$(\omega ,c)$-periodic solutions for a class of fractional integrodifferential equations

Abstract

In this paper we investigate the following fractional order in time integrodifferential problem

$$ \mathbb{D}_{t}^{\alpha}u(t)+Au(t)=f \bigl(t,u(t) \bigr)+ \int _{-\infty}^{t} k(t-s)g \bigl(s,u(s) \bigr)\,ds, \quad t \in \mathbb{R}. $$

Here, $\mathbb{D}_{t}^{\alpha}$ is the Caputo derivative. We obtain results on the existence and uniqueness of $(\omega ,c)$-periodic mild solutions assuming that −A generates an analytic semigroup on a Banach space X and f, g, and k satisfy suitable conditions. Finally, an interesting example that fits our framework is given.

1 Introduction

The aim of this paper is to investigate the existence of $(\omega ,c)$-periodic mild solutions for a class of fractional integrodifferential equations in Banach spaces. More precisely, let X be a Banach space. Our objective is to study the following problem

$$ \mathbb{D}_{t}^{\alpha}u(t)+Au(t)=f \bigl(t,u(t) \bigr)+(Ku) (t),\quad t\in \mathbb{R}. $$

(1.1)

In (1.1), $0<\alpha \leq 1$, $\mathbb{D}_{t}^{\alpha}$ denotes the Caputo fractional derivative in the t variable that is defined by

$$\begin{aligned} \mathbb{D}_{t}^{\alpha}u(t):=\frac{1}{\Gamma (1-\alpha )} \int _{0}^{t}(t- \tau )^{-\alpha}u'( \tau )\,d\tau , \end{aligned}$$

where −A generates an analytic semigroup $S(t)$ in X, and f, g are continuous functions from $\mathbb{R}\times X$ to X, and

$$ (Ku) (t):= \int _{-\infty}^{t} k(t-s)g \bigl(s,u(s) \bigr)\,ds, $$

where k is a continuous function from $\mathbb{R}\mathbbm{^{+}}$ to $\mathbb{R}$.

In many areas of science and technology, the theory of fractional differential equations and their applications is of significant importance because certain situations do not fit into classical models, see [18, 25, 26] and the references therein.

Recently, Alvarez et al. presented the concept of vector-valued $(\omega ,c)$-periodic solutions and its properties in [6]. Moreover, they proved the existence and uniqueness of $(\omega ,c)$-periodic mild solutions to the problem (1.1) with $K=0$. Then, several authors have studied related problems, see, for example, [1, 4, 5, 7, 10–15, 17, 22, 23, 27]. Also, there exist various generalizations of this kind of functions and applications to real-life problems [2, 3, 20, 21].

The problem of the existence and uniqueness of a pseudoalmost-periodic PC-mild solution for

$$\begin{aligned} \mathbb{D}_{t}^{\alpha}u(t)+Au(t)=f \bigl(t,u(t) \bigr)+ \int _{-\infty}^{t}k(t-s)g \bigl(s,u(s) \bigr)\,ds+ \sum _{j=-\infty}^{\infty}G_{j} \bigl(u(t) \bigr) \delta (t-t_{j}),\quad t\in \mathbb{R}, \end{aligned}$$

where $G_{j}$ are continuous impulsive operators, $\delta (\cdot )$ is the Dirac delta function, and $\tau _{j}$ are a sequence in $\mathbb{Z}$ was investigated by Xia in [29] for $0<\alpha <1$, and by Gu and Li in [19] for $1<\alpha <2$. The existence of almost-periodic mild solutions for the case without impulsive effects was studied in [8].

It is worth mentioning that not much seems to be known about $(\omega ,c)$-periodic mild solutions for the integrodifferential equation (1.1). This is precisely our aim in this article.

We succeed in solving this open problem using Banach fixed-point arguments and the fractional powers of operators to derive some sufficient conditions guaranteeing the existence and uniqueness of $(\omega ,c)$-periodic mild solutions to (1.1).

The paper is structured as follows. In Sect. 2, we recall the definition of $(\omega ,c)$-periodic functions, the fractional power of an operator, and the definition of Mittag–Leffler functions and their properties that will be used throughout the manuscript. In Sect. 3, we investigate the main problem where we obtain a novel regularity result related to $(\omega ,c)$-periodic mild solutions of (1.1). Finally, an interesting example is given in Sect. 4.

2 Preliminaries

Throughout this paper, $c\in \mathbb{C}\setminus \{0\}$, $\omega >0$, X will denote a Banach space with norm $\|\cdot \|_{X}$ and we will denote the set of continuous functions on $\mathbb{R}$ by

$$ C(\mathbb{R},X):=\{f:\mathbb{R}\to X: f\text{ is continuous}\}, $$

and the set of continuous functions on $\mathbb{R}\times X$ by

$$ C(\mathbb{R}\times X,X):=\{f:\mathbb{R}\times X\to X: f\text{ is continuous}\}. $$

We recall that a function $f\in C(\mathbb{R},X)$ is said to be $(\omega ,c)$-periodic if $f(t+\omega )=cf(t)$ for all $t\in \mathbb{R}$, see [6]. The collection of those functions with the same c-period ω will be denoted by $P_{\omega c}(\mathbb{R},X)$. Also, in the same article, it was proved that $P_{\omega c}(\mathbb{R},X)$ is a Banach space with the norm

$$ \Vert f \Vert _{\omega c}:=\sup_{t\in [0,\omega ]} \bigl\Vert \vert c \vert ^{\land}(-t)f(t) \bigr\Vert . $$

Definition 2.1

([28, Sect. 2.6])

Assume that −A generates an analytic semigroup $\{S(t)\}_{t\geq 0}$ in a Banach space X and $0\in \rho (A)$. For any $\beta > 0$, we define the fractional power $A^{-\beta}$ of the operator A by

$$ A^{-\beta}:=\frac{1}{\Gamma (\beta )} \int _{0}^{\infty }t^{\beta -1} S(t)\,dt. $$

We further define $A^{-0}:=I$.

Lemma 2.2

([28, Lemma 6.3])

Let the operator −A be an infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$ in the Banach space X and $0\in \rho (A)$. There exists a constant $C_{\beta}$ such that

$$ \bigl\Vert A^{-\beta}x \bigr\Vert _{X}\leq C_{\beta} \Vert x \Vert _{X}, \quad \textit{for all } x\in X, $$

where $0\leq \beta \leq 1$.

Theorem 2.3

([28, Theorem 6.13])

Let −A be an infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$. If $0\in \rho (A)$, then

• 1. $S(t): X \to D(A^{\beta})$ for all $t > 0$ and $\beta \geq 0$;

• 2. For all $x \in D(A^{\beta})$, it follows that $S(t)A^{\beta }x= A^{\beta }S(t)x$;

• 3. For all $t > 0$, the operator $A^{\beta }S(t)$ is bounded and

  $$ \bigl\Vert A^{\beta }S(t) \bigr\Vert _{\mathcal{L}(X)}\leq M_{\beta }t^{-\beta}e^{- \lambda t}, \quad M_{\beta}>0, \lambda >0, $$

  where $M_{\beta}$ is a positive constant and $\lambda >0$ satisfies that $-A+\lambda I$ remains the infinitesimal generator of the analytic semigroup $S(t)$.

