Mild Solutions of Fractional Integrodifferential Diffusion Equations with Nonlocal Initial Conditions via the Resolvent Family

Abstract

Fractional integrodifferential diffusion equations play a significant role in describing anomalous diffusion phenomena. In this paper, we study the existence and uniqueness of mild solutions to these equations. Firstly, we construct an appropriate resolvent family, through which the related equicontinuity, strong continuity, and compactness properties are studied using the convolution theorem of Laplace transform, the probability density function, the Cauchy integral formula, and the Fubini theorem. Then, we construct a reasonable mild solution for the considered equations. Finally, we obtain some sufficient conditions for the existence and uniqueness of mild solutions to the considered equations by some fixed point theorems.

1. Introduction

In this paper, we consider the existence and uniqueness of mild solutions to the following nonlinear fractional integrodifferential diffusion equations with nonlocal initial conditions:

$$\left\{\begin{array}{c}{\partial }_t^{\alpha }u(x,t)=Au(x,t)+I_t^{\beta }f(x,t,u(x,t)),(x,t)\in \Omega \times (0,T],\hfill \\ u(x,t)=0,(x,t)\in \partial \Omega \times (0,T],\hfill \\ u(x,0)+g\left(u\right)=u_0\left(x\right),x\in \Omega ,\hfill \end{array}\right.$$

(1)

where $\alpha \in (0,1)$, $\beta \in (0,\infty )$, and ${\partial }_t^{\alpha }$ and $I_t^{\beta }$ are the $\alpha$ order partial Caputo derivative and $\beta$ order partial Riemann–Liouville integral with respect to t, respectively. $u_0\left(x\right)\in L^2(\Omega )$, which is given in Section 2, $\Omega \subset {\mathbb{R}}^n$ with smooth boundary $\partial \Omega$, $T>0$, and the nonlinear term f and the nonlocal term g are given functions. The coefficient linear operator A is defined by:

$$Au(x,t)=\sum _{i=1}^n\sum _{j=1}^n{\partial }_{x_i}\left[a_{ij}\left(x\right){\partial }_{x_j}u(x,t)\right]-p\left(x\right)u(x,t),$$

(2)

where the real valued functions $a_{ij}$ satisfy:

$$a_{ij}\in C^1(\overline{\Omega }),1\le i,j\le n,$$

$$\sum _{i,j=1}^n{a}_{ij}\left(x\right){\vartheta }_i{\vartheta }_j\ge \varsigma {\left|\vartheta \right|}^2,\vartheta \in {\mathbb{R}}^n,x\in \overline{\Omega },$$

with some constants $\varsigma >0$, and p is also a real valued function satisfying

$$p\in C(\overline{\Omega }),p\left(x\right)\ge p_0>0,x\in \overline{\Omega }.$$

Fractional differential equations originated in 1695, see [1,2]. As is known, they can provide excellent descriptive models to resolve various problems in reality, and they are applied in various fields, such as control engineering [3], viscoelastic materials [4], fluid mechanics [5], electrochemistry [6], the analysis of epidemic [7] and complex networks [8], statistical mechanics [9], numerical schemes [10], etc. The relevant problems for the diffusion equations have been studied by many scholars, see [11,12]. Significantly, when $\alpha$ is used to represent the order of a fractional diffusion equation, when $\alpha \in (0,1)$, $\alpha =1$, and $\alpha \in (1,2)$, the equation describes subdiffusion, regular diffusion, and superdiffusion, respectively.

The existence results to nonlocal initial problems in Banach spaces were initiated by Byszewski and Lakshmikantham [13]. The motivation for these studies is that the nonlocal condition better describes the diffusion phenomena than using the usual local condition $u\left(0\right)=u_0$. For example, $g\left(u\right)$ can be given by

$$g\left(u\right)=\sum _{i=1}^l{c}_{i}u\left({\tau }_i\right),$$

where $c_i$$(i=1,2,\dots ,l)$ are given constants, and $0<{\tau }_1<{\tau }_2<\cdots <{\tau }_l<T$. In addition, for some applications of nonlocal conditions, please refer to [14,15,16,17,18].

Fractional equations containing only differential terms have been studied widely. For instance, in [19], Mu et al. considered the initial boundary value problem of fractional diffusion equations in the Caputo sense:

$$\left\{\begin{array}{cc}{\partial }_t^{\alpha }u(x,t)=Au(x,t)+f(x,t)\hfill & \mathrm{in}\Omega \times (0,b),\hfill \\ u=0\hfill & \mathrm{on}\partial \Omega \times (0,b),\hfill \\ u(x,0)=u_0\left(x\right)\hfill & \mathrm{in}\Omega ,\hfill \end{array}\right.$$

(3)

where f is weighted Hölder continuous [20]. The existence, uniqueness, and regularity of solutions to (3) can be established in $C([0,b],H^{-1}(\Omega ))\cap C((0,b],D(-A))\cap C^{\alpha }((0,b],H^{-1}(\Omega ))$ with some assumptions. Nevertheless, a lot of practical phenomena can be depicted via appropriate models, which include differential and integral terms. Hence, the appearance of integrodifferential equations shows their excellent applicability in some physical or engineering areas, and they have attracted the attention of many scholars, see [21,22,23]. In [23], Amin et al. obtained the solutions to an integrodifferential equation with an initial condition

$$\left\{\begin{array}{c}{\partial }_t^{\xi }w\left(t\right)=r\left(t\right)+b\left(t\right)w\left(t\right)+{\int }_0^{t}W(t,\tau )w\left(\tau \right)d\tau ,t\in [0,1],\hfill \\ w\left(0\right)=w_0,\hfill \end{array}\right.$$

where $\xi \in (0,1]$, W is the kernel of integral, and $r\left(t\right)$ and $b\left(t\right)$ are known. This integral term has certain limitations when describing some nonlocal diffusion phenomena, and the applicability of the above initial condition is also relatively weak.

The Mittag–Leffler function or the probability density function are often applied in the expression of mild solutions, see [24,25,26,27,28,29,30]. In [30], Zhou et al. obtained a mild solution, where

$${\lambda }^{q-1}{({\lambda }^{q}I-A)}^{-1}={\int }_0^{\infty }{e}^{-\lambda t}\left[{\int }_0^{\infty }{\psi }_q\left(\theta \right)T\left(\frac{t^q}{{\theta }^q}\right)d\theta \right]dt,q\in (0,1),$$

(4)

${\psi }_q\left(\theta \right)$ is the probability density function defined on $(0,\infty )$, and ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ is a $C_0$-semigroup generated by the operator A. On the other hand, if $f(x,t,u(x,t\left)\right)=g\left(u\right(x,t\left)\right)=0$, then a formal solution to (1) is

$$u(x,t)=\sum _{i=1}^{\infty }(u_0,X_i)E_{\alpha }(-{\lambda }_i{t}^{\alpha })X_i\left(x\right),$$

where $X_i$ is the eigenfunction related to the eigenvalues ${\lambda }_i$ of the corresponding problems; that is, $-A\left(X_i\right)={\lambda }_i{X}_i$, $i=1,2,\dots$; see further details in [27]. Obviously, using these techniques directly to solve problem (1) is quite difficult.

Based on the above discussions, in this paper we apply the $(\alpha ,\beta )$-resolvent family to discuss the mild solutions to (1). Resolvent families are powerful for studying solutions to fractional diffusion equations. Chen et al. [31] established the existence and controllability estimation of mild solutions for a class of evolution equations with nonlocal conditions through a resolvent family. Ponce [18] obtained properties on the behavior of mild solutions for fractional Cauchy problems by a resolvent family. Later, Chang et al. [14] proved that if the source function of a diffusion equation has vector value periodicity or almost periodicity or almost automorphism, then the diffusion equation has a mild solution through a resolvent family. Although there is also an integral of f in [14], some of the proof techniques therein are not applicable to this article due to the derivative order of u being different.

