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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1

# Weighted fractional differential equations with infinite delay in Banach spaces

Qixiang Dong
/ Can Liu
/ Zhenbin Fan
Published Online: 2016-06-08 | DOI: https://doi.org/10.1515/math-2016-0035

## Abstract

This paper is devoted to the study of fractional differential equations with Riemann-Liouville fractional derivatives and infinite delay in Banach spaces. The weighted delay is developed to deal with the case of non-zero initial value, which leads to the unboundedness of the solutions. Existence and uniqueness results are obtained based on the theory of measure of non-compactness, Schaude’s and Banach’s fixed point theorems. As auxiliary results, a fractional Gronwall type inequality is proved, and the comparison property of fractional integral is discussed.

MSC 2010: 34A08; 34K37

## 1 Introduction

In this paper, we study nonlinear functional fractional differential equation with weighted infinite delay in an abstract Banach space X, of the form $Dαy(t)= (t,y~t),t∈(o,b],$(1) $y~o=ϕ∈B,$(2)

where 0 < α ≤ 1, Dα is the Riemann-Liouville fractional derivative, (t) = t1-α y(t), f :(0, b] × 𝓑𝓑 is a given function satisfying some assumptions, and B is the phase space that will be specified later. We give the definition of solutions, and investigate the existence and continuous dependence of solutions to such equations in the space C1−α((a,b];X).

In the past several decades fractional differential equations have attracted a considerable interest in both mathematics and applications, since they have been proved to be valuable tools in modeling many physical phenomena. There has been a significant development in fractional differential equations in the past decades. See, for example, [114] and references therein. Fractional differential equations in Banach spaces are also wildly studied [1, 11–14]. Among these works, some authors study functional fractional differential equations [2, 3, 6, 8, 11]. For example, in [2], Benchohra et al. studied fractional order differential equations $Dαy(t)=f(t,yt),t∈[0,b],0<α<1,$ with infinite delay $y(t)=ϕ(t),t∈(−∞,0].$

However, it is known that the Riemann-Liouville fractional derivative of a function y is unbounded at some neighborhoods of the initial point 0, except that y(0) = 0. For this reason, when y(0) ≠ 0, the solutions to the functional fractional differential equations given in the mentioned papers may not be well-defined. An example can be found in [6]. In order to remedy this defect, in [6], the author modified the model and studied the weighted fractional differential equations with infinite delay. Existence and continuous dependence results of solutions are obtained in finite dimensional spaces.

In this paper, we continue the work of [6], to study the weighted fractional differential equations with infinite delay in Banach spaces. The difficulty is that the bounded subsets in Banach spaces are not compact in general. To get a compact subset in the space of continuous functions, we have to suppose some conditions involving the measure of non-compactness on the nonlinear term. The function space C1−α(0, b]: X) we encountered in this paper is a space of unbounded functions, which is different from C([0,b] : X). We need some further discussion on the relevant subsets. As auxiliary results, we also prove a Gronwall type inequality for fractional differential equations, and discuss the comparison property of fractional integral.

## 2 Preliminaries

In this section we collect some definitions and results needed in our further investigations.

Let (X, ∥·∥) be a Banach space. Denote by C([a, b]; X/ the space of all continuous X valued functions defined on [a, b] with the supremum norm, and ${L}_{loc}^{1}\left(\left(a,b\right);X\right)$ the space of Bochner integrable functions u : (a,b) → X with the norm $\parallel u\parallel ={\int }_{a}^{b}\parallel u\left(t\right)\parallel dt$. We also consider the space Cr((a, b]; X) consisting of all continuous functions f : (a, b] → X such that limta (ta)r f (t) exists, with the norm ∥fCr = sup{∥(ta)r f (t)∥; t ∈ (a, b]}.

([5]). Let α < 0 be a fixed number. The Riemann-Liouville fractional integral of order α > 0 of a function h : [a, b] → X is defined by $Iaαh(t)=1Γ(α)∫ta(t−s)α−1h(s)ds,t∈[a,b]$ provided the right side is pointwisely defined, where Γ(·) denotes the well-known gamma function, i.e., $\mathrm{\Gamma }\left(z\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-t}{t}^{z-1}\text{d}t$

([5]). Let α > 0 be fixed and n = [α] + 1. The Riemann-Liouville fractional derivative of order α of h : (a, b] → X at the point t is defined by $Daαh(t)=1Γ(n−α)dndtn∫ta(t−s)n−α−1h(s)ds,t∈[a,b]$ provided the right side is pointwisely defined, where Œα“ denotes the integer part of the real number α.

When 0 < α < 1, then $Daαh(t)=1Γ(1−α)ddt∫ta(t−s)α−1h(s)ds.$

For simplicity, when a = 0, we denote ${D}_{0}^{\alpha }$ and ${I}_{0}^{\alpha }$ by Dα and I α, respectively.

([5]). Let 0 < α < 1. Then the unique solution to the equation Dαh(t) = 0 is given by the formula $h(t)=Ctα−1,$ for t < 0, where CR is a constant provided $h\in C\left(\left(0,b\right]\right)\cap {L}_{loc}^{1}\left(0,b\right)$. Further, if $f\in C\left(\left(0,b\right]\right)\cap {L}_{loc}^{1}\left(a,b\right)$ such that ${D}^{\alpha }f\in C\left(\left(0,b\right]\right)\cap {L}_{loc}^{1}\left(0,b\right)$ then $IαDαf(t)=f(t)+Ctα−1$ for t > 0 and some constant CR.

