γ(α).

Furthermore, we have

|ψ(t)-ϕ1(t)|=|0t(t-s)α-1Γ(α)A(s)[ψ(s)-ϕ0(s)]𝑑s||η|ψMA2t2α+2k1+2l1𝐁(α+l1,k1+1)Γ(α)𝐁(α+l1,α+2k1+l1+1)Γ(α).

Now suppose that

|ψ(t)-ϕj-1(t)||η|ψMAjtjα+jk1+jl1i=0j-1𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α).

Then

|ψ(t)-ϕj(t)|=|0t(t-s)α-1Γ(α)A(s)[ψ(s)-ϕj-1(s)]𝑑s||η|ψMAj+1t(j+1)α+(j+1)k1+(j+1)l1i=0j𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α).

Hence

|ψ(t)-ϕn(t)||η|ψMAn+1t(n+1)α+(n+1)k1+(n+1)l1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α)|η|ψMAn+1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α)for all n=1,2,.

Similarly we have

limn+|η|ψMAn+1i=0n𝐁(α+l1,iα+(i+1)k1+il1+1)Γ(α)=0.

Then limn+ϕn(t)=ψ(t) uniformly on [0,1]. Thus ϕ(t)ψ(t). Then (3.1.5) has a unique solution ϕ. The proof is complete. ∎

Theorem 1.

Suppose that Assumption A1 holds. Then x is a solution of IVP (3.1.1) if and only if x is a solution of the integral equation (3.1.5).

Proof.

Suppose that xC[0,1] is a solution of IVP (3.1.1). Then limt0+x(t)=η and x=r<+. From Assumption A1, we have

|0t(t-s)α-1Γ(α)A(s)x(s)𝑑s|x0t(t-s)α-1Γ(α)|A(s)|𝑑s0t(t-s)α-1Γ(α)MArsk1(1-s)l1𝑑sMAr0t(t-s)α+l1-1Γ(α)sk1𝑑s=MArtα+k1+l101(1-w)α+l1-1Γ(α)wk1𝑑w=MArtα+k1+l1𝐁(α+l1,k1+1)Γ(α).

So t0t(t-s)α-1Γ(α)A(s)x(s)𝑑s is continuous on [0,1]. We have I0+αD0+αCx(t)=I0+α[A(t)x(t)+F(t)]. Thus

0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s=I0+α[A(t)x(t)+F(t)]=I0+αD0+αCx(t)=0t(t-s)α-1Γ(α)(1Γ(1-α)0s(s-w)-αx(w)𝑑w)𝑑s(interchange the order of integrals)=1Γ(α)Γ(1-α)0twt(t-s)α-1(s-w)-α𝑑sx(w)𝑑w(use s-wt-w=u)=1Γ(α)Γ(1-α)0t01(1-u)α-1u-α𝑑ux(w)𝑑w(by 𝐁(α,1-α)=Γ(α)Γ(1-α)Γ(1))=0tx(w)𝑑w=x(t)-limt0+x(t)=x(t)-η.

Then xC[0,1] is a solution of (3.1.5).

On the other hand, if x is a solution of (3.1.5), together with Claims 3.13.3, we have xC[0,1] and limt0+x(t)=η. So xC[0,1]. Furthermore,

D0+αCx(t)=1Γ(1-α)0t(t-s)-αx(s)𝑑s=1Γ(1-α)0t(t-s)-α(η+0s(s-w)α-1Γ(α)[A(w)x(w)+F(w)]𝑑w)𝑑s=1Γ(1-α)0t(t-s)-α(0s(s-w)α-1Γ(α)[A(w)x(w)+F(w)]𝑑w)𝑑s=[1Γ(2-α)0t(t-s)1-α1Γ(1-(1-α))(0s(s-w)-(1-α)[A(w)x(w)+F(w)]𝑑w)𝑑s]=1Γ(2-α)[.(t-s)1-α1Γ(1-(1-α))0s(s-w)-(1-α)[A(w)x(w)+F(w)]dw|0t+(1-α)0t(t-s)-α1Γ(1-(1-α))0s(s-w)-(1-α)[A(w)x(w)+F(w)]dwds]=1Γ(1-α)[0t(t-s)-α1Γ(1-(1-α))0s(s-w)-(1-α)[A(w)x(w)+F(w)]𝑑w𝑑s]=1Γ(1-α)1Γ(α)[0twt(t-s)-α(s-w)-(1-α)𝑑s[A(w)x(w)+F(w)]𝑑w](by changing the order of integrals)=1Γ(1-α)1Γ(α)[0t01(1-u)-αuα-1𝑑u[A(w)x(w)+F(w)]𝑑w](by s-wt-w=u)=[0t[A(w)x(w)+F(w)]𝑑w](by 𝐁(1-α,α)=Γ(1-α)Γ(α))=A(t)x(t)+F(t).

So xC[0,1] is a solution of IVP (3.1.1). ∎

Theorem 2.

Suppose that Assumption A1 holds. Then (3.1.1) has a unique solution. If there exist constants k2>-1, l20 with l2>max{-α,-α-k2}, MF0 such that |F(t)|MFtk2(1-t)l2 for all t(0,1), then the special problem

{D0+𝜶Cx(t)=λx(t)+F(t),t(0,1],limt0+x(t)=η(3.1.7)

has a unique solution

x(t)=ηEα,1(λtα)+0t(t-s)α-1Eα,α(λ(t-s)α)F(s)𝑑s,t(0,1].

Proof.

From Claims 3.13.3, Theorem 1 implies that (3.1.1) has a unique solution. From the assumption and A(t)λ, it is easy to see that Assumption A1 holds with k1=l1=0 and k2,l2 mentioned. Thus (3.1.7) has a unique solution. We get from the Picard function sequence that

ϕn(t)=η+λ0t(t-s)α-1Γ(α)ϕn-1(s)𝑑s+0t(t-s)α-1Γ(α)F(s)𝑑s=η+ηλ0t(t-s)α-1Γ(α)𝑑s+λ20t(t-s)α-1Γ(α)0s(s-w)α-1Γ(α)ϕn-2(w)𝑑w𝑑s+λ0t(t-s)α-1Γ(α)(s-w)α-1Γ(α)F(w)𝑑w𝑑s+0t(t-s)α-1Γ(α)F(s)𝑑s=η+ηλΓ(α+1)tα+λ20twt(t-s)α-1Γ(α)(s-w)α-1Γ(α)𝑑sϕn-2(w)𝑑w+λ0twt(t-s)α-1Γ(α)(s-w)α-1Γ(α)𝑑sF(w)𝑑w+0t(t-s)α-1Γ(α)F(s)𝑑s=η+ηλΓ(α+1)tα+λ20t(t-w)2α-101(1-u)α-1Γ(α)uα-1Γ(α)𝑑uϕn-2(w)𝑑w+λ0t(t-w)2α-101(1-u)α-1Γ(α)uα-1Γ(α)𝑑uF(w)𝑑w+0t(t-s)α-1Γ(α)F(s)𝑑s=η(1+λtαΓ(α+1))+λ20t(t-w)2α-1Γ(2α)ϕn-2(w)𝑑w+0t(t-s)α-1(λ(t-s)αΓ(2α)+1Γ(α))F(s)𝑑s=ηj=0n-1λjtjαΓ(jα+1)+ηλn0t(t-w)nα-1Γ(nα)𝑑w+0t(t-s)α-1(j=0n-1λj(t-s)jαΓ((j+1)α))F(s)𝑑s=ηj=0nλjtjαΓ(jα+1)+0t(t-s)α-1(j=0nλj(t-s)jαΓ((j+1)α))F(s)𝑑s

and so

ϕn(t)ηEα,1(λtα)+0t(t-s)α-1Eα,α(λ(t-s)α)F(s)𝑑s.