This paper is organized as follows. In Section 2, by selecting the appropriate space, we transform (1) into an abstract Cauchy problem and provide some necessary definitions and preliminary results that will be used in the sequel. Afterwards, we define the mild solutions for (11) using the Laplace transform. In Section 3, the existence and uniqueness of the mild solutions to (11) are established by several fixed point theorems under some assumptions. In Section 4, an example is provided to verify the reasonability of the results. In Section 5, we summarize the entire article.

2. Preliminaries

In this section, we provide some definitions and lemmas about fractional calculus and the $(\alpha ,\beta )$-resolvent family that will be used in this paper.

Definition 1

([1]). The Riemann–Liouville fractional integral of order $\alpha \in (0,1)$ with respect to t for an integrable function $f:[0,\infty )\to \mathbb{R}$ is defined as

$$I_t^{\alpha }f\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{(t-s)}^{\alpha -1}f\left(s\right)ds,t>0,$$

(5)

where $\Gamma (·)$ is the Gamma function.

Definition 2

([1]). The Riemann–Liouville fractional derivative of order $\alpha \in (0,1)$ with respect to t for an absolutely continuous function $f:[0,\infty )\to \mathbb{R}$ is defined as

$${}^{R}D_t^{\alpha }f\left(t\right)=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dt}{\int }_0^t{(t-s)}^{-\alpha }f\left(s\right)ds,t>0.$$

(6)

Definition 3

([1]). The Caputo fractional derivative of order $\alpha \in (0,1)$ with respect to t for a continuously differentiable function $f:[0,\infty )\to \mathbb{R}$ can be written as

$$D_t^{\alpha }f\left(t\right)=\frac{1}{\Gamma (1-\alpha )}{\int }_0^t{f}^{\prime }\left(s\right){(t-s)}^{-\alpha }ds=I_t^{1-\alpha }{f}^{\prime }\left(t\right).$$

(7)

If f is an abstract function with values in a Banach space, then the integrals and derivatives appearing in (5) and (7) are understood in Bochner’s sense.

Definition 4

([32]). Let $({\Omega }_1,F,{\mu }_1)$ be a measure space, $U_1$ be a Banach space, $h:{\Omega }_1\to U_1$ be a measurable function, and $A_i$ be a partition of ${\Omega }_1$; if the limitation

$$\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(t_i\right){\mu }_1\left(A_i\right)$$

exists, then

$${\int }_{{\Omega }_1}fd{\mu }_1=\underset{n\to \infty }{lim}\sum _{i=1}^{n}f\left(t_i\right){\mu }_1\left(A_i\right)$$

is called the Bochner integral of f with respect to ${\mu }_1$.

Definition 5

([33]). Let P be a metric space, and let $S\subset P$ be a bounded set. The Kuratowski measure of noncompactness is defined by:

$$\nu \left(S\right)=inf\{\delta >0|S=\bigcup _{i=1}^m{S}_i,diam{S}_i\le \delta \}.$$

Lemma 1

([34]). Let E be a Banach space, and S, $S_1$, $S_2$ be some subsets of E. Then, we have the following properties:

(i) $\nu \left(S\right)=0\iff S$ is relatively compact;

(ii) $\nu \left(S\right)=\nu (\overline{S})$;

(iii) $S_1\subset S_2\Rightarrow \nu \left(S_1\right)\le \nu \left(S_2\right)$;

(iv) $\nu (S_1+S_2)\le \nu \left(S_1\right)+\nu \left(S_2\right)$, where $S_1+S_2=\left\{x\right|x=y+z,y\in S_1,z\in S_2\}$;

(v) $\nu \left(cS\right)=\left|c\right|\nu \left(S\right)$, $c\in \mathbb{R}$, where $cS=\left\{x\right|x=cz,z\in S\}$;

(vi) $0\le \nu \left(S\right)<+\infty$.

Let $X=L^2(\Omega )$, and $J=[0,T]$. Here,

$$L^2(\Omega )=\{w:\Omega \to \mathbb{R}|w\mathrm{is}\mathrm{Lebesgue}\mathrm{measurable},\parallel w\parallel <\infty .\},$$

where

$$\parallel w\parallel ={\left({\int }_{\Omega }{\left|w\right|}^{2}dx\right)}^{\frac{1}{2}},$$

see [35]. $C(J,X)$ denotes the Banach space of all continuous functions from J into X with the norm

$${\parallel ·\parallel }_{\infty }=\underset{t\in J}{sup}\{\parallel ·\left(t\right)\parallel \}.$$

Similarly, $C(J,{\mathbb{R}}^{+})$ denotes the Banach space of all continuous functions from J into ${\mathbb{R}}^{+}$ with the norm

$${\parallel ·\parallel }_{\infty }=\underset{t\in J}{sup}\{·\left(t\right)\}.$$

We define $A:D\left(A\right)\subset X\to X$ with $D\left(A\right)=H_0^1(\Omega )\cap H^2(\Omega )$ (see [35]) and $\left(Au\right)\left(t\right)x=Au(x,t)$. Then, A generates an analytic semigroup ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ on X. Without loss of generality, we can assume that $0\in \rho \left(A\right)$, ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ is uniformly bounded, and there exists a constant $N>0$ such that

$$\parallel T\left(t\right)\parallel \le N,\mathrm{for}t>0.$$

(8)

We denote $\rho \left(A\right)$ and $R(\lambda ,A)={(\lambda I-A)}^{-1}$ as the resolvent set and resolvent operator of A, respectively, where I is the identity operator. By [36] (Theorem 5.2, p. 61), there exist $\delta \in (0,\frac{\pi }{2})$ and $C>0$ such that

$${\Sigma }_{\frac{\pi }{2}+\delta }:=\left\{\lambda |\left|\arg \lambda \right|<\frac{\pi }{2}+\delta \right\}\subset \rho \left(A\right),$$

(9)

and

$$\parallel R(\lambda ,A)\parallel \le \frac{C}{\left|\lambda \right|},\lambda \in {\Sigma }_{\frac{\pi }{2}+\delta }\setminus \left\{0\right\}.$$

(10)

We set $u\left(t\right)\left(x\right)=u(x,t)$, $f(t,u(t\left)\right)\left(x\right)=f(x,t,u(x,t\left)\right)$, then (1) can be formulated as an abstract problem with nonlocal initial conditions:

$$\left\{\begin{array}{c}{D}_t^{\alpha }u\left(t\right)=Au\left(t\right)+I_t^{\beta }f(t,u\left(t\right)),t\in (0,T],\hfill \\ u\left(0\right)+g\left(u\right)=u_0,\hfill \end{array}\right.$$

(11)

where $D_t^{\alpha }$ and $I_t^{\beta }$ denote the $\alpha$ order Caputo derivative and $\beta$ order Riemann–Liouville integral, respectively. $g:C(J,X)\to X$, $f:J\times X\to X$, and $u_0\in X$.

Lemma 2

([30]). If function $H:J\to X$ is measurable, and $\parallel H\parallel$ is Lebesgue integrable, then H is called Bochner integrable.

Definition 6.

If A generates a uniformly bounded and analytic semigroup, which satisfies (9) and (10), and for operator-valued function $S_{\alpha ,\beta }\left(t\right):X\to X$, we have

$${\lambda }^{-\beta }{({\lambda }^{\alpha }I-A)}^{-1}x={\int }_0^{\infty }{e}^{-\lambda t}{S}_{\alpha ,\beta }\left(t\right)xdt,t\ge 0,x\in X,$$

(12)

then ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is called the $(\alpha ,\beta )$-resolvent family generated by the operator A.

It can be seen from the references [19,29] that the following results can be obtained.

Remark 1.