In the literature devoted to equations with infinite delay, the selection of the state space 𝓑 plays an important role in the study of both qualitative and quantitative theory. A usual choice is a semi-normed space satisfying suitable axioms, which was introduced by Hale and Kato [15]. For a detailed discussion on the topic, we refer to the book by Hino et al [16].

([16]). A linear topological space of functions from (−∞, 0] into X, with seminorm ∥·∥𝓑, is called an admissible phase space if 𝓑 has the following properties:

(A1) There exist a positive constant H and functions K(·), M(·):[0, +∞) → [0, +∞), with K continuous and M locally bounded, such that for any a, bR and b > a, if x : (-∞,b] → X, xa𝓑 and x(·) is continuous on [a, b], then for every t ∈ [a, b], the following conditions hold:

1. xt𝓑;

2. x(t)∥ ≤ Hxt𝓑;, for some H > 0;

3. xtBK(ta supastx(s)∥ + M(ta)∥xa𝓑.

(A2) For the function x(·) in (A1), txt is a 𝓑 valued continuous function for t ∈ [a, b].

(B) The space 𝓑 is complete.

The theory of the measure of noncompactness was initiated by Kuratowski in 1930s and has played a very important role in nonlinear analysis in recent decades. It is often applied to the theories of differential and integral equations as well as to the operator theory and geometry of Banach spaces [1722]. One of the most important examples of measure of noncompactness is the Hausdorff’s measure of noncompactness βY, which is defined by $βY(B)=inf{r>0;Bcanbeconveredwithafinitenumberofballsofradiusequaltor}cov$ for bounded set B in a Banach space Y.

The following properties of Hausdorff’s measure of noncompactness are well known.

([17]). Let Y be a real Banach space and B; CY be bounded,the following properties are satisfied :

1. B is pre-compact if and only if βY(B) = 0;

2. βY(B) = βY (B) = βY (convB) where B̄ and convB mean the closure and convex hull of B respectively;

3. βY (B) ≤ βY (C) when BC;

4. βY (B + C) ≤ βY (B) + βY (C) where B + C = {x + y; xB, yC};

5. βY (B) ∪ Cmax {βY (B), βY (C)};

6. βY (λB) = |λ|βY (B) for any λR;

7. If the map Q : D(Q)YZ is Lipschitz continuous with constant k then βZ(QB) ≤ kβY (B) for any bounded subset BD(Q), where Z be a Banach space;

8. βY (B) = inf{dY (B, C) ; CY be precompact} = inf{dY (B, C) ; CY be f i nite valued}, where dY (B, C) means the nonsymmetric (or symmetric) Hausdorff distance between B and C in Y.

9. If $\left\{{W}_{n}{\right\}}_{n=1}^{+\mathrm{\infty }}$ is a decreasing sequence of bounded closed nonempty subsets of Y and limn→+∞ βY (Wn) = 0, then ${\cap }_{n=1}^{+\mathrm{\infty }}{W}_{n}$ is nonempty and compact in Y.

In this paper we denote by β the Hausdorff measure of noncompactness of X and by βc the Hausdorff measure of noncompactness of C([a, b]; X/). To discuss the existence we need the following lemmas in this paper.

([17]). If WC([a, b]; X) is bounded, then $β(W(t))≤βc(W)$ for all t ∈ [a, b], where W(t) = {u(t); uW} ⊆ X.Furthermore if W is equicontinuous on [a,b], then β(W(t)) is continuous on [a, b] and $βc(W)=sup{β(W(t)),t∈[a,b]}.$

([17]). If WC([a, b]; X) is bounded and equicontinuous, then β(W(s)) is continuous and $β(∫atW(s)ds)≤∫atβ(W(s))ds$(3) for all t ∈ [a, b], where ${\int }_{a}^{t}W\left(s\right)ds=\left\{{\int }_{a}^{t}x\left(s\right)ds:x\in W\right\}$

## 3 Auxiliary results

In the qualitative theory of differential and Volterra integral equations, especially in establishing of boundedness and stability, the Gronwall type inequalities play a very important role. See, for example, [2326]. In 1919, T. H. Gronwall proved a remarkable inequality which has attracted and continues to attract considerable attention in the literature.

Let u, p and q be real continuous functions defined on [a; b], q(t) ≥ 0 for t ∈ [a, b]. Suppose that the inequality $u(t)≤p(t)+∫taq(s)u(s)ds$ holds for all t ∈ [a, b]. Then we have $u(t)≤p(t)+∫taq(s)p(s)exp⁡[∫tsq(τ)dτ]ds$ for all t ∈ [a, b].

In 1967, S. C. Chu and F. T. Metcalf proved in [23] a generalized Gronwall inequality for convolution type integral equations as follows.