Then we get (3.1.8). The proof is complete. ∎

To get solutions of (3.1.2), we need the following assumptions.

Assumption A2.

There exist constants ki>-α and li0 with l1>max{-α,-1-k1}, l2>max{-α,-1-k2}, MA0 and MF0 such that |A(t)|MAtk1(1-t)l1 and |F(t)|MFtk2(1-t)l2 for all t(0,1).

Choose Picard function sequence as

ϕ0(t)=ηtα-1+0t(t-s)α-1Γ(α)F(s)𝑑s,t(0,1],ϕn(t)=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s,t(0,1],n=1,2,.

Claim 3.4.

One has ϕnC1-α(0,1].

Proof.

Since ϕ0C1-α(0,1], we know that ϕ1 is continuous on (0,1], and together with

t1-α|0t(t-s)α-1Γ(α)A(s)ϕ0(s)𝑑s|=t1-α|0t(t-s)α-1Γ(α)A(s)sα-1s1-αϕ0(s)𝑑s|ϕ0t1-α0t(t-s)α-1Γ(α)MA|η|sα-1sk1(1-s)l1𝑑sϕ0t1-αMA|η|0t(t-s)α+l1-1Γ(α)sα+k1-1𝑑s=MA|η|ϕ0t1+k1+l1𝐁(α+l1,α+k1)Γ(α)0as t0+,

we see that ϕ1C1-α(0,1]. By the mathematical induction method, we obtain that ϕnC1-α(0,1]. ∎

Claim 3.5.

The sequence {tt1-αϕn(t)} is convergent uniformly on (0,1].

Proof.

In fact, we have for t(0,1] similarly to Claim 3.4 that

t1-α|ϕ1(t)-ϕ0(t)|=|0t(t-s)α-1Γ(α)A(s)ϕ0(s)𝑑s|MA|η|ϕ0t1+k1+l1𝐁(α+l1,α+k1)Γ(α).

It follows that

t1-α|ϕ2(t)-ϕ1(t)|=|0t(t-s)α-1Γ(α)A(s)[ϕ1(s)-ϕ0(s)]𝑑s|t1-α0t(t-s)α-1Γ(α)MAsk1(1-s)l1(|η|ϕ0MAsk1+l1+1𝐁(α+l1,k1+1)Γ(α))𝑑s|η|ϕ0MA2t1-α0t(t-s)α+l1-1Γ(α)s2k1+l1+1𝐁(α+l1,k1+1)Γ(α)𝑑s=|η|ϕ0MA2t2k1+2l1+2𝐁(α+l1,k1+1)Γ(α)𝐁(α+l1,2k1+l1+2)Γ(α).

Again by the mathematical induction method, we get for every n=1,2, that

t1-α|ϕn(t)-ϕn-1(t)||η|Mϕ0Antnk1+nl1+ni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α)|η|MAni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α),t[0,1].

Similarly we can prove that

n=1+un=n=1+|η|MAni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α)

is convergent. Hence

t1-αϕ0(t)+t1-α[ϕ1(t)-ϕ0(t)]+t1-α[ϕ2(t)-ϕ1(t)]++t1-α[ϕn(t)-ϕn-1(t)]+,t[0,1],

is uniformly convergent. Then {tt1-αϕn(t)} is convergent uniformly on (0,1]. ∎

Claim 3.6.

The function ϕ defined on (0,1] by ϕ(t)=tα-1limn+t1-αϕn(t) is a unique continuous solution of the integral equation

x(t)=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)x(s)𝑑s,t(0,1].

Proof.

By limn+t1-αϕn(t)=t1-αϕ(t) and the uniformly convergence, we see that ϕ(t) is continuous on (0,1]. Then we know

ϕ(t)=tα-1limnt1-αϕn(t)=limn+[ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s]=ϕ0(t)+limn+t1-α0t(t-s)α-1Γ(α)A(s)ϕn-1(s)𝑑s=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ϕ(s)𝑑s.

Then ϕ is a continuous solution of (3.1.9) defined on (0,1].

Suppose that ψ defined on (0,1] is also a solution of (3.1.9). Then

ψ(t)=ϕ0(t)+0t(t-s)α-1Γ(α)A(s)ψ(s)𝑑s,t[0,1].

We need to prove that ϕ(t)ψ(t) on (0,1]. Then

t1-α|ψ(t)-ϕ0(t)|=t1-α|0t(t-s)α-1Γ(α)|A(s)ψ(s)|𝑑s||η|ψMAtk1+l1+1𝐁(α+l1,k1+1)Γ(α)+MFtk2+l2+1𝐁(α+l2,k2+1)Γ(α).

Furthermore, we have

t1-α|ψ(t)-ϕ1(t)|=t1-α|0t(t-s)α-1Γ(α)A(s)[ψ(s)-ϕ0(s)]𝑑s||η|ψMA2t2k1+2l1+2𝐁(α+l1,k1+1)Γ(α)𝐁(α+l1,2k1+l1+2)Γ(α).

By the mathematical induction method, we can get that

t1-α|ψ(t)-ϕn(t)|=t1-α|0t(t-s)α-1Γ(α)A(s)[ψ(s)-ϕn-1(s)]𝑑s||η|ψMAntnk1+nl1+ni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α).

Hence

t1-α|ψ(t)-ϕn(t)||η|ψMAni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α)for all n=1,2,.

Similarly we have limn+t1-αϕn(t)=t1-αψ(t) uniformly on (0,1]. Thus ϕ(t)ψ(t) on (0,1]. Then (3.1.9) has a unique solution ϕ. ∎

Theorem 3.

Suppose that Assumption A2 holds. Then xC1-α(0,1] is a solution of IVP (3.1.2) if and only if xC1-α(0,1] is a solution of the integral equation (3.1.9).

Proof.

Suppose that xC1-α(0,1] is a solution of IVP (3.1.2). Then tt1-αx(t)is continuous on (0,1] by defining t1-αx(t)|t=0=limt0+t1-αx(t) and x=r<+. So  by ws=u, we get

lims0+0s(s-w)-αx(w)𝑑w=lims0+0s(s-w)-αwα-1w1-αx(w)𝑑w=lims0+ξ1-αx(ξ)0s(s-w)-αwα-1𝑑w=lims0+ξ1-αx(ξ)01(1-u)-αuα-1𝑑u=η𝐁(1-α,α),

by the mean value theorem for integrals, ξ(0,s). From Assumption A2, we have

t1-α|0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s|=t1-α|0t(t-s)α-1Γ(α)[A(s)sα-1s1-αx(s)+F(s)]𝑑s|0t(t-s)α-1Γ(α)[MArsα-1sk1(1-s)l1+MFsk2(1-s)l2]𝑑st1-αMAr0t(t-s)α+l1-1Γ(α)sα+k1-1𝑑s+t1-αMF0t(t-s)α+l2-1Γ(α)sk2𝑑s=MArtα+k1+l1𝐁(α+l1,α+k1)Γ(α)+MFt1+k2+l2𝐁(α+l2,k2+1)Γ(α).