Let $\Phi \left(t\right)$ and $\Lambda \left(t\right)$ satisfy

$${\lambda }^{\alpha -1}{({\lambda }^{\alpha }I-A)}^{-1}x={\int }_0^{\infty }{e}^{-\lambda t}\Phi \left(t\right)xdt,t\ge 0,x\in X,$$

and

$${({\lambda }^{\alpha }I-A)}^{-1}x={\int }_0^{\infty }{e}^{-\lambda t}\Lambda \left(t\right)xdt,t\ge 0,x\in X.$$

(13)

Then,

$$\Lambda \left(t\right)=t^{\alpha -1}\Psi \left(t\right),\Psi \left(t\right)=\alpha {\int }_0^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)T\left(t^{\alpha }\theta \right)d\theta ,$$

$$S_{\alpha ,1-\alpha }\left(t\right)=\Phi \left(t\right)={\int }_0^{\infty }{\zeta }_{\alpha }\left(\theta \right)T\left(t^{\alpha }\theta \right)d\theta ,$$

where ${\zeta }_{\alpha }\left(\theta \right)$ is the probability density function defined on $(0,\infty )$, which satisfies

$${\zeta }_{\alpha }\left(\theta \right)\ge 0,\theta \in (0,\infty ),$$

$${\int }_0^{\infty }{\theta }^v{\zeta }_{\alpha }\left(\theta \right)d\theta =\frac{\Gamma (1+v)}{\Gamma (1+\alpha v)}\mathrm{for}v\in (-1,\infty ),$$

$${\int }_0^{\infty }{e}^{-z\theta }{\zeta }_{\alpha }\left(\theta \right)d\theta =E_{\alpha }(-z),\alpha {\int }_0^{\infty }\theta e^{-z\theta }{\zeta }_{\alpha }\left(\theta \right)d\theta =E_{\alpha ,\alpha }(-z),\mathrm{for}z\in \mathbb{C},$$

(14)

$E_{\alpha }$, $E_{\alpha ,\alpha }$ are the Mittag–Leffler functions [1].

If $u:J\to X$ is a solution to

$$\left\{\begin{array}{c}{D}_t^{\alpha }u\left(t\right)=Au\left(t\right)+f(t,u\left(t\right)),t\in (0,T],\hfill \\ u\left(0\right)=u_0,u_0\in X,\hfill \end{array}\right.$$

then

$$u\left(t\right)=S_{\alpha ,1-\alpha }\left(t\right)u_0+{\int }_0^t{(t-s)}^{\alpha -1}\Psi (t-s)f(s,u\left(s\right))ds.$$

Remark 2.

Assume that ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is the $(\alpha ,\beta )$-resolvent family generated by the operator A. By (13) and the convolution theorem of Laplace transform, we have

$$S_{\alpha ,\beta }\left(t\right)=\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -1}\Psi \left(s\right)ds.$$

Lemma 3.

If A generates a uniformly bounded and analytic semigroup, which satisfies (9) and (10), then

(i) There exist $\delta \in (0,\frac{\pi }{2})$ and a constant $C>0$ such that

$$\left\{{\lambda }^{\alpha }|\left|\arg \lambda \right|<\frac{\pi }{2}+\delta \right\}\subset \rho \left(A\right),$$

and

$$\parallel {\lambda }^{-\beta }R({\lambda }^{\alpha },A)\parallel \le \frac{C}{{\left|\lambda \right|}^{\alpha +\beta }}.$$

(15)

(ii) A generates the $(\alpha ,\beta )$-resolvent family ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$, and

$$S_{\alpha ,\beta }\left(t\right)=\frac{1}{2\pi i}{\int }_{\Gamma }{e}^{\lambda t}{\lambda }^{-\beta }R({\lambda }^{\alpha },A)d\lambda ,$$

where

$$\Gamma =\{r{e}^{-i(\frac{\pi }{2}+{\delta }_1)}|r\in [\rho ,\infty )\}\cup \{\rho e^{i{\delta }_2}||{\delta }_2|\le \frac{\pi }{2}+{\delta }_1\}\cup \{r{e}^{i(\frac{\pi }{2}+{\delta }_1)}|r\in [\rho ,\infty )\},$$

${\delta }_1\in (0,\delta )$, $\delta \in (0,\frac{\pi }{2})$ and $\rho >0$. Moreover,

$$\left|\right|S_{\alpha ,\beta }\left(t\right)\left|\right|\le M{t}^{\alpha +\beta -1},t>0,$$

(16)

and $M>0$ is a constant.

Proof. 

(i) Let $K\left(\lambda \right)={\lambda }^{\alpha }$, $\lambda =r{e}^{i(\frac{\pi }{2}+\delta )}$ with $\delta \in (0,\frac{\pi }{2})$ and $r>0$; then,

$$\begin{array}{cc}\hfill \left|\arg \right(K\left(\lambda \right)\left)\right|& =|\mathrm{Im}(\ln (K(r{e}^{i(\frac{\pi }{2}+\delta )})))|\hfill \\ \hfill & =\left|\mathrm{Im}{\int }_0^{\frac{\pi }{2}+\delta }\frac{d}{dt}\ln \left(K\left(r{e}^{it}\right)\right)dt\right|\hfill \\ \hfill & =\left|\mathrm{Im}{\int }_0^{\frac{\pi }{2}+\delta }\frac{K^{\prime }\left(r{e}^{it}\right)ir{e}^{it}}{K\left(r{e}^{it}\right)}dt\right|\hfill \\ \hfill & \le \alpha (\frac{\pi }{2}+\delta )\hfill \\ \hfill & <\frac{\pi }{2}+\delta .\hfill \end{array}$$

Thus, ${\lambda }^{\alpha }\in \rho \left(A\right)$, and

$$\parallel {\lambda }^{-\beta }R({\lambda }^{\alpha },A)\parallel \le \frac{C}{{\left|\lambda \right|}^{\alpha +\beta }},$$

where $C>0$ is a constant.

(ii) For $t>0$, and ${\delta }_1\in (0,\delta )$, we set

$$S_{\alpha ,\beta }\left(t\right)=\frac{1}{2\pi i}{\int }_{\Gamma }{e}^{\lambda t}{\lambda }^{-\beta }R({\lambda }^{\alpha },A)d\lambda ,$$

(17)

where

$$\Gamma ={\Gamma }_1\cup {\Gamma }_2\cup {\Gamma }_3,$$

$${\Gamma }_1=\{r{e}^{-i(\frac{\pi }{2}+{\delta }_1)}|r\in [\frac{1}{t},\infty )\},$$

$${\Gamma }_2=\{\frac{1}{t}{e}^{i\theta }\left|\right|\theta |\le \frac{\pi }{2}+{\delta }_1\},$$

$${\Gamma }_3=\{r{e}^{i(\frac{\pi }{2}+{\delta }_1)}|r\in [\frac{1}{t},\infty )\}$$

are oriented counterclockwise. From (15), it is easy to see that for $t>0$ the integral in (17) converges in the uniform topology. Moreover,

$$\begin{array}{cc}\hfill ∥\frac{1}{2\pi i}{\int }_{{\Gamma }_3}{e}^{\lambda t}{\lambda }^{-\beta }R({\lambda }^{\alpha },A)d\lambda )∥& \le \frac{C}{2\pi }{\int }_{\frac{1}{t}}^{\infty }\frac{e^{-rtsin{\delta }_1}}{r^{\alpha +\beta }}dr\hfill \\ \hfill & \le \frac{C{(sin{\delta }_1)}^{\alpha +\beta -1}}{2\pi }{\int }_{sin{\delta }_1}^{\infty }\frac{e^{-\tau }}{{\tau }^{\alpha +\beta }}d\tau t^{\alpha +\beta -1}\hfill \\ \hfill & \le C_1{t}^{\alpha +\beta -1}.\hfill \end{array}$$

Similarly, the integral on ${\Gamma }_1$ has the same estimation, and on ${\Gamma }_2$ we obtain

$$\begin{array}{cc}\hfill ∥\frac{1}{2\pi i}{\int }_{{\Gamma }_2}{e}^{\lambda t}{\lambda }^{-\beta }R({\lambda }^{\alpha },A)d\lambda )∥& \le \frac{C}{2\pi }{\int }_{-(\frac{\pi }{2}+\delta )}^{\frac{\pi }{2}+\delta }{e}^{cos\theta }d\theta t^{\alpha +\beta -1}\hfill \\ \hfill & \le C_2{t}^{\alpha +\beta -1}.\hfill \end{array}$$

Hence, (16) holds.