• 4. For $0 <\beta \leq 1$ and $x \in D(A^{\beta})$, there exists $C_{\beta}>0$ such that

  $$ \bigl\Vert S(t)x-x \bigr\Vert _{X}\leq C_{\beta }t^{\beta} \bigl\Vert A^{\beta }x \bigr\Vert _{X}. $$

Theorem 2.4

([28])

The space $X_{\beta}:=D(A^{\beta})\subset X$ with norm $\|x\|_{\beta}:=\|A^{\beta }x\|_{X}$ is a Banach space.

We recall that the Mittag–Leffler-type function (or the two-parameter Mittag–Leffler function) is given by

$$ E_{\alpha , \beta}(t)=\sum_{n=0}^{\infty} \frac{t^{n}}{\Gamma (\alpha n+\beta )},\quad (\alpha >0, \beta \in \mathbb{C}). $$

When $\beta =1$, we write simply $E_{\alpha}(t)$ instead of $E_{\alpha , 1}(t)$. For more details about the Mittag–Leffler function, the reader may want to consult [18].

Proposition 2.5

([25])

Let $0<\alpha <1$. If $\theta \geq 0$, the following properties are satisfied:

• (a)
  $$ M_{\alpha}(\theta )\geq 0. $$
• (b)
  $$ { \int _{0}^{\infty}} \theta ^{n} M_{\alpha}(\theta )\,d\theta =\frac{\Gamma (n+1)}{\Gamma (\alpha n+1)},\quad n\geq -1. $$
• (c)
  $$ \int _{0}^{\infty }M_{\alpha}(\theta )e^{-t\theta}\,d\theta =E_{\alpha}(-t). $$

Lemma 2.6

([25])

Let $0<\alpha <1$. If −A is an infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$ on X, $0\in \rho (A)$ and $x\in X$, then

$$ E_{\alpha} \bigl(-t^{\alpha }A \bigr)x= \int _{0}^{\infty }M_{\alpha}(\theta )S \bigl( \theta t^{\alpha} \bigr)x\,d\theta ,\quad t\geq 0 $$

and

$$ E_{\alpha ,\alpha} \bigl(-t^{\alpha }A \bigr)x= \int _{0}^{\infty}\alpha \theta M_{ \alpha}(\theta )S \bigl(\theta t^{\alpha} \bigr)x\,d\theta ,\quad t\geq 0. $$

Theorem 2.7

([9])

Let $\alpha ,\beta \in (0,1)$. If −A is the infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$ and $0\in \rho (A)$, there exists a constant $M_{E}$ such that

$$ \bigl\Vert E_{\alpha} \bigl(-t^{\alpha }A \bigr)x \bigr\Vert _{\beta}\leq M_{E} t^{- \alpha \beta} \Vert x \Vert _{X} \quad \textit{and}\quad \bigl\Vert E_{\alpha , \alpha} \bigl(-t^{\alpha }A \bigr)x \bigr\Vert _{\beta}\leq M_{E} t^{-\alpha \beta} \Vert x \Vert _{X} $$

for all $t> 0$.

Lemma 2.8

([25])

The operators $E_{\alpha ,\alpha}(-t^{\alpha }A)$ and $E_{\alpha}(-t^{\alpha }A)$ are strongly continuous, which means that for all $x\in X$ and $s,t>0$, we have that

$$ \bigl\Vert E_{\alpha ,\alpha} \bigl(-t^{\alpha }A \bigr)x-E_{\alpha ,\alpha} \bigl(-s^{\alpha }A \bigr)x \bigr\Vert _{X}\to 0 \quad \textit{and}\quad \bigl\Vert E_{\alpha} \bigl(-t^{\alpha }A \bigr)x-E_{ \alpha} \bigl(-s^{\alpha }A \bigr)x \bigr\Vert _{X}\to 0 $$

when $s\to t$.

Proposition 2.9

([26])

Let $0<\alpha <1$, $t>0$. There are two asymptotic representations set up for $E_{\alpha}(-t^{\alpha})$:

$$ E_{\alpha} \bigl(-t^{\alpha} \bigr)\sim \textstyle\begin{cases} E_{\alpha}^{0} (-t^{\alpha} ):=\exp (- \frac{t^{\alpha}}{\Gamma (1+\alpha )} ),& t\to 0; \\ E_{\alpha}^{\infty} (-t^{\alpha} ):= \frac{t^{-\alpha}}{\Gamma (1-\alpha )}= \frac{\sin (\alpha \pi )}{\pi} \frac{\Gamma (\alpha )}{t^{\alpha}},& t\to \infty . \end{cases} $$

3 $(\omega ,c)$-periodic mild solutions

In this section we prove the main result of this article. Under suitable conditions, we show the existence and uniqueness of $(\omega ,c)$-periodic mild solutions for (1.1).

Let us consider the following Cauchy problem

$$ \textstyle\begin{cases} \mathbb{D}_{t}^{\alpha}u(t)+Au(t)=f(t,u(t))+(Ku)(t), & t>t_{0}, \\ u(t_{0})=u_{0}, & t_{0}\in \mathbb{R}, u_{0}\in X, \end{cases} $$

(3.1)

where the $\mathbb{D}_{t}^{\alpha}$ denotes the fractional Caputo derivative, $0<\alpha <1$, $-A:D(-A)\subset X\to X$ generates an analytic semigroup $S(t)$ in a Banach space X, and f, g are continuous functions from $\mathbb{R}\times X$ to X and $(Ku)(t):=\int _{-\infty}^{t} k(t-s)g(s,u(s))\,ds$. Here, k is a continuous function from $\mathbb{R}\mathbbm{^{+}}$ to $\mathbb{R}$.

We assume the following:

• (H1) −A is an infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$ such that $0\in \rho (A)$ and

  $$ \bigl\Vert S (t) \bigr\Vert _{X} \leq Ce^{-\sigma t} \quad \text{for } t \geq 0, $$

  where σ and C are positive constants.

• (H2) $|k(t)|\leq C_{k} e^{-\eta t}$ for some positive constants $C_{k}$, η.