So

tt1-α0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s

is defined on (0,1] and

limt0+t1-α0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s=0.

Furthermore, we have similarly to Theorem 1 that

t0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s

is continuous on (0,1]. So

tt1-α0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s

is continuous on [0,1] by defining

t1-α0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s|t=0=limt0+t1-α0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s.

We have I0+αD0+αRLx(t)=I0+α[A(t)x(t)+F(t)]. So

0t(t-s)α-1Γ(α)[A(s)x(s)+F(s)]𝑑s=I0+α[A(t)x(t)+F(t)]=I0+αD0+αRLx(t)=0t(t-s)α-1Γ(α)[1Γ(1-α)(0s(s-w)-αx(w)𝑑w)]𝑑s=[0t(t-s)αΓ(α+1)1Γ(1-α)(0s(s-w)-αx(w)𝑑w)𝑑s]=1Γ(1-α)Γ(α+1)[(t-s)α0s(s-w)-αx(w)𝑑w|0t+α0t(t-s)α-10s(s-w)-αx(w)𝑑w𝑑s]=1Γ(1-α)Γ(α+1)[-tαlims0+0s(s-w)-αx(w)𝑑w+α0t(t-s)α-10s(s-w)-αx(w)𝑑w𝑑s]=1Γ(1-α)Γ(α+1)[α0twt(t-s)α-1(s-w)-α𝑑sx(w)𝑑w]-1Γ(1-α)Γ(α)tα-1lims0+0s(s-w)-αx(w)𝑑w=1Γ(1-α)Γ(α+1)[α0t01(1-u)α-1u-α𝑑ux(w)𝑑w]-tα-1Γ(1-α)Γ(α)lims0+0s(s-w)-αx(w)𝑑w=[0tx(w)𝑑w]=x(t)-tα-1Γ(1-α)Γ(α)η𝐁(1-α,α)=x(t)-ηtα-1,

by s-wt-w=u. Then xC1-α(0,1] is a solution of (3.1.9).

On the other hand, if xC1-α(0,1] is a solution of (3.1.9), then this together with (3.1.10) implies that limt0+t1-αx(t)=η. Furthermore, we have

D0+αRLx(t)=1Γ(1-α)(0t(t-s)-αx(s)𝑑s)=1Γ(1-α)(0t(t-s)-α(ηsα-1+0s(s-w)α-1Γ(α)[A(w)x(w)+F(w)]𝑑w)𝑑s)=1Γ(1-α)(η0t(t-s)-αsα-1𝑑s)+1Γ(1-α)(0t(t-s)-α0s(s-w)α-1Γ(α)[A(w)x(w)+F(w)]𝑑w𝑑s)=1Γ(1-α)(η01(1-u)-αuα-1𝑑u)+1Γ(1-α)(0twt(t-s)-α(s-w)α-1Γ(α)𝑑s[A(w)x(w)+F(w)]𝑑w)=1Γ(1-α)(0t01(1-u)-αuα-1Γ(α)𝑑u[A(w)x(w)+F(w)]𝑑w)=A(t)x(t)+F(t).

So xC1-α(0,1] is a solution of IVP (3.1.2). ∎

Theorem 4.

Suppose that Assumption A2 holds. Then (3.1.2) has a unique solution. If A(t)λ and there exist constants k2>-1, l20 with l2>max{-α,-1-k1} and MF0 such that |F(t)|MFtk2(1-t)l2 for all t(0,1), then the special problem

{D0+𝜶RLx(t)=λx(t)+F(t),t(0,1],limt0+t1-αx(t)=η(3.1.12)

has a unique solution

x(t)=ηΓ(α)tα-1Eα,α(λtα)+0t(t-s)α-1Eα,α(λ(t-s)α)F(s)𝑑s,t(0,1].

Proof.

From Claims 3.43.6, IVP (3.1.2) and Theorem 3 has a unique solution. From the assumption and

A(t)λ,

one sees that Assumption A2 holds with k1=l1=0 and k2,l2 mentioned. Thus (3.1.12) has a unique solution. We get from the Picard function sequence that

ϕn(t)=ηtα-1+λ0t(t-s)α-1Γ(α)ϕn-1(s)𝑑s+0t(t-s)α-1Γ(α)F(s)𝑑s=ηtα-1+ηλ0t(t-s)α-1Γ(α)sα-1𝑑s+λ20t(t-s)α-1Γ(α)0s(s-w)α-1Γ(α)ϕn-2(w)𝑑w𝑑s+λ0t(t-s)α-1Γ(α)(s-w)α-1Γ(α)F(w)𝑑w𝑑s+0t(t-s)α-1Γ(α)F(s)𝑑s=ηtα-1+ηλΓ(α)t2α-1Γ(2α)+λ20twt(t-s)α-1Γ(α)(s-w)α-1Γ(α)𝑑sϕn-2(w)𝑑w+λ0twt(t-s)α-1Γ(α)(s-w)α-1Γ(α)𝑑sF(w)𝑑w+0t(t-s)α-1Γ(α)F(s)𝑑s=ηΓ(α)tα-1j=0n-1λjtjαΓ((j+1)α)+ηλn0t(t-w)nα-1Γ(nα)𝑑w+0t(t-s)α-1(j=0n-1λj(t-s)jαΓ((j+1)α))F(s)𝑑s=ηΓ(α)tα-1j=0nλjtjαΓ((j+1)α)+0t(t-s)α-1(j=0nλj(t-s)jαΓ((j+1)α))F(s)𝑑s

and hence

ϕn(t)ηΓ(α)tα-1Eα,α(λtα)+0t(t-s)α-1Eα,α(λ(t-s)α)F(s)𝑑s.

Then we get (3.1.13). ∎

To get solutions of (3.1.3), we need the following assumptions:

Assumption A3.

There exist constants ki>-α, li0 with l1>max{-α,-α-k1}, l2>max{-α,-1+k2}, MB0 and MG0 such that |B(t)|MB(logt)k1(1-logt)l1 and |G(t)|MG(logt)k2(1-logt)l2 for all t(1,e).

Choose Picard function sequence as

ϕ0(t)=η(logt)α-1+1Γ(α)1t(logts)α-1G(s)dss,t(1,e],ϕn(t)=ϕ0(t)+1Γ(α)1t(logts)α-1B(s)ϕn-1(s)dss,t(1,e],n=1,2,.

Claim 3.7.

One has ϕnLC1-α(1,e].

Proof.

In fact, ϕ0LC1-α(1,e] and

(logt)1-α|1t(logts)α-1B(s)ϕ0(s)dss|ϕ0(logt)1-α1t(logts)α-1MB|η|(logs)α-1(logs)k1(1-logs)l1dssϕ0(logt)1-αMB|η|1t(logts)α+l1-1(logs)α+k1-1dss=MB|η|ϕ0(logt)α+k1+l1𝐁(α+l1,k1+α)0as t0+.

We know that

t1Γ(α)1t(logts)α-1B(s)ϕ0(s)dss

is continuous on (1,e] and limt0+(logt)1-αϕ1(t) exists. Then ϕ1LC1-α(1,e]. By the mathematical induction method, we obtain that ϕnLC1-α(1,e]. ∎

Claim 3.8.