Next, we fix $\lambda >0$, and we have

$$\begin{array}{cc}\hfill {\int }_0^T{e}^{-\lambda t}{S}_{\alpha ,\beta }\left(t\right)dt& =\frac{1}{2\pi i}{\int }_{\Gamma }{\mu }^{-\beta }R({\mu }^{\alpha },A){\int }_0^T{e}^{-(\lambda -\mu )t}dtd\mu \hfill \\ \hfill & ={\lambda }^{-\beta }R({\lambda }^{\alpha },A)+\frac{1}{2\pi i}{\int }_{\Gamma }{e}^{-(\lambda -\mu )T}\frac{{\mu }^{-\beta }R({\mu }^{\alpha },A)}{\mu -\lambda }d\mu ,\hfill \end{array}$$

(18)

where the Cauchy integral formula and Fubini theorem [37] are also used. Due to

$$∥{\int }_{\Gamma }{e}^{-(\lambda -\mu )T}\frac{{\mu }^{-\beta }R({\mu }^{\alpha },A)}{\mu -\lambda }d\mu ∥\le M{e}^{-T\lambda }{\int }_{\Gamma }\frac{\left|d\mu \right|}{{\left|\mu \right|}^{\alpha +\beta }|\lambda -\mu |}\to 0,T\to \infty .$$

Therefore, by taking the limit as $T\to \infty$ in (18), we have

$${\int }_0^{\infty }{e}^{-\lambda t}{S}_{\alpha ,\beta }\left(t\right)dt={\lambda }^{-\beta }R({\lambda }^{\alpha },A).$$

That is, ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is generated by the operator A. □

Remark 3.

Due to different parameters, the method in [14] (Theorem 5 and Theorem 6) cannot be directly applied to this paper. By comparing the forms of mild solutions to the studied equations, we find that (16) agrees well with $\parallel \Psi \left(t\right)\parallel \le \frac{M_{\omega }}{\Gamma \left(\alpha \right)}$ in [19].

Lemma 4.

For $t>0$, $R_{\alpha ,\beta }\left(t\right)=t^{1-\alpha -\beta }{S}_{\alpha ,\beta }\left(t\right)$ is continuous in the uniform operator topology, where ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is the $(\alpha ,\beta )$-resolvent family generated by the operator A.

Proof. 

Let $ϵ>0$ be fixed. Due to the fact that $\Psi \left(t\right)$ is continuous in the uniform operator topology for $t>0$ [38], for arbitrary $t_0>0$, there exists $\delta >0$ such that

$$\parallel \Psi \left(t_2\right)-\Psi \left(t_1\right)\parallel \le \frac{\Gamma (\alpha +\beta )}{\Gamma \left(\alpha \right)}ϵ,$$

(19)

for $t_2>t_1\ge t_0$, and $|t_2-t_1|<\delta$.

Then, owing to

$$\begin{array}{cc}\hfill \parallel R_{\alpha ,\beta }\left(t_2\right)-R_{\alpha ,\beta }\left(t_1\right)\parallel & =\frac{1}{\Gamma \left(\beta \right)}∥t_2^{1-\alpha -\beta }{\int }_0^{t_2}{(t_2-s)}^{\beta -1}{s}^{\alpha -1}\Psi \left(s\right)ds\hfill \\ \hfill & -t_1^{1-\alpha -\beta }{\int }_0^{t_1}{(t_1-s)}^{\beta -1}{s}^{\alpha -1}\Psi \left(s\right)ds∥\hfill \\ \hfill & =\frac{1}{\Gamma \left(\beta \right)}{\int }_0^1{(1-\tau )}^{\beta -1}{\tau }^{\alpha -1}\parallel \Psi \left(t_2\tau \right)-\Psi \left(t_1\tau \right)\parallel d\tau ,\hfill \end{array}$$

and

$${\int }_0^1{(1-\tau )}^{\beta -1}{\tau }^{\alpha -1}d\tau =B(\alpha ,\beta )=\frac{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}{\Gamma (\alpha +\beta )},$$

where $B(·,·)$ is the Beta function [1], we conclude that

$$\parallel R_{\alpha ,\beta }\left(t_2\right)-R_{\alpha ,\beta }\left(t_1\right)\parallel \le ϵ.$$

That is, by the arbitrariness of $t_0$, $R_{\alpha ,\beta }\left(t\right)$ is continuous in the uniform operator topology for $t>0$. □

Remark 4.

${\left\{R_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is strongly continuous. That is, for arbitrary $x\in X$ and $0\le t_1<t_2\le T$, we have

$$\parallel R_{\alpha ,\beta }\left(t_2\right)x-R_{\alpha ,\beta }\left(t_1\right)x\parallel \to 0\mathrm{as}{t}_1\to t_2.$$

Proof. 

Since ${\{\Psi \left(t\right)\}}_{t\ge 0}$ is strongly continuous [39], there exists ${\delta }_1>0$ such that

$$\parallel \Psi \left(t_2\right)x-\Psi \left(t_1\right)x\parallel \le \frac{\Gamma (\alpha +\beta )}{\Gamma \left(\alpha \right)}ϵ,$$

(20)

for $t_2>t_1\ge 0$, and $|t_2-t_1|<{\delta }_1$.

Due to

$$\parallel R_{\alpha ,\beta }\left(t_2\right)x-R_{\alpha ,\beta }\left(t_1\right)x\parallel =\frac{1}{\Gamma \left(\beta \right)}{\int }_0^1{(1-\tau )}^{\beta -1}{\tau }^{\alpha -1}\parallel \Psi \left(t_2\tau \right)x-\Psi \left(t_1\tau \right)x\parallel d\tau \le ϵ,$$

we obtain that ${\left\{R_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is strongly continuous. □

It is noteworthy that the strong continuity of ${\left\{R_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ can not be obtained immediately by Lemma 4, in which $t>0$ not $t\ge 0$.

Lemma 5.

If the analytic semigroup ${\left\{T\left(t\right)\right\}}_{t\ge 0}$ generated by the operator A is compact, then $R_{\alpha ,\beta }\left(t\right)$ is compact for $t>0$.

Proof. 

Set $B_k=\{x\in X|\parallel x\parallel \le k\}$. In order to show $R_{\alpha ,\beta }\left(t\right)$ is compact for $t>0$, we need to show that

$$\left\{R_{\alpha ,\beta }\left(t\right)x\right|x\in B_k\}$$

is relatively compact in X, for any $k>0$ and $t>0$.

