Criterion of the Existence of a Strongly Continuous Resolving Family for a Fractional Differential Equation with the Hilfer Derivative

Abstract

In the qualitative theory of differential equations in Banach spaces, the resolving families of operators of such equations play an important role. We obtained necessary and sufficient conditions for the existence of strongly continuous resolving families of operators for a linear homogeneous equation resolved with respect to the Hilfer derivative. These conditions have the form of estimates on derivatives of the resolvent of a linear closed operator from the equation and generalize the Hille–Yosida conditions for infinitesimal generators of $C_0$-semigroups of operators. Unique solvability theorems are proved for the corresponding inhomogeneous equations. Illustrative examples of the operators from the considered classes are constructed.

1. Introduction

The uniform well-posedness of the Cauchy problem for first-order differential equations is traditionally studied in terms of semigroups of operators. Briefly, a resolving $C_0$-continuous semigroup of operators $\{Z\left(t\right):t\in {\overline{\mathbb{R}}}_{+}\}$ of the equation $D^{1}z\left(t\right)=Az\left(t\right)$ exists if and only if the respective Cauchy problem $z\left(0\right)=z_0$ is uniformly well posed [1,2,3,4]. In this case, every operator $Z\left(t\right)$ maps the initial data $z_0$ to a value $z\left(t\right)$ of a solution of the problem at time $t\ge 0$. Such operators for a first-order equation form a semigroup, but for other equations, it is not so. But even in the absence of the semigroup property, the study of resolving families of operators of various differential equations allows us to obtain important qualitative results on their solutions. Therefore, the issue of the existence of a resolving family for a differential equation is very important. This issue is equivalent to the fulfillment of the criterion in the form of Hille–Yosida conditions for the operator A from a first-order equation or their generalizations for other equations. Strongly continuous resolving families and criteria of their existence have been studied for second-order equations [5,6,7], for some integro-differential equations [8], for evolution integral equations [9], for equations with the Gerasimov–Caputo derivative [10,11,12], with the Riemann–Liouville derivative [13,14], and with a distributed Gerasimov–Caputo derivative [15]. Various applications in physics of integro-differential equations and differential equations with fractional derivatives, including the Hilfer derivative, which is the focus of this work, can be found in [16,17,18].

Here, we study strongly continuous resolving families for a linear equation in a Banach space $\mathcal{Z}$:

$$D^{\alpha ,\beta }z\left(t\right)=Az\left(t\right),t\in {\mathbb{R}}_{+},$$

(1)

where $D^{\alpha ,\beta }$ is the Hilfer derivative [16] of an order $\alpha \in (m-1,m)$, $m\in \mathbb{N}$, of a kind $\beta \in [0,1]$, and A is a linear closed operator in $\mathcal{Z}$. The Cauchy-type problem

$$D^{k-(1-\beta )(m-\alpha )}z\left(0\right)=z_k,k=0,1,\dots ,m-1,$$

(2)

is considered for Equation (1). Hereafter, $D^{\gamma }z$ is the Riemann–Liouville fractional derivative for $\gamma >0$, and the Riemann–Liouville fractional integral for $\gamma \le 0$ (see below for details) is $D^{\gamma }z\left(0\right):=\underset{t\to 0+}{lim}{D}^{\gamma }z\left(t\right)$, $\gamma \in \mathbb{R}$.

In the second section, the main definitions are given, and some basic properties of operators from a resolving family for Equation (1) are proved. In the third section, it is shown that for $\alpha >2$, Equation (1) has a resolving family of operators in the case of a bounded operator A only. Hence, the only case of interest is $\alpha \in (0,2]$. The main result is proved here. It is a theorem on necessary and sufficient conditions of the existence of a resolving family of operators for Equation (1). These conditions generalize the Hille–Yosida conditions for $C_0$-continuous semigroups of operators. The most difficult issue is the sufficiency of these conditions; it was solved using Phillips-type approximations [1,19] and by applying the properties of operator A’s resolvents without assuming the presence of the Radon–Nikodym property in a considered Banach space. Due to the main theorem, problem (1) and (2) have a solution, and their uniqueness is proved. In the fourth section, we show the unique solution of existence for Cauchy-type problem (2) for the inhomogeneous equation $D^{\alpha }z\left(t\right)=Az\left(t\right)+f\left(t\right)$ under some assumptions on f. The fifth section contains the construction of a family of operators $A_{\gamma }$, $\gamma \in (-2,-\alpha ]\cup [\alpha ,2]$ such that for $\gamma \in (-2,-\alpha )\cup (\alpha ,2]$, where $\alpha \in (0,1]$ and $\beta \in [0,1]$, there exists a resolving family for Equation (1), and for $\gamma =\pm \alpha$, it is true only for $\alpha =1$ or $\beta =1$. The last section contains an application of this result for the investigation of an initial boundary value problem with a time fractional-order one-dimensional Schrödinger equation.

In the case of $\beta =1$, the obtained abstract results coincide with the corresponding results of [10,11,12]. Note the works [20,21] on equations with the Hilfer derivative and their resolving families of various classes.

2. Primary Results

Let $\mathcal{Z}$ be a Banach space. For $\delta >0$ and $h:{\mathbb{R}}_{+}\to \mathcal{Z}$, let

$$J^{\delta }h\left(t\right):={\displaystyle \frac{1}{\mathrm{\Gamma }\left(\delta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\delta -1}h\left(s\right)ds,t\in {\mathbb{R}}_{+},$$

be the Riemann–Liouville fractional integral, $J^{0}h\left(t\right):=h\left(t\right)$, $m-1<\alpha \le m\in \mathbb{N}$, $D^m$ be the usual derivative of the m-th order, and $D^{\alpha }h\left(t\right):=D^m{J}^{m-\alpha }h\left(t\right)$ be the Riemann–Liouville derivative of $h:{\mathbb{R}}_{+}\to \mathcal{Z}$. For $\alpha \le 0$, we will also denote $D^{\alpha }h\left(t\right):=J^{-\alpha }h\left(t\right)$.

Denote $D^{\gamma }h\left(0\right):=\underset{t\to 0+}{lim}{D}^{\gamma }h\left(t\right)$, $\gamma \in \mathbb{R}.$ The Hilfer fractional derivative of an order $\alpha \in (m-1,m]$, $m\in \mathbb{Z}$, and of a kind $\beta \in [0,1]$ is defined as

$$D^{\alpha ,\beta }h\left(t\right)=D^{m-\beta (m-\alpha )}\left(J^{(1-\beta )(m-\alpha )}h\left(t\right)-\sum _{k=0}^{m-1}{D}^{k-(1-\beta )(m-\alpha )}h\left(0\right){\displaystyle \frac{t^k}{k!}}\right)=$$

$$=D^m\left(J^{m-\alpha }h\left(t\right)-\sum _{k=0}^{m-1}{\displaystyle \frac{D^{k-(1-\beta )(m-\alpha )}h\left(0\right)t^{k+\beta (m-\alpha )}}{\mathrm{\Gamma }(k+\beta (m-\alpha )+1)}}\right).$$

(3)

For sufficiently smooth h, we have $D^{\alpha ,\beta }h\left(t\right):=J^{\beta (m-\alpha )}{D}^m{J}^{(1-\beta )(m-\alpha )}h\left(t\right)$. For $\alpha \le 0$, due to our designations, we obtain $D^{\alpha ,\beta }h\left(t\right):=J^{\beta (m-\alpha )}{J}^{-m}{J}^{(1-\beta )(m-\alpha )}h\left(t\right)=J^{-\alpha }h\left(t\right).$

Remark 1.

For $\beta =0$, the Hilfer fractional derivative coincides with the Riemann–Liouville derivative, and for $\beta =1$, the Hilfer derivative is the Gerasimov–Caputo fractional derivative (see [22,23,24]).

Denote the Laplace transform for $h:{\mathbb{R}}_{+}\to \mathcal{Z}$ as $\widehat{h}$ or $\mathbb{L}\left[h\right]$. Denote also ${\overline{\mathbb{R}}}_{+}:={\mathbb{R}}_{+}\cup \left\{0\right\}$. Hereafter, the principal branch of the power function will be used.

Lemma 1.

[20]. Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$, $h:{\mathbb{R}}_{+}\to \mathcal{Z}$ have the Laplace transform, and $J^{(1-\beta )(m-\alpha )}h\in C^{m-1}({\overline{\mathbb{R}}}_{+};\mathcal{Z})$. Then,

$$\widehat{D^{\alpha ,\beta }h}\left(\lambda \right)={\lambda }^{\alpha }\widehat{h}\left(\lambda \right)-\sum _{k=0}^{m-1}{D}^{k-(1-\beta )(m-\alpha )}h\left(0\right){\lambda }^{m-1-k-\beta (m-\alpha )}.$$

It is known (see, e.g., [25,26]) that

$$\widehat{J^{\alpha }h}\left(\lambda \right)={\lambda }^{-\alpha }\widehat{h}\left(\lambda \right),\widehat{D_t^{\alpha }h}\left(\lambda \right)={\lambda }^{\alpha }\widehat{h}\left(\lambda \right)-\sum _{k=0}^{m-1}{D}^{\alpha -m+k}h\left(0\right){\lambda }^{m-1-k}.$$

Let $\mathcal{L}\left(\mathcal{Z}\right)$ be the Banach space of all linear continuous operators from $\mathcal{Z}$ to $\mathcal{Z}$, and let $\mathcal{C}l\left(\mathcal{Z}\right)$ denote the set of all linear closed operators, densely defined in $\mathcal{Z}$, acting on space $\mathcal{Z}$. We supply the domain $D_A$ of an operator $A\in \mathcal{C}l\left(\mathcal{Z}\right)$ through the norm of its graph, and thus, we obtain the Banach space $D_A$.

We will use the notation $A{C}^m({\mathbb{R}}_{+};\mathcal{Z})$ for the set of all functions h that belong to $C^{m-1}({\overline{\mathbb{R}}}_{+};\mathcal{Z})$ and are absolutely continuous on every segment $[t_0,T]\subset {\overline{\mathbb{R}}}_{+}$$(m-1)$-th derivative.

Consider the Cauchy-type problem

$$D^{k-(1-\beta )(m-\alpha )}z\left(0\right)=z_k,k=0,1,\dots ,m-1,$$

(4)

for an equation

$$D^{\alpha ,\beta }z\left(t\right)=Az\left(t\right),t\in {\mathbb{R}}_{+},$$

(5)

with $\alpha \in (m-1,m]$, $m\in \mathbb{N}$, $A\in \mathcal{C}l\left(\mathcal{Z}\right)$. A solution for problem (4) and (5) is a function $z\in C({\mathbb{R}}_{+};D_A)\cap L_{1,\mathrm{loc}}({\mathbb{R}}_{+};\mathcal{Z})$ such that $J^{(1-\beta )(m-\alpha )}z\in C^{m-1}({\overline{\mathbb{R}}}_{+};\mathcal{Z})\cap A{C}^m({\mathbb{R}}_{+};\mathcal{Z})$, $D^{\alpha ,\beta }z\in C({\mathbb{R}}_{+};\mathcal{Z})$, and condition (4) and equality (5) for $t\in {\mathbb{R}}_{+}$ are valid.