• (H3) $f\in C(\mathbb{R}\times X_{\beta},X_{\beta})$ and there exists $(\omega , c)\in \mathbb{R}^{+}\times (\mathbb{C}\setminus \{0\})$ such that $f(t+\omega ,cx)=cf(t,x)$ for all $t\in \mathbb{R}$ and all $x\in X_{\beta}$. Also, there exists a positive constant $L_{f}$ such that

  $$ \bigl\Vert f(t,u)-f(t,v) \bigr\Vert _{X} \leq L_{f} \Vert u-v \Vert _{\beta}, \quad t\in \mathbb{R}, u,v\in X_{\beta}. $$

• (H4) $g\in C(\mathbb{R}\times X_{\beta},X_{\beta})$ and $g(t+\omega ,cx)=cg(t,x)$ (where ω and c are the same as given in (H3)) for all $t\in \mathbb{R}$ and all $x\in X_{\beta}$. Also, there exists a positive constant $L_{g}$ such that

  $$ \bigl\Vert g(t,u)-g(t,v) \bigr\Vert _{X} \leq L_{g} \Vert u-v \Vert _{\beta},\quad t\in \mathbb{R}, u,v\in X_{\beta}. $$

The next definition is similar to [16, Definition 3.1] and [29, Definition 3.1].

Definition 3.1

A mild solution of (3.1) is a continuous function u from $\mathbb{R}$ to X that satisfies the following integral equation:

$$ u(t)=E_{\alpha} \bigl(-(t-t_{0})^{\alpha }A \bigr)u_{0}+ \int _{t_{0}}^{t} (t-s)^{ \alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr)\,ds. $$

(3.2)

Proposition 3.2

Suppose that (H1) holds. If u is a mild solution of (3.1), then

$$ \lim_{t_{0}\to -\infty}u(t)= \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{ \alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds. $$

(3.3)

Proof

According to the definition of an improper integral, we have

$$ \begin{aligned} &\lim_{t_{0}\to -\infty} \biggl( \int _{t_{0}}^{t} (t-s)^{\alpha -1} E_{ \alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr)\,ds \biggr) \\ &\quad = \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds. \end{aligned} $$

(3.4)

On the other hand, we will prove that ${\lim_{t_{0}\to -\infty}E_{\alpha}(-(t-t_{0})^{\alpha }A)u_{0}=0}$. In fact, by Proposition 2.5 and (H1), we obtain

$$ \begin{aligned} \bigl\Vert E_{\alpha} \bigl(-(t-t_{0})^{\alpha }A \bigr)u_{0} \bigr\Vert _{X}&= \biggl\Vert \int _{0}^{\infty }M_{\alpha}(\theta )S \bigl((t-t_{0})^{\alpha}\theta \bigr)u_{0}\,d\theta \biggr\Vert _{X} \\ &{\leq} \int _{0}^{\infty }M_{\alpha}(\theta ) Ce^{-\sigma (t-t_{0})^{ \alpha}\theta } \Vert u_{0} \Vert _{X}\,d\theta \\ &{\leq} C \Vert u_{0} \Vert _{X} E_{\alpha } \bigl(- \bigl(\sigma ^{1/\alpha}(t-t_{0}) \bigr)^{ \alpha} \bigr). \end{aligned} $$

Now, by Proposition 2.9, we obtain

$$ \begin{aligned} \bigl\Vert E_{\alpha} \bigl(-(t-t_{0})^{\alpha }A \bigr)u_{0} \bigr\Vert _{X}&{\leq} C \Vert u_{0} \Vert _{X} \biggl(\frac{\sin (\alpha \pi )}{\pi}\cdot \frac{\Gamma (\alpha )}{\sigma (t-t_{0})^{\alpha}} \biggr) \xrightarrow[t_{0}\to -\infty ]{} 0, \end{aligned} $$

which shows that ${\lim_{t_{0}\to -\infty}E_{\alpha}(-(t-t_{0})^{\alpha }A)u_{0}=0}$. Using this fact, together with (3.2) and (3.4), we obtain the desired result. □

The previous proposition motivates the following definition.

Definition 3.3

A mild solution of (1.1) is a continuous function u from $\mathbb{R}$ to X that satisfies the following integral equation:

$$ u(t)= \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds, $$

(3.5)

provided that (H1) holds.

The next results are crucial for the proof of our main result.

Lemma 3.4

If (H3) and (H4) are satisfied and $u \in P_{\omega c}(\mathbb{R},X_{\beta})$, then $f_{u}=f(\cdot ,u(\cdot ))$, $g_{u}=g(\cdot ,u(\cdot ))$ lies in $P_{\omega c}(\mathbb{R},X_{\beta})$.

Proof

Let $t\in \mathbb{R}$. Then,

$$ f_{u}(t+\omega ) =f \bigl(t+\omega ,u(t+\omega ) \bigr)=f \bigl(t+ \omega ,cu(t) \bigr) =cf \bigl(t,u(t) \bigr)=cf_{u}(t). $$

By [6, Theorem 2.11] we have that $f_{u}\in P_{\omega c}(\mathbb{R},X_{\beta})$. Analogously, we can prove the claim for $g_{u}$. □

Lemma 3.5

Suppose that (H2)–(H4) are satisfied. If $u \in P_{\omega c}(\mathbb{R},X_{\beta})$, then

$$ h(\cdot ):=f \bigl(\cdot ,u(\cdot ) \bigr)+(Ku) (\cdot ) \in P_{\omega c}( \mathbb{R},X_{\beta}). $$

(3.6)

Proof

First, we will show that $h\in C(\mathbb{R},X_{\beta})$. In order to prove that h is continuous for each $t\in \mathbb{R}$, we claim that ${\lim_{\rho \to 0^{+}}}\|h(t+\rho )-h(t)\|_{\beta}=0$. Indeed, let $\rho >0$. Then,

$$ \begin{aligned} \bigl\Vert h(t+\rho )-h(t) \bigr\Vert _{\beta}={}& \bigl\Vert f \bigl(t+\rho ,u(t+\rho ) \bigr)+(Ku) (t+ \rho )-f \bigl(t,u(t) \bigr)-(Ku) (t) \bigr\Vert _{\beta} \\ \leq{}& \bigl\Vert f \bigl(t+\rho ,u(t+\rho ) \bigr)-f \bigl(t,u(t) \bigr) \bigr\Vert _{\beta} \\ &{} + \underbrace{ \int _{t}^{t+\rho} \bigl\Vert k(t+\rho -s)g \bigl(s,u(s) \bigr) \bigr\Vert _{\beta }\,ds}_{I} \\ & {}+ \underbrace{ \int _{-\infty}^{t} \bigl\Vert \bigl( k(t+\rho -s)-k(t-s) \bigr)g \bigl(s,u(s) \bigr) ) \bigr\Vert _{\beta } \,ds}_{II}. \end{aligned} $$