Let $t>0$ be fixed. For any $ϵ>0$, we define

$$R_{ϵ}\left(t\right):=\frac{\alpha }{\Gamma \left(\beta \right)}{t}^{1-\alpha -\beta }{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -1}{\int }_{ϵ}^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)T\left(t^{\alpha }\theta \right)d\theta ds.$$

Then,

$$R_{ϵ}\left(t\right)=T\left(t^{\alpha }ϵ\right)\frac{\alpha }{\Gamma \left(\beta \right)}{t}^{1-\alpha -\beta }{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -1}{\int }_{ϵ}^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)T(t^{\alpha }\theta -t^{\alpha }ϵ)d\theta ds,$$

and

$$\begin{array}{cc}& ∥\frac{\alpha }{\Gamma \left(\beta \right)}{t}^{1-\alpha -\beta }{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -1}{\int }_{ϵ}^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)T(t^{\alpha }\theta -t^{\alpha }ϵ)d\theta ds∥\hfill \\ \hfill & \le \frac{N}{\Gamma \left(\beta \right)}{\int }_0^t{(1-\tau )}^{\beta -1}{\tau }^{\alpha }d\tau ·\alpha {\int }_0^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)d\theta \hfill \\ \hfill & =\frac{N}{\Gamma \left(\beta \right)}B(\alpha ,\beta )\frac{1}{\Gamma \left(\alpha \right)}=\frac{N}{\Gamma (\alpha +\beta )},\hfill \end{array}$$

where $B(·,·)$ is the Beta function. Due to the compactness of $T\left(t^{\alpha }ϵ\right)$$(t^{\alpha }ϵ>0)$, we obtain that $\left\{R_{ϵ}\left(t\right)x\right|x\in B_k\}$ is relatively compact in X for arbitrary $ϵ>0$.

In addition, for any $x\in B_k$, we have

$$\begin{array}{c}\hfill \parallel R_{\alpha ,\beta }\left(t\right)x-R_{ϵ}\left(t\right)x\parallel \le \frac{Nk}{\Gamma (\alpha +\beta )}\alpha {\int }_0^{ϵ}\theta {\zeta }_{\alpha }\left(\theta \right)d\theta .\end{array}$$

Therefore, we obtain that $\left\{R_{\alpha ,\beta }\left(t\right)x\right|x\in B_k\}$ is relatively compact in X. □

Remark 5.

If B is the infinitesimal generator of a $C_0$-semigroup ${\left\{S\left(t\right)\right\}}_{t\ge 0}$ on a Banach space, then $S\left(t\right)$ is compact if and only if $S\left(t\right)$ is continuous in the uniform operator topology for $t>0$ and $R(\lambda ,A)$ is compact for $\lambda \in \rho \left(A\right)$ [36]. If $\alpha \in (0,1)$ and $\beta =1-\alpha$, $S_{\alpha ,\beta }$ is compact if and only if A generates a compact $C_0$-semigroup, which is also obtained in [39].

Lemma 6.

Assume that $u\in C(J,X)$, $D_t^{\alpha }u\in C((0,T],X)$, $u\left(t\right)\in D\left(A\right)$ for $t\in (0,T]$, and u satisfies (11). Then, u satisfies the formal integral equation

$$u\left(t\right)=S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))+{\int }_0^t{S}_{\alpha ,\beta }(t-s)f(s,u\left(s\right))ds.$$

Proof. 

By the definitions of the Caputo derivative and the Riemann–Liouville integral [1], we can rewrite (11) as the equivalent integral equation

$$\begin{array}{cc}\hfill u\left(t\right)& =u_0+\frac{1}{\Gamma \left(\alpha \right)}{\int }_0^t{(t-s)}^{\alpha -1}Au\left(s\right)ds\hfill \\ \hfill & +\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{(t-s)}^{\alpha +\beta -1}f(s,u\left(s\right))ds.\hfill \end{array}$$

(21)

For $\lambda \in {\Sigma }_{\frac{\pi }{2}+\delta }$, using the Laplace transform

$$\widehat{u}\left(\lambda \right)={\int }_0^{\infty }{e}^{-\lambda s}u\left(s\right)ds\mathrm{and}\widehat{f}(\lambda ,u\left(\lambda \right))={\int }_0^{\infty }{e}^{-\lambda s}f(s,u\left(s\right))ds$$

to (11), we have

$${\lambda }^{\alpha }\widehat{u}\left(\lambda \right)-{\lambda }^{\alpha -1}u\left(0\right)=A\widehat{u}\left(\lambda \right)+{\lambda }^{-\beta }\widehat{f}(\lambda ,u\left(\lambda \right)).$$

(22)

Then, (22) is equivalent to

$$\widehat{u}\left(\lambda \right)={\lambda }^{\alpha -1}{({\lambda }^{\alpha }I-A)}^{-1}u\left(0\right)+{\lambda }^{-\beta }{({\lambda }^{\alpha }I-A)}^{-1}\widehat{f}(\lambda ,u\left(\lambda \right)).$$

By the inverse Laplace transform and Definition 6, we have

$$u\left(t\right)=S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))+{\int }_0^t{S}_{\alpha ,\beta }(t-s)f(s,u\left(s\right))ds,$$

where $t\in J$. □

Consequently, we give the definition of a mild solution to (11) as follows.

Definition 7.

The function $u\in C(J,X)$ is called a mild solution of Equation (11) if

$$u\left(t\right)=S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))+{\int }_0^t{S}_{\alpha ,\beta }(t-s)f(s,u\left(s\right))ds.$$

Lemma 7

([40]). The convex closure $\overline{\mathit{conv}\left(Z\right)}$ is compact provided Z is a compact subset of a Banach space.

Lemma 8

([41]). Assume that $H:Y\to Y$ is completely continuous, where Y is a convex subset of a Banach space and $0\in Y$. Then, either there is a fixed point of H or the set $\left\{y\in Y|y=\mu H(y)\right\}$ is unbounded, where $\mu \in (0,1)$.

Lemma 9

([42]). Suppose that D is a bounded, convex, and closed subset of a Banach space, $0\in D$, and $N:D\to D$ is continuous. If $V=\overline{\mathit{conv}}N\left(V\right)$ or $V=N\left(V\right)\cup \left\{0\right\}$ can obtain $\nu \left(V\right)=0$ for every subset V of D, then N has a fixed point.

3. Main Results

In order to obtain the existence of fixed points for the solution operator, the continuity conditions, compactness conditions, and some growth conditions are given below.

(${\mathrm{H}}_1$) $f(t,z)$ is continuous with respect to $z\in X$ for almost all $t\in J$ and strongly measurable with respect to $t\in J$ for any $z\in X$;

(${\mathrm{H}}_1^{\prime }$) $f(t,z)$ is strongly measurable with respect to any $z\in X$ and almost all $t\in J$;

(${\mathrm{H}}_2$) $g:C(J,X)\to X$ is completely continuous, and there exists a constant $L>0$ such that

$$\parallel g\left(u\right)\parallel \le {L\parallel u\parallel }_{\infty }$$

for any $u\in C(J,X)$;

(${\mathrm{H}}_2^{\prime }$) For $g:C(J,X)\to X$, there exists a constant $L>0$ such that

$$\parallel g\left(u\right)\parallel \le {L\parallel u\parallel }_{\infty }$$

for any $u\in C(J,X)$;

(${\mathrm{H}}_3$) There exists a continuous function $\phi :J\to {\mathbb{R}}^{+}$ such that

$$\parallel f(t,z)\parallel \le \phi \left(t\right)\parallel z\parallel ,$$

for almost everywhere $t\in J$ and each $z\in X$;

(${\mathrm{H}}_4$) $T\left(t\right)$ is compact for $t>0$;

(${\mathrm{H}}_5$) For any bounded subset $X_1$ of X and each $t\in J$, there exists $a_1>0$ such that

$$\nu \left(g\left(X_1\right)\right)\le a_1\nu \left(X_1\right),$$

and $\nu \left(f(t,X_1)\right)\le \phi \left(t\right)\nu \left(X_1\right)$, where $\phi \left(t\right)$ is defined as in (${\mathrm{H}}_3$);

(${\mathrm{H}}_6$) There exists $L_g>0$ such that

$$\parallel g\left(y_1\right)-g\left(y_2\right)\parallel \le L_g{\parallel y_1-y_2\parallel }_{\infty },$$

and $\parallel f(t,y_1\left(t\right))-f(t,y_2\left(t\right))\parallel \le \phi \left(t\right)\parallel y_1-y_2{\parallel }_{\infty }$, where $\phi \left(t\right)$ is defined as in (${\mathrm{H}}_3$), $t\in J$, $y_1,y_2\in B_{k_2}$, and

$$B_{k_2}=\left\{u\in C(J,X)|\parallel u\parallel \le k_2,k_2=\frac{M(\parallel u_0\parallel +\parallel g\left(0\right)\parallel )(\alpha +\beta )}{(\alpha +\beta )(1-M{L}_g)-M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }}\right\}.$$

(23)

Theorem 1.