Definition 1.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$. A family of operators $\{S\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ is called resolving for Equation (5) if the following conditions are satisfied:

(i) 

$\exists K\in {\mathbb{R}}_{+}$, $\exists \omega \in {\overline{\mathbb{R}}}_{+}$$\forall t\in {\mathbb{R}}_{+}$${\parallel S\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le K{e}^{\omega t}{t}^{(1-\beta )(\alpha -m)};$

(ii) 

For every $z_0\in \mathcal{Z}$$S\left(t\right)z_0\in C({\mathbb{R}}_{+};\mathcal{Z})$, $s-\underset{t\to 0+}{lim}{J}^{(1-\beta )(m-\alpha )}S\left(t\right)=I;$

(iii) 

$S\left(t\right)\left[D_A\right]\subset D_A$, $S\left(t\right)A{z}_0=AS\left(t\right)z_0$ for all $z_0\in D_A$, $t\in {\mathbb{R}}_{+};$

(iv) 

For every $z_0\in D_A$$S\left(t\right)z_0$ is a solution of the Cauchy-type problem $D^{(1-\beta )(\alpha -m)}z\left(0\right)=z_0$, $D^{k-(1-\beta )(m-\alpha )}z\left(0\right)=0$, $k=1,2,\dots ,m-1$, for Equation (5).

Remark 2.

It is easy to show that for any $z_0,z_1,\dots ,z_{m-1}\in D_A$, problem (4) and (5) has a solution ${\displaystyle \sum _{k=0}^{m-1}}{J}^{k}S\left(t\right)z_0$.

We write $A\in C^{\alpha ,\beta }\left(\omega \right)$ if a resolving family of operators for Equation (5) with a constant $\omega \in {\overline{\mathbb{R}}}_{+}$ in condition (i) exists.

Remark 3.

In the case of $A\in \mathcal{L}\left(\mathcal{Z}\right)$, Equation (5) has a resolving family of operators that can be calculated (see, e.g., [27]):

$$S\left(t\right)=\sum _{l=0}^{\infty }{\displaystyle \frac{t^{\alpha l+(1-\beta )(\alpha -m)}{A}^l}{\mathrm{\Gamma }(\alpha l+(1-\beta \left)\right(\alpha -m)+1)}}=t^{(1-\beta )(\alpha -m)}{E}_{\alpha ,(1-\beta )(\alpha -m)+1}\left(t^{\alpha }A\right),t\in {\mathbb{R}}_{+},$$

where $E_{\alpha ,\beta }\left(z\right):={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{z^n}{\mathrm{\Gamma }(\alpha n+\beta )}}$ is the Mittag–Leffler function. Thus, $\mathcal{L}\left(\mathcal{Z}\right)\subset C^{\alpha ,\beta }\left(\omega \right)$ with some ω depending on ${\parallel A\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}$.

Definition 2.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$. An operator $A\in \mathcal{C}l\left(\mathcal{Z}\right)$ is called the operator of the class ${\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$ for a some constant $\omega \in {\overline{\mathbb{R}}}_{+}$ if the following two conditions are fulfilled:

(i) If $\mathrm{Re}\lambda >\omega$, then ${\lambda }^{\alpha }\in \rho \left(A\right):=\{\mu \in \mathbb{C}:{(\mu -A)}^{-1}\in \mathcal{L}\left(\mathcal{Z}\right)\};$

(ii) $\exists K\in {\mathbb{R}}_{+}$$\forall \lambda \in \{\mathrm{Re}\lambda >\omega \}$$\forall n\in {\mathbb{N}}_0:=\mathbb{N}\cup \left\{0\right\}$

$${\displaystyle {∥\frac{d^n}{d{\lambda }^n}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}\right)∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+n+1)}{{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+n+1}}.}$$

Remark 4.

Conditions (i) and (ii) generalize the Hille–Yosida conditions and agree with them if $\alpha =m=1$. Indeed, in this case, we can use the formula for the n-th-order derivative of the resolvent and obtain

$${∥{\displaystyle \frac{d^n}{d{\lambda }^n}}{(\lambda -A)}^{-1}∥}_{\mathcal{L}\left(\mathcal{Z}\right)}={∥{(-1)}^{n}n!{(\lambda -A)}^{-n-1}∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le {\displaystyle \frac{K\mathrm{\Gamma }(n+1)}{{(\mathrm{Re}\lambda -\omega )}^{n+1}}}={\displaystyle \frac{Kn!}{{(\mathrm{Re}\lambda -\omega )}^{n+1}}}.$$

Remark 5.

In Theorem 2.8 [11], it was proved that the equation $D^{\alpha ,1}z\left(t\right)=Az\left(t\right)$ has a resolving family of operators, where

$$D^{\alpha ,1}z\left(t\right):=D^{\alpha }\left(z\left(t\right)-\sum _{k=0}^{m-1}{D}^{k}z\left(0\right){\displaystyle \frac{t^k}{k!}}\right)$$

is the Gerasimov–Caputo derivative if and only if $A\in {\mathcal{C}}_{\alpha ,1}\left(\omega \right)$.

Lemma 2.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$. Then, ${\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)\subset {\mathcal{C}}_{\alpha ,1}\left(\omega \right)$.

Proof. 

Denote $H_{\beta }\left(\lambda \right):={\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }I-A)}^{-1}$ for $\lambda >\omega$, $\beta \in [0,1]$. For $n\in \mathbb{N}$, $\mathrm{Re}\lambda >\omega$, we have

$${\displaystyle \frac{d^n}{d{\lambda }^n}}{\lambda }^{\alpha -1}{({\lambda }^{\alpha }I-A)}^{-1}={\displaystyle \frac{d^n}{d{\lambda }^n}}{\lambda }^{(1-\beta )(\alpha -m)}{H}_{\beta }\left(\lambda \right)=\sum _{k=0}^n{C}_n^k{\displaystyle \frac{d^{n-k}}{d{\lambda }^{n-k}}}{\lambda }^{(1-\beta )(\alpha -m)}{\displaystyle \frac{d^k}{d{\lambda }^k}}{H}_{\beta }\left(\lambda \right)=$$

$$=\sum _{k=0}^n{C}_n^k(1-\beta )(\alpha -m)((1-\beta )(\alpha -m)-1)\dots ((1-\beta )(\alpha -m)-n+k+1)\times$$

$$\times {\lambda }^{(1-\beta )(\alpha -m)-n+k}{\displaystyle \frac{d^k}{d{\lambda }^k}}{H}_{\beta }\left(\lambda \right),$$

$$\begin{array}{c}\hfill {∥{\displaystyle \frac{d^n}{d{\lambda }^n}}{\lambda }^{\alpha -1}{({\lambda }^{\alpha }I-A)}^{-1}∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le K\sum _{k=0}^n{C}_n^k(1-\beta )(m-\alpha )((1-\beta )(m-\alpha )+1)\cdots \times \\ \hfill \times ((1-\beta )(m-\alpha )+n-k-1)\times {\left|\lambda \right|}^{(1-\beta )(\alpha -m)-n+k}{\displaystyle \frac{\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+k+1)}{{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+k+1}}}\le \end{array}$$

$$\le {\displaystyle \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+1)}{{(\mathrm{Re}\lambda -\omega )}^{n+1}}}\sum _{k=0}^n{C}_n^k(1-\beta )(m-\alpha )((1-\beta )(m-\alpha )+1)\cdots \times$$

$$\times \left(\right(1-\beta \left)\right(m-\alpha )+n-k-1)\left(\right(1-\beta \left)\right(\alpha -m)+k)\left(\right(1-\beta \left)\right(\alpha -m)+k-1)\cdots \times$$

$$\times ((1-\beta )(\alpha -m)+1)={\displaystyle \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+1)n!}{{(\mathrm{Re}\lambda -\omega )}^{n+1}}},$$

due to equality (10) [13] with $(1-\beta )(m-\alpha )$ instead of $\alpha$. The proof is completed. □

Remark 6.

The assertion of Lemma 2 for $\alpha \in (0,1)$, $\beta =0$ was proved in [13].

Lemma 3.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$, $A\in C^{\alpha ,\beta }\left(\omega \right)$, and let $\{S\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ be a resolving family of operators. Then, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$ and

$$\widehat{S}\left(\lambda \right)={\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }I-A)}^{-1},\mathrm{Re}\lambda >\omega .$$

(6)

Proof. 

For $\mathrm{Re}\lambda >\omega$, $z_0\in D_A$, it follows from Lemma 1 that

$$\mathbb{L}\left[D^{\alpha }S\left(t\right)z_0\right]\left(\lambda \right)z_0={\lambda }^{\alpha }\widehat{S}\left(\lambda \right)z_0-{\lambda }^{m-1-\beta (m-\alpha )}{z}_0=A\widehat{S}\left(\lambda \right)z_0=\widehat{S}\left(\lambda \right)A{z}_0$$

due to conditions (iii) and (iv) of Definition 1 and the closedness of A. This implies the invertibility of the operator ${\lambda }^{\alpha }I-A:D_A\to \mathcal{Z}$ and the fulfilment of (6). The boundedness of $\widehat{S}\left(\lambda \right)$ follows from the obtained inclusion of ${\lambda }^{\alpha }\in \rho \left(A\right)$.

If we differentiate the equality

$$\underset{0}{\overset{\infty }{\int }}S\left(t\right)e^{-\lambda t}dt={\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }I-A)}^{-1}$$

with respect to $\lambda$, we obtain, for $\mathrm{Re}\lambda >\omega$, $n\in {\mathbb{N}}_0$

$$\begin{array}{c}\hfill {∥{\displaystyle \frac{d^n}{d{\lambda }^n}}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }I-A)}^{-1}\right)∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le K\underset{0}{\overset{\infty }{\int }}{t}^{(1-\beta )(\alpha -m)+n}{e}^{(\omega -\mathrm{Re}\lambda )t}dt=\\ \hfill =K\mathbb{L}\left[t^{(1-\beta )(\alpha -m)+n}\right](\mathrm{Re}\lambda -\omega )={\displaystyle \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+n+1)}{{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+n+1}}},\end{array}$$

Consequently, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right).$ □

Corollary 1.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$. Then, $C^{\alpha ,\beta }\left(\omega \right)\subset {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$.

Lemma 4.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. Then, for every $r\in \mathbb{N}$, $z_0\in \mathcal{Z}$,

$$\underset{n\to \infty }{lim}{\left(n^{\alpha }{(n^{\alpha }-A)}^{-1}\right)}^r{z}_0=z_0.$$

Proof. 