Let the functions u and p be continuous on [0, b]; let the function K be continuous and nonnegative on the triangle 0stb. If $u(t)≤p(t)+∫t0K(t,s)u(s)ds,0≤t≤b,$ then $u(t)≤p(t)+∫t0H(t,s)p(s)ds,0≤t≤b,$ where $H\left(t,s\right)={\sum }_{i=1}^{\mathrm{\infty }}{K}_{i}\left(t,s\right),0\le s\le t\le b$ is the resolvent kernel, and the Ki(i = 1,2,…) are the iterated kernels of K.

The fractional integral is in fact a kind of convolution of functions. However, the kernel of the fractional integral $K\left(t,s\right)=\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}$ is unbounded in the triangle 0s < tb when 0 < α < 1. Observe that Lemma 3.2 holds only for continuous, and hence bounded, kernel K(t, s) on 0stb. For this reason, in 2007, H. Ye et al. proved a generalized Grownwall inequality specifically for fractional differential equations ([26], which is the generalization of [25, Lemma 7.1.1].

Suppose α > 0, u and p are nonnegative functions locally integrable on Œ0; b“ and g is a nonnegative, nondecreasing continuous function defined on [0, b], g(t)M (constant) on [0, b]. If the inequality $u(t)≤p(t)+g(t)∫t0(t−s)α−1u(s)ds$

holds for all t[0; b], then we have $u(t)≤p(t)+∫t0[∑n=1∞(g(t)Γ(α))nΓ(nα)(t−s)nα−1p(s)]ds$

for all t[0, b].

Now we suppose that $u(t)≤p(t)+∫0t(t−s)α−1Γ(α)q(s)u(s)ds$(4)

for 0tb. Let $k\left(t,s\right)=\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}q\left(s\right)$. Define $K1(t,s)=K(t,s),Ki(t,s)=∫stK(t,τ)Ki−1(τ,s)dτ,i=2,3,... .$(5)

Then we have

Let the functions u, p and q be continuous and nonnegative on [0, b]. If the inequality (4) holds, then $u(t)≤p(t)+∫0tR(t,s)p(s)ds,0≤t≤b,$(6)

where $R\left(t,s\right)={\sum }_{i=1}^{\mathrm{\infty }}{k}_{i}\left(t,s\right),0\le s, is the resolvent kernel, and the ki(i = 1, 2,…) are the iterated kernels defined by (5).

From (4), we have $u(t)≤p(t)+∫0tk(t,s)p(s)ds+∫0tk(t,s)ds∫0sk(s,τ)u(τ)d(τ)=p(t)+∫0tk1(t,s)p(s)ds+∫0tu(s)ds∫stk(t,τ)k1(τ,s)dτ=p(t)+∫0tk1(t,s)p(s)ds+∫0tk2(t,s)u(s)ds$ for all t ∈ [0, b]. By induction, we have for all nN, $u(t)≤p(t)+∫0t∑i=1nKi(t,s)p(s)ds+∫0tKn+1(t,s)u(s)ds$(7) for all t ∈ [0, b]. Now we prove that $|Kn(t,s)|≤(t−s)nα−1Γ(nα)‖q‖∞n$(8) for all nN and 0s < tb. In fact, $|K1(t,s)|=(t−s)α−1Γ(α)|q(s)|≤(t−s)α−1Γ(α)∥q∥∞$ for 0s < tb. Suppose that (8) holds for n. Then for n + 1, $|Kn+1(t−s)|=|∫stK(t,τ)Kn(τ,s)dτ|≤∥q∥∞Γ(α)∥q∥∞nΓ(nα)∫st(t−τ)α−1(τ−s)nα−1dτ=∥q∥∞n+1Γ(α)Γ(nα)(t−s)nαΓ(α)Γ(nα)Γ((n+1)α)=(t−s)nαΓ((n+1)α)∥q∥∞n+1$ for all 0s < tb. By mathematical induction, (8) holds for all nN.

We now show that $\underset{0}{\overset{t}{\int }}R\left(t,s\right)p\left(s\right)ds={\int }_{0}^{t}{\sum }_{i=1}^{\mathrm{\infty }}{k}_{i}\left(t,s\right)p\left(s\right)ds$ uniformly for t ∈ [0, b] and $\underset{i\to \mathrm{\infty }}{lim}{k}_{i}\left(t,s\right)=0$ uniformly for 0s < tb. Write $R(t,s)=∑[1/α]i=1ki(t,s)+∑∞i=[1/α]+1ki(t,s).$

For 1 ≤ i ≤ [1/α], $|∫0tki(t,s)p(s)ds|≤∥p∥∞iΓ(iα)∫0t(t−s)iα−1ds=∥p∥∞iΓ(iα+1)tiα.$(9)

While $∑i=[1/α]+1∞|ki(t,s)|≤∑i=[1/α]+1∞∥p∥∞iΓ(iα)(t−s)iα−1≤∑i=[1/α]+1∞∥p∥∞ibiα−1Γ(iα)=1b∑i=[1/α]+1∞(∥p∥∞bα)iΓ(iα).$