If ($H_1$)–($H_4$) are satisfied, then (11) has a mild solution provided

$$(\alpha +\beta )(1-ML)>M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }.$$

Proof. 

Let

$$k_1=\frac{M\parallel u_0\parallel (\alpha +\beta )}{(\alpha +\beta )(1-ML)-M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }}.$$

For any $u\in B_{k_1}$, by Lemma 3 and (${\mathrm{H}}_2$), we have

$$\parallel S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))\parallel \le M(\parallel u_0\parallel +k_{1}L).$$

(24)

Furthermore, due to (${\mathrm{H}}_1$), $f(t,u(t\left)\right)$ is a measurable function on J. By Lemma 3 and (${\mathrm{H}}_3$), we obtain

$$\begin{array}{cc}\hfill {\int }_0^t\parallel S_{\alpha ,\beta }(t-s)f(s,u\left(s\right))\parallel ds& \le M{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)\parallel u\left(s\right)\parallel ds\hfill \\ \hfill & \le M{k}_1{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)ds\hfill \\ \hfill & \le \frac{M{T}^{\alpha +\beta }{k}_1}{\alpha +\beta }{\parallel \phi \parallel }_{\infty }.\hfill \end{array}$$

(25)

Then, $\parallel S_{\alpha ,\beta }(t-s)f(s,u\left(s\right))\parallel$ is Lebesgue integrable with respect to $s\in J$ and $t\in J$, which implies that $S_{\alpha ,\beta }(t-s)f(s,u\left(s\right))$ is Bochner integrable with respect to $s\in J$ and $t\in J$ by Lemma 2.

Now, we can define an operator Q on $B_{k_1}$ as follows:

$$\left(Qu\right)\left(t\right)=S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))+{\int }_0^t{S}_{\alpha ,\beta }(t-s)f(s,u\left(s\right))ds.$$

Firstly, we prove that Q is a completely continuous operator. Suppose that

$$u_n,u\in B_{k_1}\mathrm{such}\mathrm{that}{u}_n\to u\mathrm{as}n\to \infty ,$$

(26)

then,

$$\begin{array}{cc}\hfill \parallel Q{u}_n\left(t\right)-Qu\left(t\right)\parallel & \le M\parallel g\left(u\right)-g\left(u_n\right)\parallel \hfill \\ \hfill & +M{\int }_0^t{(t-s)}^{\alpha +\beta -1}\parallel f(s,u_n\left(s\right))-f(s,u\left(s\right))\parallel ds\hfill \\ \hfill & =:I_1+I_2.\hfill \end{array}$$

Obviously, $I_1\to 0$ as $n\to \infty$ by (26) and (${\mathrm{H}}_2$). Since (${\mathrm{H}}_3$), we have

$$\begin{array}{cc}\hfill I_2& \le M{\int }_0^t{(t-s)}^{\alpha +\beta -1}(\parallel f(s,u_n\left(s\right))\parallel +\parallel f(s,u\left(s\right))\parallel )ds\hfill \\ \hfill & \le 2M{k}_1{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)ds\hfill \\ \hfill & \le \frac{2M{k}_1{T}^{\alpha +\beta }}{\alpha +\beta }{\parallel \phi \parallel }_{\infty }.\hfill \end{array}$$

Then, by the Lebesgue dominated convergence theorem [37] and (${\mathrm{H}}_1$), we obtain $I_2\to 0$ as $n\to \infty$. Therefore,

$$\parallel Q{u}_n\left(t\right)-Qu\left(t\right)\parallel \to 0\mathrm{as}n\to \infty .$$

That is, Q is continuous on $B_{k_1}$.

Next, we prove that $\left\{Qu\right|u\in B_{k_1}\}$ is relatively compact. It suffices to show that $\left\{Qu\right|u\in B_{k_1}\}$ is uniformly bounded and equicontinuous, and $\left\{Qu\left(t\right)\right|u\in B_{k_1}\}$ is relatively compact in X for any $t\in J$.

Equations (24) and (25) imply

$$\parallel \left(Qu\right)\left(t\right)\parallel \le k_1,$$

which means $\left\{Qu\right|u\in B_{k_1}\}$ is uniformly bounded. Take $u\in B_{k_1}$, and $0\le t_1<t_2\le T$; then, $\parallel Qu\left(t_2\right)-Qu\left(t_1\right)\parallel \le I_3+I_4+I_5$, where

$$\begin{array}{cc}\hfill & I_3=\parallel [S_{\alpha ,1-\alpha }\left(t_2\right)-S_{\alpha ,1-\alpha }\left(t_1\right)][u_0-g\left(u\right)]\parallel ,\hfill \\ \hfill & I_4={\int }_{t_1}^{t_2}\parallel S_{\alpha ,\beta }(t_2-s)f(s,u\left(s\right))\parallel ds,\hfill \\ \hfill & I_5={\int }_0^{t_1}\parallel [S_{\alpha ,\beta }(t_2-s)-S_{\alpha ,\beta }(t_1-s)]f(s,u\left(s\right))\parallel ds.\hfill \end{array}$$

It is easy to see that $I_3\to 0$ as $t_1\to t_2$ by Remark 4. By (${\mathrm{H}}_3$) and (16), we have

$$I_4\le M{\int }_{t_1}^{t_2}{(t_2-s)}^{\alpha +\beta -1}\phi \left(s\right)\parallel u\left(s\right)\parallel ds\le \frac{M{k}_1{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }{(t_2-t_1)}^{\alpha +\beta },$$

which implies that $I_4\to 0$ as $t_1\to t_2$.

$$\begin{array}{cc}\hfill I_5& \le k_1{\parallel \phi \parallel }_{\infty }{\int }_0^{t_1}{(t_2-s)}^{\alpha +\beta -1}\parallel R_{\alpha ,\beta }(t_2-s)-R_{\alpha ,\beta }(t_1-s)\parallel ds\hfill \\ \hfill & +M{k}_1{\parallel \phi \parallel }_{\infty }{\int }_0^{t_1}\left[{(t_1-s)}^{\alpha +\beta -1}-{(t_2-s)}^{\alpha +\beta -1}\right]ds\hfill \\ \hfill & \le k_1{\phi \parallel }_{\infty }{\int }_0^{t_1-ϵ}{(t_2-s)}^{\alpha +\beta -1}ds\underset{s\in [0,t_1-ϵ]}{sup}\parallel R_{\alpha ,\beta }(t_2-s)-R_{\alpha ,\beta }(t_1-s)\parallel \hfill \\ \hfill & +2M{k}_1{\phi \parallel }_{\infty }{\int }_{t_1-ϵ}^{t_1}{(t_2-s)}^{\alpha +\beta -1}ds\hfill \\ \hfill & +\frac{M{k}_1{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }\left[t_1^{\alpha +\beta }-t_2^{\alpha +\beta }+{(t_2-t_1)}^{\alpha +\beta }\right]\hfill \\ \hfill & \le \frac{k_1{\phi \parallel }_{\infty }}{\alpha +\beta }\left[t_2^{\alpha +\beta }-{(t_2-t_1+ϵ)}^{\alpha +\beta }\right]\underset{s\in [0,t_1-ϵ]}{sup}\parallel R_{\alpha ,\beta }(t_2-s)-R_{\alpha ,\beta }(t_1-s)\parallel \hfill \\ \hfill & +\frac{2M{k}_1{\phi \parallel }_{\infty }}{\alpha +\beta }\left[{(t_2-t_1+ϵ)}^{\alpha +\beta }-{(t_2-t_1)}^{\alpha +\beta }\right]+\frac{M{k}_1{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }\left[{(t_2-t_1)}^{\alpha +\beta }\right],\hfill \end{array}$$

where $ϵ>0$ is arbitrary. Then, $I_5\to 0$ as $t_1\to t_2$ and $ϵ\to 0$ by Lemma 4 and (16). Now, we can conclude that $\left\{Qu\right|u\in B_{k_1}\}$ is equicontinuous.

