Existence of Nonoscillatory Solutions for Fractional Functional Differential Equations

Abstract

In this paper, we develop sufficient criteria for the existence of a nonoscillatory solution to the fractional neutral functional differential equation of the form:

$$\begin{aligned} D^{\alpha }_t[x(t)+c x(t-\tau )]'+\sum ^m_{i=1}P_i(t)F_i(x(t-\sigma _i))=0,\quad t\ge t_0, \end{aligned}$$

where $D_t^{\alpha }$ is Liouville fractional derivatives of order $\alpha \ge 0$ on the half-axis, $c\in \mathbb {R}$, $\tau$, $\sigma _i\in \mathbb {R}^+$, $P_i\in C([t_0, \infty ), \mathbb {R})$, $F_i\in C(\mathbb {R}, \mathbb {R}), ~ i=1,2,\ldots ,m$, $m \ge 1$ is an integer. Our results are new and improve many known results on the integer-order functional differential equations.

1 Introduction

Fractional differential equations have gained much popularity due to their extensive applications in a variety of fields such as physics, mechanics, chemistry and engineering. The recent development on fractional-order ordinary and partial differential equations can be found in the monographs by Podlubny [1], Kilbas et al. [2], Diethelm [3], and a series of recent research articles [4,5,6,7,8] and the references therein.

Oscillation theory provides a useful platform to acquire much needed information about the qualitative properties of solutions of differential equations. This branch of mathematics has been developed for integer-order functional differential equations during the past three decades. One can find important works on the topic in the monographs by Ladde et al. [9], Györi and Ladas [10], Gopalsamy [11], Erbe et al. [12], Agarwal et al. [13].

Recently, Grace et al. [14], Bolat [15], Duan et al. [16], Harikrishnan et al. [17] investigated oscillation and forced oscillation characteristics for fractional-order delay differential equations. However, to the best of our knowledge, the nonoscillatory theory for fractional differential equations is yet to be explored.

In this paper, we discuss the nonoscillatory characteristics of solutions for the following fractional neutral functional differential equation

$$\begin{aligned} D^{\alpha }_t[x(t)+c x(t-\tau )]'+\sum ^m_{i=1}P_i(t)F_i(x(t-\sigma _i))=0, \quad t\ge t_0, \end{aligned}$$

(1)

where $D_t^{\alpha }$ is Liouville fractional derivatives of order $\alpha \ge 0$ on the half-axis, $c\in \mathbb {R}$, $\tau$, $\sigma _i\in \mathbb {R}^+$, $P_i\in C([t_0, \infty ), \mathbb {R})$, $F_i\in C(\mathbb {R}, \mathbb {R}), ~ i=1,2,\ldots ,m$, $m \ge 1$ is an integer.

Let $r=\max _{1\le i\le m}\{\tau , \sigma _i,\}$. By a solution of Eq. (1), we mean a function $x\in C([t_1-r, \infty ), \mathbb {R})$ for some $t_1\ge t_0$ such that $D_t^{\alpha }[x(t)+cx(t-\tau )]'$ exists on $[t_1, \infty )$ and that Eq. (1) is satisfied for $t\ge t_1$.

A nontrivial solution x of Eq. (1) is said to be oscillatory if it has an arbitrarily large number of zeros. Otherwise, x is said to be nonoscillatory, that is, x is nonoscillatory if there exists a $T>t_1$ such that $x(t)\ne 0$ for $t\ge T$. In other words, a nonoscillatory solution must be eventually positive or eventually negative.

For the case of $\alpha =n\in \mathbb {N}$, the existence of nonoscillatory solution to Eq. (1) has been studied extensively. The monographs [12, 13] summarize some important works on nonoscillation theory of higher-order neutral differential equations of the form

$$\begin{aligned} \frac{\mathrm{d}^n}{\mathrm{d}t^n}[x(t)+c x(t-\tau )]+P(t)x(t-\sigma )=0, \quad t\ge t_0, \end{aligned}$$

(2)

where $\tau$, $\sigma \in (0, \infty )$, $P\in C([t_0, \infty ), \mathbb {R})$.

Theorem A

[13] Assume that $c\ne -1$, $P(t)\ge 0$, and

$$\begin{aligned} \int _{t_0}^\infty t^{n-1}P(t)\mathrm{d}t <\infty ,\quad \mathrm{for}~\mathrm{some}~\mathrm{add}\quad n\in \mathbb {N}. \end{aligned}$$

Then Eq. (2) has a bounded positive solution.

Theorem B

[12, 13] Assume that $c<0$, $c\ne -1$, and

$$\begin{aligned} \int _{t_0}^\infty t^{n-1}|P(t)|\mathrm{d}t <\infty ,\quad \mathrm{for}\quad n\in \mathbb {N}. \end{aligned}$$

Then Eq. (2) has a bounded positive solution.