The sequence {t(logt)1-αϕn(t)} is convergent uniformly on (1,e].

The proof is similar to that of Claim 3.5 and is omitted.

Claim 3.9.

The function ϕ defined on (1,e] by ϕ(t)=(logt)α-1limn+(logt)1-αϕn(t) is a unique continuous solution of the integral equation

x(t)=η(logt)α-1+1Γ(α)1t(logts)α-1[B(s)x(s)+G(s)]dss,t(1,e].

Proof.

By

limn+(logt)1-αϕn(t)=(logt)1-αϕ(t)

and the uniformly convergence, we see that ϕ(t) is continuous on (1,e]. From

(logt)1-α|1t(logts)α-1[A(s)ϕn-1(s)+F(s)]dss-1t(logts)α-1[B(s)ϕm-1(s)+G(s)]dss|MBϕn-1-ϕm-1(logt)1-α1t(logts)α+l1-1(logs)k1(logs)α-1dssMBϕn-1-ϕm-1(logt)1-α1t(logts)α+l1-1(logs)α+k1-1dssMBϕn-1-ϕm-1(logt)α+k1+l1𝐁(α+l1,α+k1)MBϕn-1-ϕm-1𝐁(α+l1,α+k1)0uniformly as m,n+,

we know that

ϕ(t)=(logt)α-1limn+(logt)1-αϕn(t)=limn+[η+(logt)1-α1Γ(α)1t(logts)α-1[B(s)ϕn-1(s)+G(s)]dss]=η(logt)α-1+(logt)α-1limn+(logt)1-α1t(logts)α-1[B(s)ϕn-1(s)+G(s)]dss=η(logt)α-1+1Γ(α)1t(logts)α-1[B(s)ϕ(s)+G(s)]dss.

Then ϕ is a continuous solution of (3.1.14) defined on (1,e].

Suppose that ψ defined on (1,e] is also a solution of (3.1.14). Then

ψ(t)=η(logt)α-1+1Γ(α)1t(logts)α-1[B(s)ψ(s)+G(s)]dss,t(1,e].

We need to prove that ϕ(t)ψ(t) on (1,e]. Then

(logt)1-α|ψ(t)-ϕ0(t)|=(logt)1-α|1t(logts)α-1|B(s)ψ(s)|dss||η|ψMB(logt)k1+l1+1𝐁(α+l1,k1+1)Γ(α).

By the mathematical induction method, we can get that

(logt)1-α|ψ(t)-ϕn(t)|=(logt)1-α1Γ(α)|1t(logts)α-1B(s)[ψ(s)-ϕn-1(s)]𝑑s||η|ψMBn(logt)nk1+nl1+ni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α)|η|ψMBni=0n-1𝐁(α+l1,(i+1)k1+il1+(i+1))Γ(α)for all n=1,2,.

Similarly we have limn+(logt)1-αϕn(t)=(logt)1-αψ(t) uniformly on (1,e]. Thus ϕ(t)ψ(t) on (1,e]. Then (3.1.14) has a unique solution ϕ. ∎

Theorem 5.

Suppose that Assumption A3 holds. Then xLC1-α(1,e] is a solution of IVP (3.1.3) if and only if xLC1-α(1,e] is a solution of the integral equation (3.1.14).

Proof.

Suppose that xC1-α(0,1] is a solution of IVP (3.1.3). Then t(logt)1-αx(t)is continuous on (1,e] by defining (logt)1-αx(t)|t=1=limt1+(logt)1-αx(t) and x=r<+. So

lims1+1s(logsw)-αx(w)dww=lims1+1s(logsw)-α(logw)α-1(logw)1-αx(w)dww=lims1+(logξ)1-αx(ξ)1s(logsw)-α(logw)α-1dww=lims1+(logξ)1-αx(ξ)01(1-u)-αuα-1𝑑u(by logwlogs=u)=η𝐁(1-α,α)

by the mean value theorem for integrals, ξ(1,s). From Assumption A3, we have

(logt)1-α|1t(logts)α-1[B(s)x(s)+G(s)]dss|(logt)1-α1t(logts)α-1[MBr(logs)α-1(logs)k1(1-logs)l1+MG(logs)k2(1-logs)l2]dss(logt)1-αMBr1t(logts)α+l1-1(logs)α+k1-1dss+(logt)1-αMG1t(logts)α+l2-1(logs)k2dss=MBr(logt)α+k1+l1𝐁(α+l1,k1+α)+MG(logt)1+k1+l1𝐁(α+l2,k2+1).

It follows that

t(logt)1-α1t(logts)α-1[B(s)x(s)+G(s)]dss

is defined on (1,e] and

limt1+(logt)1-α1t(logts)α-1[B(s)x(s)+G(s)]dss=0.

Furthermore, we have similarly to Theorem 1 that

t1t(logts)α-1[B(s)x(s)+G(s)]dss

is continuous on (1,e]. So

t(logt)1-α1t(logts)α-1[B(s)x(s)+G(s)]dss

is continuous on [1,e] by defining

(logt)1-α1t(logts)α-1[B(s)x(s)+G(s)]dss|t=1=limt1+(logt)1-α1t(logts)α-1[B(s)x(s)+G(s)]dss.

We have

I1+αHD1+αRLHx(t)=I1+αH[B(t)x(t)+G(t)].

Therefore

1Γ(α)1t(logts)α-1[B(s)x(s)+G(s)]dss=I1+αH[B(t)x(t)+G(t)]=I1+αHD1+αRLHx(t)=1Γ(α)1t(logts)α-11Γ(1-α)s(1s(logsw)-αx(w)dww)dss=1Γ(α+1)1Γ(1-α)t[1t(logts)α(1s(logsw)-αx(w)dww)𝑑s]=1Γ(1-α)Γ(α+1)t[(logts)α1s(logsw)-αx(w)dww|1t+α1t(logts)α-11s(logsw)-αx(w)dwwdss]=1Γ(1-α)Γ(α+1)t[(logt)αlims1+1s(logsw)-αx(w)dww+α1twt(logts)α-1(logsw)-αdssx(w)dww]=1Γ(1-α)Γ(α+1)t[(logt)αη𝐁(1-α,α)+α1t𝐁(α,1-α)x(w)dww]=x(t)-η(logt)α-1.

Then xLC1-α(1,e] is a solution of (3.1.14).

On the other hand, if x is a solution of (3.1.14), then we deduce from (3.1.15) that

limt1+(logt)1-αx(t)=η.

Then xLC1-α(1,e]. Furthermore, we have by Definition 2.5 that

D1+αRLHx(t)=1Γ(1-α)t(1t(logts)-αx(s)dss)=1Γ(1-α)t[1t(logts)-α(η(logs)α-1+1Γ(α)1s(logsw)α-1[A(w)x(w)+F(w)]dww)dss]=1Γ(1-α)t[η1t(logts)-α(logs)α-1dss]+1Γ(1-α)t[1Γ(α)1s(logsw)α-1[A(w)x(w)+F(w)]dwwdss]=1Γ(1-α)t[η𝐁(1-α,α)]+1Γ(1-α)t[1Γ(α)1twt(logts)-α(logsw)α-1dss[A(w)x(w)+F(w)]dww]=1Γ(1-α)t[1Γ(α)1t𝐁(1-α,α)[B(w)x(w)+G(w)]dww]=B(t)x(t)+G(t).