Obviously, due to (${\mathrm{H}}_2$), $\left\{Qu\left(0\right)\right|u\in B_{k_1}\}$ is relatively compact. By (${\mathrm{H}}_2$), (${\mathrm{H}}_4$), and Remark 5, we can similarly prove the compactness of $\left\{S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))\right|u\in B_{k_1}\}$ for $t\in (0,T]$. Due to (${\mathrm{H}}_1$), (${\mathrm{H}}_4$), and Lemma 5, U is compact for $t\in (0,T]$; then, $\overline{\mathrm{conv}\left(U\right)}$ is compact for $t\in (0,T]$ by Lemma 7, where

$$U=\left\{R_{\alpha ,\beta }(t-s)f(s,u\left(s\right))\right|s\in [0,t),u\in B_{k_1}\}.$$

By the Mean-Value Theorem for the Bochner integral [43] (Corollary 8, p. 48),

$${\int }_0^t{S}_{\alpha ,\beta }(t-s)f(s,u\left(s\right))ds\in \frac{t^{\alpha +\beta }}{\alpha +\beta }\overline{\mathrm{conv}\left(U\right)},t\in (0,T].$$

As a consequence, $\overline{\left\{Qu\left(t\right)\right|u\in B_{k_1}\}}$ is compact in X for all $t\in (0,T]$. Then, $\left\{Qu\left(t\right)\right|u\in B_{k_1}\}$ is relatively compact in X for any $t\in J$.

By the Arzela–Ascoli theorem [32], $\left\{Qu\right|u\in B_{k_1}\}$ is relatively compact. By combining this with the continuity of Q, we conclude that $Q:B_{k_1}\to B_{k_1}$ is completely continuous.

We set

$$M_1=\{u\in B_{k_1}|u=\eta Qu,\eta \in (0,1)\};$$

let us prove the boundedness of $M_1$. Apparently, $0\in M_1$. For $u\in M_1$, we have

$$\begin{array}{cc}\hfill \parallel u\left(t\right)\parallel & \le \eta \left[M\right(\parallel u_0\parallel +\parallel g\left(u\right)\parallel )+M{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)\parallel u\left(s\right)\parallel ds\hfill \\ \hfill & \le \eta [M(\parallel u_0\parallel +k_{1}L)+\frac{M{T}^{\alpha +\beta }{k}_1{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }]\hfill \\ \hfill & <M(\parallel u_0\parallel +k_{1}L)+\frac{M{T}^{\alpha +\beta }{k}_1{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }\hfill \end{array}$$

for $t\in J$. Therefore, Q has a fixed point by Lemma 8. That is, (11) has a mild solution. □

Theorem 2.

Under the assumptions ($H_1$), ($H_2^{\prime }$), ($H_3$), and ($H_5$), (11) has a mild solution provided $M\left(a_1+\frac{T^{\alpha +\beta }}{\alpha +\beta }{\parallel \phi \parallel }_{\infty }\right)<1$ and $(\alpha +\beta )(1-ML)>M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }$.

Proof. 

Similar to Theorem 1, $Q:B_{k_1}\to B_{k_1}$ is continuous, and $\left\{Qu\right|u\in B_{k_1}\}$ is uniformly bounded and equicontinuous, where $k_1$ is defined as in Theorem 1. Let $V\subset B_{k_1}$ such that $V\subset \overline{Q\left(V\right)}\cup \left\{0\right\}$; then, $v\left(t\right)=\nu \left(V\right(t\left)\right)$ is continuous for any $t\in J$ because of the boundness and equicontinuity of V. Then, by Lemma 1, (${\mathrm{H}}_5$), and (16), we have

$$\begin{array}{cc}\hfill {\parallel v\parallel }_{\infty }& \le \underset{t\in J}{sup}\parallel \nu (\left(QV\right)\left(t\right)\cup \left\{0\right\})\parallel \hfill \\ \hfill & \le \underset{t\in J}{sup}\parallel \nu \left(\left(QV\right)\left(t\right)\right)\parallel \hfill \\ \hfill & \le \nu \left(g\left(V\right)\right)\underset{t\in J}{sup}\parallel S_{\alpha ,1-\alpha }\left(t\right)\parallel \hfill \\ \hfill & +\underset{t\in J}{sup}{\int }_0^t\parallel S_{\alpha ,\beta }(t-s)\parallel \phi \left(s\right)\nu \left(V\left(s\right)\right)ds\hfill \\ \hfill & \le M{a}_1{\parallel v\parallel }_{\infty }+M\underset{t\in J}{sup}{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)v\left(s\right)ds\hfill \\ \hfill & \le M(a_1+\frac{T^{\alpha +\beta }}{\alpha +\beta }{\parallel \phi \parallel }_{\infty }{)\parallel v\parallel }_{\infty }.\hfill \end{array}$$

Since $M(a_1+\frac{T^{\alpha +\beta }}{\alpha +\beta }\parallel \phi {\parallel }_{\infty })<1$, we have ${\parallel v\parallel }_{\infty }=0$; namely, $v\left(t\right)=\nu \left(V\right(t\left)\right)=0$. Consequently, $V\left(t\right)$ is relatively compact in X. By the Arzela–Ascoli theorem [32], $\left\{Qu\right|u\in B_{k_1}\}$ is relatively compact, which means $\nu \left(V\right)=0$. Thus, the proof is completed by Lemma 9. □

Theorem 3.

If ($H_1^{\prime }$), ($H_3$), and ($H_6$) are satisfied, then (11) has a unique mild solution provided $(\alpha +\beta )(1-M{L}_g)>M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }$.

Proof. 

By (${\mathrm{H}}_6$), we have

$$\parallel S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))\parallel \le M(\parallel u_0\parallel +\parallel g\left(0\right)\parallel +k_2{L}_g),$$

and g is continuous, where $k_2$ is defined as in (23). Similar to Theorem 1, for $u\in B_{k_2}$, $S_{\alpha ,1-\alpha }\left(t\right)(u_0-g\left(u\right))$ exists, and $S_{\alpha ,\beta }(t-s)f(s,u\left(s\right))$ is Bochner integrable with respect to $s\in J$ and $t\in J$. We also have that Q maps $B_{k_2}$ into itself. Apparently, we only need to prove that Q has a unique fixed point on $B_{k_2}$. For any $u,v\in B_{k_2}$ and $t\in J$, according to (${\mathrm{H}}_6$), we have

$$\begin{array}{cc}\hfill \parallel \left(Qu\right)\left(t\right)-\left(Qv\right)\left(t\right)\parallel & \le \parallel S_{\alpha ,1-\alpha }\left(t\right)[g\left(v\right)-g\left(u\right)]\parallel \hfill \\ \hfill & +{\int }_0^t\parallel S_{\alpha ,\beta }(t-s)\parallel \parallel f(s,u\left(s\right))-f(s,v\left(s\right))\parallel ds\hfill \\ \hfill & \le M{L}_g{\parallel u-v\parallel }_{\infty }+M{\parallel u-v\parallel }_{\infty }{\int }_0^t{(t-s)}^{\alpha +\beta -1}\phi \left(s\right)ds\hfill \\ \hfill & \le (M{L}_g+\frac{M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }){\parallel u-v\parallel }_{\infty }.\hfill \end{array}$$