Note that by (H3), we have $\Vert f(t+\rho ,u(t+\rho ))-f(t,u(t)) \Vert _{\beta} \xrightarrow[\rho \to 0^{+}]{}0$. Now, we estimate I and II separately. By (H2), (H4), and Lemma 3.4, we have

$$ \begin{aligned} I&= \int _{t}^{t+\rho} \bigl\Vert k(t+\rho -s)g \bigl(s,u(s) \bigr) \bigr\Vert _{\beta }\,ds \\ &\leq C_{k} \Vert g_{u} \Vert _{\omega c}e^{-\eta (t+\rho )} \int _{t}^{t+\rho} e^{s (\frac{\ln \vert c \vert +\eta \omega}{\omega} )}\,ds \\ &\leq C_{k} \Vert g_{u} \Vert _{\omega c} \biggl( \frac{\omega}{\ln \vert c \vert +\eta \omega} \biggr) \bigl( e^{(t+\rho ) \frac{\ln \vert c \vert }{\omega}}- e^{t\frac{\ln \vert c \vert }{\omega}-\rho \eta} \bigr) \xrightarrow[\rho \to 0^{+}]{} 0. \end{aligned} $$

On the other hand, by (H4) and Lemma 3.4, we obtain

$$ \begin{aligned} II&= \int _{-\infty}^{t} \bigl\Vert \bigl( k(t+\rho -s)-k(t-s) \bigr)g \bigl(s,u(s) \bigr) ) \bigr\Vert _{\beta }\,ds \\ &{\leq} \Vert g_{u} \Vert _{\omega c} \int _{-\infty}^{t} \bigl\vert k(t+ \rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k(t-s)e^{s\frac{\ln \vert c \vert }{\omega}} \bigr\vert \,ds. \end{aligned} $$

Since $k\in C(\mathbb{R}^{+},\mathbb{R})$ and $s< t+\rho$ for $\rho >0$, we have that

$$ s\mapsto k(t+\rho -s)e^{s\frac{\ln|c|}{\omega}}:(-\infty ,t+\rho )\to \mathbb{R} $$

(3.7)

is continuous. In particular,

$$ \bigl\vert k(t+\rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k({t}-s)e^{s \frac{\ln \vert c \vert }{\omega}} \bigr\vert \xrightarrow[\rho \to 0^{+}]{}0. $$

Moreover, by (H2)

$$ \begin{aligned} \bigl\vert k(t+\rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k(t-s)e^{s \frac{\ln \vert c \vert }{\omega}} \bigr\vert &\leq C_{k} \bigl( e^{-\eta (t+\rho )}+ e^{-\eta t} \bigr)e^{s (\frac{\ln \vert c \vert }{\omega}+\eta )}. \end{aligned} $$

Due to the facts that $\rho >0$ and $\eta >0$, we have

$$ e^{-\eta (t+\rho )}< e^{-\eta t}. $$

The above implies that

$$ \begin{aligned} C_{k} \bigl( e^{-\eta (t+\rho )}+ e^{-\eta t} \bigr)e^{s ( \frac{\ln|c|}{\omega}+\eta )}&< 2C_{k} e^{-\eta t}e^{s ( \frac{\ln|c|}{\omega}+\eta )}, \end{aligned} $$

and therefore,

$$ \bigl\vert k(t+\rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k(t-s)e^{s \frac{\ln \vert c \vert }{\omega}} \bigr\vert \leq 2C_{k} \bigl( e^{-\eta t} \bigr)e^{s (\frac{\ln \vert c \vert }{\omega}+\eta )}. $$

Also, the function $s\mapsto 2 C_{k} e^{-\eta t}e^{s ( \frac{\ln|c|+\eta \omega}{\omega} )}$ is integrable in $(-\infty ,t)$, since

$$ \begin{aligned} \int _{-\infty}^{t} 2 C_{k} e^{-\eta t}e^{s ( \frac{\ln \vert c \vert +\eta \omega}{\omega} )}\,ds &=2C_{k} \biggl( \frac{\omega}{\ln \vert c \vert +\eta \omega} \biggr) e^{\frac{t\ln \vert c \vert }{\omega}}< \infty . \end{aligned} $$

Hence, the criterion of comparison of improper integrals guarantees that

$$ s\mapsto \bigl\vert k(t+\rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k(t-s)e^{s \frac{\ln \vert c \vert }{\omega}} \bigr\vert $$

is integrable in $(-\infty ,t)$. By virtue of the Dominated Convergence Theorem, it follows that

$$ II \leq \Vert g_{u} \Vert _{\omega c} \int _{-\infty}^{t} \bigl\vert k(t+ \rho -s)e^{s\frac{\ln \vert c \vert }{\omega}}-k(t-s)e^{s\frac{\ln \vert c \vert }{\omega}} \bigr\vert \,ds \xrightarrow[ \rho \to 0^{+}]{}0, $$

obtaining the claim.

Analogously, we can show that ${\lim_{\rho \to 0^{-}}}\|h(t+\rho )-h(t)\|_{\beta}=0$.

Now, we will prove that $h(t+\omega )=ch(t)$ for all $t\in \mathbb{R}$. In fact, since $u\in P_{\omega c}(\mathbb{R}, X)$, by the definition of $(\omega ,c)$-periodicity, (H3), and (H4), we obtain

$$ \begin{aligned} h(t+\omega )&=f \bigl(t+\omega ,u(t+\omega ) \bigr)+(Ku) (t+\omega ) \\ &=f \bigl(t+\omega ,cu(t) \bigr)+ \int _{-\infty}^{t}k(t-r)g \bigl(r+\omega ,cu(r) \bigr) \,dr \\ &=cf \bigl(t,u(t) \bigr)+ \int _{-\infty}^{t}k(t-r)cg \bigl(r,u(r) \bigr)\,dr=ch(t). \end{aligned} $$

Consequently, $h\in P_{\omega c}(\mathbb{R},X_{\beta})$. □

Lemma 3.6

Suppose that (H1)–(H4) are satisfied. If $u \in P_{\omega c}(\mathbb{R},X_{\beta})$, then

$$ (\Theta u) (t)= \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds $$

(3.8)

lies in $P_{\omega c}(\mathbb{R},X_{\beta})$.

Proof

Define $h(s):=f(s,u(s))+(Ku)(s)$ for all $s\in \mathbb{R}$. According to Lemma 3.5, we have $h\in P_{\omega c}(\mathbb{R},X_{\beta})$.