So xLC1-α(1,e] is a solution of IVP (3.1.3). ∎

Theorem 6.

Suppose that Assumption A3 holds. Then (3.1.14) has a unique solution. If B(t)λ and there exist constants k2>-1, l20 with l2>max{-α,-1-k2} and MG0 such that |G(t)|MGtk2(1-t)l2 for all t(1,e), then the special problem

{D1+𝜶RLHx(t)=λx(t)+G(t),t(1,e],limt0+(logt)1-αx(t)=η(3.1.17)

has a unique solution

x(t)=ηΓ(α)(logt)α-1Eα,α(λ(logt)α)+1t(logts)α-1Eα,α(λ(logts)α)G(s)dss,t(1,e].

Proof.

From Claims 3.73.9, IVP (3.1.14) has a unique solution. From the assumption and B(t)λ, one sees that Assumption A3 holds with k1=l1=0 and k2,l2 mentioned in assumption. Thus (3.1.17) has a unique solution. We get from the Picard function sequence that

ϕn(t)=η(logt)α-1+λ1Γ(α)1t(logts)α-1ϕn-1(s)𝑑s+1Γ(α)1t(logts)α-1G(s)dss=η(logt)α-1+ηλΓ(α)1t(logts)α-1(logs)α-1dss+λ2Γ(α)1t(logts)α-11Γ(α)1s(logsw)α-1ϕn-2(w)dwwdss+λΓ(α)1t(logts)α-11Γ(α)(logsw)α-1G(w)dwwdss+1Γ(α)1t(logts)α-1G(s)dss=η(logt)α-1+ηλ(logt)2α-1Γ(2α)+λ2Γ(α)1twt(logts)α-11Γ(α)(logsw)α-1dssϕn-2(w)dww+λΓ(α)1twt(logts)α-1(logsw)α-11Γ(α)dssG(w)dww+1Γ(α)1t(logts)α-1G(s)dss=ηΓ(α)(logt)α-1j=0n-1λj(logt)jαΓ((j+1)α)+ηλnΓ(α)1t(logts)α-1dww+1Γ(α)1t(logts)α-1(j=0n-1λj(t-s)jαΓ((j+1)α))F(s)dss=ηΓ(α)(logt)α-1j=0nλj(logt)jαΓ((j+1)α)+1Γ(α)1t(logts)α-1(j=0nλj(ts)jαΓ((j+1)α))G(s)dss

and so

ϕn(t)ηΓ(α)(logt)α-1Eα,α(λ(logt)α)+1t(logts)α-1Eα,α(λ(logts)α)G(s)dss.

Then we get (3.1.18). The proof is complete. ∎

To get solutions of (3.1.4), we need the following assumptions:

Assumption A4.

There exist constants ki>-1 and li0 with li>max{-α,-α-ki}, MB0 and MG0 such that |B(t)|MB(logt)k1(1-logt)l1 and |G(t)|MG(logt)k2(1-logt)l2 for all t(1,e).

Choose Picard function sequence as

ϕ0(t)=η+1Γ(α)1t(logts)α-1G(s)dss,t(1,e],ϕn(t)=ϕ0(t)+1Γ(α)1t(logts)α-1B(s)ϕn-1(s)dss,t(1,e],n=1,2,.

Claim 3.10.

One has ϕnC[1,e].

The proof is similar to that of Claim 3.1 and is omitted.

Claim 3.11.

The sequence ϕn is convergent uniformly on [1,e].

The proof is similar to that of Claim 3.2 and is omitted.

Claim 3.12.

The function ϕ defined on (1,e] by ϕ(t)=limn+ϕn(t) is a unique continuous solution of the integral equation

x(t)=η+1Γ(α)1t(logts)α-1[B(s)x(s)+G(s)]dss.

The proof is similar to that of Claim 3.3 and is omitted.

Lemma 7.

Suppose that Assumption A4 holds. Then xC(1,e] is a solution of IVP (3.1.4) if and only if xC(1,e] is a solution of the integral equation (3.1.19).

Proof.

Suppose that xC(1,e] is a solution of IVP (3.1.4). Then the function tx(t) is continuous on [1,e] and x=r<+. One can see that

1t(logts)α-1(logs)k1(1-logs)l1dss1t(logts)α+l1-1(logs)k1dss(by logslogt=u)=(logt)α+k1+l101(1-u)α+l1-1uk1𝑑u(logt)α+k1+l101(1-u)α+l1-1uk1𝑑u=(logt)α+k1+l1𝐁(α+l1,k1+1).

From Assumption A4, we have for t(1,e] that

|1t(logts)α-1[B(s)x(s)+G(s)]dss|1t(logts)α-1[MBr(logs)k1(1-logs)l1+MG(logs)k2(1-logs)l2]dssMBr1t(logts)α-1(logs)k1(1-logs)l1dss+MG1t(logts)α-1(logs)k2(1-logs)l2dss=MBr(logt)α+k1+l1𝐁(α+l1,k1+1)+MG(logt)α+k2+l2𝐁(α+l2,k2+1).

It follows that

t1t(logts)α-1[B(s)x(s)+G(s)]dss

is defined on (1,e] and

limt1+1t(logts)α-1[B(s)x(s)+G(s)]dss=0.(3.1.20)

Furthermore, we have similarly to Theorem 1 that

t1t(logts)α-1[B(s)x(s)+G(s)]dss

is continuous on (1,e]. So

t1t(logts)α-1[B(s)x(s)+G(s)]dss

is continuous on [1,e] by defining

1t(logts)α-1[B(s)x(s)+G(s)]dss|t=1=0.(3.1.21)

One sees that

wt(logts)α-1(logsw)-αdss=01(1-u)α-1u-α𝑑u=Γ(1-α)Γ(α)

by logs-logwlogt-logw=u. We have by Definition 2.6 and I1+αHD1+αCHx(t)=I1+αH[B(t)x(t)+G(t)] that

1t(logts)α-1[A(s)x(s)+F(s)]dss=I1+αH[B(t)x(t)+G(t)]=I1+αHD1+αCHx(t)=1Γ(α)1t(logts)α-1[1Γ(1-α)1s(logsw)-αwx(w)dww]dss=1Γ(α)1t(logts)α-1[1Γ(1-α)0s(logsw)-αx(w)𝑑w]dss=1Γ(α)1Γ(1-α)1twt(logts)α-1(logsw)-αdssx(w)𝑑w=1tx(w)𝑑w=x(t)-limt1+x(t)=x(t)-η.

Then xC(1,e] is a solution of (3.1.19).