Due to the condition $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, for $n>\omega$, the inequalities

$$\parallel {(n^{\alpha }-A)}^{-1}{\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le {\displaystyle \frac{K}{n^{m-1-\beta (m-\alpha )}{(n-\omega )}^{(1-\beta )(\alpha -m)+1}}}$$

and $\parallel n^{\alpha }{(n^{\alpha }-A)}^{-1}{\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le C_1$ are valid. For $z_0\in D_A$,

$$\underset{n\to \infty }{lim}{n}^{\alpha }{(n^{\alpha }-A)}^{-1}{z}_0=z_0+\underset{n\to \infty }{lim}{(n^{\alpha }-A)}^{-1}A{z}_0=z_0.$$

The density in $\mathcal{Z}$ of the domain $D_A$ implies that $\underset{n\to \infty }{lim}{n}^{\alpha }{(n^{\alpha }-A)}^{-1}{z}_0=z_0$ for every $z_0\in \mathcal{Z}$.

Moreover, for $z_0\in \mathcal{Z}$,

$$\parallel ({(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^2-n^{\alpha }{(n^{\alpha }-A)}^{-1})z_0{\parallel }_{\mathcal{Z}}\le C_1{\parallel n^{\alpha }{(n^{\alpha }-A)}^{-1}{z}_0-z_0\parallel }_{\mathcal{Z}}\to 0,n\to \infty ,$$

Hence,

$$\underset{n\to \infty }{lim}{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^2{z}_0=\underset{n\to \infty }{lim}({(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^2-n^{\alpha }{(n^{\alpha }-A)}^{-1})z_0$$

$$+\underset{n\to \infty }{lim}{n}^{\alpha }{(n^{\alpha }-A)}^{-1}{z}_0=z_0.$$

Analogously, we can obtain the required statement for every $r\in \mathbb{N}$. □

Corollary 2.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. Then, for arbitrary $r\in \mathbb{N}$, the domain $D_{A^r}$ is dense in the space $\mathcal{Z}$.

Proof. 

Consider $z_0\in \mathcal{Z}$. Then, ${(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0\in D_{A^r}$ and $\underset{n\to \infty }{lim}{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0=z_0.$ □

Corollary 3.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. Then, for arbitrary $r\in \mathbb{N}$, the domain $D_{A^r}$ is dense in the space $D_A$ with the graph norm of operator A.

Proof. 

Due to Lemma 4, for $z_0\in D_A$, we obtain $\underset{n\to \infty }{lim}{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0=z_0$ and

$$\underset{n\to \infty }{lim}A{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0=\underset{n\to \infty }{lim}{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^{r}A{z}_0=A{z}_0.$$

Thus,

$$\parallel {(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0-z_0{\parallel }_{D_A}={\parallel {(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0-z_0\parallel }_{\mathcal{Z}}+$$

$$+\parallel A{(n^{\alpha }{(n^{\alpha }-A)}^{-1})}^r{z}_0-A{z}_0{\parallel }_{\mathcal{Z}}\to 0,n\to \infty .$$

□

Theorem 1.

Let $m-1<\alpha \le m\in \mathbb{N}$, $\beta \in [0,1]$. There exist a resolving family of operators $\{S\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ for Equation (5). Then, the family of operators $\{J^{(1-\beta )(m-\alpha )}S\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\overline{\mathbb{R}}}_{+}\}$ is continuous in the norm of $\mathcal{L}\left(\mathcal{Z}\right)$ at the point $t=0$ if and only if $A\in \mathcal{L}\left(\mathcal{Z}\right)$.

Proof. 

Consider $t\in {\mathbb{R}}_{+}$. Then,

$$\begin{array}{cc}\hfill \parallel J^{(1-\beta )(m-\alpha )}{S\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}& \le K\underset{0}{\overset{t}{\int }}{\displaystyle \frac{{(t-s)}^{(1-\beta )(m-\alpha )-1}}{\mathrm{\Gamma }\left(\right(1-\beta \left)\right(m-\alpha \left)\right)}}{s}^{(1-\beta )(\alpha -m)}{e}^{\omega s}ds\le \hfill \\ \hfill & \le K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)e^{\omega t}.\hfill \end{array}$$

For $\mathrm{Re}\lambda >\omega$, by Lemma 3, $\mathbb{L}\left[J^{(1-\beta )(m-\alpha )}S\right]\left(\lambda \right)={\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}$; hence,

$${\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}-{\displaystyle \frac{I}{\lambda }}=\underset{0}{\overset{\infty }{\int }}{e}^{-\lambda t}(J^{(1-\beta )(m-\alpha )}S(t)-I)dt.$$

By the assumption of the theorem, the function $\eta \left(t\right)={\parallel J^{(1-\beta )(m-\alpha )}S\left(t\right)-I\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}$ is continuous on $[0,1]$; in addition, $\eta \left(0\right)=0$. For any $\epsilon >0$, take a number $\delta >0$ such that $\eta \left(t\right)\le \epsilon$ for every $t\in [0,\delta ]$; therefore,

$${∥{\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}-{\displaystyle \frac{I}{\lambda }}∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le \underset{0}{\overset{\delta }{\int }}{e}^{-\lambda t}\eta \left(t\right)dt+\underset{\delta }{\overset{\infty }{\int }}{e}^{-\lambda t}\eta \left(t\right)dt\le {\displaystyle \frac{\epsilon }{\lambda }}+o\left({\displaystyle \frac{1}{\lambda }}\right),\mathrm{Re}\lambda \to +\infty .$$

Here, we take into account that $\eta \left(t\right)\le K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)e^{\omega t}+1$ for all $t\ge 0$. Hence, for sufficiently large $\mathrm{Re}\lambda$, we have $\parallel {\lambda }^{\alpha }{({\lambda }^{\alpha }-A)}^{-1}{-I\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}<1$, and there exists a continuous inverse operator for ${\lambda }^{\alpha }{({\lambda }^{\alpha }-A)}^{-1}$. Thus, $A\in \mathcal{L}\left(\mathcal{Z}\right)$.

Now, suppose that $A\in \mathcal{L}\left(\mathcal{Z}\right)$. Hence, the inverse Laplace transform of ${\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}$ is defined for $\left|\lambda \right|>{\parallel A\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}^{1/\alpha }$. Take ${\mathrm{\Gamma }}_R:=\{R{e}^{i\phi }\in \mathbb{C}:\phi \in (-\pi ,\pi )\}\cup \{r{e}^{i\pi }\in \mathbb{C}:r\in [R,\infty )\}\cup \{r{e}^{-i\pi }\in \mathbb{C}:r\in [R,\infty )\}$ for $R>{\parallel A\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}^{1/\alpha }$. Then, for $t\in {\mathbb{R}}_{+}$, we have equalities

$$J^{(1-\beta )(m-\alpha )}S\left(t\right)={\displaystyle \frac{1}{2\pi i}}\underset{{\mathrm{\Gamma }}_R}{\int }{\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}d\lambda =$$

$$={\displaystyle \frac{1}{2\pi i}}\underset{{\mathrm{\Gamma }}_R}{\int }\sum _{n=0}^{\infty }{\lambda }^{-\alpha n-1}{A}^n{e}^{\lambda t}d\lambda =\sum _{n=0}^{\infty }{\displaystyle \frac{t^{\alpha n}{A}^n}{\mathrm{\Gamma }(\alpha n+1)}}=E_{\alpha ,1}\left(t^{\alpha }A\right),$$

which imply that

$$\parallel J^{(1-\beta )(m-\alpha )}{S\left(t\right)-I\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le \sum _{n=1}^{\infty }{\displaystyle \frac{t^{\alpha n}{\parallel A\parallel }_{\mathcal{L}(\mathcal{Z})}^n}{\mathrm{\Gamma }(\alpha n+1)}}=t^{\alpha }{\parallel A\parallel }_{\mathcal{L}(\mathcal{Z})}{E}_{\alpha ,\alpha +1}(t^{\alpha }{\parallel A\parallel }_{\mathcal{L}(\mathcal{Z})})\to 0,$$

as $t\to 0+.$ □

3. Existence Criterion of a Resolving Family of Operators

Theorem 2.

Let $\beta \in [0,1]$ and $A\in C^{\alpha ,\beta }\left(\omega \right)$ with $\alpha >2$. Then, $A\in \mathcal{L}\left(\mathcal{Z}\right)$.

Proof. 

If $A\in C^{\alpha ,\beta }\left(\omega \right)$, then Corollary 1 implies that $F_{\alpha ,\omega }:=\{{\lambda }^{\alpha }:\mathrm{Re}\lambda >\omega ,|arg\lambda |\le {\displaystyle \frac{\pi }{\alpha }}\} \subset \rho \left(A\right)$. Therefore, the complex plane without some bounded set belongs to the resolvent set $\rho \left(A\right)$. Hence, for sufficiently large $\left|\mu \right|$, it is necessary that $\mu ={\lambda }^{\alpha }\in F_{\alpha ,\omega }$. Then, by Corollary 1,

$$\parallel \mu R_{\mu }{\left(A\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le {\displaystyle \frac{{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\left|\lambda \right|}^{\alpha -m+1+\beta (m-\alpha )}}{{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+1}}}\le$$

$$\le {\displaystyle \frac{{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\left|\lambda \right|}^{(1-\beta )(\alpha -m)+1}}{\left(\right|\lambda |cos\frac{\pi }{\alpha }{-\omega )}^{(1-\beta )(\alpha -m)+1}}}\to {\displaystyle \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+1)}{{\left(\cos \frac{\pi }{\alpha }\right)}^{(1-\beta )(\alpha -m)+1}}},\left|\mu \right|\to \infty .$$

Due to Lemma 5.2 [7], operator A is bounded. □

Let $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, for $\mathrm{Re}\lambda >\omega$$H_{\beta }\left(\lambda \right):={\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}$. For $n\in \mathbb{N}$ and $t\in {\mathbb{R}}_{+}$, define using the Phillips approximations [1] (Theorem 6.3.3) for the inverse Laplace transform as follows:

$${\mathcal{S}}_n\left(t\right):=e^{-nt}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(n(n+\omega )t\right)}^{k+1}}{k!(k+1)!}}{H}_{\beta }^{\left(k\right)}(n+\omega ).$$