First, we will show that $(\Theta u)\in C(\mathbb{R},X_{\beta})$. For this, we claim that ${\lim_{\xi \to 0^{+}}}\|(\Theta u)(t+\xi )-(\Theta u)(t) \|_{\beta}=0$. Indeed, let $\xi >0$. Then,

$$ \begin{aligned} & \bigl\Vert (\Theta u) (t+\xi )-(\Theta u) (t) \bigr\Vert _{\beta} \\ &\quad =\bigg\| \int _{-\infty}^{t+\xi}(t+\xi -s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s)\,ds \\ &\qquad {} - \int _{-\infty}^{t} ({t}-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr)h(s)\,ds\bigg\| _{\beta} \\ &\quad \leq \underbrace{ \int _{-\infty}^{t} \bigl\Vert (t+\xi -s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s)- ({t}-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr)h(s) \bigr\Vert _{\beta }\,ds}_{I} \\ &\qquad {} + \underbrace{ \int _{{t}}^{t+\xi} (t+\xi -s)^{\alpha -1} \bigl\Vert E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s) \bigr\Vert _{\beta }\,ds}_{II}. \end{aligned} $$

We will estimate I and II. Indeed, for $s\in (-\infty ,t)$, by Theorem 2.7 and Lemma 3.5, we have

$$\begin{aligned}& \bigl\Vert (t+\xi -s)^{\alpha -1} E_{\alpha ,\alpha}\bigl(-(t+ \xi -s)^{\alpha }A\bigr)h(s)- ({t}-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A\bigr)h(s) \bigr\Vert _{\beta} \\& \quad \leq \bigl\Vert (t+\xi -s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t+ \xi -s)^{ \alpha }A \bigr)h(s)-(t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s) \bigr\Vert _{\beta} \\& \qquad {} + \bigl\Vert (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s)- (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr)h(s) \bigr\Vert _{\beta} \\& \quad \leq \biggl( \frac{M_{E} C_{\beta } \Vert h \Vert _{\omega c}e^{s\frac{\ln \vert c \vert }{\omega}}}{(t+\xi -s)^{\alpha \beta}} \biggr) \biggl\vert \biggl( \frac{1}{t+\xi -s} \biggr)^{1-\alpha}- \biggl( \frac{1}{t-s} \biggr)^{1-\alpha} \biggr\vert \\& \qquad {} +(t-s)^{\alpha -1} \bigl\Vert E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s)- E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr)h(s) \bigr\Vert _{ \beta}\xrightarrow[\xi \to 0^{+} ]{}0. \end{aligned}$$

Therefore,

$$ \big\| (t+\xi -s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s)- ({t}-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr)h(s)\big\| _{\beta}\xrightarrow[\xi \to 0^{+}]{}0. $$

Again, by Theorem 2.7 and Lemma 3.5, we obtain

$$\begin{aligned} & \big\| ({t}+\xi -s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s)-({t}-s)^{ \alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{\alpha }A \bigr)h(s)\big\| _{\beta } \\ &\quad \leq \big\| ({t}+\xi -s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s) \big\Vert _{\beta}+ \big\Vert ({t}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{ \alpha }A \bigr)h(s) \big\| _{\beta} \\ &\quad \leq M_{E} C_{\beta} \Vert h \Vert _{\omega c} \biggl( \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t+\xi -s)^{1-\alpha +\alpha \beta}}+ \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} \biggr),\quad s\in (-\infty ,t). \end{aligned}$$

Due to $\xi >0$ and $0<1-\alpha +\alpha \beta <1$, we have

$$ M_{E} C_{\beta} \Vert h \Vert _{\omega c} \biggl( \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t+\xi -s)^{1-\alpha +\alpha \beta}}+ \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} \biggr)\leq M_{E} C_{\beta} \Vert h \Vert _{\omega c} \biggl( \frac{2e^{s\frac{\ln \vert c \vert }{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} \biggr), $$

and therefore,

$$\begin{aligned} & \big\| ({t+\xi}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s)-({t}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{ \alpha }A \bigr)h(s)\big\| _{\beta} \\ &\quad \leq 2M_{E} C_{\beta} \Vert h \Vert _{\omega c} \biggl( \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} \biggr), \quad s\in (-\infty ,t). \end{aligned}$$

In addition, the function ${s\mapsto 2 M_{E} C_{\beta} \Vert h \Vert _{\omega c} ( \frac{e^{s\frac{\ln |c|}{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} )}$ is integrable in $(-\infty ,t)$, since

$$ \int _{-\infty}^{t} 2 M_{E} C_{\beta} \Vert h \Vert _{\omega c} \biggl( \frac{e^{s\frac{\ln \vert c \vert }{\omega}}}{(t-s)^{1-\alpha +\alpha \beta}} \biggr)\,ds=2 M_{E} C_{\beta} \Vert h \Vert _{\omega c}e^{t \frac{\ln \vert c \vert }{\omega}} \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1- \beta )}\Gamma \bigl(\alpha (1-\beta ) \bigr). $$

Hence, the criterion of comparison of improper integrals guarantees that

$$ s\mapsto \big\| ({t+\xi}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{ \alpha }A \bigr)h(s)-({t}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{ \alpha }A \bigr)h(s)\big\| _{\beta} $$

is integrable in $(-\infty ,t)$. Thus, by the Dominated Convergence Theorem, it follows that

$$\begin{aligned} I =& \int _{-\infty}^{t} \big\| (t+\xi -s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s) \\ &{}- ({t}-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{\alpha }A \bigr)h(s) \big\| _{\beta }\,ds \\ {\xrightarrow[\xi \to 0^{+}]{}}&0. \end{aligned}$$

On the other hand, using similar arguments to those in the estimates of I, we obtain

$$ \begin{aligned} II&= \int _{{t}}^{t+\xi} (t+\xi -s)^{\alpha -1} \bigl\Vert E_{\alpha , \alpha} \bigl(-(t+\xi -s)^{\alpha }A \bigr)h(s) \bigr\Vert _{\beta }\,ds \\ &\leq M_{E} C_{\beta } \Vert h \Vert _{\omega c} \int _{{t}}^{t+\xi} (t+\xi -s)^{ \alpha -\alpha \beta -1}e^{s\frac{\ln \vert c \vert }{\omega}} \,ds \\ &\leq M_{E} C_{\beta} \Vert h \Vert _{\omega c}e^{(t+\xi ) \frac{\ln \vert c \vert }{\omega}} \int _{0}^{\xi} r^{\alpha -\alpha \beta -1}e^{-r \frac{\ln \vert c \vert }{\omega}}\,dr. \end{aligned} $$

Note that, using a change of variable and the definition of the incomplete Gamma function γ, we have

$$\begin{aligned} \int _{0}^{\xi} r^{\alpha -\alpha \beta -1}e^{-r \frac{\ln \vert c \vert }{\omega}}\,dr &= \int _{0}^{\frac{\ln \vert c \vert }{\omega}\xi} \biggl(s\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )-1}e^{-s} \frac{\omega}{\ln \vert c \vert }\,ds \\ &= \biggl(\frac{\ln \vert c \vert }{\omega} \biggr)^{\alpha (\beta -1)}\gamma \biggl(\alpha (1-\beta ), \frac{\ln \vert c \vert }{\omega}\xi \biggr). \end{aligned}$$

Thus,

$$\begin{aligned} II &\leq M_{E} C_{\beta} \Vert h \Vert _{\omega c}e^{(t+\xi ) \frac{\ln \vert c \vert }{\omega}} \biggl(\frac{\ln \vert c \vert }{\omega} \biggr)^{\alpha ( \beta -1)} \gamma \biggl(\alpha (1-\beta ), \frac{\ln \vert c \vert }{\omega}\xi \biggr) {\xrightarrow[ \xi \to 0^{+}]{}}0. \end{aligned}$$

Therefore, ${\|(\Theta u)(t+\xi )-(\Theta u)(t)\|_{\beta}\to 0}$ when $\xi \to 0^{+}$, proving the claim. In a similar way, we can show that ${\lim_{\xi \to 0^{-}}}\|(\Theta u)(t+\xi )-(\Theta u)(t) \|_{\beta}=0$.