In [18], the authors investigated the following second-order neutral functional differential equation with positive and negative coefficients

$$\begin{aligned} \frac{\mathrm{d}^2}{\mathrm{d}t^2}[x(t)+cx(t-\tau )]+P_1(t)x(t-\sigma _1)-P_2(t)x(t-\sigma _2)=0,\quad t\ge t_0. \end{aligned}$$

(3)

Obviously, Eq. (3) is a special form of Eq. (1), and the following result has been proved for this equation in [18].

Theorem C

[18] Assume that $c\not =\pm 1$, $P_1(t)\ge 0$ and $P_2(t)\ge 0$ and $aP_1(t)-P_2(t)\ge 0$, for every $t\ge T$ and $a>0$. Further, assume that

$$\begin{aligned} \int ^{\infty }_{t_0}P_1(t)\mathrm{d}t<\infty ,\quad \int ^{\infty }_{t_0}P_2(t)\mathrm{d}t<\infty . \end{aligned}$$

Then Eq. (3) has a nonoscillatory solution.

In this paper, we obtain some sufficient conditions for the existence of a nonoscillatory solution of Eq. (1) by using fixed point theorems due to Krasnoselskii and Schauder, and some new techniques. Our results are new and more general as we relax the restrictive conditions and hypotheses assumed in proving Theorems A, B and C.

2 Preliminaries

In this section, we introduce preliminary details which are used throughout this paper.

Definition 1

[2] (Liouville fractional integrals on the half-axis) The Liouville fractional integral on the half-axis is defined by

$$\begin{aligned} D_t^{-\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int _{t}^{\infty }(s-t)^{\alpha -1}f(s)\mathrm{d}s, \end{aligned}$$

where $t\in \mathbb {R}$ and $\alpha \in [0, \infty )$.

Definition 2

[2] (Liouville fractional derivatives on the half-axis) The Liouville fractional derivative on the half-axis is defined by

$$\begin{aligned} D_t^{\alpha }f(t)= & {} (-1)^n\displaystyle \frac{\mathrm{d}^n}{\mathrm{d}t^n} D_t^{-(n-\alpha )}f(t))\\= & {} \displaystyle \frac{1}{\Gamma (n-\alpha )}(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}t^n}\bigg (\int _{t}^{\infty }(s-t)^{n-\alpha -1}f(s)\mathrm{d}s\bigg ), \end{aligned}$$

where $n=[\alpha ]+1$, $\alpha \in (0, \infty )$, $[\alpha ]$ denotes the integer part of $\alpha$ and $t\in \mathbb {R}$.

In particular, if $\alpha =n\in \mathbb {N}$, then $D^{n}_tf(t)=(-1)^{n}f^{(n)}(t)$, where $f^{(n)}(t)$ is the usual derivative of f(t) of order n.

Property 1

[2] For $\alpha>0, \lambda >0$,

$$\begin{aligned} D_t^{\alpha }\left( D_t^{-\alpha }f\right) (t)=f(t);\quad D_t^{\alpha }\mathrm{e}^{-\lambda t}=\lambda ^{\alpha }\mathrm{e}^{-\lambda t}. \end{aligned}$$

Now we state fixed point theorems that we need to prove our main results.

Lemma 1

(Krasnoselskii’s fixed point theorem) Let X be a Banach space, let $\Omega$ be a bounded closed convex subset of X and let $\mathcal {A}_1$, $\mathcal {A}_2$ be maps of $\Omega$ into X such that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for every pair $x, y\in \Omega$. If $\mathcal {A}_1$ is a contraction and $\mathcal {A}_2$ is completely continuous, then the equation $\mathcal {A}_1x+\mathcal {A}_2x=x$ has a solution in $\Omega$.

Lemma 2

(Schauder’s fixed point theorem) Let $\Omega$ be a closed, convex and nonempty subset of a Banach space X. Let $\mathcal {A}:\Omega \rightarrow \Omega$ be a continuous mapping such that $\mathcal {A}\Omega$ is a relatively compact subset of X. Then $\mathcal {A}$ has at least one fixed point in $\Omega$, that is, there exists an $x\in \Omega$ such that $\mathcal {A}x=x$.

3 Main Results

We will consider the two cases: $c\ne \pm 1$ and $c=-1$. Our main results are the following theorems.

Theorem 1

Assume that $c\ne \pm 1$ and that

$$\begin{aligned} \int _{t_0}^\infty t^{\alpha } |P_i(t)|\mathrm{d}t <\infty ,\quad i=1,2,\ldots , m. \end{aligned}$$

(4)

Then (1) has a bounded nonoscillatory solution.