Using the ratio test it is easily seen that ${\sum }_{i=\left[1/\alpha \right]+1}^{\mathrm{\infty }}\frac{{\left(\parallel p{\parallel }_{\mathrm{\infty }}{b}^{\alpha }\right)}^{i}}{\mathrm{\Gamma }\left(i\alpha \right)}$ is convergent. Hence ${\sum }_{i}^{\mathrm{\infty }}{}_{=\left[1/\alpha \right]+1}{k}_{i}\left(t,s\right)$ converges absolutely and uniformly for 0s < tb. This, combined with (9), implies that ${\int }_{0}^{t}R\left(t,s\right)p\left(s\right)ds={\int }_{0}^{t}{\sum }_{i=1}^{\mathrm{\infty }}{k}_{i}\left(t,s\right)p\left(s\right)ds$ uniformly for t ∈ [0, b]. The convergence of ${\sum }_{i=\left[1/\alpha \right]+1}^{\mathrm{\infty }}{k}_{i}\left(t,s\right)$ in turn implies that limi→∞ki(t, s) = 0 uniformly for 0 ≤ s < tb.

At last, letting n → ∞ in (7), we get the desired inequality (6). ⎕

The proof of inequality (8) and the convergence of series can be found in [5, Page 145-146]. For the convenience of the readers and the completeness of the paper we repeat the proof here.

From the proof of Lemma 3.4 we can see that, if the functions p and q are locally integrable and bounded, then the result in Lemma 3.4 still holds. Further, by a slightly modification of the proof, we can get the following result, which is the generalization of [26, Theorem 1].

Let α > 0, the functions p and q be nonnegative, locally integrable and bounded on [0, b], g be nonnegative and nondecreasing continuous on [0, b]. Suppose that u is nonnegative and continuous on [0, b) satisfying $u(t)≤p(t)+g(t)∫0t(t−s)α−1Γ(α)q(s)u(s)ds$ for all t ∈ [0, b), then $u(t)≤p(t)+∫0tR~(t,s)p(s)ds,0≤t≤b.$ where $\stackrel{~}{R}\left(t,s\right)={\sum }_{i=1}^{\mathrm{\infty }}{\stackrel{~}{k}}_{i}\left(t,s\right),0\le s and i (i = 1,2, ...) are defined by $k~1(t,s)=k~(t,s)=(t−s)α−1g(t)Γ(α)q(s),k~i(t,s)=∫stk~(t,τ)k~i−1(τ,s)dτ,i=2,3....$

Next we discuss the comparison property of the fractional integral. It is well-known that if an integrable function f is nonnegative on [0, b], then ${\int }_{{t}_{1}}^{{t}_{2}}f\left(s\right)ds\ge 0$ for any t1, t2 ∈ [0, b] with t1 < t2. Equivalently, the function $\mathrm{\Phi }\left(t\right)={\int }_{0}^{t}f\left(s\right)ds$ is nondecreasing on [0, b]. For fractional case, a natural generalization of this property might be ${I}_{0}^{\alpha }f\left({t}_{2}\right)-{I}_{0}^{\alpha }f\left({t}_{1}\right)\ge 0$ for t1, t2 ∈ [0, b] with t1 < t2. Or, equivalently, the function $F\left(t\right)={I}_{0}^{\alpha }f\left(t\right)={\int }_{0}^{t}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(s\right)ds$ is nondecreasing. However, it is not always the case.

Consider f(t) = tα−1 for 0 < α < 1 and t > 0. Then f(t) > 0 for all t > 0. $F(t)=∫ts(t−s)α−1Γ(α)f(s)ds=∫t0(t−s)α−1Γ(α)sα−1ds=Γ(α)Γ(2α)t2α−1.$

So for $0<\alpha <\frac{1}{2}$, 2α−1<0, and the function F(·) is strictly decreasing for t > 0.

Fortunately, we have the following relatively weak result.

Suppose that α > 0 and fC[0, b] is nonnegative and nondecreasing. Then $F\left(t\right)={I}_{0}^{\alpha }f\left(t\right)={\int }_{0}^{t}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(s\right)ds$ is nondecreasing on [0, b].

We first suppose that fC1[0, b] is nonnegative and nondecreasing. Then f'(t) ≤ 0 for all t ∈ [0, b]. It follows that $F′(t)=∫t0(t−s)α−1Γ(α)f′(s)ds+f(0)tα−1≥0$ for all t ∈ [0, b] and hence F(·) is nondecreasing on [0, b].

For the general case that fC[0, b], we choose a sequence of nonnegative and nondecreasing functions {fn} ∈ C1[0, b] such that limn fn = f in C[0, b]. Then each ${F}_{n}={I}_{0}^{\alpha }{f}_{n}$ is nondecreasing, and the limit F = limn Fn is therefore nondecreasing. ⎕

## 4 Existence results

In this section, we study the existence of solutions to the weighted functional differential equations (1)-(2). We begin with the definition of solutions to these equations.

A function y : (−∞, b] → X is said to be a solution to (1)-(2), if $y|\left(0,b\right]\right)\in C\left(\left(0,b\right]\right){\cap }^{\text{​}}{L}_{loc}^{1}\left(0,b\right)$, 0 = ϕ and satisfies (1).