Now, we will show that $(\Theta u)(t+\omega )=c(\Theta u)(t)$ for all $t\in \mathbb{R}$. Indeed, since $h\in P_{\omega c}(\mathbb{R},X_{\beta})$, by the definition of $(\omega ,c)$-periodicity, we have

$$ \begin{aligned} (\Theta u) (t+\omega ) &= \int _{-\infty}^{t} (t-r)^{\alpha -1} E_{ \alpha , \alpha} \bigl(-(t-r)^{\alpha }A \bigr)h(r+\omega )\,dr \\ &=c \int _{-\infty}^{t} (t-r)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-r)^{ \alpha }A \bigr)h(r)\,dr=c(\Theta u) (t). \end{aligned} $$

Hence, we deduce that $(\Theta u)\in P_{\omega c}(\mathbb{R},X_{\beta})$. □

Theorem 3.7

Suppose that (H1)–(H4) are satisfied and $1<|c|<e^{\eta \omega}$. If $\delta <1$ where

$$ \delta :=M_{E} \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1- \beta )}\Gamma \bigl(\alpha (1-\beta ) \bigr) \biggl( L_{f} + \frac{C_{k}L_{g} \omega}{\eta \omega +\ln \vert c \vert } \biggr), $$

(3.9)

then (1.1) has a unique mild solution $u\in P_{\omega c}(\mathbb{R},X_{\beta})$.

Proof

Let us define the operator $\Theta :P_{\omega c}(\mathbb{R},X_{\beta})\to P_{\omega c}( \mathbb{R},X_{\beta})$ given by

$$ (\Theta u) (t)= \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds. $$

According to Lemma 3.6, we have $\Theta u\in P_{\omega c}(\mathbb{R},X_{\beta})$ for all $u\in P_{\omega c}(\mathbb{R},X_{\beta})$.

Let us see that Θ is a contraction. In fact, let $u,v\in P_{\omega c}(\mathbb{R},X_{\beta})$. By (H3) and Theorem 2.7, we have

$$\begin{aligned} & \biggl\Vert \vert c \vert ^{\wedge}(-t) \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{ \alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)-f \bigl(s,v(s) \bigr) \bigr)\,ds \biggr\Vert _{\beta } \\ & \quad \leq M_{E} L_{f} \Vert u-v \Vert _{\omega c} \int _{-\infty}^{t} (t-s)^{ \alpha (1-\beta )-1} \vert c \vert ^{\wedge} \bigl(-(t-s) \bigr)\,ds \\ &\quad \leq M_{E} L_{f} \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1- \beta )}\Gamma \bigl(\alpha (1-\beta ) \bigr) \Vert u-v \Vert _{\omega c}. \end{aligned}$$

(3.10)

On the other hand, by (H2) and (H4), we obtain

$$ \begin{aligned} \bigl\Vert (Ku) (s)-(Kv) (s) \bigr\Vert _{X} &\leq \int _{-\infty}^{s} \bigl\vert k(s-r) \bigr\vert \bigl\Vert g \bigl(r,u(r) \bigr)-g \bigl(r,v(r) \bigr) \bigr\Vert _{X} \,dr \\ &\leq C_{k}L_{g} \Vert u-v \Vert _{\omega c}e^{-\eta s} \biggl( \int _{-\infty}^{s} e^{ (\eta +\frac{\ln \vert c \vert }{\omega} ) r}\,dr \biggr) \\ &\leq C_{k}L_{g} \biggl(\frac{\omega}{\eta \omega +\ln \vert c \vert } \biggr) \Vert u-v \Vert _{\omega c} \vert c \vert ^{\wedge}(s) . \end{aligned} $$

Using this fact together with Theorem 2.7, we obtain

$$\begin{aligned} & { \biggl\Vert \vert c \vert ^{\wedge}(-t) \int _{-\infty}^{t} (t-s)^{ \alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl((Ku) (s)-(Kv) (s) \bigr)\,ds \biggr\Vert _{\beta}} \\ &\quad \leq M_{E} \int _{-\infty}^{t} \vert c \vert ^{\wedge}(-t) (t-s)^{\alpha -1- \alpha \beta} \bigl\Vert (Ku) (s)-(Kv) (s) \bigr\Vert _{X} \,ds \\ &\quad \leq M_{E} \int _{-\infty}^{t} \vert c \vert ^{\wedge}(-t) (t-s)^{\alpha -1- \alpha \beta} \biggl(C_{k}L_{g} \biggl( \frac{\omega}{\eta \omega +\ln \vert c \vert } \biggr) \Vert u-v \Vert _{ \omega c} \vert c \vert ^{\wedge}(s) \biggr)\,ds \\ &\quad \leq M_{E} C_{k}L_{g} \biggl( \frac{\omega}{\eta \omega +\ln \vert c \vert } \biggr) \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )} \Gamma \bigl(\alpha (1-\beta ) \bigr) \Vert u-v \Vert _{\omega c}. \end{aligned}$$

(3.11)

Now, by (3.10) and (3.11), we have

$$\begin{aligned} & { \bigl\Vert (\Theta u) (t)-(\Theta v) (t) \bigr\Vert _{\omega c}} \\ &\quad \leq \sup_{t\in [0,\omega ]} \biggl( \biggl\Vert \vert c \vert ^{\wedge}(-t) \int _{- \infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)-f \bigl(s,v(s) \bigr) \bigr)\,ds \biggr\Vert _{\beta} \\ & \qquad {} {} + \biggl\Vert \vert c \vert ^{\wedge}(-t) \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl((Ku) (s)-(Kv) (s) \bigr)\,ds \biggr\Vert _{\beta} \biggr) \\ &\quad \leq \sup_{t\in [0,\omega ]} \biggl( M_{E} L_{f} \biggl( \frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )}\Gamma \bigl(\alpha (1- \beta ) \bigr) \Vert u-v \Vert _{\omega c} \\ &\qquad {} {} + M_{E} C_{k}L_{g} \biggl( \frac{\omega}{\eta \omega +\ln \vert c \vert } \biggr) \biggl( \frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )}\Gamma \bigl(\alpha (1- \beta ) \bigr) \Vert u-v \Vert _{\omega c} \biggr) \\ &\quad \leq \biggl( M_{E} \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1- \beta )} \Gamma \bigl(\alpha (1-\beta ) \bigr) \biggl( L_{f} + \frac{C_{k}L_{g} \omega}{\eta \omega +\ln \vert c \vert } \biggr) \biggr) \Vert u-v \Vert _{\omega c} \\ &\quad \leq \delta \Vert u-v \Vert _{\omega c}. \end{aligned}$$

Since $\delta <1$, Θ is a contraction. Therefore, Banach’s Fixed-Point Theorem guarantees the existence of a unique fixed point $u\in P_{\omega c}(\mathbb{R},X_{\beta})$ of the operator Θ, which satisfies

$$ (\Theta u) (t)=u(t)= \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha , \alpha} \bigl(-(t-s)^{\alpha }A \bigr) \bigl(f \bigl(s,u(s) \bigr)+(Ku) (s) \bigr) \,ds. $$

This completes the proof of the theorem. □

4 An application

In this section we present an example that fits our framework.