Proof

Case I ($-1<c\le 0$). By (4), we choose a $T>t_0$ sufficiently large so that

$$\begin{aligned} \frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_1\right) \mathrm{d}s\le \frac{1+c}{3}, \end{aligned}$$

where

$$\begin{aligned} M_1=\max _{2(1+c)/3\le x\le 4/3}\{|F_i(x)|: 1\le i\le m\}. \end{aligned}$$

Let $C([t_0, \infty ), \mathbb {R})$ be the set of all continuous functions with the norm $||x||=\sup _{t\ge t_0}|x(t)|<\infty$. Then $C([t_0, \infty ), \mathbb {R})$ is a Banach space. We define a closed, bounded and convex subset $\Omega$ of $C([t_0, \infty ), \mathbb {R})$ by

$$\begin{aligned} \Omega =\left\{ x=x(t)\in C([t_0, \infty ), \mathbb {R}): \frac{2(1+c)}{3}\le x(t)\le \frac{4}{3}, t\ge t_0\right\} . \end{aligned}$$

Define two maps $\mathcal {A}_1$ and $\mathcal {A}_2:\Omega \rightarrow C([t_0, \infty ), \mathbb {R})$ as follows:

$$\begin{aligned} (\mathcal {A}_1x)(t)= & {} \left\{ \begin{array}{ll} 1+c-cx(t-\tau ), &{}t\ge T, \\ (\mathcal {A}_1x)(T),&{} t_0\le t \le T. \end{array}\right. \\ (\mathcal {A}_2x)(t)= & {} \left\{ \begin{array}{ll} \displaystyle \frac{1}{\Gamma (\alpha +1)}\displaystyle \int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}P_i(s)F_i(x(s-\sigma _i))\right) \mathrm{d}s,&{} t\ge T, \\ (\mathcal {A}_2x)(T),&{} t_0\le t \le T. \end{array} \right. \end{aligned}$$

• (i) We shall show that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

Indeed, for every $x, y\in \Omega$ and $t\ge T$, we get

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le & {} 1+c-cx(t-\tau )\\&+\frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} 1+c-\frac{4}{3}c+\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_1\right) \mathrm{d}s\\\le & {} 1+c-\frac{4}{3}c+\frac{1+c}{3}=\frac{4}{3}. \end{aligned}$$

Furthermore, we have

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\ge & {} 1+c-c x(t-\tau )\\&-\frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\ge & {} 1+c-\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_1\right) \mathrm{d}s\\\ge & {} 1+c-\frac{1+c}{3}=\frac{2(1+c)}{3}. \end{aligned}$$

From the above two inequalities, it follows that

$$\begin{aligned} \frac{2(1+c)}{3}\le (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le \frac{4}{3},\quad \mathrm{for}\quad t\ge t_0. \end{aligned}$$

Thus, $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

• (ii) We show that $\mathcal {A}_1$ is a contraction mapping on $\Omega$.

In fact, for $x, y\in \Omega$ and $t\ge T$, we have

$$\begin{aligned} |(\mathcal {A}_1x)(t)-(\mathcal {A}_1y)(t)|\le -c|x(t-\tau )-y(t-\tau )|\le -c||x-y||, \end{aligned}$$

which implies that

$$\begin{aligned} ||\mathcal {A}_1x-\mathcal {A}_1y||\le -c||x-y||. \end{aligned}$$

Since $0<-c<1$, we conclude that $\mathcal {A}_1$ is a contraction mapping on $\Omega$.

• (iii) Here we show that $\mathcal {A}_2$ is completely continuous.

First, we will show that $\mathcal {A}_2$ is continuous. Let $x_k=x_k(t)\in \Omega$ be such that $x_k(t)\rightarrow x(t)$ as $k\rightarrow \infty$. As $\Omega$ is closed, $x=x(t)\in \Omega$. For $t\ge T$, we have

$$\begin{aligned}&\left| (\mathcal {A}_2x_k)(t)-(\mathcal {A}_2x)(t)\right| \\&\quad \le \frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x_k(s-\sigma _i))-F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x_k(s-\sigma _i))-F_i(x(s-\sigma _i))|\right) \mathrm{d}s. \end{aligned}$$

Since $|F_i( x_k(t-\sigma _i))-F_i(x(t-\sigma _i))| \rightarrow 0$ as $k\rightarrow \infty$ for $i=1,2,\ldots ,m$, by applying the Lebesgue-dominated convergence theorem, we deduce that $\lim _{k\rightarrow \infty }||(\mathcal {A}_2x_k)(t)-(\mathcal {A}_2x)(t)||=0$. This means that $\mathcal {A}_2$ is continuous.

Next, we show $\mathcal {A}_2\Omega$ is relatively compact. It suffices to show that the family of functions $\{\mathcal {A}_2x:x\in \Omega \}$ is uniformly bounded and equicontinuous on $[t_0, \infty )$. The uniform boundedness is obvious. For the equicontinuity, according to Levitan’s result, we only need to show that, for any given $\varepsilon >0$, $[T, \infty )$ can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have change of amplitude less than $\varepsilon$. By (4), for any $\varepsilon >0$, take $T^*\ge T$ large enough so that

$$\begin{aligned} \frac{1}{\Gamma (\alpha +1)}\int ^{\infty }_{T^*}s^{\alpha }\left( M_1\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s<\frac{\varepsilon }{2}. \end{aligned}$$