This series is uniformly convergent on all segments $[1/T,T]$, with any $T>0$; moreover, for $t\in {\mathbb{R}}_{+}$,

$$\begin{array}{cc}\hfill & \parallel t^{(1-\beta )(m-\alpha )}{\mathcal{S}}_n{\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le \hfill \\ \hfill & \le K{e}^{-nt}{\left(nt\right)}^{(1-\beta )(m-\alpha )}\sum _{k=0}^{\infty }{\displaystyle \frac{\mathrm{\Gamma }((1-\beta )(\alpha -m)+k+1){\left((n+\omega )t\right)}^{k+1}}{k!(k+1)!}}\le \hfill \\ \hfill & \le C_1{e}^{-nt}{\left(nt\right)}^{(1-\beta )(m-\alpha )}\sum _{k=0}^{\infty }{\displaystyle \frac{{\left((n+\omega )t\right)}^{k+1}}{\mathrm{\Gamma }(k+2+(1-\beta \left)\right(m-\alpha \left)\right)}}\le \hfill \\ \hfill & \le C_1{e}^{-nt}{\left(nt\right)}^{(1-\beta )(m-\alpha )}{E}_{1,(1-\beta )(m-\alpha )+1}\left((n+\omega )t\right)\le \hfill \\ \hfill & \le C_1{e}^{\omega t}{e}^{-(n+\omega )t}{\left((n+\omega )t\right)}^{(1-\beta )(m-\alpha )}{E}_{1,(1-\beta )(m-\alpha )+1}\left((n+\omega )t\right)\le \hfill \\ \hfill & \le C_1{e}^{\omega t}\underset{x>0}{sup}{e}^{-x}{x}^{(1-\beta )(m-\alpha )}{E}_{1,(1-\beta )(m-\alpha )+1}\left(x\right),\hfill \end{array}$$

Due to the asymptotic expansion of the Euler gamma function [28] (§1.18)

$$\mathrm{\Gamma }\left(z\right)\sim {\displaystyle \frac{\sqrt{2\pi }{z}^z}{\sqrt{z}{e}^z}},\left|z\right|\to \infty ,|argz|<\pi ,$$

we have

$${\displaystyle \frac{\mathrm{\Gamma }(k+1-\delta )}{\mathrm{\Gamma }(k+1)}}\sim C_2{k}^{-\delta },{\displaystyle \frac{\mathrm{\Gamma }(k+2)}{\mathrm{\Gamma }(k+2+\delta )}}\sim C_3{k}^{-\delta },$$

$${\displaystyle \frac{\mathrm{\Gamma }(k+1-\delta )}{k!(k+1)!}}={\displaystyle \frac{\mathrm{\Gamma }(k+1-\delta )}{\mathrm{\Gamma }(k+1)\mathrm{\Gamma }(k+2)}}\le {\displaystyle \frac{C_1}{\mathrm{\Gamma }(k+2+\delta )}},\delta =(1-\beta )(m-\alpha ).$$

The asymptotics of the Mittag–Leffler function [11] (p. 12)

$$E_{\beta ,\gamma }\left(s\right)={\displaystyle \frac{1}{\beta }}{s}^{{\displaystyle \frac{1-\gamma }{\beta }}}{e}^{s^{1/\beta }}+O\left(s^{-1}\right),s\to +\infty ,$$

imply that $E_{1,(1-\beta )(m-\alpha )+1}\left(x\right)\le C_4{x}^{(1-\beta )(\alpha -m)}{e}^x$ and $\parallel t^{(1-\beta )(m-\alpha )}{\mathcal{S}}_n{\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le C_5{e}^{\omega t}.$ Consequently, for all $t\in {\mathbb{R}}_{+}$,

$$\parallel {\mathcal{S}}_n{\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le C_5{t}^{(1-\beta )(\alpha -m)}{e}^{\omega t}.$$

(7)

It is not difficult to prove the infinite differentiability of ${\mathcal{S}}_n\left(t\right)$ at every $t\in {\mathbb{R}}_{+}$.

Lemma 5.

Let $m-1<\alpha \le m\in \{1,2\}$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. Then, there exists the strong limit $s-\underset{n\to \infty }{lim}{\mathcal{S}}_n\left(t\right)$, which is uniform with respect to t on all segments of the form $[1/T,T]$, $T>0$.

Proof. 

Take in further considerations $r\in \mathbb{N}$, for which $r-1\le 1/\alpha <r$; hence, $(r-1)\alpha \le 1<r\alpha .$ If $z_0\in D_{A^r}$, then we have

$$\begin{array}{cc}\hfill H_{\beta }\left(\lambda \right)z_0& ={\lambda }^{m-1-\beta (m-\alpha )-\alpha }{z}_0+{\lambda }^{m-1-\beta (m-\alpha )-\alpha }{({\lambda }^{\alpha }-A)}^{-1}A{z}_0=\dots =\hfill \\ \hfill & =\sum _{k=1}^r{\lambda }^{m-1-\beta (m-\alpha )-k\alpha }{A}^{k-1}{z}_0+{\lambda }^{m-1-\beta (m-\alpha )-r\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^r{z}_0,\hfill \end{array}$$

Therefore,

$${∥H_{\beta }\left(\lambda \right)z_0-\sum _{k=1}^r{\lambda }^{m-1-\beta (m-\alpha )-k\alpha }{A}^{k-1}{z}_0∥}_{\mathcal{Z}}\le {\displaystyle \frac{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\parallel A^r{z}_0{\parallel }_{\mathcal{Z}}}{{\left|\lambda \right|}^{r\alpha }{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+1}}}$$

and the integral

$$\mathrm{v}.\mathrm{p}.{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{e}^{\lambda t}{H}_{\beta }\left(\lambda \right)z_{0}d\lambda =\sum _{k=1}^r{\displaystyle \frac{t^{k\alpha -m+\beta (m-\alpha )}{z}_0}{\mathrm{\Gamma }(k\alpha -m+\beta (m-\alpha )+1)}}+$$

$$+{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{e}^{\lambda t}\left(H_{\beta }\left(\lambda \right)z_0-\sum _{k=1}^r{\lambda }^{m-1-\beta (m-\alpha )-k\alpha }{A}^{k-1}{z}_0\right)d\lambda ,s>\omega ,$$

converges.

For $\mathrm{Re}\lambda >\omega$, we have

$$\begin{array}{c}{\widehat{\mathcal{S}}}_n\left(\lambda \right)=\underset{0}{\overset{\infty }{\int }}{e}^{-(n+\lambda )t}\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(n(n+\omega )t\right)}^{k+1}}{k!(k+1)!}}{H}_{\beta }^{\left(k\right)}(n+\omega )dt=\\ =\sum _{k=0}^{\infty }{\displaystyle \frac{{(-1)}^k{\left(n(n+\omega )\right)}^{k+1}{H}_{\beta }^{\left(k\right)}(n+\omega )}{k!{(n+\lambda )}^{k+2}}}=\\ ={\displaystyle \frac{n(n+\omega )}{{(n+\lambda )}^2}}\sum _{k=0}^{\infty }{\left({\displaystyle \frac{\lambda (n+\omega )}{n+\lambda }}-(n+\omega )\right)}^k{\displaystyle \frac{H_{\beta }^{\left(k\right)}(n+\omega )}{k!}}={\displaystyle \frac{n(n+\omega )}{{(n+\lambda )}^2}}{H}_{\beta }\left({\displaystyle \frac{\lambda (n+\omega )}{n+\lambda }}\right),\end{array}$$

Hence, $\underset{n\to \infty }{lim}{\widehat{\mathcal{S}}}_n\left(\lambda \right)=H_{\beta }\left(\lambda \right)$, since $H_{\beta }\left(\lambda \right)$ is analytic on $\{\lambda \in \mathbb{C}:\mathrm{Re}\lambda >\omega \}$.

If $\alpha \in (0,2]$, $\mathrm{Re}\lambda >\omega +\nu$, $\nu >0$, then for sufficiently large $n\in \mathbb{N}$,

$$\begin{array}{cc}\hfill & {∥{\widehat{\mathcal{S}}}_n\left(\lambda \right)∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le {\displaystyle \frac{{(n+\omega )}^2}{{|n+\lambda |}^2}}{∥H_{\beta }\left({\displaystyle \frac{\lambda (n+\omega )}{n+\lambda }}\right)∥}_{\mathcal{L}\left(\mathcal{Z}\right)}\le \hfill \\ \hfill & \le {\displaystyle \frac{{(n+\omega )}^2}{{|n+\lambda |}^2}}{\displaystyle \frac{K\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+1)}{{\left(\mathrm{Re}\frac{\lambda (n+\omega )}{n+\lambda }-\omega \right)}^{(1-\beta )(\alpha -m)+1}}}\le {\displaystyle \frac{C_1{(1+\frac{\omega }{n})}^2}{|\frac{1}{n}+\frac{1}{\lambda }{|}^2{\left|\lambda \right|}^2}}\le {\displaystyle \frac{C_2}{{\left|\lambda \right|}^2}}.\hfill \end{array}$$

If $s>\omega$, $z_0\in D_{A^r}$, we pass limit $n\to \infty$ in the equality

$${\mathcal{S}}_n\left(t\right)z_0=\mathrm{v}.\mathrm{p}.{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{e}^{\lambda t}{\widehat{\mathcal{S}}}_n\left(\lambda \right)z_{0}d\lambda ,t\in {\mathbb{R}}_{+}.$$

The domain $D_{A^r}$ is dense in the space $\mathcal{Z}$ due to Corollary 2. In addition, the family $\{{\mathcal{S}}_n\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ is uniformly bounded on segments $[1/T,T]$, $T>0$. Hence, the convergence of $S_n\left(t\right)z_0$ is strong and uniform with respect to t on segments $[1/T,T]$. □

Denote $Z\left(t\right):=s-\underset{n\to \infty }{lim}{\mathcal{S}}_n\left(t\right)$, $t\in {\mathbb{R}}_{+}$. Due to (7),

$${\parallel Z\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le C{t}^{(1-\beta )(\alpha -m)}{e}^{\omega t}.$$

(8)

Lemma 5 implies that $\{Z\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ is a strongly continuous family of operators. Due to the proof of Lemma 5, we have for $s>\omega$,

$$Z\left(t\right)z_0=\mathrm{v}.\mathrm{p}.{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{e}^{\lambda t}{H}_{\beta }\left(\lambda \right)z_{0}d\lambda ,z_0\in D_{A^r}.$$

Theorem 3.

Let $m-1<\alpha \le m\in \{1,2\}$, $\beta \in [0,1]$. Then, $C^{\alpha ,\beta }\left(\omega \right)={\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$.

Proof. 

The inclusion $C^{\alpha ,\beta }\left(\omega \right)\subset {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$ is proved in Corollary 1. Let $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. In this case, as it is proved above, there exists the strongly continuous family $\{Z\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$.

If $\mathrm{Re}\lambda >\omega$, then the derivatives $H_{\beta }^{\left(n\right)}\left(\lambda \right):={\displaystyle \frac{d^n}{d{\lambda }^n}}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}\right)$ are sums of natural powers of ${({\lambda }^{\alpha }I-A)}^{-1}$, which are multiplied by scalar functions. Consequently,

$${\displaystyle \frac{d^n}{d{\lambda }^n}}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}\right)z_0\in D_A,z_0\in \mathcal{Z},$$

$$A{\displaystyle \frac{d^n}{d{\lambda }^n}}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}\right)z_0={\displaystyle \frac{d^n}{d{\lambda }^n}}\left({\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}\right)A{z}_0,z_0\in D_A.$$

Due to these relations, ${\mathcal{S}}_n\left(t\right)z_0\in D_A$ and $A{\mathcal{S}}_n\left(t\right)z_0={\mathcal{S}}_n\left(t\right)A{z}_0$ for $z_0\in D_A$. If we pass the limit as $n\to \infty$, then $Z\left(t\right)z_0\in D_A$ and $AZ\left(t\right)z_0=Z\left(t\right)A{z}_0$ for $z_0\in D_A$ due to the closedness of A.