For the existence results on the problem (1)-(2), we need to transform the fractional differential equation into an integral equation. From Lemma 2.3 we can obtain that, if 0 < α < 1 and $h\in C|\left(0,b\right]\right){\cap }^{\text{​}}{L}_{loc}^{1}\left(0,b\right)$ such that I1−αh is absolutely continuous on [0, b], then the function y solves the fractional differential equation $Dαy(t)=h(t), t∈(0,b]$ if and only if y satisfies $y(t)=1Γ(α)∫0t(t−s)α−1h(s)ds+Ctα−1, t∈(0,b]$ for some constant C. See also [5, Theorem 2.23].

For the forthcoming analysis, we need the following hypothesis.

(H1) f : (0, b] × 𝓑X is continuous.

(H2) There exist a nonnegative ηLp(0, b] with p > 1/α and a continuously non-decreasing function Ω : [0,+∞) → [0,+∞) such that $∥f(t,u)∥≤η(t)Ω(‖u‖B)$(10) for t(0, b] and every u𝓑, and $β(f(t,D))≤η(t) sup−∞<θ≤0 β(D(θ))$(11) for t(0, b] and every bounded subset D ⊂ 𝓑, where D(θ) = {u(θ) : uD}.

Assume the hypotheses (H1) and (H2) hold. If $limr→+∞ sup Ω(r)r< Γ(1+α)Kbb2∥η∥p,$(12) then there exists at least a solution to (1)-(2) on (−∞,b].

From the comment above, we know that y is a solution to (1) if and only if y satisfies $y(t)={1Γ(α)∫0t(t−s)α−1f(s,y~s)ds+ϕ(0)tα−1,ϕ(t),t∈(0,b],t∈(−∞,0].$

For given ϕ : (−∞, 0] which belongs to 𝓑, let ϕ̃ be a function defined by $ϕ~(t)={0,ϕ(t), t∈(0,b],t∈(−∞,0].$

Then we have ϕ̃0 = ϕ. For zC1−α((0, b], X), where C1−α (0, b],X) is the Banach space consisting of all continuous functions f : (0, b] → X such that limt→0 t1−α f(t) exists, endowed with the norm ∥fC1−α = sup|t1−α f(t)|; t(0, b], we extend z̃ to (−∞, b], also denoted by e , defined by $z~(t)={t1−αz(t),0,t∈(0,b],t∈(−∞,0].$

It is easily seen that if y(·) satisfies the integral equation $y(t)=1Γ(α)∫0t(t−s)α−1f(s,y~s)ds+ϕ(0)tα−1,t>0,$ we can decompose y(·) as y(t) = ϕ(t) + z(t), which implies that = ϕ̃t + t for t ∈ (0, b], and the function z(·) satisfies $z(t)=1Γ(α)∫0t(t−s)α−1f(s,ϕ~s+z~s)ds+ϕ(0)tα−1,t>0.$(13)

Set W = {z : (−∞] → X;z|(0, b]; ∈. C1−α. ((0, b]; X), z0 = 0}. For zW, define ∥zW = ∥z0𝓑 + ∥zC1−α. Then (W,∥zW) becomes a Banach space. Define an operator P : WW by $(pz)(t)=1Γ(α)∫0t(t−s)α−1f(s,ϕ~s+z~s)ds+ϕ(0)tα−1,t>0.$

We will prove by the Schauder’s fixed point theorem that P has at least a fixed point z, and hence z + ϕ̃ is a solution to (1)-(2).

First note that the continuity of P can be derived by (H2) and the Lebesgue dominated convergence theorem. We now show that P maps bounded subsets in W into bounded subsets. Let Br = {zW; ∥zWr}. Then, for any zBr and t(0, b], we have $‖t1−α(pz)(t)‖≤t1−αΓ(α)∫0t(t−s)α−1∥f(s,ϕ˜s+z˜s)∥ds+∥ϕ(0)∥≤b1−αΓ(α)∫0t(t−s)α−1η(s)Ω(∥ϕ˜s+z˜s∥β)ds+∥ϕ(0)∥.$

Since $∥z˜s+ϕ˜s∥B≤K(s)sup0≤τ≤s∥z˜(τ)∥+M(s)∥z˜0∥B+K(s)sup∥ϕ˜(τ)∥+M(s)∥ϕ0˜∥B≤Kbr+Mb∥ϕ∥B,$ where Mb = sup0≤s≤b M(s). It follows from (H2) and Holder’s inequality that $∥t1−α(pz)(t)∥≤b1−αΓ(α)∫0t(t−s)α−1η(s)dsΩ(Kbr+Mb∥ϕ∥B)+∥ϕ(0)∥≤b1−αΓ(α)Ω(Kbr+Mb∥ϕ∥B)(∫0t(t−s)(α−1)qds)1/q∥η∥p+∥ϕ(0)∥.≤Ω(Kbr+Mb∥ϕ∥B)b2∥η∥pΓ(1+α)+∥ϕ(0)∥ :=l,$ where ${‖\eta ‖}_{p}={\left({\int }_{o}^{b}|\eta \left(s\right){|}^{p}ds\right)}^{1/p}$ and 1/p + 1/q = 1; (α − 1)q > −1. Therefore, ∥PzWl for every zBr, which implies that P maps bounded subsets into bounded subsets in W.