Since $M{L}_g+\frac{M{T}^{\alpha +\beta }{\parallel \phi \parallel }_{\infty }}{\alpha +\beta }<1$, by the Banach contraction principle, we conclude that Q has a unique fixed point on $B_{k_2}$, and the proof is complete. □

4. An Example

Let $X=L^2\left([0,\pi ]\right)$; we consider the following fractional integrodifferential diffusion equations

$$\left\{\begin{array}{c}{\partial }_t^{\frac{1}{3}}u(x,t)={\partial }_x^{2}u(x,t)+I_t^{\frac{1}{2}}f(x,t,u(x,t)),x\in [0,\pi ],t\in (0,T],\hfill \\ u(0,t)=u(\pi ,t)=0,t\in [0,T],\hfill \\ u(x,0)+{\displaystyle \sum _{i=0}^n}{\int }_0^{\pi }k(x,s)u(s,t_i)ds=u_0\left(x\right),x\in [0,\pi ],\hfill \end{array}\right.$$

(27)

where ${\partial }_t^{\frac{1}{3}}$ is the Caputo fractional partial derivative of order $\frac{1}{3}$, $I_t^{\frac{1}{2}}$ is the Riemann–Liouville integral of order $\frac{1}{2}$, $u_0\left(x\right)\in X$, and $k(x,s)\in L^2(\left[0,\pi \right]\times \left[0,\pi \right])$.

We set $Au=u^{″}$ with the domain

$$D\left(A\right)=\{u(·)\in X|u\mathrm{and}{u}^{\prime }\mathrm{as}\mathrm{absolutely}\mathrm{continuous},\mathrm{and} u^{″}\in X,u\left(0\right)=u\left(\pi \right)=0.\}.$$

Then, A generates a strongly continuous semigroup ${\left\{T\left(t\right)\right\}}_{t\ge 0}$, and

$$T\left(t\right)u=\sum _{n=1}^{\infty }{e}^{-n^{2}t}(u,y_n)y_n,$$

where $y_n\left(x\right)=\sqrt{\frac{2}{\pi }}\sin \left(nx\right)$ is the normalized eigenvector corresponding to the eigenvalues $n^2(n\in \mathbb{N})$ of A. Moreover, $\left({\mathrm{H}}_4\right)$ is satisfied, $T\left(t\right)$ is continuous in the uniform operator topology, and $\parallel T\left(t\right)\parallel \le e^{-t},t\ge 0$. We also found that the resolvent family $S_{\frac{1}{3},\frac{1}{2}}\left(t\right)$ is given by

$$\begin{array}{cc}\hfill S_{\frac{1}{3},\frac{1}{2}}\left(t\right)& =\frac{1}{\Gamma \left(\frac{1}{2}\right)}{\int }_0^t{(t-s)}^{-\frac{1}{2}}{s}^{-\frac{2}{3}}\left[\frac{1}{3}{\int }_0^{\infty }\theta {\zeta }_{\frac{1}{3}}\left(\theta \right)T(s^{\frac{1}{3}}\theta )d\theta \right]ds\hfill \\ \hfill & =\frac{1}{\sqrt{\pi }}{\int }_0^t{(t-s)}^{-\frac{1}{2}}{s}^{-\frac{2}{3}}\left[\sum _{n=1}^{\infty }{E}_{\frac{1}{3},\frac{1}{3}}(-n^2{s}^{\frac{1}{3}})(u,y_n)y_n\right]ds\hfill \\ \hfill & =\sum _{n=1}^{\infty }{t}^{-\frac{1}{6}}{E}_{\frac{1}{3},\frac{5}{6}}(-n^2{t}^{\frac{1}{3}})(u,y_n)y_n,t\ge 0,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \parallel S_{\frac{1}{3},\frac{1}{2}}\left(t\right)\parallel & \le \frac{1}{\Gamma \left(\frac{1}{2}\right)}{\int }_0^t{(t-s)}^{-\frac{1}{2}}{s}^{\frac{2}{3}}\frac{\frac{1}{3}}{\Gamma (1+\frac{1}{3})}ds\hfill \\ \hfill & =\frac{t^{-\frac{1}{6}}}{\Gamma \left(\frac{5}{6}\right)},t>0,\hfill \end{array}$$

where $E_{\gamma ,\eta }(·)$ is the Mittag–Leffler function for $\gamma$, $\eta >0$ [1].

We set $u\left(t\right)x=u(x,t)$, and $f(t,u(t\left)\right)x=f(x,t,u(x,t\left)\right)$, where $x\in [0,\pi ]$, $t\in [0,T]$, and $g:C\left(\right[0,T],X)\to X$ is given by

$$\left(gu\right)\left(x\right)=\sum _{i=0}^n{\int }_0^{\pi }k(x,s)u(s,t_i)ds,x\in [0,\pi ].$$

Then, (27) can be written as the following problem in X:

$$\left\{\begin{array}{c}{D}_t^{\frac{1}{3}}u\left(t\right)=Au\left(t\right)+I_t^{\frac{1}{2}}f(t,u\left(t\right)),t\in (0,T],\hfill \\ u\left(0\right)+g\left(u\right)=u_0.\hfill \end{array}\right.$$

(28)

We take $f(t,u(t\left)\right)=\sin u\left(t\right)$, $L=L_g=(n+1){\left[{\int }_0^{\pi }{\int }_0^{\pi }{k}^2(x,s)dsdx\right]}^{\frac{1}{2}}$, and note that g is completely continuous; then, $\left({\mathrm{H}}_1\right)$–$\left({\mathrm{H}}_3\right)$, $\left({\mathrm{H}}_5\right)$–$\left({\mathrm{H}}_6\right)$, $\left({\mathrm{H}}_1^{\prime }\right)$–$\left({\mathrm{H}}_2^{\prime }\right)$ are satisfied, where $\phi \left(t\right)=1$, $M_2=\frac{1}{\Gamma \left(\frac{5}{6}\right)}$, $a_1>0$ may be an arbitrary constant. Then by Theorem 1, or Theorem 2, or Theorem 3, (28) has a mild solution provided $T^{\frac{5}{6}}<\frac{5}{6}(\Gamma \left(\frac{5}{6}\right)-L)$.

5. Conclusions

This article investigates the nonlocal initial value problem for a class of fractional integrodifferential diffusion equations with Dirichlet boundary conditions. By selecting appropriate Banach spaces, we transform the original problem into an abstract Cauchy problem. At the same time, we prove that operator A can generate a resolvent family ${\left\{S_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$, by the Laplace transform, the convolution theorem, the probability density function ${\zeta }_{\alpha }\left(\theta \right)$, and semigroup ${\left\{T\left(t\right)\right\}}_{t\ge 0}$, we obtain

$$S_{\alpha ,\beta }\left(t\right)=\frac{\alpha }{\Gamma \left(\beta \right)}{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -1}{\int }_0^{\infty }\theta {\zeta }_{\alpha }\left(\theta \right)T\left(s^{\alpha }\theta \right)d\theta ds.$$

We set $R_{\alpha ,\beta }\left(t\right)=t^{1-\alpha -\beta }{S}_{\alpha ,\beta }\left(t\right)$; then, ${\left\{R_{\alpha ,\beta }\left(t\right)\right\}}_{t\ge 0}$ is strongly continuous, and $R_{\alpha ,\beta }\left(t\right)$ is continuous in the uniform topology and compact for $t>0$. It is worth noting that since A in [18] is $\omega$-sectorial ($\omega$ < 0) and the order of the equations is different from that in this paper, the proof that $\parallel S_{\alpha ,\beta }\left(t\right)\parallel$ is bounded cannot be directly applied in this paper. By the Laplace transform, the definition of a mild solution for (11) is given, and we prove the existence and uniqueness of the mild solutions through several fixed point theorems. In the future, we will further investigate the representation of solutions and the existence and regularity of solutions to fractional diffusion equations with multiple derivatives.