Then, for $x\in \Omega , ~t_2>t_1\ge T^*$, we have

$$\begin{aligned} \left| (\mathcal {A}_2x)(t_2)-(\mathcal {A}_2x)(t_1)\right|\le & {} \frac{1}{\Gamma (\alpha +1)}\int _{t_2}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\&+ \frac{1}{\Gamma (\alpha +1)}\int _{t_1}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} \frac{1}{\Gamma (\alpha +1)}\int _{t_2}^\infty s^{\alpha }\left( M_1\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\\&+ \frac{1}{\Gamma (\alpha +1)}\int _{t_1}^\infty s^{\alpha }\left( M_1\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\\< & {} \frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon . \end{aligned}$$

For $x\in \Omega$ and $T\le t_1<t_2\le T^*$, we obtain

$$\begin{aligned} \left| (\mathcal {A}_2x)(t_2)-(\mathcal {A}_2x)(t_1)\right|\le & {} \frac{1}{\Gamma (\alpha +1)}\int _{t_1}^{t_2} s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} \frac{1}{\Gamma (\alpha +1)}\int _{t_1}^{t_2} s^{\alpha }\left( M_1\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\\\le & {} \frac{1}{\Gamma (\alpha +1)}\max _{T\le s\le T^*}\Big \{ s^{\alpha }\left( M_1\sum ^m_{i=1}|P_i(s)|\right) \Big \}(t_2-t_1). \end{aligned}$$

Thus, there exists a $\delta >0$ such that

$$\begin{aligned} |(\mathcal {A}_2x)(t_2)-(\mathcal {A}_2x)(t_1)|<\varepsilon ,\quad \mathrm{if}\quad 0<t_2-t_1<\delta . \end{aligned}$$

For any $x\in \Omega ,~ t_0\le t_1<t_2\le T$, it is easy to see that

$$\begin{aligned} |(\mathcal {A}_2x)(t_2)-(\mathcal {A}_2x)(t_1)|=0<\varepsilon . \end{aligned}$$

Therefore, $\{\mathcal {A}_2x: x\in \Omega \}$ is uniformly bounded and equicontinuous on $[t_0, \infty )$, and hence $\mathcal {A}_2\Omega$ is relatively compact. In consequence, the conclusion of Lemma 1 (Krasnoselskii’s fixed point theorem) applies and there exits $x_0\in \Omega$ such that $\mathcal {A}_1x_0+\mathcal {A}_2x_0=x_0,$ that is,

$$\begin{aligned} x_0(t)=1+c-cx_0(t-\tau )+\frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}s, \end{aligned}$$

which implies that

$$\begin{aligned} x_0(t)= & {} 1+c-cx_0(t-\tau )+\frac{1}{\Gamma (\alpha )}\int _{t}^\infty \mathrm{d}s\int ^s_t(s-u)^{\alpha -1}\\&\times \left( \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}u. \end{aligned}$$

Hence

$$\begin{aligned}{}[x_0(t)+cx_0(t-\tau )]'=\frac{1}{\Gamma (\alpha )}\int _{t}^\infty (s-t)^{\alpha -1}\left( \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}s. \end{aligned}$$

By Property 1, it is easy to see that $x_0(t)$ is a nonoscillatory solution of Eq. (1).

Case II ($-\infty< c<-1$). By (4), we choose a $T>t_0$ sufficiently large such that

$$\begin{aligned} -\frac{1}{c\Gamma (\alpha +1)}\int _{T+\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_2\right) \mathrm{d}s\le -\frac{c+1}{2}, \end{aligned}$$

where

$$\begin{aligned} M_2=\max _{-(c+1)/2\le x\le -2c}\{|F_i(x)|: 1\le i \le m \}. \end{aligned}$$

Let $C([t_0, \infty ), \mathbb {R})$ be the set as in the proof of Theorem 1. We define a closed, bounded and convex subset $\Omega$ of $C([t_0, \infty ), \mathbb {R})$ as follows:

$$\begin{aligned} \Omega =\left\{ x=x(t)\in C([t_0, \infty ), \mathbb {R}): -\frac{c+1}{2}\le x(t)\le -2c, t\ge t_0\right\} . \end{aligned}$$

Define two maps $\mathcal {A}_1$ and $\mathcal {A}_2:\Omega \rightarrow C([t_0, \infty ), \mathbb {R})$ by

$$\begin{aligned} (\mathcal {A}_1x)(t)= & {} \left\{ \begin{array}{ll} -c-1-\displaystyle \frac{1}{c}x(t+\tau ), &{}t\ge T, \\ (\mathcal {A}_1x)(T),&{} t_0\le t \le T. \end{array}\right. \\ (\mathcal {A}_2x)(t)= & {} \left\{ \begin{array}{ll} \displaystyle \frac{1}{c\Gamma (\alpha +1)}\displaystyle \int _{t+\tau }^\infty (s-t-\tau )^{\alpha }\left( \sum ^m_{i=1}P_i(s) F_i(x(s-\sigma _i))\right) \mathrm{d}s, &{}t\ge T, \\ (\mathcal {A}_2x)(T),&{} t_0\le t \le T. \end{array} \right. \end{aligned}$$

In the first step, let us show that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