Next we prove that PB̃ is equicontinuous for every bounded subsets BW. Let zBr and t1, t2(0, b] with t1 < t2, then we have $∥t21−α(pz)(t2)−t11−α(pz)(t1)∥≤t21−α−t21−αΓ(α)∫0t2(t2−s)α−1∥f(s,ϕ˜s+z˜s)∥ds+t11−αΓ(α)[∫0t1((t2−s)α−1−(t1−s)α−1)∥f(s,ϕ˜s+z˜s)∥ds+∫t1t2(t2−s)α−1∥f(s,ϕ˜s+z˜s)∥ds]≤t21−α−t11−αΓ(α)∫0t2(t2−s)α−1η(s)Ω(∥ϕ˜s+z˜s∥B)ds+t11−αΓ(α)[∫0t1((t2−s)α−1−(t1−s)α−1)η(s)Ω(∥ϕ˜s+z˜s∥B)ds+∫t1t2(t2−s)α−1η(s)Ω(∥ϕ˜s+z˜s∥B)ds]≤(tz1−α−t11−α)Ω(ro)Γ(α)(∫0t2(t2−s)(α−1)qds)1/q(∫0t2ηp(s)ds)1/p+t11−αΩ(ro)Γ(α)[(∫0t1((t2−s)α−1−(t1−s)α−1)qds)1/q(∫0t1ηp(s)ds)1/p+(∫t1t2(t2−s)(α−1)qds)1/q(∫0t1ηp(s)ds)1/p≤∥η∥pΩ(ro)r1Γ(α)(br2(tz1−α−t11−α)+(2(t2−t1)r2+(t2r2−t1r2))),$ where r0 = Kbr + Mb||Φ||B, r1 = ((α − 1) q + 1)1/q and r2 = [(α−1)q+1]/q > 0. It follows that $‖{t}_{2}^{1-\alpha }\left(pz\right)\left({t}_{2}\right)-{t}_{1}^{1-\alpha }\left(pz\right)\left({t}_{1}\right)‖\to 0\text{ }\text{\hspace{0.17em}}as\text{\hspace{0.17em}}{t}_{2}\to {t}_{1}\to 0$, and the convergence is independent of zBr, which implies that the set PBr̃ is equicontinuous.

Now we have to verify that there exists a closed convex bounded subset Br0, such that PBr0Br0. We derive from inequlity (12), i.e., ${\mathrm{lim}}_{r\to +\infty }\frac{\Omega \left(r\right)}{r}<\frac{\Gamma \left(1+\alpha \right)}{{k}_{b}{b}^{2}{‖\eta ‖}_{p}}$, that there exists a constant r0 > 0 such that $b2∥η∥pΓ(1+α)Ω(Kbro+Mb∥ϕ∥B)+∥ϕ(0)∥ < ro.$

Define Br0 = {zW; ∥zWr0}. Then Br0 is closed, convex and bounded in W. Then, for every zBr0 and t(0, b], similar to the proof of P maps bounded subsets in W into bounded subset, we have $∥t1−α(p1z)(t)−t1−α(p1z*)(t)∥≤t1−αΓ(α)∫0t(t−s)α−1∥f(s,ϕ˜s+z˜s)−f(s,ϕ˜s+z˜s*∥ds≤Lt1−αΓ(α)∫0t(t−s)α−1∥z˜s−z˜s*∥Bds$

It then follows that ∥PzWr0 for all zBr0, and hence PBr0Br0.

Now we prove that there exists a compact subset MBr0 such that PMM. We first construct a series of sets {Mn} ⊂ Br0 by $M0=Bro, M1=conv¯pMo,Mn+1=conv¯pMn,n=1,2,.... .$

From the above proof it is easy to see that Mn+1Mn for n = 1, 2, … and each n is equi-continuous. Further, from Lemma 2.5, 2.6 and 2.7 we can derive that $B(M˜n+1(t))=β(t1−αMn+1(t))=β(t1−αPMn(t))=β[t1−αΓ(α)∫0t(t−s)α−1f(s,ϕ˜s+M˜ns)ds+ϕ(0)]≤β[t1−αΓ(α)∫0t(t−s)α−1f(s,ϕ˜s+M˜ns)ds]≤t1−αΓ(α)∫0t(t−s)α−1f(s,ϕ˜s+M˜ns))ds]≤t1−αΓ(α)∫0t(t−s)α−1η(s)sup−∞<θ≤0β(ϕ˜s(θ)+M˜ns(θ))ds≤t1−αΓ(α)∫0t(t−s)α−1η(s)sup−∞<θ≤0[β(ϕ˜s(θ)+βM˜ns(θ))]ds≤t1−αΓ(α)∫0t(t−s)α−1η(s)sup−∞<θ≤0β(M˜ns(θ))ds=t1−αΓ(α)∫0t(t−s)α−1η(s)sup−∞<θ≤0β(s1−α(θ+s)α−1(θ+s)1−αMn(θ+s))ds=t1−αΓ(α)∫0t(t−s)α−1η(s)sup−∞<θ≤0β(s1−α(θ+s)α−1(θ+s)1−αMn(θ+s))ds$(14)