On the other hand, if xC(1,e] is a solution of (3.1.19), then this together with (3.1.20) implies limt1+(t)=η. Furthermore, we have that

D1+αCHx(t)=1Γ(1-α)1t(logts)-αsx(s)dss=1Γ(1-α)1t(logts)-α(η+1Γ(α)1s(logsw)α-1[B(w)x(w)+G(w)]dww)𝑑s=1Γ(1-α)1t(logts)-α(1Γ(α)1s(logsw)α-1[B(w)x(w)+G(w)]dww)𝑑s=tΓ(2-α)(1t(logts)1-α(1Γ(α)1s(logsw)α-1[B(w)x(w)+G(w)]dww)𝑑s)=tΓ(2-α)[(logts)1-α1Γ(α)1s(logsw)α-1[B(w)x(w)+G(w)]dww|1t+(1-α)1s1t(logts)-α1Γ(α)1s(logsw)α-1[B(w)x(w)+G(w)]dwwds]=tΓ(1-α)1Γ(α)[1s1t(logts)-α1s(logsw)α-1[B(w)x(w)+G(w)]dww𝑑s]=tΓ(1-α)1Γ(α)[1twt(logts)-α(logsw)α-1dss[B(w)x(w)+G(w)]dww](by logslogt=u)=tΓ(1-α)1Γ(α)[1t01(1-u)-αuα-1𝑑u[B(w)x(w)+G(w)]dww](by 𝐁(1-α,α)=Γ(1-α)Γ(α))=B(t)x(t)+G(t).

So xC(1,e] is a solution of IVP (3.1.4). ∎

Theorem 8.

Suppose that Assumption A4 holds. Then (3.1.4) has a unique solution. If B(t)λ and there exist constants k2>-1, l20 with l2>max{-α,-α-k2} and MG0 such that |G(t)|MGtk2(1-t)l2 for all t(1,e), then the special problem

{D0+𝜶CHx(t)=λx(t)+G(t),t(1,e],limt1+x(t)=η(3.1.22)

has a unique solution

x(t)=ηEα,1(λ(logt)α)+1t(logts)α-1Eα,α(λ(logts)α)G(s)dss,t(1,e].

Proof.

From Claims 3.103.12 and Lemma 7, IVP (3.1.4) has a unique solution. From the assumption and

A(t)λ,

one sees that Assumption A4 holds with k1=l1=0 and k2,l2 mentioned. Thus (3.1.22) has a unique solution. We get from the Picard function sequence that

ϕn(t)=η+λ1Γ(α)1t(logts)α-1ϕn-1(s)dss+1Γ(α)1t(logts)α-1G(s)dss=η+ηλ1Γ(α)1t(logts)α-1dss+λ21Γ(α)1t(logts)α-11s(logsw)α-1ϕn-2(w)dwwdss+λ1Γ(α)1t(logts)α-11s(logsw)α-1G(w)dwwdss+1Γ(α)1t(logts)α-1G(s)dss=η+ηλ(logt)αΓ(α+1)+λ21Γ(α)1twt(logts)α-1(logsw)α-1dssϕn-2(w)dww+λ1Γ(α)0twt(logts)α-1(logsw)α-1dssF(w)dww+1Γ(α)1t(logts)α-1F(s)dss=η(1+λ(logt)αΓ(α+1))+λ21Γ(2α)1t(logts)2α-1ϕn-2(w)dww+λ1Γ(2α)0t(logts)2α-1G(w)dww+1Γ(α)1t(logts)α-1G(s)dss=ηj=0nλj(logt)jαΓ(jα+1)+1t(logts)α-1(j=0nλj(logts)jαΓ((j+1)α))G(s)𝑑s

and so

ϕn(t)ηEα,1(λ(logt)α)+1t(logts)α-1Eα,α(λ(logts)α)G(s)dss.

Then we get (3.1.23). The proof is complete. ∎

Theorem 9 (Schaefer’s fixed point theorem).

Let E be a Banach spaces and let T:EE be a completely continuous operator. If the set E(T)={x=θ(Tx):for some θ[0,1],xE} is bounded, then T has at least a fixed point in E.

3.2 Exact piecewise continuous solutions of impulsive FDEs

In this subsection, we present exact piecewise continuous solutions of the following fractional differential equations, respectively

D0+𝜶Cx(t)=λx(t)+F(t),t(ti,ti+1],i0,(3.2.1)D0+𝜶RLx(t)=λx(t)+F(t),t(ti,ti+1],i0,(3.2.2)D0+𝜶RLHx(t)=λx(t)+G(t),t(ti,ti+1],i0,(3.2.3)D0+𝜶CHx(t)=λx(t)+G(t),t(ti,ti+1],i0,(3.2.4)

where λ, 0=t0<t1<<tm<tm+1=1 in (3.2.1) and (3.2.2) and 1=t0<t1<<tm<tm+1=e in (3.2.3) and (3.2.4). We say that x:(0,1]R is a piecewise solution of (3.2.1) (or (3.2.2) if xPmC(0,1] (or PmC1-α(0,1] and satisfies (3.2.1) or (3.2.2). We say that x(1,e] is a piecewise continuous solutions of (3.2.3) (or (3.2.4)) if xLPmC1-α(1,e] (or LPmC(1,e]) and x satisfies all equations in (3.2.3) (or (3.2.4)).

Theorem 1.

Suppose that F is continuous on (0,1) and there exist constants k>-1 and l(-α,-α-k,0] such that |F(t)|tk(1-t)l for all t(0,1). Then x is a piecewise solution of (3.2.1) if and only if x and there exists constants ciR (i0) such that

x(t)=v=0jcv𝐄α,1(λ(t-tv)α)+0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s,t(tj,tj+1],j0.(3.2.5)

Proof.

Firstly, we have for t(ti,ti+1] that

|0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s|0t(t-s)α-1𝐄α,α(λ(t-s)α)|F(s)|𝑑s0t(t-s)α-1𝐄α,α(λ(t-s)α)sk(1-s)l𝑑s=j=0+λj𝚪((j+1)α)0t(t-s)α-1(t-s)αjsk(1-s)l𝑑sj=0+λj𝚪((j+1)α)0t(t-s)α+l-1(t-s)αjsk𝑑s=j=0+λj𝚪((j+1)α)tα+αj+k+l01(1-w)α+αj+l-1wk𝑑wj=0+λjtαj𝚪((j+1)α)tα+k+l01(1-w)α+l-1wk𝑑w=tα+k+l𝐄α,α(λtα)𝐁(α+l,k+1).

Then 0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s is convergent and is continuous on [0,1]. If x is a piecewise continuous solution of (3.2.5), then we know that xPmC(0,1] and limtti+x(t) (i0) exist. Now we prove that x satisfies differential equation in (3.2.1). In fact, for t(ti,ti+1] (i0), we have