Indeed, for every $x, y\in \Omega$ and $t\ge T$, we get

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le & {} -c-1-\frac{1}{c} x(t+\tau )\\&- \frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{t+\tau }^\infty (s-t-\tau )^{\alpha }\\&\times \left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} -c-1+2-\frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{T+\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_2\right) \mathrm{d}s\\\le & {} -c+1-\frac{c+1}{2}\le -2c \end{aligned}$$

and

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\ge & {} -c-1-\frac{1}{c} x(t+\tau )\\&+ \frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{t+\tau }^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)||F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\ge & {} -c-1+\frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_2\right) \mathrm{d}s\\\ge & {} -c-1+\frac{c+1}{2}=-\frac{c+1}{2}, \end{aligned}$$

which imply that

$$\begin{aligned} -\frac{c+1}{2}\le (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le -2c, \quad \mathrm{for}\quad t\ge t_0. \end{aligned}$$

Thus, $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

Next we show that $\mathcal {A}_1$ is a contraction mapping on $\Omega$.

For $x, y\in \Omega$ and $t\ge T$, we have

$$\begin{aligned} |(\mathcal {A}_1x)(t)-(\mathcal {A}_1y)(t)|\le -\frac{1}{c}|x(t+\tau )-y(t+\tau )|\le -\frac{1}{c}||x-y||, \end{aligned}$$

which implies that

$$\begin{aligned} ||\mathcal {A}_1x-\mathcal {A}_1y||\le -\frac{1}{c}||x-y||. \end{aligned}$$

In view of the condition $0<-1/c<1$, it follows that $\mathcal {A}_1$ is a contraction mapping on $\Omega$.

As in the proof of Case I, we can obtain that the mapping $\mathcal {A}_2$ is completely continuous. Therefore, all the conditions of Lemma 1 are satisfied. Hence there exists $x_0\in \Omega$ such that $\mathcal {A}_1x_0+\mathcal {A}_2x_0=x_0$. Clearly, $x_0=x_0(t)$ is a bounded positive solution of Eq. (1).

Case III ($0\le c<1$). By (4), we choose a $T>t_0$ sufficiently large so that

$$\begin{aligned} \frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_3\right) \mathrm{d}s\le 1-c, \end{aligned}$$

where

$$\begin{aligned} M_3=\max _{2(1-c)\le x\le 4}\{F_i(x): 1\le i \le m \}. \end{aligned}$$

Let $C([t_0, \infty ), \mathbb {R})$ be the set defined in the proof of Theorem 1. We define a closed, bounded and convex subset $\Omega$ of $C([t_0, \infty ), \mathbb {R})$ as follows:

$$\begin{aligned} \Omega =\{x=x(t)\in C([t_0, \infty ), \mathbb {R}): 2(1-c)\le x(t)\le 4, t\ge t_0\}, \end{aligned}$$

and consider two maps $\mathcal {A}_1$ and $\mathcal {A}_2:\Omega \rightarrow C([t_0, \infty ), \mathbb {R})$ defined by

$$\begin{aligned} (\mathcal {A}_1x)(t)=\left\{ \begin{array}{ll} 3+c-cx(t-\tau ), &{}t\ge T, \\ (\mathcal {A}_1x)(T),&{} t_0\le t \le T, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} (\mathcal {A}_2x)(t)=\left\{ \begin{array}{ll} \displaystyle \frac{1}{\Gamma (\alpha +1)}\displaystyle \int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}P_i(s)F_i(x(s-\sigma _i))\right) \mathrm{d}s, &{}t\ge T, \\ (\mathcal {A}_2x)(T),&{} t_0\le t \le T. \end{array}\right. \end{aligned}$$

As before, for any $x, y\in \Omega$ and $t\ge T$, we have

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le & {} 3+c-c x(t-\tau )\\&+\frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} 3+c+\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_3\right) \mathrm{d}s\\\le & {} 3+c+1-c=4, \end{aligned}$$

and

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\ge & {} 3+c-c x(t-\tau )\\&-\frac{1}{\Gamma (\alpha +1)}\int _{t}^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\ge & {} 3+c-4c-\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_3\right) \mathrm{d}s\\\ge & {} 3+c-4c-(1-c)=2(1-c). \end{aligned}$$

In consequence, we get

$$\begin{aligned} 2(1-c)\le (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le 4,\quad \mathrm{for}\quad t\ge t_0. \end{aligned}$$

This shows that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

Proceeding as in the proof of Case I, we can establish that the mapping $\mathcal {A}_1$ is a contraction mapping on $\Omega$ and the mapping $\mathcal {A}_2$ is completely continuous. By Lemma 1, there exists $x_0\in \Omega$ such that $\mathcal {A}_1x_0+\mathcal {A}_2x_0=x_0$. Clearly, $x_0=x_0(t)$ is a bounded positive solution of (1).