From the proof above we know that {(θ + s)1−α Mn(θ+s)} is equicontinuous at θ and θ+s)1−α Mn(θ + s)|θ=−s = {ϕ(0)}. So from Lemma 2.5 we know that the function θβ((θ+s)1−α Mn(θ+s))is continuous, and β((θ+s)1−α Mn. (θ+s))|θ=−s = 0. Therefore, if the supermum supsθ0 β(s1−α(θ + s)α−1(θ + s1−α Mn (θ + s)) is taken at θ = −s, then supsθ≤0 β(s1−α(θ + s)α−1 (θ + s)1−α Mn (θ + s)) ≡ 0. Otherwise, there exist δn > 0, such that $sup−s≤θ≤0β(s1−α(θ+s)α−1(θ+s)1−αMn(θ+s))=sup−s+δn≤θ≤0β(s1−α(θ+s)α−1((θ+s)1−αMn(θ+s))=supδn≤τ≤sβ(s1−ατα−1τ1−αMn(τ))≤supδn≤τ≤sβ(s1−αδnα−1τ1−αMn(τ))=s1−αδnα−1sup0≤τ≤sβ(M˜n(τ)).$(15)

Subsitting (15) into (14) we obtain that $β(M˜n+1(t))≤t1−αΓ(α)∫0t(t−s)α−1η(s)s1−αδnα−1sup0≤τ≤sβ(M˜n(τ))ds$(16) for n = 1, 2, …. Define the functions fn(t) = supτ ∈ [0, t] β(n(τ)) for n = 1,2,… and take supermum on both sides of (16), We get due to Lemma 3.9 that $fn+1(t)≤t1−αΓ(α)∫0t(t−s)α−1η(s)s1−αδnα−1fn(s)ds$(17) for n = 1, 2, …. The fact that Mn+1Mn implies that fn+1(t)fn(t) for all t ∈ [0, b] and n = 1, 2, …, and each fn is continuous on [0, b]. Therefore, the limit limn fn(t) = f(t) exists for t ∈ [0, b]. We claim that f(t) = 0 for all t ∈ [0, b]. In fact, take infn≤1 δn = δ. If δ > 0, then inequality (17) implies that $fn+1(t)≤t1−αΓ(α)∫0t(t−s)α−1η(s)s1−αδα−1f(s)s$(18) for all t ∈ [0,b] and n = 1, 2, …. Taking limit as n → ∞ in (18) we get $f(t)≤t1−αΓ(α)∫0t(t−s)α−1η(s)1−α δα−1f(s)ds$ for all t ∈ [0, b]. An application of Lemma 3.7 yields f(t) = 0 for all t ∈ [0, b].

If δ = 0, then there are two cases. In case when there exists an n0 such that δn0 = 0, fn0(t) = 0 for all t ∈ [0, b] according to the definition of δn, the inequality (17) implies that fn(t) = 0 for all n > n0 and t ∈ [0, b], and hence f(t) = limn fn(t) = 0 for all t ∈ [0, b]. In case when δn ≠ 0 for all n ≥ 1, limn δn = 0, the definition of fn and δn imply that limn fn(t) = 0 = f(t) for all t ∈ [0, b]. So we have f(t) = 0 for all t ∈ [0, b] in each case, as claimed. Therefore, ${\cap }_{n=1}^{\infty }{M}_{n}=M$ is nonempty and compact in W due to Lemma 2.5, and PMM by the definition of Mn.

Up to now we have verified that there exists a nonempty bounded convex and compact subset MW such that PMM. An employment of Schauder’s fixed point theorem shows that there exists at least a fixed point z of P in M. Then y = z + ϕ̃ is the solution to (1)-(2) on [0, b], which completes the proof. ⎕

Below we consider the existence result which is based on the Lipschitz condition. We need the following hypothesis.

(H3) There exists a constant L > 0 such that $∥f(t,u)−f(t,u)∥≤L∥u−v∥B$ for t(0, b] and every u, v ∈ 𝓑.

Assume that (H1) and (H3) hold. Then there exists a unique solution to (1)-(2) on (−∞, b].

As in the proof of Theorem 4.2, we define the operator P : WW by $(PZ)(t)=1Γ(α)∫0t(t−s)α−1f(s,ϕ˜s+z˜s)s+ϕ(0)tα−1,t >0.$(19)

If zW is a fixed point of P, then y = z + ϕ̃ is a solution of (1)-(2).

Let K(b) = sup{K(t); t(0, b]}, where K(·) is the function that appeared in Definition 2.4. Let N = [b(2LKbb1−α=Γ(1+α))1/α], and hi = ib/N. Then 0 < h1 < h2 < … < hN = b and $LKbb1−α(hi+1−hi)αΓ(1+α)<12$(20) for i = 1, 2, …, N:

We first focus on the interval(0, h1]. Let W1 = {z : (−∞, h1] → X ; z |0,h1] ∈ C1−α((0, h1]; X), z|0 = 0} and define ∥zW1 = ∥z0𝓑 + sup{|t1−αz(t)|; 0 < t < h1} = sup{|t1−αz(t)|; 0 < t < h1} for zW1.Then (W1, ∥zW1) is a Banach space. Define the operator P1 : W1W1 by $(p1z)(t)=1Γ(α)∫0t(t−s)α−1f(s,ϕ~s+z~s)ds+ϕ(0)tα−1.t∈(0,h1].$(21)