If $z_0\in D_{A^r}$, we have

$$\begin{array}{cc}\hfill \mathbb{L}\left[J^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0\right]& ={\lambda }^{\alpha -1}{({\lambda }^{\alpha }-A)}^{-1}{z}_0={\lambda }^{-1}{z}_0+{\lambda }^{-1}{({\lambda }^{\alpha }-A)}^{-1}A{z}_0=\hfill \\ \hfill & ={\lambda }^{-1}{z}_0+{\lambda }^{-1-\alpha }A{z}_0+{\lambda }^{-1-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^2{z}_0=\hfill \\ \hfill & =\sum _{k=0}^{r-1}{\lambda }^{-1-k\alpha }{A}^k{z}_0+{\lambda }^{-1-(r-1)\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^r{z}_0,\hfill \end{array}$$

$$\parallel {\lambda }^{-1-(r-1)\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^r{z}_0{\parallel }_{\mathcal{Z}}\le {\displaystyle \frac{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\parallel A^r{z}_0{\parallel }_{\mathcal{Z}}}{{\left|\lambda \right|}^{m+(r-1)\alpha -\beta (m-\alpha )}{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+1}}},$$

(9)

$m+(r-1)\alpha -\beta (m-\alpha )\ge r\alpha >1,$

$$J^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0=\sum _{k=0}^{r-1}{\displaystyle \frac{t^{k\alpha }{A}^k{z}_0}{\mathrm{\Gamma }(k\alpha +1)}}+{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-1-(r-1)\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^r{z}_{0}d\lambda .$$

From (9), it follows that

$${∥\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-1-(r-1)\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^r{z}_{0}d\lambda ∥}_{\mathcal{Z}}\le C_1{t}^{m+(r-1)\alpha -\beta (m-\alpha )-1}.$$

Hence, $\left(J^{(1-\beta )(m-\alpha )}Z\right)\left(0\right)z_0=z_0$.

If $\alpha \in (1,2]$, then $m=2$ and $r=1$, and for $z_0\in D_{A^2}$, we have

$$\begin{array}{cc}\hfill \mathbb{L}\left[J^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0\right]& ={\lambda }^{-1}{z}_0+{\lambda }^{-1}{({\lambda }^{\alpha }-A)}^{-1}A{z}_0=\hfill \\ \hfill & ={\lambda }^{-1}{z}_0+{\lambda }^{-1-\alpha }A{z}_0+{\lambda }^{-1-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^2{z}_0,\hfill \end{array}$$

$$\parallel {\lambda }^{-1-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^2{z}_0{\parallel }_{\mathcal{Z}}\le {\displaystyle \frac{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\parallel A^2{z}_0{\parallel }_{\mathcal{Z}}}{{\left|\lambda \right|}^{2+\alpha -\beta (2-\alpha )}{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+1}}},$$

$${∥\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-1-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^2{z}_{0}d\lambda ∥}_{\mathcal{Z}}\le C_1{t}^{1+\alpha -\beta (2-\alpha )},t\to 0+,$$

$$J^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0=z_0+{\displaystyle \frac{t^{\alpha }A{z}_0}{\mathrm{\Gamma }(\alpha +1)}}+{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-1-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^2{z}_{0}d\lambda ,$$

$$D^1{J}^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0={\displaystyle \frac{t^{\alpha -1}A{z}_0}{\mathrm{\Gamma }\left(\alpha \right)}}+{\displaystyle \frac{1}{2\pi i}}\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^2{z}_{0}d\lambda ,$$

$$\parallel {\lambda }^{-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{A}^2{z}_0{\parallel }_{\mathcal{Z}}\le {\displaystyle \frac{K\mathrm{\Gamma }((1-\beta )(\alpha -m)+1)\parallel A^2{z}_0{\parallel }_{\mathcal{Z}}}{{\left|\lambda \right|}^{1+\alpha -\beta (2-\alpha )}{(\mathrm{Re}\lambda -\omega )}^{(1-\beta )(\alpha -m)+1}}},$$

$${∥\underset{s-i\infty }{\overset{s+i\infty }{\int }}{\lambda }^{-\alpha }{({\lambda }^{\alpha }-A)}^{-1}{e}^{\lambda t}{A}^2{z}_{0}d\lambda ∥}_{\mathcal{Z}}\le C_1{t}^{\alpha -\beta (2-\alpha )},$$

$D^1{J}^{(1-\beta )(2-\alpha )}Z\left(0\right)z_0=0$, since $\alpha -\beta (2-\alpha )>0$.

For any $z_0\in \mathcal{Z}$, choose $v_0={({\lambda }_0^{\alpha }-A)}^{-r}{z}_0\in D_{A^r}$, where $\mathrm{Re}{\lambda }_0>\omega$ and the denotation ${({\lambda }_0^{\alpha }-A)}^{-r}:={({({\lambda }_0^{\alpha }-A)}^{-1})}^r$ is used. Since the operator $Z(t)$ commutes with ${({\lambda }_0^{\alpha }-A)}^{-1}$ by the construction, we have

$${({\lambda }_0^{\alpha }-A)}^{-r}(J^{(1-\beta )(m-\alpha )}Z)\left(0\right)z_0=(J^{(1-\beta )(m-\alpha )}Z)\left(0\right)v_0=v_0={({\lambda }_0^{\alpha }-A)}^{-r}{z}_0.$$

(10)

If $\alpha \in (1,2]$, then we choose $v_0={({\lambda }_0^{\alpha }-A)}^{-2}{z}_0\in D_{A^2}$ and have

$${({\lambda }_0^{\alpha }-A)}^{-2}(D^1{J}^{(1-\beta )(2-\alpha )}Z)\left(0\right)z_0=(D^1{J}^{(1-\beta )(2-\alpha )}Z)\left(0\right)v_0=0.$$

(11)

After using ${({\lambda }_0^{\alpha }-A)}^r$ on the left and right sides of equality (10) and ${({\lambda }_0^{\alpha }-A)}^2$ on both sides of (11), we obtain $(J^{(1-\beta )(m-\alpha )}Z)(0)z_0=z_0$, and for $\alpha \in (1,2]$, $z_0\in \mathcal{Z}$, we have $(D^1{J}^{(1-\beta )(2-\alpha )}Z)(0)z_0=0$.

If $z_0\in D_{A^r}$, then

$$\begin{array}{c}\mathbb{L}\left[Z\left(t\right)A{z}_0\right]={\lambda }^{m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}A{z}_0=\\ ={\lambda }^{\alpha +m-1-\beta (m-\alpha )}{({\lambda }^{\alpha }-A)}^{-1}{z}_0-{\lambda }^{m-1-\beta (m-\alpha )}{z}_0={\lambda }^{\alpha }{H}_{\beta }\left(\lambda \right)z_0-{\lambda }^{m-1-\beta (m-\alpha )}{z}_0,\end{array}$$

$${\lambda }^{-\alpha }\mathbb{L}\left[Z\left(t\right)A{z}_0\right]=H_{\beta }\left(\lambda \right)z_0-{\lambda }^{(1-\beta )(m-\alpha )-1}{z}_0,$$

$$\begin{array}{c}\hfill J^{\alpha }Z\left(t\right)A{z}_0={\mathbb{L}}^{-1}\left[{\lambda }^{-\alpha }\mathbb{L}\left[Z\left(t\right)A{z}_0\right]\right]={\mathbb{L}}^{-1}[H_{\beta }\left(\lambda \right)z_0-{\lambda }^{(1-\beta )(m-\alpha )-1}{z}_0]=\\ \hfill =Z\left(t\right)z_0-{\displaystyle \frac{t^{(1-\beta )(\alpha -m)}{z}_0}{\mathrm{\Gamma }\left(\right(1-\beta \left)\right(\alpha -m)+1)}},\end{array}$$

$$AZ\left(t\right)z_0=D^m{J}^{m-\alpha }{J}^{\alpha }Z\left(t\right)A{z}_0=D^m{J}^{m-\alpha }Z\left(t\right)z_0-{\displaystyle \frac{t^{\beta (m-\alpha )-m}{z}_0}{\mathrm{\Gamma }(1+\beta (m-\alpha )-m)}}=$$

$$=J^{\beta (m-\alpha )}{D}^m{J}^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0=D^{\alpha ,\beta }Z\left(t\right)z_0.$$

(12)

In this case, for $z_0\in D_A$, $t\in {\mathbb{R}}_{+}$,

$${\parallel AZ\left(t\right)z_0\parallel }_{\mathcal{Z}}={\parallel Z\left(t\right)A{z}_0\parallel }_{\mathcal{Z}}\le {\parallel Z\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}{\parallel z_0\parallel }_{D_A}\le K{t}^{(1-\beta )(\alpha -m)}{e}^{\omega t}{\parallel z_0\parallel }_{D_A},$$

Hence, $Z\left(t\right)A\in \mathcal{L}(D_A;\mathcal{Z})$. Due to Corollary 3, the domain $D_{A^r}$ is dense in $D_A$; therefore, equality (12) can be extended on $z_0\in D_A$. Indeed,

$$\begin{array}{cc}\hfill & D^{\alpha ,\beta }Z\left(t\right)z_0=\underset{n\to \infty }{lim}{D}^{\alpha ,\beta }Z\left(t\right){[n^{\alpha }{(n^{\alpha }-A)}^{-1}]}^{r-1}{z}_0=\hfill \\ \hfill & =\underset{n\to \infty }{lim}Z\left(t\right)A{[n^{\alpha }{(n^{\alpha }-A)}^{-1}]}^{r-1}{z}_0=\underset{n\to \infty }{lim}{[n^{\alpha }{(n^{\alpha }-A)}^{-1}]}^{r-1}Z\left(t\right)A{z}_0=Z\left(t\right)A{z}_0\hfill \end{array}$$

for $n>\omega$, $z_0\in D_A$. Therefore, for $z_0\in D_A$, $D^{\alpha ,\beta }Z\left(t\right)z_0=Z\left(t\right)A{z}_0\in C({\mathbb{R}}_{+};\mathcal{Z})$ and $J^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0\in C({\overline{\mathbb{R}}}_{+};\mathcal{Z})\cap C^1({\mathbb{R}}_{+};\mathcal{Z})$, and for $\alpha \in (1,2]$, $D^1{J}^{(1-\beta )(m-\alpha )}Z\left(t\right)z_0\in C({\overline{\mathbb{R}}}_{+};\mathcal{Z})\cap C^1({\mathbb{R}}_{+};\mathcal{Z})$.

Thus, $\{Z\left(t\right)\in \mathcal{L}\left(\mathcal{Z}\right):t\in {\mathbb{R}}_{+}\}$ is a resolving family of operators for Equation (5). □

Remark 7.