D0+αCx(t)=1𝚪(1-α)0t(t-s)-αx(s)𝑑s=1𝚪(1-α)[j=0i-1tjtj+1(t-s)-αx(s)𝑑s+tit(t-s)-αx(s)𝑑s]=1𝚪(1-α)[j=0i-1tjtj+1(t-s)-α(κ=0jcκ𝐄α,1(λ(s-tκ)α)+0s(s-v)α-1𝐄α,α(λ(s-v)α)F(v)dv)ds+tit(t-s)-α(κ=0icκ𝐄α,1(λ(s-tκ)α)+0s(s-v)α-1𝐄α,α(λ(s-v)α)F(v)dv)ds]=1𝚪(1-α)j=0i-1tjtj+1(t-s)-α[κ=0jcκ𝐄α,1(λ(s-tκ)α)]𝑑s+1𝚪(1-α)tit(t-s)-α[κ=0icκ𝐄α,1(λ(s-tκ)α)]𝑑s+1𝚪(1-α)t0t(t-s)-α[0s(s-v)α-1𝐄α,α(λ(s-v)α)F(v)𝑑v]𝑑s=1𝚪(1-α)j=0i-1κ=0jcκtjtj+1(t-s)-α[m=0+λm(s-tκ)mα𝚪(mα+1)]𝑑s+1𝚪(1-α)κ=0icκtit(t-s)-α[m=0+λm(s-tκ)mα𝚪(mα+1)]𝑑s+1𝚪(1-α)m=0+λm𝚪(α(m+1))0t(t-s)-α[0s(s-v)α+mα-1F(v)𝑑v]𝑑s=1𝚪(1-α)j=0i-1κ=0jcκm=1+λmmα𝚪(mα+1)tjtj+1(t-s)-α(s-tκ)mα-1𝑑s+1𝚪(1-α)m=1+mαλm𝚪(mα+1)κ=0icκtit(t-s)-α(s-tκ)mα-1𝑑s+m=0+λmD0+αI0+α(m+1)F(t)=1𝚪(1-α)m=1+λmmα𝚪(mα+1)j=0i-1κ=0jcκ(t-tκ)mα-αtj-tκt-tκtj+1-tκt-tκ(1-w)-αwmα-1𝑑w+1𝚪(1-α)m=1+mαλm𝚪(mα+1)κ=0icκ(t-tκ)αm-αti-yκt-tκ1(1-w)-αwmα-1𝑑w+m=0+λmI0+αmF(t)=1𝚪(1-α)m=1+λmmα𝚪(mα+1)κ=0i-1cκ(t-tκ)mα-αj=κi-1tj-tκt-tκtj+1-tκt-tκ(1-w)-αwmα-1𝑑w+1𝚪(1-α)m=1+mαλm𝚪(mα+1)κ=0i-1cκ(t-tκ)αm-αti-yκt-tκ1(1-w)-αwmα-1𝑑w+ci𝚪(1-α)m=1+mαλm𝚪(mα+1)01(1-w)-αwmα-1𝑑w+f(t)+λ0tm=1+(t-s)α-1λm-1(t-s)α(m-1)𝚪(αm)F(s)ds=1𝚪(1-α)m=1+λmmα𝚪(mα+1)κ=0i-1cκ(t-tκ)mα-α01(1-w)-αwmα-1𝑑w+ci𝚪(1-α)m=1+mαλm𝚪(mα+1)01(1-w)-αwmα-1𝑑w+f(t)+λ0tm=1+(t-s)α-1λm-1(t-s)α(m-1)𝚪(αm)F(s)ds=λx(t)+F(t).

We have done that x satisfies (3.2.1) if x satisfies (3.2.5).

Now, we suppose that x is a solution of (3.2.1). We will prove that x satisfies (3.2.5) by the mathematical induction method. Since x is continuous on (ti,ti+1] and the limit limtti+x(t) (i0) exists, it follows that xPmC(0,1]. For t(t0,t1], we know from Theorem 2 that there exists c0 such that

x(t)=c0𝐄α,1(λtα)+0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s,t(t0,t1].

Then (3.2.5) holds for j=0. We suppose that (3.2.5) holds for all j=0,1,,i. We derive the expression of x on (ti+1,ti+2]. Suppose that

x(t)=Φ(t)+j=0icj𝐄α,1(λ(t-tj)α)+0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s,t(ti+1,ti+2].

By

D0+αCx(t)-λx(t)=f(t),t(ti+1,ti+2],

we get

F(t)+λx(t)=D0+αCx(t)=1𝚪(1-α)0t(t-s)-αx(s)𝑑s=1𝚪(1-α)(j=0itjtj+1(t-s)-αx(s)𝑑s+ti+1t(t-s)-αx(s)𝑑s)=1𝚪(1-α)[j=0itjtj+1(t-s)-α(v=0jcv𝐄α,1(λ(s-tv)α)+0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)du)ds+ti+1t(t-s)-α(Φ(s)+v=0icv𝐄α,1(λ(s-tv)α)+0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)du)ds]=Dti+1+αCΦ(t)+1𝚪(1-α)[j=0itjtj+1(t-s)-α(v=0jcv𝐄α,1(λ(s-tv)α))ds+ti+1t(t-s)-α(v=0icv𝐄α,1(λ(s-tv)α))𝑑s+0t(t-s)-α(0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)du)ds]=Dti+1+αCΦ(t)+1𝚪(1-α)[j=0iv=0jcvtjtj+1(t-s)-α(ι=0+λι(s-tv)ιαΓ(αι+1))ds+v=0icvti+1t(t-s)-α(ι=0+λι(s-tv)ιαΓ(αι+1))ds]+1Γ(2-α)[0t(t-s)1-α(0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)𝑑u)𝑑s]=Dti+1+αCΦ(t)+1𝚪(1-α)[j=0iv=0jcvtjtj+1(t-s)-α(ι=1+(ια)λι(s-tv)ια-1Γ(αι+1))ds+v=0icvti+1t(t-s)-α(ι=1+(αι)λι(s-tv)ια-1Γ(αι+1))ds]+1Γ(2-α)[(t-s)1-α(0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)du)|0t+(1-α)0t(t-s)-α0s(s-u)α-1𝐄α,α(λ(s-u)α)F(u)duds]=Dti+1+αCΦ(t)+1𝚪(1-α)[j=0iv=0jcvι=1+λιΓ(αι)tjtj+1(t-s)-α(s-tv)ια-1ds+v=0icvι=1+λιΓ(αι)ti+1t(t-s)-α(s-tv)ια-1ds]+1Γ(2-α)[(1-α)ι=0+λιΓ(α(j+1))0tut(t-s)-α(s-u)αj+α-1𝑑sF(u)𝑑u]=Dti+1+αCΦ(t)+1𝚪(1-α)[v=0icvj=viι=1+λι(t-tv)α(ι-1)Γ(αι)tj-tvt-tvtj+1-tvt-tv(1-w)-αwια-1dw+v=0icvι=1+λι(t-tv)α(ι-1)Γ(αι)ti+1-tvt-tv1(1-w)-αwια-1dw]+1Γ(2-α)[(1-α)ι=0+λιΓ(α(j+1))0t(t-u)αj01(1-w)-αwαj+α-1𝑑wF(u)𝑑u]=Dti+1+αCΦ(t)+1𝚪(1-α)[v=0icvι=1+λι(t-tv)α(ι-1)Γ(αι)0ti+1-tvt-tv(1-w)-αwια-1dw+v=0icvι=1+λι(t-tv)α(ι-1)Γ(αι)ti+1-tvt-tv1(1-w)-αwια-1dw]+[ι=0+λιΓ(αj+1)0t(t-u)αjF(u)𝑑u]=Dti+1+αCΦ(t)+1𝚪(1-α)[v=0icvι=1+λι(t-tv)α(ι-1)Γ(αι)01(1-w)-αwια-1𝑑w]+[ι=0+λιΓ(αj+1)0t(t-u)αjF(u)𝑑u]=Dti+1+αCΦ(t)+v=0icvι=1+λι(t-tv)α(ι-1)Γ(α(ι-1)+1)+ι=1+λιΓ(αj)0t(t-u)αj-1F(u)𝑑u=F(t)+λx(t)+Dti+1+αcΦ(t)-λΦ(t).