Case IV ($1< c< \infty$). Again, by (4), we can choose a $T>t_0$ sufficiently large so that

$$\begin{aligned} \frac{1}{c\Gamma (\alpha +1)}\int _{T+\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_4\right) \mathrm{d}s\le c-1, \end{aligned}$$

where

$$\begin{aligned} M_4=\max _{2(c-1)\le x\le 4c}\{F_i(x): i=1,2,\ldots ,m\}. \end{aligned}$$

Let $C([t_0, \infty ), \mathbb {R})$ be the set considered in the proof of Theorem 1. Let $\Omega$ be a closed, bounded and convex subset of $C([t_0, \infty ), \mathbb {R})$ defined by

$$\begin{aligned} \Omega =\{x=x(t)\in C([t_0, \infty ), \mathbb {R}): 2(c-1)\le x(t)\le 4c, t\ge t_0\}. \end{aligned}$$

Define two maps $\mathcal {A}_1$ and $\mathcal {A}_2:\Omega \rightarrow C([t_0, \infty ), \mathbb {R})$ as follows:

$$\begin{aligned} (\mathcal {A}_1x)(t)=\left\{ \begin{array}{ll} 3c+1-\displaystyle \frac{1}{c}x(t+\tau ), &{}t\ge T, \\ (\mathcal {A}_1x)(T),&{} t_0\le t \le T, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} (\mathcal {A}_2x)(t)=\left\{ \begin{array}{ll} \displaystyle \frac{1}{c\Gamma (\alpha +1)}\int _{t+\tau }^\infty (s-t-\tau )^{\alpha }\left( \sum ^m_{i=1}P_i(s)F_i(x(s-\sigma _i))\right) \mathrm{d}s, &{}t\ge T, \\ (\mathcal {A}_2x)(T),&{} t_0\le t \le T. \end{array} \right. \end{aligned}$$

In order to show that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$ and $t\ge T$, we consider

$$\begin{aligned} (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\le & {} 3c+1-\frac{1}{c} x(t+\tau )\\&+ \frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{t+\tau }^\infty (s-t-\tau )^{\alpha }\\&\times \left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} 3c+1+\frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{T+\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}(|P_i(s)|M_4\right) \mathrm{d}s\\\le & {} 3c+1+(c-1)=4c,\quad (\mathcal {A}_1x)(t)+(\mathcal {A}_2y)(t)\\\ge & {} 3c+1-\frac{1}{c} x(t+\tau )\\&- \frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{t+\tau }^\infty (s-t)^{\alpha }\left( \sum ^m_{i=1}|P_i(s)| |F_i(y(s-\sigma _i))|\right) \mathrm{d}s\\\ge & {} 3c+1-4-\frac{1}{c}\frac{1}{\Gamma (\alpha +1)}\int _{T}^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_4\right) \mathrm{d}s\\\ge & {} 3c-3-(c-1)=2(c-1). \end{aligned}$$

Hence, we obtain

$$\begin{aligned} 2(c-1)\le \mathcal {A}_1x(t)+\mathcal {A}_2y(t)\le 4c,\quad \mathrm{for}\quad t\ge t_0, \end{aligned}$$

which shows that $\mathcal {A}_1x+\mathcal {A}_2y\in \Omega$ for any $x, y\in \Omega$.

As in the proof of Case I, it can be shown that the mapping $\mathcal {A}_1$ is a contraction mapping on $\Omega$ and the mapping $\mathcal {A}_2$ is completely continuous. In consequence, the conclusion of Lemma 1 applies, and there exists $x_0\in \Omega$ such that $\mathcal {A}_1x_0+\mathcal {A}_2x_0=x_0$. By Property 1, it is easy to see that $x_0=x_0(t)$ is a bounded positive solution of Eq. (1). The proof is complete. $\square$

Remark 1

We emphasize that Theorem 1 is a new result in the context of fractional functional differential equations. In particular, for $\alpha =n\in \mathbb {N}$, Theorem 1 improves essentially Theorems A, B and C by removing the restrictive conditions: $P(t)\ge 0$ in Theorem A, $aP_1(t)-P_2(t)\ge 0$ in Theorem C, and relaxing the hypothesis that $c<0$ in Theorem B.

Remark 2

Minor adjustments are only necessary to discuss the neutral functional differential equation of the form

$$\begin{aligned} D^{\alpha }_t[x(t)+C(t)x(t-\tau )]'+F(t, x(\sigma _{1}(t)),\ldots , x(\sigma _{m} (t)))=f(t),\quad t\ge t_0, \end{aligned}$$

where $\tau \in \mathbb {R}^+=[0, \infty )$, $\sigma _i (t)\rightarrow \infty ~(i=1,2,\ldots ,m)$ as $t\rightarrow \infty$, $m\ge 1$ is an integer, and $F: [t_0, \infty ) \times \mathbb {R}\times \cdots \times \mathbb {R}\rightarrow \mathbb {R}$ is continuous and bounded, $C, f\in C([t_0, \infty ), \mathbb {R})$. So we omit the details.

Theorem 2

Assume that $c=-1$ and that

$$\begin{aligned} \sum ^{\infty }_{j=0}\int _{t_0+j\tau }^\infty t^{\alpha } |P_i(t)|\mathrm{d}t <\infty ,\quad i=1,2,\ldots , m. \end{aligned}$$

(5)

Then Eq. (1) has a bounded positive solution.