Due to Lemma 3, the resolving family of operators for Equation (5) is unique.

Denote

$$C^{\alpha ,\beta }:=\bigcup _{\omega \ge 0}{C}^{\alpha ,\beta }\left(\omega \right),{\mathcal{C}}_{\alpha ,\beta }:=\bigcup _{\omega \ge 0}{\mathcal{C}}_{\alpha ,\beta }\left(\omega \right).$$

Remark 8.

We can affirm that for $\alpha >2$, ${\mathcal{C}}_{\alpha ,\beta }=C^{\alpha ,\beta }=\mathcal{L}\left(\mathcal{Z}\right)$. Indeed, due to Lemma 2 and Theorem 2, $C^{\alpha ,\beta }\subset {\mathcal{C}}_{\alpha ,\beta }\subset \mathcal{L}\left(\mathcal{Z}\right)$, and by means of Remark 3, $\mathcal{L}\left(\mathcal{Z}\right)\subset C^{\alpha ,\beta }$.

Corollary 4.

Let $\alpha \in (0,1]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, and $z_0\in D_A$. Then, the Cauchy-type problem $J^{(1-\beta )(1-\alpha )}z\left(0\right)=z_0$ for Equation (5) has a unique solution. It has a form $z\left(t\right)=Z\left(t\right)z_0$.

Proof. 

Due to Theorem 3, it remains to prove that the problem solution is unique.

Use $J^{1-\beta (1-\alpha )}$ in Equation (5) and obtain

$$J^{1-\beta (1-\alpha )}{J}^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}z\left(t\right)=g_{(1-\beta )(1-\alpha )}\ast z\left(t\right)-z_0=g_{1-\beta (1-\alpha )}\ast Az\left(t\right).$$

(13)

Here, $g_{\beta }\left(t\right):=t^{\beta -1}/\mathrm{\Gamma }\left(\beta \right)$, and the Laplace convolution * has a form

$$(f\ast g)\left(t\right):=\underset{0}{\overset{t}{\int }}f(t-s)g\left(s\right)ds.$$

Equality (13) is true for every solution of the Cauchy-type problem $J^{(1-\beta )(1-\alpha )}z\left(0\right)=z_0\in D_A$ for Equation (5). In particular, for $Z\left(t\right)z_0$, the equality

$$(g_{(1-\beta )(1-\alpha )}\ast Z\left(t\right)-g_{1-\beta (1-\alpha )}\ast AZ\left(t\right))z_0\equiv z_0,$$

is valid; therefore, for another solution z,

$$\begin{array}{cc}\hfill & 1\ast z=1\ast (g_{(1-\beta )(1-\alpha )}\ast Z-g_{1-\beta (1-\alpha )}\ast AZ)z=(g_{(1-\beta )(1-\alpha )}\ast Z-g_{1-\beta (1-\alpha )}\ast AZ)\ast z\hfill \\ \hfill & =Z\ast (g_{(1-\beta )(1-\alpha )}\ast z-g_{1-\beta (1-\alpha )}\ast Az)=Z\ast z_0=1\ast Z{z}_0.\hfill \end{array}$$

Differentiating by t, we obtain $z\left(t\right)=Z\left(t\right)z_0$, $t\in {\mathbb{R}}_{+}$. □

Corollary 5.

Let $\alpha \in (1,2]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, and $z_0,z_1\in D_A$. Then, the Cauchy-type problem

$$D^{-(1-\beta )(2-\alpha )}z\left(0\right)=z_0,D^{1-(1-\beta )(2-\alpha )}z\left(0\right)=z_1$$

(14)

for Equation (5) has a unique solution. It has a form $z\left(t\right)=Z_0\left(t\right)z_0+J^1{Z}_0\left(t\right)z_1$.

Proof. 

As in the previous proof, we can prove that for a solution of problem (5) and (14) with $z_1=0$, $z_0\in D_A$,

$$J^{2-\beta (2-\alpha )}{J}^{\beta (2-\alpha )}{D}^2{J}^{(1-\beta )(2-\alpha )}z\left(t\right)=g_{(1-\beta )(2-\alpha )}\ast z\left(t\right)-z_0=g_{2-\beta (2-\alpha )}\ast Az\left(t\right).$$

Here, $J^{(1-\beta )(2-\alpha )}z\in A{C}^2({\mathbb{R}}_{+};\mathcal{Z})$ by the definition of a solution. Taking the solution $Z\left(t\right)z_0$, we have $(g_{(1-\beta )(2-\alpha )}\ast Z\left(t\right)-g_{2-\beta (2-\alpha )}\ast AZ\left(t\right))z_0\equiv z_0;$ therefore, for another solution z,

$$\begin{array}{c}\hfill 1\ast z=(g_{(1-\beta )(2-\alpha )}\ast Z-g_{2-\beta (2-\alpha )}\ast AZ)\ast z=\\ \hfill =Z\ast (g_{(1-\beta )(2-\alpha )}\ast z-g_{2-\beta (2-\alpha )}\ast Az)=Z\ast z_0=1\ast Z{z}_0\end{array}$$

and $z\left(t\right)=Z_0\left(t\right)z_0$, $t\in {\mathbb{R}}_{+}$.

For a solution z for problem (5) and (14) with $z_0,z_1\in D_A$, the function $z\left(t\right)-J^{1}Z\left(t\right)z_1$ solves this problem with $z_1=0$. Hence, $z\left(t\right)-J^{1}Z\left(t\right)z_1=Z_0\left(t\right)z_0$. □

4. Unique Solvability of Inhomogeneous Equations

Consider the Cauchy-type problem

$$J^{(1-\beta )(1-\alpha )}z\left(0\right)=z_0,$$

(15)

$$D^{\alpha ,\beta }z\left(t\right)=Az\left(t\right)+f\left(t\right),t\in (0,T],$$

(16)

with $\alpha \in (0,1]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, and $f:(0,T]\to \mathcal{Z}$. A solution to problem (15) and (16) is called a function $z\in C((0,T];D_A)\cap L_1(0,T;\mathcal{Z})$ such that

$$J^{(1-\beta )(1-\alpha )}z\in C([0,T];\mathcal{Z})\cap A{C}^1((0,T];\mathcal{Z}),$$

condition (15) is valid, and equality (16) is fulfilled for $t\in (0,T]$.

Lemma 6.

Let $\alpha \in (0,1]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, and $f\in C^1([0,T];D_A)$. Then, the Cauchy-type problem $J^{(1-\beta )(1-\alpha )}z\left(0\right)=0$ for Equation (16) has a unique solution. It has a form

$$z_f\left(t\right)=\underset{0}{\overset{t}{\int }}{D}^{\beta (1-\alpha )}Z(t-s)f\left(s\right)ds=J^{1-\beta (1-\alpha )}Z\left(t\right)f\left(0\right)+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f\left(s\right)ds.$$

Proof. 

From the proof of Theorem 3, it follows that for $t>0$,

$$(J^{(1-\beta )(1-\alpha )}Z)(0)=I,J^{1-\beta (1-\alpha )}Z(t)=J^{\alpha }{J}^{(1-\beta )(1-\alpha )}Z(t),(J^{1-\beta (1-\alpha )}Z)(0)=0,$$

$$\begin{array}{cc}\hfill z_f(t)& =-\underset{0}{\overset{t}{\int }}{D}_s^1{J}^{1-\beta (1-\alpha )}Z(t-s)f(s)ds=\hfill \\ \hfill & =-(J^{1-\beta (1-\alpha )}Z)(0)f(t)+J^{1-\beta (1-\alpha )}Z(t)f(0)+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f(s)ds=\hfill \\ \hfill & =J^{1-\beta (1-\alpha )}Z(t)f(0)+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f(s)ds,\hfill \end{array}$$

$$J^{(1-\beta )(1-\alpha )}{z}_f(t)=J^{(1-\beta )(1-\alpha )}{J}^{1-\beta (1-\alpha )}Z(t)f(0)+\underset{0}{\overset{t}{\int }}{J}^{(1-\beta )(1-\alpha )}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f(s)ds,$$

and

$${∥\underset{0}{\overset{t}{\int }}{J}^{(1-\beta )(1-\alpha )}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f\left(s\right)ds∥}_{\mathcal{Z}}\le C\underset{0}{\overset{t}{\int }}{\parallel D^{1}f\left(s\right)\parallel }_{\mathcal{Z}}ds.$$

Hence, $\left(J^{(1-\beta )(1-\alpha )}{z}_f\right)\left(0\right)=0.$ So, we obtain

$${∥\underset{0}{\overset{t}{\int }}Z(t-s)Af\left(s\right)ds∥}_{\mathcal{Z}}\le {\parallel f\parallel }_{C([0,T];D_A)}K{e}^{\omega t}\underset{0}{\overset{t}{\int }}{s}^{(1-\beta )(\alpha -1)}\le C{e}^{\omega t}{\parallel f\parallel }_{C([0,T];D_A)}.$$

Due to the closedness of the operator A and its commutation with operators $Z\left(t\right)$, for $f\in C^1([0,T];D_A)$,

$$\underset{0}{\overset{t}{\int }}Z(t-s)Af\left(s\right)ds=A{z}_f\left(t\right),$$

which implies $z_f\left(t\right)\in D_A$ for all $t>0$.

Further, for $t>0$,

$$\begin{array}{cc}\hfill D^{\alpha ,\beta }{z}_f(t)& =J^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}{z}_f(t)=J^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}{J}^{1-\beta (1-\alpha )}Z(t)f(0)+\hfill \end{array}$$

$$+J^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}Z(t-s)D^{1}f(s)ds=$$

$$=J^{1-\beta (1-\alpha )}{J}^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}Z(t)f(0)+{\displaystyle \frac{J^{\beta (1-\alpha )}{t}^{-\beta (1-\alpha )}}{\mathrm{\Gamma }(1-\beta (1-\alpha ))}}(J^{(1-\beta )(1-\alpha )}Z)(0)f(0)+$$

$$+J^{\beta (1-\alpha )}{D}^1\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}{J}^{(1-\beta )(1-\alpha )}Z(t-s)D^{1}f(s)ds=$$

$$=J^{1-\beta (1-\alpha )}AZ(t)f(0)+f(0)+J^{\beta (1-\alpha )}(J^{1-\beta (1-\alpha )}{J}^{(1-\beta )(1-\alpha )}Z)(0)D^{1}f(t)+$$

$$+\underset{0}{\overset{t}{\int }}{J}^{\beta (1-\alpha )}{D}^1{J}^{1-\beta (1-\alpha )}{J}^{(1-\beta )(1-\alpha )}Z(t-s)D^{1}f(s)ds=$$

$$=J^{1-\beta (1-\alpha )}AZ(t)f(0)+f(0)+\underset{0}{\overset{t}{\int }}{\displaystyle \frac{J^{\beta (1-\alpha )}{(t-s)}^{-\beta (1-\alpha )}}{\mathrm{\Gamma }(1-\beta (1-\alpha ))}}(J^{(1-\beta )(1-\alpha )}Z)(0)D^{1}f(s)ds+$$

$$+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (1-\alpha )}AZ(t-s)D^{1}f(s)ds=A{z}_f(t)+f(t).$$

□

Theorem 4.