Let $X=(L^{2}[0,1],\|\cdot \|_{L^{2}})$. Consider the following problem:

$$ \textstyle\begin{cases} \partial ^{\alpha}_{t} w(t,x)+\partial _{x}^{2} w(t,x)=a(t)\cos (b(t)w(t,x))+Kw(t,x), & t\in \mathbb{R},x\in (0,1), \\ w(t,0)=w(t,1)=0, & t\in \mathbb{R}, \end{cases}$$

(4.1)

where $0<\alpha < 1$, $\partial _{t}^{\alpha}$ denotes the Caputo fractional derivative with respect to t and

$$ Kw(t,x)= \int _{-\infty}^{t} k(t-s) \bigl(a(s)\sin \bigl( b(s)w(s,x) \bigr) \bigr)\,ds. $$

The functions k and a, b will be specified later.

We define the linear operator −A on X by

$$ \begin{aligned} &D(-A)= \bigl\{ u\in X: u,u'\in X\text{ are absolutely continuous},u''\in X\text{ and } u(0)=u(1)=0 \bigr\} , \\ &-Au(x)=u''(x),\quad \forall x\in (0,1), u\in D(-A). \end{aligned} $$

It is well known that −A is an infinitesimal generator of an analytic semigroup $\{S(t)\}_{t\geq 0}$ on X (see, for example, [24, Example 4.1.7] with a little modification). In addition, −A has a discrete spectrum, namely, the eigenvalues $-\lambda _{n}=-n^{2}$, $n\in \mathbb{N}$. The associated normalized eigenfunctions are given by $e_{n}(x)=\sqrt{2}\sin (n\pi x)$, $n\in \mathbb{N}$. Moreover, the semigroup is

$$ S(t)u(x)=\sum_{n=1}^{\infty }e^{-n^{2}\pi ^{2} t} \langle u,e_{n} \rangle _{L^{2}}e_{n}(x). $$

Also, $\|S(t)\|_{L^{2}}\leq e^{-\pi ^{2} t}$ for $t\geq 0$. This shows that (H1) holds. This in turn implies that the fractional powers of A can be defined as in Sect. 2. More precisely, since A has a compact resolvent, we have that

$$\begin{aligned} A^{\beta}u=\sum_{n=1}^{\infty}\langle u,e_{n}\rangle _{L^{2}}e_{n} \lambda _{n}^{\beta}=\sum_{n=1}^{\infty} \langle u,e_{n}\rangle _{L^{2}}e_{n} n^{2\beta}, \end{aligned}$$

with domain

$$\begin{aligned} \Biggl\{ u\in X:\sum_{n=1}^{\infty} \bigl\vert \langle u,e_{n}\rangle _{L^{2}} \bigr\vert ^{2} n^{4\beta}< \infty \Biggr\} . \end{aligned}$$

Now, let $k(t)=e^{-\pi ^{2} t}$. Then, ${|e^{-\pi ^{2} t}|\leq (2/3)e^{-9 t}}$. Thus, (H2) holds with $C_{k}=\frac{2}{3}$ and $\eta =9$.

Let $a\in P_{\omega ,c}(\mathbb{R},X_{\beta})$ and $b\in P_{\omega ,\frac{1}{c}}(\mathbb{R},X_{\beta})$ with $1<|c|<e^{9\omega}$.

Let us define $f(t,x)=a(t)\cos (b(t)x)$ and $g(t,x)=a(t)\sin (b(t)x)$. Then, the problem (4.1) can be reformulated as (1.1) with A, k, f, and g defined as above.

Next, we will show that (H3) and (H4) hold. Indeed,

$$ \begin{aligned} f(t+\omega ,cx)&=a(t+\omega )\cos \bigl(b(t+\omega )cx \bigr) \\ &=ca(t)\cos \biggl( \frac{1}{c}b(t)cx \biggr) \\ &=ca(t)\cos \bigl(b(t)x \bigr)= cf(t,x). \end{aligned} $$

Since $a\in P_{\omega c}(\mathbb{R},X_{\beta})$, $b\in P_{\omega \frac{1}{c}}(\mathbb{R},X_{\beta})$, we have $f\in C(\mathbb{R}\times X_{\beta}, X_{\beta})$. Also, for $x, y\in X_{\beta}$, we obtain

$$ \begin{aligned} \bigl\Vert f(t,x)-f(t,y) \bigr\Vert _{L^{2}} &\leq \Vert a \Vert _{L^{2}} \bigl\Vert \cos \bigl(b(t)x \bigr)-\cos \bigl(b(t)y \bigr) \bigr\Vert _{L^{2}} \\ &\leq \Vert a \Vert _{L^{2}} 2 \biggl\Vert \sin \biggl( \frac{b(t)x-b(t)y}{2} \biggr) \biggr\Vert _{L^{2}} \biggl\Vert \sin \biggl(\frac{b(t)x+b(t)y}{2} \biggr) \biggr\Vert _{L^{2}} \\ &\leq \Vert a \Vert _{L^{2}} 2 \biggl\Vert \frac{b(t)x-b(t)y}{2} \biggr\Vert _{L^{2}} \cdot 1 \\ &\leq \Vert a \Vert _{L^{2}} \Vert b \Vert _{L^{2}} \Vert x-y \Vert _{L^{2}} \\ &\leq \Vert a \Vert _{L^{2}} \Vert b \Vert _{L^{2}} C_{\beta } \Vert x-y \Vert _{\beta},\quad \forall t\in \mathbb{R}, x,y\in X_{\beta}, \end{aligned} $$

obtaining $(H3)$. The proof for (H4) is analogous. More precisely,

$$ \bigl\Vert g(t,x)-g(t,y) \bigr\Vert _{L^{2}}\leq \Vert a \Vert _{L^{2}} \Vert b \Vert _{L^{2}} C_{ \beta } \Vert x-y \Vert _{\beta},\quad \forall t\in \mathbb{R}, x,y\in X_{\beta}. $$

From the estimated

$$ \begin{aligned} \bigl\Vert A^{-\beta}x \bigr\Vert _{L^{2}} &\leq \frac{ \Vert x \Vert _{L^{2}}}{\Gamma (\beta )} \int _{0}^{\infty } \biggl(\frac{s}{\pi ^{2}} \biggr)^{\beta -1} e^{-s} \frac{ds}{\pi ^{2}} \\ &\leq \frac{1}{\pi ^{2\beta}} \Vert x \Vert _{L^{2}},\quad x\in X_{\beta}, \end{aligned} $$

we see that the constant $C_{\beta}$ can be chosen as ${\frac{1}{\pi ^{2\beta}}}$ (see Lemma 2.2).