Thus

Dti+1+αCΦ(t)-λΦ(t)=0

for all t(ti+1,ti+2]. By Theorem 2, we know that there exists a constant ci+1 such that

Φ(t)=ci+1𝐄α,1(λ(t-ti+1)α)

for t(ti+1,ti+2]. Substituting Φ into (3.2.6), we get that (3.2.5) holds for j=i+1. Now suppose that (3.2.5) holds for all j0. By the mathematical induction method, we know that x satisfies (3.2.5) and x|(ti,ti+1] is continuous and limtti+x(t) exists. ∎

Theorem 2.

Suppose that F is continuous on (0,1) and there exist constants k>-1 and l(-α,-1-k,0] such that |F(t)|tk(1-t)l for all t(0,1). Then x is a solution of (3.2.2) if and only if there exists constants ciR (i0) such that

x(t)=v=0jcv(t-tv)α-1𝐄α,α(λ(t-tv)α)+0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s,t(tj,tj+1],j0.(3.2.7)

Proof.

For t(tj,tj+1] (j0), similarly to the beginning of the proof of Theorem 1 we know that

t1-α|0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s|0t(t-s)α-1𝐄α,α(λ(t-s)α)|F(s)|𝑑st1-α0t(t-s)α-1𝐄α,α(λ(t-s)α)sk(1-s)l𝑑s=t1-αj=0+λj𝚪((j+1)α)0t(t-s)α-1(t-s)αjsk(1-s)l𝑑st1-αj=0+λj𝚪((j+1)α)0t(t-s)α+l-1(t-s)αjsk𝑑s=t1-αj=0+λj𝚪((j+1)α)tα+αj+k+l01(1-w)α+αj+l-1wk𝑑wt1-αj=0+λjtαj𝚪((j+1)α)tα+k+l01(1-w)α+l-1wk𝑑w=t1+k+l𝐄α,α(λtα)𝐁(α+l,k+1).

So t1-α0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s is convergent and is continuous on [0,1].

If x is a solution of (3.2.7), we have

xPmC1-α(0,1].

It follows for t(ti,ti+1] that

D0+αRLx(t)=1𝚪(1-α)[0t(t-s)-αx(s)𝑑s]=1𝚪(1-α)[j=0i-1tjtj+1(t-s)-α(κ=0jcκ(s-tκ)α-1𝐄α,α(λ(s-tκ)α)+0s(s-v)α-1𝐄α,α(λ(s-v)α)f(v)𝑑v)𝑑s]+1𝚪(1-α)[tit(t-s)-α(κ=0icκ(t-tκ)α-1𝐄α,α(λ(s-tκ)α)+0s(s-v)α-1𝐄α,α(λ(s-v)α)F(v)dv)ds]=1𝚪(1-α)[j=0i-1κ=0jcκtjtj+1(t-s)-α(s-tκ)α-1m=0+λm(s-tκ)αm𝚪(α(m+1))ds]+1𝚪(1-α)[κ=0icκtit(t-s)-α(t-tκ)α-1m=0+λm(s-tκ)αm𝚪(α(m+1))ds]+1𝚪(1-α)[0t(t-s)-α0s(s-v)α-1m=0+λm(s-v)αm𝚪(α(m+1))F(v)dvds]=1𝚪(1-α)[m=0+λm𝚪(α(m+1))j=0i-1κ=0jcκtjtj+1(t-s)-α(s-tκ)α+αm-1𝑑s]+1𝚪(1-α)[κ=0icκm=0+λm𝚪(α(m+1))tit(t-s)-α(t-tκ)α+αm-1𝑑s]+1𝚪(1-α)[m=0+λm0t(t-s)-α0s(s-v)α+αm-1𝚪(α(m+1))F(v)𝑑v𝑑s]=1𝚪(1-α)[m=0+λm𝚪(α(m+1))κ=0i-1cκ(t-tκ)αmj=κi-1tj-tκt-tκtj+1-tκt-tκ(1-w)-αwα+αm-1𝑑w]+1𝚪(1-α)[m=0+κ=0icκ(t-tκ)αmλm𝚪(α(m+1))ti-tκt-tκ1(1-w)-αwα+αm-1𝑑w]+1𝚪(1-α)[m=0+λm0tvt(t-s)-α(s-v)α+αm-1𝚪(α(m+1))𝑑sF(v)𝑑v]=1𝚪(1-α)[m=0+λm𝚪(α(m+1))κ=0i-1cκ(t-tκ)αm0ti-tκt-tκ(1-w)-αwα+αm-1𝑑w]+1𝚪(1-α)[m=0+κ=0i-1cκ(t-tκ)αmλm𝚪(α(m+1))ti-tκt-tκ1(1-w)-αwα+αm-1dw+m=0+ci(t-ti)αmλm𝚪(α(m+1))01(1-w)-αwα+αm-1dw]+1𝚪(1-α)[m=0+λm0t(t-v)αm01(1-w)-αwα+αm-1𝚪(α(m+1))𝑑wF(v)𝑑v]=1𝚪(1-α)[κ=0i-1cκm=0+(t-tκ)αmλm𝚪(α(m+1))01(1-w)-αwα+αm-1𝑑w]+1𝚪(1-α)[cim=0+(t-ti)αmλm𝚪(α(m+1))01(1-w)-αwα+αm-1𝑑w]+1𝚪(1-α)[m=0+λm0t(t-v)αm01(1-w)-αwα+αm-1𝚪(α(m+1))𝑑wF(v)𝑑v]=1𝚪(1-α)[κ=0i-1cκm=0+(t-tκ)αmλm𝚪(α(m+1))Γ(1-α)Γ(α(m+1))Γ(αm)]+1𝚪(1-α)[cim=0+(t-ti)αmλm𝚪(α(m+1))Γ(1-α)Γ(α(m+1))Γ(αm)]+1𝚪(1-α)[m=0+λm𝚪(α(m+1))Γ(1-α)Γ(α(m+1))Γ(αm)0t(t-v)αmF(v)𝑑v]=κ=0i-1cκm=1(t-tκ)αm-1+(αm)λm𝚪(αm)+cim=1+(t-ti)αm-1(αm)λm𝚪(αm)+m=1+(αm)λm𝚪(αm)0t(t-v)αm-1F(v)𝑑v+F(t)=F(t)+λκ=0icκ(t-tκ)α-1Eα,α(λ(t-tκ)α)+m=1+(αm)λm𝚪(α(m+1))0t(t-v)αm-1F(v)𝑑v=F(t)+λκ=0icκ(t-tκ)α-1𝐄α,α(λ(t-tκ)α)+λ0t(t-v)α-1𝐄α,α(λ(t-v)α)F(v)𝑑v=λx(t)+F(t),t(ti,ti+1].

It follows that x is a solution of (3.2.2).

Now we prove that if x is a solution of (3.2.2), then x satisfies (3.2.7) and xPmC1-α(0,1] by the mathematical induction method. By Theorem 4 we know that there exists a constant c0 such that

x(t)=c0tα-1𝐄α,α(λtα)+0t(t-s)α-1𝐄α,α(λ(t-s)α)F(s)𝑑s,t(t0,t1].

Hence (3.2.7) holds for j=0. Assume that (3.2.7) holds for j=0,1,2,,im, we will prove that (3.2.7) holds for j=i+1. Suppose that

x(t)=Φ(t)+j=0icj(t-tj)α-1𝐄α,α(λ(t-tj)α)+