Let $\alpha \in (0,1]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, $f\in C^1([0,T];D_A)$, and $z_0\in D_A$. Then, Cauchy-type problem (15) and (16) has a unique solution. It has a form

$$z\left(t\right)=Z\left(t\right)z_0+\underset{0}{\overset{t}{\int }}{D}^{\beta (1-\alpha )}Z(t-s)f\left(s\right)ds.$$

Now, consider the Cauchy-type problem

$$D^{-(1-\beta )(2-\alpha )}z\left(0\right)=z_0,D^{1-(1-\beta )(2-\alpha )}z\left(0\right)=z_1,$$

(17)

for a linear inhomogeneous equation

$$D^{\alpha ,\beta }z\left(t\right)=Az\left(t\right)+f\left(t\right),t\in (0,T],$$

(18)

where $\alpha \in (1,2]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, and $f:(0,T]\to \mathcal{Z}$. A solution to problem (17) and (18) is called a function $z\in C((0,T];D_A)\cap L_1(0,T;\mathcal{Z})$ such that $D^{-(1-\beta )(2-\alpha )}z\in C^1([0,T];\mathcal{Z})\cap A{C}^2((0,T];\mathcal{Z})$, condition (17) is valid, and (18) is fulfilled for all $t\in (0,T]$.

Lemma 7.

Let $\alpha \in (1,2]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, $f\in C((0,T];D_A)$, and

$$\exists \gamma >\beta (2-\alpha )-1\exists C_f>0\forall t\in (0,T]{\parallel f\left(t\right)\parallel }_{D_A}\le C_f{t}^{\gamma }.$$

(19)

Then, the Cauchy-type problem $D^{-(1-\beta )(2-\alpha )}z\left(0\right)=0$, $D^{1-(1-\beta )(2-\alpha )}z\left(0\right)=0$ for Equation (18) has a unique solution. It has a form

$$z_f\left(t\right)=\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}Z(t-s)f\left(s\right)ds.$$

Proof. 

For $t>0$, due to (8),

$$\parallel J^{1-\beta (2-\alpha )}{Z\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le K\underset{0}{\overset{t}{\int }}{\displaystyle \frac{{(t-s)}^{-\beta (2-\alpha )}{s}^{(1-\beta )(\alpha -2)}}{\mathrm{\Gamma }(1-\beta (2-\alpha \left)\right)}}{e}^{\omega s}ds\le C_1{t}^{\alpha -1}{e}^{\omega t},$$

$$\parallel z_f{\left(t\right)\parallel }_{\mathcal{Z}}\le C_1{e}^{\omega t}\underset{0}{\overset{t}{\int }}{(t-s)}^{\alpha -1}{\parallel f\left(s\right)\parallel }_{\mathcal{Z}}ds\le C_2{t}^{\alpha -1}{e}^{\omega t}\underset{0}{\overset{t}{\int }}{\parallel f\left(s\right)\parallel }_{\mathcal{Z}}ds,$$

$$\parallel J^{(1-2\beta )(2-\alpha )+1}{Z\left(t\right)\parallel }_{\mathcal{L}\left(\mathcal{Z}\right)}\le K\underset{0}{\overset{t}{\int }}{\displaystyle \frac{{(t-s)}^{(1-2\beta )(2-\alpha )}{s}^{(1-\beta )(\alpha -2)}}{\mathrm{\Gamma }\left(\right(1-2\beta \left)\right(2-\alpha )+1)}}{e}^{\omega s}ds\le C_3{t}^{1-\beta (2-\alpha )}{e}^{\omega t}.$$

Since $1-\beta (2-\alpha )>0$, we have $(J^{(1-2\beta )(2-\alpha )+1}Z)(0)=0$. Hence,

$$\begin{array}{cc}\hfill \parallel J^{(1-\beta )(2-\alpha )}{z}_f{\left(t\right)\parallel }_{\mathcal{Z}}& ={∥\underset{0}{\overset{t}{\int }}{J}^{(1-2\beta )(2-\alpha )+1}Z(t-s)f\left(s\right)ds∥}_{\mathcal{Z}}\le C_4{t}^{1-\beta (2-\alpha )}{e}^{\omega t}\underset{0}{\overset{t}{\int }}{\parallel f\left(s\right)\parallel }_{\mathcal{Z}}ds,\hfill \end{array}$$

and $(J^{(1-\beta )(2-\alpha )}{z}_f)(0)=0.$ Further,

$$D^1{J}^{(1-2\beta )(2-\alpha )+1}Z\left(t\right)z_0=J^{1-\beta (2-\alpha )}{D}^1{J}^{(1-\beta )(2-\alpha )}Z\left(t\right)z_0+{\displaystyle \frac{t^{-\beta (2-\alpha )}{z}_0}{\mathrm{\Gamma }(1-\beta (2-\alpha \left)\right)}},$$

$${\parallel D^1{J}^{(1-2\beta )(2-\alpha )+1}Z\left(t\right)z_0\parallel }_{\mathcal{Z}}\le {\displaystyle \frac{C_5{t}^{-\beta (2-\alpha )}{\parallel z_0\parallel }_{\mathcal{Z}}}{\mathrm{\Gamma }(1-\beta (2-\alpha \left)\right)}},$$

and

$$\begin{array}{cc}\hfill D^{1-(1-\beta )(2-\alpha )}{z}_f(t)& =D^1\underset{0}{\overset{t}{\int }}{J}^{(1-2\beta )(2-\alpha )+1}Z(t-s)f(s)ds=\hfill \\ \hfill & =(J^{(1-2\beta )(2-\alpha )+1}Z)(0)f(t)+\underset{0}{\overset{t}{\int }}{D}^1{J}^{(1-2\beta )(2-\alpha )+1}Z(t-s)f(s)ds=\hfill \\ \hfill & =\underset{0}{\overset{t}{\int }}{D}^1{J}^{(1-2\beta )(2-\alpha )+1}Z(t-s)f(s)ds,\hfill \end{array}$$

Therefore,

$$\parallel D^{1-(1-\beta )(2-\alpha )}{z}_f{\left(t\right)\parallel }_{\mathcal{Z}}\le C_6{e}^{\omega t}\underset{0}{\overset{t}{\int }}{(t-s)}^{-\beta (2-\alpha )}{s}^{\gamma }ds\le C_7{e}^{\omega t}{t}^{\gamma -\beta (2-\alpha )+1}.$$

Since $\gamma >\beta (2-\alpha )-1$, we have $(D^{1-(1-\beta )(2-\alpha )}{z}_f)\left(0\right)=0$.

Further,

$${∥\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}Z(t-s)Af\left(s\right)ds∥}_{\mathcal{Z}}\le C{e}^{\omega t}\underset{0}{\overset{t}{\int }}{\parallel Af\left(s\right)\parallel }_{\mathcal{Z}}ds.$$

The closedness of the operator A and its commutation with operators $Z\left(t\right)$ and the condition $f\left(t\right)\in D_A$ for $t\in (0,T]$ imply that

$$\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}Z(t-s)Af\left(s\right)ds=\underset{0}{\overset{t}{\int }}A{J}^{1-\beta (2-\alpha )}Z(t-s)f\left(s\right)ds=A{z}_f\left(t\right),$$

Therefore, $z_f\left(t\right)\in D_A$ for $t\in (0,T]$.

Finally,

$$D^{\alpha ,\beta }{z}_f(t)=J^{\beta (2-\alpha )}{D}^2{J}^{(1-\beta )(2-\alpha )}{z}_f(t)=D^2{J}^{2-\alpha }\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}Z(t-s)f(s)ds-$$

$$-{\displaystyle \frac{t^{\beta (2-\alpha )-2}}{\mathrm{\Gamma }(\beta (2-\alpha )-1)}}(J^{(1-\beta )(2-\alpha )}{z}_f)(0)-{\displaystyle \frac{t^{\beta (2-\alpha )-1}}{\mathrm{\Gamma }(\beta (2-\alpha ))}}(D^{1-(1-\beta )(2-\alpha )}{z}_f)(0)=$$

$$\begin{array}{cc}\hfill & =D^2\underset{0}{\overset{t}{\int }}{J}^{1+(1-\beta )(2-\alpha )}Z(t-s)f(s)ds=\hfill \\ \hfill & =D^1(J^{1+(1-\beta )(2-\alpha )}Z)(0)f(t)+D^1\underset{0}{\overset{t}{\int }}{D}^1{J}^{1+(1-\beta )(2-\alpha )}Z(t-s)f(s)ds=\hfill \\ \hfill & =(J^{(1-\beta )(2-\alpha )}Z)(0)f(t)+\underset{0}{\overset{t}{\int }}{D}^2{J}^{1+(1-\beta )(2-\alpha )}Z(t-s)f(s)ds=\hfill \\ \hfill & =f(t)+\underset{0}{\overset{t}{\int }}\left(J^1{D}^2{J}^{(1-\beta )(2-\alpha )}Z(t-s)+(D^{1-(1-\beta )(2-\alpha )}Z)(0)\right)f(s)ds=\hfill \\ \hfill & =f(t)+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}{D}^{\alpha ,\beta }Z(t-s)f(s)ds=\hfill \\ \hfill & =f(t)+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}AZ(t-s)f(s)ds=A{z}_f(t)+f(t),\hfill \end{array}$$

since $D^{1-(1-\beta )(2-\alpha )}Z\left(0\right)z_0=0$ for every $z_0\in \mathcal{Z}$, as it is proved in Theorem 3. □

Corollary 5 and the linearity of Equation (18) imply the next statement.

Theorem 5.

Let $\alpha \in (1,2]$, $\beta \in [0,1]$, $A\in {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$, $z_0,z_1\in D_A$, and $f\in C((0,T];D_A)$ satisfy condition (19). Then, Cauchy-type problem (17) and (18) has a unique solution. It has a form

$$z\left(t\right)=Z\left(t\right)z_0+J^{1}Z\left(t\right)z_1+\underset{0}{\overset{t}{\int }}{J}^{1-\beta (2-\alpha )}Z(t-s)f\left(s\right)ds.$$

5. Simple Examples of Operators from $C^{\alpha ,\beta }$

Take the Banach space $l_1$, which consists of sequences $z={\left\{z_n\right\}}_{n=1}^{\infty }$, where $z_n\in \mathbb{C}$ for $n\in \mathbb{N}$, such that the series ${\parallel z\parallel }_{l_1}={\sum }_{n=1}^{\infty }\left|z_n\right|$ is convergent.