The constant $M_{\beta}$ of Theorem 2.3 can be taken as ${\frac{1}{\pi ^{2(1-\beta )}}}$. In fact, note that $\|S(t) x(\cdot )\|_{L^{2}}\leq e^{-\pi ^{2}t}\|x\|_{L^{2}}$ for $t\geq 0$ and $\|AS(t)x(\cdot )\|_{L^{2}}\leq t^{-1}e^{-\pi ^{2}t}\|x\|_{L^{2}}$ for $t> 0$. Moreover, for $x\in X_{\beta}$, we have

$$ \begin{aligned} \bigl\Vert A^{\beta}S(t) x \bigr\Vert _{L^{2}} &= \bigl\Vert A^{-(1-\beta )}AS(t) x \bigr\Vert _{L^{2}} \\ &\leq \frac{1}{\Gamma (1-\beta )} \int _{0}^{\infty }s^{-\beta}(t+s)^{-1}e^{- \pi ^{2}(t+s)} \Vert x \Vert _{L^{2}}\,ds \\ &\leq \frac{1}{\pi ^{2(1-\beta )}} t^{-\beta} e^{-\pi ^{2} t} \Vert x \Vert _{L^{2}}. \end{aligned} $$

Finally, the constant $M_{E}$ of Theorem 2.7 can be taken as $M_{E}=\frac{1}{\pi ^{2(1-\beta )}} ( \frac{\Gamma (1-\beta )}{\Gamma (\alpha (1-\beta ))} )$. Indeed,

$$ \begin{aligned} \bigl\Vert E_{\alpha ,\alpha} \bigl(-t^{\alpha }A \bigr)x \bigr\Vert _{\beta }&= \biggl\Vert A^{\beta} \int _{0}^{\infty}\alpha \theta M_{\alpha}(\theta )S \bigl( \theta t^{\alpha} \bigr)x\,d\theta \biggr\Vert _{L^{2}} \\ &\leq \int _{0}^{\infty}\alpha \theta M_{\alpha}(\theta ) \bigl\Vert A^{ \beta }S \bigl(\theta t^{\alpha} \bigr)x \bigr\Vert _{L^{2}}\,d\theta \\ &\leq \int _{0}^{\infty}\alpha \theta M_{\alpha}(\theta ) \biggl( \frac{1}{\pi ^{2(1-\beta )}} \bigl(\theta t^{\alpha} \bigr)^{-\beta}e^{-\pi ^{2} \theta t^{\alpha}} \Vert x \Vert _{L^{2}} \biggr)\,d\theta \\ &\leq \frac{\alpha}{\pi ^{2(1-\beta )}} \biggl( \int _{0}^{\infty }M_{ \alpha}(\theta )\theta ^{1-\beta}\,d\theta \biggr)t^{-\alpha \beta} \Vert x \Vert _{L^{2}}. \end{aligned} $$

Due to the Proposition 2.5, it is fulfilled that ${\int _{0}^{\infty}} \theta ^{n} M_{\alpha}(\theta )\,d\theta =\frac{\Gamma (n+1)}{\Gamma (\alpha n+1)}$, for $n\geq -1$ and by the definition of the Gamma function one has that $\Gamma (\theta +1)=\theta{\Gamma (\theta )}$.

Then,

$$ \begin{aligned} \bigl\Vert E_{\alpha ,\alpha} \bigl(-t^{\alpha }A \bigr)x \bigr\Vert _{\beta }&\leq \frac{\alpha}{\pi ^{2(1-\beta )}} \biggl( \frac{\Gamma (1-\beta +1)}{\Gamma (\alpha (1-\beta )+1)} \biggr)t^{- \alpha \beta} \Vert x \Vert _{L^{2}} \\ &\leq \frac{1}{\pi ^{2(1-\beta )}} \biggl( \frac{\Gamma (1-\beta )}{\Gamma (\alpha (1-\beta ))} \biggr)t^{- \alpha \beta} \Vert x \Vert _{L^{2}}. \end{aligned} $$

Consequently $M_{E}=\frac{1}{\pi ^{2(1-\beta )}} ( \frac{\Gamma (1-\beta )}{\Gamma (\alpha (1-\beta ))} )$.

Now, by (3.9), we have

$$ \delta =M_{E} \biggl(\frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )} \Gamma \bigl(\alpha (1-\beta ) \bigr) \biggl( L_{f} + \frac{C_{k}L_{g} \omega}{\eta \omega +\ln \vert c \vert } \biggr), $$

and therefore,

$$\begin{aligned} \delta =& \biggl( \frac{\Gamma (1-\beta )}{\pi ^{2(1-\beta )}\Gamma (\alpha (1-\beta ))} \biggr) \\ &{}\times \biggl( \frac{\omega}{\ln \vert c \vert } \biggr)^{\alpha (1-\beta )} \Gamma \bigl(\alpha (1-\beta ) \bigr) \cdot \Vert a \Vert _{L^{2}} \Vert b \Vert _{L^{2}}\cdot \frac{1}{\pi ^{2\beta}} \biggl( 1 + \frac{(2/3) \omega}{9\omega +\ln \vert c \vert } \biggr) \\ =&\frac{\Gamma (1-\beta )}{\pi ^{2}}\cdot \Vert a \Vert _{L^{2}} \Vert b \Vert _{L^{2}} \biggl( \frac{\omega ^{\alpha (1-\beta )}}{(\ln \vert c \vert )^{\alpha (1-\beta )}} \biggr) \biggl(1+\frac{(2/3) \omega}{9\omega +\ln \vert c \vert } \biggr). \end{aligned}$$

According to Theorem 3.7, the fractional problem (4.1) has a unique $(\omega ,c)$-periodic mild solution whenever $\delta <1$. Moreover, the solution is given by

$$\begin{aligned} u(t) =& \int _{-\infty}^{t} (t-s)^{\alpha -1} E_{\alpha ,\alpha} \bigl(-(t-s)^{ \alpha }A \bigr) \\ &{}\times \biggl(a(s)\cos \bigl(b(s)u(s) \bigr)+ \int _{-\infty}^{s} e^{\pi ^{2}(s-r)}a(s) \sin \bigl(b(s)u(s) \bigr)\,dr \biggr)\,ds. \end{aligned}$$