For z, z*W1 and t ∈ (0, h1], we have $∥t1−α(p1z)(t)−t1−α(p1z*)(t)∥≤t1−αΓ(α)∫0t(t−s)α−1∥f(s,ϕ˜s+z˜s)−f(s,ϕ˜s+z˜s*∥ds≤Lt1−αΓ(α)∫0t(t−s)α−1∥z˜s−z˜s*∥Bds$

Since $∥z˜s−z˜s*∥B ≤K(s) sup0≤τ≤s {∥z˜(τ)−z˜*(τ)∥}+M(s)∥z˜0−z˜0*∥B≤Kb sup0≤τ≤s {∥τ1−αz(τ)−τ1−αz*(τ)∥}≤Kb∥z−z*∥W1,$ we have $∥t1−α(p1z)(t)−t1−α(p1z*)(t)∥≤LKbt1−αΓ(α)∫0t(t−s)α−1ds∥z−z*∥W1≤LKbb1−αh1αΓ(1+α)∥z−z*∥W1,$ and hence $∥ p1z − p1z∗∥w1≤LKbb1−αh1αΓ(1+α)∥z−z∗∥W1.$(22)

From (20) and the Banach contraction principle we know that there exists a unique zW1 satisfying $z(t)=1α∫0t(t−s)α−1f(s,φs~+z~s)ds +φ(0)tα−1$(23) for t ∈(0, h1], which is the unique solution to the integral equation (13) on the interval (0, h1].

Next we consider the interval (h1, h2]. Restrict the function zW on the interval (h1, h2] to construct W2 and define ∥zW2 = ∥z0𝓑 + sup{|t1−α z(t)|; h1 < th2} = sup{|t1−α z(t);| h1 < t ≤ h2} for zW2. Then (W2, ∥zW2) is Banach space. For t ∈ (h1, h2], rewrite equation (13) as $z(t)=1Γ(α)∫h1t(t−s)α−1f(s,ϕ~s+z~s)ds+1Γ(α)∫0h1(t−s)α−1f(s,ϕ~s+z~s)ds+ϕ(0)tα−1$(24)

Since the function z is uniquely defined on (0, h1], the second integral can be considered as a known function. Using the same arguments as above, we can obtain that there exists a unique function zW2 satisfying $z(t)=1Γ(α)∫h1t(t−s)α−1f(s,ϕ~s+z~s)ds+1Γ(α)∫0h1(t−s)α−1f(s,ϕ~s+z~s)ds+ϕ(0)tα−1$(25) for t ∈ (h1, h2], which is the unique solution to the integral equation (13) on the interval (h1, h2]. Taking the next interval (h2, h3], repeating this process, we conclude that there exists a unique solution to the integral equation (13) on the interval (0, hN] = (0, b]. Set y = z + ϕ̃, then y is the unique solution to the fractional differential equation (1)-(2). ⎕

In the most of similar results, there is a restriction on the Lipschitz constant. Here we do not have any restriction on the Lipschitz constant L.

## 5 An example

In this section, we discuss an example to illustrate our results. For any real constant γ > 0 we set $Cγ={φ∈C((−∞,0];R):limx→−∞eγθφ(θ)exists in R},$ and set $|φ|γ=sup{eγθ|φθ|:−∞<θ≤0}$ for φCγ. Then by [16, Theorem 3.7], Cγ satisfies (A1), (A2) and (B) in Definition 2.4 with H = 1, K(t) = max{1, eγt} and M(t) = eγt, Cγ is a phase space.

Let us consider the weighted fractional differential equation with infinite delay $Dαy(t)=ce−yt+t|y~t|γ(et+e−t)(1+|y~t|γ),t∈(0,b], α∈(0,1),$(26) $y(t)=φ(t),t∈(−∞,0],$(27) where φCγ and c > 0 can be any constant. Set $f(t,x)=ce−yt+tx(et+e−t)(1+x), (t,x)∈(o,b]×R+.$

Then for any x, yR+, one has $|f(t,x)−f(t,y)|=ce−yt+t|x−y|(et+e−t)(1+x)(1−y)≤cet|x−y|(et + e−t)≤c|x−y|.$ i.e., f is Lipschitz with respect to the second variable with Lipschitz constant c. Then by Theorem 4.3 the problem (26)-(27) has a unique solution on. (−∞, b].

In [2], the authors discussed an example similar to the problem (26)- (27) without weight. They have to suppose that φ(0) = 0 since otherwise the solution y may be unbounded at the right neighbourhood of 0. Here we do not need this restriction. Furthermore, we don’t have any restriction on the constant c.

## Acknowledgement

This research was supported by the National Natural Science Foundation of China (11271316 and 11571300) and TAPP of Jiangsu Higher Education Institutions(No. PPZY2015B109). Zhenbin Fan’s work was also supported by Qing Lan Project of Jiangsu Province of China and High-Level Personal Support Project of Yangzhou University.

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Qixiang Dong E-mail:

Accepted: 2016-05-23

Published Online: 2016-06-08

Published in Print: 2016-01-01

Citation Information: Open Mathematics, Volume 14, Issue 1, Pages 370–383, ISSN (Online) 2391-5455,

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