Consider the Cauchy-type problem

$$J^{(1-\beta )(1-\alpha )}{z}_n\left(0\right)=z_{0n},n\in \mathbb{N},$$

(20)

$$D^{\alpha ,\beta }{z}_n\left(t\right)=n{e}^{i\pi \gamma /2}{z}_n\left(t\right),t\in (0,T],n\in \mathbb{N},$$

(21)

where $0<\alpha \le 1$, $0\le \beta \le 1$, $-2<\gamma \le 2$, and $z_0={\left\{z_{0n}\right\}}_{n=1}^{\infty }\in l_1$. Set $\mathcal{Z}=l_1$ and $A_{\gamma }z={\left\{n{e}^{i\pi \gamma /2}{z}_n\right\}}_{n=1}^{\infty }$ for $z={\left\{z_n\right\}}_{n=1}^{\infty }\in l_1$ such that ${\left\{n{e}^{i\pi \gamma /2}{z}_n\right\}}_{n=1}^{\infty }\in l_1$, which make up the domain $D_{A_{\gamma }}$. It is not difficult to prove that for all $\gamma \in (-2,2]$, $A_{\gamma }\in \mathcal{C}l\left(\mathcal{Z}\right)$ and $\sigma \left(A_{\gamma }\right)=\{n{e}^{i\pi \gamma /2}:n\in \mathbb{N}\}$. Let ${\mathrm{\Pi }}_{\omega }^{\alpha }:=\{{\mu }^{\alpha }\in \mathbb{C}:\mathrm{Re}\mu >\omega \}.$ Then, for every $\gamma \in (-\alpha ,\alpha )$ and $\omega \ge 0$, we have $\sigma \left(A_{\gamma }\right)\cap {\mathrm{\Pi }}_{\omega }^{\alpha }\ne \varnothing$ and $A\notin {\mathcal{C}}_{\alpha ,\beta }\left(\omega \right)$. So, we will consider $\gamma \in (-2,-\alpha ]\cup [\alpha ,2]$ only.

Solving problem (20) and (21) for every $n\in \mathbb{N}$ individually, we obtain that the resolving family of equation has a form $S_{\gamma }\left(t\right)z_0={\left\{t^{(1-\beta )(\alpha -1)}{E}_{\alpha ,(1-\beta )(\alpha -1)+1}\left(t^{\alpha }n{e}^{i\pi \gamma /2}\right)z_{0n}\right\}}_{n=1}^{\infty },$ where $z_0={\left\{z_{0n}\right\}}_{n=1}^{\infty }\in l_1$ (see Remark 3). Since $arg\left(n{e}^{i\pi \gamma /2}\right)=\pi \gamma /2$, we have for a fixed $t>0$, due to the asymptotic expansion of the Mittag–Leffler function ([11], p. 13) for $\gamma \in (-2,-\alpha )\cup (\alpha ,2]$,

$$|t^{(1-\beta )(\alpha -1)}{E}_{\alpha ,(1-\beta )(\alpha -1)+1}(t^{\alpha }n{e}^{i\pi \gamma /2})| \le C_1{t}^{(1-\beta )(\alpha -1)}{(t^{\alpha }n)}^{-1}={\displaystyle \frac{C_1{t}^{\beta (1-\alpha )-1}}{n}},n\to \infty .$$

Therefore, $S_{\gamma }\left(t\right)\in \mathcal{L}\left(l_1\right)$. Since for $z_0\in l_1$ $J^{(1-\beta )(1-\alpha )}{S}_{\gamma }\left(t\right)z_0={\left\{E_{\alpha ,1}\left(t^{\alpha }n{e}^{i\pi \gamma /2}\right)z_{0n}\right\}}_{n=1}^{\infty },$ the equality $J^{(1-\beta )(1-\alpha )}{S}_{\gamma }\left(0\right)=I$ holds, and for every $z_0\in D_{A_{\gamma }}$, we have

$$\begin{array}{cc}\hfill J^{\beta (1-\alpha )}{D}^1{J}^{(1-\beta )(1-\alpha )}{S}_{\gamma }\left(t\right)z_0& ={\left\{n{e}^{i\pi \gamma /2}{t}^{(1-\beta )(\alpha -1)}{E}_{\alpha ,(1-\beta )(\alpha -1)+1}\left(t^{\alpha }n{e}^{i\pi \gamma /2}\right)z_{0n}\right\}}_{n=1}^{\infty }\hfill \\ \hfill & =A{S}_{\gamma }\left(t\right)z_0=S_{\gamma }\left(t\right)A{z}_0\in C({\mathbb{R}}_{+};l_1).\hfill \end{array}$$

Thus, $A_{\gamma }\in C^{\alpha ,\beta }\left(0\right)$ for every $\gamma \in (-2,-\alpha )\cup (\alpha ,2]$.

Now, take $\gamma =\alpha$. Then, $arg\left(n{e}^{i\pi \alpha /2}\right)=\pi \alpha /2$, and for a fixed $t>0$ using the asymptotic expansion of the Mittag–Leffler function ([11], p. 12), we obtain

$$|t^{(1-\beta )(\alpha -1)}{E}_{\alpha ,(1-\beta )(\alpha -1)+1}(t^{\alpha }n{e}^{i\pi \alpha /2})| \sim C{t}^{(1-\beta )(\alpha -1)}{(t^{\alpha }n)}^{{\displaystyle \frac{(1-\beta )(1-\alpha )}{\alpha }}}|e^{i{n}^{1/\alpha }t}|=C{n}^{{\displaystyle \frac{(1-\beta )(1-\alpha )}{\alpha }}},$$

as $n\to \infty$. So, $S_{\alpha }\left(t\right)\notin \mathcal{L}\left(l_1\right)$ for $\beta \in [0,1)$, $\alpha \in (0,1)$. If $\alpha =1$ or $\beta =1$, (20) is the Cauchy problem for the equation $D^{1,\beta }{z}_n\left(t\right)=n{e}^{i\pi \alpha /2}{z}_n\left(t\right)$, $n\in \mathbb{N}$, or $D^{\alpha ,1}{z}_n\left(t\right)=n{e}^{i\pi \alpha /2}{z}_n\left(t\right)$, $n\in \mathbb{N}$, $S_{\alpha }\left(t\right)z_0={\left\{E_{\alpha ,1}\left(t^{\alpha }n{e}^{i\pi \alpha /2}\right)z_{0n}\right\}}_{n=1}^{\infty },$ and $A_1\in C^{1,\beta }\left(0\right)$ or $A_{\alpha }\in C^{\alpha ,1}\left(0\right)$.

Analogously, we can consider the case $\gamma =-\alpha$.

6. Application to Initial Boundary Value Problem

For $d\in \mathbb{N}$, consider in a bounded domain $\mathrm{\Omega }\subset {\mathbb{R}}^d$ with a smooth boundary $\partial \mathrm{\Omega }$, the initial boundary value problem

$$J_t^{(1-\beta )(1-\alpha )}w(\xi ,0)=w_0\left(\xi \right),\xi \in \mathrm{\Omega },$$

(22)

$$w(\xi ,t)=0,(\xi ,t)\in \partial \mathrm{\Omega }\times {\overline{\mathbb{R}}}_{+},$$

(23)

for the linear-time fractional Schrödinger equation

$$D_t^{\alpha ,\beta }w(\xi ,t)=i\mathrm{\Delta }w(\xi ,t),(\xi ,t)\in \mathrm{\Omega }\times {\overline{\mathbb{R}}}_{+},$$

(24)

where $\alpha \in (0,1)$, $\beta \in [0,1]$, $J_t^{(1-\beta )(1-\alpha )}$ is the Riemann–Liouville fractional integral with respect to the time variable t, $D_t^{\alpha ,\beta }$ is the Hilfer fractional derivative with respect to t, i is the imaginary unit, $\mathrm{\Delta }={\displaystyle \sum _{j=1}^d}{\displaystyle \frac{{\partial }^2}{\partial {\xi }_j^2}}$ is the Laplace operator, and $\xi =({\xi }_1,{\xi }_2,\dots ,{\xi }_d)$. In setting $\mathcal{Z}=L_2(\mathrm{\Omega })$, linear operator $A\in \mathcal{C}l\left(\mathcal{Z}\right)$ is defined as $Av:=i\mathrm{\Delta }v$, $D_A:=\{v\in H^2(\mathrm{\Omega }):v\left(\xi \right)=0,\xi \in \partial \mathrm{\Omega }\}$, $w_0\in D_A$. Thus, problem (22)–(24) is a partial case of problem (4) and (5).

It is known that $\sigma \left(A\right)=\{i{\lambda }_n:n\in \mathbb{N}\}$, where ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n\ge \dots$ are real negative eigenvalues of the Laplace operator, which are numbered in non-decreasing order, taking into account their multiplicities. Let $\left\{{\phi }_n\right\}$ be the corresponding eigenfunctions of the operator A, which form an orthonormal basis in $L_2(\mathrm{\Omega })$. Decomposing all the vectors in $L_2(\mathrm{\Omega })$ according to this basis, we present problem (22)–(24) as follows:

$$J^{(1-\beta )(1-\alpha )}{z}_n\left(0\right)=z_{0n},n\in \mathbb{N},$$

(25)

$$D^{\alpha ,\beta }{z}_n\left(t\right)=-i\left|{\lambda }_n\right|z_n\left(t\right),t\in (0,T],n\in \mathbb{N}.$$

(26)

Here, $z_n\left(t\right)={⟨w(·,t),{\phi }_n⟩}_{L_2(\mathrm{\Omega })}$ and $z_{0n}\left(t\right)={⟨w_0(·),{\phi }_n⟩}_{L_2(\mathrm{\Omega })}$, $n\in \mathbb{N}$. Thus, we obtained problem (20) and (21) in the $l_2$ space, since the Fourier coefficient sequences ${\left\{z_n\left(t\right)\right\}}_{n=1}^{\infty }$ for every $t\ge 0$ and ${\left\{z_{0n}\right\}}_{n=1}^{\infty }$ belong to $l_2$.

For simplicity, let $d=1$ and $\mathrm{\Omega }=(0,\pi )$; then, ${\lambda }_n=-n^2$, ${\phi }_n\left(\xi \right)=\sqrt{\pi /2}sinn\xi$, $n\in \mathbb{N}$. Arguing in a similar manner as in the previous section, we obtain that this operator is analogous to operator $A_{-1}$ ($\gamma =-1$) and that there exists a family of resolving operators $S_{-1}\left(t\right)\in \mathcal{L}\left(l_2\right)$ for every $\alpha \in (0,1)$ and $\beta \in [0,1]$.

7. Conclusions

A Cauchy-type problem for a linear homogeneous differential equation in a Banach space with the Hilfer derivative and with a closed operator A at the unknown function is considered. A generalization of the Hille–Yosida theorem on generation using the operator A of a strongly continuous resolving family of operators for such equations is obtained. It allows for the investigation of the unique solvability of the Cauchy-type problem for the corresponding inhomogeneous equations. The obtained results are illustrated by examples of specially constructed operators A. We plan to continue studying the conditions for the existence of strongly continuous resolving families of operators for equations with other fractional derivatives and integro-differential operators.