Proof

By the condition (5), we can choose a sufficiently large $T>t_0$ so that

$$\begin{aligned} \frac{1}{\Gamma (\alpha +1)}\sum ^{\infty }_{j=1}\int _{T+j\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_5\right) \mathrm{d}s\le 1, \end{aligned}$$

where $M_5=\max _{0\le x\le 1}\{F_i(x): 1\le i\le m\}$.

We consider a closed, bounded and convex subset $\Omega$ of $C([t_0, \infty ), \mathbb {R})$ given by

$$\begin{aligned} \Omega =\{x=x(t)\in C([t_0, \infty ), \mathbb {R}): 2\le x(t)\le 4, t\ge t_0\} \end{aligned}$$

and define a mapping $\mathcal {A}:\Omega \rightarrow C([t_0, \infty ), \mathbb {R})$ as follows:

$$\begin{aligned} (\mathcal {A}x)(t)=\left\{ \begin{array}{ll} 3-\displaystyle \frac{1}{\Gamma (\alpha +1)}\displaystyle \sum ^\infty _{j=1}\displaystyle \int _{t+j\tau }^{\infty }(s-t-j\tau )^{\alpha }\\ \quad \times \left( \displaystyle \sum ^m_{i=1}P_i(s)F_i(x(s-\sigma _i))\right) \mathrm{d}s, &{}t\ge T, \\ (\mathcal {A}x)(T),&{} t_0\le t \le T. \end{array}\right. \end{aligned}$$

We first show that $\mathcal {A}\Omega \subset \Omega$. Indeed, for every $x\in \Omega$ and $t\ge T$, we get

$$\begin{aligned} (\mathcal {A}x)(t)\le & {} 3+\frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t+j\tau }^{\infty }(s-t-j\tau )^{\alpha }\left( \sum ^m_{i=1}|P_i(s)||F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\\le & {} 3+\frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{T+j\tau }^{\infty }s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_5\right) \mathrm{d}s\\\le & {} 4, \end{aligned}$$

and

$$\begin{aligned} (\mathcal {A}x)(t)\ge & {} 3-\frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t+j\tau }^{\infty }(s-t-j\tau )^{\alpha }\left( \sum ^m_{i=1}|P_i(s)||F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\\ge & {} 3-\frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{T+j\tau }^{\infty }s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|M_5\right) \mathrm{d}s\\\ge & {} 2. \end{aligned}$$

Hence $\mathcal {A}\Omega \subset \Omega$.

We now show that $\mathcal {A}$ is continuous. Let $x_k=x_k(t)\in \Omega$ be such that $x_k(t)\rightarrow x(t)$ as $k\rightarrow \infty$. Since $\Omega$ is closed, $x=x(t)\in \Omega$. For $t\ge T$, we have

$$\begin{aligned}&|(\mathcal {A}x_k)(t)-(\mathcal {A}x)(t)|\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\sum ^{\infty }_{j=1}\int _{T+j\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x_k(s-\sigma _i))-F_i(x(s-\sigma _i))|\right) \mathrm{d}s. \end{aligned}$$

Noting that $|F_i( x_k(t-\sigma _i))-F_i(x(t-\sigma _i))| \rightarrow 0$ as $k\rightarrow \infty$ for $i=1,2,\ldots ,m$, and applying the Lebesgue-dominated convergence theorem, we conclude that $\lim _{k\rightarrow \infty }||(\mathcal {A}x_k)(t)-(\mathcal {A}x)(t)||=0$. This means that $\mathcal {A}$ is continuous.

In what follows, we show that $\mathcal {A}\Omega$ is relatively compact. By (5), for any $\varepsilon >0$, take $T^*\ge T$ large enough so that

$$\begin{aligned} \frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int ^{\infty }_{T^*+j\tau }s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s<\frac{\varepsilon }{2}. \end{aligned}$$

Then, for $x\in \Omega , ~t_2>t_1\ge T^*$, we get

$$\begin{aligned}&|(\mathcal {A}x)(t_2)-(\mathcal {A}x)(t_1)|\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t_2+j\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\&\qquad + \frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t_1+j\tau }^\infty s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t_2+j\tau }^\infty s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\\&\qquad + \frac{1}{\Gamma (\alpha +1)}\sum ^\infty _{j=1}\int _{t_1+j\tau }^\infty s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\\&\quad <\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon . \end{aligned}$$

For $T\le t_1<t_2\le T^*$, we choose a sufficiently large $J\in \mathbf{N}^+$ such that $T+j\tau \ge T^*$ as $j\ge J$. For $x\in \Omega$, we obtain

$$\begin{aligned}&|(\mathcal {A}x)(t_2)-(\mathcal {A}x)(t_1)|\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\sum _{j=1}^\infty \int _{t_1+j\tau }^{t_2+j\tau } s^{\alpha }\left( \sum ^m_{i=1}|P_i(s)|| F_i(x(s-\sigma _i))|\right) \mathrm{d}s\\&\quad \le \frac{1}{\Gamma (\alpha +1)}\left[ \sum _{j=1}^{J} \int _{t_1+j\tau }^{t_2+j\tau } s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\right. \\&\qquad + \left. \sum _{j=J+1}^\infty \int _{t_1+j\tau }^{t_2+j\tau } s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\right] \\&\quad \le \frac{1}{\Gamma (\alpha +1)}\left[ \max _{T+\tau \le s\le T^*+j\tau }\Big \{ s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \Big \}J(t_2-t_1)\right. \\&\qquad + \left. \sum ^\infty _{j=1}\int _{T^*+j\tau }^\infty s^{\alpha }\left( M_5\sum ^m_{i=1}|P_i(s)|\right) \mathrm{d}s\right] . \end{aligned}$$

Then there exists a $\delta >0$ such that

$$\begin{aligned} |(\mathcal {A}x)(t_2)-(\mathcal {A}x)(t_1)|<\varepsilon ,\quad \mathrm{if}\quad 0<t_2-t_1<\delta . \end{aligned}$$

For any $x\in \Omega ,~ t_0\le t_1<t_2\le T$, it is easy to see that

$$\begin{aligned} |(\mathcal {A}x)(t_2)-(\mathcal {A}x)(t_1)|=0<\varepsilon . \end{aligned}$$

Therefore, $\{\mathcal {A}x: x\in \Omega \}$ is uniformly bounded and equicontinuous on $[t_0, \infty )$, and hence, $\mathcal {A}\Omega$ is relatively compact. By Lemma 2 (Schauder’s fixed point theorem), there exists $x_0\in \Omega$ such that $\mathcal {A}x_0=x_0$, that is,

$$\begin{aligned} x_0(t)=\left\{ \begin{array}{ll} 3-\displaystyle \frac{1}{\Gamma (\alpha +1)}\displaystyle \sum ^\infty _{j=1}\displaystyle \int _{t+j\tau }^{\infty }(s-t-j\tau )^{\alpha }\\ \quad \times \left( \displaystyle \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}s, &{}t\ge T, \\ x_0(T),&{} t_0\le t \le T. \end{array}\right. \end{aligned}$$

Then, we have

$$\begin{aligned} x_0(t)-x_0(t-\tau )=\frac{1}{\Gamma (\alpha +1)}\int ^\infty _{t}(s-t)^{\alpha }\left( \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}s, \quad t\ge T, \end{aligned}$$

which implies that

$$\begin{aligned}{}[x_0(t)-x_0(t-\tau )]'=\frac{1}{\Gamma (\alpha )}\int ^\infty _{t}(s-t)^{\alpha -1}\left( \sum ^m_{i=1}P_i(s)F_i(x_0(s-\sigma _i))\right) \mathrm{d}s, \quad t\ge T. \end{aligned}$$

By Property 1, it follows that $x_0=x_0(t)$ is a bounded positive solution of Eq. (1). This completes the proof. $\square$

Remark 3

When $F_i(x)\equiv x, P_i(t)\equiv p_i\in \mathbb {R}, i=1,2,\ldots ,m$, Eq. (1) reduces to

$$\begin{aligned} D^{\alpha }_t[x(t)+c x(t-\tau )]'+\sum ^m_{i=1}p_ix(t-\sigma _i)=0, \quad t\ge t_0. \end{aligned}$$

(6)

In this case, (4) and (5) cannot be satisfied. So, we provide an alternative sufficient condition for existence of nonoscillatory solutions of Eq. (1).

Theorem 3

Assume that $\alpha >0$, c, $\tau$, $p_i$, $\sigma _i\in \mathbb {R}$, $i=1,2,\ldots ,m$. If the characteristic equation of (6):

$$\begin{aligned} \lambda ^{\alpha +1} +c\lambda ^{\alpha +1} \mathrm{e}^{-\lambda \tau }=\sum _{i=1}^mp_i\mathrm{e}^{-\lambda \sigma _i} \end{aligned}$$

(7)

has a positive real root, then Eq. (6) has a bounded positive solution.

Proof

Let $\lambda _0 >0$ be a real root of (7). Set $y(t)=\mathrm{e}^{-\lambda _0 t}$. By using Property 1 and Eq. (7), we get

$$\begin{aligned} \begin{array}{rl} D^{\alpha }_t[y(t)+cy(t-\tau )]'&{}=-(\lambda _0^{\alpha +1}+c\lambda _0^{\alpha +1} \mathrm{e}^{-\lambda _0\tau })\mathrm{e}^{-\lambda _0 t}\\ &{}=-\bigg (\displaystyle \sum _{i=1}^mp_i\mathrm{e}^{-\lambda _0\sigma _i}\bigg )\mathrm{e}^{-\lambda _0 t}\\ &{}=-\displaystyle \sum ^m_{i=1}p_iy(t-\sigma _i). \end{array} \end{aligned}$$

Clearly, y(t) is a bounded positive solution of Eq. (6). The proof is complete. $\square$